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190100 | https://www.geeksforgeeks.org/python/how-to-evaluate-an-algebraic-expression-in-sympy/ | How to evaluate an algebraic expression in Sympy?
Last Updated :
23 Jul, 2025
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SymPy is a symbolic mathematics Python package. Its goal is to develop into a completely featured computer algebra system while keeping the code as basic as possible to make it understandable and extendable. An algebraic expression is an expression or statement that has been combined using operations like addition, subtraction, division, multiplication, modulus, etc... . for example 10x+2, etc... Let's demonstrate how to evaluate algebraic expressions in sympy with a few examples. evalf() function and subs() function from sympy is used to evaluate algebraic expressions.
Example 1:
In this example, we import symbols from sympy package. An expression is created and evalf() function is used to evaluate the expression. subs is a parameter in the function, where we pass in a dictionary of symbol mappings to values. In sympy float, precision is up to 15 digits by default. precision can be overridden up to 100 digits.
Python3
````
import packages
from sympy.abc import x, y ,z
creating an expression
expression = 4x+5y+6z
evaluating the expression
print(expression.evalf(subs={x:1,y:2,z:3}))
````
Output:
32.0000000000000
In this code, precision is set to 3 digits.
Python3
````
import packages
from sympy.abc import x, y ,z
creating an expression
expression = 4x+5y+6z
evaluating the expression
print(expression.evalf(3,subs={x:1,y:2,z:3}))
````
Output:
32.0
Example 2:
sympy also has built-in values like pi, which helps us solve numerical problems such as finding the circumference of a circle etc... . precision is set to 10 digits in this example.
Python3
````
import packages
from sympy.abc import
from sympy import pi
finding the circumference of a circle
expression = 2pir
evaluating the expression
print(expression.evalf(10,subs={r:2}))
````
Output:
12.56637061
Example 3:
In this example, we use the subs() function to substitute symbols with values, and it evaluates the algebraic expression given. 24 -42 +1 == 16-8+1 ==9.
Python3
````
import packages
from sympy.abc import
creating an expression
expression = 2x - 4y + z
substituting values in the expression
print(expression.subs([(x, 4), (y, 2), (z, 1)]))
````
Output:
9
S
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Detonation vs Pre-Ignition
April 13th, 2016
5 min read
By Alan Jackson
There are many ways to ruin a perfectly good engine, but today, I just want to talk about two of the most violent ones. Detonation and Pre-Ignition, often interchanged and/or used to describe the same thing, are actually quite different things that lead to similar outcomes. They are both referred to as abnormal combustion, and they are extremely bad for your engine. In order to better explain both Detonation and Pre-Ignition, I need to also explain normal combustion.
Normal Combustion:
Normal combustion is the burning of an air/fuel mixture in the combustion chamber. Normal combustion begins with the flame front originating at the spark plug and traveling outward evenly and steadily throughout the combustion chamber. It is very much like blowing up a balloon. As you blow, the balloon expands away from the source of air in a very controlled and even fashion. In a perfect world, the combustion event burns up all the air and fuel in the cylinder leaving none behind (this happens with a stoichiometric mixture of Lambda 1). The heat from the combustion event is transferred from the flame front to the piston, and from the piston to the cylinder wall, and from there into the coolant system. A common misconception about combustion is that there is an explosion involved. This simply isn’t true… Ideally, once the spark plug ignites the mixture, the flame fills the cylinder in a very quick, yet very controlled manner.
Detonation:
Definition:
Spontaneous combustion of the end-gas, or remaining air/fuel mixture in the chamber.
Detonation always comes after normal combustion has been started by the spark plug. Normal combustion expands out, but the gasses on the edge of the flame front compress and begin combusting spontaneously. This is likely caused by excess heat and pressure. The most important thing to remember about detonation though is that it occurs after the flame front has been initiated by the spark plug.
There are many factors that come together to create an ideal scenario for detonation to occur. While engine design and fuel octane plays a significant role, the most common causal factor is too much spark advance. Overly advanced ignition timing sets the burn off too soon allowing pressure to increase too quickly. This is a very high/very sharp pressure spike, and this is what often brings about engine damage.
As you can see in the image, the graph on the top has a smooth pressure profile and would be considered normal combustion. However, the bottom graph shows the pressure rising normally until the detonation even occurs after the spark. Then you see the large pressure spike from the abnormal combustion. This pressure spike causes the structure of the engine to resonate as if it were a tuning fork. This resonance is what gets picked up by the knock sensor and reported to the ECU.
Knock sensors are a big cause for concern among many enthusiast. Being able to see what they see with a monitoring device like the Cobb Tuning AccessPort gives people a window into what is going on with their engine at all times. This window allows you to see things that you normally wouldn’t ever notice or care about. I am asked very frequently about ongoing detonation that occurs at part throttle. Fortunately, detonation is not always destructive. Low levels of detonation occur all the time and can even be sustained for long periods without effecting damage. Part throttle knock is very normal in a variety of cars and, while sometimes it can be caused by actual detonation, in most cases it is just noise as the engine resonates at certain rpms. It will also manifest itself at shift points where the engine moves significantly between gear changes at WOT, and this should not be cause for concern. However, detonation does become a big cause for concern once you start operating at higher loads. If you are seeing significant knock at Wide Open Throttle (WOT), you should contact your tuner.
Damage:
There are a few main failures brought on by detonation. The lesser of the points of failure is pitting or abrasion on the piston crown. You will also see this pitting on exhaust valves as well since this is the hotter side of the cylinder with the air/fuel mixture cooling the intake side. The pitting looks as if the piston was hit with bird shot from a shot gun.
Another point of damage with detonation is the ring lands. Often times under the sharp pressure spikes, you will end up with crushed or broken ring lands. In less severe cases, you will still see broken rings. This is often the case with cast pistons as they were never meant to withstand those kind of cylinder pressures, especially such sudden and violent changes in pressure.
With detonation also comes heat. The pressure spikes break down the boundary layer of gas that quenches the flame front and keeps the relatively cool piston protected from the relatively hot combustion. As this boundary layer is broken down and more and more heat is absorbed into the piston, you will see deformation of the piston and cylinder wall scoring which inevitably leads to the need for a rebuilt engine. You will also see higher coolant temps because of this as well since the coolant system is having to do more work with the excess heat.
Indicators:
Higher levels of detonation are audible and will sound like quarters tapping on glass. You can’t really hear it in most new cars due to the insulated cabins, so when you do hear it, it is most likely a higher level of knock. If you have a tuning device which monitors knock retard, such as the Cobb Tuning Accesport, you are only seeing the engines response to perceived noise. It is a good idea to pay attention though as the engine, for whatever reason, is trying to save itself from something by reducing spark advance.
Pre-Ignition:
Definition:
Pre-Ignition is the ignition of the air/fuel mixture before the spark plug fires.
A pre-ignition combustion event looks something like this…
The fuel air mixture enters the combustion chamber as the piston is on its downward intake stroke. The piston then returns upward for the compression stroke. The more compressed the mixture is, the more difficult it is to ignite it, therefore, with the piston on the low side of the compression stroke, the mixture is actually easier to ignite than when it gets closer to Top Dead Center (TDC). A hot spot in the cylinder such as a glowing spark plug tip can ignite this low compression mixture very early, before the spark plug fires. Now the upward motion of the piston is fighting the combustion force that is expanding. This adds a tremendous amount of heat and load to the engine, and for this reason pre-ignition is responsible for much higher cylinder pressures than detonation. The pressure from pre-ignition is not rapid like it is with detonation. Instead, the pressure is very high and has a much longer duration.
Damage:
Damage from pre-ignition is much more severe and instantaneous than that from detonation. Typically, with pre-ignition, you will see holes melted in pistons, spark plugs melted away, and engine failure happens pretty much immediately.
Due to the longer duration of the heat and pressure bought on by pre-ignition, you will notice a lot more melted parts, whereas, with detonation, you get more parts that are just blown apart.
Indicators:
There are not really any early warning signs with pre-ignition. The best thing you can do to prevent it, is to make sure the engine is set up as well as it can be to minimize potential hot spots. From the OEM, cars come with appropriate spark plug heat ranges and everything setup to minimize/eliminate pre-ignition. So making sure you have the right spark plugs and the right gaps is important when doing plug swaps and adding more boost and therefore higher cylinder temps to your combustion chamber.
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George
9/23/2021, 3:19:48 AM
Very nicely explained in simple and understandable terms. Very much appreciated. This is the first site after much searching that I feel accurately explained the 2 scenarios.
11/6/2021, 8:43:57 AM
This article was a perfect tool to help explain these issues to a coworker who is learning how to diagnose small engines thru failure analysis. (We finally have a backpack blower that has detonation happening) The explanations here are easy to understand and follow, and show the clear difference between the two.
Nick Reynolds
2/15/2022, 12:33:30 PM
I've been reading "The Secret Horsepower Race" Highly recommended. The author used the terms preignition and detonation frequently I could not understand teh difference.
Now I do - many thanks.
Vince |
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190103 | https://mathworld.wolfram.com/NaturalNumber.html | Natural Number
The term "natural number" refers either to a member of the set of positive integers 1, 2, 3, ... (OEIS A000027) or
to the set of nonnegative integers 0, 1, 2,
3, ... (OEIS A001477; e.g., Bourbaki 1968,
Halmos 1974). Regrettably, there seems to be no general agreement about whether to
include 0 in the set of natural numbers. In fact, Ribenboim (1996) states "Let
be a set of natural numbers; whenever
convenient, it may be assumed that ."
The set of natural numbers (whichever definition is adopted) is denoted N.
Due to lack of standard terminology, the following terms and notations are recommended in preference to "counting number,"
"natural number," and "whole number."
| | | |
---
| set | name | symbol |
| ..., , , 0, 1, 2, ... | integers | Z |
| 1, 2, 3, 4, ... | positive integers | Z-+ |
| 0, 1, 2, 3, 4, ... | nonnegative integers | Z- |
| 0, , , , , ... | nonpositive integers | |
| , , , , ... | negative integers | Z-- |
See also
Counting Number, Integer, N, Positive, Z,
Z--, Z-+, Z-
Explore with Wolfram|Alpha
More things to try:
is 1 a natural number
sum of all natural numbers from 1 to 100
1111
References
Bourbaki, N. Elements of Mathematics: Theory of Sets. Paris, France: Hermann, 1968.Courant,
R. and Robbins, H. "The Natural Numbers." Ch. 1 in What
Is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed. Oxford,
England: Oxford University Press, pp. 1-20, 1996.Halmos, P. R.
Naive
Set Theory. New York: Springer-Verlag, 1974.Ribenboim, P. "Catalan's
Conjecture." Amer. Math. Monthly 103, 529-538, 1996.Sloane,
N. J. A. Sequences A000027/M0472
and A001477 in "The On-Line Encyclopedia
of Integer Sequences."Welbourne, E. "The Natural Numbers."
Referenced on Wolfram|Alpha
Natural Number
Cite this as:
Weisstein, Eric W. "Natural Number." From
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Quantum Numbers and their types?
Aug 11, 2022
Quantum Numbers
An atom can consist of a large number of orbitals. Based on the quantity, these orbitals can be differentiated by their shape, size, and direction of the electron or their spin, i.e., orientation.
The smaller the size of an orbital, the more the chances of finding the electrons near the nucleus. Likewise, the shape and orientation of an atom inform the probability of locating the electron in a fixed direction more than others. Atomic orbitals are accurately distinguished by some numbers, which are called quantum numbers.
This article will inform you about quantum numbers and how to calculate them.
Quantum Numbers and Their Types
Quantum numbers are the integral numbers used to describe the precise information of an electron in an atom. Simply put, quantum numbers are used to explain the correct address of an electron.
A quantum number narrates
The energy of an electron in orbit.
The position or distance of an electron from the nucleus.
The shape and number of orientations of an electron cloud around its axis.
The direction of the spinning of the electron around its axis.
These are the four quantum numbers, i.e., principal quantum number (n), magnetic quantum number (m), azimuthal quantum number (l), and spin quantum number (s). Therefore, it can say that to describe an electron in an atom completely, four quantum numbers are required.
Principal Quantum Number (n)
The principal quantum number gives the following information:
It determines the energy of an electron and the average distance of an electron from the nucleus. The principal quantum number tells the principal energy level of the electron. It can have any whole number value such as 1, 2, 3, 4, etc.
The energy levels or shells corresponding to these numbers are designated as K, L, M, N, etc. As the value of n increases, the electron gets farther from the nucleus, increasing its energy. The higher the value of n, the higher the electronic energy of the electron. Therefore, it is less tightly held with the nucleus.
The energy of the electron in a hydrogen atom is related to the principal quantum number by the following relation:
Eₙ = – k² [2𝜋²me⁴Z²] / [n² h²]
Where m = mass of an electron
e = charge on electron
h = Planck’s Constant
Eₙ = energy of the electron in nth principle shell in Joule
n = Principal quantum number used to designate the principal shell
k = Coulomb’s law constant
This relation is similar to the expression given by Bohr.
The principal quantum number, n, also gives the maximum number of electrons that may accommodate a given shell. The maximum number of electrons in the nth shell = 2n².
Azimuthal Quantum Number (l)
This quantum number is also called the angular quantum number. It decides the angular momentum of an electron and is denoted by l. The value of l gives the sublevel or subshell in which the electron is located.
This quantum number also decides the shape of the orbital in which the electron is located. The number of subshells with a principal shell is determined by the value of n for that principal energy level. Therefore, l may have all possible whole number values from 0 to (n – 1) for each principal energy level.
This quantum number directs the magnitude of the orbital angular momentum. Therefore, it is also called orbital angular momentum quantum number. The maximum number of electrons that may accommodate a given subshell equals 2(2l + 1).
Azimuthal quantum number rules:
For n = 1, l will be 0. It means that an electron in the first energy level can be present only in an s -shell. So, the first energy level has solely one subshell, i.e., 1s. It is spherical.
For n = 2, l will have values 0 and 1. It means that the electron in the second principal energy level may be either in a spherical s-subshell or p-subshell. So, the second energy level has only two subshells, i.e., 2s & 2p. The p-subshells are dumb-bell.
For n = 3, l can have values 0, 1, and 2. An electron in the third principal energy level may either be in the s-subshell or p-subshell. So the third energy level has three subshells that are 3s, 3p, & 3d. The d-subshells are leaf-like.
Similarly, the fourth energy level can have four subshells, i.e., 4s, 4p, 4d, & 4f.
The relation between orbital angular momentum and azimuthal quantum number, l is
Orbital angular momentum = [h/ 2𝜋] √[l (l + 1)]
Magnetic Quantum Number (mₗ)
This quantum number refers to the different orientations of the electron cloud in a particular subshell. In other words, the direction of the angular momentum is determined by the magnetic quantum number (mₗ).
The number of orbitals in a specific subshell within a principal energy level is stated by the number of values sanctioned to mₗ, which in turn hangs on the values of l. The possible values of mₗ range +l to -l, including 0 values. The total number of mₗ values gives the total number of orbitals within a given subshell.
Magnetic quantum number rules:
For l = 0, i.e., s-subshell, mₗ = 0. Therefore, the s-subshell consists solely of one orbital.
For l = 1, i.e., p-subshell, mₗ can be -1, 0, and +1. Therefore, the p-subshell consists of three orbitals.
For l = 2, i.e., d-subshell, mₗ can be -2, -1, 0, +1, and +2. Therefore, the d-subshell consists of five orbitals.
For l = 3, i.e., f-subshell, mₗ can be -3, -2, -1, 0, +1, +2, and +3. Therefore, the f-subshell consists of seven orbitals.
Spin Quantum Number (mₛ)
This quantum number is denoted by mₛ. It does not arise from the mechanical wave treatment but the spectral proof that an electron, in addition to its revolving motion about the nucleus, also spins about its axis.
The spin quantum number tells the direction of spin of the electron. The spin can either be anticlockwise or clockwise. The spin quantum number can have only two values which are +1/2 and -1/2. If the positive value indicates clockwise spin, i.e. ↑, upward arrow, and is known as 𝛼-spin. The negative value indicates anti-clockwise spin, i.e. ↓, downward arrow, known as 𝛽-spin.
Quantum Numbers Chart
The relationship between all quantum numbers is represented with the help of a quantum numbers chart.
| | | | | | | | |
--- --- --- --- |
| n | l | mₗ | mₛ | Number of Orbitals | Orbital Name | Number of Electrons | Total Electrons |
| (K shell) | 0 | 0 | +½, -½ | 1 | 1s | 2 | 2 |
| (L shell) | 0 | 0 | +½, -½ | 1 | 2s | 2 | 8 |
| | 1 | -1, 0, +1 | +½, -½ | 3 | 2p | 6 | |
| (M shell) | 0 | 0 | +½, -½ | 1 | 3s | 2 | 18 |
| | 1 | -1, 0, +1 | +½, -½ | 3 | 3p | 6 | |
| | 2 | -2, -1, 0, +1, +2 | +½, -½ | 5 | 3d | 10 | |
Information Provided by Four Quantum Numbers
The four quantum numbers give the following information :
n defines the shell number and determines the size of the orbital and the orbital’s energy to a large extent.
There are many subshells in the nth shell. In which subshell the electron lies and decides the shape of the orbital is pointed out by l. There are (2l+1) types of orbitals in a subshell, i.e., the value of l for a one s orbital is 0, for three p orbitals is 1, and for five d orbitals is 2 per subshell. To some limit, l also decides the energy of the orbital in a numerous electrons atom.
mₗ appoints the orientation of the orbital. For a given value of l, mₗ has (2l+1) values, the same as the orbitals’ number per subshell. It suggested that the orbitals’ number equals a lot of ways in which they are positioned.
mₛ refers to the orientation of the spin of the electron.
Conclusion
From the above discussion, it is concluded that four types of quantum numbers allocate the position of an electron in an atom. When the Schrodinger wave equation is calculated for the wave function, 𝛹, the values of three quantum numbers, i.e., n, l, and m, become easy to calculate. Spin quantum number (s) arise due to the electron’s spinning about its axis.
With the help of a quantum numbers chart, you can understand the values of all quantum numbers and memorize them most simply.
Frequently Asked Questions
1. Why is it impracticable to get a principal quantum number as zero or a negative value?
The principal quantum number describes an electron’s electron shell or energy level. It is not feasible for an electron to possess the value of the principal quantum number, n, as zero or a negative value. It is because an atom can’t have a negative value or no value for a principal shell.
2. What are the assigned values of all the four quantum numbers?
The assigned values of all four quantum numbers are:
A principal quantum number, i.e., ‘n’, can take all the natural number values.
Magnetic quantum number, i.e., ‘mₗ’, can take values from -l to +l, depending upon ‘l’ and hence, ‘n’ too.
Azimuthal quantum numbers, i.e., ‘l’, can take values from 0 to (n – 1).
The spin quantum number for an electron is either -1/2 or 1/2.
3. Which quantum number decides the magnetic property of an atom?
Due to spin, the electron behaves as a tiny magnet. The electron’s spin is responsible for the magnetic properties of atoms, and molecules of a substance are paired. It behaves as diamagnetic. It is because the magnetic field weakly repels them.
Whereas if atoms or molecules of a substance have one or more unpaired or odd electrons, it behaves as a paramagnetic substance. It is because the magnetic field weakly attracts them.
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190105 | https://www.quora.com/How-can-unknown-quantities-in-geometric-problems-be-solved | Something went wrong. Wait a moment and try again.
Problem Solving
Unknown Numbers
Mathematical Problems
Mathematical Concepts
Geometric Equations
Geometry Word Problems
Geometric Mathematics
SPATIAL PROBLEMS
5
How can unknown quantities in geometric problems be solved?
·
Solving for unknown quantities in geometric problems typically involves a combination of geometric principles, algebraic techniques, and sometimes trigonometric relationships. Here’s a structured approach to tackling these problems:
Understand the Problem
Carefully read the problem statement to identify the given information and what you need to find.
Sketch the geometric figure if it’s not provided, labeling all known quantities.
Identify Relevant Geometric Principles
Use properties of shapes (triangles, circles, polygons, etc.) and their relationships.
Apply theorems such as:
Pythagorean Theo
Solving for unknown quantities in geometric problems typically involves a combination of geometric principles, algebraic techniques, and sometimes trigonometric relationships. Here’s a structured approach to tackling these problems:
Understand the Problem
Carefully read the problem statement to identify the given information and what you need to find.
Sketch the geometric figure if it’s not provided, labeling all known quantities.
Identify Relevant Geometric Principles
Use properties of shapes (triangles, circles, polygons, etc.) and their relationships.
Apply theorems such as:
Pythagorean Theorem for right triangles: a2+b2=c2
Area and perimeter formulas for different shapes.
Properties of similar and congruent figures.
Set Up Equations
Translate the geometric relationships into algebraic equations. This might involve:
Setting up equations based on the properties of similar triangles (e.g., ratios of corresponding sides).
Using the area or perimeter formulas where necessary.
Employing trigonometric ratios (sine, cosine, tangent) for angles and sides in right triangles.
Solve the Equations
Use algebraic techniques to solve for the unknowns. This can include:
Combining like terms.
Factoring or using the quadratic formula if necessary.
Isolating variables to find their values.
Check Your Work
Verify that the solution makes sense in the context of the problem.
Check that all dimensions and units are consistent.
If applicable, substitute back into original equations to ensure they hold true.
Example Problem
Problem: A right triangle has one leg measuring 6 cm, and the hypotenuse measures 10 cm. What is the length of the other leg?
Solution Steps:
1. Understand the Problem: We need to find the length of the unknown leg (let’s call it b).
2. Identify Relevant Geometric Principles: Use the Pythagorean Theorem: a2+b2=c2.
3. Set Up the Equation:
- Given a=6 cm and c=10 cm, the equation becomes:
62+b2=102
4. Solve the Equation:
36+b2=100b2=100−36b2=64b=8 cm
5. Check Your Work: The dimensions satisfy the Pythagorean theorem, confirming the solution is correct.
This systematic approach can be adapted to a wide range of geometric problems involving unknown quantities.
Steve Sparling
BSc in Mathematics, The University of British Columbia
·
Author has 365 answers and 280.2K answer views
·
4y
The beauty of geometry is that there are so many different interesting problems. Some problems involve a number of strategies.
For starters look at what you are given and what you want to find. What you want is a numeric path from the given to the end so look for obvious consequences of the given. Now go to the end and work backwards - I can figure out this number if I only knew that number. I find working backwards is a great strategy.
In order to work the above you need to look at your toolbox. If there are right triangles will you use Pythagoras or trigonometry or something else. Can you draw
The beauty of geometry is that there are so many different interesting problems. Some problems involve a number of strategies.
For starters look at what you are given and what you want to find. What you want is a numeric path from the given to the end so look for obvious consequences of the given. Now go to the end and work backwards - I can figure out this number if I only knew that number. I find working backwards is a great strategy.
In order to work the above you need to look at your toolbox. If there are right triangles will you use Pythagoras or trigonometry or something else. Can you draw a line to make a right triangle? If there is a circle then thing of your circle rules. Think of your other tools: parallel lines; vertical angles etc.
If it is a good problem nothing is obvious. To see what needs to be done get some paper and coloured pencils and start drawing. At some point a potential solution will seem to emerge. If that solution doesn’t work then you can eliminate it but don’t forget it as changing it a little may yield a solution.
I hope this helps. The best way to become a geometry problem solver is to work lots of problems and develop your own strategies.
Sachin Tiwary
4y
Geometry is a branch of mathematics which serves as a connecting link between the real world and our theoretical understanding of mathematics which we study in our mathematics textbook. If you have a good understanding of geometry ( it should be at your fingertips ) and algebra, then using the combined concepts from the two ( i.e. geometry and algebra ) , you can convert an unknown quantity in geometry ( or from a real world situation ) into an algebric equations. Then you can obtain the value of the unknown quantity by solving the equation by using various methods of solving.
Related questions
How can real life problems involving unknown quantities in geometric problems be formulated and solved?
How can challenging problems involving geometric figures be analyzed and solved?
How can I solve geometric problems mentally?
How can I solve the geometric problem?
How do you solve a difficult problem in geometry?
Herbert Levinson
Former Retired Maintenance Tech
·
Author has 225 answers and 118.1K answer views
·
4y
That's what Co-ordinated Geometry and Trigonometry is all about. The geometry is Co-ordinated on graphs, on paper and then in your mind. Along with proven theorems.
Bernd Leps
Former scientific official; retired
·
Author has 5.7K answers and 1.2M answer views
·
2y
Originally Answered: What is the formula for finding and solving for an unknown in geometry?
·
Geometry is not algebra and not arithmetics.
There are ways to transfer geometrical problems into arithmetical ones - especially coordinate systems and vectors -, but in general geometry will not work with formulas.
That will not say, that for standard problems there are not standard ways of solution, for which you can get a “recipe”.
For example, in Euclidean geometry (compass and ruler without divisions only, no coordinate system, planar field of work) nearly any given problem can be solved by connecting and coupling single steps or building bricks like
construct a straight line out of two point
Geometry is not algebra and not arithmetics.
There are ways to transfer geometrical problems into arithmetical ones - especially coordinate systems and vectors -, but in general geometry will not work with formulas.
That will not say, that for standard problems there are not standard ways of solution, for which you can get a “recipe”.
For example, in Euclidean geometry (compass and ruler without divisions only, no coordinate system, planar field of work) nearly any given problem can be solved by connecting and coupling single steps or building bricks like
construct a straight line out of two points,
construct a circle around a given center with a given radius or through a given second point,
construct a straight line orthogonal to a given straight line and through a given point (not being at the given line),
construct a stright line parallel to a given line through a given point (not being at the given line),
transport a given distance as to start now from a given point on a given line,
construct a rectangular triangle, if the longest and another side are given,
multiply a given distance by an integer factor,
divide a given distance in an integer number of equal parts,
transport a given angle now starting with a given point and having a given line as side,
multiply an angle by an integer factor,
divide a given angle by two,
transfer an area given as rectangle into a square,
transfer the area of a given square into an rectangle, of which one side is given,
transfer the area of two given squares into one square,
construct a regular hexagon and a regular triangle, if a side is given,
costruct triangles out of three given sides or two given sides and the angle between them or one side and the two angles at its ends or one side and two angles, of which one is not attached to the given side or two sides and an angle not between them,
and the like (here given with increasing difficulty) and try to solve the given problem by dismembering it into such steps.
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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Michael Paglia
Former Journeyman Wireman IBEW
·
Author has 33.3K answers and 5.1M answer views
·
1y
Originally Answered: Can you suggest an interesting method for solving unknown values in geometric problems?
·
You don't find any values in plane geometry
It's all.on a compass
Polar you can use algebra because it's on a grid and there are values there
Your response is private
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Related questions
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Trevor
B.A. with MMath in Mathematics, University of Cambridge (Graduated 2023)
·
Author has 1.3K answers and 5.5M answer views
·
Updated 8y
Related
How can I solve the geometric problem?
Finally! The answer is 5√3 !
This is a pure geometric approach, but it still requires a clear mindset and knowledge of quite a few important theorems.
Let us denote P as the intersection of AM and DN, and Q as the intersection of AN and EM. X will be the intersection of the three lines. S will be the intersection of the altitude and MN.
An important dimension that you might take note of is that BC is divided by the altitude into length of 1 and 4 respectively. This can be proven by cosine law or simply the Pythagorean theorem.
Step 1 Identify the two right angled triangles are similar
This sho
Finally! The answer is 5√3 !
This is a pure geometric approach, but it still requires a clear mindset and knowledge of quite a few important theorems.
Let us denote P as the intersection of AM and DN, and Q as the intersection of AN and EM. X will be the intersection of the three lines. S will be the intersection of the altitude and MN.
An important dimension that you might take note of is that BC is divided by the altitude into length of 1 and 4 respectively. This can be proven by cosine law or simply the Pythagorean theorem.
Step 1 Identify the two right angled triangles are similar
This should be easy because the two angles (right angle) and another angle (∠ADM and ∠ANE) are given to be equal. So we have
[math]\displaystyle \boxed {\frac {AD}{AN}=\frac {AM}{AE}} \tag{}[/math]
Step 2 Identify another pair of similar triangle
This is a difficult step. Since we have established the ratios in Step 1, we can conclude another pair of similar triangle based on an additional infornation:
[math]\angle DAN = \angle CAB+90^\circ = \angle EAM \tag{}[/math]
So we also have
[math]\displaystyle \boxed {\triangle DAN \sim \triangle MAE} \tag{}[/math]
Step 3 Recognize the two hidden circles
This step is extremely difficult. But from the similarity, we have
[math]\angle ADN=\angle AME \text { and } \angle AEM=\angle AND \tag{}[/math]
Because of converse of angle in the same segment we have ADMX and AXNE to be cyclic quadrilateral.
Step 4 Recognize the right angles
This step follows from the cyclic quadrilateral conclusion. Since we have angle in the same segment we also have
[math]\boxed {\angle DXM=\angle EXN=90^\circ} \tag{}[/math]
Step 5 See that we have yet another pair of similar triangles
Or alternatively, if you already know how to construct the geometric mean of two numbers using straight edge and compass, you should know that [math]SX^2=MS\times SN[/math], but you can recognize the pair of similar triangles of
[math]\triangle XMS \sim \triangle NXS \sim \triangle NMX \tag{}[/math]
to obtain that. Since we know that [math]BS=0.5[/math] and [math]SC=2[/math], we have [math]SX=\sqrt {0.5\times 2}=1[/math]
Therefore,
[math]\displaystyle \boxed {XN=\sqrt 5}\tag{}[/math]
[math]\displaystyle \boxed {XM=\frac {\sqrt 5}2} \tag{}[/math]
Step 6 The final pairs of similar triangles
Since we have proved the cyclic quadrilateral, so we also have [math]\angle XAN=\angle XEN[/math] and [math]\angle XDM=\angle XAM[/math]. Together with the right angle we have discovered earlier, we have found the final pairs of similar triangles:
[math]\begin {align} \triangle DXM&\sim \triangle ASM\ \triangle EXN&\sim \triangle ASN \end {align}[/math]
With the values we have found for [math]XM[/math] and [math]XN[/math], we can find [math]XD[/math] and [math]XE[/math] using the ratios.
[math]XD= 2\sqrt {15} \text { and } XE=\sqrt {15}[/math]
[math]DE=\sqrt {5}\sqrt {15}=\boxed {5\sqrt 3}[/math]
The typing is a bit of work…
[math]\huge \ddot \smile \tag{}[/math]
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Dean Rubine
Programming since 1975
·
Author has 10.5K answers and 23.4M answer views
·
5y
Related
How would I solve this geometry problem?
I promised Paul I’d look at this one two or three weeks ago and I’m just getting to it. Sorry Paul.
I pasted Paul’s figure on the grid; I reflected and rotated to get it in position.
In the figure we have [math]p=\tan 45^\circ=1[/math], [math]q=\tan 30^\circ = 1/\sqrt{3}[/math] but let’s do it in general to start. Let’s say [math]BC=s[/math] in general, [math]s=12[/math] at the end. Our unknowns are the slope [math]m, [/math] lengths [math]b[/math] and [math] d.[/math]
[math]A(2b,2mb)[/math] is on [math]y=p(x+d)[/math]
[math]2mb = p(2b+d) = 2pb + pd[/math]
That’s one equation.
[math]B (b,mb)[/math] is on [math]y=q(x+d)[/math]
[math]mb=q(b+d) = qb + qd[/math]
That’s a second equation. The final one is [math]BC^2=s^2[/math]
[math]b^2 + m^2b^2 = s^2[/math]
mathb^2 = s^2[/math]
OK, let’s try to solve th
I promised Paul I’d look at this one two or three weeks ago and I’m just getting to it. Sorry Paul.
I pasted Paul’s figure on the grid; I reflected and rotated to get it in position.
In the figure we have [math]p=\tan 45^\circ=1[/math], [math]q=\tan 30^\circ = 1/\sqrt{3}[/math] but let’s do it in general to start. Let’s say [math]BC=s[/math] in general, [math]s=12[/math] at the end. Our unknowns are the slope [math]m, [/math] lengths [math]b[/math] and [math] d.[/math]
[math]A(2b,2mb)[/math] is on [math]y=p(x+d)[/math]
[math]2mb = p(2b+d) = 2pb + pd[/math]
That’s one equation.
[math]B (b,mb)[/math] is on [math]y=q(x+d)[/math]
[math]mb=q(b+d) = qb + qd[/math]
That’s a second equation. The final one is [math]BC^2=s^2[/math]
[math]b^2 + m^2b^2 = s^2[/math]
mathb^2 = s^2[/math]
OK, let’s try to solve this non-linear system. Eliminate [math]d[/math] first,
[math]mpb= pqb + pqd[/math]
[math]2mqb = 2pqb + pqd[/math]
[math]mb(2q-p) = pqb[/math]
We can assume [math]b\ne 0[/math] and eliminate [math]b[/math] at the same time.
[math]m = \dfrac{pq}{2q-p}[/math]
Interesting. We get the slope independently of the length of AB=BC.
[math]b^2 = \dfrac{s^2}{1+m^2}[/math]
[math]2mb = 2pb + pd[/math]
[math]d = 2b(m-p)/p = 2b(m/p - 1)[/math]
The thing we’re after is [math]AD^2 = (2b + d)^2 + (2mb)^2[/math]
I won’t try to simplify.
OK, let’s substitute in [math]p=\tan 45^\circ=1[/math], [math]q=\tan 30^\circ=1/\sqrt{3}, s=12[/math]
[math]m = \dfrac{pq}{2q-p} = \dfrac{1/\sqrt{3}}{2/\sqrt{3}-1} = \dfrac{1}{2 -\sqrt{3}} \dfrac{2+\sqrt{3}}{2+\sqrt{3}} [/math]
[math]m = 2+\sqrt{3}[/math]
Rather than use the rest of the solution we just worked out, we have the shortcut through the trig tunnel. It happens that [math]\tan 75^\circ = m;[/math] let’s show that.
[math]\sin 75^\circ=\sin (45^\circ+30^\circ) = \sin 45 \cos 30 + \cos 45 \sin 30 \\quad = \dfrac{\sqrt{2}}{2} \left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\right)=\frac 1 4(\sqrt{6}+\sqrt{2})[/math]
[math]\cos 75 =\cos (45+30) = \cos 45 \cos 30 - \sin 45 \sin 30 \\quad = \dfrac{\sqrt{2}}{2} \left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}\right)=\frac 1 4(\sqrt{6}-\sqrt{2})[/math]
[math]\tan 75^\circ = \dfrac{ \sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}} \cdot \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} = \dfrac{8+2\sqrt{12}}{4} = 2 + \sqrt{3}[/math]
So
[math]\arctan m = 75^\circ[/math]
exactly. We didn’t really need to find the angle to find
[math]\sin C = \frac 1 4(\sqrt{6}+\sqrt{2})[/math]
which we could have (rather tediously) calculated from the tangent [math]2 + \sqrt{3}[/math].
The Law of Sines gives our answer,
[math]\dfrac{\sin(180^\circ - 75^\circ)}{AD} = \dfrac{\sin 45^\circ}{24}[/math]
[math]AD = \dfrac{ 24 \sin 75 }{\sin 45 } [/math]
[math]AD = \dfrac{24 \cdot \frac 1 4 (\sqrt{2} + \sqrt{6})}{1/\sqrt{2}} = 6(2 + \sqrt{12})[/math]
[math]AD = 12 + 12\sqrt{3}[/math]
I’ll refrain from the pointless adding up of the numbers.
Hugh Bothwell
Studied Electrical Engineering
·
Author has 2.6K answers and 1.5M answer views
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5y
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How do you solve a difficult problem in geometry?
Always start by sketching a diagram.
I like to think of the initial model as being “floppy” - the lines and intersections can shift around. As you add constraints - lengths, angles, relationships - it “stiffens” your model, reducing the ways it can move.
Now you need to think about how different pieces of the model interact. When I change this angle, what effect does it have on that one? If I shorten this line, does it make that one longer? Look for invariants - is this angle always the same as that one, or the complement of it?
Finally, think about your desired final result. What value do I need
Always start by sketching a diagram.
I like to think of the initial model as being “floppy” - the lines and intersections can shift around. As you add constraints - lengths, angles, relationships - it “stiffens” your model, reducing the ways it can move.
Now you need to think about how different pieces of the model interact. When I change this angle, what effect does it have on that one? If I shorten this line, does it make that one longer? Look for invariants - is this angle always the same as that one, or the complement of it?
Finally, think about your desired final result. What value do I need? What does that value depend on? How can I get the inputs I need to produce that output?
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Aaron Briseno
B.S in Mathematics & Teaching, University of California, Los Angeles (Graduated 2010)
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Author has 1.3K answers and 3.1M answer views
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4y
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How would this geometry problem be solved?
With similar triangles, drop a perpendicular line from where the transversal crosses the first horizontal line to the second horizontal line(as shown), then drop another perpendicular from the second horizontal line to the third. It’ll look like:
Note the triangles formed are similar and so:
[math]\dfrac{x}{12}=\dfrac{6}{x+14}[/math]
Cross multiply
[math]x(x+14)=72[/math]
This might seem odd to you but I’d list the factors of 72: (1, 72) (2, 36) (3, 24) (4, 18) DONE the last one!
Do you see it? To be fair I’ve done this a bit; you see, the equation to me is read “you need two numbers who are 14 units apart that multiply to 7
With similar triangles, drop a perpendicular line from where the transversal crosses the first horizontal line to the second horizontal line(as shown), then drop another perpendicular from the second horizontal line to the third. It’ll look like:
Note the triangles formed are similar and so:
[math]\dfrac{x}{12}=\dfrac{6}{x+14}[/math]
Cross multiply
[math]x(x+14)=72[/math]
This might seem odd to you but I’d list the factors of 72: (1, 72) (2, 36) (3, 24) (4, 18) DONE the last one!
Do you see it? To be fair I’ve done this a bit; you see, the equation to me is read “you need two numbers who are 14 units apart that multiply to 72.” we found the two numbers 4 and 18, so
[math]x=4[/math]
Edit: I drew my perpendiculars different than I said them so I corrected the words to match the picture :)
Michael Grady
Former OS Software, Tutor, Math/Phys/Chem/Lang Teacher at Apple (company) (1995–2022)
·
Author has 1.3K answers and 1.1M answer views
·
Updated 10mo
Related
Can you provide examples of geometric problems that can be solved using vectors?
(A) A basic, but fun one, is the proof of the MidPoint Theorem: The line segment between Midpoints of 2 sides of a triangle has the length of half the 3rd side, and is parallel to it.
The work is easy and fun, but a picture from my old Calculus book is easier.
DE = (1/2)AC means that DE and AC are parallel, and |DE| = (1/2)|AC|
(B) Another fun exercise would be…
Show that the Midpoints of the sides of any Quadrilateral are the vertices of a Parallelogram…
[Update]
OK, (B) was an exercise in my old Calculus book, so I messed around with it for fun. Here we go…
A = vector OA, B = OB, etc. B, D, H, F
(A) A basic, but fun one, is the proof of the MidPoint Theorem: The line segment between Midpoints of 2 sides of a triangle has the length of half the 3rd side, and is parallel to it.
The work is easy and fun, but a picture from my old Calculus book is easier.
DE = (1/2)AC means that DE and AC are parallel, and |DE| = (1/2)|AC|
(B) Another fun exercise would be…
Show that the Midpoints of the sides of any Quadrilateral are the vertices of a Parallelogram…
[Update]
OK, (B) was an exercise in my old Calculus book, so I messed around with it for fun. Here we go…
A = vector OA, B = OB, etc. B, D, H, F are midpoints… “bold” means vector.
B = A + (1/2)(C - A) = (1/2)(A + C)
Similarly, D = (1/2)(C + E), F = (1/2)(E + G), H = (1/2)(A + G)
B - D = (1/2)(A + C) - (1/2)(C + E) = (1/2)(A - E)
H - F = (1/2)(A + G) - (1/2)(E + G) = (1/2)(A - E)
Therefore Vectors (B - D) = (H - F)
Similarly, we have (D - F) = (1/2)(C - G) = (B - H)
Conclusion: Opposite sides of quadrilateral BDFH, formed from midpoints of arbitrary quadrilateral ACEG, are parallel and have equal magnitudes ==> BDFH is a parallelogram !
The OP question was Geometric problems solved by Vectors. I believe these 2 results are a good introduction.
Peter Wengert
Author has 73 answers and 111K answer views
·
Updated 5y
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How do I solve this problem? It is a problem related to geometry.
I’ve drawn out a diagram of the problem for you here.
The task is to prove that the area shaded pink is the same as the area shaded blue.
Now let x be the distance of D from AB, let y be the distance of C from AB, let z be the distance of F from AB, and let l be the length of EB like so:
The area of a triangle is just its base multiplied by its height.
The area of ∆EBC is [math]\frac{Iy}{2}[/math], the area of ∆EAD is [math]\frac{Ix}{2}[/math], and the area of both ∆EBF and ∆EAF is [math]\frac{Iz}{2}[/math].
The thing to notice is that z is just the average of x and y, so the area of both ∆EBF and ∆EAF can be rewritten as [math]\frac{I\frac{x[/math]
I’ve drawn out a diagram of the problem for you here.
The task is to prove that the area shaded pink is the same as the area shaded blue.
Now let x be the distance of D from AB, let y be the distance of C from AB, let z be the distance of F from AB, and let l be the length of EB like so:
The area of a triangle is just its base multiplied by its height.
The area of ∆EBC is [math]\frac{Iy}{2}[/math], the area of ∆EAD is [math]\frac{Ix}{2}[/math], and the area of both ∆EBF and ∆EAF is [math]\frac{Iz}{2}[/math].
The thing to notice is that z is just the average of x and y, so the area of both ∆EBF and ∆EAF can be rewritten as [math]\frac{I\frac{x+y}{2}}{2}[/math] = [math]\frac{I(x+y)}{4}[/math]
The blue region therefore has an area of [math]2\frac{I(x+y)}{4}[/math] - (Area of ∆EBH + Area of ∆EAG)
The red region has a total area of [math]\frac{Iy}{2} + \frac{Ix}{2}[/math] - (Area of ∆EBH + Area of ∆EAG)
They both have ∆EBH and ∆EAG added to their areas, so proportionately speaking, they can be ignored.
The area of the blue region, [math]\frac{2I(x+y)}{4}[/math] simplifies into [math]\frac{I(x+y)}{2}[/math]
The area of the red region, [math]\frac{Iy}{2} + \frac{Ix}{2}[/math] also simplifies into [math]\frac{I(x+y)}{2}[/math]
Kranti Kumar Udayasingh
5+ years in IT Services & 3+ yrs in Professional Photography
·
8y
Related
Can someone solve this geometry problem?
This is otherwise called as
Langley’s Adventitious Angles
or 80° - 80° - 20° triangle problem. Found the solution from :—
The 80-80-20 Triangle Problem, Solution #1.
Calculate some known angles:
ACB = 180-(10+70)-(60+20) = 20°
AEB = 180-60-(50+30) = 40°
Draw a line from point E parallel to AB, labeling the intersection with AC as a new point F and conclude:
FCE ACB
CEF = CBA = 50+30 = 80°
FEB = 180-80 = 100°
AEF = 100-40 = 60°
CFE = CAB = 60+20 = 80°
EFA = 180-80 = 100°
Draw a line FB labeling the intersection with AE as a new point G and conclude:
AFE BEF
AFB = BEA = 40°
BFE = AEF = 60°
FGE = 180-6
This is otherwise called as Langley’s Adventitious Angles or 80° - 80° - 20° triangle problem. Found the solution from :— The 80-80-20 Triangle Problem, Solution #1.
Calculate some known angles:
ACB = 180-(10+70)-(60+20) = 20°
AEB = 180-60-(50+30) = 40°
Draw a line from point E parallel to AB, labeling the intersection with AC as a new point F and conclude:
FCE ACB
CEF = CBA = 50+30 = 80°
FEB = 180-80 = 100°
AEF = 100-40 = 60°
CFE = CAB = 60+20 = 80°
EFA = 180-80 = 100°
Draw a line FB labeling the intersection with AE as a new point G and conclude:
AFE BEF
AFB = BEA = 40°
BFE = AEF = 60°
FGE = 180-60-60 = 60° = AGB.
ABG = 180-60-60 = 60°
Draw a line DG. Since AD=AB (leg of isosceles) and AG=AB (leg of equilateral), conclude:
AD = AG.
DAG is isosceles
ADG = AGD = (180-20)/2 = 80°
Since DGF = 180-80-60 = 40°, conclude:
FDG (with two 40° angles) is isosceles, so DF = DG
With EF = EG (legs of equilateral) and DE = DE (same line segment) conclude:
DEF DEG by side-side-side rule
DEF = DEG = x
FEG = 60 = x+x
Answer X = 30 degree .
Source : World's Hardest Easy Geometry Problem
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190106 | https://www.facebook.com/groups/2692404370981189/posts/4225709047650706/ | Mathematics for Learning | For what values of m and n ,limit of the function (m.e^x -n.cosx)/x is finite non zero number, | Facebook
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Values of m and n for finite limit of function?
Summarized by AI from the post below
Mathematics for Learning
Sharafat Rasool
· August 10 ·
For what values of m and n ,limit of the function (m.e^x -n.cosx)/x is finite non zero number,
as x--> 0
For what values of m and n ,limit of the function (m.e^x -n.cosx)/x is finite non zero number,
as x--> 0
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Hans Karl Abele
Assuming m = n:
lim n(e^x - cos(x))/x = (L'H) = lim n(e^x + sin(x))/1 = n
7w
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Jonathan Burros
Attempting to take the limit gives
(m - n)/0
So m = n…
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190107 | https://www.bhg.com/homekeeping/house-cleaning/tips/how-to-clean-a-thermometer/ | Clean Your Thermometer After Each Use—Here’s How to Do It Safely
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Clean Your Thermometer After Each Use—Here’s How to Do It Safely
Learn how to disinfect digital and infrared thermometers to prevent the spread of germs between household members.
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When someone in your family feels sick, one of the first things you typically do is take their temperature. Fever is a common symptom of many illnesses, including influenza and Coronavirus, making a thermometer an essential household tool. Because temperature readings often involve direct contact with a sick person, thermometers should be cleaned before and after each use. If multiple people are using the same device, to avoid passing germs between people, you should wash and disinfect the thermometer between each and every use. To make sure you’re sanitizing it safely, follow these instructions for cleaning digital and infrared versions.
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How to Clean a Digital Thermometer
Typically the least expensive option, a digital thermometer uses an electronic heat sensor to measure a person’s body temperature in less than a minute. These can typically be used in the mouth, armpit, or rectum, but each device should be limited to only one type of use in order to prevent cross-contamination. To clean a thermometer, you’ll need rubbing alcohol with at least 60 percent alcohol to effectively kill germs, according to the Centers of Disease Control and Prevention (CDC). Then follow these instructions to clean and sanitize, before and after use:
Soak a cotton ball or pad in rubbing alcohol and use it to thoroughly coat the entire device. Use a cotton swab dipped in alcohol to clean the inside any small crevices.
Allow the alcohol to air-dry on the thermometer to effectively kill germs.
You can rinse the device under cool water to remove any traces of alcohol, taking care not to wet the electronic elements, such as the display.
Let the thermometer air-dry completely before using or storing.
Alternatively, you can wash with soap and water, but take care not to submerge the electronic components, which can damage the device. You should also avoid using hot water to clean a thermometer, as this could damage the sensor that reads the temperature.
How to Clean an Infrared or Forehead Thermometer
Infrared thermometers use an infrared sensor to measure body temperature in a few seconds through a forehead scan. Although many are designed to be used without touching skin, the device should be sanitized before and after each use, in case contact has occurred. Refer to the manufacturer’s instructions for guidance on cleaning your specific model, but follow these basic steps for cleaning a forehead thermometer:
Clean the thermometer’s sensor with a cotton ball or pad soaked in at least 60 percent rubbing alcohol. You can also use a bleach wipe or alcohol pad.
Let the thermometer air-dry completely before using or storing.
Cleaning a thermometer is vital for avoiding the spread of germs between members of your household. With these tips, you’ll be ready to take a safe, effective temperature reading whenever you need it.
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190108 | https://atozmath.com/example/StatsUG.aspx?q=1&q1=E2 | Median Example for ungrouped data
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Support us (New) All problem can be solved using search box ~~I want to sell my website www.AtoZmath.com with complete code~~Powered by Translate HomeWhat's newCollege AlgebraGamesFeedbackAbout us AlgebraMatrix & VectorNumerical MethodsStatistical MethodsOperation ResearchWord ProblemsCalculusGeometryPre-Algebra
Home>Statistical Methods calculators>Mean, Median and Mode for ungrouped data example Median Example for ungrouped data ( Enter your problem ) ( Enter your problem ) 1. Formula & Example 2. Mean Example 3. Median Example 4. Mode ExampleOther related methods 1. Mean, Median and Mode 2. Quartile, Decile, Percentile, Octile, Quintile 3. Population Variance, Standard deviation and coefficient of variation 4. Sample Variance, Standard deviation and coefficient of variation 5. Population Skewness, Kurtosis 6. Sample Skewness, Kurtosis 7. Geometric mean, Harmonic mean 8. Mean deviation, Quartile deviation, Decile deviation, Percentile deviation 9. Five number summary 10. Box and Whisker Plots 11. Construct an ungrouped frequency distribution table 12. Construct a grouped frequency distribution table 13. Maximum, Minimum 14. Sum, Length 15. Range, Mid Range 16. Stem and leaf plot 17. Ascending order, Descending order 2. Mean Example (Previous example)4. Mode Example (Next example) ### 3. Median Example Median : After arranging the observations in ascending (or descending) order, then the middle number is called median. It is denoted by M 1. If n is odd, then M= value of (n+1 2)t h observation 2. If n is even, then M=Value of (n 2)t h observation+Value of (n 2+1)t h observation 2 1. Find the Median of 4,14,12,16,6,3,1,2,3 Solution: Median : Observations in the ascending order are : 1,2,3,3,4,6,12,14,16 Here, n=9 is odd. M= value of (n+1 2)t h observation = value of (9+1 2)t h observation = value of 5 t h observation =4 2. Find the Median of 69,66,67,69,64,63,65,68,72 Solution: Median : Observations in the ascending order are : 63,64,65,66,67,68,69,69,72 Here, n=9 is odd. M= value of (n+1 2)t h observation = value of (9+1 2)t h observation = value of 5 t h observation =67 3. Find the Median of 3,23,13,11,15,5,4,2 Solution: Median : Observations in the ascending order are : 2,3,4,5,11,13,15,23 Here, n=8 is even. M=Value of (n 2)t h observation+Value of (n 2+1)t h observation 2 =Value of (8 2)t h observation+Value of (8 2+1)t h observation 2 =Value of 4 t h observation+Value of 5 t h observation 2 =5+11 2 =8 This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then Submit Here 2. Mean Example (Previous example)4. Mode Example (Next example) Share this solution or page with your friends.
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Liu Hui (fl. third century)
Liu Hui (fl. third century) was the founder of ancient China’s classical mathematical theories. He lived in the Wei Kingdom in the third century, but the dates of his birth and death are unknown. Most of his greatest achievements were related to Commentary on Nine Chapters on Mathematical Art, a famous Chinese classic of mathematics. Liu first read Nine Chapters on Mathematical Art when he was still a child. As an adult, he conducted in-depth research on its theories and problems. The research bore great fruit in 263, when he finished a ten-volume commentary, in which he analyzed the methods and equations in the classic, corrected its mistakes, and advanced many original theories. His algorithm for calculating square roots, which resembles the modern method, laid essential foundations for improving the accuracy of pi (π) and inventing decimal fractions. He also improved the solutions for linear equations with an ingenious method that is quite similar to the modern one, and created a formula for calculating the sum of a finite arithmetic progression. When analyzing a classic mathematical problem known as “five families sharing one well,” he noted it was actually an indeterminate equation (an equation for which there is more than one solution), making him the first Chinese mathematician to note this concept. He defined many other mathematical concepts as well, including “power,” “equation,” and “positive/negative numbers.” Moreover, he advanced a large number of axioms for proving new theorems. His rigorous, logical approach greatly solidified the factual foundations of both Nine Chapters on Mathematical Art and his own theories. Although he never wrote a systematic treatise, Liu had in fact created a distinctive theoretical system that included definitions, judgments, and most importantly, mathematical proofs. His factual reasoning, independent thinking, and dialectical philosophy became invaluable to later generations of mathematicians.
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190110 | https://web.math.princeton.edu/~smorel/adic_notes.pdf | Adic spaces Sophie Morel April 22, 2019 Conventions : - Every ring is commutative.
- N is the set of nonnegative integers.
2 Contents I The valuation spectrum 7 I.1 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 I.1.1 Valuations and valuation rings . . . . . . . . . . . . . . . . . . . . . . .
7 I.1.2 Some properties of valuations and valuation rings . . . . . . . . . . . . .
10 I.1.3 Rank of a valuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11 I.1.4 Comparing valuation subrings of a field . . . . . . . . . . . . . . . . . .
12 I.1.5 Valuation topology and microbial valuations . . . . . . . . . . . . . . .
13 I.1.6 The Riemann-Zariski space of a field . . . . . . . . . . . . . . . . . . .
17 I.2 The valuation spectrum of a ring . . . . . . . . . . . . . . . . . . . . . . . . . .
19 I.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 I.2.2 Some topological notions . . . . . . . . . . . . . . . . . . . . . . . . . .
23 I.2.3 The constructible topology . . . . . . . . . . . . . . . . . . . . . . . . .
25 I.2.4 The constructible topology on a spectral space . . . . . . . . . . . . . .
27 I.2.5 A criterion for spectrality . . . . . . . . . . . . . . . . . . . . . . . . . .
29 I.2.6 Spectrality of Spv(A) . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 I.3 The specialization relation in Spv(A) . . . . . . . . . . . . . . . . . . . . . . .
33 I.3.1 Specialization in a spectral space . . . . . . . . . . . . . . . . . . . . . .
34 I.3.2 Vertical specialization . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 I.3.3 Horizontal specialization . . . . . . . . . . . . . . . . . . . . . . . . . .
36 I.3.4 Putting things together . . . . . . . . . . . . . . . . . . . . . . . . . . .
40 I.3.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 I.4 Valuations with support conditions . . . . . . . . . . . . . . . . . . . . . . . . .
46 I.4.1 Supports of horizontal specializations . . . . . . . . . . . . . . . . . . .
47 I.4.2 The subspace Spv(A, J) . . . . . . . . . . . . . . . . . . . . . . . . . .
51 II Topological rings and continuous valuations 55 II.1 Topological rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 II.1.1 Definitions and first properties . . . . . . . . . . . . . . . . . . . . . . .
55 II.1.2 Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 II.1.3 Bounded sets and continuous maps . . . . . . . . . . . . . . . . . . . .
60 II.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 II.2 Continuous valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 II.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 II.2.2 Spectrality of the continuous valuation spectrum . . . . . . . . . . . . .
64 II.2.3 Specializations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66 3 Contents II.2.4 Analytic points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66 II.2.5 Tate rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69 II.3 Constructions with f-adic rings . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 II.3.1 Completions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 II.3.2 Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
76 II.3.3 Rings of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78 II.3.4 Localizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83 II.4 The Banach open mapping theorem . . . . . . . . . . . . . . . . . . . . . . . .
85 II.4.1 Statement and proof of the theorem . . . . . . . . . . . . . . . . . . . .
86 II.4.2 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88 III The adic spectrum 91 III.1 Rings of integral elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
91 III.2 The adic spectrum of a Huber pair . . . . . . . . . . . . . . . . . . . . . . . . .
95 III.3 Perturbing the equations of a rational domain . . . . . . . . . . . . . . . . . . .
98 III.4 First properties of the adic spectrum . . . . . . . . . . . . . . . . . . . . . . . . 100 III.4.1 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 III.4.2 Spa and completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 III.4.3 Rational domains and localizations . . . . . . . . . . . . . . . . . . . . 102 III.4.4 Non-emptiness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 III.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 III.5.1 Completed residue fields and adic Points . . . . . . . . . . . . . . . . . 107 III.5.2 The closed unit ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 III.5.3 Formal schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 III.6 The structure presheaf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 III.6.1 Universal property of rational domains . . . . . . . . . . . . . . . . . . 112 III.6.2 Definition of the structure presheaf . . . . . . . . . . . . . . . . . . . . . 113 III.6.3 Stalks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 III.6.4 The category V pre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 III.6.5 Adic spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 IV When is the structure presheaf a sheaf ?
121 IV.1 The main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 IV.1.1 Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 IV.1.2 Examples of strongly Noetherian Tate rings . . . . . . . . . . . . . . . . 122 IV.1.3 Examples of stably uniform Tate rings . . . . . . . . . . . . . . . . . . . 123 IV.2 Some preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 IV.2.1 Strictness and completion . . . . . . . . . . . . . . . . . . . . . . . . . 126 IV.2.2 The ˇ Cech complex for a special cover . . . . . . . . . . . . . . . . . . . 126 IV.2.3 Refining coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) . . . . . . . . . . . . . . . . . 132 IV.3.1 A local criterion for power-boundedness . . . . . . . . . . . . . . . . . . 132 IV.3.2 Calculation of the ˇ Cech cohomology . . . . . . . . . . . . . . . . . . . . 133 4 Contents IV.3.3 The strongly Noetherian case . . . . . . . . . . . . . . . . . . . . . . . . 136 IV.3.4 Cases (c) and (d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 IV.3.5 Case (a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 V Perfectoid algebras 141 V.1 Perfectoid Tate rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 V.1.1 Definition and basic properties . . . . . . . . . . . . . . . . . . . . . . . 141 V.1.2 Tilting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 V.1.3 Witt vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 V.1.4 Untilting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 V.1.5 A little bit of almost mathematics . . . . . . . . . . . . . . . . . . . . . 156 V.1.6 Tilting and the adic spectrum . . . . . . . . . . . . . . . . . . . . . . . . 164 V.2 Perfectoid spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 V.3 The almost purity theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 V.3.1 Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 V.3.2 The trace map and the trace pairing . . . . . . . . . . . . . . . . . . . . 173 V.3.3 From almost finite ´ etale to finite ´ etale . . . . . . . . . . . . . . . . . . . 174 V.3.4 The positive characteristic case . . . . . . . . . . . . . . . . . . . . . . . 176 V.3.5 Deforming almost finite ´ etale extensions . . . . . . . . . . . . . . . . . . 178 V.3.6 The case of perfectoid fields . . . . . . . . . . . . . . . . . . . . . . . . 179 V.3.7 Proof of theorems V.3.1.3 and V.3.1.9 . . . . . . . . . . . . . . . . . . . 181 5 I The valuation spectrum I.1 Valuations I.1.1 Valuations and valuation rings Notation I.1.1.1. If R is a local ring, we denote its maximal ideal by mR.
Definition I.1.1.2. If A ⊂B are local rings, we say that B dominates A if mA ⊂mB (which is equivalent to mA = A ∩mB).
Remark I.1.1.3. If K is a field, domination is an order relation on local subrings of K.
Proposition I.1.1.4. Let K be a field and R ⊂K be a subring. The following are equivalent : (a) R is local and it is maximal for the relation of domination (amnog local subrings of K); (b) for every x ∈K×, we have x ∈R or x−1 ∈R; (c) Frac(R) = K, and the set of ideals of R is totally ordered for inclusion; (c) Frac(R) = K, and the set of principal ideals of R is totally ordered for inclusion.
If these conditions are satisfied, we say that R is a valuation subring of K.
Remark I.1.1.5. We use the convention that K is a valuation subring of itself.
Now we define valuations.
Definition I.1.1.6. An ordered abelian group is an abelian group (Γ, +) with an order relation ≤such that for all a, b, c ∈Γ, a ≤b ⇒a + c ≤b + c.
We will only be interested in totally ordered abelian groups. Here are some examples.
Example I.1.1.7.
- (R, +) with its usual order relation; more generally, any subgroup of (R, +), for example (Z, +).
- (R>0, ×) with its usual order relation. Note that this is isomorphic (as an ordered group) to (R, +) by the map log.
- (R × R, +) with the lexicographic order; more generally, (Rn, +) with the lexicographic order, for any positive integer n.
7 I The valuation spectrum Remark I.1.1.8. Let (Γ, ×) be a totally ordered abelian group .
(1) Γ is torsionfree : Indeed, if γ ∈Γ is a torsion element, then there is some positive integer n such that γn = 1. If γ ≤1, then 1 ≤γ ≤γn = 1, so γ = 1. If γ ≤1, then 1 ≥γ ≥γ2 = 1, so γ = 1.
(2) Suppose that Γ is not the trivial group. Then, for every γ ∈Γ, there exists γ′ ∈Γ such that γ′ < γ. Indeed, if γ > 1 then γ−1 < γ, if γ < 1 then γ2 < γ, and if γ = 1 then there exists some δ ∈Γ −{1} (because Γ is not trivial), and then either δ < 1 or δ−1 < 1.
Notation I.1.1.9. Let Γ be a totally ordered abelian group . We will want to add an element to Γ that is bigger or smaller than all the elements of Γ. More precisely : (a) If the group law of Γ is written additively, we denote the unit element of Γ by 0, and we write Γ ∪{∞} for the union of Γ and of an element ∞, and we extend + and ≤to this set by the following rules : for every a ∈Γ, • a + ∞= ∞+ a = ∞; • a ≤∞.
(b) If the group law of Γ is written multiplicatively, we denote the unit element of Γ by 1, and we write Γ ∪{0} for the union of Γ and of an element 0, and we extend × and ≤to this set by the following rules : for every a ∈Γ, • a × 0 = 0 × a = 0; • 0 ≤a.
Definition I.1.1.10. Let R be a ring.
(i) An additive valuation on R is a map v : R →Γ ∪{∞}, where (Γ, +) is a totally ordered abelian group , satisfying the following conditions : • v(0) = ∞, v(1) = 0; • ∀x, y ∈R, v(xy) = v(x) + v(y); • ∀x, y ∈R, v(x + y) ≥min(v(x), v(y)).
The value group of v is the subgroup of Γ generated by Γ ∩v(R). The kernel (or support) of v is Ker(v) = {x ∈R | v(x) = ∞}; it is a prime ideal of R.
(ii) A multiplicative valuation (or non-Archimedean absolute value) on R is a map |.| : R →Γ ∪{0}, where (Γ, ×) is a totally ordered abelian group , satisfying the fol-lowing conditions : • |0| = 0, |1| = 1; • ∀x, y ∈R, |xy| = |x||y|; • ∀x, y ∈R, |x + y| ≤max(|x|, |y|).
8 I.1 Valuations The value group of |.| is the subgroup of Γ generated by Γ ∩|R|. The kernel (or support) of |.| is Ker(|.|) = {x ∈R | |x| = 0}; it is a prime ideal of R.
Remark I.1.1.11. Of course, whether we write the group law of a given totally ordered abelian group Γ additively or multiplicatively is an arbitrary choice. Fix Γ. Note that a map v : R →Γ ∪{∞} is an additive valuation if and only if −v is a multiplicative valuation (with the obvious convention that minus the biggest element is a new smallest element). This is a bit unfortunate, as both expressions “additive valuation” and “multiplicative valuation” are often shortened to “valuation”, but hopefully the meaning is always clear from context. Modulo this sign issue, both definitions are equivalent, and the notions of value group and kernel are the same on both sides.
In these notes, we will eventually take all valuations to be multiplicative (as this is the usual convention for adic spaces), and “valuation” will mean “multiplicative valuation”. Note however that some commutative algebra references, Matsumura’s or Bourbaki’s for example, use additive valuations.
Example I.1.1.12.
(1) Let Γ = ({1}, ×). Suppose that R is an integral domain. Then there is a valuation |.|triv : R →Γ ∪{0}, called the trivial valuation, defined by : |0| = 0 and |x| = 1 for every x ̸= 0. Its value group is {1} and its kernel is {0}.
(2) More generally, if R is a ring and ℘is a prime ideal of R, then composing the quotient map R →R/℘with the trivial valuation on R/℘gives a valuation on R with value group {1} and kernel ℘. We call this the trivial valuation on R with kernel ℘. We will often denote it by |.|℘,triv.
(3) For every prime number ℓ, the usual ℓ-adic valuation vℓ: Q →Z ∪{∞} is an additive valuation on Q (and on all its subrings), and the ℓ-adic absolute value |.|ℓ: Q →R≥0 is a multiplicative valuation on Q. They are related by |.|ℓ= ℓ−vℓ, and for our purposes they are interchangeable.
(4) By Ostrowski’s theorem, every nontrivial valuation on Q is of the form |.|s ℓ, for some prime number ℓand some s ∈R>0. It is easy to deduce from this that the valuations on Z are the trivial valuation, the valuation described in point (2) (one for each prime number ℓ), and the valuation |.|s ℓ, for ℓa prime number and s ∈R>0.
Definition I.1.1.13. Let R be a ring and |.|1, |.|2 be two valuations on R. We denote by Γ1 and Γ2 their respective value groups. We say that |.|1 and |.|2 are equivalent if there exists an isomorphism of ordered groups ϕ : Γ1 ∼ →Γ2 such that ϕ ◦|.|1 = |.|2 (with the convention that ϕ(0) = 0).
We compare the notions of valuations on a field and of valuation subrings of this field.
Proposition I.1.1.14. Let K be a field.
(i) If v : K →Γ ∪{0} is a valuation, then R := {x ∈K | |x| ≤1} is a valuation subring of K, and its maximal ideal is given by the formula mR = {x ∈K | |x| < 1}.
9 I The valuation spectrum (ii) Let R be a valuation subring of K. Let Γ = K×/R×; for a, b ∈K×, we write aR× ≤bR× if ab−1 ∈R. Then this makes Γ into a totally ordered abelian group , and the map |.| : K →Γ ∪{0} defined by |0| = 0 and |a| = aR× if a ̸= 0 is a valuation on K.
(iii) The constructions of (i) and (ii) induce inverse bijections between the set of valuation subrings of K and the set of equivalence classes of valuations on K.
Notation I.1.1.15. If K is a field and R ⊂K is a valuation subring, we denote by ΓR = K×/R× the value group of the corresponding valuation on K, and we call it the value group of R. We also denote by |.|R : K →ΓR ∪{0} the valuation defined by R.
Example I.1.1.16. The valuation subring of Q corresponding to the ℓ-adic absolute value |.|ℓis Z(ℓ), with maximal ideal ℓZ(ℓ).
I.1.2 Some properties of valuations and valuation rings Proposition I.1.2.1. Let K be a field and R ⊂K be a valuation subring. Then : (i) R is integrally closed in K.
(ii) Every finitely generated ideal of R is principal (in other words, R is a B´ ezout domain).
(iii) If I ⊂R is a finitely generated ideal, then ℘= √ I is a prime ideal of R, and it is minimal among prime ideals of R containing I.
Theorem I.1.2.2. Let K be a field and A be a subring of K.
(i) ( Theorem 10.2) Let ℘be a prime ideal of A. Then there exists a valuation subring R ⊃A of K such that ℘= A ∩mR.
(ii) ( Theorem 10.4) Let B be the integral closure of A in K. Then B is the intersection of all the valuation subrings of K containing A.
Proposition I.1.2.3. ( §2 No3 prop. 1 and prop. 2, and No4 prop. 4.) Let K be a field, |.| : K →Γ ∪{0} be a multiplicative valuation and K′ be an extension of K. Then there exists a multiplicative valuation |.|′ : K′ →Γ′ ∪{0} such that |.|′ |K is equivalent to |.|.
Moreover, if x1, . . . , xn ∈K′ are algebraically independent over K and γ1, . . . , γn ∈Γ, then we can find such a |.|′ such that |xi|′ = γi for every i ∈{1, . . . , n}.
Corollary I.1.2.4. (Proposition 2.25 of .) Let K′/K be a field extension, let R′ be a valu-ation subring of K′, and set R = K ∩R′. Then R is a valuation subring of K, and the map S′ 7− →K ∩S′ induces surjections {valuation subrings R′ ⊂S′ ⊂K′} →{valuation subrings R ⊂S ⊂K} and {valuation subrings S′ ⊂R′ ⊂K′} →{valuation subrings S ⊂R ⊂K}.
Moreover, if the extension K′/K is algebraic, then these surjections are actually bijections.
10 I.1 Valuations I.1.3 Rank of a valuation We first define the height of a totally ordered abelian group Γ.
Definition I.1.3.1. ( §4 No2 D´ efinition 1 p. 108) Let Γ be a totally ordered abelian group . A convex subgroup (or isolated subgroup) of Γ is a subgroup ∆of Γ such that, for all a, b, c ∈Γ, if a ≤b ≤c and a, c ∈∆, then b ∈∆.
Remark I.1.3.2. The condition in definition I.1.3.1 is also equivalent to the following condition (using additive conventions for Γ) : () for all a, b ∈Γ such that 0 ≤a ≤b, if b ∈∆, then a ∈∆.
Indeed, the condition of definition I.1.3.1 is obviously stronger than (). Conversely, suppose that ∆satisfies (), and let a, b, c ∈Γ such that a ≤b ≤c and a, c ∈∆; then 0 ≤b −a ≤c −a and c −a ∈∆, so b −a ∈∆by (), hence b = (b −a) + a ∈∆.
Example I.1.3.3.
(1) {0} and Γ are convex subgroups of Γ.
(2) If Γ = R × R with the lexicogrpahic order, then {0} × R is a convex subgroup of Γ.
Proposition I.1.3.4. ( §4 No4 p. 110) The set of all convex subgroups of Γ, ordered by inclu-sion, is a well-ordered set. Its ordinal is called the height of Γ and denoted by ht(Γ).
Example I.1.3.5.
(1) If ht(Γ) = 0, then Γ has only one convex subgroup. As {0} and Γ are always convex subgroups, this means that Γ is the trivial group.
(2) The condition ht(Γ) = 1 means that Γ is nontrivial and that the only convex subgroups of Γ are the trivial subgroup and Γ itself. For example, if Γ is a nontrivial subgroup of (R, +), then Γ has height 1. (We will see in proposition I.1.3.6 that the converse is true.) (3) Γ = Rn (with the lexicographic order) has height n. (See proposition I.1.4.1.) In these notes, we mostly care about the distinction “height ≤1” versus “height > 1”.
Proposition I.1.3.6. ( Theorem 10.6 or §4 No5 proposition 8 p. 112) Let (Γ, +) be a nontrivial totally ordered abelian group . The following are equivalent : (i) ht(Γ) = 1; (ii) for all a, b ∈Γ such that a > 0 and b ≥0, there exists n ∈N such that b ≤na; (iii) there exists an injective morphism of ordered groups Γ →(R, +).
Proof.
(i)⇒(ii) Consider the smallest convex subgroup ∆of Γ containing a. Condition (i) means that ∆= {0} or ∆= Γ, so ∆= Γ. This obviously implies (ii).
11 I The valuation spectrum (ii)⇒(iii) We will just indicate the construction of the map ϕ : Γ →R. Fix some a ∈Γ such that a > 0. Let b ∈Γ, and let n0 = min{n ∈Z | na ≤b} (this makes sense by (ii)) and b1 = b −n0a. We define sequences (ni)i≥1 of N and (bi)i≥1 of Γ by induction on i in the following way : Suppose that we have defined b1, . . . , bi and n1, . . . , ni−1, for some i ≥1. Then we set ni = min{n ∈N | nia ≤10bi} and bi+1 = 10bi −nia. Then we take ϕ(b) = n0 + P i≥1 ni10−i.
Definition I.1.3.7. The rank of a valuation (resp. of a valuation ring) if the height of its value group.
I.1.4 Comparing valuation subrings of a field Proposition I.1.4.1. ( §4 No2 proposition 3 p. 108 and §4 No4 Exemples p. 111) Let Γ be a totally ordered abelian group .
(i) If ϕ : Γ →Γ′ is a morphism of ordered groups, then Ker ϕ is a convex subgroup of Γ.
(ii) Let ∆be a convex subgroup of Γ. We define a relation ≤on Γ/∆in the following way : if c, c′ ∈Γ/∆, then c ≤c′ if and only if there exists x ∈c and x′ ∈c′ such that x ≤x′ in Γ.
Then this makes Γ/∆into an ordered group (necessarily totally ordered) if and only if ∆ is a convex subgroup.
(iii) In the situation of (ii), we have ht(Γ) = ht(∆) + ht(Γ/∆).
Theorem I.1.4.2. ( theorem 10.1 or §4 No1 p. 106) Let K be a field.
(i) Let B ⊂A ⊂K be two subrings of K, and suppose that B is a valuation subring of K.
Then : (a) A is also a valuation subring of K.
(b) mA ⊂mB, with equality if and only if A = B.
(c) mA is a prime ideal of B, and we have A = BmA.
(d) B/mA is a valuation subring of the field A/mA.
(ii) Conversely, let A ⊂K be a valuation subring, let B be a valuation subring of A/mA, and denote by B the inverse image of B in A. Then B is a valuation subring of K (in particular, K = Frac(B)), its maximal ideal is the inverse image of the maximal ideal of B, and we have an exact sequence of ordered groups : 1 →ΓB →ΓB →ΓA →1, where the maps ΓB = (A/mA)×/(B/mA)× ≃A×/B× →ΓB = K×/B× and ΓB = K×/B× →ΓA = K×/A× are the obvious ones.
12 I.1 Valuations Example I.1.4.3. Take K = k((u))((t)) and A = k((u)) (the valuation subring correspond-ing to the t-adic valuation on K). Then A/mA = k((u)) has the u-adic valuation, and the corresponding valuation subring is B = k. Its inverse image in A is B = {f = X n≥0 fntn ∈k((u)) | f0 ∈k}.
This is a valuation subring of rank 2 of K, and its value group is Z × Z (with the lexicographic order).
Corollary I.1.4.4. ( §4 No1 proposition 1 p. 106 and §4 No3 proposition 4 p. 109) Let K be a field and B ⊂K be a valuation subring.
(i) The map ℘7− →B℘is an order-reversing bijection from the set of prime ideals of B to the set of subrings of K containing B; its inverse is A 7− →mA.
(ii) The map A 7− →Ker(ΓB →ΓA) is an order-preserving bijection from the set of subrings A ⊃B of K to the set of convex subgroups of ΓB. Its inverse sends a convex subgroup H of ΓB to the subring A := B℘, where ℘= {x ∈B | ∀δ ∈H, |x|B < δ}.
Corollary I.1.4.5. Let K be a field and R ⊂K be a valuation subring. Then the rank of R is equal to the Krull dimension of R.
Corollary I.1.4.6. ( §4 No5 proposition 6 p. 111) Let K be a field and R ⊂K be a valuation subring. Then R has rank 1 if and only if it is maximal among all the subrings of K distinct from K.
Note also the following result : Proposition I.1.4.7. ( §3 No6 proposition 9 p. 105) Let K be a field and R ⊂K be a valuation subring. The following are equivalent : (i) R is a principal ideal domain.
(ii) R is Noetherian.
(iii) The ordered group ΓR is isomorphic to (Z, +, ≤).
(iv) The ideal mR is principal and T n≥0 Rn = (0).
If these conditions are satisfied, we say that R is a discrete valuation ring and that the corre-sponding valuation on K is discrete.
I.1.5 Valuation topology and microbial valuations Definition I.1.5.1. Let R be a ring and |.| : R →Γ ∪{0} be a valuation on R.
The valuation topology on R associated to |.| is the topology given by the base of open subsets B(a, γ) = {x ∈R | |x −a| < γ}, for a ∈R and γ ∈Γ.
13 I The valuation spectrum This makes R into a topological ring, and the map |.| is continuous if we put the discrete topology on Γ ∪{0}. (See §5 No1 for the case of a field, the general case is similar.) Remark I.1.5.2. Let R and |.| be as in definition I.1.5.1.
(1) The valuation topology is Hausdorff if and only Ker |.| = {0} (which implies that R is a domain).
(2) If the value group of |.| is trivial, then the valuation topology is the discrete topology.
(3) For every a ∈R and every γ ∈Γ, let B(a, γ) = {x ∈R | |x −a| ≤γ}. This is an open subset in the valuation topology; indeed, for every x ∈B(a, γ), we have B(x, γ) ⊂B(a, γ).
Suppose that the value group of |.| is not trivial.
Then the sets B(a, γ) also form a base of the valuation topology. Indeed, for every a ∈R and every γ ∈Γ, we have B(a, γ) = S δ<γ B(a, δ). (This uses remark I.1.1.8(2), and it is not true in general if the value group is trivial, as then all the sets B(a, γ) are equal to R, so they cannot form a base of the discrete topology unless R = {0}.) Definition I.1.5.3. Let R be a topological ring. An element x ∈R is called topologically nilpotent if 0 is a limit of the sequence (xn)n≥0 Theorem I.1.5.4. Let K be a field, let |.| : K →Γ ∪{0} be a valuation, and let R be the corresponding valuation ring. We put the valuation topology on K.
Then the following are equivalent : (i) The topology on K coincides with the valuation topology defined by a rank 1 valuation on K.
(ii) There exists a nonzero topologically nilpotent element in K.
(iii) R has a prime ideal of height 1.1 If these conditions are satisfied, we say that the valuation |.| is microbial. Moreover, if the valuation is microbial, then : (a) If ϖ ∈K× is topologically nilpotent, we have K = R[ϖ−1], and, if ϖ ∈R, then the subspace topology on R coincides with the ϖR-adic topology and also with the ϖ-adic topology.
(b) If ℘is a prime ideal of height 1 of R, then the valuation subring R℘has rank 1, and the corresponding valuation defines the same topology as |.|.
Note that, as the ideals of R are totally ordered by inclusion, R has at most one prime ideal of height 1.
1That is, a prime ideal that is minimal among the set of nonzero prime ideals of R.
14 I.1 Valuations Example I.1.5.5. We use the notation of example I.1.4.3. We claim that the valuation subrings A and B define the same topology on K = k((u))((t)). Indeed, the valuation corresponding to A is the t-adic valuation, which has rank 1, so we can try to apply theorem I.1.5.4 to B. We know that B has Krull dimension 2, so any prime ideal that is neither (0) not maximal has height 1; this is the case for I := tB, which is prime and not maximal because B/I = k by definition of B. Also, we see easily that I = mA, so we are done by point (b) of the theorem.
Lemma I.1.5.6. Let K be a field and |.| be a valuation on K with valuation ring R. Then R and mR are open for the valuation topology.
Proof. In the notation of definition I.1.5.1 and remark I.1.5.2, we have R = B(0, 1) and mR = B(0, 1).
Lemma I.1.5.7. Let R be a ring and |.| be a valuation on R. If x ∈R is topologically nilpotent for the valuation topology, then |x| < 1. If moreover |.| has rank 1, then the converse is true.
Proof. Suppose that |x| ≥1. Then, for every integer n ≥1, we have |xn| ≥1, so xn ̸∈B(0, 1), which implies that 0 is not a limit of the sequence (xn)n≥0.
Conversely, suppose that the valuation has rank 1 and that |x| < 1. Let Γ be the value group of |.|, and let γ = |x|−1 > 1. Let δ ∈Γ. By proposition I.1.3.6, there exists n ∈N such that δ−1 < γm for every m ≥n, and then we have : for every m ≥n, |xm| = γ−m < δ, that is, xm ∈B(0, δ). This shows that 0 is a limit of the sequence (xn)n≥0.
Remark I.1.5.8. The converse in lemma I.1.5.7 is not true for a valuation of rank > 1.
For example, take K = k((u))((t)) with the rank 2 valuation defined by the valuation ring B of example I.1.4.3. Then u ∈mB so it has valuation < 1, but it is not topologically nilpotent.
Indeed, by example I.1.5.5, the valuation topology on K coincides with the the t-adic topology, and the sequence (un)n≥0 does not tend to 0 in the t-adic topology.
Lemma I.1.5.9. Let K be a field and |.| be a valuation on K with valuation ring R. If ϖ ∈K× is topologically nilpotent for the valuation topology, then K = R[ϖ−1]. Moreover, there exists a positive integer r such that ϖr ∈R (or even ϖn ∈mR), and then the subspace topology on R coincides with the ϖrR-adic topology.
Proof. By lemma I.1.5.6, R and mR are open neighborhoods of 0 in K. As the sequence (ϖn)n≥0 converges to 0, we have ϖn ∈mR for n big enough. So we may assume that ϖ ∈R.
We first show that K = R[ϖ−1]. Let x ∈K. We want to prove that xϖn ∈R for n big enough. This is obvious if x = 0, so we may assume x ̸= 0. By definition of R, we have xϖn ∈R if and only if |xϖn| ≤1, i.e. if and only if |ϖn| ≤|x|−1. As ϖ is topologically nilpotent, we have ϖn ∈B(0, |x|−1) for n big enough, which implies the desired result.
15 I The valuation spectrum Now we prove that the subspace topology on R is the ϖR-adic topology. First, as ϖ ∈K×, multiplication by ϖ is a homeomorphism of K. So the subsets ϖnR are open for every n ∈Z.
It remains to show that the family (ϖnR)n≥0 is a basis of neighborhoods of 0. Let γ ∈Γ, where γ is the value group of |.|. We want to find some n ≥0 such that ϖnR ⊂B(0, γ). Let n ∈N.
Then ϖnR ⊂B(0, γ) if and only if, for every a ∈R, |aϖn| < γ. As every a ∈R has valuation ≤1, this will hold if |ϖn| < γ. But ϖn →0 as n →+∞, so |ϖn| < γ for n big enough.
Proof of theorem I.1.5.4. Note that point (a) follows from lemma I.1.5.9.
(i)⇒(ii) Let |.|′ be a rank 1 valuation on K defining the same topology as |.|. Then any x ∈K× with |x|′ < 1 is topologically nilpotent by lemma I.1.5.7. (Such an element exists because the value group of |.|′ has rank 1, so it is not trivial, so it at least one element < 1, and this element must be the image of a nonzero element of K.) (ii)⇒(iii) Let ϖ ∈K× be a topologically nilpotent element. By lemma I.1.5.9, we may assume that ϖ ∈mR, and the subspace topology on R is the I-adic topology, where I = ϖR. Let ℘= √ I. By proposition I.1.2.1, ℘is the smallest prime ideal of R containing I. We show that ℘has height 1. Let q ⊊℘be another prime ideal of R; we want to show that q = (0).
Note that q does not contain I (otherwise it would contain ℘), and so ϖ ̸∈q. As the ideals of R are totally ordered by inclusion (see proposition I.1.1.4, we must have q ⊂I. We show by induction on n that q ⊂In for every n ≥1 : • We just did the case n = 1.
• Suppose that q ⊂In = ϖnR. Let x ∈q. As q ⊂ϖR, we can write x = ϖy, with y ∈R.
As q is prime and ϖ ̸∈q, this implies that y ∈q, and so x = ϖy ∈ϖIn = In+1.
But we have seen that the topology on R is the I-adic topology, and it is Hausdorff because Ker |.| = {0} (remember that Ker |.| is an ideal of K). So T n≥1 In = (0), and q = (0).
(iii)⇒(i) Let ℘be a height 1 prime ideal of R, and let R′ = R℘. Then R′ is also a valuation subring of R. Also, the Krull dimension of R′ is the height of ℘, i.e. 1, so the rank of the valuation corresponding to R′ is 1 by corollary I.1.4.5. So we need to show that the valuation topol-ogy corresponding to R′ coincides with the valuation topology corresponding to R (this will also prove point (b)). Let |.| (resp. |.|′) be the valuation corresponding to R (res. R′), and Γ = K×/R× (resp. Γ′ = K×/R′×) be its value. We denote the obvious projection Γ →Γ′ by π; this is order-preserving by theorem I.1.4.2(ii). Note that Γ is not trivial (because R has a prime ideal of height 1, so it cannot be equal to K).
Let a ∈R and γ ∈Γ. If |x −a| ≤γ, then |x −a|′ ≤π(γ) (because π is order-preserving).
So B(a, γ) ⊂{x ∈R | |x −a|′ ≤π(γ)}.
16 I.1 Valuations Also, if |x −a| ≥γ, then |x −a|′ ≥π(γ). So B(a, γ) ⊃{x ∈R | |x −a|′ < π(γ)}.
Thanks to remark I.1.5.2(3), this implies that the two valuation topologies on K are equal.
I.1.6 The Riemann-Zariski space of a field This is a simple example of valuation spectrum, and it will also be useful to understand the general case.
Let K be a field and A ⊂K be a subring.
Definition I.1.6.1. We say that a valuation subring R ⊂K has a center in A if A ⊂R; in that case, the center of R in A is the prime ideal A ∩mR of A.
Definition I.1.6.2. The Riemann-Zariski space of K over A is the set RZ(K, A) of valuation subrings R ⊃A of K. We put the topology on it with the following base of open subsets (sometimes called Zariski topology) : the sets U(x1, . . . , xn) = RZ(K, A[x1, . . . , xn]) = {R ∈RZ(K, A) | x1, . . . , xn ∈R}, for x1, . . . , xn ∈K. (If |.| is the valuation on K corresponding to R, the condition that R ⊃A becomes |a| ≤1 for every a ∈A, and the condition that R ∈RZ(K, A) becomes |xi| ≤1 for 1 ≤i ≤n.) If A is the image of Z, we write RZ(K, Z) = RZ(K) and we call it the Riemann-Zariski space of K; this is the set of all valuation subrings of K.
Note that U(x1, . . . , xn) ∩U(y1, . . . , ym) = U(x1, . . . , xn, y1, . . . , ym), so this does define a topology on RZ(K, A).
Example I.1.6.3. Let k be a field, X/k be a smooth projective geometrically connected curve and K be the function field of X. Then RZ(K, k) is canonically isomorphic to X as a topological space.
More generally, if K/k is a finitely generated field extension, then RZ(K, k) is isomorphic (as a topological space) to the inverse limit of all the projective integral k-schemes with function field K.
See also example I.2.1.6 for a description of RZ(Q).
Remark I.1.6.4. Let R, R′ ∈RZ(K, A). Then R is a specialization of R′ in RZ(K, A) (i.e. R is in the closure of {R′}) if and only if R ⊂R′.
Indeed, R is a specialization of R′ if and only every open set of RZ(K, A) that contains R also contains R′. This means that, for all a1, . . . , an ∈K such that a1, . . . , an ∈R, we must also have a1, . . . , an ∈R′, so it is equivalent to the fact that R ⊂R′.
17 I The valuation spectrum Here is a particular case of the topological results to come.
Proposition I.1.6.5. ( theorem 10.5) RZ(K, A) is quasi-compact.
Proof. We write X = RZ(K, A). Let A be a family of closed subsets of X having the finite intersection property. We must show that T F∈A F ̸= ∅. Using Zorn’s lemma, we can replace with a family of closed subsets of X having the finite intersection property and maximal among all families of closed subsets of X having the finite intersection property. Then the following hold (otherwise, we could enlarge A without losing the finite intersection property) : (a) if F1, . . . , Fr ∈A , then F1 ∩. . . ∩Fr ∈A ; (b) if Z1, . . . , Zr are closed subsets of X and Z1 ∪. . . ∪Zr ∈A , then at least one of the Zi is in A ; (c) if F ∈A and Z ⊃F is a closed subset of X, then Z ∈A .
We claim that \ F∈A F = \ a∈K|X−U(a)∈A (X −U(a)).
Indeed, it is obvious that the right hand side is contained in the left hand side. Conversely, let x ∈X, and suppose that x ̸∈T F∈A F. Then there exists F ∈A such that x ̸∈F. As X −F is open, there exists a1, . . . , an ∈K such that x ∈U(a1, . . . , an) and U(a1, . . . , an)∩F = ∅. Then F ⊂X −U(a1, . . . , an) = Sn i=1(X −U(ai)), so, by (b) and (c), there exists i ∈{1, . . . , n} such that X −U(ai) ∈A . As x ∈U(a1, . . . , an) ⊂U(ai), we have x ∈X −U(ai). This finishes the proof of the claim.
Now let C = {a ∈K× | X −U(a−1) ∈A }. If a point x ∈X corresponds to a valuation subring A ⊂R ⊂K, then, for every a ∈K× : x ∈X −U(a−1) ⇔a−1 ̸∈R ⇔a ∈mR.
So : \ F∈A F = \ a∈C (X −U(a−1)) = {R ∈RZ(K, A) | mR ⊃C}.
Let I be the ideal of A[C] generated by C. If 1 ∈I, then there exist a1, . . . , an ∈C such that 1 ∈Pn i=1 aiA[C], and then Tn i=1(X −U(a−1 i )) = ∅, which contradicts the fact that A has the finite intersection property. So 1 ̸∈I, hence I is contained in a prime ideal of A, so, by theorem I.1.2.2(i), there exists a valuation subring A ⊂R ⊂K such that I ⊂mR, and then R ∈T F∈A F.
18 I.2 The valuation spectrum of a ring I.2 The valuation spectrum of a ring I.2.1 Definition Definition I.2.1.1. Let A be a commutative ring. The valuation spectrum Spv(A) of A is the set of equivalence classes of valuations on A, equipped for the topology generated by the subsets U f1, . . . , fn g = {|.| ∈Spv(A) | |f1|, . . . , |fn| ≤|g| ̸= 0}, for all f1, . . . , fn, g ∈A. (Note that this subset is empty if g = 0.) Note that U f1, . . . , fn g ∩U f ′ 1, . . . , f ′ m g′ = U f1g′, . . . , fng′, f ′ 1g, . . . , f ′ mg gg′ .
In particular, we can use the subsets U f g , f, g ∈A, to generate the topology of Spv(A).
Remark I.2.1.2. Let X be the set of pairs (℘, R), where ℘is a prime ideal of A and R is a valuation subring of Frac(A/℘). Then Spv(A) and X are naturally in bijection. Indeed, if (℘, R) ∈X, then we get a valuation on A by composing the quotient map A →A/℘with the valuation on Frac(A/℘) defined by R as in proposition I.1.1.14(ii). Conversely, if |.| is a valuation on A, then ℘:= Ker |.| is a prime ideal of A, and |.| defines a valuation on the domain A/℘, hence also on its fraction field; we take for R the valuation subring of that valuation, as in proposition I.1.1.14(i). It follows easily from proposition I.1.1.14(iii) that these maps are inverse bijections.
For f1, . . . , fn, g ∈A, the image of the open subset U f1,...,fn g by this bijection is {(℘, R) ∈X | g ̸∈℘and ∀i ∈{1, . . . , n}, (fi + ℘)(g + ℘)−1 ∈R}, where (fi + ℘)(g + ℘)−1 is an element of Frac(A/℘) (this makes sense because the image of g in A/℘is nonzero by the first condition).
In particular, we get a canonical map supp : Spv(A) →Spec(A) that sends a valuation to its kernel or support.
Notation I.2.1.3. If x ∈Spv(A) corresponds to the pair (℘, R), we write ℘x = ℘, Rx = R, Γx = ΓR, K(x) = Frac(A/℘x), and we denote by |.|x : A →Γx ∪{0} the composition of A →A/℘x and of the valuation corresponding to Rx on K(x). For f ∈A, we often write f(x) for the image of f in A/℘x, and |f(x)| for the image of f(x) in Γx.
Proposition I.2.1.4. Let A be a commutative ring.
19 I The valuation spectrum (i) If A is a field, then Spv(A) = RZ(A) (as topological spaces).
(ii) In general, the map supp : Spv(A) →Spec(A) is continuous and surjective. For every ℘∈Spec(A), the fiber of this map over ℘is isomorphic (as a topological space) to RZ(Frac(A/℘)).
Proof.
(i) This is a particular case of (ii).
(ii) For every ℘∈Spec(A), we have (℘, Frac(A/℘)) ∈supp−1(℘), so supp is surjective.
Next, for every f ∈A, we have supp−1(D(f)) = {(℘, R) ∈Spv(A) | f ̸∈℘} = U f f (where D(f) = {℘∈Spec(A) | f ̸∈℘}), so supp is continuous.
Finally, fix ℘∈Spec(A).
Then supp−1(℘) = {(℘, R) | R ∈Frac(A/℘)} is canonically in bijection with RZ(Frac(A/℘)). Moreover, if f1, . . . , fn, g ∈A, then the intersection U f1,...,fn g ∩supp−1(℘) is empty if g ∈℘, and, if g ̸∈℘, it is equal to {(℘, R) ∈supp−1(℘) | (f1 + ℘)(g + ℘)−1, . . . , (fn + ℘)(g + ℘)−1 ∈R}, which corresponds by the bijection supp−1(℘) ≃ RZ(Frac(A/℘)) to the open subset U((f1 + ℘)(g + ℘)−1, . . . , (fn + ℘)(g + ℘)−1).
So the bijection supp−1(℘) ≃RZ(Frac(A/℘)) is a homeomorphism.
Remark I.2.1.5. The map supp : Spv(A) →Spec(A) has an obvious section, sending a prime ideal ℘of A to the pair (℘, Frac(A/℘)) (in terms of valuations, this is the composition of the quotient map A →A/℘and of the trivial valuation on A/℘). This section is also a continuous map, because the inverse image of the open set U f1,...,fn g is {℘∈Spec(A) | g ̸∈℘}, which is open in Spec(A).
Example I.2.1.6.
(1) We have (see example I.1.1.12(4) Spv(Q) = RZ(Q) = {|.|triv, |.|ℓ, ℓ∈Z prime}.
By remark I.1.6.4, the trivial valuation |.|triv is the generic point of Spv(Q), and each |.|ℓ is a closed point. In fact, for all f1, . . . , fn, g ∈Z such that g ̸= 0, we have U f1, . . . , fn g = {|.|triv} ∪{|.|ℓ, for ℓnot dividing g}.
So Spv(Q) is isomorphic to Spec(Z) as a topological space.
20 I.2 The valuation spectrum of a ring (2) We have Spv(Z) = Spv(Q) ∪{|.|ℓ,triv, ℓ∈Z prime}, where |.|ℓ,triv is the composition of Z →Z/ℓZ and of the trivial valuation on Z/ℓZ. Note that Spv(Q) = supp−1((0)) and {|.|ℓ,triv} = supp−1(ℓZ). The points |.|ℓ,triv are all closed, we have {|.|ℓ} = {|.|ℓ, |.||ell,triv} (so |.|ℓspecializes to |.|ℓ), and the point |.|triv is generic.
Indeed, note that, for x, y in a topological space X, y specializes to x if and only if ev-ery open subset of X that contains x also contains y. Let f1, . . . , fn, g ∈Z, and let U = U f1,...,fn g . As U = ∅if g = 0, we assume that g ̸= 0. Then |.|triv is always in U, |.|ℓ,triv is in U if and only if g ̸∈ℓZ and |.|ℓis in U if and only if f1g−1, . . . , fng−1 ∈Z(ℓ) (which is automatically true if g ̸∈ℓZ).
Definition I.2.1.7. Let ϕ : A →B be a morphism of rings. We denote by Spv(ϕ) the map Spv(B) →Spv(A), |.| 7− →|.| ◦ϕ.
Note that this is a continuous map, because, for all f1, . . . , fn, g ∈A, we have Spv(ϕ)−1 U f1, . . . , fn g = U ϕ(f1), . . . , ϕ(fn) ϕ(g) .
(This follows immediately from the definitions.) Remark I.2.1.8. We get a commutative square Spv(B) / Spv(A) Spec(B) / Spec(A) This square is cartesian if B is a localization or a quotient of A, but not in general.
The goal of this section is to prove that Spv(A) is a spectral space (see definition I.2.2.7) and that the continuous maps Spv(ϕ) are spectral (i.e. quasi-compact).
We will outline the strategy of the proof. First “remember” that a topological space X is called spectral if it satisfies the following conditions : (i) X is quasi-compact.
(ii) X has a collection of quasi-compact open subsets that is stable by finite intersections and generates its topology.
(iii) X is quasi-separated (see definition I.2.2.1(iv)).
(iv) X is T0 (see definition I.2.2.4(v)).
(v) X is sober, that is, every closed irreducible subset of X has a unique generic point.
21 I The valuation spectrum We spelled out all these conditions because they are all important, but they are not indepen-dent; indeed, (iii) follows from (ii) (see remark I.2.2.3(1)) and (iv) follows from (v) (see remark I.2.2.5).
The main example of a spectral space (and the reason for the name) is Spec(A) with its Zariski topology, for A a ring.2 We quickly indicate how to check conditions (i), (ii) and (v) in this case : (i) Remember that the closed subsets of Spec(A) are the V (I) = {℘∈Spec(A) | I ⊂℘}, for I an ideal of A (and that we have a canonical isomorphism V (I) = Spec(A/I)). If (Ij)j∈J is a family of ideals of A, then \ j∈J V (Ij) = ∅⇔1 ∈ sX j∈J Ij ⇔1 ∈ X j∈J Ij, and this happens if and only if there is a finite subset J′ of J such that 1 ∈P j∈J′ Ij. So Spec(A) is quasi-compact.
(ii) We take as our generating family of open subsets the subsets D(f1, . . . , fn) := Spec(A) −V ((f1, . . . , fn)), for f1, . . . , fn ∈A.
The fact that D(f1, . . . , fn) is quasi-compact is proved in example I.2.2.2. (If n = 1, it follows from the obvious isomorphism D(f) = Spec(A[f −1]).) (v) It is easy to see that a closed subset V (I) of Spec(A) is irreducible if and only if ℘:= √ I is a prime ideal of A. Then V (I) = V (℘) = Spec(A/℘) has a unique generic point, given by the prime ideal (0) of A/℘.
In the case of Spv(A), the quasi-compactness is still not too hard to check directly (see propo-sition I.1.6.5 for a particular case), but we cannot apply the same strategy afterwards : we want to use the subsets U f1,...,fn g as our generating family of quasi-compact open subsets, but they are not isomorphic in general to valuation spectra even when n = 1, and neither are the closed subsets of Spv(A). Instead, we take a more indirect route that uses the constructible topology of Spv(A).
If X is a quasi-compact and quasi-separated space, the collection of constructible subsets of X is the smallest collection of subsets of X that is stable by finite unions, finite intersec-tions and complements and that contains all quasi-compact open subsets of X. (In general, we need to replace “quasi-compact” with a different condition called “retrocompact”, see definitions I.2.2.1(iii) and I.2.3.1). Constructible subsets form the base of a new topology on X, called the constructible topology, in which quasi-compact open subsets of X are open and closed. The constructible topology is Hausdorff and quasi-compact if X is spectral (proposition I.2.4.1). The main technical tool is a partial converse of this result, due to Hochster (theorem I.2.5.1) : If X′ is a quasi-compact topological space and if U is a collection of open and closed subsets of X′, 2Indeed, Hochster has proved that every spectral space is homeomorphic to the spectrum of a ring, see theorem 6 in section 7 of .
22 I.2 The valuation spectrum of a ring it gives a criterion for the topology generated by U to be spectral (and then the topology of X′ will be the corresponding constructible topology).
To apply Hochster’s spectrality criterion to Spv(A), we still need to be able to check its hy-potheses. The key point is the following (see the proof of theorem I.2.6.1 for details): Each equivalence class of valuations on A defines a binary relation on A (“divisibility with respect to the corresponding valuation ring”). This gives an injective map from Spv(A) to the set {0, 1}A×A of binary relations on A, and if we put the product topology of {0, 1}A×A (which is Hausdorff and quasi-compact by Tychonoff’s theorem), then this topology induces the constructible topology on Spv(A) and every set U f1,...,fn g is open and closed.
I.2.2 Some topological notions First we fix some vocabulary : Let X be a topological space and U be a collection of open subsets of X. We say that U is a base of the topology of X if every open subset of X is a union of elements of U ; we say that U generates the topology of X or that U is a subbase of the topology of X if the collection of finite intersections of elements of U is a base of the topology (or, in other words, if the topology of X is the coarsest topology for which all the elements of U are open subsets).
Definition I.2.2.1.
(i) We say that a topological space X is quasi-compact if every open cov-ering of X has a finite refinement.
(ii) We say that a continuous map f : X →Y is quasi-compact if the inverse image of any quasi-compact open subset of Y is quasi-compact.
(iii) We say that X is quasi-separated if the diagonal embedding X →X × X is quasi-compact. This means that, for any quasi-compact open subspaces U and V of X, the intersection U ∩V is still quasi-compact.
Example I.2.2.2. Let A be a ring. Then an open subset U of Spec(A) is quasi-compact if and only if there exists a finitely generated ideal I of A such that U = Spec(A) −V (I).
Indeed, suppose that U = Spec(A)−V (I), with I finitely generated, say I = (f1, . . . , fr). Let (Ij)j∈J be a family of ideals of A such that U ∩T j∈J V (Ij) = ∅, that is, T j∈J V (Ij) ⊂V (I).
Then I ⊂ qP j∈J Ij, so there exists N ≥1 such that f N 1 , . . . , f N r ∈P j∈J Ij. Choose a finite subset J′ of J such that f N 1 , . . . , f N r ∈P j∈J′ Ij. Then V (I) = V (f N 1 , . . . , f N r ) ⊃T j∈J′ V (Ij), so U ∩T j∈J′ V (Ij) = ∅.
Conversely, suppose that U is quasi-compact, and let I be an ideal of A such that U = Spec(A) −V (I). We have U = S f∈I D(f), so, by the quasi-compactness assumption, there exist f1, . . . , fr ∈I such that U = Sr i=1 D(fi) = Spec(A) −V (f1, . . . , fr).
Remark I.2.2.3. To check that a continuous map f : X →Y is quasi-compact, it suffices to check that f −1(U) is quasi-compact for U in a base of quasi-compact open subsets of Y . In 23 I The valuation spectrum particular, if the topology of X has a basis of quasi-compact open subsets which is stable by finite intersections, then X is quasi-separated.
Definition I.2.2.4. Let X be a topological space.
(i) We say that X is irreducible if, whenever X = Y ∪Z with Y and Z closed subsets of X, we have X = Y or X = Z.
(ii) If x, y ∈X, we say that x is a specialization of y (or that y is a generization3 of x, or that y specializes to x) if x ∈{y}.
(iii) We say that a point x ∈X is closed if {x} is closed.
(iv) We say that a point x ∈X is generic if {x} is dense in X.
(v) We say that X is a Kolmogorov space (or a T0 space if, for all x ̸= y in X, there exists an open subset U of X such that x ∈U and y ̸∈U, or that x ̸∈U and y ∈U.
(vi) We say that X is sober if every irreducible closed subset of X has a unique generic point.
Remark I.2.2.5. A topological space X is T0 if and only every irreducible closed subset of X has at most one generic point.
Indeed, note that X is T0 if and only, for all x, y ∈X such that x ̸= y, there exists a closed subset of X that contains exactly one of x and y, which means that x ̸∈{y} or y ̸∈{x}. If X is T0, let Z be an irreducible closed subset of X and let x, y ∈Z be two generic points. Then x ∈{y} = Z and y ∈{x} = Z, so x = y. Conversely, suppose that every irreducible closed subset of X has at most one generic point, and let x, y ∈X such that x ̸= y. As {x} and {y} are irreducible closed and have distinct generic points, we have {x} ̸= {y}. So we have {x} ̸⊂{y}, and then x ̸∈{y}, or we have {y} ̸⊂{x}, and then y ̸∈{x}.
Example I.2.2.6. For every ring, Spv(A) is T0.
Indeed, let x, y ∈Spv(A) such that x ̸= y. If A is a field, then Spv(A) = RZ(A) and x, y correspond to distinct valuation subrings Rx, Ry ⊂K. We may assume without loss of generality that there exists a ∈Rx −Ry, and then the open subset U(a) of RZ(A) contains x but not y.
In the general case, we have either supp(x) = supp(y) or supp(x) ̸= supp(y).
If supp(x) = supp(y), then x and y are both in the same fiber of supp, which is a Riemann-Zariski space, and so, by the first case treated above and proposition I.2.1.4(ii), we can find an open subset of Spv(A) that contains exactly one of x and y. If supp(x) ̸= supp(y), as Spec(A) is T0, we may assume without loss of generality that there exists an open subset U of Spec(A) such that supp(x) ∈U and supp(y) ̸∈U. Then supp−1(U) is an open subset of Spv(A) that contains x and not y.
3EGA says “g´ en´ erisation”, so I translated it as “generization”, but some English-language references (for example the stacks project) use “generalization”.
24 I.2 The valuation spectrum of a ring Definition I.2.2.7. A topological space X is called spectral if it satisfies the following conditions : (a) X is quasi-compact and quasi-seperated; (b) the topology of X has a base of quasi-compact open subsets; (c) X is sober.
We say that X is a locally spectral space if it has an open covering by spectral spaces.
Example I.2.2.8. Any affine scheme is a spectral space, and any scheme is a locally spectral space. (In fact, Hochster has proved that these are the only examples, see theorems 6 and 9 of .) Proposition I.2.2.9. ( remark 3.13) (i) A sober space is T0.
(ii) A locally spectral space is sober.
(iii) A locally spectral space is spectral if and only if it is quasi-compact and quasi-separated.
(iv) Let X be a sober space. Then every locally closed subspace of X is sober.
(v) Let X be a spectral space. Then every quasi-compact open subspace (resp. every closed subspace) of X is spectral.
(vi) Let X be a locally spectral space. Then every open subspace of X is locally spectral, and the topology of X has a base consisting of spectral open subspaces.
Proof. (i) follows from remark I.2.2.5. (ii) follows from the fact that a space that has a covering by sober open subspaces is sober (see (3) of [25, Lemma 06N9]). (iii) is obvious from the definitions. (iv) is (3) of [25, Lemma 0B31]. (v) is a particular case of [25, Lemma 0902]. (vi) is clear.
I.2.3 The constructible topology We will only define the constructible topology on a quasi-compact and quasi-separated topolog-ical space, because that is the only case when we will need it (at least for now). For the general definition, see for example [25, Section 04ZC].
Definition I.2.3.1. Let X be a quasi-compact and quasi-separated topological space and Y be a subset of X.
(i) We say that Y is constructible if it is a finite union of subsets of the form U ∩(X −V ), with U and V quasi-compact open subsets of X.
25 I The valuation spectrum (ii) We say that Y is ind-conctructible (resp. pro-constructible) if it a union (resp. an inter-section) of consrtuctible subsets of X.
Any finite union of quasi-compact open subsets is quasi-compact, and, if X is quasi-separated, so is any finite intersection of quasi-compact open subsets. This immediately implies the follow-ing lemma.
Lemma I.2.3.2. Let X be a quasi-compact and quasi-separated topological space. The collec-tion of constructible subsets of X is closed under finite unions, finite intersections and taking complements.
So we could also have defined the collection of constructible subsets of X as the smallest collection of subsets of X that is stable by finite unions, finite intersections and complements, and contains the quasi-compact open subsets of X.
Definition I.2.3.3. Let X be a quasi-compact and quasi-separated topological space. The con-structible topology on X is the topology with base the collection of the constructible subsets of X. Equivalently, it is the topology generated by the quasi-compact open subsets of X and their complements.
We denote by Xcons the space X equipped with its constructible topology.
It is clear from the definitions that the open (resp. closed) subsets for the constructible topology are the ind-constructible (resp. pro-constructible) subsets of X.
Example I.2.3.4. Let X be a finite T0 space. Then X is spectral and every subset of X is constructible.
Proof. Every subset of X is quasi-compact (because it is finite), so X is quasi-compact and quasi-separated. As X is T0, we just need to prove that every closed irreducible subset of X has at least one generic point. Let Z be a closed irreducible subset of X. Then Z is the union of the finite family of closed subsets ({z})z∈Z, so there exists z ∈Z such that Z = {z}. Finally, we proved that every subset of X is constructible. It suffices to prove that {x} is constructible for every x ∈X. So let x ∈X. As X if T0, for every y ∈X −{x}, we can choose Yy open or closed (and in particular constructible) such that x ∈Yy and y ̸∈Yy. So {x} = T y∈X−{x} Yy is constructible.
Proposition I.2.3.5. Let X be a quasi-compact and quasi-separated topological space. Then any constructible subset of X is quasi-compact.
In particular, if Y is an open (resp. closed) subset of X, then Y is constructible if and only if it is quasi-compact (resp. if and only if X −Y is quasi-compact).
26 I.2 The valuation spectrum of a ring Proof. Let Y be a constructible subsets of X.
We write Y = Sn i=1(Ui −Vi), with U1, . . . , Un, V1, . . . , Vn quasi-compact open subsets of X. For every i ∈{1, . . . , n}, Ui −Vi is a closed subset of the quasi-compact space Ui, so it is also quasi-compact. Hence Y is quasi-compact.
I.2.4 The constructible topology on a spectral space Proposition I.2.4.1. Let X be a spectral space. Then Xcons is Hausdorff, totally disconnected and quasi-compact.
Proof. (See [25, Lemma 0901].) Let x, y ∈X such that x ̸= y. As X is T0, we may assume that there exists an open subset U of X such that x ∈U and y ̸∈U. As quasi-compact open subsets form a base of the topology of X, we may assume that U is quasi-compact. Then U and X −U are constructible, hence open and closed in the constructible topology, and we have x ∈U and y ∈X −U. So Xcons is Hausdorff and totally disconnected.
We show that X is quasi-compact. Let B be the collection of quasi-compact open subsets of X and of their complements. Then B generates the topology of X, so, by the Alexander subbase theorem ([25, Lemma 08ZP]), it suffices to show that every covering of X by elements of B has a finite refinement. As the collection B is stable by taking complements, it also suffices to show that every subcollection of B that has the finite intersection property has nontrivial intersection.
So let B′ ⊂ B, and suppose that B′ has the finite intersection property and that T B∈B′ B = ∅. Using Zorn’s lemma, we may assume that B′ is maximal among all subcol-lections that the finite intersection property and empty intersection. Let B′′ ⊂B′ be the set of B ∈B′ that are closed, and let Z = T B∈B′′ B; this is a closed subset of X, and it is not empty because X is quasi-compact.
Suppose first that Z is reducible. Then we can write Z = Z′ ∪Z′′, with Z′, Z′′ ̸= Z. In particular, we have Z′ ̸⊂Z′′ and Z′′ ̸⊂Z′, so we can find quasi-compact open subsets U ′ and U ′′ of X such that U ′ ⊂X −Z′′, U ′′ ⊂X −Z′, U ′ ∩Z′ ̸= ∅and U ′′ ∩Z′′ ̸= ∅.
Let B′ = X −U ′ ⊃Z′′ and B′′ = X −U ′′ ⊃Z′. We want to show that B′ ∪{B′} or B′ ∪{B′′} has the finite intersection property, which will contradict the maximality of B′.
Suppose that this does not hold, then there exist there exist B′ 1, . . . , B′ n, B′′ 1, . . . , B′′ m ∈B′ such that B′ ∩B′ 1 ∩. . .∩B′ n = B′′ ∩B′′ 1 ∩. . .∩B′′ m = ∅, then Z ∩B′ 1 ∩. . .∩B′ n ∩B′′ 1 ∩. . .∩B′′ m = ∅.
As B′ 1∩. . .∩B′ n∩B′′ 1 ∩. . .∩B′′ m is quasi-compact and Z = T B∈B′′ B with every B ∈B′′ closed, we can find B1, . . . , Br ∈B′′ such that B1 ∩. . . ∩Br ∩B′ 1 ∩. . . ∩B′ n ∩B′′ 1 ∩. . . ∩B′′ m = ∅; but this contradicts the fact that B′ has the finite intersection property. So Z cannot be reducible.
Suppose now that Z is irreducible. As X is sober, Z has a unique generic point, say η. Let U ∈B′ −B′′. Then U is quasi-compact and the family of closed subsets (B ∩U)B∈B′′ has the finite intersection property, so their intersection Z ∩U is nonempty; as Z ∩U is an open 27 I The valuation spectrum subset of Z, it contains its generic point η. But then η ∈T B∈B′ B, which contradicts the fact that T B∈B′ B = ∅.
Corollary I.2.4.2. (See proposition 3.23 of .) Let X be a spectral space. Then : (i) The constructible topology is finer than the original topology on X.
(ii) A subset of X is constructible if and only if it is open and closed in the constructible topology (i.e. if and only if it is both ind-constructible and pro-constructible).
(iii) If U is an open subspace of X, then the map Ucons →Xcons is open and continuous (i.e. the subspace topology on U induced by the constructible topology on X is the constructible topology on U).
Proof.
(i) Every open subset of X is the union of its quasi-compact open subsets, hence is ind-constructible.
(ii) We already know that constructible subsets of X are open and closed in Xcons. Conversely, let Y ⊂X be open and closed in Xcons. Then Y is ind-constructible, so we can write Y = S i∈I Yi, with the Yi constructible. Also, by I.2.4.1, Y is quasi-compact for the constructible topology, so there exists a finite subset I′ of I such that Y = S i∈I′ Yi, which proves that Y is constructible.
(iii) Let U be an open subset of X. By [25, Lemma 005J], the map Ucons →Xcons is continuous.
Conversely, let E be a constructible subset of U; we want to show that E is ind-constructible in X. We can write U = S i∈I Ui, with the Ui quasi-compact. For every i ∈I, the set E ∩Ui is constructible in Ui, hence constructible in X by [25, Lemma 09YD]. So E = S i∈I(E ∩Ui) is ind-constructible in X.
Corollary I.2.4.3. ([25, Lemma 0902].) Let X be a spectral space and Y ⊂X be closed in the constructible topology. Then Y is a spectral space for the subspace topology.
Definition I.2.4.4. A continuous map f : X →Y of locally spectral spaces is called spectral if, for every open spectral subspace V of Y and every open spectral subspace U of f −1(V ), the induced map U →V is quasi-compact.
Proposition I.2.4.5. (Proposition 3.27 of .) Let f : X →Y be a continuous map of spectral spaces. Then the following are equivalent : (i) f is spectral.
(ii) f : Xcons →Ycons is continuous.
(iii) f is quasi-compact.
28 I.2 The valuation spectrum of a ring (iv) The inverse image by f of every constructible subset of Y is a constructible subset of X.
If these conditions are satisfied, then f : Xcons →Ycons is proper.
Proof. The implications (i)⇒(iii)⇒(iv)⇒(v)⇔(ii) are clear. By proposition I.2.3.5 and corollary I.2.4.2(ii), an open subset of X is quasi-compact if and only if it is open and closed in the constructible topology, so (ii) implies (iii). It remains to show that (iii) implies (i). Suppose that f is quasi-compact. Let V ⊂Y and U ⊂f −1(V ) be open spectral subspaces, and let W be a quasi-compact open subset of V . We want to show that f −1(W) ∩U is quasi-compact, but this follows from the fact that f −1(W) and U are quasi-compact and that X is quasi-separated.
Finally, as Xcons and Ycons are Hausdorff and quasi-compact by proposition I.2.4.1, if f : Xcons →Ycons is continuous, then it is proper.
I.2.5 A criterion for spectrality The goal of this section is to give a criterion for a topological space to be spectral, due to Hochster.
Theorem I.2.5.1. (Proposition 3.31 of .) Let X′ = (X0, T ′) be a quasi-compact topological space, let U ⊂T ′ be a collection of open and closed subsets of X′, let T be the topology on X0 generated by U , and set X = (X0, T ).
If X is T0, then it is spectral, every element of U is a quasi-compact open subset of X, and X′ = Xcons.
Proof. After replacing U by the collection of finite intersections of sets of U , we may assume that U is stable by finite intersections. Note that the topology of X is coarser than the topology of X′. In particular, every quasi-compact subset of X′ is also quasi-compact as a subset of X; in particular, X itself is quasi-compact. As elements of U are closed in X′, the previous sentence also applies to them, and we see that they are all quasi-compact as subsets of X. So the topology of X has a basis of quasi-compact open subsets which is stable by finite intersections. Let T ′′ be the topology on X generated by the quasi-compact open subsets and their complements. By lemma I.2.5.2, it suffices to show that T ′′ = T ′.
First note that T ′′ is coarser than T ′, because T ′′ is generated by elements of U and their complements, and every element of U is open and closed for T ′. We also claim that T ′′ is Hausdorff : Indeed, let x, y ∈X such that x ̸= y; as X is T0, we may assume (up to switching x and y) that there exists U ∈U such that x ∈U and y ∈X −U, and U and X −U are both open in T ′′ by definition of T ′′. So the identity map from X′ to X′′ := (X0, T ′′) is a continuous map from a quasi-compact space from a Hausdorff space, which implies that it is a homeomorphism (see chapitre I §9 No4 corollaire 2 du th´ eoreme 2, p. 63).
29 I The valuation spectrum Lemma I.2.5.2. (Lemma 3.29 of .) Let X be a quasi-compact T0 space, and suppose that its topology has a basis consisting of quasi-compact open subsets which is stable under finite intersections. Let X′ be the topological space with the same underlying set as X, and whose topology is generated by the quasi-compact open subsets of X and their complements. Then the following are equivalent : (i) X is spectral.
(ii) X′ is Hausdorff and quasi-compact, and its topology has a basis consisting of open and closed subsets.
(iii) X′ is quasi-compact.
Also, if these conditions are satisfied, then X′ = Xcons.
Proof.
(i)⇒(ii) If X is spectral, then it is quasi-compact and quasi-separated, so X′ = Xcons by defini-tion of the constructible topology, and the fact that X′ is Hausdorff and quasi-compact is proposition I.2.4.1. Also, the topology of X′ is generated by the consrtuctible subsets of X, which are open and closed as subsets of X′ corollary I.2.4.2(ii).
(ii)⇒(iii) Obvious.
(iii)⇒(i) We already know that X is quasi-separated and has a basis of quasi-compact open subsets.
So we just need to show that X is sober. By definition of the topology of X′, this topology is finer than the topology of X.
Let Z be a closed irreducible subset of X. As X is T0, this subset has at most one generic point, and we want to show that it has at least one. Let Z′ be the set with the topology induced by the topology of X′. As X′ is quasi-compact and Z′ is closed in X′ (because the topology of X′ is finer than that of X), Z′ is also quasi-compact. Suppose that Z has no generic point. For every z ∈Z, we have {z} ⊊Z, so there exists an open quasi-compact subset Uz fo Z such that Uz ∩{z} = ∅. We have Z = S z∈Z(Z \Uz), and each Z \Uz is open in Z′ by definition of the topology of X′. As Z′ is quasi-compact, there exists z1, . . . , zn ∈Z such that Z = Sn i=1(Z\Uzi); but this contradicts the irreducibility of Z.
30 I.2 The valuation spectrum of a ring I.2.6 Spectrality of Spv(A) Theorem I.2.6.1. (Proposition 4.7 of .) Let A be a commutative ring. Then Spv(A) is spectral. The open subsets U f1, . . . , fn g = {|.| ∈Spv(A) | |f1|, . . . , |fn| ≤|g| ̸= 0}, for f1, . . . , fn, g ∈A, are quasi-compact, and they and their complements generate the con-structible topology of Spv(A).
Moreover, if ϕ : A → B is a morphism of rings, then the induced map Spv(ϕ) : Spv(B) →Spv(A) is spectral.
Remark I.2.6.2. The continuous map supp : Spv(A) →Spec(A) is also spectral, since supp−1(D(g)) = U g g for every g ∈A.
Proof. Let X = Spv(A), and denote its topology by T . We want to apply theorem I.2.5.1 to the family of subsets U f1,...,fn g , for f1, . . . , fn, g ∈A.
(A) For every x ∈X, if (℘x, Rx) is the pair corresponding to x as in remark I.2.1.2, we define a relation |x on A by : f|xg if there exists an element a of Rx such that a(f +℘x) = g+℘x.
(In other words, if |.|x is a valuation in the equivalence class corresponding to x, we have f|xg if and only |f|x ≥|g|x.) This defines a map ρ from X to the set {0, 1}A×A of relations on A. It follows directly from the definition that, for every x ∈X, we have supp(x) is the set of f ∈A such that 0|xf; so ρ(x) determines supp(x).
(B) Claim : The map ρ : X →{0, 1}A×A is injective.
Indeed, let x, y ∈X be such that ρ(x) = ρ(y), and let (℘x, Rx) and (℘y, Ry) be the pairs corresponding to x and y. By the remark at the end of (A), we have ℘x = ℘y. Let K = Frac(A/℘x). Let a ∈K. Then we can find f, g ∈A such that a = (f+℘x)(g+℘x)−1, and we have : a ∈Rx ⇔g|xf ⇔g|yf ⇔a ∈Ry.
So Rx = Ry.
(C) Claim : The image of ρ is the set of relations | on A satisfying the following conditions : for all f, g, h ∈A, (a) f|g or g|f; (b) if f|g and g|h, then f|h; (c) if f|g and f|h, then f|(g + h); 31 I The valuation spectrum (d) if f|g, then fh|gh; (e) if fh|gh and 0 ̸ |h, then f|g; (f) 0 ̸ |1.
Let’s prove this claim. It is easy to check that every element in the image of ρ satisfies (a)-(f).
Conversely, let | be a relation on A satisfying (a)-(f). Note that we have f|0 for every f ∈R by (d). Let ℘= {f ∈A such that 0|f}. Then ℘is an ideal of A by (c) and (d), it does not contain 1 by (f), and it is a prime ideal by (e). Next we note that, for all f, f ′ ∈A such that f + ℘= f ′ + ℘, we have f|f ′ and f ′|f; indeed, if f ′ −f ∈℘, then 0|(f ′ −f), so f|(f ′ −f) by (b) (and the fact that f|0), hence f|f ′ by (c), and similarly f ′|f. Let’s check that | induces a relation on A/℘. Let f, f ′, g, g′ ∈A such that f −f ′, g −g′ ∈℘. We need to show that f|g ⇒f ′|g′; but we have just seen that f ′|f and g|g′, so this follows from (b). Now let R be the set of a ∈Frac(A/℘) that can be written a = (f + ℘)(g + ℘)−1, with f, g ∈A and g|f; by (d) and (e), this condition is actually independent of the choice of f and g. We see easily that R is a subring of Frac(A/℘), and it is a valuation subring thanks to condition (a). So the pair (℘, R) defines a point x of X, and ρ(x) is equal to | by definition of ℘and R.
(D) We put the discrete topology on {0, 1} and the product topology on {0, 1}A×A. By Ty-chonoff’s theorem, the space {0, 1}A×A is compact. Let T ′ be the topology on X induced by the topology of {0, 1}A×A via ρ, and let X′ = (X, T ′). Then X′ is also compact.
Indeed, for f, g, h ∈A fixed, each of the conditions (a)-(f) of (X) defines a closed subset of {0, 1}A×A, so, by the conclusion (C), the subset ρ(X) of {0, 1}A×A is closed.
(E) Let f1, . . . , fn, g ∈A. Then an element | of ρ(X) is in ρ U f1,...,fn g if and only g|fi for every i ∈{1, . . . , n} and 0 ̸ |g. Each of these conditions defines an open and closed subset on {0, 1}A×A (which is the pullback by one of the projection maps {0, 1}A×A of a subset of {0, 1}), so U f1,...,fn g is open and closed in the topology T ′. Also, the space X is T0 by remark I.2.2.6. Applying theorem I.2.5.1 to X′ and to the collection of subsets U f1,...,fn g , for f1, . . . , fn, g ∈A, we get that X is spectral, that the U f1,...,fn g form a basis of quasi-compact open subsets of X, and that X′ = Xcons.
(F) Finally, we prove the last statement. Let ϕ : A →B be a morphism of rings. By propo-sition I.2.4.5, to show that the continuous map Spv(f) : Spv(B) →Spv(A) is spectral, it suffices to show that it is quasi-compact; but this follows from the fact that Spv(ϕ)−1 U f1, . . . , fn g = U ϕ(f1), . . . , ϕ(fn) ϕ(g) for all f1, . . . , fn, g ∈A.
32 I.3 The specialization relation in Spv(A) Remark I.2.6.3. It follows from the theorem that the collection of consrtuctible subsets of Spv(A) is the smallest collection of subsets of Spv(A) that is stable by finite unions, finite intersections and complements and that contains all subsets of the form U f g , for f, g ∈A. It is also the smallest collection of subsets of Spv(A) that is stable by finite unions, finite intersections and complements and that contains all subsets of the form U ′(f, g) := {|.| ∈Spv(A) | |f| ≤|g|}, for f, g ∈A.
Indeed, we have for all f, g ∈A : U f g = U ′(f, g) ∩(Spv(A) −U ′(g, 0)) and U ′(f, g) = U f g ∪ Spv(A) − U 0 g ∪U 0 f .
I.3 The specialization relation in Spv(A) The goal of this section is to study the specialization relation in Spv(A). Here are the main points : (1) Specialization is an order relation on Spv(A) (and more generally on any T0 space), and it tells us which constructible subsets are open or closed.
(2) Specialization in Spv(A) breaks into two simpler cases, horizontal specialization and ver-tical specialization. More precisely (see theorem I.3.4.3), if x ∈Spv(A), every specializa-tion of x is a horizontal specialization of a vertical specialization of x; moreover, in many cases (for example if |A|x ̸⊂Γx,≥1), every specialization of x is also a vertical specializa-tion of a horizontal specialization of x.
(3) We say that a specialization y of x (in Spv(A)) is a vertical specialization if ℘y = ℘x. So vertical specializations of x are parametrized by valuation subrings of K(x) contained in Rx (i.e.
by valuation subrings of of Rx/mRx), and vertical generizations of x are parametrized by valuation subrings of K(x) containing Rx (i.e. by prime ideals of Rx).
See theorem I.1.4.2.
(4) Horizontal specialization changes the support of a valuation. Here are some facts about it.
Let x ∈Spv(A).
(a) If H is any subgroup of Γx, we define a map |.|x|H : A →H ∪{0} by |a|x|H = |a|x if |a|x ∈H 0 otherwise.
If this is a valuation, it is a specialization of x, and we call this a horizontal specialization of x. Also, |.|x|H is a valuation if and only if H is a convex subgroup of Γx and contains the convex subgroup cΓx generated by |A|x ∩Γx,≥1. (Note that |.|x|H does not determine H in general.) 33 I The valuation spectrum (b) Horizontal specializations of x are totally ordered (by specialization), and the mini-mal horizontal specializaton of x is x|cΓx.
(c) A horizontal specialization y of x is uniquely determined by its support, and it is the generic point of {x} ∩supp−1(℘y).
(d) The possible supports of horizontal specializations of x are the x-convex prime ideals of A (where ℘∈Spec(A) is x-convex if 0 ≤|a|x ≤|b|x and b ∈℘implies that a ∈℘).
(e) If |A|x ̸⊂Γx,≤1, then every fiber of supp : {x} →Spec(A) contains a horizontal specialization of x, which is its generic point.
I.3.1 Specialization in a spectral space Notation I.3.1.1. Let X be a topological space, and let x, y ∈X. Remember from definition I.2.2.4(ii) that we say that y specializes to x (or that x is a specialization of y, or that y is a generization of x) if x ∈{x}.
If this is the case, we write y ⇝x or x ⇝ y.
Lemma I.3.1.2. If X is T0, then specialization is an order relation on X.
Proof. Specialization is clearly reflexive and transitive (without any condition on X), so we just need to check that it is antisymmetric. Let x, y ∈X such that x ⇝y and y ⇝x. Then x ∈{y} and y ∈{x}, so x and y are both generic points of the irreducible closed subset {x}. By remark I.2.2.5, this implies that x = y.
The following result is one of the reasons that it is useful to understand the specialization relation in a spectral space.
Proposition I.3.1.3. (Proposition 3.30 of and [25, Lemma 0903]). Let X be a spectral space and Z ⊂X be a subspace.
(i) The following are equivalent : (a) Z is pro-constructible; (b) Z is spectral, and the inclusion Z →X is spectral.
(ii) (a) Suppose that Z is pro-constructible. If x ∈Z, then x is the specialization of a point of Z. (In other words, Z is the set of specializations of points of Z.) (b) If Z is pro-constructible, then Z is closed if and only if it is stable under specializa-tion.
34 I.3 The specialization relation in Spv(A) (c) If Z is ind-constructible, then Z is open if and only if it is stable under generization.
Proof.
(i) Suppose that (b) holds. Then Z is closed in Xcons by proposition I.2.4.5, which means that it is pro-constructible.
Suppose that (a) holds. Then Z is spectral by corollary I.2.4.3. We want to show that the inclusion Z →X is spectral. By proposition I.2.4.5, this is equivalent to the fact that it is quasi-compact. Let U be a quasi-compact open subsets of X. Then U ∩Z is pro-constructible. As Xcons is quasi-compact by proposition I.2.4.1, its closed subset U ∩Z is also quasi-compact, and so U ∩Z is quasi-compact in the topology induced by X (which is coarser than the constructible topology).
(ii) Note that (b) follows immediately from (a), and that (c) is just (b) applied to X −Z. So we just need to prove (a). Let x ∈Z. Let (Ui)i∈I be the family of all quasi-compact open neighborhoods of x in X. Then (Ui)i∈I is cofinal in the set of all open neighborhoods of x and stable by finite intersection, because X is spectral, and Ui ∩Z ̸= ∅for every i ∈I because x ∈Z. Also, for every i ∈I, the intersection Z ∩Ui is pro-constructible, hence closed in Xcons. As Xcons is quasi-compact (proposition I.2.4.1), we have Z∩T i∈I Ui ̸= ∅.
Let z ∈Z ∩T i∈U Ui. Then z is contained in every open neighborhood of x, so x ∈{z}.
We easily see that this implies the following corollary.
Corollary I.3.1.4. ([25, Lemma 09XU].) Let f : X →Y be a bijective continuous spectral map of spectral spaces. If generizations (resp. specializations) lift along f, then f is a homeomor-phism.
I.3.2 Vertical specialization We now fix a commutative ring A.
Remark I.3.2.1.
(1) Let ℘, ℘′ ∈Spec(A). Then ℘is a specialization of ℘′ if and only if ℘⊃℘′. Indeed, the smallest closed subset of Spec(A) containing ℘′ is V (℘′), that is, {℘′} = V (℘′).
(2) Let x, y ∈Spv(A). If x is a specialization of y, then supp(x) ⊃supp(y). (This just follows from the fact that supp : Spv(A) →Spec(A) is continuous.) Definition I.3.2.2. Let x, y ∈Spv(A). We say that x is a vertical specialization of y (or that y is a vertical generization of x) if x is a specialization of y and supp(x) = supp(y).
35 I The valuation spectrum Remember that supp−1(℘) = RZ(Frac(A/℘)) for every ℘∈Spec(A) (proposition I.2.1.4), and that the specialization relation in RZ(K), for K a field, just corresponds to the inclusion of valuation subrings (remark I.1.6.4). So points (i) and (ii) of the following proposition are a direct consequence of corollary I.1.4.4, and point (ii) follows immediately from theorem I.1.4.2.
Proposition I.3.2.3. Let x ∈Spv(A). Then : (i) There is a canonical order-reversing bijection from the set of vertical generizations of x (ordered by specialization, where we think of the specialization as the “smaller” element) to Spec(Rx) (ordered by inclusion). If y = (℘x, Ry) is a vertical generization of x, then the corresponding element of Spec(Rx) is mRy; conversely, if ℘is a prime ideal of Rx, the corresponding vertical generization of x is (℘x, (Rx)℘).
(ii) There is a canonical order-preserving bijection from the set of vertical generizations of x (ordered by specialization) to the set of convex subgroups of Γx (ordered by inclusion). If y = (℘x, Ry) is a vertical generization of x, the corresponding convex subgroup of Γx is Ker(Γx →ΓRy).
(iii) There is a canonical order-preserving bijection from the set of vertical specializations of x (ordered by specialization) to Spv(Rx/mRx) = RZ(Rx/mRx) (also ordered by special-ization).
Notation I.3.2.4. If x ∈Spv(A) and H is a convex subgroup of Γx, we denote by x/H the vertical generization of x corrseponding to H.
I.3.3 Horizontal specialization Definition I.3.3.1. Let x ∈Spv(A). The characteristic group of x is the convex subgroup cΓx of Γx generated by Γx,≥1 ∩|A|x.
Example I.3.3.2.
(1) We have cΓx = {1} if and only if |a|x ≤1 for every a ∈A. This does not hold in general.
(2) If A = K is a field, then |K|x = {0} ∪Γx, so cΓx = Γx.
(3) If K is a field, A is a valuation subring of K and x ∈Spv(A) is the corresponding valua-tion, then |a|x ≤1 for every a ∈A, so cΓx = {1}.
Definition I.3.3.3. Let x ∈Spv(A) and let H be a subgroup of Γx.
We define a map |.|x|H : A →Γx ∪{0} by setting |f|x|H = |f|x if |f|x ∈H 0 otherwise.
If |.|x|H is a valuation on A, we denote the corresponding point of Spv(A) by x|H.
36 I.3 The specialization relation in Spv(A) Proposition I.3.3.4. (Remarks 4.15 and 4.16 of .) Let x ∈Spv(A) and let H be a convex subgroup of Γx. We denote the map A →A/℘x ⊂Frac(A/℘x) by π. Then the following are equivalent : (i) |.|x|H is a valuation on A.
(ii) H ⊃cΓx.
(iii) If we denote by ℘H = {a ∈Rx | ∀δ ∈H, |a|x < δ} the corresponding prime ideal of Rx (see corollary I.1.4.4), then π(A) = A/℘x ⊂Rx,℘H.
Proof.
(i)⇒(ii) Suppose that |.|x|H is a valuation on A. To show that H contains cΓx, it suffices to show that it contains Γx,≥1 ∩|A|x. So let a ∈A be such that |a|x ≥1. If |a|x = 1, we have |a|x ∈H, so assume that |a|x > 1. Then |a + 1|x = |a|x. If |a|x ̸∈H, then 0 = |a + 1|x|H = max{|a|x|H, 1} = 1, contradiction. So |a|x ∈H.
(ii)⇒(iii) Note that Rx,℘H is the valuation subring of Frac(A/℘x) corresponding to the composition of |.|x : Frac(A/℘x) →Γx ∪{0} and of the quotient map Γx ∪{0} →(Γx/H)∪{0}, so an element a of Frac(A/℘x) is in Rx,℘H if and only if there exists γ ∈H such that |a|x ≤γ.
Now suppose that H ⊃cΓx, and let a ∈A. We want to show that there exists γ ∈H such that |a|x ≤γ. If |a|x ≤1, then we can take γ = 1. If |a|x > 1, then |a|x ∈cΓx ⊂H, so we can take γ = |a|x.
(iii)⇒(i) We have seen that (iii) implies that, for every a ∈A, there exists γ ∈H such that |a|x ≤γ; in particular, if |a|x ≥1, then |a|x ∈H, because H is convex. (This proves (ii).) We check that |.|x|H satisfies the conditions of definition I.1.1.10(ii).
It is clear that |0|x|H = 0 and |1|x|H = 1. Let a, b ∈A. The only way we can have |ab|x|H ̸= |a|x|H|b|x|H is if |ab|x ∈H but |a|x, |b|x ̸∈H ∪{0}. So suppose that |a|x, |b|x ̸∈H ∪{0}. This implies that |a|x, |b|x < 1, and then |ab|x ≤|a|x ≤1 and |ab|x ≤|b|x ≤1, so the convexity of H implies that |ab|x cannot be in H. Finally, we prove that |a + b|x|H ≤max{|a|x|H, |b|x|H}.
If |a + b|x ̸∈H, this obviously holds, so suppose that |a + b|x ∈H. We may assume that |a|x ≤|b|x. If |b|x ∈H, we are done, so we may assume that |b|x ̸∈H. But then |a + b|x ≤|b|x ≤1 and H is convex, so this impossible.
Proposition I.3.3.5. (Remark 4.16 of .) We use the notation of proposition I.3.3.4, and we denote the quotient map A →A/℘x by π. If the conditions of this proposition hold, then : (i) x|H is a specialization of x.
(ii) Ker(x) ⊂Ker(x|H) = π−1(℘HRx,℘H) = {a ∈A | ∀γ ∈H, |a|x < γ}, with equality if and only if H = Γx (which is also equivalent to ℘H = 0, or to x = x|H).
37 I The valuation spectrum (iii) Rx/℘H is a valuation subring of the field Rx,℘H/℘H, and |.|x|H is the composition of A π →Rx,℘H →Rx,℘H/℘H and of the valuation corresponding to Rx/℘H.
Note that it is not easy in general to describe the valuation group of x|H (beyond the obvious fact that it is included in H).
Proof.
(i) Let f1, . . . , fn, g ∈A such that x||H ∈U := U f1,...,fn g . We want to show that x ∈U. We have |fi|x|H ≤|g|x|H ̸= 0 for every i, so in particular |g|x ∈H and |g|x = |g|x|H. Suppose that we |fi|x > |g|x for some i. Then |fi|x ̸∈H, so |fi|x < 1 (because H ⊃cΓx), so |g|x < |fi|x < 1, which implies that |fi|x ∈H because H is convex, contradiction. So x ∈U.
(ii) The inclusion Ker(x) ⊂Ker(x|H) is obvious.
As ℘HRx,℘H is the maximal ideal of Rx,℘H and Rx,℘H is the valuation subring of Frac(A/℘x) corresponding to the valuation Frac(A/℘x) |.|x →Γx ∪{0} →(Γx/H) ∪{0}, we have ℘HRx,℘H = {a ∈Frac(A/℘x) | ∀γ ∈H, |a|x < γ}.
Let a ∈A. We have a ∈Ker(x|H) if and only if |a|x ̸∈H. As H ⊃cΓx ⊃Γx,≥1 ∩|A|x, this implies that |a|x < 1. So we have |a|x ̸∈H if and only if |a|x < γ for every γ ∈H.
(Indeed, if |a|x ̸∈H and |a|x > γ for some γ ∈H, then γ < |a|x < 1, contradicting the convexity of H). This proves the formula for Ker(x|H).
We obviously have Ker(x) = Ker(x|H) if Γx = H.
Conversely, suppose that Ker(x) = Ker(x|H), and let’s prove that H = Γx. Let γ ∈Γx. Write γ = |a|x/|b|x, with a, b ∈A. Then |a|x, |b|x ̸= 0, so |a|x,H, |b|x,H ̸= 0, so |a|x, |b|x ∈H, so γ ∈H.
(iii) In Frac(A/℘x), we have the valuation subring with valuation group Γx, and the bigger ring Rx,℘H with valuation group Γx/H and maximal ideal ℘H.. So we are in the situation of theorem I.1.4.2, and we see that Rx/℘H is a valuation subring of Rx,℘H/℘H with valuation group Ker(Γx →Γx/H) = H. The second statement also follows easily from this theorem and from (ii).
Definition I.3.3.6. Let x, y ∈Spv(A). We say that y is a horizontal specialization of x (or that x is a horizontal generization of y) if y is the form x|H, for H a convex subgroup of Γx containing cΓx.
Remark I.3.3.7. Note that x is a horizontal specialization of itself (corresponding to the convex subgroup H = Γx).
We now give a decription of the possible kernels of horizontal specializations of x.
Definition I.3.3.8. Let x ∈Spv(A). A subset T of A is called x-convex if for all a1, a2, b ∈A, if |a1|x ≤|b|x ≤|a2|x and a1, a2 ∈T, then b ∈T.
38 I.3 The specialization relation in Spv(A) For example, if 0 ∈T, then T is x-convex if and only if, for all a ∈A and b ∈T such that |a|x ≤|b|x, we have a ∈T.
Proposition I.3.3.9. (Proposition 4.18 of .) Let x ∈Spv(A). Let C be the set of x-convex prime ideals of A, order by inclusion, and let S be the set of horizontal specializations of x, ordered by specialization. Then y 7− →Ker(y) induces an order-reversing bijection S ∼ →C, and the sets S and C are totally ordered.
Remark I.3.3.10. So we have a commutative diagram : S / ≀ {x} / Spv(A) supp C / Spec(A/℘x) / Spec(A) We will see in the next subsection (see corollary I.3.4.5) that, if |A|x ̸⊂Γx,≤1 (which occurs for example if A contains a subfield on which the valuation |.|x is nontrivial), then C is actually the image of {x} in Spec(A), so every x-convex prime ideal q of A lifts to a unique horizontal specialization y of x, and moreover this y is the unique generic point of supp−1(q) ∩{x}.
Proof of proposition I.3.3.9. The kernel of any horizontal specialization of x is x-convex by proposition I.3.3.5(ii), so supp : S →C is well-defined, and it is clearly order-reversing.
We have seen in proposition I.3.3.4 that S is in bijection to the set P of prime ideals ℘of Rx such that A/℘x ⊂Rx,℘, and this bijection is order-reversing. As the set of ideals of a valuation ring is totally ordered (proposition I.1.1.4), this implies that S is totally ordered. If we identify S and P, the map supp : S →C becomes ℘7− →π−1(℘Rx,℘), where π : A →Frac(A/℘x) is the quotient map (see proposition I.3.3.5(ii)).
Now we construct the inverse map. If q is a x-convex prime ideal of A, then we clearly have q ⊃℘x, and |a|x < |b|x for every a ∈q and every b ∈A −q; in particular, |q|x ⊂Γx,<1.
So, if H is the convex subgroup of Γx generated by |A −q|x, we have H ⊃cΓx. We want to prove that supp(x|H) = q, or equivalently that q = π−1(℘Rx,℘).
But we know from proposition I.3.3.5(ii) that Ker(x|H) = {a ∈A|∀γ ∈H, |a|x < γ}, and this is equal to {a ∈A|∀b ∈A −q, |a|x < |b|x} by definition of H.
Proposition I.3.3.11. (Proposition 4.4.2 of .) Let x, y, z ∈Spv(A). If y is a horizontal specialization of x and z is a horizontal specialization of y, then z is a horizontal specialization of x.
Proof. Let H be a convex subgroup of Γx and G be a convex subgroup of Γy such that y = x|H and z = y|G. In particular, we have Γy ⊂H. We denote by G′ the smallest convex subgroup of H containing G. Then : 39 I The valuation spectrum (a) G = G′ ∩Γy : Indeed, we obviously have G ⊂G′ ∩Γy. Conversely, if δ ∈G′, then, by definition of G′, there exist γ1, γ2 ∈G such that γ1 ≤δγ2. If moreover δ ∈Γy, then this implies that δ ∈G, because G is convex in Γy.
(b) G′ ⊃cΓx : As G′ is convex, it suffices to show that G′ ⊃|A|x ∩Γx,≥1. Let a ∈A such that |a|x ≥1. Then |a|x ∈H, so |a|y = |a|x|H = |a|x ≥1, so |a|x = |a|y ∈G ⊂G′.
(c) x|G′ (which makes sense by (b)) is equal to z : Indeed, this follows immediately from definition I.3.3.3 and from (a).
I.3.4 Putting things together Proposition I.3.4.1. (Lemma 4.19 of .) (i) If x ⇝y is a horizontal specialization and y ⇝z is a vertical specialization, then there exists a vertical specialization x ⇝y′ that admits z as a horizontal specialization.
x / ∃ y y′ / z (ii) If x ⇝y is a horizontal specialization and x ⇝y′ is a vertical specialization, then there exists a unique horizontal specialization y′ ⇝z such that z is a vertical specialization of y.
x / y ∃!
y′ / z Proof.
(i) Let H ⊃ cΓx be a convex subgroup of Γx such that y = x|H, and ℘H be the corresponding prime ideal of Rx.
Then |.|y is the composition of A →A/℘x →Rx,℘H →Rx,℘H/℘HRx,℘H and of the valuation on Rx,℘H/℘H corre-sponding to the valuation subring Rx/℘H, so ℘y = Ker(A →Rx,℘H/℘H) and we have an extension of fields K(y) ⊂Rx,℘H/℘H such that Ry = K(y) ∩(Rx/℘H). Let Rz ⊂Ry be the valuation subring of K(y) corresponding to the vertical specialization y ⇝z. Then, by corollary I.1.2.4, there exists a valuation subring B ⊂Rx/℘H of Rx,℘H/℘H such that Rz = K(y) ∩B. Let B′ be the inverse image of B in Rx,℘H. Then B′ is a valuation subring of K(x) by theorem I.1.4.2(ii), and B′ ⊂Rx by definition. Let y′ be the valuation on A given by the composition of A →A/℘x ⊂K(x) and of |.|B′ (so that ℘y′ = ℘x and Ry′ = B′). Then y′ is a vertical specialization of x.
40 I.3 The specialization relation in Spv(A) It remains to show that z is a horizontal specialization of y′. Let q = Ry′∩℘H ∈Spec(Ry′).
Then Ry′,q = Rx,℘H by theorem I.1.4.2(i)(c), so Ry′,q contains the image of A/℘y′ = A/℘x, so it defines a horizontal specialization of y′ by proposition I.3.3.4.
By proposition I.3.3.5, this horizontal specialization is the composition of A →A/℘y′ ⊂Ry′,q = Rx,℘H →Rx,℘H/℘HRx,℘H and of the valuation defined by the valuation subring Ry′,q/q = B; in other words, it is the valuation z.
(ii) We have ℘z = ℘y for every vertical specialization z and y, and horizontal valuations of y′ are uniquely determined by their kernel (by proposition I.3.3.9), so z is unique if it exists.
Let H be the convex subgroup of Γy′ such that x = y′/H (so we have Γx = Γy′/H), let G ⊃cΓx be a convex subgroup of Γx such that y = x|G, and denote by G′ the inverse image of G in Γy′. Then G′ is a convex subgroup of Γy′, and we have G′ ⊃cΓy′ (indeed, if a ∈A is such that |a|y′ ≥1, then |a|x ≥1 because |a|x is the image of |a|y′ by the quotient map Γy′ →Γx, so |a|x ∈G, and finally |a|y′ ∈G′). Let z = y′ |G′. Then it is easy to check (from the formulas for |.|z and |.|y given in definition I.3.3.3) that z = y/H.
Corollary I.3.4.2. (Corollary 4.20 of .) Let x ∈Spv(A), and let ℘be a generization of ℘x in Spec(A) (i.e. ℘⊂℘x). Then there exists a horizontal generization y of x such that ℘y = ℘.
Proof. Let R be the localization (A/℘)℘/℘x. Then R is a local ring and Frac(R) = Frac(A/℘), so, by theorem I.1.2.2, there exists a valuation subring B of Frac(A/℘) such that mR = R ∩mB.
Let y′ be the corresponding valuation on A (i.e. such that ℘y′ = ℘and Ry′ = B). By def-inition of y′, the image of A in K(y′) is included in Ry′, so we can construction a horizontal specialization z of y′ using the maximal ideal m of Ry′. The valuation corresponding to z is the composition of the map A →A/℘→Ry′ →Ry′/m and of the trivial valuation on Ry′/m. As mR = R ∩m and R also contains the image of A in A/℘, this valuation is also the composition of A →R →R/mR = Frac(A/℘x) and of the trivial valuation on Frac(A/℘x). In particular, we have ℘z = ℘x, so x is a vertical specialization of z. By proposition I.3.4.1(i), there exists a vertical specialization y of y′ such that x is a horizontal specialization of y; then ℘y = ℘y′ = ℘, so we are done.
Theorem I.3.4.3. (Proposition 4.21 of .) Let x, y ∈Spv(A) such that y is a specialization of x. Then : (i) There exists a vertical specialization x ⇝x′ such that y is a horizontal specialization of x′.
(ii) There exists a vertical generization y′ of y such that one of the following conditions holds : (a) y′ is a horizontal specialization of x; 41 I The valuation spectrum (b) |A|x ⊂Γx,≤1, y′ induces the trivial valuation on K(y′) = K(y), and ℘y′ = ℘y contains ℘x|{1}.
Note that the condition |A|x ⊂Γx,≤1 is equivalent to the fact that cΓx = {1}.
Proof.
(ii) We first assume that cΓx = {1} and that |A −℘y|x = {1}. Let H be the triv-ial subgroup of Γx. Then we can form the horizontal specialization x|H, and we have Ker(x|H) = {a ∈A | |a|x < 1} ⊂℘y. Let y′ be the composition of the map A →A/℘y and of the trivial valuation on Frac(A/℘y). Then y is a vertical specialization of y′ and ℘y′ ⊃Ker(x|H), so we are done. (And we are in case (b).) Now assume that cΓx ̸= {1} or |A −℘y|x ̸= {1}. We claim that the following relation holds : (∗) ∀a ∈A −℘y, ∀b ∈℘y, |a|x ≤|b|x ⇒|a|x = |b|x ̸= 0.
Indeed, the hypothesis that a ∈A −℘y and b ∈℘y means that |b|y = 0 ≤|a|y ̸= 0, which implies that y ∈U b a . As x is a generization of y, we must then also have x ∈U b a , i.e.
|b|x ≤|a|x ̸= 0. So, if |a|x ≤|b|x, we get |a|x = |b|x ̸= 0.
We prove that the ideal ℘y is x-convex. Let a, b ∈A such that b ∈℘y and |a|x ≤|b|x; we need to prove that a ∈℘y. Suppose that a ̸∈℘y; then, by (), we have |a|x = |b|x ̸= 0.
There are two cases: (1) If |A|x ̸⊂Γx,≤1 : Then there exists c ∈A such that |c|x > 1. We have |a|x < |bc|x, a ∈A −℘y and bc ∈℘y, which contradicts ().
(2) If |A|x ⊂Γx,≤1 and |A −℘y|x ̸= {1} : Then there exists c ∈A −℘y such that |c|x < 1. We have |ac|x < |b|x, ac ∈A −℘y and b ∈℘y, which again contradicts ().
So both cases are impossible, and this finishes the proof that ℘y is x-convex.
By proposition I.3.3.9, there exists a horizontal specialization y′ of x such that ℘y′ = ℘y.
To finish the proof, it suffices to show that y is a vertical specialization of y′; as ℘y = ℘y′, it suffices to show that y is a specialization of y′. So let f, g ∈A such that y ∈U f g ; we want to show that y′ ∈U f g . We have |f|y ≤|g|y ̸= 0. As x is a generization of y, this implies that |f|x ≤|g|x. We know that |a|y′ = |a|x or 0 for every a ∈A, by the formula in definition I.3.3.3. As ℘y = ℘y′, we have g ̸∈℘y′, so |g|y′ ̸= 0, hence |g|y′ = |g|x; so |f|y′ ≤|f|x ≤|g|y′, and we are done.
(i) Let y′ be the vertical generization of y given by (ii). If we are in case (ii)(a), then we get (i) by proposition I.3.4.1(i). So we may assume that we are in case (ii)(b), that is, that cΓx = {1}, that |.|y′ is trivial on K(y) and that ℘y ⊃Ker(x|H), where H = {1} ⊂Γx. By corollary I.3.4.2, there exists a horizontal generization z of y such that ℘z = Ker(x|H). As x|H induces the trivial valuation on K(x|H), it is generic in the fiber supp−1(Ker(x|H)), and so z is a vertical specialization of x|H. By proposition I.3.4.1(i), there exists a vertical 42 I.3 The specialization relation in Spv(A) specialization x′ of x such that z is a horizontal specialization of x′. Finally, by proposition I.3.3.11, y is a horizontal specialization of x′.
x / x|H y′ x′ / _ supp z / _ supp y _ supp ℘x / ℘x|H / ℘y Corollary I.3.4.4. Let x ∈Spv(A). If |A|x ̸⊂Γx,≤1, then x|cΓx has only vertical specializations.
Proof. Let y be a specialization of x|cΓx. Then y is also a specialization of x, so, by theorem I.3.4.3(ii), there exists a vertical generization y′ of y such that y′ is a horizontal specialization of x. But x|cΓx is the minimal horizontal specialization of x (remember that the set of horizontal specializations of x is totally ordered by proposition I.3.3.9), so it is a horizontal specialization of y′. As ℘y′ = ℘y ⊃℘x|cΓx (because y is a specialization of x|cΓx), we must have ℘y′ = ℘x|cΓx, hence y′ = x|cΓx by proposition I.3.3.9 again, so y is a vertical specialization of x|cΓx.
Corollary I.3.4.5. Let x ∈Spv(A). If |A|x ̸⊂Γx,≤1, then supp({x}) ⊂Spec(A) coincides with the set of x-convex prime ideals q of A, and, for every such q, the intersection supp−1(q) ∩{x} has a unique generic point, which is the horizontal specialization of x corresponding to q by proposition I.3.3.9.
Proof. Let y be a specialization of x. We want to show that ℘y is x-convex. By theorem I.3.4.3(ii), there exists a horizontal specialization y′ of x such that y is a vertical specialization of y′, and then ℘y = ℘y′, so ℘y is x-convex.
Now let q be a x-convex prime ideal of A, and let z be the unique horizontal specialization of x such that ℘z = q. and let y be a specialization of x such that ℘y = q. Then, by theorem I.3.4.3(ii), there exists a vertical specialization of y′ that is also a horizontal specialization of x.
As ℘y′ = q, we must have z = y′. So z is dense in {x} ∩supp−1(q).
Remark I.3.4.6. Let x ∈Spv(A). Suppose that A has a subfield k such that |.|x is not trivial on k. Then |A|x ̸⊂Γx,≤1. Indeed, let a ∈k be such that |a|x ̸= 1. Then either |a|x > 1 and we are done, or |a|x < 1 and then |a−1|x > 1.
43 I The valuation spectrum I.3.5 Examples Note the following useful remark to recognize specializations : Remark I.3.5.1. If x, y ∈Spv(A), then y is a specialization of x if and only if, for every f, g ∈A : (|f|y ≤|g|y ̸= 0) ⇒(|f|x ≤|g|x ̸= 0).
Indeed, y is a specialization of x if and only every open subset of Spv(A) that contains y also contains x, and the sets U f g generate the topology of Spv(A).
I.3.5.1 A = Qℓ[T]: For every r ∈(0, +∞), we define a valuation |.|r on A by the following formula : if f = P n≥0 anT n ∈A, then |f|r = max n≥0 {|an|ℓrn}.
It is not too hard to prove that |f|r = sup x∈Qℓ, |x|ℓ≤r |f(x)|ℓ.
We have Ker |.|r = (0) and cΓ|.|r = Γ|.|r = ⟨ℓ, r⟩⊂R>0. In particular, the valuation |.|r has no proper horizontal specialization.
Claim: The map (0, +∞) →Spv(A), r 7− →|.|r is continuous exactly at the points of (0, +∞) −ℓQ.
For every a ∈Qℓ, we define a valuation |.|a on A by |f|a = |f(a)|ℓ. The kernel of this valuation is (Pa), where Pa is the minimal polynomial of a over Qℓ; its valuation group is the subgroup of R>0 generated by ℓand |a|ℓ(so it is a subgroup of ℓQ), and we have cΓ|.|a = Γ|.|a.
We can also consider the T-valuation on A. Take Γ = γZ, with the convention that γ < 1, and define |.|T by f|T = γord0(f), where ord0(f) is the order of vanishing of f at 0 (and with the convention that ord0(0) = +∞and γ+∞= 0). We have Ker |.|T = (0), Γ|.|T = γZ and cΓ|.|T = {1}. So we can form the horizontal specialization |.|T|{1}. It sends f ∈A to 1 if ord0(f) = 0 and to 0 otherwise, so it is the trivial valuation on A with support (T).
These are all rank 1 valuations.
The rank 0 valuations are all of the form |.|℘,triv : A →A/℘ |.|triv →{0, 1}, where ℘∈Spec(A). For example, if ℘= (0), we get the trivial valuation of A, which is the generic point of Spv(A). We have seen in the previous paragraph that |.|(T),triv is a horizontal specialization of |.|T and that |.|T is a vertical specialization of |.|triv.
Does there exist a horizontal specialization of |.|triv that has |.|(T),triv as a vertical specialization 44 I.3 The specialization relation in Spv(A) ?
|.|triv / ?
|.|T / |.|(T),triv Let’s construct some rank 2 valuations. Consider the group Γ = R+ >0 × {1−}Z with the lexicographic order (here “1−” is just a symbol that we use to denote a generator of the second factor); in other words, we have r < 1−< 1 for every r ∈(0, 1), where we abbreviate (r, 1) to r for every r ∈R>0. Let r ∈(0, +∞), set r−= r · 1−∈Γ (so we have s < r−< r for every s ∈(0, r)), and define a valuation |.|r−on A by the following formula : if f = P n≥0 anT n ∈A, then |f|r = max n≥0 {|an|ℓ(r−)n}.
We have Ker |.|r−= (0) and cΓ|.|r−= Γ|.|r−= ⟨ℓ, r−⟩⊂Γ. It is easy to see that |.|r−is a vertical specialization of |.|r.
I.3.5.2 A = Z[T]: We can restrict all the valuations of the previous example to Z[T] (and we will use the same notation for them). But note that the groups cΓx can change.
For example, if r ∈(0, 1], we now have have cΓ|.|r = {1}, so we can form the horizontal specialization |.|′ r of |.|r corresponding to H = {1}. This is the trivial valuation on A with kernel {f ∈A | |f|r < 1}. If for example r = 1, this kernel is ℓA. If r = ℓ−1, then Ker |.|′ r is equal to { X n≥0 anT n | ∀n ∈N, |an|ℓ< ℓn} = (ℓ, T).
By proposition I.3.4.1(ii), there exists a unique horizontal specialization of |.|r−that is also a vertical specialization of |.|′ r. What is it ?
|.|r / |.|′ r |.|r− / ?
Remember that we also have the T-adic valuation |.|T : A →γZ ∪{0}. We can consider the mod ℓT-adic valuation |.|ℓ,T defined by |f|ℓ,T = γord0(f mod ℓ). This is a rank 1 valuation with kernel ℓA. It is a vertical specialization of the trivial valuation with kernel ℓA, which is a horizontal specialization of |.|r for r = 1 (it is equal to |.|′ 1). So there exists a vertical specialization of |.|1 that has |.|ℓ,T as a horizontal specialization. Question : what is this valuation 45 I The valuation spectrum ?
|.|1 / |.|′ 1 = |.|(ℓ),triv ?
/ |.|ℓ,T I.4 Valuations with support conditions Let A be commutative ring and J be an ideal of A such that Spec(A) −V (J) is quasi-compact.
The goal of this section is to study a subset of Spv(A) that we will denote by Spv(A, J). This subset will turn out to be the set of valuations on A that either have support in V (J), or that have support in Spec(A) −V (J) and have every proper horizontal specialization with support in V (J).
Our reason for studying this subset is that, if A is a topological ring of the type used by Huber (in Huber’s terminology, a f-adic ring), and if A00 is the set of topologically nilpotent elements of A, then the set Cont(A) of continuous valuations on A is a closed subset of Spv(A, A00 · A); in fact, it is the subset of valuations x such that |A00|x ⊂Γx,<1. So we will be able to deduce properties of Cont(A) from properties of the Spv(A, J). (Our reason for studying Cont(A) is that the space we are really interested in, the adic spectrum Spa(A, A+) of an affinoid ring, is a pro-constructible subset of Cont(A).) In this section, we will do the following things : - Give a more explicit definition of Spv(A, J).
- Prove that Spv(A, J) is spectral, and give an explicit base of quasi-compact open subsets (they are of the form Spv(A, J) ∩U f1,...,fn g for well-chosen sets {f1, . . . , fn} ⊂A).
- Construct a continuous and spectral retraction from Spv(A) onto Spv(A, J).
From now on, we fix a ring A an ideal J of A such that Spec(A) −V (J) is quasi-compact; remember that this last condition is equivalent to the fact that √ J is equal to the radical of a finitely generated ideal of A (see remark I.2.2.2).
Here is a summary of the section. In the first subsection, we want to construct a (somewhat explicit) retraction r : Spv(A) →Spv(A, J) such that r(x) is a horizontal specialization of x for evey x ∈Spv(A). Let x ∈Spv(A). There are three possibilities : (1) If supp(x) ∈V (J) (i.e. if |J|x = {0}), then we take r(x) = x.
(2) If supp(x) ̸∈V (J) and if x has no horizontal specialization with support in V (J), we take r(x) = x|cΓx (the minimal horizontal specialization of x; note that this is Spv(A, J), because it has no proper horizontal specializations). This happens if and only if |J|x ∩cΓx ̸= ∅. (See proposition I.4.1.1.) 46 I.4 Valuations with support conditions (3) If supp(x) ̸∈V (J) but x has at least one horizontal specialization with support in V (J), we want to take for r(x) the minimal horizontal specialization of x with support in Spec(A) −V (J). Proposition I.4.1.3 (and its corollaries) shows that such a specializa-tion exists and gives a construction of a convex subgroup HJ of Γx such that r(x) = x|HJ.
This is where we need the hypothesis on √ J : if √ J = p (a1, . . . , an), then HJ is the convex subgroup of Γx generated by max1≤i≤n{|ai|x}.
In the second subsection, we show that Spv(A, J) is spectral and give an explicit base of quasi-compact open subsets of its topology. This is a pretty straightforward application of Hochster’s spectrality criterion (theorem I.2.5.1) and of the spectrality of Spv(A) (theorem I.2.6.1).
I.4.1 Supports of horizontal specializations Proposition I.4.1.1. 4 ( Lemma 9.1.11 and Remark 9.1.12.) Let x ∈Spv(A). Then the following are equivalent : (i) |J|x ∩cΓx ̸= ∅.
(ii) |J|x ∩Γx,≥1 ̸= ∅.
(iii) Every horizontal specialization of x has support in Spec(A) −V (J).
(iv) The valuation x|cΓx has support in Spec(A) −V (J).
Proof. As x|cΓx is the horizontal specialization of x with minimal support (for the specialization relation in Spec(A)), (iii) and (iv) are equivalent. Also, as cΓx contains |A|x ∩Γx,≥1, (ii) implies (i).
Let’s prove that (i) implies (ii). Assume that (i) holds, and choose a ∈J such that |a|x ∈cΓx.
If |a|x ≥1, then (ii) holds, so we assume that |a|x < 1. By definition of cΓx, we can find b, b′ ∈A such that |b|x, |b′|x ≥1 and |b|x|b′|−1 x ≤|a|x < 1. Then 1 ≤|b|x ≤|ab′|x and ab′ ∈J, so |J|x ∩Γx,≥1 ̸= ∅.
Finally, we prove that (ii) and (iv) are equivalent. If |J|x ∩cΓx ̸= ∅, then there exists a ∈J such that |a|x ∈cΓx, and then we have |a|x|cΓx = |a|x ̸= 0, so a ̸∈supp(x|cΓx), i.e.
supp(x|cΓx) ̸∈V (J). Conversely, if supp(x|cΓx) ̸∈V (J), then J ̸⊂supp(x|cΓx), so there exists a ∈J such that |a|x|cΓx ̸= 0, and then we have |a|x = |a|x|cΓx ∈cΓx.
We now turn to the case where |J|x ∩cΓx = ∅. The following definition will be useful.
Definition I.4.1.2. If (Γ, ×) is a totally ordered abelian group and H is a subgroup of Γ, we say that γ ∈Γ ∪{0} is cofinal for H if for all h ∈H there exists n ∈N such that γn < h.
4This result holds for an ideal J of A, without the extra condition on √ J.
47 I The valuation spectrum Proposition I.4.1.3. ( Proposition 9.1.13.) Let x ∈Spv(A), and suppose that |J|x∩cΓx = ∅.
Then : (i) The set of convex subgroups H of Γx such that cΓx ⊂H and that every element of |J|x is cofinal for H is nonempty, and it has a maximal element (for the inclusion), which we will denote by HJ.
(ii) If moreover |J|x ̸= {0} (i.e.
if supp(x) ∈Spec(A) −V (J)), then HJ ̸= cΓx, |J|x ∩HJ ̸= ∅, and HJ is contained in every convex subgroup H of Γx satisfying |J|x ∩H ̸= ∅. In particular, x|HJ has support in Spec(A) −V (J), and it is the mini-mal horizontal specialization of x with that property.
Proof. If |J|x = {0}, then the elements of |J|x are cofinal for every subgroup of Γx, so every convex subgroup H ⊃cΓx satisfies the conditions of (i), and we can take HJ = Γx.
From now on, we assume that |J|x ̸= {0}, i.e. that supp(x) ̸∈V (J). By lemma I.4.1.4, none of the statements change if we replace J by its radical, so we may assume that J is finitely generated, say J = (a1, . . . , an). Let δ = max{|ai|x, 1 ≤i ≤n}. As |J|x ̸= {0}, we have δ ̸= 0, i.e. δ ∈Γx; also, as |J|x ∩cΓx = ∅, we have δ ̸∈cΓx and δ < 1. Let HJ be the convex subgroup of Γx generated by δ, that is, HJ = {γ ∈Γx|∃n ∈N, δn ≤γ ≤δ−n}.
Note that δ ∈HJ, and so we have HJ ∩|J|x ̸= ∅. We will show that this group HJ satisfies the properties of (i) and (ii).
First we show that HJ strictly contains cΓx. As convex subgroups of Γx are totally ordered by inclusion (see proposition I.1.3.4), we have cΓx ⊂HJ or HJ ⊂cΓx. As δ ∈HJ −cΓx, the second case is impossible, so cΓx ⊊HJ.
Next we show that every element of |J|x is cofinal for HJ.
Let I = {a ∈A | |a|x is cofinal for HJ}.
Note that a1, . . . , an ∈I.
Indeed, for every i ∈{1, . . . , n}, we have |ai|x ≤δ, and δ is cofinal for HJ by definition of HJ. By lemma I.4.1.4, I is a radical ideal of A, and in particular I ⊃J.
We have shown that HJ satisfies the properties of (i). Let’s show that it is maximal for these properties. Let H be a subgroup of Γx such that every element of |J|x is cofinal for H. In particular, the generator δ of HJ (which is an element of J|x) is cofinal for H. Let γ ∈H. There exists n ∈N such that δn < γ; as δ < 1, this means that δn < γ for every n big enough. As γ−1 ∈H, a similar property holds for γ−1, so we can find n ∈N such that δn < γ and δn < γ−1, and then we have δn < γ < δ−n, hence γ ∈HJ.
Finally, we prove (ii). We have already shown that cΓx ⊊HJ, and we have |J|x ∩HJ ̸= ∅by definition of HJ. Let H be a convex subgroup of Γx such that |J|x ∩H ̸= ∅. As before, using the fact that convex subgroups of Γx are totally ordered by inclusion, we see that cΓx ⊊H. To prove that HJ ⊂H, it suffices to show that δ ∈H. Let a ∈J such that |a|x ∈H. We write a = Pn i=1 biai, with b1, . . . , bn ∈A. Then |a|x ≤max{|bi|x|ai|x, 1 ≤i ≤n}, and we choose 48 I.4 Valuations with support conditions i ∈{1, . . . , n} such that |a|x ≤|bi|x|ai|x; note that |ai|x ≤δ by definition of δ. If |bi|x ≤1, then H ∋|a|x ≤|ai|x ≤δ < 1, so δ ∈H because H is convex. If |bi|x ≥1, then |bi|x ∈cΓx, so H ∋|bi|−1 x |a|x ≤|ai|x ≤δ < 1, and again we deduce that δ ∈H.
The last sentence of (ii) follows from the fact that, for any convex subgroup H ⊃cΓx of Γx, the valuation x|H has support in Spec(A) −V (J) if and only if |J|x ∩H ̸= ∅. (See for example the end of the proof of proposition I.4.1.1.) Lemma I.4.1.4. Let x ∈Spv(A).
(i) For every subgroup H of Γx, we have : |J|x ∩H = ∅⇔| √ J|x ∩H = ∅.
(ii) For every subgroup H of Γx, the following are equivalent : (a) every element of |J|x is cofinal for H; (b) every element of | √ J|x is cofinal for H.
Proof.
(i) We obviously have |J|x ∩H ⊂| √ J|x ∩H, so |J|x ∩H = ∅if | √ J|x ∩H = ∅.
Conversely, suppose that | √ J|x ∩H ̸= ∅, and let a ∈ √ J such that |a|x ∈H. There exists N ≥1 such that aN ∈J, and then |a|N x ∈|J|x ∩H, so |J|x ∩H ̸= ∅.
(ii) Obviously (b) implies (a). If (a) holds, let a ∈ √ J. Then aN ∈I for some N ≥1. Let γ ∈H. Then there exists n ∈N such that |aN|n x < γ, i.e. |a|nN x < γ. So |a|x is cofinal for H.
Lemma I.4.1.5. Let x ∈Spv(A), and let H be a subgroup of Γx such that cΓx ⊊H. Then I = {a ∈A | |a|x is cofinal for H} is a radical ideal of A.
Proof. Let a, b ∈I. As |a + b|x ≤max(|a|x, |b|x), and as both |a|x and |b|x is cofinal for H, so is |a + b|x, hence a + b ∈I.
Let a ∈I and c ∈A. If |c|x ≤1, then |ca|x ≤|a|x, so |ca|x is cofinal for H, and ca ∈I.
Suppose that |c|x > 1, then |c|x ∈cΓx ⊂H. Let γ ∈H −cΓx. As cΓx is convex, γ is either smaller than all the elements of cΓx, or bigger than all the elements of cΓx; replacing γ by γ−1 if necessary, we may assume that we are in the first case. So δ < γ−1 for every δ ∈cΓx, and in particular |c|n x < γ−1 for every n ∈N. Let n ∈N such that |a|n x < γ. Then we have, for every N ∈N, |ca|n+N x = |c|n+N x |a|n x|a|N x < γ−1|a|n x|a|N x < |a|N x , so |ca|x is cofinal for H (because |a|x is), and finally ca ∈I.
49 I The valuation spectrum So we have shown that I is an ideal. The fact that I is radical follows immediately from (ii) of lemma I.4.1.4.
Definition I.4.1.6.
(i) Let x ∈Spv(A). We define a convex subgroup cΓx(J) of Γx by the following formula cΓx(J) = HJ if |J|x ∩cΓx = ∅; cΓx if |J|x ∩cΓx ̸= ∅.
(Where HJ is defined in (i) of proposition I.4.1.3.) (ii) We define a map r : Spv(A) →Spv(A) by r(x) = x|cΓx(J).
Remark I.4.1.7. For x ∈Spv(A), the subgroup cΓx(J) of Γx depends only on √ J. Hence the map r depends only on √ J. (We actually showed this at the beginning of the proof of proposition I.4.1.3.) These objects satisfy the following properties.
Corollary I.4.1.8. ( Proposition 9.2.2.) Let x ∈Spv(A). Then : (i) cΓx(J) is a convex subgroup of Γx, and cΓx ⊂cΓx(J).
(ii) cΓx(J) = Γx if and only if every proper horizontal specialization of x has support in V (J).
(iii) If |J|x ̸= {0}, then cΓx(J) is minimal among all the convex subgroups H of Γx such that H ⊃cΓx and H ∩|J|x ̸= ∅.
(iv) If |J|x ∩cΓx = ∅, then cΓx(J) is maximal among all the convex subgroups H of Γx such that H ⊃cΓx and that every element of |J|x is cofinal for H.
(v) We have r(x) = x if and only if cΓx(J) = Γx.
Proof.
(i) This follows immediately from the definition of cΓx(J).
(iii) This is point (ii) of proposition I.4.1.3.
(iv) This is point (i) of proposition I.4.1.3.
(ii) If |J|x = {0}, then cΓx(J) = HJ = Γx and x has support in V (J) (hence all its special-izations also do).
Suppose that |J|x ̸= {0}. If cΓx(J) = Γx, let y be a proper horizontal specialization of x, and write y = x|H, with cΓx ⊂H ⊊Γx = cΓx(J); by (iii), we have H ∩|J|x = ∅, hence supp(y) ̸∈V (J). Conversely, suppose that supp(y) ̸∈V (J) for every proper horizontal specialization y of x (in particular, supp(x) ̸∈V (J)). Then H ∩|J|x = ∅for every proper convex subgroup H ⊃cΓx of Γx, so Γx is the only convex subgroup of Γx containing cΓx and meeting |J|x, hence Γx = cΓx(J) by (iii).
50 I.4 Valuations with support conditions (v) If cΓx(J) = Γx, then obviously r(x) = x. Conversely, suppose that r(x) = x. Then |A|x ⊂cΓx(J). As |A|x generates Γx, this implies that cΓx(J) = Γx.
I.4.2 The subspace Spv(A, J) Definition I.4.2.1. We define a subset Spv(A, J) of Spv(A) by Spv(A, J) = {x ∈Spv(A) | r(x) = x} = {x ∈Spv(A) | cΓx(J) = Γx}.
Remark I.4.2.2.
(1) We have supp−1(V (J)) = {x ∈Spv(A) | |J|x = {0}} ⊂Spv(A, J).
(2) If J = A, then Spv(A, A) is the set of x ∈Spv(A) having no proper horizontal specializa-tions. (This follows from corollary I.4.1.8(ii).) (3) By remark I.4.1.7, we have Spv(A, J) = Spv(A, √ J) (in other words, Spv(A, J) only depends on √ J).
(4) By definition, the map r is a retraction from Spv(A) onto Spv(A, J) (i.e.
Im(r) = Spv(A, J) and r2 = r).
Lemma I.4.2.3. ( Lemma 9.2.4) Let a1, . . . , an ∈J such that p (a1, . . . , an) = √ J, and let x ∈Spv(A). The following are equivalent : (i) x ∈Spv(A, J).
(ii) Γx = cΓx, or |a|x is cofinal for Γx for every a ∈J.
(iii) Γx = cΓx, or |ai|x is cofinal for Γx for every i ∈{1, . . . , n}.
Proof. If (i) holds and cΓx ̸= Γx, then cΓx(J) ̸= cΓx, so Γx = cΓx(J) = HJ, and the second part of (ii) holds by proposition I.4.1.1(i). Suppose that (ii) holds. If cΓx = Γx, then cΓx(J) = Γx. If every element of |J|x is cofinal for Γx, then |J|x ∩Γx,≥1 = ∅, so |J|x ∩cΓx = ∅by proposition I.4.1.1. So, by corollary I.4.1.8(iv), cΓx(J) is maximal among all the convex subgroup H ⊃cΓx of Γx such that every element of |J|x is cofinal for H; as Γx itself satisfies these properties by assumption, we have cΓx(J) = Γx.
As (ii) obviously implies (iii), it remains to show that (iii) implies (ii). Suppose that (iii) holds, and that Γx ̸= cΓx. By lemma I.4.1.5, the set of elements of a such that |a|x is cofinal for Γx is a radical ideal of A, so, if it contains a1, . . . , an, it also contains J.
We come to the main theorem of this section.
Theorem I.4.2.4. (Proposition 9.2.5 of , lemma 7.5 of .) 51 I The valuation spectrum (i) Spv(A, J) is a spectral space (for the topology induced by the topology of Spv(A)).
(ii) A base of quasi-compact open subsets for the topology of Spv(A, J) is given by the sets UJ f1, . . . , fn g = {x ∈Spv(A, J) | ∀i ∈{1, . . . , n}, |fi|x ≤|g|x ̸= 0}, for nonempty finite sets {f1, . . . , fn} such that J ⊂ p (f1, . . . , fn).
(iii) The retraction r : Spv(A) →Spv(A, J) is a continuous and spectral map.
(iv) If x ∈Spv(A) has support in Spec(A) −V (J), so does r(x).
Note that the inclusion Spv(A, J) →Spv(A) is not spectral in general.
Proof. We may assume that J is finitely generated.
We proceed in several steps.
(1) Let U be the family of subsets defined in (ii). First, as the elements of U are the intersec-tion with Spv(A, J) of open subsets of Spv(A), they are all open in Spv(A, J). Also, if f1, . . . , fn ∈A are such that J ⊂ p (f1, . . . , fn) and g ∈A, then clearly UJ f1, . . . , fn g = UJ f1, . . . , fn, g g .
(2) We show that U is stable by finite intersections. Let T, T ′ be two finite subsets of A such that J ⊂ p (T) and J ⊂ p (T ′), and let g, g′ ∈A.
We want to show that UJ T g ∩UJ T ′ g′ ∈U .
By (1), we may assume that g ∈T and g′ ∈T ′.
Let T ′′ = {ab, a ∈T, b ∈T ′}. Then J ⊂ p (T ′′), and we have UJ T g ∩UJ T ′ g′ = UJ T ′′ gg′ .
(3) We show that U is a base of the topology of Spv(A, J). Let x ∈Spv(A, J), and let U be an open neighborhood of x in Spv(A). We want to find an element of U that is contained in U. Choose f1, . . . , fn, g ∈A such that x ∈U f1,...,fn g ⊂U.
Suppose that Γx = cΓx. Then there exists a ∈A such that |g|−1 x ≤|a|x, i.e., |ag|x ≥1, and then x ∈UJ af1, . . . , afn, 1 ag ⊂U f1,...,fn g .
Suppose that Γx ̸= cΓx. Let a1, . . . , am be a set of generators of J. By lemma I.4.2.3, there exists r ∈N such that |ai|r x < |g|x for every i ∈{1, . . . , m}, and then x ∈UJ f1, . . . , fn, ar 1, . . . , ar m g ⊂U f1,...,fn g .
52 I.4 Valuations with support conditions (4) Let T be a finite subset of A such that J ⊂ p (T), and let g ∈A. We write V = UJ T g and W = U T g We claim that r−1(V ) = W.
We obviously have V ⊂W. As every point of r−1(V ) is a (horizontal) generization of a point of V and as W is open, this implies that r−1(V ) ⊂W. Conversely, let x ∈W; we want to show that y := r(x) ∈V . If |J|x = {0}, then cΓx(J) = Γx, so r(x) = x ∈V .
So we may assume that |J|x ̸= {0}, i.e. that supp(x) ̸∈V (J). Then |J|x ∩cΓx(J) ̸= ∅ (indeed, if |J|x ∩cΓx = ∅, then cΓx(J) = HJ, and HJ ∩|J|x ̸= ∅by definition); in particular, supp(y) ̸∈V (J), so (iv) holds. Let H = cΓx(J), so that y = x|H. Suppose that g ∈Ker(y). As Ker(y) is x-convex, this implies that a ∈Ker(y) for every a ∈T, so Ker(y) ⊃ p (T) ⊃J, which contradicts the fact that |J|x ∩H ̸= ∅. So g ̸∈Ker(y), and in particular |g|y = |g|x ̸= 0. As |a|y ≤|a|x for every a ∈A, we deduce that y ∈W, hence that y ∈V = W ∩Spv(A, J).
(5) Let C be the smallest collection of subsets of Spv(A, J) that contains U and is stable by finite unions, finite intersections and complements, and let X′ be Spv(A, J) with the topology generated by C . By (4), for every Y ∈C , r−1(Y ) is a constructible subset of Spv(A). Hence r : Spv(A)cons →X′ is a continuous map. Since Spv(A)cons is quasi-compact (proposition I.2.4.1) and r is surjective, X′ is also quasi-compact. By definition of the topology of X′, every element of U is open and closed in X′. Also, Spv(A, J) is T0, because it is a subspace of the T0 space Spv(A), and U is a base of the topology of Spv(A, J) by (3). So Hochster’s spectrality criterion (theorem I.2.5.1) implies that Spv(A, J) is spectral, that X′ = Spv(A, J)cons and that U is a base of quasi-compact open subsets of Spv(A, J). This shows (i) and (ii), and (iii) follows from (4).
53 II Topological rings and continuous valuations II.1 Topological rings II.1.1 Definitions and first properties Definition II.1.1.1. Let A be a topological ring.
(i) We say that A is non-Archimedean if 0 has a basis of neighborhoods consisting of sub-groups of the underlying additive group of A.
(ii) We say that A is adic if there exists an ideal I of A such that (In)n≥0 is a fundamental system of neighborhoods of 0 in A. In that case, we call the topology on A the I-adic topology and we say that I is an ideal of definition.
If M is a A-module, the topology on M for which (InM)n≥0 is a fundamental system of neighborhoods of 0 is also called the I-adic topology on M.
(iii) We say that A if a f-adic ring (or a Huber ring) if there exists an open subring A0 of A and a finitely generated ideal I of A0 such that (In)n≥0 is a fundamental system of neighborhoods of 0 in A0. In that case, we say that A0 is a ring of definition (for the topology of A), that I is an ideal of definitin of A0 and that (A0, I) is a couple of definition.
(iv) We say that A is a Tate ring if it is a f-adic ring and has a topologically nilpotent unit.
Note that A0 and I in point (iii) are far from unique in general. (See for example corollary II.1.1.8.) Remark II.1.1.2.
(1) Note that we are not assuming that A is separated and/or complete for the I-adic topology.
(2) If I and J are two ideals of A, then J-adic topology on A is finer than the I-adic topology if and only if there exists a positive integer n such that Jn ⊂I.
Definition II.1.1.3. Let A be a topological ring. A subset E of A is called bounded (in A) if for every neighborhood U of 0 in A there exists an open neighborhood V of 0 such that ax ∈U for every a ∈E and x ∈V .
55 II Topological rings and continuous valuations Remark II.1.1.4. Let A be a topological ring and A0 be an open adic subring of A. Then A0 is bounded in A. Indeed, let I be an ideal of A0 such that the topology on A0 is the I-adic topology.
As A0 is open in A, the family (In)n≥0 is a fundamental system of neighborhoods of 0 in A. Let U be an open subset of A such that 0 ∈U. Then there exists n ≥0 such that In ⊂U, so, if we take V = In, then ax ∈U for every a ∈A0 and every x ∈V .
Notation II.1.1.5. Let A be a ring. If U, U1, . . . , Ur are subsets of A (with r ≥2) and n is a positive integer, we write U1 · . . . · Ur for the set of finite sums of products a1 . . . ar with ai ∈Ui, and U(n) = {a1 . . . an, a1, . . . , an ∈U}. If U1 = . . . = Ur = U, we write U r instead of U · . . . · U.
Proposition II.1.1.6. (Proposition 6.1 of .) Let A be a topological ring. Then the following are equivalent : (i) A is a f-adic ring.
(ii) There exists an additive subgroup U of A and a finite subset T of U such that (U n)n≥1 is a fundamental system of neighborhoods of 0 in A and such that T · U = U 2 ⊂U.
The proof rests on the following lemma.
Lemma II.1.1.7. (Lemma 6.2 of .) Let A be a topological ring and A0 be a subring of A (with the subspace topology). The following are equivalent : (a) A is f-adic and A0 is a ring of definition.
(b) A is f-adic, and A0 is open in A and adic.
(c) A satisfies condition (ii) of proposition II.1.1.6, and A0 is open in A and bounded.
Proof. Note that (a) trivially implies (b). Also, (b) implies (c) by remark II.1.1.4. So it remains to show that (c) implies (a).
Let U and T be as in condition (ii) of proposition II.1.1.6, and suppose that A0 is open and bounded in A. Then there exists a positive integer r such that U r ⊂A0, and we have T(r) ⊂U r ⊂A0. Let I be the ideal of A0 generated by T(r); note that I is finitely generated, because T(r) is finite. For every n ≥1, we have In = T(nr)A0 ⊃T(nr)U r = U r+nr, so In is an open neighborhood of 0 in A0. Let U be any open neighborhood of 0 in A0. As A0 is bounded and (U m)m≥1 is a fundamental system of neighborhoods of 0, there exists a positive integer m such that U mA0 ⊂U, hence Im ⊂U. So (In)n≥1 is a fundamental system of neighborhoods of 0 in A0, i.e. the topology on A0 is the I-adic topology. As A0 is open in A, we are done.
56 II.1 Topological rings Proof of proposition II.1.1.6. It is easy to see that (i) implies (ii) (take U = I and take for T a finite system of generators of I).
Conversely, suppose that A satisfies (ii). Let A0 = Z + U. This is open in A because U is, and it is a subring because U n ⊂U for every n ≥1. If we show that A0 is bounded, we will be done thanks to lemma II.1.1.7. Let V be a neighborhood of 0 in A. By assumption, there exists a positive integer m such that U m ⊂V . As U m is open and A0 · U m = U m + U m+1 ⊂U m ⊂V , we are done.
Corollary II.1.1.8. (Corollary 6.4 of .) Let A be a f-adic ring.
(i) If A0 and A1 are two rings of definition of A, then so are A0 ∩A1 and A0 · A1.
(ii) Every open subring of A is f-adic.
(iii) If B ⊂C are subrings of A with B bounded and C open, then there exists a ring of definition A0 of A such that B ⊂A0 ⊂C.
(iv) A is adic if and only if it is bounded (in itself).
Proof.
(i) If A0 and A1 are rings of definition, they are open and bounded, hence so are A0 ∩A1 and A0 · A1, so A0 ∩A1 and A0 · A1 are also rings of definition by lemma II.1.1.7.
(ii) Let B be an open subring of A, and let (A0, I) be a couple of definition in A. Then there exists a positive integer n such that In ⊂B, and (B ∩A0, In) is a couple of definition in B.
(iii) By (ii), we may assume that C = A. Let A0 be a ring of definition of A. Then A0 · B is open and bounded, hence is a ring of definition by lemma II.1.1.7.
(iv) If A is bounded, then it is a ring of definition of itself by lemma II.1.1.7, so it is adic.
Conversely, if A is adic, then it is bounded by remark II.1.1.4.
Remark II.1.1.9. Suppose that A is f-adic and that (A0, I0) and (A1, I1) are couples of definition.
Then we know that A0 · A1 is a ring of definition, and it is easy to see that I0 · A1 and I1 · A0 both are ideals of definition in it. On the other hand, we also know that A0 ∩A1 is a ring of definition, but there is no reason for I0 ∩I1 to be an ideal of definition (because we don’t know if it is finitely generated).
II.1.2 Boundedness Recall that bounded subsets of topological rings are introduced in definition II.1.1.3.
57 II Topological rings and continuous valuations Definition II.1.2.1. Let A be a topological ring. We say that a subset E of A is power-bounded if the set S n≥1 E(n) is bounded, where (as in notation II.1.1.5 E(n) = {e1 . . . en, e1, . . . , en ∈E}.
We say that E is topologicall nilpotent if, for every neighborhood W of 0 in A, there exists a positive integer N such that E(n) ⊂W for n ≥N.
If E is a singleton {a}, we say that a is power-bounded (resp. topologically nilpotent) if E is.
1 Notation II.1.2.2. Let A be a topological ring. We denote by A0 the subset of its power-bounded elements, and by A00 the subset of its topologically nilpotent elements.
Lemma II.1.2.3. (Remark 5.26 of .) Let A be an adic ring. If x ∈A, the following are equivalent : (i) x is topologically nilpotent.
(ii) There exists an ideal of definition I such that the image of x in A/I is nilpotent.
(iii) There exists an ideal of definition I such that x ∈I.
In particular, A00 is an open radical ideal of A and it is the union of all the ideals of definition.
Moreover, A00 itself is an ideal of definition if and only if there exists an ideal of definition I such that the nilradical of A/I is nilpotent (and then this condition holds for all ideals of definition).
Proof. We first prove the equivalence of (i), (ii) and (iii), for x ∈A. It is clear that (iii) implies (i). Suppose that (i) holds. Let I be an ideal of definition of A. As I is a neighborhood of 0, there exists N ∈N such that xn ∈I for n ≥N; so the image of x in A/I is nilpotent, and (ii) holds.
Finally, suppose that (ii) holds, and let I be an ideal of definition such that x + I is nilpotent in A/I. Let n be a positive integer such that xn ∈I, and let J = I + xA. Then J is an open ideal of A, I ⊂J, and Jn ⊂I, so the I-adic and J-adic topologies on A coincide, which means that J is an ideal of definition; this shows that (iii) holds.
We now prove the rest of the lemma. The fact that A00 is the union of all the ideals of definition follows from the equivalence of (i) and (iii); in particular, as ideals of definition are open, A00 is also open; it is clear that A00 is radical. If A00 is an ideal of definition, then there exists an ideal of definition I such that the nilradical of A/I is nilpotent (just take I = A00, and observe that the nilradical of A/A00 is (0)). Conversely, suppose that there exists an ideal of definition I such that the nilradical of A/I is nilpotent. Then there exists a positive integer n such that (A00)n ⊂I, so the I-adic and A00-adic topologies on A coincide, i.e., A00 is an ideal of definition.
Proposition II.1.2.4. (Corollary 6.4 of .) Let A be a f-adic ring. Then the set of power-bounded elements A0 is an open and integrally closed subring of A, and it is the union of all the rings of definition of A. Moreover, A00 is a radical ideal of A0.
1Note that this agrees with definition I.1.5.3.
58 II.1 Topological rings Note that A00 is not an ideal of A in general.
Remark II.1.2.5. (See proposition 5.30 of , or adapt the proof below.) All the assertions remain true for a non-Archimedean ring A, except the fact that A0 is open and the union of all the rings of definition of A.
Proof of the proposition. By (iii) of corollary II.1.1.8, every bounded subring of A is contained in a ring of definition; by (ii) of lemma II.1.2.6, this implies that every power-bounded element of A is contained in a ring of definition, so A0 is contained in the union of all the rings of definition of A. Conversely, if A0 is a ring of definition of A, then it is bounded by remark II.1.1.4, so all its elements are power-bounded by lemma II.1.2.6(ii), so A0 ⊂A0. This proves that A0 is the union of all the rings of definition of A, so it is a subring of A; as rings of definition are open, A0 is open.
We show that A0 is integrally closed in A. Let a ∈A be integral over A0. By the previous paragraph, there exists a ring of definition A0 such that a is integral over A0; in particular, A0 is bounded. So there exists n ∈N such that A0[a] = A0 + A0a + . . . + A0an, hence A0[a] is bounded, which implies that a is power-bounded, i.e., a ∈A0.
We prove that A00 is a radical ideal of A0. Let a, a′ ∈A00 and b ∈A0. We prove that a + a′, ab ∈A00. Let U be a neighborhood of 0 in A; as A is non-Archimedean, we may assume that U is an additive subgroup of A. Let V ⊂U be a neighborhood of 0 such that bnV ⊂U for every n ≥1, and let N be a positive integer such that an, (a′)n ∈V for n ≥N. Then (a+a′)n ∈U for n ≥2N by the binomial formula, and (ab)n = bnan ∈bnV ⊂U for n ≥N. It remains to show that A00 is a radical ideal of A0. Let a ∈A0, and suppose that we have ar ∈A00 for some positive integer r. Let U be a neighborhood of 0 in A, and let V be a neighborhood of 0 such that anV ⊂U for every n ≥1. Choose a positive integer N such that (ar)n ∈V for every n ≥N. Then we have an ∈U for every n ≥rN. This shows that a ∈A00.
Lemma II.1.2.6. (Proposition 5.30 of .) Let A be a non-Archimedean topological ring.
(i) Let T be a subset of A, and let T ′ be the subgroup generated by T. Then T ′ is bounded (resp. power-bounded, resp. topologically nilpotent) if and only if T is.
(ii) Let T be a subset of A. Then T is power-bounded if and only if the subring generated by T is bounded.
Proof.
(i) We prove the non-obvious direction. Suppose that T is bounded. Let U be a neigh-borhood of 0, and let V be a neighborhood of 0 such that ax ∈U for every a ∈T and x ∈V . As A is non-Archimedean, we may assume that U and V are additive subgroups of A, and then we have T ′ · V ⊂U. So T ′ is bounded. The proofs are similar for T power-bounded and T topologically nilpotent.
(ii) Let B be the subring generated by T. Then B is the subgroup generated {1} ∪S n≥1 T(n), so, by (i), it is bounded if and only if {1} ∪S n≥1 T(n) is bounded; we see easily that this 59 II Topological rings and continuous valuations equivalent to the fact that S n≥1 T(n) is bounded, i.e. that T is power-bounded.
Note the following useful lemma characterizing open ideals of A.
Lemma II.1.2.7. Let A be a f-adic ring and J be an ideal of A. Then J is open if and only if A00 ⊂ √ J.
Proof. Suppose that J is open. Then it is a neighborhood of 0 in A, so, for every a ∈A00, we have an ∈J for n big enough, which means that A00 ⊂ √ J.
Conversely, suppose that A00 ⊂ √ J. Let (A0, I) be a couple of definition of A. Then I ⊂A00 by lemma II.1.2.3, so I ⊂ √ J. Write I = (a1, . . . , ar) with a1, . . . , ar ∈A0, and choose N ∈N such that aN 1 , . . . , aN r ∈J. Then J contains IrN, and IrN is open, so J is open.
II.1.3 Bounded sets and continuous maps A continuous map of f-adic rings does not necessarily send bounded sets to bounded sets. We want to introduce a condition that will guarantee this property.
Definition II.1.3.1. Let A and B be f-adic rings. A morphism of rings f : A →B is called adic if there exist a couple of definition (A0, I) of A and a ring of definition B0 of B such that f(A0) ⊂B0 and that f(I)B0 is an ideal of definition of B.
Example II.1.3.2.
(1) A continuous, surjective and open morphism of f-adic rings is adic.
(2) Let A be Qℓwith the discrete topology and B be Qℓwith the topology given by the ℓ-adic valuation. Then the identity f : A →B is continuous, and A is bounded in A, but f(A) is not bounded in B. By proposition II.1.3.3, this implies that f is not adic.
(3) Let f : Zℓ→Zℓ be the inclusion, where Zℓhas the ℓ-adic topology and Zℓ has the (ℓ, X)-adic topology. Then f is not adic.
Proposition II.1.3.3. Let A and B be f-adic rings and f : A →B be an adic morphism of rings.
Then : (i) f is continuous.
(ii) If A0 and B0 are rings of definition of A and B such that f(A0) ⊂B0, then, for every ideal of definition I of A0, the ideal f(I)B0 is an ideal of definition of B.
(iii) For every bounded subset E of A, the set f(E) is bounded in B.
Proof. Let (A0, I) and B0 be as in definition II.1.3.1, and write J = f(I)B0.
60 II.1 Topological rings (i) For every n ≥1, we have f −1(Jn) = f −1(f(In)B0) ⊃In. So f is continuous.
(ii) Let (A′ 0, I′) be a couple of definition of A and B′ 0 be a ring of definition of B such that f(A′ 0) ⊂B′ 0. We want to show that J := f(I′)B′ 0 is an ideal of definition of B′ 0.
(iii) Let U be a neighborhood of 0 in B. We may assume that U = Jn for some n ≥1. Let V be a neighborhood of 0 in A such that ax ∈In for every a ∈E and every x ∈V ; we may assume that V = Im for some m ≥1. Then f(E) · f(I)m = f(E · Im) ⊂f(In) ⊂Jn, so f(E)Jm ⊂Jn.
Proposition II.1.3.4. (Proposition 6.25 of .) Let A and B be f-adic rings and f : A →B be a continuous morphism of rings. Suppose that A is a Tate ring. Then B is a Tate ring, f is adic, and, for every ring of definition B0 of B, we have f(A) · B0 = B.
Proof. Let B0 be a ring of definition of B. By lemma II.1.3.5, we can find a ring of definition A0 of A such that f(A0) ⊂B0. Let ϖ ∈A be a topologically nilpotent unit. After replacing replacing ϖ by some ϖr, we may assume that ϖ ∈A0. As f is a continuous morphism of rings, f(ϖ) ∈B0 is a topologically nilpotent unit of B. In particular, B is a Tate ring. By proposition II.2.5.2, I := ϖA0 is an ideal of definition of A0, and f(I)B0 = f(ϖ)B0 is an ideal of definition of B0. So f is adic. Also, by the same lemma, we have B = B0[f(ϖ)−1], so B = f(A) · B.
Lemma II.1.3.5. Let f : A →B be a continuous morphism of f-adic rings. For every ring of definition B0 of B, there exists a ring of definition A0 of A such that f(A0) ⊂B0.
Proof. Let A′ 0 and B0 be rings of definition of A and B. Then f −1(B0) is an open subring of A and A′ 0 ∩f −1(B0) is a bounded subring, so, by corollary II.1.1.8(iii), there exists a ring of definition A0 of A such that A′ 0 ∩f −1(B0) ⊂A0 ⊂f −1(B0), and we clearly have f(A0) ⊂B0.
II.1.4 Examples Some of these will be particular cases of constructions that we will see later, I’ll add references later for the others.
1. Any ring is a topological ring for the discrete topology. It is f-adic but not Tate.
2. The rings R and C (with the usual topology) are topological rings. They are not non-Archimedean. A subset of C (or R) is bounded if and only if it is bounded in the usual sense. We have C0 = {z ∈C | |z| ≤1} 61 II Topological rings and continuous valuations and C00 = {z ∈C | |z| < 1}.
Note that these are not additive subgroups.
3. Zℓwith the ℓ-adic topology is an adic topological ring (any power of ℓZℓis an ideal of definition), and Qℓwith the ℓ-adic topology is an f-adic topological ring (with Zℓas a ring of definition). We have Q0 ℓ= Zℓand Q00 ℓ= ℓZℓ.
4. Qℓ(with the topology coming from the unique extension of the ℓ-adic valuation on Qℓ) is a f-adic ring. We have Q 0 ℓ= Zℓ, the integral closure of Zℓin Qℓ; this is a ring of definition, and it is adic with ideal of definition ℓZℓ(for example). Also, Q 00 ℓis the maximal ideal of Zℓ; it is not an ideal of definition, because it is equal to its own square (also, it is not finitely generated).
Note that Z + ℓZℓis also a ring of definition of Qℓ, because it is open and bounded.
5. Let A be a Noetherian ring and I an ideal of A. The I-adic topology on A is Hausdorff if I is contained in the Jacobson radical of A (for example if A is local and I ̸= A), or if A is a domain and I ̸= A.
6. Let A be a ring with the topology induced by a rank 1 valuation |.|, and let Γ be the valuation group of |.|. Then : - a subset E of A is bounded if and only if there exists γ ∈ Γ such that E ⊂{a ∈A | |a| ≤γ}.
- A0 = {a ∈A | |a| ≤1}; - A00 = {a ∈A | |a| < 1}.
These statements are all false if |.| has rank ≥2. Indeed, in that case Γ has a proper convex subgroup ∆, and an element of A that has valuation in ∆cannot be topologically nilpotent, because there exists γ ∈Γ such that γ < δ for every δ ∈∆.
7. Let A = k((t))((u)), with the valuation |.| corresponding to the valuation subring R := {f = P n≥0 anun | an ∈k((t)) and a0 ∈k}. More explicitly, take Γ = Z × Z; if f = P n≥r anun ∈A with an ∈k((t)) and ar ̸= 0, and if ar = P m≥s bmtm with bm ∈k and bs ̸= 0, then |f| = (−r, −s).
2 Note that the valuation topology on A coincides with the topology defined by the u-adic valuation (see example I.1.5.5). So t ∈A is not topologically nilpotent, even though it has valuation < (0, 0).
8. Let k be a field with the topology induced by a rank 1 valuation |.|.
Then k0 = {x ∈k | |x| ≤1} is a ring of definition of k (often called the ring of integers of k); it is a local ring with maximal ideal k00 = {x ∈k | |x| < 1}. A nonzero topologi-cally nilpotent in k, i.e. an element of k00 −{0} is called a pseudo-uniformizer. k is a Tate ring; a ring of definition is k0, and any pseudo-uniformizer generates an ideal of definition 2We put the minus signs so that |.| will be a multiplicative valuation; see remark I.1.1.11.
62 II.1 Topological rings of k0.
The Tate algebra in n indeterminates over k is the subalgebra Tn = Tn,k = l⟨X1, . . . , Xn⟩ of k whose elements are power series f = P ν∈Nn aνXν such that |aν| →0 as ν1 + . . . + νn →+∞(where ν = (ν1, . . . , νn)).
The Gauss norm ∥.∥on Tn is defined by ∥P ν aνXν∥= supν∈Nn |aν|.
With the topology induced by this norm, Tn is a Tate ring, and contains k[X1, . . . , Xn] as a dense subring.
We have T 0 n = k0⟨X1, . . . , Xn⟩:= Tn ∩k0, and this is a ring of defi-nition.
Any pseudo-uniformizer of k generates an ideal of definition of T 0 n.
Also, T 00 n = k00⟨X1, . . . , Xn⟩.
9. We keep the notation of the previous example, and we suppose that k is complete.3 The algebra k⟨X1, . . . , Xn⟩ is a Banach k-algebra, it is Noetherian and all its ideals are closed.
Alors, if Bn(k) is the closed unit ball in k n (i.e.
Bn(k) = {(x1, . . . , xn) ∈k n | ∀i ∈{1, . . . , n}, |xi| ≤1}, where we denote by |.| the uniaue extension of the valuation |.| to k), then a formal power series f ∈k is in k⟨X1, . . . , Xn⟩if and only if, for every x ∈Bn(k), the series f(x) converges in the completion of k.
An affinoid k-algebra is a quotient of an algebra k⟨X1, . . . , Xn⟩. These are also called topologically finitely generated k-algebras. . Such an algebra is also a Tate ring, it is Noetherian, and all its ideals are closed. Also, if A is an affinoid k-algebra, then A0 is a ring of definition of A (i.e. bounded) if and only if A is reduced. (The fact that A is reduced if A0 is bounded is proved in remark IV.1.1.3, and the converse follows immediately from Theorem 1 of section 6.2.4 of .) For example, A = Qℓ[T]/(T 2) is an affinoid Qℓ-algebra (note that A is also equal to Qℓ⟨T⟩/(T 2)), but A0 = Zℓ⊕QℓT is not bounded (neither is A00 = ℓZℓ⊕QℓT).
10. Another important class of examples are perfectoid algebras. These are always Tate rings.
For example, Cℓis a perfectoid field, and the ℓ-adic completion of S n≥1 Cℓ⟨X1/ℓn⟩is a perfectoid Cℓ-algebra.
11. A = Zℓ with the (ℓ, T)-adic topology is an adic and f-adic ring, but it is not a Tate ring. This type of f-adic ring is also very useful, because their adic spectra will be formal schemes.
3Some of the results are still true under weaker conditions.
63 II Topological rings and continuous valuations II.2 Continuous valuations II.2.1 Definition Definition II.2.1.1. Let A be a topological ring. We say that a valuation |.| on A is continuous if the valuation topology on A is coarser than its original topology.
In other words, a valuation |.| : A →Γ ∪{0} is continuous on A if and only if, for every γ ∈Γ, the set {a ∈A | |a| < γ} is an open subset of A. By remark I.1.5.2(3), if the value group of |.| is not trivial, then |.| is continuous if and only if, for every γ ∈Γ, the set {a ∈A | |a| ≤γ} is an open subset of A.
Definition II.2.1.2. If A is a topological ring, the set of continuous valuations of A is called the continuous valuation spectrum of A and denoted by Cont(A). We see it as a topological space with the topology induced by the topology of Spv(A).
II.2.2 Spectrality of the continuous valuation spectrum Theorem II.2.2.1. (Theorem 7.10 of .) Let A be a f-adic ring. Then Cont(A) = {x ∈Spv(A, A00 · A) | ∀a ∈A00, |a|x < 1}.
If I is an ideal of definition of a ring of definition of A, we also have Cont(A) = {x ∈Spv(A, I · A) | ∀a ∈I, |a|x < 1}.
Proof. By lemma II.2.2.2 and remark I.4.2.2(3), we have {x ∈Spv(A, A00 · A) | ∀a ∈A00, |a|x < 1} = {x ∈Spv(A, I · A) | ∀a ∈I, |a|x < 1}.
Let x ∈Cont(A).
Let a ∈A00 and let γ ∈Γ.
As a is topologically nilpotent and {b ∈A | |b|x < γ} is a neighborhood of 0 in A, there exists a positive integer n such that |an|x = |a|n x < γ. This shows that every element of |A00|x is cofinal for Γx, and so, by lemma I.4.2.3, x ∈Spv(A, A00 · A). Also, we have |a|x < 1 for every a ∈A00 by lemma I.1.5.7.
Conversely, let x ∈{x ∈Spv(A, A00 · A) | ∀a ∈A00, |a|x < 1}. If cΓx ̸= Γx, then |a|x is cofinal for Γx for every a ∈A00 by lemma I.4.2.3. Suppose that cΓx = Γx. Let a ∈A00 and γ ∈Γx. As Γx = cΓx, there exists b ∈A such that |b|x ̸= 0 and |b|−1 x ≤γ. We can find n ≥1 such that ban ∈A00 (because A00is open in A), and then |ban|x < 1 by the assumption on x, so |a|n x < |b|−1 x ≤γ. So we see again that every element of |A00|x is cofinal for Γx.
We finally show that x is continuous. Write I = (a1, . . . , ar) with a1, . . . , ar ∈A0, and set δ = max{|ai|x, 1 ≤i ≤n}. Let γ ∈Γx. By the previous paragraph, there exists n ≥1 such that 64 II.2 Continuous valuations δn < γ. As |a|x < 1 for every a ∈I, this implies that |a|x < δn < γ for every a ∈In · I = In+1, so the open neighborhood In+1 of 0 is included in {a ∈A | |a|x < γ}. This implies that |.|x is continuous.
Lemma II.2.2.2. Let A be a f-adic ring, and let I be an ideal of definition of a ring of definition of A. Then √ A00 · A = √ I · A.
Also, if x ∈Spv(A), the following are equivalent : (a) |a|x < 1 for every a ∈I; (b) |a|x < 1 for every a ∈A00.
Proof. We prove both statements at the same time. Let x ∈Spv(A). Let A0 be the ring of definition in which I is an ideal of definition. We have A00 ∩A0 ⊃I by lemma II.1.2.3. So (b) implies (a), and also A00·A ⊃I·A, hence √ A00 · A ⊃ √ I · A. Conversely, if a ∈A00, then there exists r ≥1 such that ar ∈I. This shows that (a) implies (b), and also that A00 · A ⊂ √ I · A, and hence that √ A00 · A ⊂ √ I · A.
Corollary II.2.2.3. For every f-adic ring A, the continuous valuation spectrum Cont(A) is a spectral space.
Moreover, the sets Ucont f1, . . . , fn g = {x ∈Cont(A) | ∀i ∈{1, . . . , n}, |fi|x ≤|g|x ̸= 0}, for f1, . . . , fn, g ∈A such that A00 ⊂ p (f1, . . . , fn), form a base of quasi-compact open subsets of Cont(A).
Note that the condition A00 ⊂ p (f1, . . . , fn) is equivalent to saying that the ideal (f1, . . . , fn) is open. (See lemma II.1.2.7.) Proof. Let J = A00 · A. We have Cont(A) = Spv(A, J) − [ g∈A00 UJ 1 g , so Cont(A) is a closed subset of Spv(A, J). As √ J is the radical of the ideal of A generated by an ideal of definition of a subring of definition, which is finitely generated, the theorem follows from theorem I.4.2.4 and corollary I.2.4.3.
65 II Topological rings and continuous valuations II.2.3 Specializations The subset Cont(A) of Spv(A) is not stable by general specializations (or generizations), but we do have the following result.
Proposition II.2.3.1. Let A be a f-adic ring and x ∈Cont(A). Then : (i) Every horizontal specialization of x is continuous.
(ii) Every vertical generization y of x such that Γy ̸= {1} is continuous.
Remark II.2.3.2. If Γy = {1}, then y is the trivial valuation with support ℘y, and it is continuous if and only if ℘y is an open ideal of A. This might or might not be the case in general.
Proof of the proposition.
(i) Let y be a horizontal specialization of x.
Then we have |a|y ≤|a|x for every a ∈A (by the formula of definition I.3.3.3). So, if γ ∈Γ, the subgroup {a ∈A | |a|y < γ} of (A, +) contains the open subgroup {a ∈A | |a|x < γ}; this implies that {a ∈A | |a|y < γ} is open.
(ii) Let y be a vertical generization of x. By proposition I.3.2.3(ii), there exists a convex subgroup H of Γx such that Γy = Γx/H and |.|y is the composition of |.|x and of the quotient map π : Γx ∪{0} →Γy ∪{0}. Let γ ∈Γx. Then {a ∈A | |a|x ≤γ} ⊂{a ∈A | |a|y = π(|a|x) ≤π(γ)}.
As both these sets are additive subgroups of A, and as the smaller one is open, the bigger one is also open. By remark I.1.5.2(3), if Γy ̸= {1}, this implies that y is a continuous valuation.
II.2.4 Analytic points In this section, A is a f-adic ring.
Definition II.2.4.1. A point x ∈Cont(A) is called analytic if ℘x is not open.
We denote by Cont(A)an the subset of analytic points in Cont(A).
Proposition II.2.4.2. Let x ∈Cont(A). The following are equivalent : (i) x is analytic.
(ii) |A00|x ̸= {0}.
(iii) For every couple of definition (A0, I) of A, we have |I|x ̸= 0.
(iv) There exists a couple of definition (A0, I) of A such that we have |I|x ̸= 0.
66 II.2 Continuous valuations Proof. (ii) implies (iii) because A00 contains every ideal of definition of a ring of definition of A, and (iii) obvisouly implies (iv). Suppose that (i) holds. As ℘x is not open, it cannot contain the open additive subgroup A00 of A, so (ii) holds. Suppose that (iv) holds, and let a ∈I such that |a|x ̸= 0. Then |an|x ̸= 0 for every n ≥1, so ℘x does not contain any of the sets In, so it cannot be open.
Remark II.2.4.3. Remember that Cont(A) = {x ∈Spv(A, A00 · A) | ∀a ∈A00, |a|x < 1} by theorem II.2.2.1. So, by proposition II.2.4.2, Cont(A)an is the set of points of x of Spv(A) such that - the support of x is not in V (A00 · A); - every proper horizontal specialization of x has support in V (A00 · A); - |a|x < 1 for every a ∈A00.
Remark II.2.4.4. It is easy to show (see lemma 6.6 of ) that an ideal a of A is open if and only √a contains the ideal A00 · A.
Corollary II.2.4.5. Let I be an ideal of definition of a ring of definition of A, and let f1, . . . , fn be generators of I. Then Cont(A)an = n [ i=1 Ucont f1, . . . , fn fi .
In particular, Cont(A)an is a quasi-compact open subset of Cont(A).
Proof. Let x ∈Cont(A). For i ∈{1, . . . , n}, we have x ∈Ucont f1,...,fn fi if and only if 0 ̸= |fi|x = max1≤j≤n{|fj|x}.
So x is in Sn i=1 Ucont f1,...,fn fi if and only if there exists i ∈{1, . . . , n} such that |fi|x ̸= 0. This is equivalent to the fact that |I|x ̸= {0}, so it is equivalent to x ∈Cont(A)an by proposition II.2.4.2.
Proposition II.2.4.6. Let x ∈Cont(A)an. Then x has rank ≥1, and the valuation |.|x on K(x) is microbial.
Proof. If x ∈Cont(A) and Γx = {1}, then ℘x = {a ∈A | |a|x < 1} is open, so x cannot be analytic. So analytic points of Cont(A) must have positive rank.
We prove the second statement. Let x ∈Cont(A)an. By proposition II.2.4.2, there exists a ∈A00 such that |a|x ̸= 0. So the image of a in Frac(A/℘x) ⊂K(x) is topologically nilpotent 67 II Topological rings and continuous valuations (for the valuation topology on K(x)) and invertible. By theorem I.1.5.4, this implies that the valuation |.|x on K(x) is microbial.
Finally, we show that specializations among analytic points are particularly simple.
Proposition II.2.4.7. Every specialization inside Cont(A)an is vertical.
In particular, if A is a Tate ring, then every specialization in Cont(A) is vertical.
The second part follows from the first and from remark II.2.5.7.
Proof. Let x, y ∈Cont(A)an such that y is a specialization of x. Let y′ be the vertical generiza-tion of y (in Spv(A)). given by theorem I.3.4.3(ii). If we were in case (b) of theorem I.3.4.3(ii), then we would have ℘y ⊃℘x|{1} ⊃{a ∈A | |a|x < 1}; but this would imply that ℘y is open and contradict the condition y ∈Cont(A)an. So we are in case (a) of theorem I.3.4.3(ii), which means that y′ is a horizontal specialization of x. In particular, y′ is continuous by proposition II.2.3.1(i), and it is analytic because ℘y′ = ℘y is not open.
Let H ⊃cΓx be a convex subgroup of Γx such that y′ = x|H. We want to show that H = Γx, which will imply that y is a vertical specialization of x = y′. Suppose that H ̸= Γx. Then we can find γ ∈Γx −H such that γ < 1. Let a ∈A such that |a|x < γ. Then |a|x ̸∈H (otherwise γ would be in H, because H is convex), so |a|y′ = 0. This shows that ℘y′ contains the open subset {a ∈A | |a|x < γ} and contradicts the fact that y′ is analytic.
Corollary II.2.4.8. For every x ∈Cont(A)an, the set of generizations of x in Cont(A)an is to-tally ordered and admits an order-preserving bijection with the set of proper convex subgroups of Γx. In particular, the continuous rank 1 valuations are exactly the maximal points of Cont(A)an for the order given by specialization (i.e. the x ∈Cont(A)an such that {x} is an irreducible component of Cont(A)an).
Moreover, every x ∈Cont(A)an has a unique rank 1 generization, which is its maximal gener-ization.
Proof. All generizations of x in Cont(A)an is vertical by proposition II.2.4.7. As every nontrivial vertical generization of x is continuous by proposition II.2.3.1(ii), and as vertical generizations of x have the same support as x, we see that generizations of x in Cont(A)an are exactly the non-trivial vertical generizations of x. By proposition I.3.2.3, these are in order-preserving bijection with proper convex subgroups of Γx, and in order-reversing bijection with the nonzero prime ideals of Rx. In particular, x is maximal if and only Γx has no nonzero proper convex subgroups, i.e. if and only if Γx has height 1.
To finish the proof, we must show that x has a maximal vertical generization. By proposition II.2.4.6, the valuation |.|x on K(x) is microbial, so Rx has a prime ideal of height 1. As the ideals 68 II.2 Continuous valuations of Rx are totally ordered by inclusion, this implies that Rx has a unique prime ideal of height 1.
The corresponding generization of x is the maximal generization of x in Cont(A), and also the unique rank 1 generization of x in Cont(A).
II.2.5 Tate rings In this section, we gather some results that are specific to Tate rings. In general, Tate rings behave more nicely than general f-adic rings.
Definition II.2.5.1. If A is a Tate ring, a topologically nilpotent unit of A is called a pseudo-uniformizer.
Proposition II.2.5.2. Let A be a Tate ring, let A0 be a ring of definition of A, and let ϖ be a topologically nilpotent unit of A. Suppose that ϖ ∈A0. Then ϖA0 is an ideal of definition of A0, and A = A0[ϖ−1].
Proof. Let I = ϖA0. As ϖ is a unit in A, multiplication by ϖ is continuous, so In = ϖnA0 is an open subset of A0 for every n ≥1. So we just need to show that every neighborhood of 0 contains some In. Let U be an open neighborhood of 0 in A. As A0 is bounded, there exists an open neighborhood of 0 such that ax ∈U for every a ∈A0 and every x ∈V . As ϖ is topologically nilpotent, there exists a positive integer n such that ϖn ∈V . Then we have In = ϖnA0 ⊂U.
We show the last statement. Let a ∈A. As multiplication by a is continuous and ϖ is topologically nilpotent, 0 is a limit of the sequence (aϖn)n≥0. So there exists N ∈N such that aϖn ∈A0 for every n ≥N, and a ∈A0[ϖ−1].
Conversely, if a f-adic ring A has a pair of definition (A0, I) with I principal, then any gen-erator of I is topologically nilpotent in A, so A is a Tate ring if I is generated by a unit of A.
Remark II.2.5.3. (Proposition 6.2.6 of .) If A0 is an adic ring with a principal ideal of definition I := ϖA0, then A := A0[ϖ−1] is a Tate ring for the topology for which the image of A0 is a ring of definition and the image of I an ideal of definition.
Proof. Let u : A0 →A be the canonical map. We check that the subgroups (u(ϖnA0))n≥0 of A satisfy the conditions of lemma II.3.3.8, hence are a fundamental system of neighbor-hood for a topological ring structure on A. Conditions (a) and (c) of the lemma are clear. Let a ∈A and n ∈N. We write a = u(b)u(ϖ)−r, with b ∈A0 and r ∈N. Then we have au(ϖn+rA0) ⊂u(ϖnA0).
69 II Topological rings and continuous valuations Example II.2.5.4. Take A0 = Z[X] with the X-adic topology. Then A = A0[X−1] is a Tate ring. Note that this ring does not contain a field.
Corollary II.2.5.5. Let A be a Tate ring, let A0 be a ring of definition of A, and let ϖ be a topologically nilpotent unit of A. Then a subset E of A is bounded if and only there exists n ∈Z such that E ⊂ϖnA0.
Remember proposition II.1.3.4.
Proposition II.2.5.6. Let f : A →B be a continuous ring morphism between f-adic ring. If A is a Tate ring, then so is B, and f is adic.
Remark II.2.5.7. If A is a Tate ring, then Contan(A) = Cont(A).
This follows immediately from the definition and from the next lemma.
Lemma II.2.5.8. Let A be a Tate ring. Then the only open ideal of A is A itself.
Proof. Let ϖ be a topologically nilpotent unit of A, and let J be an open ideal of A. Then there exists r ≥1 such that ϖr ∈J, so J contains a unit and J = A.
So proposition II.2.4.7 implies that, if A is a Tate ring, every specialization in Cont(A) is vertical. Also, corollary II.2.4.8 says that each point of Cont(A) has a unique rank 1 generization in Cont(A), which is also its maximal generization in Cont(A).
Definition II.2.5.9. A non-Archimedean field is a topological field K whose topology is given by a rank 1 valuation.
Note that we do not assume that K is complete.
The following result is an immediate consequence of theorem I.1.5.4.
Corollary II.2.5.10. Let K be a topological field whose topology is given by a valuation. Then K is a non-Archimedean field if and only if it is a Tate ring.
Corollary II.2.5.11. Let K be a non-Archimedean field, and let |.| be a rank 1 valuation defining its topology.
(i) The ring K0 is local with maximal ideal K00.
(ii) Let x ∈Cont(A). Then |.|x is microbial and its valuation topology coincides with the original topology of K; moreover, we have K00 ⊂Rx ⊂K0.
(iii) If R is a valuation subring of K such that K00 ⊂R ⊂K0, then the corresponding valuation is continuous.
70 II.3 Constructions with f-adic rings In other words, we get a canonical bijection Cont(K) ∼ →Spv(K0/K00).
Proof.
(ii) It suffices to show that every element of K0 −K00 is invertible in K0. This follows immediately from the fact that K0 = {a ∈K | |a| ≤1}, K00 = {a ∈K | |a| < 1} and (K0)× = {a ∈K | |a| = 1}.
(ii) Any topologically nilpotent unit of K is also topologically nilpotent for the valuation topology, so |.|x is microbial by theorem I.1.5.4, that is, it admits a rank 1 generiza-tion y ∈Cont(K) such that |.|x and |.|y define the same topology. We obviously have K00 ⊂Rx ⊂Ry, so it suffices to prove the result for y. In that case, the statement is equivalent to the fact that |.|y and |.| are equivalent. As y has rank 1, we must have |a|y ≤1 for every power-bounded element a ∈K, so we have K0 ⊂Ry. This means that y is a vertical generization of |.|; but, as |.|y and |.| have the same rank, they must be equivalent.
(iii) The ring K0 is maximal among all proper valuation subrings of K containing R, so, by corollary I.1.4.4, its maximal ideal K00 is a height 1 prime ideal of R. Now theorem I.1.5.4 implies that |.|R and |.| define the same topology on K, and in particular |.|R is continuous.
II.3 Constructions with f-adic rings II.3.1 Completions Remember the following definitions from general topology.
Definition II.3.1.1. Let X be a set. A filter of subsets of X (or filter on X) is a nonempty family F of subsets of X that is stable by finite intersection and such that, if A ∈F and B ⊃A, then B ∈F.
If X is a topological space and x ∈X, we say that x is a limit of the filter F if every neighborhood of x is in F.
In a metric space (or more generally in a first-countable topological space), we can characterize many topological properties using sequences. This does not work in a general topological space, but we can use filters (or their cousins nets) instead, and everything adapts quite easily.
Remark II.3.1.2. A topological space X is Hausdorff if and only if every filter on X has at most one limit.
Remark II.3.1.3. If (xn)n≥0 is a sequence in X, the associated filter is F := {E ∈A | ∃n ∈N, xm ∈E for m ≥n}.
Then x is a limit of (xn)n≥0 if and only if it is a limit of F.
71 II Topological rings and continuous valuations Remark II.3.1.4. Any f-adic ring is a first-countable topological space, so we are being somewhat pedantic here.
Definition II.3.1.5. Let A be a commutative topological group (for example the additive sub-group of a topological ring).
(i) We say that a filter F on A is a Cauchy filter if, for every neighborhood U of 0, there exists E ∈F such that x −y ∈U for all x, y ∈E.
(ii) We say that A is complete if it is Hausdorff and if every Cauchy filter on A has a limit.
Remark II.3.1.6. Note that Bourbaki does not require complete commutative topological groups to be Hausdorff.
Remark II.3.1.7. If (xn)n≥0 is a sequence in A, we say that it is a Cauchy sequence if, for every neighborhood U of 0, there exists n ∈N such that xm −xp ∈U for all m, p ≥n. So (xn)n≥0 is a Cauchy sequence if and only if the associated filter (see remark II.3.1.3) is a Cauchy filter.
Completions of abelian topological groups and topological rings always exist, and they satisfy the obvious universal property. (See for example Chapitre III §3 No5 Th´ eor eme 2 and §6 No5 Th´ eoreme 1.) For f-adic rings, these completions take a more explicit form, thanks to the following theorem.
Theorem II.3.1.8. Let A0 be a ring and I be an ideal of A0. For every A0-module M, we set c M = lim ← −n≥0 M/InM and denote the obvious map M →c M by f.
Suppose that the ideal I is finitely generated. Then : (i) The abelian group c M is Hausdorff and complete for the f(I)c M-adic topology.
(ii) For every n ≥0, the map f induces an isomorphism M/InM ∼ →c M/f(I)nc M.
(iii) If A0 is Noetherian, then b A0 is a flat A0-algebra.
Points (i) and (ii) are proved in [25, Lemma 05GG], and point (iii) in [25, Lemma 00MB].
Note that (i) and (ii) are false in general if I is not finitely generated (see [25, Section 05JA]), and that (iii) is false in general if A0 is not Noetherian, even for a finitely generated ideal (see [25, Example 0BNU] and [25, Section 0AL8]).
In particular, if A0 is an adic ring and I is a finitely generated ideal of definition of A0, then b A0 is the completion of A0. It is easy to see that b A0 does not depend on the choice of the ideal of definition (see remark II.1.1.2(2)).
Corollary II.3.1.9. Let A be a f-adic ring, let (A0, I) be a couple of definition of A, and set b A = lim ← −n≥0 A/In (as an abelian group; note that we take the quotient of A by the ideal In of A0 and not by the ideal that In generates in A). Then : 72 II.3 Constructions with f-adic rings (i) The canonical map b A0 →b A is injective, and the square A0 / b A0 A / b A is cartesian.
(ii) If we put the unique topology on b A for which b A0 is an open subgroup, then the abelian topological group b A is complete.
(iii) There is a unique ring structure on b A that makes the canonical map A →b A continuous, and b A is a topological ring.
(iv) The ring b A is f-adic and ( b A0, I b A0) is a couple of definition of b A. Moreover, the canonical map A →b A is adic.
(v) The canonical map b A0 ⊗A0 A →b A is an isomorphism.
(vi) If A0 is Noetherian, then b A is a flat A-algebra.
(vii) If A0 is Noetherian and A is a finitely generated A0-algebra, then b A is Noetherian.
It is easy to see that b A does not depend on the choice of the pair of definition (A0, I).
Proof.
(i) For every n ≥0, the map A0/In →A/In is injective. As projective limits are left exact, the morphism b A0 →b A is injective.
Let i : A →b A be the canonical map. To prove the second statement, we must show that i(A) ∩b A0 = i(A0). The fact that i(A0) ⊂i(A) ∩b A0 is obvious. Conversely, let a ∈A such that i(a) ∈b A0. Then, for every n ≥1, there exists bn ∈A0 such that a ∈bn + In; in other words, a is in the closure of A0 in A. As A0 is an open subgroup of A, it is also closed, so a ∈A0.
(ii) It is easy to see that b A is Hausdorff (because 0 has a Hausdorff neighborhood, i.e. b A0). Let F be a Cauchy filter on b A. As b A0 is a neighborhood of 0 in b A, there exists F ∈F such that x −y ∈b A0 for all x, y ∈F. Let x0 ∈F, and define a family F0 of subsets of b A0 by : G ∈F0 ⇔x0 + G ∈F. Then F0 is not empty because F −x0 ∈F0, and it is clearly a Cauchy filter on b A0. As b A0 is complete, F0 has a limit a, and then a + x0 is a limit of F.
(iii) As A0 is dense in b A0, A is dense in b A. This implies uniqueness. The existence of the product on b A follows from Th´ eor eme 1 of Chapitre III §6 No5.
(iv) b A0 is an open subring of b A by (i), and it has the I b A0-adic topology by theorem II.3.1.8, so b A is f-adic. The fact that the map A →b A follows immediately from the definition.
(v) This is lemma 1.6 of . Let us explain the proof. Consider the commutative diagram 73 II Topological rings and continuous valuations (where all the maps are the obvious ones) : A0 i / _ b A0 _ g { b A0 ⊗A0 A j $ A f : i / b A We want to show that the map j is an isomorphism. We will do this by constructing an inverse.
First note that, by proposition II.3.2.1, b A0 ⊗A0 A has a natural structure of f-adic ring and that f, g and j are continuous. Indeed, the maps A0 →b A0 and A0 →A are adic.
We now turn to the construction of an inverse h of j : b A0 ⊗A0 A →b A. Let a ∈b A. As i(A) is dense in b A, we can find a0 ∈b A0 and b ∈A such that a = a0 + i(b), and we want to set h(a) = f(b) + g(a0). We have to check that this does not depend on the choices. Suppose that a = a0 + i(b) = a′ 0 + i(b′), with a0, a′ 0 ∈b A0 and b, b′ ∈A. Then a0 −a′ 0 = i(b′ −b), so b′ −b ∈A0 by (i), and f(b) + g(a0) = f(b′) + f(b −b′) + g(a0) = f(b′) + g(i(b −b′)) + g(a0) = f(b′) + g(a′ 0).
So h is well-defined, and it is clear that h is additive and that f = h ◦i and g = h| b A0.
The last property implies that h is continuous in a neighborhood of 0, hence that h is continuous. As f = h ◦i, i has dense image and f is a morphism of rings, h is also a morphism of rings. By construction of h, we have h ◦j = id. Also, if a ∈f(A) or a ∈g( b A0), then j(h(a)) = a, also by construction of h; as h is a morphism or rings, this implies that j ◦h = id.
(vi) and (vii) These follow immediately from (v) (and from theorem II.3.1.8(iii) for (vi)).
Definition II.3.1.10. If A is a f-adic ring, the f-adic ring b A defined in corollary II.3.1.9 is called the completion of A.
Lemma II.3.1.11. Let A be an abelian topological group and i : A →b A be its completion.
Then there is a bijection between the set of open subgroups of A and the set of open subgroups of b A; it sends an open subgroup G of A to i(G) = b G, and its inverse sends an open subgroup H of b A to i−1(H).
Proof. If Y is a subset of A, then i(Y ) is canonically isomorphic to the completion of Y by Chapitre II §3 No9 corollaire 1 de la proposition 18.
74 II.3 Constructions with f-adic rings Note also that Ker i = {0} and that i(A) is dense in b A (for example by Chapitre II §3 No7 proposition 12). If G is an open subgroup of A, it is also closed, hence contains Ker i, and so we have G = i−1(i(G)). Conversely, if H is an open subgroup of b A, then H ∩i(A) is dense in H, so H is the closure of i(i−1(H)).
Proposition II.3.1.12. Let A be a f-adic ring.
(i) We have b A0 = c A0 and b A00 = d A00. (That is, A0 (resp. A00 is sent to b A0 (resp. b A00) by the bijection of lemma II.3.1.11.) (ii) If we have open sugroups of G and H of A and b A that correspond to each other by the bijection of lemma II.3.1.11, then G is a ring of definition of A if and only if H is a ring of definition of b A.
(iii) The map i : A →b A induces a bijective map Cont( b A) →Cont(A).
In fact, the bijection of (iii) is a homeomorphism, and this is not so obvious and quite impor-tant. We will prove this later, after we introduce adic spectra. (See corollary III.4.2.2.) Proof.
(i) Let i : A →b A be the obvious map. By lemma II.3.1.11, a subset E of A is bounded if and only if i(E) is bounded in b A, and an element x ∈A is topologically nilpotent if and only if i(x) ∈b A is topologically nilpotent. In particular, i−1( b A0) (resp. i−1( b A00)) is contained in A0 (resp. A00), so b A0 ⊂c A0 (resp. b A00 ⊂d A00).
Conversely, as i(A0) ⊂b A0 ⊂c A0, b A0 is dense in c A0; but b A0 is open in b A, hence closed, so b A0 = c A0. The case of b A00 is similar.
(ii) As i(G) is dense in H and i(A) is a subring of b A, H is a subring if and only if G is a subring. Also, we have seen in (i) that G is bounded if i(G) is, so G is a ring of definition if H is. Conversely, suppose that G is a bounded subring of A. Then i(G) is bounded. As b A has a fundamental system of open bounded neighborhoods of 0 (for example the powers of an ideal of definition of a ring of definition), we can find an open bounded subgroup U ⊂H. We have H = i(G) + U because i(G) is dense in U, and so H is bounded.
(iii) Let |.| : A →Γ ∪{0} be a continuous valuation, and let F be a Cauchy filter on A. We claim that : (a) either, for every γ ∈Γ, there exists F ∈F such that |a| < γ for every a ∈F; (b) otherwise there exists F ∈F such that |.| is constant on F.
Indeed, suppose that (a) does not hold. Then there exists γ0 ∈Γ such that, for every F ∈F, there exists a ∈F with |a| ≥γ0. As |.| is continuous, the set {a ∈A | |a| < γ0} is an open neighborhood of 0. So, as F is a Cauchy filter, there exists F ∈F such that |a −b| < γ0 for all a, b ∈F. Fix a0 ∈F such that |a0| ≥γ0. Then, for every a ∈F, we have |a −a0| < γ0 ≤|a0|, so the strong triangle inequality implies that |a| = |a|0.
75 II Topological rings and continuous valuations Let i : A →b A be the canonical map. Applying the result of the previous paragraph to b A and using the fact that i(A) is dense in b A, we see that a continuous valuation on b A is uniquely determined by its restriction to i(A), so the map Cont( b A) →Cont(A) is injective.
We now show that Cont( b A) →Cont(A) is surjective. Let |.| : A →Γ ∪{0} be a continuous valuation. Let a ∈Ker i, and let F be the filter of neighborhoods of a in A; this is clearly a Cauchy filter. If it satisfies condition (a) above, then |a| < γ for every γ ∈Γ, so |a| = 0. Otherwise, F satisfies condition (b), so there exists F ∈F such that |.| is constant on F; but we have a ∈F and 0 ∈F (because i(A) is the maximal Hausdorff quotient of A), so again |a| = 0. This shows that |.| factors through i(A), so we may assume that i is injective.
We now extend |.| to a map |.|′ : b A →Γ ∪{0}. Let a ∈b A. Then there exists a Cauchy filter F on A that converges to a. If F satifies condition (a) above, then we set |a|′ = 0.
Otherwise, we choose F ∈F such that |.| is constant on F, and we set |a|′ = |b|, for any b ∈F. It is easy to check that this does not depend on the choices and defines a valuation on b A. We finally show that |.|′ is continuous. Let γ ∈Γ, and let G = {a ∈A | |a| < γ}.
This is an open subgroup of A and, by the definition of |.|′, its closure in b A is contained in the group {a ∈b A | |a|′ < γ}; so {a ∈b A | |a|′ < γ} is open.
II.3.2 Tensor products We have to be a bit careful with tensor products of f-adic rings, because they don’t make sense in general. This corresponds to the fact that fiber products of adic spaces don’t always exist, and has a simple geometric explanation, that is given in the remark below. However, if we assume that all the maps are adic, then there is no problem; in particular, we can always define tensor products of Tate rings.
Proposition II.3.2.1. (See theorem 5.5.4 of .) Let f : A →B and f : A →C be two adic morphisms of f-adic rings. Choose a couple of definition (A0, I) of A and rings of definition B0 and C0 of B and C such that f(A0) ⊂B0 and g(A0) ⊂C0. Let D0 be the image of B0 ⊗A0 C0 in D := B ⊗A C, and let J be the ideal of D0 generated by the image of I.
We put the J-adic topology on D0 and equip D with the unique structure of topological group that makes D0 an open subgroup. Then D is a f-adic ring with couple of definition (D0, J), and the obvious ring morphisms u : B →D and v : C →D are continuous and adic.
Moreover, for every non-Archimedean topological ring D′ and every pair of continuous ring morphisms (u′ : B →D′, v′ : C →D′) such that u ◦f = v ◦g, there exists a unique continuous ring morphism ϕ : D →D′ such that u′ = ϕ ◦u and v′ = ϕ ◦v. If D′ if f-adic and u′ and v′ are adic maps, then ϕ is also an adic map.
76 II.3 Constructions with f-adic rings Proof. The ideal J of D0 if of finite type because I is, so the only thing we need to prove to get the first statement is that D is a topological ring. By Chapitre III §6 No3 Remarque, it suffices to prove that the multiplication of D is continuous in a neighborhood of 0 and that the map x 7− →ax is continuous for every a ∈D. The first statement follows from the fact that D0 is a topological ring, and it suffices to prove the second statement for pure tensors. So let b ∈B and c ∈C, and let U be a neighborhood of 0 in D. We may assume that U is of the form Jr, for some r ≥0. Let s ≥0 such that b(f(I)B0)s ⊂(f(I)B0)r and c(g(I)C0)s ⊂(g(I)C0)r. If x ∈(f(I)B0)s ⊗A0 (g(I)C0)s, then we have (b ⊗c)x ∈Jr; as J2s = (f(I)B0)s ⊗A0 (g(I)C0)s, this shows that (b ⊗c)J2s ⊂Jr.
We now turn to the second statement. By the usual property of the tensor product, there exists a unique morphism of rings ϕ : D →D′ such that u′ = ϕ ◦u and v′ = ϕ ◦v. We want to show that ϕ is continuous. Let U be an open subgroup of D′. As u′ and v′ are continuous, there exists r ≥1 such that u′−1(U) ⊃(f(I)B0)r and v′−1(U) ⊃(f(I)C0)r. Then ϕ−1(U) ⊃Jr, so ϕ−1(U) is open. The last statement is easy.
Example II.3.2.2. (See example 5.5.5 of .) Here is an example where things don’t work. Take A = A0 = Zℓwith the ℓ-adic topology, B = B0 = Zℓ with the (ℓ, X)-adic topology and C = Qℓ⊃C0 = Zℓwith the ℓ-adic topology. Note that the obvious map A →C is adic, but the obvious map A →B is not. We have D = Zℓ[ℓ−1] and D0 = Zℓ.
Suppose that there is an ideal J of D0 such that D is f-adic with couple of definition (D0, J) and such that the canonical maps B →D and C →D are continuous. Suppose that 1 ̸∈J. In particular, the map B0 →D0 is continuous, so J must contain a power of the ideal (ℓ, X). As Zℓ is a local ring with maximal ideal (ℓ, X), we must also have J ⊂(ℓ, X). So the topology on D0 is the (ℓ, X)-adic topology. But then there is no structure of topological ring on D that makes D0 an open subring; indeed, ℓis invertible in D, so multiplication by ℓwould have to be a homeomorphism, and this not possible because ℓD0 is not an open subset of D0 (for example because it contains no power of X even though X is topologically nilpotent for the (ℓ, X)-adic topology).
So we must have 1 ∈J, which means that the only open subsets of D0 are ∅and D0. But then there can be no topological ring structure on D that makes D0 an open subring, for the same reason as before : ℓis invertible in D, so ℓD0 would have to be an open subset of D0.
Remark II.3.2.3. We keep the notation of example II.3.2.2. We temporarily write Spa(R) for “the affinoid adic space of R”, even though that is not quite correct because we need an extra piece of data to define this space. The geometric interpretation of the previous example is that Spa(Zℓ[ℓ−1]), if it made sense, would be the fiber product Spa(Zℓ) ×Spa(Zℓ) Spa(Qℓ), that is, the generci fiber of of the “formal affine line” Spa(Zℓ). But this generic fiber should be the open unit disc, which is not an affinoid space.
77 II Topological rings and continuous valuations II.3.3 Rings of polynomials If A is a f-adic ring, we want to define topologies on rings of polynomials over A that make them f-adic rings. The model is the Tate algebra over a non-Archimedean field k. For example, if k = Qℓ, we defined in section II.1.4 the Tate algebra Tn = k⟨X1, . . . , Xn⟩and saw (or at least claimed) that it is complete and contains the polynomial ring k[X1, . . . , Xn]. So another way to define Tn would be to say that it is the completion of k[X1, . . . , Xn] for the Gauss norm.
Remember also that Tn is the ring of convergent power series on the closed unit ball in Cn ℓ. Of course, the choice of 1 as the radius of the ball was arbitrary, and in general we will want to allow a different radius for each indeterminate (this makes a real difference if the radius is not in ℓQ).
This will modify the norm on k[X1, . . . , Xn] for which we take the completion. In the case of a general f-adic ring A, it does not make sense to talk of radii in R≥0, and they will be replaced by the family (Ti)i∈I in the next proposition.
We start with the case of general non-Archimedean rings and specialize to the case of f-adic rings at the end.
Proposition II.3.3.1. (Remark 5.47 of .) Let A be a non-Archimedean topological ring, let X = (Xi)i∈I be a family of indeterminates (not necessarily finite) and let T = (Ti)i∈I be a family of subsets of A. Suppose that, for every i ∈I, every n ∈N and every neighborhood U of 0 in A, the subgroup T n i U is open. 4 For every function ν : I →N with finite support, we set T ν = Q i∈I T ν(i) i . For every open subgroup U of A, we set U[X,T] = { X ν∈N(I) aνXν ∈A[(Xi)i∈I] | aν ∈T νU for all ν ∈N(I)}.
Then : (i) For every open neighborhood U of 0 in A and every ν ∈N(I), T νU is an open subgroup of A.
(ii) There is a unique structure of topological ring on A[X] := A[(Xi)i∈I] for which the sub-groups U[X,T], for U running through the open subgroups of A, form a fundamental system of neighborhoods of 0.
We denote the resulting topological ring by A[X]T.
(iii) The inclusion ι : A →A[X]T is continuous and the set {ι(t)Xi, i ∈I, t ∈Ti} is power-bounded.
(iv) For every non-Archimedean topological ring B, every continuous ring morphism f : A →B and every family (xi)i∈I of elements of B such that {f(t)xi, i ∈I, t ∈Ti} is power-bounded, there exists a unique continuous ring morphism g : A[X]T →B such that f = g ◦ι and g(Xi) = xi for every i ∈I.
4Where we denote by T n i U the subgroup generated by products of n elements of Ti and of one element of U.
78 II.3 Constructions with f-adic rings Proof.
(i) Let ν ∈N(I). Let i1, . . . , ir be the elements of I on which ν takes nonzero values, and let U be an open neighborhood of 0 in A. We know that T n i ◦V is an open subgroup of A for every i ∈I, every n ≥0 and every open subgroup V of A, so T n ◦U = T νi1 i1 · . . . · T νir ir · U is an open subgroup of A.
(ii) If U and V are two open subgroups of A, then U[X,T] ∩V[X,T] = (U ∩V )[X,T] and U[X,T] · V[X,T] = (U · V )[X,T]. So the family of the U[X,T], for U an open subgroup of A, satisfies all the conditions of lemma II.3.3.8.
(iii) The map ι is continuous by definition on the topology on A[X]T. Let E be the subgroup generated by all the products ι(t1)Xi1 . . . ι(tn)Xin, for n ∈N, i1, . . . , in ∈I and tis ∈Tis, 1 ≤s ≤n. We want to show that E is bounded in A[X]T. Let G be an open subgroup of A[X]T. Then there exists an open subgroup U of A such that U[X,T] ⊂G. But we have E · U[X,T] = U[X,T], so E · U[X,T] ⊂G.
(iv) By the universal property of the polynomial ring, there exists a unique morphism of rings g : A[X]T →B such that g ◦i = f and that g(Xi) = xi for every i ∈I. So we just need to show that this g is continuous. Let E be the subgroup of B generated by all the products f(t1)Xi1 . . . f(tn)Xin, for n ∈N, i1, . . . , in ∈I and tis ∈Tis, 1 ≤s ≤n. We know that E is bounded. Let H be an open subgroup of B. Then there exists an open subgroup G of B such that E ◦G ⊂H. As f is continuous, U := f −1(G) is an open subgroup of A.
Then U[X,T] ⊂g−1(H); as g−1(H) is a subgroup of A[X]T, this implies that it is open.
Proposition II.3.3.2. (Remark 5.38 of .) Let A, X and T be as in proposition II.3.3.1 (with the same condition on T). We denote by A the formal power series ring A. Then : (i) The set A⟨X⟩T := { X ν∈N(I) aνXν ∈A | for all open subgroups U of A, aν ∈T νU for almost all ν} (here “almost all” means “all but a finite number”) is a subring of A.
(ii) There is a unique structure of topological ring on A⟨X⟩T for which the subgroups U⟨X,T⟩= { X ν∈N(I) aνT ν ∈A⟨X⟩T | aν ∈T νU for all ν ∈N(I)}, for U running through all the open subgroups of A, form a fundmental system of neigh-borhoods of 0.
Proof.
(i) First, it is easy to see that A⟨X⟩T is stable by multiplication by all the elements of A and all the Xi, i ∈I : Let f = P ν∈N(I) aνXν ∈A⟨X⟩T, let a ∈A and i ∈I. Let U be an open subgroup of A.
79 II Topological rings and continuous valuations (a) Choose an open subgroup V of A such that aV ⊂U. Then aνT ν · V for almost all ν, so (aaν) ∈T ν · U for almost all ν.
(b) By assumption, Ti · U is an open subgroup of A, so aν ∈T ν · Ti · U for almost all ν.
So A⟨X⟩T is a A[X]-submodule of A.
Let f, g ∈A⟨X⟩T. We want to check that fg ∈A⟨X⟩T. Write fg = P ν cνXν. Let U be an open subgroup of A, and choose an open subgroup V of A such that V · V ⊂U.
As f, g ∈A⟨X⟩T, we can write f = f0 + f1 and g = g0 + g1, with f0, g0 ∈A[X] and f1, g1 ∈V⟨X,T⟩. If f0g0+f1g0+f0g1 = P ν aνXν and f1g1 = P ν bνXν, then cν = aν +bν.
We now that f0g0 + f1g0 + f0g1 by the first paragraph, so aν ∈T ν · U for almost all ν. On the other hand, bν ∈T ν · U for every ν by the choice of V . So cν ∈T ν · U for almost all ν.
(ii) Again, we have U⟨X,T⟩∩V⟨X,T⟩= (U ∩V )⟨X,T⟩and U⟨X,T⟩∩V⟨X,T⟩⊂(U · V )⟨X,T⟩for all open subgroups U, V ⊂A, so lemma II.3.3.8 applies.
Proposition II.3.3.3. (See proposition 5.49 of .) Let A, X and T be as in propositions II.3.3.1 and II.3.3.2. Then : (i) A[X]T is a dense subring of A⟨X⟩T, and the topology on A[X]T is the one induced by the topology on A⟨X⟩T.
(ii) If A is Hausdorff and Ti is bounded for every i ∈I, the topological rings A[X]T and A⟨X⟩T are Hausdorff.
(iii) If A is complete and Ti is bounded for every i ∈I, the topological ring A⟨X⟩T is complete (so it is the completion of A[X]T).
Proof.
(i) As we already noted in the proof of proposition II.3.3.2(i), for every open subgroup U of A, every element of A⟨X⟩T is the sum of a polynomial and of an element of U⟨X,T⟩ (by the very definition of A⟨X⟩T). So A[X]T is dense in A⟨X⟩T. The second statement just follows from the fact that A[X]T ∩U⟨X,T⟩= U[X,T] for every open subgroup U of A.
(ii) By (i), it suffices to show that A⟨X⟩T is Hausdorff. Let ν ∈N(I). As all the Ti are bounded, T ν is bounded. So the intersection of all the T ν · U, for U ⊂A an open subgroup, is equal to the intersection of all the open neighborhoods of 0 in A, i.e. {0} because A is Hausdorff.
This shows that T U U⟨X,T⟩= {0}.
(iii) It suffices to show that A⟨X⟩T is complete. If E is a subset of A⟨X⟩T, let Eν ⊂A be the set of all the ν-coefficients of elements of E. Let F be a Cauchy filter on A⟨X⟩T. For every ν ∈N(I), let Fν = {Eν, E ∈F}. As each T ν is bounded, all the Fν are Cauchy filters on A, so they converge because A is complete. Let aν be the limit of Fν. We want to show that f := P ν∈N(I) aνXν ∈A is in A⟨X⟩T, and that F converges to f.
Let U be an open subgroup of A. Choose E ∈F such that g −h ∈U⟨X,T⟩for all g, h ∈E. In other words, if g = P ν∈N(I) gνXν and h = P ν∈N(I) hνXν are in E, we 80 II.3 Constructions with f-adic rings have gν −hν ∈T ν ◦U for every ν; fixing ν and going to the limit on h, we get that gν −aν ∈T ν ◦U; in particular, this implies that aν ∈T ν ◦U for almost every ν. As U was arbitrary, this shows first that f ∈A⟨X⟩T, and then that, for every open subgroup U of A, we can find E ∈F such that g −f ∈U⟨X,T⟩for every g ∈E, i.e. that f is the limit of F.
Corollary II.3.3.4. Suppose that A is complete and Ti is bounded for every i ∈I. For every complete non-Archimedean topological ring B, every continuous ring morphism f : A →B and every family (xi)i∈I of elements of B such that {f(t)xi, i ∈I, t ∈Ti} is power-bounded, there exists a unique continuous ring morphism g : A⟨X⟩T →B such that f = g ◦ι and g(Xi) = xi for every i ∈I.
Proof. We already know that there exists a unique continuous ring morphism g : A[X]T →B satisfying the two conditiions of the statement (by proposition II.3.3.1(iv)). By Chapitre III §3 No5 corollaire de la proposition 8, g has a unique extension to a continuous morphism of topological groups from A⟨X⟩T to B, and this extension is clearly a morphism of rings.
We now specialize to the case of interest of us, i.e. that of f-adic rings.
First we note the following useful fact.
Lemma II.3.3.5. (Lemma 6.20 of .) Let A be a f-adic ring and T be a subset of A. If T generates an open ideal of A, then, for any open subgroup U of A and any n ≥0, the subgroup T n · U is open.
Proof. Let U and n be as in the statement, and let (A0, I) be a couple of definition of A. By assumption, the ideal J of A generated by T is open, so it contains some power of I. Hence Jn also contains a power of I. After changing the ideal of definition, we may assume that I ⊂Jn = T n · A. Let L be a finite set of generators of I, and let M be a finite set such that L ⊂T n ·M. As M is finite, it is bounded, so we can find an integer r ∈N such that M ·Ir ⊂U.
Then we have Ir+1 = L · Ir ⊂T n · M · Ir ⊂T n · U, so T n · U is open.
Proposition II.3.3.6. (Proposition 6.21 of .) Let A be a non-Archimedean topological ring, let X = (Xλ)λ∈L be a family of indeterminates (not necessarily finite) and let T = (Tλ)λ∈L be a family of subsets of A.
Let (A0, I) be a couple of definition of A. Suppose that, for every λ ∈L, the subset Tλ generates an open ideal of A. Then : (i) The ring A[X]T is f-adic, with couple of definition (A0[X,T], I[X,T]). In particular, the canonical map A →A[X]T is adic. Moreover, if A is a Tate ring, then A[X]T is also a Tate ring.
81 II Topological rings and continuous valuations (ii) Suppose that the family of indeterminates (Xλ)λ∈L is finite.
5 Then the ring A⟨X⟩T is f-adic, with couple of definition (A0⟨X,T⟩, I⟨X,T⟩), and the canonical map A →A⟨X⟩T is adic. Moreover, if A is a Tate ring, then A⟨X⟩T is also a Tate ring.
Proof.
(i) It is clear that A0[X,T] is an open subring of A[X]T, and that I[X,T] = I · A0[X,T] is a finitely generated ideal of A0[X,T]. As (I[X,T])n = (In)[X,T] for every n ≥1, we see that the topology on A0[X,T] is the I[X,T]-adic topology, so A[X]T is adic. The map A →A[X]T is clearly adic, and the last statement is also clear.
(ii) First note that A0⟨X,T⟩is an open subring of A⟨X⟩T.
Let J be an ideal of definition of A0; we claim that J⟨X,T⟩= JA0⟨X,T⟩. The inclusion J⟨X,T⟩⊃JA0⟨X,T⟩is clear. Conversely, let f = P ν aνXν ∈J⟨X,T⟩. Let x1, . . . , xr ∈A0 be generators of J. We write NL = S+∞ k=1 Nk, with all the Nk finite and with aν ∈T ν · Jk for every k ∈N and every ν ∈Nk. If k ∈N and aν ∈Nk, we write aν = Pr i=1 xiaν,i, with the aν,i ∈T ν · Ik−1. Let fi = P ν∈NL aν,iXν, for 1 ≤i ≤s. Then f1, . . . , fs ∈A0⟨X,T⟩, and so f = x1f1 + . . . xsfs ∈JA0⟨X,T⟩.
In particular, we have I⟨X,T⟩ = IA0⟨X,T⟩.
Hence, for every n ≥ 1, ap-plying the previous paragraph to the ideal of definition In of A0, we get that (I⟨X,T⟩)n = InA0⟨X,T⟩= (In)⟨X,T⟩.
This shows that the topology on A0⟨X,T⟩is the IA0⟨X,T⟩-adic topology, and so A⟨X⟩T is f-adic. The map A →A⟨X⟩T is clearly adic, and the last statement is also clear.
Example II.3.3.7. Take A = Zℓ, I = {1} and T1 = {ℓ}. We get a f-adic ring Zℓ⟨X⟩T = { X n≥0 anXn ∈Zℓ | ℓ−nan →0 as n →+∞}, with ring of definition Zℓ⟨X,T⟩= { X n≥0 anXn ∈Zℓ⟨X⟩T | ∀n ∈N, ℓ−nan ∈Zℓ}.
The topology on Zℓ⟨X,T⟩is the ℓZℓ⟨X,T⟩-adic topology. Note in particular that Zℓ⟨X⟩T is strictly bigger than Zℓ⟨X,T⟩, and that the ring Zℓ⟨X⟩T is not adic, even though Zℓis. For example, ℓis topologically nilpotent in Zℓ⟨X⟩T, but ℓX is not, because ℓnXn ̸∈ℓZℓ⟨X,T⟩for n ≥0. (This could not happen in an adic ring, in which all elements are power-bounded.) Lemma II.3.3.8. (Remark 5.24 of .) Let A be a ring and G be a set of additive subgroups of A. Then requiring G to be a fundamental system of neighborhoods of 0 makes A a topological ring if and only the following conditions hold : (a) For all G, G′ ∈G , there exists H ∈G such that H ⊂G ∩G′.
5We would get the same conclusion if we assumed instead that A is Hausdorff and that all the Tλ are bounded.
82 II.3 Constructions with f-adic rings (b) For every a ∈A and every G ∈G , there exists H ∈G such that aH ⊂G.
(c) For every G ∈G , there exists H ∈G such that H · H ⊂G.
Proof. The conditions are obviously necessary. Conversely, suppose that they hold. Condition (a) says that G does give a fundamental system of neighborhoods for a topological group struc-ture on A (see Chapitre III §1 No2 proposition 1). Conditions (b) and (c) now say that A is a topological ring (see Chapitre III §6 No3 Remarque).
Notation II.3.3.9. If Ti = {1} for every i ∈I, we write A[X]T = A[X] and A⟨X⟩T = A⟨X⟩.
II.3.4 Localizations Proposition II.3.4.1. (Proposition 5.51 of .) Let A be a non-Archimedean topological ring and T = (Ti)i∈I be a family of subsets of A satisfying the condition of proposition II.3.3.1.
Let S = (si)i∈I be a family of elements of A, and denote by R the multiplicative subset of A generated by {si, i ∈I}.
Then there exists a unique non-Archimedean topological ring structure on R−1A, making it into a topological ring that we will denote by A T S = A Ti si | i ∈I , satisfying the following properties : (i) The canonical morphism ϕ : A →A T S is continuous and the set { ϕ(t) ϕ(si), i ∈I, t ∈Ti} is power-bounded in A T S .
(ii) For every non-Archimedean topological ring B and every continuous map f : A →B such that f(si) is invertible in B for every i ∈I and that the set { f(t) f(si), i ∈I, t ∈Ti} is power-bounded in B, there exists a unique continuous ring morphism g : A T S →B such that f = gϕ.
Proof. Let D be the subring of R−1A generated by all the ϕ(t) ϕ(si), for i ∈I and t ∈Ti. Then the family of subsets D · ϕ(U) ⊂R−1A, for U an open subgroup of A, satisfies the conditions of lemma II.3.3.8, which implies that there is a unique structure of topological ring on R−1A that makes this family a fundamental system of neighborhoods of 0. It is clear that ϕ : A →R−1A is continuous for this topology, because ϕ−1(D · ϕ(U)) ⊃U.
As D is a subring of R−1A, and as it is bounded by definition of the topology, the se t{ ϕ(t) ϕ(si), i ∈I, t ∈Ti} is power-bounded.
We check that ϕ : A →R−1A satisfies the universal property of (ii) (which will also im-ply uniqueness). Let f : A →B be as in (ii). By the universal property of the localization, there exists a unique ring morphism g : R−1A →B such that f = g ◦ϕ, so we just need to 83 II Topological rings and continuous valuations check that this g is continuous. Let E be the subring of B generated by the power-bounded set { f(t) f(si), i ∈I, t ∈Ti}; then E is bounded by lemma II.1.2.6(ii), and we have g(D) ⊂E.
Let U be an open subgroup of B. As E is bounded, there exists an open subgroup V of B such that E · V ⊂U. As f is continuous, W := f −1(V ) is an open subgroup of A. As g(ϕ(W) · D) = f(W) · f(D) ⊂V · E ⊂U, g−1(U) contains the open subgroup ϕ(W) · D, so it is open.
Remark II.3.4.2. (Remark 5.52 of .) (1) We have A Ti si | i ∈I = A Ti ∪{si} si | i ∈I , so we can always assume that si ∈Ti for every i ∈I.
(2) Let J be the ideal of A[X]T generated by the set {1 −siXi, i ∈I}. Then A[X]T/J, with the quotient topology, satifies the same universal property as A T S , so we have a canonical isomorphism A[X]T/J = A T S .
Proposition II.3.4.3. (Proposition 6.21 of .) Let A, T = (Ti)i∈I and S = (si)i∈I be as in proposition II.3.4.1, and suppose that A is f-adic.
Then A T S is also f-adic, and the canonical map A →A T S is adic.
If I = {1} is a singleton and T1 = {t1, . . . , tn} is finite, we also write A T S = A t1,...,tn s0 .
Proof. This follows immediately from remark II.3.4.2 and from proposition II.3.3.6(i).
Remark II.3.4.4. Let us give an explicit ring of definition of B := A T S (the notation is that of proposition of II.3.4.3). Let A0 be a ring of definition of A. Then A0[X,T] is a ring of definition of A[X]T, so its image by the surjective map A[X]T →B is a ring of definition of B. By definition, we have A0[X,T] = { X ν∈N(I) aνXν ∈A[(Xi)i∈I] | aν ∈T νA0 for all ν ∈N(I)}.
So its image B0 in B is the A0-submodule of B generated by the sets Q i∈I T νi i s−νi i , for (νi)i∈I ∈NI. In other words, B0 is the A0-subalgebra of B generated by the elements ts−1 i , for i ∈I and t ∈Ti.
In particular, if I = {1} is a singleton, s = s1 and T1 = {t1, . . . , tn}, then B0 = A0[t1s−1, . . . , tns−1].
84 II.4 The Banach open mapping theorem Definition II.3.4.5. If A is a f-adic ring and T and S are as in proposition II.3.4.1, we denote the completion of A T S by A T S . This is also a f-adic ring, and the canonical map A →A⟨T S ⟩is adic.
If A is complete, we can also see A⟨T S ⟩as the quotient of A⟨X⟩T by the closure of the ideal generated by {1 −siXi, i ∈I}.
Putting the universal properties of proposition II.3.4.1 and of the completion together, we get the following result : Proposition II.3.4.6. Let A be a f-adic ring and T and S be as in proposition II.3.4.1. Then, for every complete non-Archimedean topological ring B and every continuous map f : A →B such that f(si) is invertible in B for every i ∈I and that the set { f(t) f(si), i ∈I, t ∈Ti} is power-bounded in B, there exists a unique continuous ring morphism g : A⟨T S ⟩→B such that f = gϕ.
Example II.3.4.7. Take A = A0 = Zℓ, with ideal of definition J = (ℓ, X). Take I = {1} and T1 = {ℓ, u}.
Let B = A[X]T and B0 = A[X,T]. Note that B0 is strictly contained in B, because a poly-nomial P n≥0 anXn is in B0 if and only if an ∈(ℓ, u)nA for every n ≥0, so for example X ∈B −B0. We have b B = A⟨X⟩T and b B0 = A⟨X,T⟩, and again b B0 ⊊b B.
We now consider the localizations A ℓ,u ℓ and A ℓ,u u as f-adic rings, and in particular we want to write down rings of definition. As rings, we have A T s = Zℓ[ℓ−1] and A T s = Zℓ[u−1].
We get rings of definition by using the description of remark II.3.4.4. As A is a ring of definition of itself, this remark shows that A[ u ℓ] is a ring of definition of A ℓ,u ℓ and A[ ℓ u] is a ring of definition of A ℓ,u u . Note that ℓis not invertible in A[ u ℓ], even though it is of course invertible in A ℓ,u ℓ ; similarly, u is not invertible in the ring of definition A[ ℓ u] of A ℓ,u u . Note also that the completed localizations are not adic rings, even though we started from an adic ring.
In A⟨ℓ,u ℓ⟩, a ring of definition is A⟨X,T⟩/(1 −ℓX)A⟨X,T⟩, which is isomorphic to A⟨u ℓ⟩:= { X n≥0 an u ℓ n , an ∈A, an →0 as n →+∞}.
II.4 The Banach open mapping theorem The reference for this section is Henkel’s note , where Henkel explains how to adapt the proof of chapitre I §3 No3 Th´ eoreme 1.
85 II Topological rings and continuous valuations II.4.1 Statement and proof of the theorem Theorem II.4.1.1. (Theorem 1.6 of ) Let A be a topological ring that has a sequence of units converging to 0. Let M and N be Hausdorff topological A-modules that have countable topological systems of open neighborhoods of 0, and let u : M →N be a continuous A-linear map. Suppose that M is complete. Then the following properties are equivalent : (i) N is complete and u is surjective; (ii) N is complete and u(M) is open in N; (iii) for every neighborhood U of 0 in M, u(U) is a neighborhood of 0 in N; (iv) u is open.
Proof. Note that the conditions on M and N imply that their topology is defined by translation-invariant metrics (see theorem II.4.1.7).
Condition (i) clearly implies (ii), and (iii) implies (iv) by theorem II.4.1.7 and lemma II.4.1.4.
We show that (ii) implies (iii). Choose a sequence of units (an)n≥0 of A that converges to 0.
Let U be a neighborhood of 0 in M, and let V be a neighborhood of 0 in M such that V −V ⊂U.
By lemma II.4.1.2, we have M = S n≥0 a−1 n · V , so u(M) = [ n≥0 a−1 n · u(V ) ⊂ [ n≥0 a−1 n · u(V ).
As the elements an are units, multiplication by an and a−1 n is a homeomorphism of N. In par-ticular, all the sets a−1 n · u(V ) are closed in N. As u(M) is open, the Baire category theo-rem implies that at least one of the a−1 n · u(V ) has nonempty interior, so u(V ) has nonempty interior.
Let y be an interiot point of u(V ).
Then 0 = y −y is an interior point of u(V ) −u(V ) ⊂u(V −V ) ⊂u(U), which means that u(U) is a neighborhood of 0, as de-sired.
Finally, we show that (iv) implies (i). If u is open, then u(M) is open, and this implies that u is surjective by lemma II.4.1.3. As u is open, it induces an isomorphism of topological A-modules M/ Ker(u) ∼ →N (i.e. an isomorphism of A-modules that is also a homeomorphism). As N is Hausdorff, Ker(u) is a closed subgroup of M by chapitre III §2 No6 proposition 18, and then M/ Ker(u) is complete by chapitre IX §3 No1 proposition 4.
Lemma II.4.1.2. (Lemma 1.7 of .) Let A be a topological ring, let (ai)i∈I be a family of elements of A whose closure contains 0, let M be a topological A-module, and let U be a neighborhood of 0 in M. Then, for every x ∈M, there exists i ∈I such that ai · x ∈U.
Proof. Let x ∈M. As the action map A × M →M is continuous, there exists a neighborhood V of 0 in A such that V · x ⊂U. If i ∈I is such that ai ∈V , we have ai · x ∈U.
86 II.4 The Banach open mapping theorem Lemma II.4.1.3. (Lemma 1.13 of .) Let A be a topological ring in which each neighborhood of 0 contains a unit, let N be a topological A-module, and let N ′ be a A-submodule of N. Then N ′ = N if and only if N ′ contains a nonempty open subset of N.
Proof. Suppose that N ′ contains a nonempty open subset U of N. Translating U by an element of N ′, we may assume that 0 ∈U. Applying lemma II.4.1.2 to the family of all units of A, we get N = S a∈A× a−1 · U ⊂N ′.
Lemma II.4.1.4. (Proposition 1.12 of .) Let M and N be commutative topological groups whose topology comes from a translation-invariant metric, and let u : M →N be a continuous morphism of groups such that, for every neighborhood U of 0 in M, u(U) is a neighborhood of 0 in N. Suppose that M is complete. Then u is open.
Proof. We fix translation-invariant metrics giving the topologies of M and N, and denote by BM(x, r) (resp. BN(x, r)) the open ball with center x and radius r for both the metric on M (resp. N).
The hypothesis says that, for every r > 0, there exists ρ(r) > 0 such that BN(0, ρ(r)) ⊂u(BM(0, r)). Using the fact that the metrics are translation-invariant, we eas-ily get BN(u(x), ρ(r)) ⊂u(BM(x, r)) for every x ∈M and every r > 0.
Fix r > 0 and a > r. We want to show that BN(0, ρ(r)) ⊂u(BM(0, a)). (This will clearly finish the proof.) The argument is that of chapitre I §3 No3 lemme 2. Choose a sequence (rn)n≥1 of positive real numbers such that r1 = r and P n≥1 rn = a, and a sequence of positive real numbers (ρn)n≥1 such that ρn ≤ρ(rn) and that limn→+∞ρn = 0. Let y ∈BN(0, r). We want to show that y ∈u(BM(0, a)). We define a sequence (xn)n≥0 of elements of M such that xn ∈BM(xn−1, rn) and u(xn) ∈BN(y, ρn+1) for n ≥1 in the following way : - x0 = 0; - if n ≥1 and x0, . . . , xn−1 have been chosen to satisfy the two required conditions, then we have y ∈BN(u(xn−1), ρn) ⊂u(BM(xn−1, rn)), so BN(y, ρn+2) ∩u(BM(xn−1, rn)) is not empty, and we choose xn in this set.
The sequence (xn)n≥1 is a Cauchy sequence because P n≥1 rn converges, so P n≥N rn tends to 0 as N →+∞. As M is complete, (xn)n≥1 has a limit x. We have x ∈BM(0, a) by the triangle inequality, and u(x) = limn→+∞u(xn) = y because u is continuous. So we are done.
Corollary II.4.1.5. (Theorem 1.17 of .) Let A be a topological ring that has a sequence of units converging to 0. Let M and N be topological A-modules such that M is finitely generated, Hausdorff, complete and has a countable fundamental system of open neighborhoods of 0. Then any A-linear map u : M →N is continuous.
87 II Topological rings and continuous valuations Proof. Let π : An →M be a surjective A-linear map. Then the u ◦π : An →N is given by the formula (u ◦π)(a1, . . . , an) = n X i=1 ai · u(π(ei)), where (e1, . . . , en) is the canonical basis of An, so it is continuous. Similarly, π is continuous.
By theorem II.4.1.1, π is open, so, for open subset U of N, the subset u−1(U) = π((u◦π)−1(U)) of M is open. This shows that u is continuous.
Corollary II.4.1.6. Let A be a complete Tate ring, let M and N be topological A-modules, and let u : M →N be a surjective A-linear map. Suppose that M and N are quotients of finite free A-modules by closed submodules. Then u is open.
Proof. If ϖ is a topologically nilpotent unit of A, then (ϖn)n≥0 is a sequence of units of 0 converging to 0. Also, as A has a countable system of neighborhoods of 0, its topology comes from a translation-invariant metric by theorem II.4.1.7. By chapitre IX §3 No1 proposition 4, the A-modules M and N are Hausdorff, complete and metrizable. Also, u is continuous by corollary II.4.1.5. So we can apply theorem II.4.1.1 to get the conclusion.
Theorem II.4.1.7. ( chapitre IX §3 No1 propositions 1 and 2) Let G be a commutative topo-logical group. Then the topology of G is given by a translation-invariant pseudometric if and only if G has a countable fundamental system of neighborhoods of 0.
II.4.2 Applications Proposition II.4.2.1. (Proposition 2.11 of .) Let A be a complete Tate ring, and let M be a complete topological A-module that has a countable fundamental system of neighborhoods of 0.
Then the following conditions are equivalent : (i) M is a Noetherian A-module; (ii) every A-submodule of M is closed.
In particular, the ring A is Noetherian if and only if every ideal of A is closed.
Proposition II.4.2.2. (Theorems 1.17 and 2.12 of .) Let A be a complete Noetherian Tate ring. Then : (i) Every finitely generated A-module has a unique topology that makes it a Hausdorff com-plete topological A-module having a countable fundamental system of neighborhoods of 0.
88 II.4 The Banach open mapping theorem (ii) Let M and N be finitely generated A-modules endowed with the topology of (i).
If u : M →N is a A-linear map, then u is continuous, Im u is closed in N and u : M →Im u is open.
We call the topology of point (i) of the proposition the canonical topology on M.
89 III The adic spectrum In all this chapter, A is a f-adic ring.
III.1 Rings of integral elements We will be interested in subsets of Cont(A) defined by conditions of the type |a|x ≤1, for a in some fixed subset Σ of A. We will see in this section that we can only assume that Σ is an open an integrally closed subring of A, and that the case of most interest is when Σ ⊂A0.
Remark III.1.1. Let x ∈Cont(A) and let a ∈A0 (i.e. a is power-bounded). If x is a rank 1 valuation, then we necessarily have |a|x ≤1. In general, this is not true. See the next example.
Example III.1.2. Let k a field, and let A = k((t))((u)), with the rank 2 valuation x of example I.1.4.3 (so we have Rx = {f = P n≥0 fntn ∈k((t)) | f0 ∈k}). Let y be the u-adic valuation on A. Then we have seen in example I.1.5.5 that |.|x and |.|y define the same topology on A. If we put this topology on A, then x and y are in Cont(A). As y has rank 1, we have A0 = {f ∈A | |f|x ≤1} = k((t)) and A00 = {f ∈A | |f|x < 1} = uA0.
If a is any element of A0 −Rx (for example a = 1 t), then a is power-bounded but |a|x > 1.
Definition III.1.3. Let Σ be a subset of A. We write Spa(A, Σ) = {x ∈Cont(A) | ∀a ∈Σ, |a|x ≤1}.
Note that we obviously have Spa(A, Σ) ⊃Spa(A, Σ′) if Σ ⊂Σ′.
Proposition III.1.4. (Lemma 3.3 of .) Let Σ be a subset of A. We denote by AΣ the smallest open and integrally closed subring of A containing Σ. Then : (i) Spa(A, Σ) is a pro-constructible subset of Cont(A); (ii) we have AΣ = {f ∈A | ∀x ∈Spa(A, Σ), |f|x ≤1} and Spa(A, Σ) = Spa(A, AΣ).
91 III The adic spectrum Proof.
(i) If a ∈A, then Spa(A, {a}) = {x ∈Cont(A) | |a|x ≤1} = Ucont 1, a 1 is a quasi-compact open subset of Cont(A), and in particular it is constructible.
So Spa(A, Σ) = T a∈Σ Spa(A, {a}) is pro-constructible.
(ii) Let A′ = {f ∈A | ∀x ∈Spa(A, Σ), |f|x ≤1}.
Then A′ is clearly a subring of A, and it is open because it contains the open subgroup A00.
We claim that A′ is integrally closed in A. Indeed, let f ∈A, and suppose that we have an equation f n + a1f n−1 + . . . + an = 0, with n ≥1 and a1, . . . , an ∈A′. Then, for every x ∈Spa(A, Σ), we have |f|n x ≤max 1≤i≤n |ai|x|f|n−1 x ≤max(1, |f|x, . . . , |f|n−1 x ), and this is only possible if |f|x ≤1.
As A′ contains Σ, it also contains AΣ. Note also that Spa(A, Σ) = Spa(A, A′) by definition of A′. So it just remains to show that A′ ⊂AΣ.
So suppose that we have an element a ∈A −AΣ. We want to construct a continuous valuation x on A such that |a|x > 1. Consider the element a−1 of AΣ[a−1] ⊂A[a−1]; this is not a unit, because otherwise a would be an element of AΣ[a−1], so it would be integral over AΣ, which is impossible because AΣ is integrally closed in A and a ̸∈AΣ.
So there exists a prime ideal ℘of AΣ[a−1] such that a−1 ∈℘. Let q ⊂℘be a mini-mal prime ideal of AΣ[a−1]. Then B := (AΣ[a−1]/q)℘/q is a local subring of the field K := Frac(AΣ[a−1/q]), so, by theorem I.1.2.2(i), there exists a valuation subring R ⊃B of K such that mB = B ∩mR. This valuation subring defines a valuation |.|R on K, hence a valuation |.| on AΣ[a−1] via the obvious map AΣ[a−1] →AΣ[a−1]/q ⊂K, and we have Ker |.| = q. Also, by the choice of R, we have |f| ≤1 for every f ∈AΣ[a−1] and |f| < 1 for every f ∈℘, and in particular |a−1| < 1.
Let S = AΣ[a−1] −q.
As S−1AΣ[a−1] is flat over AΣ[a−1], the map S−1AΣ[a−1] →S−1A[a−1] is injective, so S−aA[a−1] ̸= {0}, and there exists a prime ideal q′ of A[a−1] such that q ∩S = ∅. In particular, we have q′ ∩AΣ[a−1] ⊂q, which implies that q′ ∩AΣ[a−1] = q because q is a minimal prime ideal, and we get a field extension K ⊂K′ := Frac(A[a−1]/q′). By proposition I.1.2.3, there exists a valuation subring R′ of K′ such that R′ ∩K = R. Let v ∈Spv(A) correspond to the composition of A →A[a−1]/q′ ⊂K′ and of |.|R′. Then we have |f|v ≤1 for every f ∈AΣ and |a|v > 1. Note also that |f|v < 1 for every f ∈A00. Indeed, if f ∈A00, then there exists an integer r ≥1 such that f ra ∈AΣ (because AΣ is an open subring of A), and then |f|r v|a|v = |f ra|v ≤1 and |a|v > 1, which implies that |f|v < 1.
We would be done if the valuation v was continuous, but this has no reason to be true.
So let w = v|cΓv be the minimal horizontal specialization of v.
We obviously have 92 III.1 Rings of integral elements w ∈Spv(A, A00 · A). Also, |f|w ≤|f|v for every f ∈A, so we have |f|w ≤1 for every f ∈AΣ and |f|w ≤|f|v < 1 for every f ∈A00. In particular, v ∈Cont(A) by theorem II.2.2.1. Also, we have |a|v > 1, so |a|v ∈cΓv, hence |a|w = |a|v > 1. Finally, we have found w ∈Cont(A) such that w ∈Spa(A, AΣ) and |a|w > 1, which means that a ̸∈A′.
Proposition III.1.5. (Lemma 3.3 of .) Let A+ ⊂A be an open and integrally closed subring.
(i) If A+ ⊂A0, then Spa(A, A+) is dense in Cont(A) and contains all the trivial valuations in Cont(A) and all the rank 1 points of Cont(A). More precisely, every x ∈Cont(A) is a vertical specialization of a point y of Spa(A, A+), and we can choose y to be of rank 1 if x is analytic.
(ii) Suppose that A is a Tate ring and has a Noetherian ring of definition. If Spa(A, A+) is dense in Cont(A), then A+ ⊂A0.
Proof.
(i) We first prove that the continuous trivial valuations and the continuous rank 1 val-uations are in Spa(A, A+). Let x ∈Cont(A). If Γx = {1}, then obviously |a|x ≤1 for every A+, so x ∈Spa(A, A+). If x has rank 1, then |a|x ≤1 for every x ∈A0 ⊃A+, so again x ∈Spa(A, A+).
Now we prove the last sentence. Let x ∈Cont(A). If supp(x) is open, then every vertical generization (or specialization) of x is continuous, and in particular the maximal vertical generization y = x/Γx of x is in Cont(A). As |f|y ∈{0, 1} for every f ∈A, we clearly have y ∈Spa(A, A+).
Suppose that supp(x) is not open, i.e. that x ∈Cont(A)an. By corollary II.2.4.8, x has a rank 1 generization y ∈Cont(A)an. We have already seen that such a y has to be in Spa(A, A+).
(ii) Suppose that A+ ̸⊂A0, and choose a ∈A+ −A0. By proposition III.1.4(ii), there exists x ∈Spa(A, A0) such that |a|x > 1.
Let L be as in lemma III.1.6, and let S = {x ∈L | |a|x > 1}; by what we just wrote, we have S ̸= ∅. As S is the intersec-tion of L and of a constructible subset of Spv(A), lemma III.1.6 implies that there exists a maximal element y ∈Cont(A) such that |a|y > 1. In particular, y is not in Spa(A, A+) and, as it is maximal, it cannot be a proper specialization of a point of Spa(A, A+). As Spa(A, A+) is pro-constructible in Cont(A), proposition I.3.1.3(ii) implies that y is not in the closure of Spa(A, A+). So Spa(A, A+) is not dense in Cont(A).
Lemma III.1.6. Let A be a Tate ring which has a Noetherian ring of definition. Let L = {x ∈Spv(A) | ∀a ∈A0, |a|x ≤1 and ∀a ∈A00, |a|x < 1}, and let Cont(A)max be the set of maximal points of Cont(A) (for the order given by specializa-tion).
93 III The adic spectrum Then Cont(A)max is the set of rank 1 points of Cont(A), Cont(A)max ⊂L, and Cont(A)max is dense in L for the constructible topology of Spv(A).
Proof. As A is a Tate ring, we have Cont(A) = Cont(A)an (see remark II.2.5.7). So, by II.2.4.8, Cont(A)max is the set of rank 1 valuations in Cont(A), and in particular Cont(A)max ⊂L.
Fix a pair of definition (A0, I) of A with A0 Noetherian, and let ϖ be a topologically nilpotent unit in A. To show that Cont(A)max is dense in L for the constructible topology, we must show that every ind-constructible subset of Spv(A) that intersects L also intersects Cont(A)an. So let E be a ind-constructible subset of Spv(A) such that E ∩L ̸= ∅. It suffices to treat the case where there exist a1, . . . , an, b1, . . . , bn, c1, . . . , cm, d1, . . . , dm ∈A such that E = {x ∈Spv(A) | ∀i ∈{1, . . . , n}, |ai|x ≤|bi|x and ∀j ∈{1, . . . , m}, |cj|x < |dj|x} (these sets form a base of the constructible topology). Let x ∈E ∩L, and denote the canonical map A →A/℘x ⊂K(x) by f. We may assume that there exists r ∈{0, . . . , n} such that f(b1), . . . , f(br) ̸= 0 and f(br+1) = . . . = f(bn) = 0. Let B be the subring of K(x) generated by f(A0), the f(ai) f(bi) for 1 ≤i ≤r and the f(cj) f(dj) for 1 ≤j ≤m. Then B is Noetherian, B ⊂Rx and B ∩mRx contains the f(cj) f(dj) for 1 ≤j ≤m and f(I) (because I ⊂A00). As the prime ideal ℘x is not open (because A is Tate, so its only open ideal is A itself), it does not contain I, so f(I) ̸= {0} and so ℘:= B ∩mRx ̸= {0}. The Noetherian local ring B℘is not a field, and K is a finitely generated extension of Frac(B℘) (it is generated by f(ϖ−1), by proposition II.2.5.2), so, by EGA II 7.1.7, there exists a discrete valuation subring R of K such that B℘⊂R and mR ∩B℘= ℘B℘. Let y be the corresponding valuation on A (i.e. the composition of f and of |.|R). We have ℘y = ℘x, and Ry = R contains B, so |ai|y ≤|bi|y for 1 ≤i ≤r, |ai|y = |bi|y = 0 for r + 1 ≤i ≤n, |cj|y < |dj|y for 1 ≤j ≤m and |f|y ≤1 for every f ∈A0. In particular, y ∈E. We want to show that y ∈Cont(A)max. As y has rank 1, it suffices to show that y is continuous by the first paragraph of the proof. As y is discrete, it is continuous if and only if {a ∈A | |a|y ≤1} is open; but this subring contains A0 and A0 is open, so we are done.
Propositions III.1.4 and III.1.5 suggest that it reasonable to consider the sets Spa(A, Σ) when Σ is an open and integrally closed subring of A contained in A0 (because then Spa(A, Σ) is dense in Cont(A) and determines Σ). We give a special name to these rings.
Definition III.1.7. A ring of integral elements in A is an open and integrally closed subring A+ of A such that A+ ⊂A0. We also say that (A, A+) is an affinoid ring (in Huber’s terminology) or a Huber pair (in other people’s terminology). A morphism of Huber pairs ϕ : (A, A+) →(B, B+) is a continuous ring morphism ϕ : A →B such that ϕ(A+) ⊂B+; it is called adic if the morphism A →B is adic.
We say that the Huber pair (A, A+) is Tate (resp. adic, resp. complete) if A is.
94 III.2 The adic spectrum of a Huber pair Example III.1.8.
(1) The biggest ring of integral elements is A0, and the smallest one in the integral closure of Z · 1 + A00. More generally, if A′ ⊂A0 is an open subring of A, then its integral closure (in A) is a ring of integral elements.
(2) If K is a topological field whose topology is defined by a valuation x, then, for any vertical specialization y of x, Ry is a ring of integral elements in K. Indeed, Ry is integrally closed in K because it is a valuation subring, and it contains the maximal ideal of Rx by theorem I.1.4.2(i), so it is open.
For example, if we take K = k((t))((u)) with the topology given by the u-adic valuation, then {f ∈k((t)) | f(0) ∈k} is a ring of integral elements in K.
III.2 The adic spectrum of a Huber pair Definition III.2.1. Let (A, A+) be a Huber pair. Then its adic spectrum is the topological space Spa(A, A+).
A rational subset of Spa(A, A+) is a subset of the form R f1, . . . , fn g = {x ∈Spa(A, A+) | ∀i ∈{1, . . . , n}, |fi|x ≤|g|x ̸= 0}, with f1, . . . , fn, g ∈A such that f1, . . . , fn generate an open ideal of A.
We also write Spa(A, A+)an = Spa(A, A+) ∩Cont(A)an for the set of analytic points of Spa(A, A+).
Remark III.2.0.1. Let f1, . . . , fn, g, f ′ 1, . . . , f ′ m, g′ ∈A such that the ideals (f1, . . . , fn) and (f ′ 1, . . . , f ′ m) are open. Then Example III.2.2. If A+ is the integral closure of Z · 1 + A00, then Spa(A, A+) = Cont(A).
Remark III.2.3. Suppose that (A, A+) is a Huber pair, with A a Tate ring, and let R f1,...,fn g be a rational subset of Spa(A, A+).
As the only open ideal of A is A itself, there exist a1, . . . , an ∈A such that a1f1 + . . . + anfn = 1. In particular, for every x ∈Spv(A), we have 1 = |1|x = max1≤i≤n |ai|x|fi|x, so there exists i ∈{1, . . . , n} such that |fi|x ̸= 0. So we have R f1, . . . , fn g = {x ∈Spa(A, A+) | ∀i ∈{1, . . . , n}, |fi|x ≤|g|x}.
(That is, we can delete the condition “|g|x ̸= 0” from the definition, because it follows from the other conditions.) Corollary III.2.4. Let (A, A+) be a Huber pair. Then Spa(A, A+) is a spectral space, and the rational subsets are open quasi-compact and form a base of the topology of Spa(A, A+).
95 III The adic spectrum Proof. By proposition III.1.4(i), Spa(A, A+) is a pro-constructible subset of Cont(A). By corol-lary II.2.2.3, Cont(A) is spectral, and it has base of quasi-compact open subsets given by the Ucont f1, . . . , fn g = {x ∈Cont(A) | ∀i ∈{1, . . . , n}, |fi|x ≤|g|x ̸= 0}, for f1, . . . , fn, g ∈ A such that f1, . . . , fn generate an open ideal of A.
By propo-sition I.3.1.3(i), Spa(A, A+) is spectral, and the inclusion Spa(A, A+) → Cont(A) is spectral (i.e.
quasi-compact).
The result follows from this and from the fact that R f1,...,fn g = Spa(A, A+) ∩Ucont f1,...,fn g .
Let ϕ : (A, A+) →(B, B+) be a morphism of Huber pairs. Then the continuous map Spv(B) →Spv(A) restricts to a continuous map Spa(B, B+) →Spa(A, A+), which we will denote by Spa(ϕ). The basic properties of these maps are given in the next proposition.
Proposition III.2.5. (Proposition 3.8 of .) Let ϕ : (A, A+) →(B, B+) be a morphism of Huber pairs. Then (i) If x ∈Spa(B, B+) is not analytic, then Spa(ϕ)(x) is not analytic.
(ii) If ϕ is adic, then Spa(ϕ) sends Spa(B, B+)an to Spa(A, A+)an.
(iii) If B is complete and Spa(ϕ) sends Spa(B, B+)an to Spa(A, A+)an, then ϕ is adic.
(iv) If ϕ is adic, then the inverse image by Spa(ϕ) of any rational domain of Spa(A, A+) is a rational domain of Spa(B, B+). In particular, Spa(ϕ) is spectral.
Proof. We write f = Spa(ϕ).
(i) We have supp(f(x)) = ϕ−1(supp(x)), so supp(f(x)) is open if supp(x) is open.
(ii) Let (A0, I) be a couple of definition of A and B0 be a ring of definition of B such that ϕ(A0) ⊂B0 and ϕ(I)B0 is an ideal of definition of B0. Let x ∈Spa(B, B+). If f(x) is not analytic, then supp(f(x)) = ϕ−1(supp(x)) is an open prime ideal of A, so it contains I, so supp(x) contains f(I), which implies that supp(x) is an open ideal of B and that x is not analytic.
(iii) Suppose that ϕ is not adic. Choose a couple of definition (A0, I) and (B0, J) of A and B such that ϕ(A0) ⊂B0 and ϕ(I) ⊂J. As ϕ is not adic, we have p ϕ(I)B0 ̸= √ J, so there exists a prime ideal ℘of B0 such that ϕ(I) ⊂℘and J ̸⊂℘. Since B0 is J-adically complete, J is contained in the Jacobson radical of B0 by Chapitre III §2 No13 lemma 3 (the idea is that every a ∈J is topologically nilpotent, so 1 −a is in-vertible with inverse P n≥0 an). So there exists a prime ideal q of B0 containing both J and ℘(take for example any maximal ideal of B0).
Let R be a valuation subring of Frac(B0/℘) dominating the local subring (B0/℘)q/℘, and let x be the corresponding 96 III.2 The adic spectrum of a Huber pair valuation on B0; in particular, we have supp(x) = ℘and |a|x < 1 for every a ∈J.
Let r : Spv(B0) →Spv(B0, J) be the retraction introduced in definition I.4.1.6. As J ̸⊂supp(x), we have J ̸⊂supp(r(x)) by theorem I.4.2.4(iv). Also, as r(x) is a horizon-tal specialization of x, we have |a|r(x) ≤|a|x < 1 for every a ∈J. So r(x) ∈Cont(B0) by theorem II.2.2.1. Let y be the unique point of Cont(B)an such that the restriction of |.|y to B0 is |.|r(x) (see lemma III.2.6). Let z ∈Cont(B)an be the unique rank 1 vertical generiza-tion of y (see corollary II.2.4.8). Then z ∈Spa(B, B+) because z has rank 1. On the other hand, supp(z) = supp(y) ⊃supp(r(x)) ⊃supp(x) = ℘, so supp(f(z)) = ϕ−1(℘) ⊃I is open and so f(z) is not analytic.
(iv) Let f1, . . . , fn, g ∈A such that f1, . . . , fn generate an open ideal a of A.
Then a contains any ideal of definition of a ring of definition of B, and, as ϕ is adic, ϕ(a) = (ϕ(f1), . . . , ϕ(fn)) is an open ideal of B. As we clearly have f −1 R f1,...,fn g = R ϕ(f1),...,ϕ(fn) ϕ(g) , this proves the result.
Lemma III.2.6. (Lemma 3.7 of , lemma 7.44 of .) Let B be an open subring of A.
Remember that B is f-adic by corollary II.1.1.8(ii). We consider the commutative square Spv(A) g / Spv(B) Spec(A) f / Spec(B) where the horizontal maps are induced by the inclusion B ⊂A.
(i) If T ⊂Spec(B) is the subset of open prime ideals, then f −1(T) ⊂Spec(A) is the subset of open prime ideals, and f induces a homeomorphism Spec(A) −f −1(T) ∼ →Spec(B) −T.
(ii) Cont(A) = g−1(Cont(B)).
(iii) The restriction of g to Cont(A)an induces a homeomorphism Cont(A)an ∼ →Cont(B)an, and, for every x ∈Cont(A)an, the canonical injection Γg(x) →Γx is an isomorphism.
Proof.
(i) Let ℘∈Spec(A). As B is an open subring of A, ℘is open in A if and only if ℘∩B is open (in B or A). This proves the first statement.
Let q ∈Spec(B) −T. Then q does not contains B00; we fix s ∈B00 −q. As B is open in A, for every a ∈A, we have sna ∈B for n big enough. So the injective ring morphism Bs →As is also surjective, and the map Spec(Bs) →Spec(As) is a homeomorphism. As Spec(B) −T = S s∈B00 Spec(Bs) and Spec(A) −f −1(T) = S s∈B00 Spec(As) (as B is open in A, so is B00, so it cannot be contained in a non-open prime ideal of A), we get the second statement.
97 III The adic spectrum (iii) If x ∈Cont(A)an, then g(x) ∈Cont(B), and the support of g(x) is not open by (i), so g(x) ∈Cont(B)an.
Let x ∈Cont(A)an and y ∈Spv(A) such that g(x) = g(y) (that is, such that |.|x and |.|y re-strict to the same valuation on B). As x is analytic and B00 is an open subgroup of A (hence not contained in supp(x)), there exists b ∈B00 such that |b|x = |b|y ̸= 0. Let a ∈A. Then there exists n ≥1 such that bna ∈B, and we get |a|x = |bna|x|b|−n x = |bna|y|b|−n y = |a|y.
So x = y.
Let x ∈ Cont(B)an.
By (i), there exists a non-open prime ideal ℘of A such that ℘∩B = ℘x.
By proposition I.1.2.3, we can extend the valuation |.|x on K(x) = Frac(B/℘x) to a valuation on Frac(A/℘); by composing with A →A/℘, we get an element y of Spv(A) such that g(y) = x. Also, it follows from the proof of (i) that B℘x →A℘is an isomorphism, so K(x) →K(y) is an isomorphism and the injection Γx →Γy is an isomorphism. Also, for every γ ∈Γy, the group {a ∈A | |a|y < γ} con-tains the open subgroup {b ∈B | |a|x < γ}, so it is open; this shows that y ∈Cont(A).
Finally, we have constructed an element y ∈Cont(A)an such that g(y) = x.
We have shown that g : Cont(A)an →Cont(B)an is bijective and continuous. Also, if R is a rational subset of Cont(B), then g(R) is a rational subset of Cont(A) (because, if f1, . . . , fn ∈B generate an open ideal of B, they also generate an open ideal of A); so g is open, hence g : Cont(A)an →Cont(B)an is a homeomorphism.
(ii) If x ∈Cont(A), then we clearly have g(x) ∈Cont(B). Let x ∈Spv(A) such that y := g(x) ∈Cont(B). We want to show that x ∈Cont(A). If y ∈Cont(B)an, then there exists x′ ∈Cont(A)an such that g(x′) = y (by (ii)), and we have in the beginning of the proof of (ii) that this implies that x = x′ ∈Cont(A)an. If y ̸∈Cont(B)an, then supp(y) is open, so supp(x) ⊃supp(y) is also open, and x is continuous (because {a ∈A | |a|x < γ} contains supp(x), hence is open, for every γ ∈Γx).
III.3 Perturbing the equations of a rational domain The goal of this section is to show that a rational domain R f1,...,fn g is not affected by a small perturbation of f1, . . . , fn, g. This result is very important in many parts of the theory (for exam-ple the proof that rational domain are preserved when we complete the Huber pair or, if it is a perfectoid Huber pair, when we tilt it).
Theorem III.3.1. (Lemma 3.10 of .) Let A be a complete f-adic ring and let f1, . . . , fn, g be elements of A such that f1, . . . , fn generate an open ideal of A. Then there exists a neighborhood V of 0 in A such that, for all f ′ 1, . . . , f ′ n, g′ ∈A, if f ′ i ∈fi +U for i ∈{1, . . . , n} and g′ ∈g +U, 98 III.3 Perturbing the equations of a rational domain then the ideal of A generated by f ′ 1, . . . , f ′ n is open and Ucont f1, . . . , fn g = Ucont f ′ 1, . . . , f ′ n g′ .
Proof. Let U = Ucont f1,...,fn g .
Let A0 be a ring of definition of A, and let J be the ideal of A generated by f1, . . . , fn. As A0 ∩J is open, it contains an ideal of definition I of A0. Choose generators a1, . . . , ar of I. By lemma III.3.2, if a′ 1 ∈a1 + I, . . . , a′ r ∈ar + I2, then I = a′ 1A0 + . . . + a′ rA0. For j ∈{, . . . , r}, write aj = Pn i=1 aijfi, with a1j, . . . , anj ∈A. Let W be a neighborhood of 0 in A such that, for every b ∈W and every (i, j) ∈{1, . . . , n} × {1, . . . , r}, we have aijb ∈I2. Then, if f ′ 1, . . . , f ′ n ∈A are such that f ′ i ∈fi + W for every i, the elements Pn i=1 aijf ′ i, 1 ≤j ≤r, of A are in A0 and generate the ideal I; in particular, the ideal of A generated by f ′ 1, . . . , f ′ n contains I, so it is open.
Set f0 = g. For every i ∈{0, . . . , n}, we write Ui = Ucont f0,...,fn fi ; note that U0 = U, that U0, . . . , Un are open quasi-compact, and that, for every i ∈{0, . . . , n} and every x ∈Ui, we have |fi|x ̸= 0. By lemma III.3.3, for every i ∈{0, . . . , n}, there exists a neighborhood Vi of 0 in A such that, for every x ∈Ui and every f ∈Vi, we have |f|x < |fi|x.
Let V = A00 ∩W ∩V0 ∩. . .∩Vn. This is a neighborhood of 0 in A. Let f ′ 1, . . . , f ′ n, g ∈A such that f ′ i ∈fi + V for every i ∈{0, . . . , n} and g′ ∈g + V . We have already seen that f ′ 1, . . . , f ′ n generate an open ideal of A. Let U ′ = Ucont f′ 1,...,f′ n g′ . We want to show that U0 = U ′. Write f ′ 0 = g′.
We first prove that U0 ⊂U ′. Let x ∈U0. For i ∈{0, . . . , n}, we have f ′ i −fi ∈V0, so |f ′ i −fi|x < |f0|x. In particular, taking i = 0 we get |f ′ 0|x = |f0|x. If i ∈{1, . . . , n}, then |fi|x ≤|f0|x = |f ′ 0|x and |f ′ i −fi|x < |f0|x imply that |f ′ i|x ≤max(|fi|x, |f ′ i −fi|x) ≤|f ′ 0|x. So x ∈U ′.
Conversely, we prove that U ′ ⊂U0.
Let x ∈Cont(A) −U0.
If |fi|x = 0 for every i ∈{0, . . . , n}, then supp(x) ⊃(f1, . . . , fn) is open, so it contains A00, so f ′ 0 −f0 ∈supp(x), so |f ′ 0|x = 0 and x ̸∈U ′. From now on, we assume that there exists i ∈{0, . . . , n} such that |fi|x ̸= 0. Choose j ∈{0, . . . , n} such that |fj|x = max0≤i≤n |fi|x; in particular, we have |fj|x ̸= 0. As x ̸∈U0, we have |f0|x < |fj|x. Note also that x ∈Uj. For every i ∈{0, . . . , n}, we have f ′ i −fi ∈Vj, so |f ′ i −fi|x < |fj|x. In particular, |f ′ j −fj|x < |fj|x, so |fj|x = |f ′ j|x. On the other hand, |f ′ 0|x ≤max(|f0|x, |f ′ 0 −f0|x) < |fj|x = |f ′ j|x, so x ̸∈U ′.
Lemma III.3.2. Let A0 be a complete adic ring, let I be an ideal of definition of A0, and suppose that we have elements a1, . . . , ar ∈I such that I = (a1, . . . , ar). Then, for all a′ 1, . . . , a′ r ∈A such that a′ i −ai ∈I2 for every i ∈{1, . . . , r}, we have I = (a′ 1, . . . , a′ r).
99 III The adic spectrum Proof. As I2 ⊂I, we have a′ 1, . . . , a′ r ∈I, so (a′ 1, . . . , a′ r) ⊂I. Consider the A-linear map u′ : Ar →I, (b1, . . . , br) 7− →b1a′ 1 + . . . + bra′ r. We want to show that u′ is surjective. For every n ≥0, we have u′(InAr) ⊂In+1. So, by Chapitre III §2 No8 corollaire 2 du th´ eor eme 1, it suffices to show that, for every n ≥0, the map grn(u′) : (InAr)/(In+1Ar) →In+1/In+2 induced by u′ is surjective. But we have grn(u′) = grn(u), where u is the map Ar →I, (b1, . . . , br) 7− →b1a1 + . . . + brar, and all the grn(u) are surjective because I = (a1, . . . , ar).
Lemma III.3.3. (Lemma 3.11 of .) Let (A, A+) be a Huber pair, X be quasi-compact subset of Spa(A, A+) and s be an element of A such that |s|x ̸= 0 for every x ∈X. Then there exists a neighborhood V of 0 in A such that, for every x ∈X and a ∈U, we have |a|x < |s|x.
In particular, there exists a finite subset T of A such that the ideal T · A is open and that, for every t ∈T, we have |t|x < |s|x.
Proof. Let T be a finite subset of A00 such that T ◦A00 is open (for example a set of generators of an ideal of definition of a ring of definition of A). For every n ≥1, the group T n ◦A00 is open by lemma II.3.3.5, so Xn = {x ∈Spa(A, A+) | ∀t ∈T n, |t|x ≤|s|x ̸= 0} is a rational subset of Spa(A, A+), and in particular it is open and quasi-compact. Also, as every element of T is topologically nilpotent and T is finite, for every x ∈Spa(A, A+) such that |s|x ̸= 0, we have x ∈Xn for n big enough. In particular, we have X ⊂S n≥1 Xn. But X is quasi-compact, so there exists n ≥1 such that X ⊂Xn. If we take V = T n · A00, then V is an open subgroup of A and |a|x < |s|x for every x ∈Xn ⊃X.
We prove the second statement. Let (A0, I) be a couple of definition of A. As V is open, it contains some power of I. Replacing I by this power, we may assume that I ⊂V . Then we can take for T a finite set of generators of I.
III.4 First properties of the adic spectrum III.4.1 Quotients Notation III.4.1.1. Let (A, A+) be a Huber pair, and let a be an ideal of A. The quotient Hu-ber pair (A, A+)/a = (A/a, (A/a)+) is defined by taking (A/a)+ to be the integral closure of A+/(A+ ∩a) in A/a.
100 III.4 First properties of the adic spectrum Proposition III.4.1.2. Let (A, A+) be a Huber pair, and let a be an ideal of A. We denote the canonical map (A, A+) →(A, A+)/a by ϕ. Then Spa(ϕ) induces a homeomorphism from Spa((A, A+)/a) to the closed subset Spa(A, A+) ∩supp−1(V (a)) = {x ∈Spa(A, A+) | supp(x) ⊃a}.
A closed subset of Spa(A, A+) as in the proposition is called a Zariski closed subset of Spa(A, A+). Not every closed subset of Spa(A, A+) is Zariski closed.
Proof. It is easy to see that Spa(ϕ) is injective with image Spa(A, A+)∩supp−1(V (a)). Also, if f1, . . . , fn, g are such that f1, . . . , fn generate an open ideal J of A and if f 1, . . . , f n, g are their images in A/a, then f 1, . . . , f n generate the ideal (J + a)/a of A/a, which is clearly open, and the image by Spa(ϕ) of R f1,...,fn g is R f1,...,fn g ∩supp−1(V (a)). This finishes the proof.
III.4.2 Spa and completion Definition III.4.2.1. Let (A, A+) be a Huber pair. Its completion is the pair ( b A, b A+), where b A+ is the closure of the image of A+ in b A.
By lemma III.4.2.3, the completion of a Huber pair is a Huber pair.
Corollary III.4.2.2. (Proposition 3.9 of .) Let (A, A+) be a Huber pair. Then the canonical map Spa( b A, b A+) →Spa(A, A+) is a homeomorphism, and a subset of Spa( b A, b A+) is a rational domain if and only if its image in Spa(A, A+) is a rational domain.
Proof. Let ϕ : A →b A be the canonical map. By proposition II.3.1.12(iii), ϕ induces a bijection Cont( b A) →Cont(A). It follows immediately from the definition of b A+ that ϕ induces a mor-phism of Huber pairs (A, A+) →( b A, b A+) and that Spa(ϕ) : Spa( b A, b A+) →Spa(A, A+) is a bijection. Also, as ϕ is adic, the inverse image by Spa(ϕ) of a rational domain of Spa(A, A+) is a rational domain by proposition III.2.5(iv). It remains to show that Spa(ϕ) maps rational subsets to rational subsets.
Let R be a rational subset of Spa( b A, b A+). As ϕ(A) is dense in b A, by theorem III.3.1, there exist f1, . . . , fn, g ∈A such that ϕ(f1), . . . , ϕ(fn) generate an open ideal of b A and R = R ϕ(f1), . . . , ϕ(fn) ϕ(g) .
We would be done if we knew that f1, . . . , fn generate an open ideal of A, but we don’t. On the other hand, as Spa(ϕ)(R) is a quasi-compact subset of Spa(A, A+) and |g|x ̸= 0 for every 101 III The adic spectrum x ∈Spa(ϕ)(R), we know by lemma III.3.3 that there exists f ′ 1, . . . , f ′ m ∈A generating an open ideal of A and such that |f ′ j|x < |g|x for every j ∈{1, . . . , m}. Then Spa(ϕ)(R) = R f1, . . . , fn, f ′ 1, . . . , f ′ m g .
Lemma III.4.2.3. Let A be a f-adic ring. If we have open sugroups of G and H of A and b A that correspond to each other by the bijection of lemma II.3.1.11, then G is a ring of integral elements of A if and only if H is a ring of integral elements of b A.
Proof. Let i : A →b A be the canonical map. As i(A) is dense in b A, it is easy to see that G is a subring of A if and only if H is a subring of b A. By proposition II.3.1.12(i), we have G ⊂A0 if and only H ⊂b A0.
Suppose that G is an open and integrally closed subring of A.
We want to prove that H is integrally closed in b A.
Let x ∈ b A be integral over H, and write xd + a1xd−1 + . . . + ad = 0, with d ≥1 and a1, . . . , ad ∈H. As H is an open neigh-borhood of 0 in b A, we can find x′ ∈A and a′ 1, . . . , a′ d ∈G such that x −i(x′) ∈H and (xd + a1xd−1 + . . . + ad) −(i(x′)d + i(a′ 1)i(x′)d−1 + . . . + i(a′ d)) ∈H.
But then x′d+a′ 1x′d−1+. . .+ad ∈G, so x′ ∈G as G is integrally closed, and x = (x−i(x′))+i(x′) ∈H.
Conversely, suppose that H is an open and integrally closed subring of b A. Then, if x ∈A is integral over G, then i(x) is integral over H, so i(x) ∈H, so x ∈G. Hence G is integrally closed.
III.4.3 Rational domains and localizations Notation III.4.3.1. Let (A, A+) be a Huber pair, let X = (Xi)i∈I be a family of indeterminates, and let T = (Ti)i∈I be a family of subsets of A such that Ti generates an open ideal of A for every i ∈I.
(1) Remember that A[X]T is f-adic by proposition II.3.3.6. We denote by A[X]+ T the integral closure in A[X]T of the open subring A+ [X,T]. So we get a Huber pair (A[X]T, A[X]+ T ).
(2) If A is complete and I is finite, we denote by A⟨X⟩+ T the integral closure in A⟨X⟩T of the open subring A+ ⟨X,T⟩. We get a complete Huber pair (A⟨X⟩T, A⟨X⟩+ T ).
Now suppose that I is a singleton, so T is a subset of A that generates an open ideal, and let s ∈A.
102 III.4 First properties of the adic spectrum (3) Denote by A T s + the integral closure in A T s of the A+-subalgebra generated by all the ts−1, for t ∈T (this A+-algebra is open in A T s by definition of the topology in the proof of proposition II.3.4.1). Then (A T s , A T s +) is a Huber pair.
It is also canonically isomorphic to the Huber pair (A[X]T, A[X]+ T )/(1 −sX) (see III.4.1.1).
(4) By combining (3) and definition III.4.2.1, we get a Huber pair (A⟨T s ⟩, A⟨T s ⟩+).
Corollary III.4.3.2. Let (A, A+) be a Huber pair, and let f1, . . . , fn, g ∈A such that f1, . . . , fn generate an open ideal of A. Let ϕ : (A, A+) →(A f1,...,fn g , A f1,...,fn g + ) be the canonical map.
Then Spa(ϕ) induces a homeomorphism Spa(A f1,...,fn g , A f1,...,fn g + ) ∼ →R f1,...,fn g , and a subset R of Spa(A f1,...,fn g , A f1,...,fn g + ) is a rational domain if and only if Spa(ϕ)(R) is a rational domain (in Spa(A, A+)).
Proof. We write f = Spa(ϕ), (B, B+) = (A f1,...,fn g , A f1,...,fn g + ) and U = R f1,...,fn g .
As ϕ is spectral, the inverse image by f of a rational domain of Spa(A, A+) is a rational domain of Spa(B, B+) (proposition III.2.5(iv)). So it suffices to prove that f sends rational domain to rational domains and induces a bijection from Spa(B, B+) to U.
As the underlying ring of B is just A[g−1], if we have two valuations on B that coincide on the image of A, then they coincide on B. So f is injective.
Let x ∈Spa(B, B+), and let y = f(x). Then |g|x = |g|y ̸= 0 because g is invertible in B. For every i ∈{1, . . . , n}, we have fig−1 ∈B+, so |fig−1|x ≤1, and |fi|y = |fi|x ≤|g|x = |g|y ̸= 0.
So y ∈U.
Let y ∈U. Then |g|y ̸= 0, so |.|y extends to a valuation on A[g−1], hence gives a point x of Spv(B); note that Γx = Γy. We want to show that x ∈Spa(B, B+). Let D be the subring of B generated by f1g−1, . . . , fng−1; as y ∈U, we have |b|x ≤1 for every b ∈D. Let γ ∈Γx. Then V := {a ∈A | |a|y < γ} is an open subgroup of A, and V · D ⊂{b ∈B | |b|x < γ}; as V · D is open by definition of the topology on B (see the proof of proposition II.3.4.1), this shows that x is continuous. Also, B+ is the integral closure of A+ · D, and |b|x ≤1 for every b ∈A+ · D, so x ∈Spa(B, B+) (see proposition III.1.4(ii)).
It remains to show that f sends rational domains of Spa(B, B+) to rational domains of Spa(A, A+). Let t1, . . . , tm, s ∈B such that t1, . . . , tm generate an open ideal of B, and let E = R t1,...,tm s ⊂Spa(B, B+). After multiplying t1, . . . , tm, s by a high enough power of g (which does not affect the condition on (t1, . . . , tm) because g is a unit in B), we may assume that t1, . . . , tm, s ∈A. Of course, we don’t know that t1, . . . , tm generate an open ideal, because it has no reason to be true. But, as g(E) ⊂Spa(A, A+) is quasi-compact and |s|x ̸= 0 for every x ∈g(E), we can find by lemma III.3.3 elements t′ 1, . . . , t′ p ∈A generating an open ideal and 103 III The adic spectrum such that |t′ j|x < |s|x for every j ∈{1, . . . , p}. Then g(E) = R t1, . . . , tm, t′ 1, . . . , t′ p s , so g(E) is a rational domain of Spa(A, A+).
Using corollary III.4.2.2, we also get : Corollary III.4.3.3. We can replace (A f1,...,fn g , A f1,...,fn g + ) with (A⟨f1,...,fn g ⟩, A⟨f1,...,fn g ⟩+) in the statement of corollary III.4.3.2.
III.4.4 Non-emptiness In this section, we give a crierion for Spa(A, A+) to be non-empty, and some consequences.
Proposition III.4.4.1. Let (A, A+) be a Huber pair.
(i) The following are equivalent : (a) Spa(A, A+) = ∅; (b) Cont(A) = ∅; (c) A/{0} = {0}.
(ii) The following are equivalent : (a) Spa(A, A+)an = ∅; (b) Cont(A)an = ∅; (c) A/{0} has the discrete topology.
Proof. Note that the support of a continuous valuation x is always a closed prime ideal of A (because it is the intersection of the open and closed subgroups {a ∈A | |a|x < γ}, for γ ∈Γx).
In particular, for every x ∈Cont(A), we have {0} ⊂supp(x).
(ii) If (c) holds, then {0} is open, so supp(x) is open for every x ∈Cont(A), and so (b) holds. Also, Spa(A, A+)an is dense in Cont(A)an by proposition III.1.5(i), so (a) and (b) are equivalent.
Suppose that (b) holds. We want to prove (c). Let (A0, I) be a couple of definition of A.
We claim that, if ℘⊂q are prime ideals of A0 and I ⊂q, then I ⊂℘. Indeed, assume that I ̸⊂℘. Let x ∈Spv(A0) such that supp(x) = ℘and that Rx dominates the local subring 104 III.4 First properties of the adic spectrum (A0/℘)q/℘of K(x) = Frac(A0,x/℘). Let r : Spv(A0) →Spv(A0, I) be the retraction of definition I.4.1.6. Then r(x) ∈Spv(A0, I) and |a|r(x) < 1 for every a ∈q ⊃I, so r(x) ∈Cont(A0) by theorem II.2.2.1. As I ̸⊂supp(x), we have I ̸⊂supp(r(x)) by theorem I.4.2.4(iv), so supp(r(x)) is not open, which means that r(x) is analytic. By lemma III.2.6(iii), x extends to an analytic point y of Cont(A), which contradicts the assumption that Cont(A)an = ∅.
Now we prove that (c) holds. Let S = 1 + I, let B = S−1A0 and let ϕ : A0 →B be the canonical map. Then ϕ(I) is contained in the Jacobson radical of B, so, if ℘∈Spec(B), then we can find q ⊃℘in Spec(B) such that ϕ(I) ⊂q (just take q to be a maximal ideal containing ℘). By the claim, this implies that every prime ideal of B contains ϕ(I), i.e.
that ϕ(I) is contained in the nilradical of B. As I is finitely generated, there exists n ≥1 such that ϕ(I)n = {0}. So there exists a ∈I such that (1 + a)In = {0} (in A0 this time).
This implies that In ⊂In+1, hence that In = In+r for every r ≥0. So {0} = In and the topology on A/{0} is discrete.
(i) We again have that (a) and (b) are equivalent because Spa(A, A+) is dense in Cont(A) (proposition III.1.5). Also, (c) clearly implies (b). Assume that (b) holds. Then, by (ii), the topology on A/{0} is discrete. If {0} ̸= A, then we can find a prime ideal ℘of A containing {0}, and then ℘is open and the trivial valuation with support ℘is an element of Cont(A), contradicting (b). So A = {0}.
Corollary III.4.4.2. Let (A, A+) be a complete Huber pair. If Spa(A, A+)an = ∅, then A is a discrete ring.
Corollary III.4.4.3. Let (A, A+) be a complete Huber pair, and let T be a subset of A. Then the following are equivalent : (a) The ideal generated by T is A.
(b) For every x ∈Spa(A, A+), there exists t ∈T such that |t|x ̸= 0.
If these conditions are satisfied and T is finite, then (R T t )t∈T is an open covering of Spa(A, A+).
Proof. If (a) holds, we can write 1 = a1t1 + . . . + antn with a1, . . . , an ∈A and t1, . . . , tn ∈T.
For every x ∈Spa(A, A+), we have 1 = |1|x ≤max1≤i≤n |ai|x|ti|x, so there exists i ∈{1, . . . , n} such that |ti|x ̸= 0.
Suppose that (a) does not holds, and let m be a maximal ideal of A such that T ⊂m. By lemma III.4.4.5, m is a closed ideal of A, hence A/m is Hausdorff, so Spa((A, A+)/m) ̸= ∅ by proposition III.4.4.1(i). As Spa((A, A+)/m) ≃Spa(A, A+) ∩supp(V (m)) by proposition III.4.1.2, there exists x ∈Spa(A, A+) such that supp(x) = m. As T ⊂m, we have |t|x = 0 for every t ∈T. So (b) does not hold.
105 III The adic spectrum We prove the last statement.
Let x ∈ Spa(A, A+).
Choose t0 ∈ T such that |t0|x = maxt∈T |t|x. Then |t0|x ̸= 0 by condition (b), so x ∈R T t0 .
Corollary III.4.4.4. Let (A, A+) be a Huber pair, and let f ∈A.
(i) We have f ∈A+ if and only if |f|x ≤1 for every x ∈Spa(A, A+).
(ii) If A is complete, we have f ∈A× if and only if |f|x ̸= 0 for every x ∈Spa(A, A+).
(iii) If A is a Tate ring, then f is topologically nilpotent if and only if |f|n x →0 as n →+∞1 for every x ∈Spa(A, A+).
Proof. Point (i) is just proposition III.1.4(ii), and point (ii) is corollary III.4.4.3 applied to T = {f}.
We prove (iii). Assume that f ∈A00, and let x ∈Spa(A, A+). Let γ ∈Γx. As 0 is a limit of (f n)n≥0, there exists n ∈N such that, for every m ≥n, f m is in the open neighborhood {a ∈A | |a|x < γ} of 0. Conversely, suppose that |f|n x →0 for every x ∈Spa(A, A+). Let ϖ ∈A be a topologically nilpotent unit. We have Spa(A, A+) = S n≥0 R fn,ϖ ϖ by hypothesis.
As Spa(A, A+) is quasi-compact, there exists n ∈N such that Spa(A, A+) = R fn,ϖ ϖ . This means that |f n/ϖ|x ≤1 for every x ∈Spa(A, A+), hence f n ∈ϖA+ by (i). As A+ ⊂A0, we get that f n = ϖa for some power-bounded a ∈A, and so f is topologically nilpotent.
Lemma III.4.4.5. Let A be a complete f-adic ring. Then A× is open in A and every maximal ideal of A is closed.
Proof. For every a ∈A00, we have 1 −a ∈A× (because P n≥0 an is an inverse of a, see Chapitre III §2 No13 lemma 3). So 1 + A00 ⊂A×. As A00 is open, this implies that A× is open in A (if a ∈A×, then multiplication by a is a homeomorphism of A, so a(1 + A00) is an open neighborhood of a).
Let m be a maximal ideal of A. Then m is contained in the closed subset A−A×, so its closure is also contained in A −A×, and in particular it does not contains 1. But the closure of m is an ideal of A and m is maximal, so m is equal to its closure.
Example III.4.4.6. Let k be a non-Archimedean field whose topology is given by the rank 1 valuation |.|, and let (A, A+) = (k[X], k0[X]), where k[X] = k[X]{1}. Let ϖ ∈k such that 0 < |ϖ| < 1. Then f = 1 + ϖX ∈A satisfies |f|x ̸= 0 for every x ∈Spa(A, A+) (see section III.5.2), but f ̸∈A×. So some hypothesis is necessary in corollary III.4.4.4(ii). (We can get away with less than completeness, see proposition 7.3.10(6).) 1That is, for every γ ∈Γx, there exists n ∈N such that |f|m x < γ for m ≥n.
106 III.5 Examples III.5 Examples III.5.1 Completed residue fields and adic Points Definition III.5.1.1. Let (A, A+) be a Huber pair and let x ∈Spa(A, A+). Remember that we denote by K(x) the fraction field of A/ supp(x). If x is not analytic, we put the discrete topology on K(x) and we set κ(x) = K(x) and K(x)+ = κ(x)+ = Rx. If x is analytic, we put the topology defined by the valuation |.|x on K(x), we denote by κ(x) the completion of K(x) and by κ(x)+ the completion of K(x)+ := Rx.
We call κ(x) the completed residue field of Spa(A, A+) at x.
We also denote by |.|x the valuation induced by |.|x on K(x) (resp. κ(x)). It is a continuous valuation, with valuation subring K(x)+ (resp. κ(x)+), and it defines the topology of K(x) and κ(x) if x is analytic.
Note that (κ(x), κ(x)+) is a Huber pair, and that κ(x)+ is a valuation subring of κ(x).
Proposition III.5.1.2. Let x ∈Spa(A, A+). Then : (i) x is analytic if and only if κ(x) is microbial.
(iii) The map Spa(κ(x), κ(x)+) → Spa(A, A+) (coming from the canonical map (A, A+) →(κ(x), κ(x)+)) induces a homeomorphism between Spa(κ(x), κ(x)+) and the set of vertical generizations of x.
Proof.
(i) If x is not analytic, then κ(x) is discrete, so is is not microbial. Conversely, suppose that x is analytic. Then, by corollary II.2.4.8, Rx has a prime ideal of height 1, so K(x) is microbial, and so is its completion κ(x).
(ii) Suppose that x is analytic.
Then, by corollary III.4.2.2, the map Spa(κ(x), κ(x)+) →Spa(K(x), Rx) is a homeomorphism.
So, in both cases, we have to show that the map Spa(K(x), Rx) →Spa(A, A+) induces a homeomorphism from Spa(K(x), Rx) to the set of vertical generizations of x in Spa(A, A+), where we put the valuation topology on K(x) to define Spa(K(x), Rx). We already know that the set of valuations rings of R of K(x) such that Rx ⊂R (i.e. RZ(K(x), Rx)) is homeomorphic to the set of vertical generizations of x in Spv(A), and every vertical generization of x is au-tomatically ≤1 on A+, so we just need to check that, if R ∈RZ(K(x), Rx) corresponds to y, then |.|R is continuous if and only if |.|y is. As |.|R and |.|y have the same valuation group and as vertical generizations with nontrivial valuation group of a continuous valua-tion are automatically continuous (proposition II.2.3.1(ii)), the only nontrivial case is the case R = K(x). In that case, |.|y is not continuous because {a ∈A | |a|y < 1} = supp(x) is not open in A, and neither is |.|R because {a ∈K(x) | |a|R < 1} = {0} is not open in K(x).
107 III The adic spectrum If x is not analytic, then κ(x) has the discrete topology, so Spa(κ(x), κ(x)+) = RZ(K(x), Rx).
On the other hand, every vertical generization of x is continuous because it has open kernel. So we also get the result.
Here is a good reason to use the completed residue field κ(x) instead of K(x).
Proposition III.5.1.3. Let U := R T s be a rational domain of X := Spa(A, A+), and let f : Y := Spa(A⟨T s ⟩, A⟨T s ⟩+) ∼ →U be the homeomorphism of corollary III.4.3.2. We fix a point y ∈Y and let x = f(y). The canonical morphism ϕ : (A, A+) →(A⟨T s ⟩, A⟨T s ⟩+) induces a morphism (K(x), K(x)+) → (K(y), K(y)+), and this gives an isomorphism (κ(x), κ(x)+) ∼ →(κ(y), κ(y)+) on the completions of these Huber pairs.
Note that the inclusion K(x) ⊂K(y) is strict in general.
Proof. We have |.|x = |.|y ◦ϕ, so ϕ−1(supp(y)) = supp(x), which gives the first statement, and also the fact that K(x)+ is the inverse image of K(y)+ in K(x). To prove the second statement, we must show that the image of K(x) in K(y) is dense.
First note that, as the canonical map A →A⟨T s ⟩is adic, x is analytic if and only if y is analytic (see proposition III.2.5).
As |s|x = |s|y ̸= 0, K(x) is also the fraction field of A[s−1]/(A[s−1] ∩supp(y)). Also, we know that A[s−1] = A T s is dense in A′ := A⟨T s ⟩(by definition of the second ring), and that the valuation topology on A induced by |.|y is weaker than the original topology of A′; so A[s−1] is dense in A′ for the valuation topology, and this implies that K(x) is dense in K(y) is x and y are analytic.
If x and y are not analytic, then the open subgroup supp(y) of A′ is the completion of supp(y)∩A[s−1], and the morphism of discrete rings A[s−1]/(A[s−1]∩supp(y)) →A′/ supp(y) is an isomorphism. So the morphism K(x) →K(y) is an isomorphism.
Definition III.5.1.4. An affinoid field is a pair (k, k+), where k is a complete non-Archimedean field and k+ ⊂k0 is an open valuation subring of k.
Example III.5.1.5. If (A, A+) is a Huber pair and x ∈Spa(A, A+)an, then (κ(x), κ(x)+) is an affinoid field.
Remark III.5.1.6. Let (k, k+) be an affinoid field. Then k+ is a ring of integral elements and a ring of definition of k. In particular, (k, k+) is a Huber pair.
Indeed, the open subring k+ of k is also integrally closed because it is a valuation subring (see proposition I.1.2.1(i)). So it is a ring of integral elements. On the other hand, k+ is open and 108 III.5 Examples bounded in k (it is bounded because it is contained in k0, and k0 is bounded because the topology of k is defined by a rank 1 valuation), so it is a ring of definition by lemma II.1.1.7.
Note also that the topology defined by the valuation corresponding to k+ is equal to the original topology on k. Indeed, the maximal ideal k00 is a height 1 prime ideal in k+, so we can apply theorem I.1.5.4.
Definition III.5.1.7. An adic Point is the adic spectrum of an affinoid field.
Note that an adic Point is not a point in general. In fact : Lemma III.5.1.8. Let (k, k+) be an affinoid field. Then Spa(k, k+) is totally ordered by special-ization. It has a unique closed point (the minimal element) corresponding to k+, and a unique generic point (the maximal element) corresponding to k0.
Proof. By remark III.5.1.6, any valuation subring R ⊂k0 of k defines a continuous valuation on k, this valuation is in Spa(k, k+) if and only if k+ ⊂R. So Spa(k, k+) is in order-reversing bijection with the set of valuation subrings R of k such that k+ ⊂R ⊂k0 (a locally closed subset of RZ(k)). By theorem I.1.4.2, it is also in order-preserving bijection with the set of prime ideals ℘of k+ such that ℘⊂k00. (Note that k00 is the prime ideal of k+ corresponding to the valuation ring k0, see theorem I.1.5.4.) As the set of ideals of k+ is totally ordered by inclusion, this shows the first statement. The second statement is clear.
Proposition III.5.1.9. Let (A, A+) be a Huber pair. Consider the following sets : (a) Σ is the set of maps of Huber pairs ϕ : (A, A+) →(k, k+), where (k, k+) is an affi-noid field and Frac(ϕ(A)) is dense in k, modulo the following equivalence relation : ϕ1 : (A, A+) →(k1, k+ 1 ) and ϕ2 : (A, A+) →(k2, k+ 2 ) are equivalent if there exists an isomorphism of Huber pairs u : (k1, k+ 1 ) ∼ →(k2, k+ 2 ) such that u ◦ϕ1 = ϕ2.
(b) Σ′ is the set of equivalence classes of continuous valuations x ∈Spa(A, A+) such that κ(x) is microbial.
Then the map Σ′ →Σ sending x to the canonical map (A, A+) →(κ(x), κ(x)+) and the map Σ →Spa(A, A+)an sending ϕ : (A, A+) →(k, k+) to the image of the closed point of Spa(k, k+) by Spa(ϕ) are both well-defined and bijective.
Note that we have a similar statement for an affine scheme Spec(A), where we use actual points (i.e. Spec(k)) and we get all the points of Spec(A).
Proof. If x ∈Spa(A, A+) and κ(x) is microbial, then (κ(x), κ(x)+) is an affinoid field; so the first map is well-defined. We show that it is a bijection by constructing an inverse. Let ϕ : (A, A+) →(k, k+) be an element of Σ.
Then composing the valuation on k corre-sponding to k+ with ϕ gives a continuous valuation x on A such that supp(x) = Ker(ϕ) and 109 III The adic spectrum A+/ supp(x) ⊂Rx; note that x is the image of the closed point of Spa(k, k+). We get a map K(x) →k such that Rx = K(x) ∩k+, and that has dense image by assumption, hence induces an isomorphism (κ(x), κ(x)+) ∼ →(k, k+). It is easy to check that this defined an inverse of the map Σ′ →Σ.
Now we consider the second map. By the first paragraph, we just need to show that an element x ∈Spa(A, A+) is analytic if and only if κ(x) is microbial; but this is proposition III.5.1.2(i).
III.5.2 The closed unit ball Let k be a complete and algebraically closed non-Archimedean field, and denote its rank 1 val-uation by |.| : k →R≥0. Remember that k00 is the maximal ideal of k0 (this is true for any non-Archimedean field). We denote the residue field k0/k00 by κ; it is also algebraically closed.
Let A = k⟨t⟩(see II.3.3.9), and let A+ = A0 = k0⟨t⟩. The points of X := Spa(A, A+) are usually divided into 5 types : (1) Classical points : Let x ∈k0 (i.e. a point of the closed unit disk in k). Then the map A →R≥0, f 7− →|f(x)| is an element of Spa(A, A+), and its support is the maximal ideal (t −x) of A. We will often denote this point by x.
Note that every maximal ideal of A is of the form (t −x) for x ∈k0 (cf. section 2.2 corollary 13), so classical points are in bijection with the maximal spectrum of A.
(2),(3) Let r ∈[0, 1] and let x ∈k0. Let xr be the point of Spv(A) corresponding to the valuation f = X n≥0 an(t −x)n 7− →sup n≥0 |an|rn = sup y∈k0, |y−x|≤r |f(y)|.
Then xr ∈Spa(A, A+), and it only depends on D(x, r) := {y ∈k0 | |x −y| ≤r}. If r = 0 then xr = x is a classical point, and if r = 1 then xr is independent of x and is called the Gauss norm.
If x ∈k0 is fixed, then the map [0, 1] →Spa(A, A+), r 7− →xr is continuous precisely at the points of [0, 1] −|k×|.
If r ∈|k×|, then we say that the pointn xr is of type (2); otherwise, we say that it is of type (3).
(4) Let D1 ⊃D2 ⊃. . . be an infinite sequence of closed disks in k0 such that T n≥1 Dn = ∅.
2 Then the valuation f 7− →inf n≥1 sup x∈Dn |f(x)| 2Such sequences exist if and only if k is not spherically complete. For example, Cℓ= b Qℓis not spherically complete.
110 III.5 Examples defines a rank 1 point of X, which is not of type (1), (2) or (3).
(5) Rank 2 valuations : Let x ∈k0 and r ∈(0, 1]. We denote by Γ0 × γZ, with the unique order such that r′ < γ < r for every r′ < r. (With the notation of section I.3.5.1, this is the group Γ and γ = r−.) Denote by x<r the point of Spv(A) corresponding to the valuation f = X n≥0 an(t −x)n 7− →max n≥0 |an|γn ∈Γ<r ∪{0}.
Then x<r is a point of Spa(A, A+), and it only depends on D0(x, r) := {y ∈k0 | |x −y| < r}.
Similarly, if r ∈(0, 1), let Γ>r the abelian group R>0 × γZ, with the unique order such that r′ > γ > r for every r′ > r. Denote by x>r the point of Spv(A) corresponding to the valuation f = X n≥0 an(t −x)n 7− →max n≥0 |an|γn ∈Γ>r ∪{0}.
Then x>r is a point of Spa(A, A+), and it only depends on D(x, r). (So x>r = x′ >r if xr = x′ r.) If r ̸∈|k×|, then xr = xr. But if r ∈|k×|, we get two new points of Spa(A, A+), which are called points of type (5).
If we think of Spa(A, A+) as a tree, then points of (1) are end points, points of type (2) and (3) are points on the limbs of the tree (and type (2) points are exactly the branching points), points of type (4) are “dead ends”. Points of type (5) are in the closure of points of type (2) (so they are less easy to visualize).
Points of type (1), (3), (4) and (5) are closed. If x ∈k0 and r ∈|k|× ∩(0, 1], then the closure of the corresponding point of type (2) xr is {xr, xr} (where x>r appears only if r < 1).
Remark III.5.2.1. We could also have defined a point x>1 of Spv(A), for x ∈k0. This is a continuous valuation on A, but it is not a point of Spa(A, A+), because it is not ≤1 on A+. In fact, if A′+ is the integral closure of k0 + A00 in A, then Spa(A, A′+) = Spa(A, A+) ∪{x>1}.
Remark III.5.2.2. We get the Berkovich space of A by identifying the points of type (5) xr with xr. Note that this is a Hausdorff space. In general, if k is a complete non-Archimedean field, the Berkovich space of an affinoid k-algebra A is the maximal Hausdorff quotient of Spa(A, A0).
III.5.3 Formal schemes Let A be an adic ring with a finitely generated ideal of definition I. Let X = Spa(A, A) = {x ∈Cont(A) | ∀a ∈A, |a|x ≤1}.
111 III The adic spectrum Remember that a valuation is called trivial if it has rank 0, i.e. if its value group is the trivial group. A trivial valuation is continuous if and only if it has open support, so the subset Xtriv of trivial valuation is in bijection with the set of open prime ideal ideals of A, i.e. Spf(A). It is easy to see that this is a homeomorphism.
We also have a retraction Spa(A, A) →Spa(A, A)triv given by x 7− →x|cΓx, and it is a spectral map. If OX is the structure presheaf of X to be defined shortly, then we have an isomorphism of locally ringed spaces Spf(A) ≃(Xtriv, r∗OX).
This will give a fully faithful functor from the category of Noetherian formal affine schemes to the category of adic spaces.
III.6 The structure presheaf In this section, (A, A+) is a Huber pair.
III.6.1 Universal property of rational domains Proposition III.6.1.1. (Lemma 8.1 and proposition 8.2 of .) Let T be a finite subset of A such that the ideal T · A is open, let s ∈A, and let U = R T s ⊂Spa(A, A+).
(i) The canonical map ι : (A, A+) →(A⟨T s ⟩, A⟨T s ⟩+) induces a spectral homeomorphism Spa(A⟨T s ⟩, A⟨T s ⟩+) ∼ →U sending rational domains to rational domains.
(ii) For every continuous morphism ϕ : (A, A+) →(B, B+) to a complete Huber pair such that Spa ϕ : Spa(B, B+) →Spa(A, A+) factors through U, there is a unique continuous ring morphism ψ : A⟨T s ⟩→B such that ψ ◦ι = ϕ, and we have ψ(A⟨T s ⟩+) ⊂B+.
(iii) Let T ′ be another finite subset of A such that the ideal T ′ · A is open, let s′ ∈A, and let U ′ = R T ′ s′ ⊂Spa(A, A+). If U ′ ⊂U, then there exists a unique continuous ring morphism ρ : A⟨T s ⟩→A⟨T ′ s′ ⟩such that ι′ = ρ ◦ι, where ι′ : A →A⟨T ′ s′ ⟩is the canonical map.
Proof.
(i) This is corollary III.4.3.3.
(ii) The assumption on Spa(ϕ) means that, for every x ∈Spa(B, B+) and every t ∈T, we have |ϕ(t)|x ≤|ϕ(s)|x ̸= 0. By points (ii) and (i) of corollary III.4.4.4, this implies that ϕ(s) ∈B×, and then that ϕ(t)ϕ(s)−1 ∈B+ for every t ∈T. In particular, ϕ(t)ϕ(s)−1 is power-bounded for every t ∈T, so the existence and uniqueness of ψ follow from the universal property of A⟨T s ⟩(proposition II.3.4.6), and the fact that ψ preserves the rings of integral elements follows immediately from the fact that ψ(t)ψ(s)−1 ∈B+ for every t ∈T.
112 III.6 The structure presheaf (iii) This follows immediately from (i) and (ii).
Corollary III.6.1.2. Let T, T ′ ⊂A be finite subsets such that the ideals T · A and T ′ · A are open, and let s, s′ ∈A. If R T s = R T ′ s′ , then there is a canonical isomorphism of Huber pairs (A⟨T s ⟩, A⟨T s ⟩+) ∼ →(A⟨T ′ s′ ⟩, A⟨T ′ s′ ⟩+) making the following diagram commute : A / A⟨T s ⟩ ≀ A⟨T ′ s′ ⟩ In other words, the rational domain R T s uniquely determines the Huber pair (A⟨T s ⟩, A⟨T s ⟩+) as a Huber pair over (A, A+).
III.6.2 Definition of the structure presheaf Definition III.6.2.1. Let (A, A+) be a Huber pair, and let X = Spa(A, X+). The structure presheaf OX on X is the presheaf with values in the category of complete topological rings and continuous ring morphisms defined by the following formulas : - if U = R T s is a rational domain of X, then OX(U) = A⟨T s ⟩; - if U is an arbitrary open subset of X, then OX(U) = lim ← − U′⊂U OX(U ′), where U ′ ranges over rational domains of X contained in U and the transition maps are given by proposition III.6.1.1(iii), and where we put the projective limit topology on OX(U).
Note that the definition of OX(U) for U a rational domain makes sense because U determines the topological A-algebra A⟨T s ⟩by corollary III.6.1.2. Also, the presheaf does take its value in the category of complete topological rings, because a projective limit of complete topological rings is complete by Chapitre II §3 No9 corollaires 1 et 2 de la proposition 18.
Remark III.6.2.2. In particular, we have OX(X) = b A.
Definition III.6.2.3. We use the notation of definition III.6.2.1. We define a subpresheaf O+ X of OX by the formula O+ X(U) = {f ∈OX(U) | ∀x ∈U, |f|x ≤1}, for every open subset U of X.
113 III The adic spectrum This is also a presheaf of complete topological rings.
Lemma III.6.2.4. If U = R T s is a rational domain, then (OX(U), O+ X(U)) = (A⟨T s ⟩, A⟨T s ⟩+).
Proof. The formula for OX(U) is just its definition.
For O+ X(U), we use the fact that U = Spa(A⟨T s ⟩, A⟨T s ⟩+) (corollary III.4.3.2) and proposition III.1.4(ii).
Let ϕ : (A, A+) → (B, B+) be a morphism of Huber pairs, and let f = Spa(ϕ) : Y := Spa(B, B+) →X := Spa(A, A+). If U ⊂X and V ⊂Y are ratio-nal domains such that f(V ) ⊂U, then proposition III.6.1.1 (and III.4.3.2) gives a continuous ring morphism OX(U) →OY (V ). So, if U ⊂X is an open subset, we get a morphism of rings f ♭ U : OX(U) →OY (f −1(U)) = f∗OX(U), and this is clearly a morphism of presheaves. It also follows immediately from the definitions that f ♭sends O+ X to f∗O+ Y .
Lemma III.6.2.5. Let T ⊂A is a finite subset generating an open ideal, s ∈A, and let ϕ : (A, A+) →(A⟨T s ⟩, A⟨T s ⟩+) be the obvious morphism. We get as before a continuous spec-tral map f : U := Spa(A⟨T s ⟩, A⟨T s ⟩+) →X := Spa(A, A+) and a morphism of presheaves f ♭: OX →f∗OU.
Then, for every open subset V of X such that V ⊂f(U), the map f ♭ V : OX(V ) →OU(f −1(V )) is an isomorphism.
Proof. We know that f is a homeomorphism from U to R T s by corollary III.4.3.2, and that an open subset V ⊂f(U) of X is a rational domain if and only if f −1(V ) is a rational domain of U. Moreover, if V is a rational domain, it is easy to see that f ♭ V : OX(V ) →OU(f −1(X)) is an isomorphism (using the explicit formulas for these rings). The lemma follows immediately from this.
III.6.3 Stalks Let (A, A+) be a Huber pair, and let X = Spa(A, A+). If x ∈X, we consider the stalk OX,x = lim − → U∋x open OX(U) = lim − → U∋x rational OX(U) (the equality follows from the fact that rational domains are a base of the topology of X) as an abstract ring without a topology. For every rational domain U ∋x of X, the valuation |.|x extends to a unique continuous valuation on OX(U) (by corollary III.4.3.2). So we get a valuation |.|x : OX,x →Γx ∪{0}.
114 III.6 The structure presheaf Proposition III.6.3.1.
(i) The ring OX,x is local, with maximal ideal mx := {f ∈OX,x | |f|x = 0}.
We denote by k(x) the residue field of OX,x, and we still write |.|x for the valuation induced on k(x) by |.|x. Let k(x)+ be the valuation subring of k(x).
(ii) The stalk O+ X,x of O+ X at x is given by the formula O+ X,x = {f ∈OX,x | |f|x ≤1}.
In other words, O+ X,x is the inverse image of k(x)+ in OX,x.
(iii) The ring O+ X,x is also local, with maximal ideal m+ x := {f ∈OX,x | |f|x < 1}. In particular, we have a canonical isomorphism between the residue fields of O+ X,x and k(x)+.
(iv) Let u : A → OX,x be the morphism coming from the restriction morphisms A →b A = OX(X) →OX(U), for U ∋x an open subset of X. Then we have |.|x◦u = |.|x, so u gives a morphism (K(x), K(x)+) →(k(x), k(x)+), which induces an isomorphism on the completions. In other words, the completion of (k(x), k(x)+) is canonically iso-morphic to (κ(x), κ(x)+).
(v) If ϕ : (A, A+) → (B, B+) is a morphism of Huber pair and y is a point of Y := Spa(B, B+) such that Spa(ϕ)(y) = x, then the morphism of rings Spa(ϕ)♭ x : OX,x →OY,y induced by Spa(ϕ)♭is such that |.|x ◦Spa(ϕ)♭ x = |.|y. In par-ticular, Spa(ϕ)♭ x is a morphism of local rings, it sends O+ X,x to O+ Y,y and also induces a morphism of local rings O+ X,x →O+ Y,y.
Proof.
(i) It is clear that mx is an ideal of OX,x, so it suffices to show that every element of OX,x −mx is a unit. Let U ∋x be an open subset of X and let f ∈OX(U) such that |f|x ̸= 0. We want to show that the image of f in OX,x is invertible. After shrinking U, we may assume that U is a rational domain of X. Then we have U = Spa(OX(U), OX(U)+) and x defines a continuous valuation on OX(U). As |f|x ̸= 0, there exists by lemma III.3.3 a finite subset T of OX(U) such that the ideal T · OX(U) is open and that |t|x < |f|x for every t ∈T. Let V be the rational domain R T f of Spa(OX(U), OX(U)+). Then x ∈V , V is open in X, and f is invertible in OX(V ) = OU(V ).
(ii) We have an injective morphism O+ X,x → OX,x induced by the injections OX(U)+ →OX(U), and it is obvious that its image is contained in {f ∈O+ X,x | |f|x ≤1} (because, for every open subset U of X, OX(U)+ = {f ∈OX(U) | ∀y ∈U, |f|y ≤1}).
Conversely, let U ∋X be an open subset of X, and let f ∈OX(U) such that |f|x ≤1.
We want to show that the image of f in OX,x is in O+ X,x. By lemma III.6.3.2, the set V := {y ∈U | |f|y ≤1} is an open subset of X, and we obviously have x ∈X and f|V ∈OX(V )+; this implies the desired result.
(iii) As m+ x is clearly an ideal of O+ X,x, it suffices to show that every element of O+ X,x −m+ x is 115 III The adic spectrum invertible. Let f ∈O+ X,x, and suppose that f ̸∈mx, i.e. that |f|x = 1. Then f ∈O× X,x by (i), and |f −1|x = 1, so f −1 ∈O+ X,x.
(iv) The fact that |.|x ◦u = |.|x follows immediately from the definition of the valuation |.|x on OX,x. By (i), we have k(x) = lim − → U∋x OX(U)/{f ∈OX(U) | |f|x = 0} and k(x)+ = lim − → U∋x OX(U)+/{f ∈OX(U)+ | |f|x = 0}, where U runs through all rational domains of X containing x. By proposition III.5.1.3, if U ′ ⊂U are two rational domains of X containing x, then the restriction maps OX(U)/{f ∈OX(U) | |f|x = 0} →OX(U ′)/{f ∈OX(U ′) | |f|x = 0} and OX(U)+/{f ∈OX(U)+ | |f|x = 0} →OX(U ′)+/{f ∈OX(U ′)+ | |f|x = 0} have dense image. So the image of K(x) (resp. K(x)+) in k(x) (resp. k(x)+) is dense, which implies the result.
(v) The fact that |.|x ◦Spa(ϕ)♭ x = |.|y follows immediately from the definitions, and the other statements follow from this.
Lemma III.6.3.2. (Remark 8.12 of .) Let U be an open subset of X := Spa(A, A+) and f, g ∈OX(U). Then V := {x ∈U | |f|x ≤|g|x ̸= 0} is an open subset of X.
Proof. We know that U is a union of rational domains of X, and it suffices to show that the intersection of V with each of these rational domains is open. So we may assume that U is a rational domain, U = R T s = Spa(A⟨T s ⟩, A⟨T s ⟩+). It also suffices to show that V is open in U, so we may assume that U = X, i.e. f, g ∈A. Then V is open by definition of the topology on Spa(A, A+) as the topology induced by that of Spv(A). 3 If A is a Tate ring, then we can show that the categories of finite ´ etale covers of O+ X,x and k(x)+ are equivalent. First we need a definition.
Definition III.6.3.3. (See [25, Definition 09XE].) Let R be a ring and I be an ideal of R. We say that the pair (R, I) is henselian (or I-adically henselian) if : 3Note that V is not a rational domain in general, because we did not assume that f generates an open ideal of A.
116 III.6 The structure presheaf (a) I is contained in the Jacobson radical of A; (b) for any monic polynomial f ∈A[X] and any factorization f = g0h0 in A/I[X], where f is the image of f in A/I[X], if g0 and h0 are monic and generate the unit ideal of A/I[X], then there exists a factorization f = gh in A[X] with g, h monic and such that g0 = g mod I and h0 = g mod I.
If R is local and I is its maximal ideal, we also say that R is henselian.
Theorem III.6.3.4. ([25, Lemma 0ALJ].) Let R be a ring and I be an ideal of R. If R is I-adically complete, then the pair (R, I) is henselian.
Theorem III.6.3.5. ([25, Lemma 09XI].) Let R be a ring and I be an ideal of R. The following are equivalent : (i) The pair (R, I) is henselian.
(ii) For every ´ etale map of rings R →R′ and every map of R-algebras σ : R′ →R/I, there exists an map of R-algebras R′ →R lifting σ.
R′ σ !
R O / R/I (iii) For any finite R-algebra S, the map S →S/IS induces a bijection on idempotent ele-ments.
(iv) For any integral R-algebra S, the map S →S/IS induces a bijection on idempotent elements.
(v) I is contained in the Jacobson radical of R and every monic polynomial f ∈R[X] of the form f(X) = Xn(X −1) + anXn + . . . + a1X + a0, with a0, . . . , an ∈I and n ≥1, has a root in 1 + I.
Moreover, if these conditions hold, then the root in point (v) is unique.
Theorem III.6.3.6. ([25, Lemma 09ZL].) Let (R, I) be a henselian pair.
Then the functor S 7− →S/IS induces an equivalence between the category of finite ´ etale R-algebras and the category of finite ´ etale R/I-algebras.
Proposition III.6.3.7. (See proposition 7.5.5 of .) We use the notation of proposition III.6.3.1, and we suppose that A is a Tate ring. Let ϖ ∈A be a topologically nilpotent element.
(i) The ring O+ X,x is ϖ-adically henselian, and the map O+ X,x →k(x)+ induces an isomor-phism on ϖ-adic completions.
117 III The adic spectrum (ii) The pairs (OX,x, mx) and (O+ X,x, mx) are henselian.
Proof.
(i) An inductive limit of henselian pairs is henselian by [25, Lemma 0A04]. So it suffices to show that OX(U)+ is ϖ-adically henselian for every rational domain U of X.
Let U be a rational domain of X. Then (B, B+) := (OX(U), OX(U)+) is a complete Huber pair by lemma III.6.2.4, and it is a Tate pair because A is a Tate ring. Let B0 ⊂B+ be a ring of definition of B. By proposition II.2.5.2, ϖB0 is an ideal of definition of B0, so B0 is ϖ-adically complete, hence ϖ-adically henselian by theorem III.6.3.4. As B+ is the union of all the rings of definition of B contained in it (by corollary II.1.1.8(iii)), anotehr application of [25, Lemma 0A04] shows that B+ is ϖ-adically henselian.
We prove the second statement. Note that mx ⊂O+ X,x and that O+ X,x/mx = k(x)+. Also, we have ϖmx = mx because ϖ is a unit in A, so mx = ϖnmx ⊂ϖnO+ X,x for every n ∈N.
This implies that the map O+ X,x →k(x)+ induces an isomorphism on ϖ-adic completions.
(ii) As mx = ϖmx ⊂ϖO+ X,x, the fact that (O+ X,x, mx) is henselian follows from the first statement of (i) and from [25, Lemma 0DYD].
To prove that (OX,x, mx) is henselian, it suffices to note that mx is contained in the Jacobson readical of OX,x, and that it satisfies the property of theorem III.6.3.5(v) (because it does in O+ X,x).
III.6.4 The category V pre Definition III.6.4.1. We denote by V pre the category of triples (X, OX, (|.|x)x∈X), where : - X is a topological space; - OX is a presheaf of complete topological rings on X such that, for every x ∈X, the stalk OX,x (seen as an abstract ring) is a local ring; - for every x ∈X, |.|x is an equivalence class of valuations on OX,x whose support is equal to the maximal ideal of OX,x.
A morphism (X, OX, (|.|x)x∈X) →(Y, OY , (|.|y)y∈Y ) is a pair (f, f ♭), where f : X →Y is a continuous map and f ♭: OX →f∗OY is a morphism of presheaves of topological rings such that, for every x ∈X, the morphism f ♭ x : OX,x →OY,f(x) induced by f ♭is compatible with the valuations (i.e. |.|f(x) ◦f ♭ x = |.|x). Note that this implies that f ♭ x is a local morphism.
Example III.6.4.2. If (A, A+) is a Huber pair, then Spa(A, A+) is an object of V pre, and any morphism of Huber pairs ϕ : (A, A+) → (B, B+) induces a morphism Spa(ϕ) : Spa(B, B+) →Spa(A, A+) in V pre.
Corollary III.6.4.3. Let (A, A+) be a Huber pair, and let ϕ : (A, A+) →( b A, b A+) be the canonical morphism. Then Spa(ϕ) is an isomorphism in V pre.
118 III.6 The structure presheaf Proof. By corollary III.4.2.2, the map Spa(ϕ) : Spa( b A, b A+) →Spa(A, A+) is a homeomor-phism preserving rational domains. Also, if T ⊂A is a finite subset such that T · A is open and if s ∈A, we have (A⟨T s ⟩, A⟨T s ⟩+) = ( b A⟨T s ⟩, b A⟨T s ⟩+). This implies the result.
Proposition III.6.4.4. Let (A, A+) and (B, B+) be Huber pairs, and suppose that B is complete.
Then ϕ 7− →Spa(ϕ) induces a bijection Hom((A, A+), (B, B+)) →HomV pre(Spa(B, B+), Spa(A, A+)), where the first Hom is taken in the category of Huber pairs. The inverse of this bijections sends a morphism (f, f ♭) : Y := Spa(B, B+) →X := Spa(A, A+)) to the morphism (A, A+) →( b A, b A+) = (OX(X), OX(X)+) f♭ X →(OY (Y ), OY (Y )+) = (B, B+).
Proof. If ϕ : (A, A+) →(B, B+) is a morphism of Huber pairs, then the composition of Spa(ϕ)♭ and of the canonical map A →b A is ϕ by definition of Spa(ϕ)♭.
Conversely, let (f, f ♭) : Y →X be a morphism in V pre, and let ϕ : A →B be the composition of f ♭ X and of A →b A. We want to show that (f, f ♭) = (Spa(ϕ), Spa(ϕ)♭). Let U = R T s be a rational domain in X, and let V = f −1(U). We have V = {y ∈Y | ∀t ∈T, |t|f(y) ≤|s|f(y) ̸= 0} = {y ∈Y | ∀t ∈T, |ϕ(t)|y ≤|ϕ(s)|y ̸= 0}, where the second equality comes from the fact that |a|f(y) = |f ♭ y(a)|y = |ϕ(a)|y for every a ∈A.
If W is a quasi-compact open subset of V , then, by lemma III.3.3, we can find a finite subset TW of B generating an open ideal of B and such that |t|y ≤|ϕ(s)|y for every y ∈W and t ∈TW.
Then we have W ⊂W ′ := R ϕ(T),TW ϕ(s) ⊂V , so the map f ♭ U : OX(U) →OY (W) factors through OY (W ′) = B⟨ ϕ(T),T ′ W ϕ(s) ⟩. We know that f ♭ U is equal to ϕ on the image of A, that ϕ is continuous and that the rings A⟨T s ⟩and B⟨ ϕ(T),T ′ W ϕ(s) ⟩are completions of localizations of A and B, so this implies that f ♭ U : OX(U) →OX(W) is equal to Spa(ϕ)♭ U. Going to the limit on W ⊂V open quasi-compact, we see that f ♭ U : OX(U) →OY (V ) is also equal to Spa(ϕ)♭. This implies the analogous statement for an arbitrary open subset U of X by the definition of the presheaf OX.
III.6.5 Adic spaces Definition III.6.5.1.
An open immersion in V pre is a morphism (f, f ♭) : (X, OX, (|.|x)x∈X) → (Y, OY , (|.|y)y∈Y ) such that f : X → Y is a homeomorphism onto an open subset U of Y and that the induced morphism (X, OX, (|.|x)x∈X) →(U, OY |U, (|.|y)y∈U) is an isomorphism in V pre.
119 III The adic spectrum Example III.6.5.2. Let (A, A+) be a Huber pair. If T ⊂A generates an open ideal and s ∈A, and if ϕ : (A, A+) →(A⟨T s ⟩, A⟨T s ⟩+) is the obvious map, then Spa(ϕ) is an open immersion in V pre by lemma III.6.2.5.
Definition III.6.5.3. We denote by V the full subcategory of V pre whose objects are the triples (X, OX, (|.|x)x∈X) such that OX is a sheaf.
An affinoid adic space is an object of V that is isomorphic to Spa(A, A+), for (A, A+) a Huber pair.
An adic space is an object (X, OX, (|.|x)x∈X) of V such that there exists an open covering (Ui)i∈I such that, for every i ∈I, the triple (Ui, OX|Ui, (|.|x)x∈Ui) such that is an affinoid adic space.
A morphism of adic spaces is a morphism of V .
The next natural question is : which Huber pairs give rise to affinoid adic spaces ? (It is not true that the structural presheaf of Spa(A, A+) is always a sheaf.) We will give some criteria in the next chapter.
120 IV When is the structure presheaf a sheaf ?
The goal of this chapter is to give sufficient conditions on the Huber pair (A, A+) for X = Spa(A, A+) to be an adic space, i.e. for the structure presheaf OX to be a sheaf. As we will see, these conditions also imply that the cohomology of OX on any rational domain of X is concentrated in degree 0, as we would expect from the cohomology of the structural sheaf of an affinoid adic space. 1 IV.1 The main theorem IV.1.1 Statement Before we can state the main theorem of this chapter, we need some definitions.
Definition IV.1.1.1. Let A be a Tate ring. We say that A is strongly Noetherian if b A⟨X1, . . . , Xn⟩ 2 is Noetherian for every n ≥0.
Definition IV.1.1.2. A non-Archimedean topological ring A is called uniform if A0 is bounded in A.
Note that, if A is f-adic, this is equivalent to the fact that A0 is a ring of definition of A.
Remark IV.1.1.3. Any Hausdorff uniform Tate ring is reduced. Indeed, let A be a Tate ring, and let ϖ ∈A be a topologically nilpotent unit. Suppose that A is uniform, so that A0 is a ring of definition of A. Let a ∈A be a nilpotent element. For every n ∈N, the element ϖ−na is nilpotent, hence power-bounded, so a ∈ϖnA0. As the topology of A0 is the ϖA0-adic topology by proposition II.2.5.2, and as A0 is Hausdorff by hypothesis, we have T n≥0 ϖnA0 = {0}. So a = 0.
Definition IV.1.1.4. Let (A, A+) be a Huber pair. Then we say that (A, A+) is stably uniform if, for every rational subset U of Spa(A, A+), the f-adic ring OSpa(A,A+)(U) is uniform.
1Note however that, unlike the case of schemes, there is no cohomological characterization of affinoid adic spaces.
2See notation II.3.3.9.
121 IV When is the structure presheaf a sheaf ?
Theorem IV.1.1.5. (Theorem 2.2 of , theorem 8.27 of , theorem 7 of .) Let (A, A+) be a Huber pair, and let X = Spa(A, A+). Suppose that (A, A+) satisfies one of the following conditions : (a) the completion b A is discrete; (b) the completion b A has a Noetherian ring of definition; (c) A is a strongly Noetherian Tate ring; (d) the Huber pair (A, A+) is Tate and stably uniform.
Then OX is a sheaf, and, for every rational domain U of X and every i ≥1, we have Hi(U, OX) = 0.
See section IV.3 for the proof in cases (c) and (d).
Remark IV.1.1.6. Case (a) of the theorem applies to discrete rings A, and we get that Spv(A) and Riemann-Zariski spaces are adic spaces.
Case (b) applies for example to a complete Noetherian adic ring A, and gives (with some more work) a fully faithful embedding of the category of locally Noetherian formal schemes over Spf(A) into the category of adic spaces over Spa(A, A). Note that adic rings are not Tate, so we cannot apply (c) or (d).
Case (c) applies for example to affinoid algebras over a complete non-Archimedean field k, and gives a fully faithful embedding of the category of rigid analytic varieties over k into the category of adic spaces over Spa(k, k0). Note that these affinoid algebras (even k itself) do not have a Noetherian ring of definition unless the valuation defining the topology of k is discrete, so we cannot apply case (b) in general.
Finally, case (d) typically applies to perfectoid algebras. Note that, if k is a complete non-Archimedean field and A is an affinoid k-algebra, then A is uniform if and only if it is reduced (see section II.1.4 for a reference); in particular, case (d) is not sufficient if we are interested in non-reduced rigid analytic varieties.
IV.1.2 Examples of strongly Noetherian Tate rings Proposition IV.1.2.1. (Proposition 6.29 of .) Let ϕ : A →B be a morphisms of ring.
Suppose that A and B are f-adic rings, and that B is complete. Then the following conditions are equivalent : (i) There exists a positive integer n, finite subsets T1, . . . , Tn such that Ti · A is open in A for every i and a surjective continuous open A-algebra morphism π : A⟨X1, . . . , Xn⟩T1,...,Tn →B.
122 IV.1 The main theorem (ii) The morphism ϕ is adic, there exists a finite subset M of B such that the A-subalgebra ϕ(A)[M] of B is dense in B, and there exist rings of definition A0 of A and B0 of B and a finite subset N of B0 such that ϕ(A0) ⊂B0 and that ϕ(A0)[N] is dense in B0.
(iii) There exists rings of definition A0 of A and B0 of B such that : (a) ϕ(A0) ⊂B0; (b) B is finitely generated (as an algebra) over ϕ(A) · B0; (c) there exists a surjective continuous open A0-algebra morphism b A0⟨X1, . . . , Xn⟩→B0, for some n ∈N.
(iv) For every open subring A0 of A, there exists an open subring B0 of B such that conditions (a), (b) and (c) of (iii) hold.
Definition IV.1.2.2. If a morphism ϕ : A →B satisfies the equivalent conditions of proposition IV.1.2.1, we say that the A-algebra B is topologically of finite type.
Proposition IV.1.2.3. (Propositions 6.33, 6.35 and 6.36 of .) Let A be a strongly Noetherian f-adic ring, and let B be a A-algebra that is topologically of finite type (so B is a complete f-adic ring). Then : (i) B is strongly Noetherian (in particular, it is Noetherian).
(ii) If A has a Noetherian ring of definition, so does B.
Theorem IV.1.2.4. (Theorem 1 of section 5.2.6 of .) Any complete non-Archimedean field is strongly Noetherian.
IV.1.3 Examples of stably uniform Tate rings We fix a prime number ℓ.
Definition IV.1.3.1. We say that a ring A is of characteristic ℓif ℓ·1A = 0. If A is of characteristic ℓ, then the map FrobA : A →A, a 7− →aℓis a ring endormophism called the Frobenius endomorphism of A; we say that A is perfect (resp. semiperfect) if FrobA is bijective (resp.
injective).
Notation IV.1.3.2. If A is perfect, we often write a 7− →a1/ℓfor the inverse of FrobA.
Remark IV.1.3.3. Unfortunately, there is another definition of perfect and semiperfect rings (for example, a not necessarily commutative ring R is called left perfect if every left module has a projective cover); it is totally unrelated to the previous definition. We will only use definition IV.1.3.1 in these notes.
Remark IV.1.3.4. Let A be a ring of characteristic ℓ. If A is reduced, then Ker(FrobA) = {0}, so A is perfect if and only if it is semiperfect.
123 IV When is the structure presheaf a sheaf ?
Theorem IV.1.3.5. (Lemma 7.1.6 of .) Let A be a complete Tate ring of characteristic ℓ.
Suppose that A is perfect. Then A has a perfect ring of definition and is stably uniform.
We will see later that these rings are exactly the perfectoid Tate rings of characteristic ℓ.
Proof. Suppose that we have shown that A has a perfect ring of definition. By lemmas IV.1.3.8 and IV.1.3.9, for every finite subset T of A generating an open ideal and every s ∈A, the ring A⟨T s ⟩is also a complete and perfect Tate ring. So it suffices to prove that A has a perfect ring of definition and is uniform.
Let A0 be a ring of definition of A and ϖ ∈A be a topologically nilpotent unit. For every n ≥1, let An = A1/ℓn 0 , and let A∞= S n≥0 An. Then A∞is a perfect subring of A and A0 ⊂A∞⊂A0.
We first show that A∞is bounded, hence a ring of definition of A. Note that FrobA : A →A is continuous and surjective, and it is A-linear if we put the obvious A-module structure on its source and the A-module structure given by a · b = aℓb (a, b ∈A) on its target. So, by the Banach open mapping theorem (see corollary II.4.1.6), FrobA is open. In particular, the subring FrobA(A0) of A is open, so there exists r ∈N such that ϖrA0 ⊂FrobA(A0). Applying Frob−1 A , we see that we have s ∈N such that ϖsA1 ⊂A0 (any s ≥rℓ−1 will do). As in the proof of lemma IV.1.3.8, this implies that ϖs+s/ℓ+...+s/ℓn−1An ⊂A0 for every n ≥1, hence that ϖ2sA∞⊂A0. So A∞is bounded.
Now we show that ϖA0 ⊂A∞, which will imply that A0 is bounded. Let a ∈A0. As a is power-bounded, there exists r ∈N such that {ϖran, n ∈N} ⊂A∞. As A∞is closed in A under taking ℓth roots, this implies that ϖr/ℓna ∈A∞for every n ≥0. In particular, taking n such that r ≤ℓn, we get ϖa ∈A∞.
Lemma IV.1.3.6. Let A be a ring of characteristic ℓand S ⊂A be a multiplicative system. If A is perfect (resp. semiperfect), so is S−1A.
Proof. Let B = S−1A. Suppose that A is semiperfect. Let b ∈B, and write b = as−1, with a ∈A and s ∈S. As A is semiperfect, we can find c, t ∈A such that cℓ= a and tℓ= s. Then ctℓ−1s−1 ∈B, and (ctℓ−1s−1)ℓ= b. So B is semiperfect.
We now assume that A is perfect, and we want to show that Ker(FrobB) = {0}. So let a ∈A and s ∈S such that (as−1)ℓ= 0 in B. This means that there exists t ∈S such that taℓ= 0 in A.
Then ta ∈Ker(FrobA), so ta = 0 in A, so as−1 = 0 in B.
Lemma IV.1.3.7. Let A be a topological ring, N be a positive integer and T be a finite subset of A. Then T is power-bounded if and only the set {tN, t ∈T} is power-bounded.
124 IV.1 The main theorem Proof. Write T ′ = {tN, t ∈T}.
Remember that T is power-bounded is and only if the set S n≥1 T(n) is bounded, where, for every n ≥1, T(n) = {t1 . . . tn, t1, . . . , tn ∈T}. As S n≥1 T(n) ⊃S n≥1(T ′)(n), T ′ is power-bounded if T is. On the other hand, we have [ n≥1 T(n) = [ (mt)t∈T ∈{0,...,N−1}T Y t∈T tnt/N ! [ n≥1 (T ′)(n) !
, so S n≥1 T(n) is a finite union of translates of S n≥1(T ′)(n). This shows that T is power-bounded if T ′ is.
Lemma IV.1.3.8. Let A be a f-adic ring, T be a finite subset of A that generates an open ideal and s ∈A. Suppose that A is perfect. Then we have a canonical isomorphism of A-algebras A T s = A T 1/ℓ s1/ℓ .
If moreover A is a Tate ring and has a perfect ring of definition, then A T s also has a perfect ring of definition.
Proof. Write B = A T s . As an abstract ring, the f-adic ring B is isomorphic to A[s−1]. So, by lemma IV.1.3.6, we know that B is perfect.
The first statement follows immediately from the universal property of the localization (see proposition II.3.4.1) and from lemma IV.1.3.7 : the topological rings A T s and A T 1/ℓ s1/ℓ satisfy the same universal property.
Let A0 be a ring of definition of A, and let B0 be the A0-subalgebra of B generated by the elements ts−1, t ∈T. Then B0 is a ring of definition of B by remark II.3.4.4.
Suppose that A is a Tate ring, let ϖ ∈A be a topologically nilpotent unit, and suppose that A0 is perfect. Then B1/ℓ 0 is the A0-subalgebra of B generated by the t1/ℓs−1/ℓ, t ∈T, so it is also a ring of definition of B by the first statement of the lemma and the previous paragraph. In particular, B1/ℓ 0 is bounded, so there exists a positive integer r such that ϖrB1/ℓ 0 ⊂B0. An easy induction on n then shows that ϖr+r/ℓ+...+r/ℓn−1B1/ℓn 0 ⊂B0 for every n ≥1. In particular, we have ϖ2B1/ℓn 0 ⊂B0 for every n ≥1. Let B′ 0 = S n≥1 B1/ℓn 0 . This is a perfect subring of B, it is open because it contains B0, and it is bounded because ϖ2rB′ 0 ⊂B0. So B′ 0 is a perfect ring of definition of B, and we are done.
Lemma IV.1.3.9. Let A be a Tate ring with a perfect ring of definition. Then A and b A are perfect, and b A also has a perfect ring of definition.
Proof. Let A0 be a perfect ring of definition of A, and let ϖ ∈A0 a topologically nilpotent unit.
Then A = A0[ϖ−1] by proposition II.2.5.2, so A is perfect by lemma IV.1.3.6.
125 IV When is the structure presheaf a sheaf ?
Let b A0 = lim ← −n≥1 A0/ϖnA0. As b A = b A0[ϖ−1] by (v) of corollary II.3.1.9, it suffices to prove that b A0 is perfect. As ϖ1/ℓis also a topologically nilpotent element of A0, the canonical map lim ← −n≥1 A0/ϖnℓA0 →b A0 (sending (xn) ∈lim ← −n≥1 A0/ϖn/ℓA0 to (xℓn)n≥1) is an isomorphism, and this implies immediately that b A0 is perfect.
IV.2 Some preliminary results In this section, we fix a Huber pair (A, A+) with A a Tate ring, and we set X = Spa(A, A+).
IV.2.1 Strictness and completion Definition IV.2.1.1. ( chapitre III §2 No8 d´ efinition 1.) Let ϕ : M →M ′ be a continuous morphism of topological groups. We say that ϕ is strict if the following two topologies on ϕ(M) coincide : - the quotient topology given by the isomorphism ϕ(M) ≃M/ Ker ϕ; - the subspace topology given by the inclusion ϕ(M) ⊂M ′.
Proposition IV.2.1.2. ( chapitre III §2 No12 lemme 2.) Let M, M ′, M ′′ be abelian topological groups that have countable fundamental systems of neighborhoods of 0, and let ϕ : M →M ′ and ψ : M ′ →M ′′ be continuous group morphisms. Suppose that : - the sequence M ϕ →M ′ ψ →M ′′ is exact as a sequence of abstract group, i.e. Ker ψ = Im ϕ; - the morphisms ϕ and ψ are strict.
Then the sequence c M b ϕ →c M ′ b ψ →c M ′′ is exact and b ϕ and b ψ are strict.
IV.2.2 The ˇ Cech complex for a special cover Let t ∈A. We consider the rational domains U = R 1,t 1 = {x ∈X | |t|x ≤1} and V = R 1 t = {x ∈X | |t|x ≥1}. Note that U ∩V = {x ∈X | |t|x = 1} is the rational domain R 1,t,t2 t .
Let B = A 1,t 1 , C = A 1 t and D = A 1,t,t2 t . We have canonical adic maps ϕB : A →B, ϕC : A →C, ψB : B →D and ψC : C →D such that ψB ◦ϕB = ψC ◦ϕC is the canonical map 126 IV.2 Some preliminary results from A to D. Note that ϕB and ψC are continuous and bijective maps of topological rings, but they are not homeomorphisms in general. For example, B is just A as an abstract ring, but with a topology that makes t power-bounded.
We denote the map ϕB ⊕ϕC : A →B ⊕C by ε and the map ψB −ψC : B ⊕C →D by δ.
Then A ε →B ⊕C δ →D is a complex, whose completion is the ˇ Cech complex OX(X) →OX(U) ⊕OX(V ) →OX(U ∩V ) of the open cover (U, V ) of X. We want to know when this ˇ Cech complex is exact. This is the goal of the following proposition.
Proposition IV.2.2.1. (See Lemma 2 of and the discussion preceding it.) (i) The complex 0 →A ε →B ⊕C δ →D →0 is exact as a complex of abstract commutative groups.
(ii) The map δ : B ⊕C →D is strict.
(iii) The map OX(U)⊕OX(V ) →O(U∩V ) sending (f, g) to f|U∩V −g|U∩V (i.e. the completion of δ) is surjective.
(iv) The following conditions are equivalent : (a) the map ε : A →B ⊕C is strict; (b) the complex 0 →OX(X) →OX(U) ⊕OX(V ) →OX(U ∩V ) is exact; (c) the complex 0 →OX(X) →OX(U) ⊕OX(V ) →OX(U ∩V ) →0 is exact; (d) there exists rings of definition A0 ⊂A, B0 ⊂B and C0 ⊂C, a topologically nilpotent unit ϖ ∈A and n ∈N such that ϖn(ϕ−1 B (B0) ∩ϕ−1 C (C0)) ⊂A0.
Remember that a morphism of topological groups u : G →H is called strict if the quotient topology on u(G) ≃G/ Ker u coincides with the subspace topology induced by the topology of H.
Proof. Let A0 be a ring of definition of A, and let ϖ ∈A be a topologically nilpotent unit such that ϖ ∈A0. Then, by remark II.3.4.2, A0[t] is a ring of definition of B, A0[t−1] is a ring of definition of C and A0[t, t−1] is a ring of definition of D. Also, as the f-adic rings A, B, C and D are Tate, the topology on these rings of definition is the ϖ-adic topology by proposition II.2.5.2.
(i) The fact that ε is injective and δ surjective follows from the fact that ϕB and ψC are iso-morphisms of abstract rings. Let (b, c) ∈B ⊕C such that δ(b, c) = 0, i.e. b −c = 0 in A[t−1]. We have b ∈A, so (b, c) = ε(b).
(ii) The map δ is surjective by (i), so we just need to check that δ is open. This follows from the obvious fact that δ(ϖnA0[t] ⊕ϖnA0[t−1]) = ϖnA0[t, t−1] for every n ∈N.
(iii) This follows from (ii) and from chapitre III §2 No12 lemme 2.
(iv) The map ε is strict if and only, for all rings of definition B0 ⊂B and C0 ⊂C0, the groups ϖn(ϕ−1 B (B0) ∩ϕ−1 C (C0)) = ε−1(ϖnB0 ⊕ϖnC0), n ≥0, form a fundamental system of 127 IV When is the structure presheaf a sheaf ?
neighborhoods of 0 in A. This shows that (a) and (d) are equivalent. Also, (c) obviously implies (b), and (b) implies (c) by (iii).
Suppose that (a) holds. Then all the maps in the exact sequence of (i) are strict, so its completion is still exact by chapitre III §2 No12 lemme 2. This shows that (a) implies (c).
It remains to show that (c) implies (d). Fix rings of definition B0 ⊂B and C0 ⊂C such that A0 ⊂ϕ−1 B (B0) ∩ϕ−1 C (C0) := A′ 0, and let A′ be the ring A with the topology for which (ϖnA′ 0)n≥0 is a fundamental system of neighborhoods of 0. (This does define a structure of topological ring on A′ by lemma II.3.3.8.) The identity A →A′ is continuous (because A′ 0 is an open subring of A), and (d) is equivalent to the fact that it is an open map. Note that A′ is isomorphic as a topological ring to ε(A) with the subspace topology, so the obvious sequence 0 →A′ →B ⊕C →D →0 is exact and all the maps in it are strict. Using chapitre III §2 No12 lemme 2 again, we see that the sequence 0 →b A′ →b B ⊕b C →b D →0 is exact. As we are assuming that 0 →b A′ →b B ⊕b C →b D →0 is exact, this implies that the canonical map b A →b A′ is bijective. By the open mapping theorem (theorem II.4.1.1), the map b A →b A′ is open. Using lemma II.3.1.11, we see that this implies the openness of A →A′, hence condition (d).
IV.2.3 Refining coverings In this section, we fix a Huber pair (A, A+), with A a f-adic ring. We want show the existence of enough manageable covers of X = Spa(A, A+).
Definition IV.2.3.1.
(i) Let t1, . . . , tn ∈ A generating the ideal (1) of A.
The stan-dard rational covering of X generated by t1, . . . , tn is the covering (Ui)1≤i≤n, where Ui = R t1,...,tn ti . We say that this standard rational covering is generated by units if t1, . . . , tn ∈A×.
(ii) Let t1, . . . , tn ∈A. For every I ⊂{1, . . . , n}, let VI = {x ∈X | |ti|x ≤1 ∀i ∈I and |ti|x ≥1 ∀i ̸∈I}; note that VI is the rational domain R TI sI , where sI = Q j̸∈I ti and TI = {1} ∪{tisI, i ∈I} ∪{ Y j̸∈I∪{i} ti, i ̸∈I}.
The family (VI)I⊂{1,...,n} is an open covering of X, which we call the standard Laurent covering generated by t1, . . . , tn. Again, we say that the covering is generated by units if t1, . . . , tn ∈A×.
128 IV.2 Some preliminary results (iii) A simple Laurent covering is a standard Laurent covering generated one element.
(iv) A rational covering of X is a covering of X by rational domains.
Note that the covering of proposition IV.2.2.1 is a simple Laurent. Note also that we diverge from the vocabulary of when defining rational coverings and follow instead.
Remark IV.2.3.2. Note that, if t1, . . . , tn ∈A are such that (t1, . . . , tn) = A, the standard rational covering generated by t1, . . . , tn is a covering of X. Indeed, we have n [ i=1 R t1,...,tn ti = n [ i=1 {x ∈X| max 1≤j≤n |tj|x = |ti|x ̸= 0} = {x ∈X | max 1≤j≤n |tj|x ̸= 0}, which is also the set of x ∈X such that at least one of the |ti|x is nonzero; but this is all of X, because 1 ∈Pn i=1 Ati (and |1|x = 1 ̸= 0 for every x ∈X). The same observation also shows that, for every i ∈{1, . . . , n}, R t1, . . . , tn ti = {x ∈X | ∀j ∈{1, . . . , n}, |tj|x ≤|ti|x}.
(If we have |tj|x ≤|ti|x for every j and |ti|x = 0, then |t1|x = . . . = |tn|x = 0, and we have just seen that this implies |1|x = 0, which is impossible.) Proposition IV.2.3.3. (Lemma 8 of and lemma 2.4.19 of .) (i) For every open covering U of X, there exists a rational covering refining U . If moreover A is complete, then, for every open covering U of X, there exists a standard rational covering V of X refining U .
(ii) Suppose that A is a Tate ring. Then, for every standard rational covering U of X, there exists a standard Laurent covering V of X such that for every V ∈V , the covering (V ∩U)U∈U of V is a standard rational covering generated by units.
(iii) For every standard rational covering generated by units U of X, there exists a standard Laurent covering V of X generated by units refining U .
Proof.
(i) This is lemma 2.6 of . First we construct a rational covering refining U . Let x ∈X, and let y = x|cΓx (so y is the minimal horizontal specialization of x). Note that y is still in Spa(A, A+) : y is continuous by proposition II.2.3.1(i), and, for every a ∈A+, we have |a|y ≤|a|x ≤1. Let U ∈U such that y ∈U. As y has no proper horizontal specialization, we have y ∈Spv(A, A) (see remark I.4.2.2(2)). So, by point (3) in the proof of theorem I.4.2.4, there exists a finite subset Tx of A such that Tx · A = A and sx ∈A such that y ∈R Tx sx ⊂U; after replacing Tx by Tx ∪{sx}, we may assume that sx ∈Tx. As x is a generization of y, we also have x ∈R Tx sx . So X = S x∈X R Tx sx .
As X is quasi-compact, we can find x1, . . . , xn ∈X such that X = S x∈X R Txi sxi . This is a finite rational covering of X refining U .
129 IV When is the structure presheaf a sheaf ?
We now assume that A is complete and refine this rational covering forther to a standard rational covering. We write Txi = Ti and sxi = si. Let T = {t1 . . . tn, ti ∈Ti ∀i ∈{1, . . . , n}} and S = {t1 . . . tn, ti ∈Ti ∀i ∈{1, . . . , n} and ∃i ∈{1, . . . , n} such that ti = si}.
As each Ti · A = A for every i, we have S · A = s1A + . . . snA. As X is the union of the R Ti si , for every x ∈X, there exists i ∈{1, . . . , n} such that |si|x ̸= 0. By corollary III.4.4.3, this implies that S · A = A.
We want to show that the standard rational covering generated by S refines the cover-ing (R Ti si )1≤i≤n, hence U . Let s ∈S, and write s = t1 . . . tn, with ti ∈Ti. Pick j ∈{1, . . . , n} such that tj = sj. We claim that R S s ⊂R Tj sj . Indeed, let x ∈R S s , and let t ∈Tj. Then |t1 . . . ti−1tti+1 . . . tn|x ≤|s|x ̸= 0, and this implies |t|x ≤|sj|x ̸= 0.
(ii) Let t1, . . . , tn ∈A such that (t1, . . . , tn), and let (U1, . . . , Un) be the standard rational covering generated by t1, . . . , tn. Let ϖ ∈A be a topologically nilpotent unit. We choose a1, . . . , an ∈A such that a1t1 + . . . + antn = 1. As A+ is open, there exists N ∈N such that ϖNai ∈A+ for every i ∈{1, . . . , n}. Then, if x ∈X = Spa(A, A+) and we have |ϖNai|x ≤1 for every i, so |ϖN|x = |ϖNa1t1 + . . . + ϖNantn|x ≤max1≤i≤n |ti|x, and finally |ϖN+1|x < max1≤i≤n |ti|x (because |ϖ|x < 1). Let (VI)I⊂{1,...,n} be the standard Laurent covering generated by ϖ−(N+1)t1, . . . , ϖ−(N+1)tn. We will show that this covering works.
Let I ⊂{1, . . . , n}. We have VI = {x ∈X | |ti|x ≤|ϖN+1|x ∀i ∈I and |ti|x ≥|ϖN+1|x ∀i ̸∈I}.
In particular, by the choice of N, V{1,...,n} = ∅. Suppose that I ⊊{1, . . . , n}. By the description of VI as a rational subset in definition IV.2.3.1, we have ti ∈OX(VI)× for i ̸∈I. If x ∈VI, max i∈I |ti|x ≤|ϖN+1|x < max 1≤i≤n |ti|x, so max 1≤i≤n |ti|x = max i̸∈I |ti|x.
In particular, VI ∩Ui = ∅if i ∈I, and VI ∩Ui = VI ∩R tj, j̸∈I ti if i ̸∈I. This shows the statement.
(iii) Let t1, . . . , tn ∈A×, and let (U1, . . . , Un) be the standard rational covering generated by t1, . . . , tn. Let I = {(i, j) ∈{1, . . . , n} | i < j}, and let t(i,j) = tit−1 j for (i, j) ∈I.
130 IV.2 Some preliminary results We claim that the standard Laurent covering generated by the family (t(i,j))(i,j)∈I works.
Denote this covering by (VJ)J⊂I. For J ⊂I, we have VJ = {x ∈X | |ti|x ≤|tj|x if (i, j) ∈J and |ti|x ≥|tj|x if (i, j) ̸∈J}.
Choose a finite sequence (i1, . . . , ir) of elements of {1, . . . , n} such that (is, is+1) ∈J for 1 ≤s ≤r −1 and of maximal length for that property. (Such a chain exists, be-cause the condition implies that i1 < . . . < ir, so we must have r ≤n.) Then, if i ∈{1, . . . , n} −{ir}, we cannot have (ir, i) ∈J because this would contradict the maximality of (i1, . . . , ir), so |ti|x ≤|tir|x for every x ∈VJ. This shows that VJ ⊂Uir.
Corollary IV.2.3.4. (Proposition 2.4.20 of .) Suppose that A is a complete Tate ring. Let P be a property of rational coverings of rational domains of X. Suppose that P satisfies the following conditions : (a) P is local, i.e. if it holds for a refinement of a covering then it also holds for the original covering.
(b) P is transitive : let U be a rational domain of X, (Ui)i∈I be a rational covering of U and (Uij)j∈Ji be a rational covering of Ui for every i ∈I; if P holds for the covering (Ui)i∈I of U and for each covering (Uij)j∈Ji of Ui, i ∈I, then it holds for the covering (Uij)i∈I,j∈Ji of U.
(c) P holds for every simple Laurent covering of a rational domain of X.
Then P holds for any rational covering of every rational domain of X.
We will see examples of properties P satisfies (a), (b) and (c) in corollary IV.3.2.1.
Proof.
(1) If U is a rational domain of X and U is a standard Laurent covering of U of OX(U), then P holds for U : We prove this by induction on the number n of elements generating the Laurent covering. If n = 1, this is condition (c). Suppose that n ≥2 and that we know the result for n −1 (and for every rational domain of X). Let t1, . . . , tn ∈OX(U), and consider the standard Laurent covering U = (VI)I⊂{1,...,n} of U that they generate. We also set W = {x ∈U | ||t1|x ≤1} and W ′ = {x ∈U | ||t1|x ≥1}; this is a simple Laurent covering of U, so P holds for this covering by (c). Then (W ∩VI)1∈I (resp.
(W ′ ∩VI)i̸∈I) is the standard Laurent covering of W (resp. W ′) generated by t2, . . . , tn.
By the assumption hypothesis, P holds for these two coverings, so it holds for U by condition (b).
(2) By (1), property (a) and proposition IV.2.3.3(iii), P holds for any standard rational cover-ing generated by units of a rational domain of X.
(3) Let U be a standard rational covering of a rational domain U of X. By proposition IV.2.3.3(ii), there exists a standard Laurent covering V of U such that, for every V ∈V , 131 IV When is the structure presheaf a sheaf ?
(V ∩U)U∈U is a standard rational covering generated by units of V . By (2), property P holds for each covering (V ∩U)U∈U , and by (1), it holds for V . So, by property (b), P holds for U .
(4) Let U be a rational covering of a rational domain U of X. By proposition IV.2.3.3(i), there exists a standard rational covering V of U refining U . Property P holds for V by (3), so it holds for U by (a).
IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) In this section, we fix a Huber pair (A, A+) such that A is a Tate ring, and we write X = Spa(A, A+). We also fix a ring of definition A0 of A and a topologically nilpotent unit ϖ of A such that ϖ ∈A0.
IV.3.1 A local criterion for power-boundedness Proposition IV.3.1.1. (Lemma 3 of .) Let t1, . . . , tn ∈A, and suppose that the ideal (t1, . . . , tn) is A itself. For every i ∈{1, . . . , n}, let ϕi : A →Ai := A t1,...,tn ti be the canonical map. Let Ai,0 = A0[ t1 ti , . . . , tn t0 ] ⊂Ai; this is a ring of definition of Ai.
Then A0 ⊃ \ 1≤i≤n ϕ−1 i (Ai,0).
In other words, an element a ∈A such that ϕi(a) ∈Ai,0 for every i is power-bounded.
Proof. Let a ∈T 1≤i≤n ϕ−1 i (Ai,0). For each i ∈{1, . . . , n}, the image of a in A[t−1 i ] is in the subring A0[t1t−1 i , . . . , tnt−1 i ].
Choose a homogeneous polynomial fi ∈A0[X1, . . . , Xn] such that ai = t−deg(fi) i fi(t1, . . . , tn) in A[t−1 i ].
Then we can find ci ∈ N such that tci i (tdeg(fi) i a −fi(t1, . . . , tn)) = 0 in A. If we set gi = Xci i fi ∈A0[X1, . . . , Xn], then gi is homo-geneous of degree di := ci + deg(fi) and tdi i a −gi(t1, . . . , tn) = 0 in A for every i ∈{1, . . . , n}.
Let N = d1 + . . . + dn, and choose A ∈N such that ϖAti ∈A0 for every i ∈{1, . . . , n}.
We show by induction on m that ϖNAh(t1, . . . , tn)am ∈A0 for every homogeneous polynomial h ∈A0[X1, . . . , Xn] of degree N and every m ∈N. The statement is clear if m = 0, because then ϖNAh(t1, . . . , tn)rm = ϖNAh(t1, . . . , tn) is a polynomial in ϖAt1,..., ϖAtn with coefficients in A0. Suppose that m ≥1 and that we know the result for m −1. It suffices to prove the statement for h a monomial of degree N, i.e. h = Xe1 1 . . . Xen n with e1 + . . . + en = N. Since N = d1 + . . . + dn, there is at least one i such that ei ≥di, and we may assume that i = 1. Then ϖNAte1 1 . . . ten n am = ϖNAte1−d1 1 te2 2 . . . ten n g1(t1, . . . , tn)am−1, 132 IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) and the right hand side is in A0 by the induction hypothesis.
Now we show that a ∈A0. Choose a1, . . . , an ∈A such that a1t1 + . . . + antn = 1, and choose B ∈ N such that ϖBai ∈ A0 for every i ∈ {1, . . . , n}.
Then h = (ϖBa1X1 + . . . + ϖBanXn)N ∈A0[X1, . . . , Xn] is homogeneous of degree N, so, by the previous paragraph, ϖNAh(t1, . . . , tn)am = ϖN(A+B)am is in A0 for every m ∈N. This shows that {am, m ∈N} is bounded.
Corollary IV.3.1.2. We keep the notation of proposition IV.3.1.1, and we suppose that A is uni-form. Then the morphism ϕ : A →Qn i=1 Ai, a 7− →(ϕi(a))1≤i≤n is strict.
Proof. As A is uniform, we may assume that A0 = A0. The subspace topology on ϕ(A) has the sets ϕ(A) ∩(Qn i=1 ϖNAi,0), N ∈N, as a fundamental system of neighborhoods of 0, and the quotient topology has the sets ϕ(ϖNA0) as a fundamental system of neighborhoods of 0. We already know that the quotient topology is finer than the subspace topology because ϕ is contin-uous. On the other hand, proposition IV.3.1.1 says that ϕ(ϖNA0) ⊃ϕ(A) ∩(Qn i=1 ϖNAi,0) for every N ∈N, so the subspace topology is finer than the quotient topology, and we are done.
Corollary IV.3.1.3. (Corollary 4 of .) Suppose that A is uniform. Let t ∈A, and consider the open cover (U = R 1,t 1 , V = R 1 t of X. Then the ˇ Cech complex 0 →OX(X) →OX(U) ⊕OX(V ) →OX(U ∩V ) →0 is exact.
Proof. We are in the situation of proposition IV.2.2.1, so it suffices to check that condition (iv)(d) of this proposition holds. Applying proposition IV.3.1.1 with t1 = 1 and t2 = t shows that, with the notation of proposition IV.2.2.1, ϕ−1 B (B0) ∩ϕ−1 C (C0) ⊃A0, where B0 = A0[t] and C0 = A0[t−1]. As A is uniform, A0 is bounded, so there exists n ∈N such that ϖnA0 ⊂A0.
This shows condition (iv)(d) of proposition IV.2.2.1.
IV.3.2 Calculation of the ˇ Cech cohomology If F is a presheaf of abelian groups on X, U is an open subset of X and U is an open covering of U, we denote by ˇ C •(U , F) the associated ˇ Cech complex and by ˇ H i(U , F) its cohomology, i.e., the ˇ Cech cohomology groups of F on U for the covering U . (See [25, Definition 01EF].) Corollary IV.3.2.1. (Proposition 2.4.21 of .) Let F be a presheaf of abelian groups on X. Consider the following property P of rational coverings U of a rational domain U of 133 IV When is the structure presheaf a sheaf ?
X : for every rational domain V of U, if V ∩U = (V ∩W)W∈U , then the canonical map F(V ) →ˇ H 0(V ∩U , F is an isomorphism and ˇ H i(V , F) = 0 for i ≥1.
Then property P satisfies conditions (a) and (b) of corollary IV.2.3.4, so in particular, it holds for every rational covering of every rational domain of X if and only if it satisfies property (c) of that corollary.
Proof. We check (a). Suppose that U = (Ui)i∈I and V = (Vj)j∈J are rational coverings of a ratinal domain U of X, that V refines U , and that P holds for V . We want to apply corollary IV.3.2.3 to show that ˇ H •(U , OX) ≃ˇ H •(V , OX) (which will imply that P holds for U ), so we need to check that the hypotheses of this corollary hold. For all i0, . . . , ip, the fact that the map OX(Ui0 ∩. . . ∩Uip) →ˇ C •(Ui0 ∩. . . ∩Uip ∩V , OX) is a quasi-isomorphism follows immediately from the fact that property P holds for V . Let j0, . . . , jq, and let V ′ = Vj0∩. . .∩Vjq. We want to show that the map OX(V ′) →ˇ C •(V ′∩U , OX) is a quasi-isomorphism. But this follows from proposition IV.3.2.4 and from the fact that the coverings V ′ ∩U and {V ′} of V ′ refine each other.
We now check (b). Let U be a rational domain of X, consider a rational covering U = (Ui)i∈I of U and a rational covering Ui = (Uij)j∈Ji of Ui for every i ∈I. Suppose that P holds for U and for every covering Ui, i ∈I.
We want to show that it holds for the covering V = (Uij)i∈I,j∈Ji of U. First note that, for every i ∈I, the covering Ui of Ui refines Ui ∩V .
As P satisfies condition (a), this implies that P holds for Ui ∩V . As before, we want to ap-ply corollary IV.3.2.3 to show that ˇ H •(U , OX) ≃ˇ H •(V , OX) (which will imply that P holds for V ). By the beginning of the paragraph, we already know that, for all i0, . . . , ip ∈I, the map F(Ui0 ∩. . . ∩Uip) →ˇ C •(Ui0 ∩. . . ∩Uip ∩V , F) is a quasi-isomorphism. Moreover, if V ∈V , then the coverings V ∩U and {V } of V refine each other, so they have isomorphic ˇ Cech complexes by proposition IV.3.2.4. So the hypotheses of corollary IV.3.2.3 are satisfied.
In the end of this section, we give some auxiliary results that are used in the proof of corollary IV.3.2.1. We need a way to compare ˇ Cech cohomology for two different covers. This is done in section 8.1.4 of . We review their construction.
Let X be a topological space, F be a presheaf of abelian groups on X and U = (Ui)i∈I, V = (Vj)j∈J be two open coverings of X. 3 For all i0, . . . , ip ∈I (resp. j0, . . . , jq ∈J), we write Ui0,...,ip = Ui0 ∩. . . ∩Uip (resp. Vj0,...,jq = Vj0 ∩. . . ∩Vjq). We define a double complex ˇ C •,•(U , V ; F) by ˇ C p,q(U , V ; F) = Y i0,...,ip∈I j0,...,jq∈J F(Ui0,...,ip ∩Vj0,...,jq), 3This would work just as well for X a site and F a presheaf with values in an abelian category.
134 IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) with differentials ′dp,q : ˇ C p,q(U , V ; F) →ˇ C p+1,q(U , V ; F) ′′dp,q : ˇ C p,q(U , V ; F) →ˇ C p,q+1(U , V ; F) such that, if f ∈ˇ C p,q(U , V ; F), then the (i0, . . . , ip+1, j0, . . . , jq)-component of ′dp,q(f) is given by p+1 X r=0 (−1)r+qfi0,...,ir−1,ir+1,...,ip,j0,...,jq|Ui0,...,ip+1∩Vj0,...,jq and the (i0, . . . , ip, j0, . . . , jq+1)-component of ′′dp,q(f) is given by q+1 X s=0 (−1)r+qfi0,...,ip,j0,...,js−1,js+1,...,jq|Ui0,...,ip∩Vj0,...,jq+1.
We also denote by ˇ C •(U , V ; F) the associated simple complex.
The obvious maps F(Vj0,...,jq) →ˇ C 0(Vj0,...,jq ∩U , F) induce a morphism of complexes ˇ C •(V , F) →ˇ C •(U , V ; F).
The following result is Lemma 1 of 8.1.4.
Proposition IV.3.2.2. Suppose that, for all j0, . . . , jq ∈J, the obvious morphism F(Vj0,...,jq) →ˇ C •(Vj0,...,jq ∩U , F) is a quasi-isomorphism. Then the morphism ˇ C •(V , F) →ˇ C •(U , V ; F) defined above is a quasi-isomorphism.
Proof. This is a general result for double complexes. See for example corollary 12.5.5 of .
We obviously have a similar result if we switch the roles of U and V , so we get the following corollary.
Corollary IV.3.2.3. ( 8.2 Theorem 2.) Assume, for i0, . . . , ip ∈I (resp. j0, . . . , jq ∈J), the obvious morphism F(Ui0,...,ip) →ˇ C •(Ui0,...,ip ∩V , F) (resp.
F(Vj0,...,jq) →ˇ C •(Vj0,...,jq ∩U , F)) 135 IV When is the structure presheaf a sheaf ?
is a quasi-isomorphism. Then the two morphisms ˇ C •(V , F) →ˇ C •(U , V ; F) ←ˇ C •(U , F) defined above are quasi-isomorphisms.
In particular, we get canonical isomorphisms ˇ H r(U , F) ≃ˇ H r(V , F), for all r ∈N.
We note another and simpler way to compare the ˇ Cech complexes under extra hypotheses.
Suppose that V refines U . For every j ∈J, we choose c(j) ∈I such that Vj ⊂Uc(j). Then we get restriction morphisms F(Uc(j0),...,c(jq)) →F(Vj0,...,jq), for all j0, . . . , jq ∈J, and these induce a morphism of complexes ˇ C •(U , F) →ˇ C •(V , F).
This morphism of complexes depends on the choice of c : J →I, but the maps it induces on cohomology do not (see for example [25, Section 09UY] for details). This immediately implies the following result.
Proposition IV.3.2.4. If U refines V and V refines U , then the maps ˇ C •(V , F) →ˇ C •(U , F) and ˇ C •(U , F) →ˇ C •(V , F) are quasi-isomorphisms quasi-inverse of each other.
IV.3.3 The strongly Noetherian case In this section, we explain what happens in the strongly Noetherian case. For now, we assume that A is a Tate ring.
Definition IV.3.3.1. Let M be a finitely generated A-module, endowed with its canonical topol-ogy (see proposition II.4.2.2). We denote by M⟨X⟩the A⟨X⟩-submodule of M of elements f = P ν≥0 mνXν such that, for every neighborhood U of 0 in M, we have mν ∈U for all but finitely many ν.
Proposition IV.3.3.2. ( remark 8.28 and lemma 8.30) Suppose that A is complete and Noetherian.
(i) For every finitely generated A-module M, if we put the canonical topology on M, then the morphism M ⊗A A⟨X⟩→M⟨X⟩, m ⊗a 7− →ma is an isomorphism of A⟨X⟩-modules.
(ii) The ring A⟨X⟩is faithfully flat over A.
(iii) For every f ∈A, the ring A⟨X⟩/(f −X) and A⟨X⟩/(1 −fX) are flat over A.
136 IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) Corollary IV.3.3.3. (Proposition 8.29 of .) Suppose that A is strongly Noetherian, and let U ⊂V be two rational domains of X = Spa(A, A+).
Then the restriction map OX(V ) →OX(U) is flat.
Corollary IV.3.3.4. (Corollary 8.31 of .) Suppose that A is strongly Noetherian, and let (Ui)1≤i≤n be a finite rational covering of X. Then the morphism OX(X) → n Y i=1 OX(Ui) is faithfully flat.
Corollary IV.3.3.5. (Lemma 8.32 of .) Suppose that A is strongly Noetherian. Then, for every simple Laurent covering (U, V ) of X, the sequence 0 →OX(X) →OX(U) ⊕OX(V ) →OX(U ∩V ) →0 of proposition IV.2.2.1 is exact.
IV.3.4 Cases (c) and (d) We now finish the proof of theorem IV.1.1.5 in cases (c) and (d). We assume that A is a Tate ring and fix a topologically nilpotent unit ϖ of A. By corollary III.6.4.3, we can (and will) assume that A is complete.
Consider property P of corollary IV.3.2.1 for the presheaf OX. If A is stably uniform (resp.
strongly Noetherian), then, by corollary IV.3.1.3 (resp. IV.3.3.5), P holds for every simple Laurent covering of every rational domain of X. By corollary IV.3.2.1, this implies that P holds for every rational covering of every rational domain of X. Let B be the set of rational domains of X; this is a base of the topology of X, and we have just shown that OX is a sheaf of abelian groups, hence of abstract rings, on B (i.e. it satisfies the sheaf condition for every covering of a rational domain by rational domains).
We want to prove that OX is a sheaf of topological rings. We use the criterion of EGA I chapitre 0 (3.2.2); if we combine it with the observations of EGA I chapitre 0 (3.1.4), it says that it suffices to check that, for every U ∈B and every rational covering (Ui)i∈I of U, the sequence (∗) 0 →OX(U) → Y i∈I OX(Ui) → Y i,j∈I OX(Ui ∩Uj) is exact (as a sequence of abelian groups) and the morphism ϕ : OX(U) → Y i∈I OX(Ui) is strict. We already know that () is exact.
137 IV When is the structure presheaf a sheaf ?
We show the strictness of ϕ. For A stably uniform, this is corollary IV.3.1.2, but this does not work for A stronly Noetherian. Here is an argument that works in both cases : First, as X is quasi-compact, we may assume that I is finite. By the exactness of (), the image of ϕ is the kernel of the continuous map Q i∈I OX(Ui) →Q i,j∈I OX(Ui ∩Uj), so it is closed in Q i∈I OX(Ui), and in particular it is a complete OX(U)-module. So the fact that ϕ : OX(U) →Im ϕ is open just follows from the open mapping theorem (theorem II.4.1.1).
Finally, we need to prove that Hi(U, OX) = 0 for every rational domain U of X and every i ≥1. But this follows immediately from the similar property for ˇ Cech cohomology (with respect to rational coverings) and from [25, Lemma 01EW].
Let us outline the proof given in that referece. We see OX as a sheaf of abstract rings in this paragraph. Consider the category A of sheaves of (abstract) abelian groups F such that, for every rational domain U of X and every rational covering U of U, the canonical map F(U) →ˇ C •(U , F) is a quasi-isomorphism (i.e. the augmented ˇ Cech complex associated to U is acyclic). By the beginning of the proof, OX is an object of A . We claim that, for every object F of A , every rational domain U of X and every i ≥1, we have Hi(U, F) = 0. We prove this claim by induction on i. So suppose that i ≥1 and that we know the claim for every j < i. Let F be an object of A . Choose an injective morphism of F into an injective abelian sheaf I . Let U be a rational covering of a rational domain U of X. Note that I|U is still injective as an abelian sheaf on U. As the forgetful functor from sheaves to presheaves admits a lft adjoint (the sheafification functor), it preserves injective objects, so I|U is still injective when seen as a presheaf of abelian groups on U. As the ˇ H i(U , .) are the right derived functors of ˇ H 0(U , .) on the category of presheaves of abelian groups on U (see [25, Lemma 01EN]), we have ˇ H i(U , I ) = 0 for every i ≥1. In particular, I is an object of A .
Let G be the cokernel of the map F →I . It is easy to see (cf. [25, Lemma 01EU]) that the vanishing of ˇ H 1(U , I ) for every rational covering of a rational domain of X im-plies that, for every rational domain U of X, the I (U) →G (U) is surjective. Also, using the long exact sequence of ˇ Cech cohomology coming from the exact sequence of presheaves 0 →F →I →G →0, and using the fact that F and I are in A , we see that G is also an object of A .
Fix a rational domain U of X and consider the long exact sequence of cohomology groups : 0 →F(U) →I (U) →G (U) →H1(U, F) →H1(U, I ) →H1(U, G ) →. . .
. . . →Hi−1(U, G ) →Hi(U, F) →Hi(U, I ) →Hi(U, G ) →. . .
As I is injective, we have Hp(U, I ) = 0 for every p ≥1. So, if i ≥2, then the exact sequence above and the induction hypothesis applied to G imply that Hi(U, F) = 0. Suppose that i = 1. As I (U) →G (U), the map H1(U, F) →H1(U, I ) = 0 is injective, so we also get H1(U, F) = 0.
138 IV.3 Proof of theorem IV.1.1.5 in cases (a), (c) and (d) IV.3.5 Case (a) We explain how to prove theorem IV.1.1.5 in case (a), i.e., when the topology on b A is discrete.
Again, by corollary III.6.4.3, we can (and will) assume that A is complete, so that A is a discrete ring. Then A+ can be any integrally closed subring of A, and Spa(A, A+) = {x ∈Spv(A) | ∀a ∈A+, |a|x ≤1}.
Remember that we have a continuous and spectral map supp : Spv(A) → Spec(A), x 7− →Ker(|.|x). We also denote by supp its restriction to the subset Spa(A, A+). Note that the map supp : Spa(A, A+) →Spec(A) is surjective because, for every ℘∈Spec(A), the trivial valuation with support ℘is in Spa(A, A+). For the same reason, for every finite subset T of A and every s ∈A, the image of R T s ⊂Spa(A, A+) by supp is the principal open subset D(s) of Spec(A).
Let T ⊂A be a finite subset and s ∈A. Then A T s is the ring A[s−1] with the discrete topology, so OX(R T s ) = A[s−1] with the discrete topology. Also, as supp(R T s ) = D(s) is open, we have (supp∗OSpec(A))(R T s ) = OSpec(A)(D(s)) = OX(R T s ).
This shows that the presheaves OX and supp∗OSpec(A) coincide on the family of rational domain of X, and in particular that OX is a sheaf on this family and that its augmented ˇ Cech complex for any rational covering of a rational domain is exact.
We now get the result by applying the criterion of EGA I chapitre 0 (3.2.2) (to show that OX is a sheaf) and [25, Lemma 01EU] (to show that its higher cohomology vanishes on rational domains) as in section IV.3.4.
139 V Perfectoid algebras In this chapter, we will study the main example of stably uniform Tate rings, i.e. perfectoid Tate rings. The main references are Scholze’s papers and , as well as Fontaine’s Bourbaki seminar and Bhatt’s notes .
In all this chapter, we fix a prime number ℓ.
V.1 Perfectoid Tate rings V.1.1 Definition and basic properties Definition V.1.1.1. (Section 1.1 of , definition 3.1 of ) Let A be a Tate ring. We say that A is perfectoid if A is complete and uniform, and if there exists a pseudo-uniformizer ϖ of A such that (a) ϖℓdivides ℓin A0; (b) the Frobenius map Frob : A0/ϖ →A0/ϖℓ, a 7− →aℓis bijective.
If A is a perfectoid Tate ring and a field, we say that A is a perfectoid field.
Note that condition (a) implies that A0/ϖ is a ring of characteristic ℓ, so the Frobenius map in (b) is a morphism of rings.
Remark V.1.1.2. It follows from proposition V.1.1.3 that, if A is a perfectoid Tate ring, then every pseudo-uniformizer ϖ of A such that ϖℓdivides ℓin A0 satisfies condition (b) of definition V.1.1.1.
Note also that, if A is perfectoid of characteristic 0, then ℓis topologically nilpotent (because it is a multiple in A0 of some uniformizer), so, if ℓis invertible in A, then the topology on A0 is the ℓ-adic topology and A = A0[ 1 ℓ] (by proposition II.2.5.2).
Finally, remember that perfectoid Tate ring, like all complete uniform Tate rings, are reduced.
(See remark IV.1.1.3.) Proposition V.1.1.3. (Lemma 3.9 of .) Let A be a Tate ring, and let ϖ be a pseudo-uniformizer of A such that ϖℓdivides ℓin A0.
141 V Perfectoid algebras (i) The map Frob : A0/ϖ →A0/ϖℓ, a 7− →aℓis injective.
(ii) If A is complete and uniform, then the following conditions are equivalent : (a) every element of A0/ℓϖA0 is a ℓth power; (b) every element of A0/ℓA0 is a ℓth power; (c) every element of A0/ϖℓA0 is a ℓth power.
Moreover, if these conditions holds, then there exist units u and v in A0 such that uϖ and vℓadmit compatible systems of ℓ-power roots in A0.
Proof.
(i) Let a ∈A0 such that aℓ∈ϖℓA0. Then (aϖ−1)ℓ∈A0, so aϖ−1 ∈A0, and a ∈ϖA0.
(ii) As ϖℓdivides ℓand ℓdivides ℓϖ in A0, it is clear that (a) implies (b) and (b) implies (c).
We want to show that (c) implies (a).
Suppose that (c) holds, and let a ∈A0.
By lemma V.1.1.5, there exists a sequence (an)n≥0 of elements of A0 such that a = P n≥0 aℓ nϖℓn. By lemma V.1.1.6, this implies that a −(P n≥0 anϖn)ℓ∈ϖℓA0, which gives (a).
We prove the last statement. By lemma V.1.1.7 (applied to A0 and to A0/πℓA0), the canon-ical map lim ← −a7− →aℓA0 →lim ← −a7− →aℓA0/ϖℓA0 is an isomorphism. In particular, we can find ω = (ω(n)) ∈lim ← −a7− →aℓsuch that ω(0) = ϖ mod. ϖℓA0 (resp. ω(0) = ℓmod. ϖℓA0). In other words, there exists a ∈A0 such that ω(0) = ϖ(1 + ℓa) (resp. ω(0) = ℓ(1 + ϖa)).
The claim now follows from the fact that, for every a ∈A0, 1 + ϖa and 1 + ℓa are units in A0 (because ϖa and ℓa are topologically nilpotent).
The following corollary is an immediate consequence of the proposition, but it is convenient when we want to prove that a Tate ring is perfectoid.
Corollary V.1.1.4. Let A be a complete uniform Tate ring. Then the following conditions are equivalent : (i) A is a perfectoid; (ii) every element of A0/ℓA0 is a ℓth power, and A has a pseudo-uniformizer ϖ such that ϖℓ divides ℓin A0.
Lemma V.1.1.5. We use the notation of proposition V.1.1.3. Suppose that every element of A0/ϖℓA0 is a ℓth power, and let a ∈A0.
Then there exists a sequence (an)n≥0 of elements of A0 such that, for every n ∈N, a −Pn i=0 aℓ iϖℓi ∈ϖℓ(n+1)A0.
142 V.1 Perfectoid Tate rings Proof. We construct the elements an by induction on n. The assumption immediately implies that there exists a0 ∈A0 such that a −aℓ 0 ∈ϖℓA0. Suppose that n ≥1 and that we have found a0, . . . , an−1 such that a −Pn−1 i=0 aℓ iϖℓi ∈ϖℓnA0. Let b ∈A0 such that a −Pn−1 i=0 aℓ iϖℓi = ϖℓnb, and choose an ∈A0 such that b −aℓ n ∈ϖℓA0. Then a −Pn i=0 aℓ iϖℓi ∈ϖℓ(n+1)A0.
The following lemma is an easy consequence of the binomial formula.
Lemma V.1.1.6. We use the notation of proposition V.1.1.3.
For all a, b ∈A0, we have (a + ϖb)ℓ−aℓ−(ϖb)ℓ∈ϖℓA0.
Lemma V.1.1.7. (Lemma 3.4(i) of .) Let S be a ring and ϖ ∈S. Suppose that S is ϖ-adically complete (and Hausdorff) and that ϖ divides ℓin S. Then the canonical map lim ← − a7− →aℓ S →lim ← − a7− →aℓ S/ϖS is an isomorphism of topological monoids (where the monoid operations are given by the multi-plications of the rings).
Proof. Let S1 = lim ← −a7− →aℓS and S2 = lim ← −a7− →aℓS/ϖS. We prove the statement by constructing a continuous multiplicative inverse of the canonical map S1 →S2.
First, we construct a continuous multiplicative map α : S2 →S such that α((sn)n≥0) = s0 mod. πS. Let (sn)n≥0 ∈S2. Choose representatives sn ∈S of the sn. We claim that : (i) limn→+∞sℓn n exists; (ii) the limit in (i) is independent of the choice of the representatives sn.
To prove (i), note that, for every n ∈N, we have sℓ n+1 −sn ∈ϖS. Applying lemma V.1.1.6 repeatedly and using the fact that ϖ divides ℓ, we deduce that sℓn+1 n+1 −sℓn n ∈ϖn+1S. This implies that (sℓn n )n≥0 is a Cauchy sequence in S, so it has a limit. To prove (ii), choose some other lifts s′ n of the sn. Then, for every n ∈N, we have s′ n −sn ∈ϖS, so as before we get (s′ n)ℓn −sℓn n ∈ϖn+1S. This implies that limn→+∞sℓn n = limn→+∞(s′ n)ℓn.
By claims (i) and (ii), we can define a map α : S2 →S by sending (sn)n≥0 to limn→+∞sℓn n , where the sn ∈S are any lifts of the sn. It is clear that α is multiplicative, and it is also easy to see that it is continuous.
Note that, if s = (sn)n≥0 ∈S2, then it has a canonical ℓth root, which is its shift s′ = (sn+1)n≥0, and we have α(s′)ℓ= α(s) by definition of α.
We now get the desired map S2 →S1 by sending (sn)n≥0 ∈S2 to the sequence (α((sr+n)n≥0))r≥0, which is clearly an element of S1.
143 V Perfectoid algebras We now look at two particular cases : perfectoid fields and perfectoid rings of characteristic ℓ.
Proposition V.1.1.8. (Proposition 3.5 of .) Let A be a Tate ring of characteristic ℓ. Then the following are equivalent : (i) A is perfectoid.
(ii) A is complete and perfect.
Proof. Suppose that A is complete and perfect. Then it is also uniform by theorem IV.1.3.5. So, by corollary V.1.1.4, it suffices to check that every element of A0 is a ℓth power, which follows immediately from the fact that A is perfect.
Conversely, suppose that A is perfectoid. Then it is complete by assumption, and we want to show that it is perfect. As A is a localization of A0, it suffices to show that A0 is perfect. As ℓ= 0 in A, we have A0/ℓA0 = A0, so the conclusion follows from proposition V.1.1.3(ii).
For fields, we have the following result of Kedlaya.
Theorem V.1.1.9. (Theorem 4.2 of .) Let A be a perfectoid Tate ring that is also a field.
Then the topology of A is given by a rank 1 valuation; in other words, A is a complete non-Archimedean field.
This implies that the definition of perfectoid field given here is equivalent to their original definition (definition 3.1 of ).
Proposition V.1.1.10. Let K be a complete topological field. Then the following conditions are equivalent : (i) K is a perfectoid field; (ii) the topology of K is given by a rank 1 valuation |.| satisfying the following conditions : (a) |.| is not discrete (i.e., its valuation group is not isomorphic to Z); (b) |ℓ| < 1; and the ℓth power map on K0/ℓK0 is surjective.
Proof. Suppose that K is perfectoid. Then we know that its topology is given by a rank 1 valu-ation |.| by theorem V.1.1.9. By corollary V.1.1.4, the ℓth power map on K0/ℓK0 is surjective, and K has a pseudo-uniformizer ϖ such that ϖℓdivides ℓin K0. In particular, ℓis topologically nilpotent in K, so |ℓ| < 1.
Now suppose that K satisfies the conditions of (ii). Then K0 = {a ∈K | |a| ≤1}, so K0 is bounded in K, i.e. K is uniform. We check that K satisfies the conditions of corollary 144 V.1 Perfectoid Tate rings V.1.1.4(ii). The first condition is part of the assumption on K. For the second condition, note that, as |.| is not discrete, there exists ϖ ∈K −{0} such that |ϖ|ℓ≤|ℓ|. In particular, |ϖ| < 1, so ϖ is topologically unipotent, hence a pseudo-uniformizer of K; moreover, as |ℓϖ−ℓ| ≤1, we have ℓϖ−ℓ∈K0, which means that ϖℓdivides ℓin K0.
Example V.1.1.11. (See 3.3, 3.4 of .) (1) The field Qℓis not perfectoid because its topology is given by a discrete valuation.
(2) The field Cℓis perfectoid; more generally, any algebraically closed complete non-archimedean field is perfectoid.
(3) Let Qcycl ℓ be the completion of Qℓ[µ1/ℓ∞] := S n≥0 Qℓ[µ1/ℓn] for the unique valuation |.| extending the ℓ-adic valuation on Qℓ, and let Zcycl ℓ = (Qcycl ℓ )0. Then Qcycl ℓ is perfec-toid. Indeed, it is complete non-Archimedean. For every r ≥1, the cyclotomic exten-sion Qℓ[µ1/ℓr]/Qℓis of degree ℓr−1(ℓ−1), and, if ω is a primitive ℓrth root of 1, then NQℓ[ω]/Qℓ(1 −ω) = ℓ, hence |ω|ℓr−1ℓ= |ℓ|. This shows that |.| is not a discrete valuation.
We obviously have |ℓ| < 1. Finally, let a ∈Zcycl ℓ /ℓZcycl ℓ . We can find a lift a of a in the ring of integers of Qℓ[µ1/ℓn] for some n ≥0. Pick a primitive ℓn+1th root of unity ω. Then we can write a = Pℓn−1 i=0 aiωℓi, with the ai in Zℓ. Then, if b = Pℓn−1 i=0 aiωi, then bℓ= a modulo ℓZcycl ℓ .
(4) Similarly, the completion L of Qℓ[ℓ1/ℓ∞] := S n≥0 Qℓ[ℓ1/ℓn] for the unique valuation |.| extending the ℓ-adic valuation on Qℓis a perfectoid field. Indeed, K is complete and non-Archimedean, and the valuation group of |.| is not isomorphic to Z because its element |ℓ| < 1 is divisible by ℓr for every r ∈N. Also, we have K0/ℓK0 = Fℓ, so every element of K0/ℓK0 is a ℓth power.
(5) As in (3), we can show that the completion Fℓ((t1/ℓ∞)) of S n≥0 Fℓ((t1/ℓn)) for the unique valuation extending the t-adic valuation on Fℓ((t)) is a perfectoid field. This construction still makes sense if we replace Fℓby any ring A, and it gives a perfectoid field if A is a perfect field of characteristic ℓ.
(6) Let K be a perfectoid field of characteristic 0.
If A0 is the ℓ-adic completion of S n≥0 K0[T 1/ℓn], then K⟨T 1/ℓ∞⟩:= A0[ 1 ℓ] is a perfectoid Tate ring. (See proposition V.1.2.8.) Note that, if n ≥0, then the Gauss norms on K⟨T 1/ℓn+1⟩and on its subring K⟨T 1/ℓn⟩coincide, so we get a norm on S n≥0 K⟨T 1/ℓn⟩. Then K⟨T 1/ℓ∞⟩is the comple-tion of S n≥0 K⟨T 1/ℓn⟩for this norm.
Note that the construction of A⟨T 1/ℓ∞⟩makes sense for any f-adic ring A.
(6) Finally, we give an example of a perfectoid Tate alegbra that does not contain a field. First, consider the f-adic ring A0 = Zcycl ℓ (see (3) and (5)); this is an adic ring, with the T-adic topology. Let B = A⟨(ℓ/T)1/ℓ∞⟩:= A⟨X1/ℓ∞⟩/(ℓ−XT). This is still an adic ring with the T-adic topology. Finally, the f-adic ring B[ 1 T ] is a Tate ring, and it is perfectoid.
145 V Perfectoid algebras A pseudo-uniformizer satisfying the conditions of definition V.1.1.1 is ϖ = T 1/ℓ(note that ϖℓ= T divides ℓin B). The ring B[ 1 T ] doesn’t contain a field, because ℓis nonzero and not invertible in it.
V.1.2 Tilting Proposition V.1.2.1. (Section 1.3 of and lemma 3.10 of .) Let A be a perfectoid Tate ring. We consider the set A♭= lim ← − a7− →aℓ A := {(a(n)) ∈AN | ∀n ∈N, (a(n+1))ℓ= a(n)}, with the projective limit topology, the pointwise multiplication and the addition defined by (a(n)) + (b(n)) = (c(n)), with c(n) = lim r→+∞(a(n+r) + b(n+r))ℓr.
We denote the map A♭→A, (a(n)) 7− →a(0) by f 7− →f ♯.
Then : (i) The addition is well-defined (i.e. the limit above always exists) and makes A♭into a per-fectoid Tate ring of characteristic ℓ.
(ii) The subring A♭0 of power-bounded elements in A♭is given by A♭0 = lim ← − a7− →aℓ A0.
If ϖ is a pseudo-uniformizer of A that divides ℓin A0, then the canonical map A♭0 →lim ← − a7− →aℓ A0/ϖ is an isomorphism of topological rings.
(iii) There exists a pseudo-uniformizer ϖ of A such that ϖℓdivides ℓin A0 and that ϖ is in the image of the map (.)♯: A♭→A. Moreover, if ϖ♭is an element of A♭such that ϖ = (ϖ♭)♯, then ϖ♭is a pseudo-uniformizer of A♭, the map f 7− →f ♯induces a ring isomorphism A♭0/ϖ♭≃A0/ϖ, and A♭= A♭0[ 1 ϖ♭].
Note that, if A is a perfectoid ring of characteristic ℓ, then FrobA : A →A is an isomorphism of topological rings (by the open mapping theorem, i.e. theorem II.4.1.1), so A♭is canonically isomorphic to A (via the map (.)♯).
146 V.1 Perfectoid Tate rings Proof.
(1) Note that the image of the map (.)♯: A♭→A is the set of elements of A that have compatible systems of ℓth power roots. So, by proposition V.1.1.3, there exists a pseudo-uniformizer ϖ of A such that ϖℓdivides ℓin A0 and that ϖ is in the image of (.)♯. Choose ϖ♭∈A♭such that (ϖ♭)♯.
(2) We show that the addition of A♭is well-defined on elements of A♭0. Let (a(n)), (b(n)) be elements of A♭0. Fix n ∈N. For every r ∈N, we have (a(n+r+1) + b(n+r+1))ℓ= (a(n+r+1))ℓ+ (b(n+r+1))ℓ= a(n+r) + b(n+r) mod ℓA0 ⊂ϖA0, so repeated applications of lemma V.1.1.6 give (a(n+r+1) + b(n+r+1))ℓr+1 == (a(n+r) + b(n+r))ℓr mod ϖr+1A0.
This implies that ((a(n+r) + b(n+r))ℓr)r≥0 is a Cauchy sequence, so it admits a limit c(n) in A0. The fact that (c(n+1))ℓ= c(n) for every n (i.e. that (c(n)) is an element of A♭) follows immediately from the definition of c(n).
(3) By lemma V.1.1.7, the canonical map A♭0 = lim ← − a7− →aℓ A0 →lim ← − a7− →aℓ A0/ϖA0 is an isomorphism of topological monoids. We show that it is also compatible with addi-tion. Let (a(n)), (b(n)) be elements of A♭0, and let (c(n)) = (a(n)) + (b(n)). Fix n ∈N.
Then c(n) = lim r→+∞(a(n+r) + b(n+r))ℓr.
But we have seen in the proof of (2) that the sequence ((a(n+r) + b(n+r))ℓr)r≥0 is constant modulo ϖA0, so c(n) = a(n) + b(n) modulo ϖA0. This shows the claim. In particular, we get that A♭0 is a non-Archimedean topological ring of characteristic ℓ.
(4) Write ϖ♭= (ϖ1/ℓn)n≥0. (In other words, we choose a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots of ϖ.) For every a = (a(n)) ∈A♭and every r ∈N, we have (ϖ♭)ra = (ϖr/ℓna(n))n≥0 by definition of the multiplication of A♭. In particular, taking r = ℓ, we see that the isomorphism os topological rings A♭0 ∼ →lim ← −a7− →aℓA0/ϖA0 induces an isomorphism of topological rings A♭0/(ϖ♭)ℓA♭0 ∼ →lim ← − a7− →aℓ A0/ϖℓm−nA0 ≃A0/ϖℓA0, where the last isomorphism follows from condition (b) in definition V.1.1.1, applied to the pseudo-uniformizers ϖ1/ℓn, n ≥0. As the square A♭0/ϖ♭A♭0 Frob (.)♯ ∼ / A0/ϖA0 Frob A♭0/(ϖ♭)ℓA♭0 (.)♯ ∼ / A0/ϖℓA0 is clearly commutative, this shows that Frob : A♭0/ϖ♭A♭0 →A♭0/(ϖ♭)ℓA♭0 is bijective.
147 V Perfectoid algebras (5) We show that A♭0 is ϖ♭-adically Hausdorff and complete. It suffices to show that the ϖ♭-adic topology on A♭0 is finer than the product topology coming from the isomorphism A♭0 ∼ →lim ← −a7− →aℓA0/ϖA0. But, by the formula of (4) for (ϖ♭)r(a(n)), if m ∈N, then the projections of (ϖ♭)ℓmA♭0 on the last m factors of Q n≥0 A0/ϖA0 are 0, which implies the desired result.
(6) We show that addition is well-defined on A♭. Let (a(n)) ∈A♭. We choose N ∈N such that ϖNa(0) ∈A0. Then, for every n ≥0, (ϖN/ℓna(n))ℓn = ϖNa(0) ∈A0, so ϖN/ℓna(n) ∈A0.
This shows that (ϖ♭)N(a(n)) ∈A♭0.
Let (b(n)) be another element of A♭.
Up to increasing N, we may assume that ϖN/ℓnb(n) ∈A0 for every n ∈N. Fix n ∈N. By (2), we know that the sequence (ϖN/ℓn(a(n+r)+b(n+r))ℓr)r≥0 converges in A0. This implies immediately that the sequence ((a(n+r) + b(n+r))ℓr)r≥0 converges in A.
(7) Using the fact that A♭0 is a topological ring and the calculation of (6), it is easy to check that A♭is a topological ring and that A♭= A♭0[ 1 ϖ♭]. By remark II.2.5.3, A♭is a Tate ring having A♭0 as a ring of definition.
(8) We show that (A♭)0 = A♭0. (This implies that A♭is perfectoid and finishes the proof of the proposition.) As A♭0 is a bounded subring of A♭, we clearly have A♭0 ⊂(A♭)0.
Conversely, let a = (a(n)) ∈(A♭)0. As a is power-bounded, there exists N ∈N such that (ϖ♭)Nan ∈A♭0 for every n ≥0. This implies that ϖN(a(0))n ∈A0 for every n ≥0, i.e. that a(0) is power-bounded. As (a(n))ℓn = a(n) for every n ∈N, a(n) is also power-bounded, and we have shown that a ∈A♭0.
Definition V.1.2.2. The perfectoid Tate ring A♭of proposition V.1.2.1 is called the tilt of A.
Remark V.1.2.3. Let A and A′ be perfectoid Tate algebras, with A′ of characteristic ℓ, and let ϖ and ϖ′ be pseudo-uniformizers of A and A′ such that ϖℓdivides ℓin A0 and that ϖ = (ϖ♭)♯for some pseudo-uniformizer ϖ♭of A♭.
Then any isomorphism ϕ : A0/ϖℓA0 ∼ →A′0/ϖ′ℓA′0 sending ϖ to ϖ′ induces an isomorphism A♭ ∼ →A′.
Indeed, by the second formula in proposition V.1.2.1(ii), we have A♭0 = lim ← −a7− →aℓA0/ϖℓA0 and A′♭0 = lim ← −a7− →aℓA′0/ϖ′ℓA′0, so ϕ induces an isomorphism ψ : A♭0 ∼ →A′♭0 = A′0, and ψ(ϖ♭) = ϖ′ modulo ϖ′ℓA′0. This implies that ψ(ϖ♭) is also a pseudo-uniformizer of A′0, so ψ extends to an isomorphism A♭ ∼ →A′.
Remark V.1.2.4. Let A be a perfectoid Tate ring. Then the map (.)♯: A♭0 →A0 induces an isomorphism of rings A♭0/A♭00 ∼ →A0/A00.
Proof. First note that A00 (resp. A♭00) is an ideal of A0 (resp. A♭0), so the statement makes sense.
By proposition V.1.2.1, we can choose a pseudo-uniformizer ϖ♭of A♭such that ϖ := (ϖ♭)♯is 148 V.1 Perfectoid Tate rings a pseudo-unfiformizer of A, and then (.)♯induces an isomorphism A♭0/ϖ♭A♭0 ∼ →A0/ϖA0. We have A00 ⊃ϖA0, and, by lemma II.1.2.3, A00/ϖA0 is the nilradical of A0/ϖA0. We have a similar result for A♭, and this implies the statement.
Proposition V.1.2.5. Let A be a perfectoid Tate ring. If A+ is a ring of integral elements in A (i.e. an open and integrally closed subring of A contained in A0), then A♭+ := lim ← −a7− →aℓA+ is a ring of integral elements in A♭, and, for every pseudo-uniformizer ϖ♭of A♭such that ϖ := (ϖ♭)♯ is a pseudo-uniformizer of A, the isomorphism A♭0/ϖ♭≃A0/ϖ sends A♭+/ϖ♭to A+/ϖ.
Moreover, this induces a bijection between rings of integral elements in A and A♭.
Proof. Choose pseudo-uniformizers ϖ and ϖ♭of A and A♭such that ϖ = (ϖ♭)♯.
If A+ is a ring of integral elements in A, then ϖA0 ⊂A+ ⊂A0 (the first inclusion comes from the fact that ϖA0 ⊂A00), and A+/ϖA0 is an integrally closed subring of A0/ϖA0. Conversely, if S is an integrally closed subring of A0/ϖA0, then its inverse image in A0 is an open and integrally closed subring of A0, i.e. a ring of integral elements in A. So rings of integral elements in A are in natural bijection with integrally closed subrings of A0/ϖA0. We have a similar result for A♭. As A0/ϖA0 ≃A♭0/ϖ♭A♭0, this gives a bijection betweem rings of integral elements in A and A♭.
It remains to show that this bijection is given by the formula of the proposition. Let A+ be a ring of integral elements in A, and let A♭+ be the corresponding ring of integral elements in A♭.
Let a = (a(n)) ∈A♭0 = lim ← −a7− →aℓA0. We want to check that a ∈A♭+ if and only if a(n) ∈A+ for every n ∈N. First, as A+ is integrally closed in A0, we have a(n) ∈A+ for every n ∈N if and only if a(0) ∈A+. Then, as ϖA0 ⊂A+, we have a(0) ∈A+ if and only if a(0)+ϖA0 ∈A+/ϖA0.
But a(0) + ϖA0 is the image of a + ϖ♭A♭0 by the isomorphism A0/ϖA0 ≃A♭0/ϖ♭A♭0, and this isomrophism sends A+/ϖA0 to A♭+ϖ♭A♭0, so we get the desired equivalence.
Definition V.1.2.6. If A is a perfectoid Tate ring and A+ ⊂A is a ring of integral elements, we say that (A, A+) is a perfectoid Huber pair and we call (A♭, A♭) its tilt.
Proposition V.1.2.7. (Proposition 3.6 of .) Let A be a perfectoid Tate ring. Then, for every continuous valuation |.| : A →Γ ∪{0} on A, the map |.|♭: A♭→Γ ∪{0}, a 7− →|a♯| is a continuous valuation on A♭.
Moreover, if A = K is a perfectoid field, then this induces a bijection Cont(K) ∼ →Cont(K♭).
Proof. Let |.| : A →Γ ∪{0} be a continuous valuation on A. Then |.|♭obviously satisfies all the properties of a valuation on A♭, except maybe for the strong triangle inequality. We check this last property. Let a = (a(n)) and b = (b(n)) be elements of A♭. By definition of the addition on A♭, we have (a + b)♯= lim n→+∞(a(n) + b(n))ℓn = lim n→+∞((a1/ℓn)♯+ (b1/ℓn)♯)ℓn, 149 V Perfectoid algebras so |a + b|♭= |(a + b)♯| ≤sup n∈N (max(|(a1/ℓn)♯|ℓn, |(b1/ℓn)♯|ℓn)) = max(|a|♭, |b|♭).
We show that |.|♭is continuous. Let γ ∈Γ. By definition |.|♭, {a ∈A♭| |a|♭< γ} is the inverse image by the continuous map (.)♯: A♭→A of the open subset {a ∈A | |a| < γ} of A, so it is an open subset of A♭.
Now assume that K is a perfectoid field. We want to show that |.| 7− →|.|♭induces a bijection Cont(K) ∼ →Cont(K♭). By corollary II.2.5.11, Cont(K) is canonically in bijection with the set of valuation subrings K+ of K such that K00 ⊂K+ ⊂K0, hence with the set of valuation subrings of the field K0/K00. We have a similar statement for K♭. So the result follows from remark V.1.2.4.
We fnish with some examples of tilting.
Proposition V.1.2.8. (Proposition 5.20 of .) Let K be a perfectoid field, and let ϖ be a pseudo-uniformizer of K and K+ be a ring of integral elements in K. As in example V.1.1.11(6), we denote by K+⟨X1/ℓ∞⟩the ϖ-adic completion of the ring S n≥0 K+[X1/ℓn], and we set K⟨X1/ℓ∞⟩= K+⟨X1/ℓ∞⟩[ 1 ϖ] (note that this last ring does not depend on the choice of K+).
Then K⟨X1/ℓ∞⟩ is perfectoid, with ring of power-bounded elements (K⟨X1/ℓ∞⟩)0 = K0⟨X1/ℓ∞⟩, and its tilt is canonically isomorphic to K♭⟨X1/ℓ∞⟩.
Moreover, K+⟨X1/ℓ∞⟩is a ring of integral elements in K⟨X1/ℓ∞⟩and the tilt of (K⟨X1/ℓ∞⟩, K+⟨X1/ℓ∞⟩) is (K♭⟨X1/ℓ∞⟩, K♭+⟨X1/ℓ∞⟩).
Proof. Set A = K⟨X1/ℓ∞⟩and A+ = K+⟨X1/ℓ∞⟩.
We know that A is a complete Tate ring and that A0 := K0⟨X1/ℓ∞⟩is a ring of definition of A by remark II.2.5.3. In particu-lar, we have A0 ⊂A0. Also, if we set A′ 0 = S n≥0 K0[X1/ℓn] with the ϖ-adic topology and A′ = A′ 0[ 1 ϖ], then, by the same remark, A′ is a Tate ring with ring of definition A′ 0 and, by corollary II.3.1.9(v), A is the completion of A′. So, by proposition II.3.1.12(i), to show that A0 = A0, it suffices to show that (A′)0 = A′ 0. Let f ∈(A′)0. Then there exists n ∈N such that f ∈K0[X1/ℓn][ 1 ϖ] = K[X1/ℓn], and we want to show that f is in K0[X1/ℓn]. Without loss of generality, we may assume that n = 0, so that f ∈K[X]. Denote by |.| the rank 1 valuation giving the topology of K. Suppose that f ̸∈K0[X], write f = P s≥0 asXs, and let r be the smallest integer such that |ar| = maxs≥0 |as|. Then |as| ≤|ar| for every s, and |as| < |ar| if s < r; also, for every n ∈Z, we have |n · 1K| ≤1. So, if n ∈N, then the coefficient α of Xrn in f n satisfies |α| = |ar|n. This shows that the set {f n, n ∈N} is not bounded, so f is not power-bounded.
We have shown that A is uniform. As K is perfectoid, we may assume that ϖ ∈K+ and that ϖℓdivides ℓin K+; then ϖ is a pseudo-uniformizer of A, and ϖℓdivides ℓin A+, and in 150 V.1 Perfectoid Tate rings particular in A0 ⊃A+. It remains to show that every element of A0/ϖA0 is a ℓth power. But this is obvious, because A0/ϖA0 = S n≥0(K0/ϖK0)[X1/ℓn].
We may assume that ϖ = (ϖ♭)♯, with ϖ♭∈K♭+ a pseudo-uniformizer of K♭. Then we have a canonical isomorphism A+/ϖℓA+ ≃A♭+/(ϖ♭)ℓA♭+. By remark V.1.2.3, this extends to an isomorphism A♭ ∼ →K♭⟨X1/ℓ∞⟩sending A♭+ to K♭+⟨X1/ℓ∞⟩.
Proposition V.1.2.9. (Lemma 5.21 of .) Let A be a perfectoid Tate algebra. Then A is a perfectoid field if and only if A♭is a perfectoid field.
Proof. Let a = (a(n)) ∈A♭= lim ← −c7− →cℓA. As A is reduced (see remark IV.1.1.3), a = 0 if and only if a(0) = 0. On the other hand, as (a(n))ℓn = a(0) for every n ∈N, we have a ∈(A♭)× if and only if a(0) ∈A×. This shows that A♭is a field if and ony if A is a field. By theorem V.1.1.9, this finishes the proof.
Example V.1.2.10. Let L be the perfectoid field of example V.1.1.11(4).
Then (Qcycl ℓ )♭= L♭= Fℓ((t1/ℓ∞)). (We can prove this using remark V.1.2.3.) V.1.3 Witt vectors We will explain Fontaine’s approach to untilting using Witt vectors, so we need a few reminders about their construction and properties. A reference for this is chapitre II §6.
Consider the Witt polynomials W0, W1, . . . ∈Z[X0, X1, . . .], defined by Wm = m X n=0 Xℓm−n n ℓn = Xℓm 0 + ℓXℓm−1 1 + . . . + ℓm−1Xℓ m−1 + ℓmXm.
Theorem V.1.3.1. ( chapitre II §6, Th´ eoreme 5.) For every Φ ∈Z[X, Y ], there exists a unique sequence (ϕn)n≥0 of elements of the polynomial ring Z[Xn, Yn, n ∈N] such that, for every m ∈N : Wm(ϕ0, ϕ1, . . .) = Φ(Wm(X0, X1, . . .), Wm(Y0, Y1, . . .)).
Applying this theorem to the polynomials X + Y and XY , we get sequences of polynomials (Sn)n≥0 and (Pn)n≥0.
Theorem V.1.3.2. ( chapitre II §6, Th´ eor eme 6.) Let A be a commutative ring. We write W(A) = AN, and define an addition and multiplication on W(A) by a + b = (S0(a, b), S1(a, b), . . .) 151 V Perfectoid algebras and a · b = (P0(a, b), P1(a, b), . . .), for a, b ∈AN. Then this makes W(A) into a commutative ring, called the ring of Witt vectors of A.
We denote by W∗the map W(A) →AN sending a to the sequence (W0(a), W1(a), . . .). The idea is that this map is an isomorphism of rings if ℓis invertible in A, and that we can reduce to this case by functoriality.
The ring morphism Wm : W(A) →A is called the mth ghost component map.
Definition V.1.3.3. ( chapitre II §5.) A strict ℓ-ring is a commutative ring A such that ℓis not a zero divisor in A, A is complete (and Hausdorff) for the ℓ-adic topology and the ring A/ℓA is perfect.
Proposition V.1.3.4. ( chapitre II §5 proposition 8.) If A is a strict ℓ-ring, there exists a unique multiplicative section of the reduction map A →A/ℓA.
We denote this section by a 7− →[a] and call it the Teichm¨ uller representative.
Theorem V.1.3.5. ( chapitre II §5 th´ eoreme 5 and §6 th´ eor eme 7.) Let A be a perfect ring of characteristic ℓ. Then there exists a unique (up to unique isomorphism) strict ℓ-ring B such that B/ℓB = A, and this ring is canonically isomorphic to W(A).
Moreover, the isomorphism from W(A) to B sends a = (an)n≥0 to P m≥0[a1/ℓm m ]ℓm.
In particular, if A is a perfect ring of characteristic ℓ, then we have a Teichm¨ uller representative [.] : A →W(A), and every element of W(A) can be written as P n≥0[an]ℓn, with the an ∈A uniquely determined.
V.1.4 Untilting We will use the ring of Witt vectors to “untilt” a perfectoid Tate ring of characteristic ℓ. First we need a definition.
Definition V.1.4.1. (Section 1.4 of and definition 3.15 of .) Let (A, A+) be a perfectoid Huber pair of characteristic ℓ. An ideal I of W(A+) is called primitive of degree 1 if it is generated by an element ξ of the form ξ = ℓ+ [ϖ]α, where ϖ is a pseudo-uniformizer of A and α ∈W(A+). We also say that the element ξ of W(A+) is primitive of degree 1.
Lemma V.1.4.2. (Lemma 3.16 of .) Any element ξ of W(A+) that is primitive of degree 1 is torsionfree in W(A+).
152 V.1 Perfectoid Tate rings Proof. Write ξ = ℓ+ [ϖ]α, where ϖ is a pseudo-uniformizer of A and α ∈W(A+). Let b ∈W(A+) such that bξ = 0. We want to show that b = 0. By theorem V.1.3.5, we can write b = P n≥0[an]ℓn, with the an ∈A+ uniquely determined, and it suffices to show that all the an are 0. We show by induction on r that an ∈ϖrA+ for every n, r ∈N, which implies that an = 0 because A+ is ϖ-adically separated. The result is obvious for r = 0. Suppose that r ≥1 and that we have an ∈ϖr−1A+ for every n. Then we can write an = ϖr−1a′ n with a′ n ∈A+.
Let b′ = P n≥0[a′ n]ℓn ∈W(R+). Then we have 0 = bξ = [ϖ]r−1b′ξ, so b′ξ = 0 because [ϖ] is not a zero divisor in W(A+). Reducing the euqlity (ℓ+ [ϖ]α)b′ = 0 modulo [ϖ] gives P n≥0[a′ n]ℓn+1 ∈[ϖ]W(A+), hence a′ n ∈ϖA+ for every n ∈N, and we are done.
Theorem V.1.4.3. (Lemma 3.14 of .) Let (A, A+) be a perfectoid Huber pair, and let (A♭, A♭+) be its tilt.
(i) The map θ : W(A♭+) → A+ P n≥0[an]ℓn 7− → P n≥0 a♯ nℓn is a surjective morphism of rings.
(ii) The kernel of θ is primitive of degree 1.
Note that, if char(A) = ℓ(so that A♭= A), then Ker θ is the ideal generated by ℓ(which is primitive of degree 1).
Proof. Choose a pseudo-uniformizer ϖ♭of A♭such that ϖ := (ϖ♭)♯is a pseudo-uniformizer of A and that ϖℓdivides ℓin A+. (We have seen that we can choose ϖ♭and ϖ satisfying all these properties, except that ϖ♭may not be in A♭+ and ϖℓonly divides ℓin A0. But, as A♭+ and A+ are open and ϖ♭and ℓϖ1−ℓare topologically nilpotent (for the second one, because it is in ϖA0), there exists some integer N ≥1 such that (ϖ♭)N ∈A♭+ and ℓNϖN−Nℓ∈A+, and we get the last property if we replace ϖ♭by (ϖ♭)N.) (i) We first check that θ is a morphism of rings. It suffices to check that its composition with each projection A+ →A+/ϖmA+ is a morphism of rings.
Fix m ≥1.
Re-member that we have the mth ghost component map Wm : W(A+) →A+, sending (an)n≥0 ∈W(A+) to Pm n=0 aℓm−n n ℓn. The composition W(A+) Wm →A+ →A+/ϖmA+ sends a family (an)n≥0 of elements of ϖA+ to 0, hence it factors through a mor-phism of rings W(A+/ϖA+) →A+/ϖmA+.
We can compose this with the map A♭+ = lim ← −A+ →A+/ϖA+, (a(n)) 7− →a(m) + ϖA+ to get a morphism of rings θ′ : W(A♭+) →W(A+/ϖA+) →A+/ϖmA+. We claim that θ′ is the composition of θ with the projection A+ →A+/ϖmA+. Indeed, let a = (an)n≥0 ∈W(A♭+), and write an = (a(i) n )i≥0. Then a = P n≥0[a1/ℓn n ]ℓn, so θ(a) = X n≥0 (a1/ℓn n )♯ℓn = X n≥0 a(n) n ℓn.
153 V Perfectoid algebras On the other hand, the image of a in W(A+/ϖA+) is the sequence (a(m) n + ϖA+)n≥0. A lift of this in W(A+) is (a(m) n )n≥0. Applying the map Wm : W(A+) →A+/ϖmA+, we get θ′(a) = m X n=0 (a(m) n )ℓm−nℓn = m X n=0 a(n) n ℓn.
Now we show that θ is surjective.
We have θ([ϖ♭]) = ϖ, and the map W(A♭+)/[ϖ♭]W(A♭+) → A+/ϖA+ induces by θ sends P n≥0[an]ℓn to (a0)♯+ ϖA+, so it is the composition of the canonical maps W(A♭+) → A♭+ and A♭+ → A♭+/ϖ♭A♭+ ≃ A+/ϖA+, and in particular it is surjec-tive.
As W(A♭+) is [ϖ♭]-adically complete (because, for every r ∈ N, [ϖ♭]rW(A♭+) = {P n≥0[an]ℓn, an ∈(ϖ♭)rW ♭+) and A+ is ϖ-adically complete, this implies that θ is surjective by 25, Lemma 0315.
(ii) First we show that there exists f ∈ϖ♭A♭+ such that f ♯= ℓmodulo ℓϖA+. Indeed, as ϖℓdivides ℓin A+, α := ℓϖ−1 ∈A+. As every element of A+/ℓA+ is a ℓth power, there exists β ∈A♭+ such that β♯= α modulo ℓA+. Take f = ϖ♭β. Then f ♯= ϖβ♯is equal to ϖα = ℓmodulo ℓϖA+.
By the claim proved in the previous paragraph the surjectivity of θ, we can write ℓ= f ♯+ ℓ(ϖ♭)♯X n≥0 a♯ nℓn for some f ∈ϖ♭A♭+ and an ∈A♯+. Let ξ = ℓ−[f] −[ϖ♭] X n≥0 [an]ℓn+1 ∈W(A♭+).
Then ξ is primitive of degree 1 (because ϖ♭divides f in A♭+, so [ϖ♭] divides [f] in W(A♭+)), and θ(ξ) = 0 by the choice of f and the an.
It remains to show that Ker(θ) = ξW(A♭+).
Note that ξ ∈ℓ+ [ϖ♭]W(A♭+), so W(A♭+)/(ξ, [ϖ♭]) = W(A♭+)/(ℓ, [ϖ♭]) ≃ A♭+/ϖ♭A♭+, and the map W(A♭+)/(ξ, [ϖ♭]) →A+/ϖA+ induced by θ is the map (.)♯: A♭+/ϖ♭A♭+ →A+/ϖA+, which is an isomorphism. To conclude that θ induces an isomorphism, it suffices (by lemma V.1.4.4) to check that W(A♭+)/(ξ) is [ϖ♭]-adically complete.
To show that W(A♭+)/(ξ) is [ϖ♭]-adically complete, consider the short exact sequence of W(A♭+)-modules 0 →ξW(A♭+) →W(A♭+) →W(A♭+)/(ξ) →0.
As ξ is not a zero divisor (lemma V.1.4.2), the W(A♭+)-module ξW(A♭+) is flat, and so, by lemma 25, Lemma 0315, the sequence of [ϖ♭]-adic completions is still exact. But the first two modules are [ϖ♭]-adically complete, so the third must also be [ϖ♭]-adically complete.
154 V.1 Perfectoid Tate rings Lemma V.1.4.4. Let R be a ring, let ϖ ∈R, and let θ : M →N be a morphism of R-modules.
Suppose that M and N are ϖ-adically complete and Hausdorff, that N is ϖ-torsionfree, and that θ induces an isomorphism M/ϖM ∼ →N/ϖN. Then θ is an isomorphism.
Proof. We already know that θ is surjective, by 25, Lemma 0315. To show that θ is injective, it suffices to show that, for every n ≥1, the morphism M/ϖnM →N/ϖnN induces by θ is injective. We show this by induction on n. The case n = 1 is the assumption, so suppose that n ≥2 and that we know the result for every n′ < n. We have a commutative diagram with exact rows : M/ϖn−1M ϖ·(.) / θ M/ϖnM / θ M/ϖM θ / 0 0 / N/ϖn−1N ϖ·(.) / N/ϖnN / N/ϖN / 0 Indeed, the map N/ϖn−1N →N/ϖnN, x 7− →ϖx is injective because N is ϖ-torsionfree. By the induction hypothesis, the first and third vertical maps are injective. So the injectivity of the middle vertical arrow follows from the five lemma (or an easy diagram chase).
Corollary V.1.4.5. (Theorem 3.17 of , see also proposition 1.1 of .) There is an equiva-lence of categories between : (a) perfectoid Huber pairs (S, S+); (b) triples (R, R+, I), where (R, R+) is a perfectoid Huber pair of characteristic ℓand I ⊂W(R+) is an ideal that is primitive of degree 1.
This equivalence is given by the functors that send a pair (S, S+) as in (a) to (S♭, S♭+, Ker(θ : W(S♭+) → S+)), and a triple (R, R+, I) as in (b) to ((W(R+)/I)[ 1 [ϖ]], W(R+)/I), where ϖ is any pseudo-uniformizer of R.
Corollary V.1.4.6. (See th´ eoreme 1.2 of .) Fix a perfectoid Huber pair (A, A+). Then there is an equivalence of categories between perfectoid Huber pairs over (A, A+) and over (A♭, A♭+).
This equivalence is given by the functor that send a morphism of perfectoid Huber pairs (A, A+) →(B, B+) to its tilt (A♭, A♭+) →(B♭, B♭+). In the other direction, suppose that we have a morhism of perfectoid Huber pairs (A♭, A♭+) →(B♭, B♭+), then we get a mor-phism α : W(A♭+) →W(B♭+).
By the definition of a primitive ideal of degree 1, if I = Ker(θ : W(A♭+) →A+), then J := α(I) is a primitive ideal of degree 1 of W(B♭+), so we get a morphism of perfectoid Huber pairs (A, A+) ≃((W(A♭+)/I)[ 1 [ϖ♭]], W(A♭+)/I) →((W(B♭+)/J)[ 1 [ϖ♭]], W(B♭+)/J) (where ϖ♭is any pseudo-uniformizer of A♭).
155 V Perfectoid algebras V.1.5 A little bit of almost mathematics We will introduce as little almost mathematics as possible. For a complete treatment, see .
Definition V.1.5.1. Let (A, A+) be a perfectoid Huber pair.
(i) We say that a A+-module M is almost zero if A00 · M = 0.
(ii) We say that a morphism of A+-modules u : M →N is almost injective (resp. almost surjective) if ker u (resp. Coker u) is almost zero; we say that it is almost an isomorphism if it is both almost injective and almost surjective.
(iii) If M ⊂N are A+-modules, we say that they are almost equal if N/M is almost zero.
Example V.1.5.2. A+ and A0 are almost equal.
Remark V.1.5.3. If u : M →N is almost injective (resp. almost surjective, resp. almost an isomorphism), then u[ 1 ϖ] : M[ 1 ϖ] →N[ 1 ϖ] is injective (resp. surjective, resp. an isomorphism).
Proposition V.1.5.4. Let (A, A+) be a perfectoid Huber pair and M be a A+-module. Choose a pseudo-uniformizer ϖ of A that has a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots. (In other words, ϖ is in the image of (.)♯: A♭→A.) Then M is almost zero if and only if, for every x ∈M and for every n ∈N, we have ϖ1/ℓnx = 0.
Proof. We have ϖ1/ℓn ∈A00 for every n ∈N, so, if M is almost zero, then ϖ1/ℓnx = 0 for every x ∈M and every n ∈N.
Conversely, suppose that, for every x ∈M and every n ∈N, we have ϖ1/ℓnx = 0.
Let a ∈A00.
As ϖA+ is open in A, there exists n ∈N such that aℓn ∈ϖA+.
Then (ϖ−1/ℓna)ℓn ∈A+, so ϖ−1/ℓna ∈A+ because A+ is integrally closed in A. So, for every x ∈M, ax = ϖ1/ℓn((ϖ−1/ℓna)x) = 0.
Corollary V.1.5.5. If (A, A+) →(B, B+) is a morphism of perfectoid Huber pairs and M is a B+-module, then M is almost zero as a B+-module if and only if it is almost zero as a A+-module.
In particular, for perfectoid Huber pairs over a perfectoid Huber pair (K, K0) with K a field, we recover the same definition as in definition 4.1 of .
Definition V.1.5.6. Let (A, A+) be a perfectoid Huber pair and M be a A+-module. We set M∗= HomA+(A00, M). This is also a A+-module, and it is called the module of almost elements of M.
By the general properties of Hom, the functor M 7− →M∗commutes with projective limits (in particular, it is left exact).
156 V.1 Perfectoid Tate rings Remark V.1.5.7. If M1, M2, N are A+-modules and µ : M1 × M2 →N is a A+-bilinear map, then we have a canonical A+-linear map M1∗⊗A+ M2∗→N∗, induced by the A+-bilinear map HomA+(A00, M1) × HomA+(A00, M2) → HomA+(A00, N) (f, g) 7− → (a 7− →µ(f(a) ⊗g(a))).
In particular, if B+ is a A+-algebra and M is a B+-module, then (B+)∗is a A+-algebra and M∗is a (B+)∗-module. Moreover, if M1, M2, N are B+-modules and the map µ : M1×M2 →N is actually B+-bilinear, then it is easy to see from the construction given above that the induced map M1∗× M2∗→N∗is (B+)∗-bilinear; so, if for example C+ is a B+-algebra, then (C+)∗is a (B+)∗-algebra.
Proposition V.1.5.8. Let (A, A+) be a perfectoid Huber pair and M be a A+-module.
(i) The canonical map M →M∗, x 7− →(a 7− →ax) is almost an isomorphism.
(ii) Choose a pseudo-uniformizer ϖ of A that has a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots. If M is ϖ-torsionfree (so that the canonical map M →A ⊗A+ M is injective), then M∗= {x ∈A⊗A+ M | ∀n ∈N, ϖ1/ℓnx ∈M} = {x ∈A⊗A+ M | ∀a ∈A00, ax ∈M}.
(iii) If M is almost zero, then M∗= 0.
(iv) If u : M →N is an almost injective (resp. surjective) map of A+-modules, then the map u∗: M∗→N∗is injective (resp. surjective).
Proof.
(i) This follows from the exact sequence 0 →HomA+(A+/A00, M) →M →M∗→Ext1 A+(A+/A00, M).
(ii) Let v : M∗→M be evaluation at ϖ. If left multiplication by ϖ is an isomorphism on M, then the map ϖ−1v : M∗→M is an inverse of the canonical map M →M∗, so M = M∗.
Now suppose that M is only ϖ-torsionfree. As (A ⊗A+ M)∗= A ⊗A+ M by the pre-vious paragraph and M →(A ⊗A+ M) is injective by hypothesis, we get injections M ⊂M∗⊂A ⊗A+ M. Let M ′ = {x ∈A ⊗A+ M | ∀n ∈N, ϖ1/ℓnx ∈M} and M ′′ = {x ∈A ⊗A+ M | ∀a ∈A00, ax ∈M}. We obviously have M ′′ ⊂M ′. Con-versely, if x ∈M ′, then (M + A+x)/M is almost zero by proposition V.1.5.4, so x ∈M ′′.
This shows that M ′ = M ′′. Now we show that M∗= M ′′. If x ∈M ′′, then the map A00 →M, a 7− →ax is in M∗, and its image by ϖ−1v is x. So M ′′ ⊂M∗. Conversely, let c : A00 →M be an element of M∗, and let x = ϖ−1c(ϖ) ∈A ⊗A+ M. Then, for every n ∈N, we have ϖ1/ℓnx = ϖ1/ℓnϖ−1c(ϖ1−1/ℓnϖ1/ℓn) = ϖ−1+1/ℓnϖ1−1/ℓnc(ϖ1/ℓ) = c(ϖ1/ℓn) ∈M, so x ∈M ′ = M ′′.
157 V Perfectoid algebras (iii) Let u : A00 →M be an element of M∗. Let a ∈A00. Then we can find n ∈N such that ϖ−1/ℓna ∈A00 (just choose n such that aℓn ∈ϖA00), so u(a) = ϖ1/ℓnu(ϖ−1/ℓna) = 0.
This shows that u = 0.
(iv) Suppose that u is almost injective.
As the functor (.)∗is right exact, the kernel of u∗: M∗→N∗is equal to the image of (Ker u)∗→M∗. But (Ker u)∗= 0 by (iii), so u∗is injective.
Suppose that u is almost surjective.
Then, applying the right exact functor (.)∗ to the exact sequence M → N → Coker u → 0, we get an exact sequence M∗→N∗→(Coker u)∗→0. But we know that (Coker u)∗= 0 by (iii), so u∗is surjective.
Proposition V.1.5.9. (See lemma 4.4.1 of .) Let (A, A+) be a perfectoid Huber pair and M be a A+-module. Choose a pseudo-uniformizer ϖ of A that has a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots.
Let M be a A+-module that is ϖ-torsionfree and ϖ-adically separated and complete. Then the A+-module M∗is also ϖ-torsionfree and ϖ-adically separated and complete.
Moreover, for every a ∈ A+ such that M is a-torsionfree, the canonical morphism aM∗→(aM)∗is an isomorphism, and the canonical morphism M∗/aM∗→(M/aM)∗is injective, with image equal to that of (M/ϖ1/ℓnaM)∗for every n ∈N.
Proof. First we show that M∗is ϖ-torsionfree. This follows from the fact that the functor (.)∗is left exact, so Ker(ϖ : M∗→M∗) = (Ker(ϖ : M →M))∗= 0.
Let a ∈A+ such that M is a-torsionfree.
Then, applying (.)∗to the exact sequence 0 →M a·(.) →M →M/aM →0, we get an exact sequence 0 →M∗ a·(.) →M∗→(M/aM)∗→Ext1 A+(A00, M).
This shows in particular that aM∗= (aM)∗and that M∗/aM∗→(M/aM)∗is injective. Let n ∈N, and let ε = ϖ1/ℓn. We have a commutative diagram with exact rows 0 / M a·(.) / M / M/aM / 0 0 / M/εM a·(.) / M/εaM / M/aM / 0 158 V.1 Perfectoid Tate rings Applying the functor (.)∗, we get a commutative diagram with exact rows M∗ a·(.) / M∗ b1 / (M/aM)∗ / Ext1 A+(A00, M) c (M/εM)∗ a·(.) / (M/εaM)∗ b2 / (M/aM)∗ / Ext1 A+(A00, M/εM) We want to show that the maps b1 and b2 have the same image.
It suffices to prove that the map c is injective, which is equivalent to the fact that multiplication by ε is equal to zero on Ext1 A+(A00, M).
Applying the functor HomA+(., M) to the exact sequence 0 → A00 → A+ → A+/A00 → 0, we get an isomorphism Ext1 A+(A00, M) ≃Ext2 A+(A+/A00, M), which shows that Ext1 A+(A00, M) is almost zero and implies the desired result.
We finally prove that A∗is ϖ-adically separated and complete.
By the assumption on M, we have M ∼ → lim ← −M/ϖnM.
As the functor (.)∗commutes with projective lim-its, the canonical map M∗ → lim ← −(M/ϖnM)∗is an isomorphism.
But this map fac-tors as M∗→lim ← −M∗/ϖnM∗→lim ← −(M/ϖnM)∗, and the second map is injective be-cause M∗/ϖnM∗→(M/ϖnM)∗is injective for every n ∈N.
So the canonical map M∗→lim ← −M∗/ϖnM∗is also an isomorphism. (We could also have used the previous para-graph to show directly that lim ← −M∗/ϖnM∗→lim ← −(M/ϖnM)∗is an isomorphism.) Proposition V.1.5.10. (See proposition 5.2.6 and theorem 6.2.5 of .) Let (A, A+) be a per-fectoid Huber pair, and choose a pseudo-uniformizer ϖ of A that has a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots and that divides ℓin A+. Let ϕ : A →B be a morphism of f-adic rings and B0 be an open subring of B containing ϕ(A+).
Suppose that : (a) B is complete; (b) B0 is ϕ(ϖ)-adically separated and complete; (c) the map B0/ϕ(ϖ1/ℓ) →B0/ϕ(ϖ), b 7− →bℓis almost an isomorphism.
Then B is perfectoid, B0 is a ring of definition of B and B0 = (B0)∗.
Proof. We know that B is a Tate ring and that ϕ(ϖ) is a pseudo-uniformizer of B (see proposi-tion II.1.3.4). To simplify the notation, we will write ϖ instead of ϕ(ϖ). By lemma V.1.5.11, B0 is a ring of definition of B.
We write B′ = (B0)∗. We know that B′ is ϖ-adically separated and complete by proposition V.1.5.9, and that B′ = {b ∈B | ∀n ∈N, ϖ1/ℓnb ∈B0} 159 V Perfectoid algebras by proposition V.1.5.8(ii). In particular, lemma V.1.5.11 implies that B′ is a bounded subring of B, i.e. a ring of definition. By proposition V.1.5.8(iv), the map (B0/ϖ1/ℓB0)∗→(B0/ϖB0)∗ induced by b 7− →bℓis injective, so the map B′/ϖ1/ℓB′ →B′/ϖB′, b 7− →bℓis also injective, because B′/ϖ1/ℓB′ (resp. B′/ϖB′) injects in (B0/ϖ1/ℓB0)∗(resp. (B0/ϖB0)∗) by proposition V.1.5.9.
We prove that B′ is closed by taking ℓth roots. Let b ∈B such that bℓ∈B′. We want to prove that b ∈B′. Choose an integer r ∈N such that ϖr/ℓb ∈B′ (this is possible because B = B′[1/ϖ]). If r = 0, then b ∈B′ and we are done. Suppose that r ≥1. We have (ϖr/ℓb)ℓ= ϖrbℓ∈ϖrB′ ⊂ϖB′. By the injectivity of the map B′/ϖ1/ℓB′ →B′/ϖB′, b 7− →bℓ, this implies that ϖr/ℓb ∈ϖ1/ℓB′, hence that ϖ(r−1)/ℓb ∈B′. If r −1 ≥1, we can apply this process again to show that ϖ(r−2)/ℓb ∈B′, etc. In the end, we get that b ∈B′.
Note that the previous paragraph implies that B00 ⊂B′. Indeed, if b ∈B00, then there exists n ∈N such that bℓn ∈B′ (because B′ is open), so b ∈B′ by what we just proved.
We show that the map B′/ϖ1/ℓB′ →B′/ϖB′, b 7− →bℓis surjective. Let b ∈B′. As the map B0/ϖ1/ℓB0 →B0/ϖB0, c 7− →cℓis surjective, there exists c0 ∈B0 such that ϖ1/ℓb ∈cℓ+ϖB0.
Let d = ϖ−1/ℓ2c. Then dℓ= ϖ−1/ℓcℓ∈b + ϖ1−1/ℓB0 ⊂B′, so, as B′ in closed under taking ℓth roots, d ∈B′. Write b = dℓ+ ϖ1−1/ℓb′, with b′ ∈B0. Then there exists d′ ∈B0 such that ϖ1−1/ℓb′ ∈(d′)ℓ+ ϖB0, and we finally get b ∈(d + d′)ℓ+ ϖB′.
To finish the proof, we just need to show that B′ = B0. The inclusion B′ ⊂B0 follows from the fact that B′ is bounded and from proposition II.1.2.4. Conversely, let b ∈B0. Then, for every n ∈N, we have ϖ1/ℓnb ∈B00 ⊂B′; so b ∈B′.
Lemma V.1.5.11. Let A be a complete Tate ring and ϖ be a pseudo-uniformizer of A. Let B be an open subring of A that is ϖ-adically separated and complete. Then B is bounded in A.
Proof. By corollary II.1.1.8(iii), we can choose a ring of definition A0 of A such that A0 ⊂B.
We may assume, after replacing ϖ by a power, that ϖ is in A0. Note that ϖ is not a zero divisor in B, because it is invertible in A. By remark II.2.5.3, the ring A′ = B[ 1 ϖ], with the topology for which B is a ring of definition, is a complete Tate ring with pseudo-uniformizer ϖ. Of course, as a ring, A′ is canonically isomorphic to A. Moreover, the obvious isomorphism A ∼ →A′ is continuous, because B is open in A. By the open mapping theorem (theorem II.4.1.1), this map is open, which means that A ∼ →A′ is an isomorphism of topological rings, hence that B is a ring of definition of A.
Using the proposition, we can (almost) generalize proposition V.1.2.8.
Corollary V.1.5.12. . Let (A, A+) be a perfectoid Huber pair, and let ϖ ∈A+ be a pseudo-uniformizer of A. We denote by A+⟨X1/ℓ∞⟩the ϖ-adic completion of the ring S n≥0 A+[X1/ℓn], and we set A⟨X1/ℓ∞⟩= A+⟨X1/ℓ∞⟩[ 1 ϖ].
160 V.1 Perfectoid Tate rings Then A⟨X1/ℓ∞⟩is perfectoid, with ring of power-bounded elements (A⟨X1/ℓ∞⟩)0 al-most equal to A+⟨X1/ℓ∞⟩, and its tilt is canonically isomorphic to A♭⟨X1/ℓ∞⟩.
Moreover, A+⟨X1/ℓ∞⟩is a ring of integral elements in A⟨X1/ℓ∞⟩and the tilt of (A⟨X1/ℓ∞⟩, A+⟨X1/ℓ∞⟩) is (A♭⟨X1/ℓ∞⟩, A♭+⟨X1/ℓ∞⟩).
Proof. We may assume that there is a pseudo-uniformizer ϖ♭∈A♭+ of A♭such that ϖ = (ϖ♭)♯, and that ϖℓdivides ℓin A+. Then we may apply proposition V.1.5.10 to B = A⟨X1/ℓ∞⟩and to its open subring B0 = A+⟨X1/ℓ∞⟩, and we get that B is perfectoid and B0 = (B0)∗.
As B0/ϖℓB0 = S n≥0(A+/ϖℓA+)[X1/ℓn], we have a canonical isomorphism B0/ϖℓB0 ≃S n≥0 A♭+/(ϖ♭)ℓA♭+[X1/ℓn].
By remark V.1.2.3, this extends to an isomor-phism B♭ ∼ →A♭⟨X1/ℓ∞⟩sending B♭+ to A♭+⟨X1/ℓ∞⟩.
Corollary V.1.5.13. (See lemma 9.2.3 of .) Let (A, A+) be a perfectoid Huber pair of char-acteristic ℓ, let ϖ be a pseudo-uniformizer of A, let f1, . . . , fn, g ∈A+ with fn = ϖN (N ∈N) and U = R f1,...,fn g . Then : (i) Let A+⟨ fi g 1/ℓ∞ ⟩be the ϖ-adic completion of the subring A+[ fi g 1/ℓ∞ ] of A[ 1 g]. Then A+⟨ fi g 1/ℓ∞ ⟩is ϖ-torsionfree, ϖ-adically separated and complete, and perfect.
(ii) The canonical A+-algebra map ψ : A+[X1/ℓ∞ i ] → A+[ fi g 1/ℓ∞ ] sending X1/ℓm i to fi g 1/ℓm is surjective with kernel containing and almost equal to I := (g1/ℓmX1/ℓm i −f 1/ℓm i , 1 ≤i ≤n, m ∈N).
(iii) Let X = Spa(A, A+). The Tate ring OX(U) is perfectoid, and OX(U)0 = A+⟨ fi g 1/ℓ∞ ⟩∗ (in particular, these subrings of OX(U) are almost equal).
Proof. Remember that A and A+ are perfect (see proposition V.1.1.8).
(i) This ring is the ϖ-adic completion of the perfect and ϖ-torsionfree ring A+[ fi g 1/ℓ∞ ].
(ii) It is clear that ψ is surjective and that I ⊂Ker ψ. We have A+[X1/ℓ∞ i ][ 1 ϖ] = A[X1/ℓ∞ i ] and A+[ fi g 1/ℓ∞ ][ 1 ϖ] = A[ 1 g] (because fn = ϖn).
We claim that Ker ψ[ 1 ϖ] = I[ 1 ϖ].
Indeed, let f ∈ Ker ψ[ 1 ϖ].
We may assume that f ∈ A[X1, . . . , Xn], and we want to show that f ∈(gXi −fi) = (gX1 −f1, . . . , gXn−1 −fn−1, ϖ−NgXn −1).
We have an isomorphism A[X1, . . . , Xn]/(ϖ−NgXn −1) ≃A[g−1][X1, . . . , Xn−1] and the ideal (gXi −fi) to I′ := (X1 −f1g−1, . . . , Xn−1 −fn−1g−1). This ideal I′ is clearly the 161 V Perfectoid algebras kernel of the map A[g−1][X1, . . . , Xn−1] →A[g−1] induced by ψ, so the image of f in A[g−1][X1, . . . , Xn−1] is in I′, which is what we wanted.
Now let B = A+[X1/ℓ∞ i ]/I. It is clear from the definition of I that B is perfect, and ψ induces a surjective map ψ : B →A+[ fi g 1/ℓ∞ ] that is an isomorphism after inverting ϖ.
In particular, Ker(ψ) is ϖ∞-torsion, so it is almost zero by lemma V.1.5.14, and we are done.
(iii) By (i) and proposition V.1.1.3(i), the open subring A+⟨ fi g 1/ℓ∞ ⟩of OX(U) satisfies the conditions of proposition V.1.5.10. The result follows immediately from this proposition.
Lemma V.1.5.14. Let (A, A+) be a perfectoid Huber pair of characteristic ℓand B a perfect A+-algebra. Then the ϖ∞-torsion in B is almost zero.
Proof. Let N = {b ∈B | ∃n ∈Z≥1, ϖnb = 0}. We want to show that N is almost zero. Let b ∈N, and let n ∈Z≥1 such that ϖnb = 0. Let r ∈N. As B is perfect, there exists c ∈B such that b = cℓr, and then (ϖn/ℓrc)ℓr = 0, so, using the fact that B is perfect again, ϖn/ℓrc = 0, which finally gives ϖn/ℓrb = 0. By proposition V.1.5.4, this implies that N is almost 0.
Corollary V.1.5.15. (Lemma 9.2.5 of .) Let (A, A+) be a perfectoid Huber pair, and X = Spa(A, A+). Choose a pseudo-uniformizer ϖ of A that is of the form (ϖ♭)♯for ϖ♭a pseudo-uniformizer of A♭, and such that ϖ divides ℓin A+. Let f1, . . . , fn, g ∈A+ such that fn = ϖN for some N ∈N. Suppose that we have a1, . . . , an, b ∈A♭with a♯ i = fi, b♯= g and an = (ϖ♭)N. Let U = R f1,...,fn g ⊂X and U ♭= R a1,...,an b ⊂X♭, so that U is the preimage of U ♭by the map X →X♭, x 7− →x♭. Then : (i) Let B0 := A+⟨ fi g 1/ℓ∞ ⟩be the ϖ-adic completion of the subring A+[ fi g 1/ℓ∞ ] of A[ 1 g].
Then B0 is ϖ-torsionfree, ϖ-adically separated and complete, and the map B0/ϖ1/ℓB0 →B0/ϖB0, b 7− →bℓis bijective.
(ii) The canonical A+-algebra map ψ : A+[X1/ℓ∞ i ] → A+[ fi g 1/ℓ∞ ] sending X1/ℓm i to fi g 1/ℓm is surjective with kernel containing and almost equal to I := (g1/ℓmX1/ℓm i −f 1/ℓm i , 1 ≤i ≤n, m ∈N).
(iii) The Tate ring OX(U) is perfectoid, and OX(U)0 = A+⟨ fi g 1/ℓ∞ ⟩∗(in particular, these subrings of OX(U) are almost equal).
(iv) The tilt of (OX(U), OX(U)+) is canonically isomorphic to (OX♭(U ♭), OX♭(U ♭)+).
162 V.1 Perfectoid Tate rings Proof. We show (i). The ring B0 is ϖ-torsionfree and ϖ-adically separated and complete by definition. Also, if B′ 0 := A♭+⟨ ai b 1/ℓ∞ ⟩, then we have B0/ϖB0 ≃B′ 0/ϖB′ 0, and B′ 0 is perfect; this gives the last statement of (i).
It is clear that ψ is surjective and I ⊂ Ker ψ.
Let P0 = A+[X1/ℓ∞ i ]/I and a0 : P0 →A+[ fi g 1/ℓ∞ ] be the map induced by ψ. By definition of O+ X(U) (and lemma IV.1.3.8), we have a canonical morphism b0 : A+[ fi g 1/ℓ∞ ] →O+ X(U). Let (S, S+) be the per-fectoid Huber pair over (A, A+) that we get by untilting the perfectoid pair (OX♭(U ♭), O+ X♭(U ♭)) over (A♭, A♭+) (see corollary V.1.4.6). If i ∈{1, . . . , n}, then b divides ai in O+ X♭(U ♭), so b♯= g divides a♯ i = fi in S+, so the map Spa(S, S+) →Spa(A, A+) factors through U, and the univer-sal property of U (see proposition III.6.1.1(ii)) implies that the map (A, A+) →(S, S+) extends to a map (OX(U), O+ X(U)) →(S, S+). We denote the map O+ X(U) →S+ by c, and we write d0 = c ◦b0. The maps a0 : P0 →A+[ fi g 1/ℓ∞ ] and d0 : A+[ fi g 1/ℓ∞ ] →S+ extends by continuity to maps a : P →B0 = A+⟨ fi g 1/ℓ∞ ⟩and d : B0 →S+ between the ϖ-adic com-pletions (we use the fact that S+ is complete). As a0 is surjective, a is also surjective (see 25, Lemma 0315). By corollary V.1.5.13, the map d0 ◦a0 modulo ϖ is almost an isomorphism.
By lemma V.1.5.16, Ker a0 is almost zero, which proves (ii).
Moreover, as d0 ◦a0 and d ◦a are equal modulo ϖ, d ◦a modulo ϖ is also almost an isomor-phism. On the other hand, as d modulo ϖ is surjective, so is d (see 25, Lemma 0315), so d ◦a is also surjective. By lemma V.1.5.16 again, we get that d ◦a is almost an isomorphism.
As a is surjective and Ker a ⊂Ker(d ◦a), a is also almost an isomorphism, hence so is d. In particular (see remark V.1.5.3), the map d[ 1 ϖ] : B0[ 1 ϖ] →OX(U) →S is an isomorphism. As the first map is injective by definition of B0, both maps are isomorphisms. Now (iii) follows immediately from proposition V.1.5.10.
We finally prove (iv).
We have already seen that we have a canonical map (OX(U), O+ X(U)) →(S, S+), and we just proved that OX(U) →S is an isomorphism; this isomorphism then identifies O+ X(U) to a subring of S+. So it suffices to construct a contin-uous morphism of A+-algebras S+ →O+ X(U). Let (A♭, A♭+) →(T, T +) be the tilt of the morphism (A, A+) →(OX(U), O+ X(U)). As O+ X(U) ⊃B0, we have fi g ∈O+ X(U) for every i, so ai b ∈T + for every i. (This is clear on the formula for the untilting given after corol-lary V.1.4.6.) So Spa(T, T +) →Spa(A♭, A♭+) factors through U ♭, and the universal prop-erty of U ♭gives a morphism (OX♭(U ♭), O+ X♭(U ♭)) →(T, T +), which untilts to a morphism (S, S+) →(OX(U), O+ X(U)).
Lemma V.1.5.16. We use the notation of corollary V.1.5.15. Let α : M →N be a A+-module map. Suppose that α is almost surjective, that the induced map M/ϖM →N/ϖN is almost an isomorphism, that M is ϖ-adically separated and that N is ϖ-torsionfree. That α is almost an isomorphism.
163 V Perfectoid algebras Proof. We want to check that Coker α and Ker α are almost zero. This is true for Coker α by assumption.
Let L = Ker α.
As N is ϖ-torsionfree, L/ϖL is the kernel of the map M/ϖM →N/ϖN induced by α, so it is almost zero. This implies that, for every n ∈N, the A+-module L/ϖnL is almost zero. Let x ∈L. If r ∈N, then the image of ϖ1/ℓrx in L/ϖnL is almost zero for every n ∈N, which means that ϖ1/ℓrx ∈T n≥0 ϖnL. As M is ϖ-adically separated, we have T n≥0 ϖnL ⊂T n≥0 ϖnM = 0, so ϖ1/ℓrx = 0. This shows that L is almost zero.
V.1.6 Tilting and the adic spectrum Let (A, A+) be a perfectoid Huber pair.
By proposition V.1.2.7, we have a map (.)♭: Cont(A) →Cont(A♭) sending a continuous valuation |.| on A to the continuous valu-ation a 7− →|a|♭= |a♯| on A♭. By the formula for A♭+ in proposition V.1.2.5, this map sends Spa(A, A+) to Spa(A, A♭+).
Theorem V.1.6.1. (See corollary 6.7(ii),(iii) of , theorem 3.12 of .) Let (A, A+) be a perfectoid Huber pair. Then the map X := Spa(A, A+) →X♭:= Spa(A♭, A♭+), x 7− →x♭is a homeomorphism identifying rational domains of X and X♭, and, for every rational domain U of X, the Huber pair (OX(U), O+ X(U)) is perfectoid with tilt (OX♭(U ♭), O+ X♭(U ♭)).
Moreover, for every x ∈X, the completed residue field (κ(x), κ(x)+) is perfectoid, with tilt (κ(x♭), κ(x♭)+).
Applying theorem IV.1.1.5, we immediately get the following corollary.
Corollary V.1.6.2. Let (A, A+) be a perfectoid Huber pair, and let X = Spa(A, A+). Then OX is a sheaf, and, for every rational domain U of X and every i ≥1, we have Hi(U, OX) = 0.
Proof of theorem V.1.6.1. We choose pseudo-uniformizers ϖ and ϖ♭of A and A♭such that ϖ = (ϖ♭)♯and that ϖ divides ℓin A+. We write f for the map (.)♭: X →X♭.
(1) Let t1, . . . , tn, s ∈A♭such that (t1, . . . , tn) = A♭, and let V = R t1,...,tn s . By lemma V.1.6.6, V does not change if we add a power of ϖ♭to t1, . . . , tn, so we may assume that tn = (ϖ♭)N for some N ∈N. Then t♯ n = ϖN, so (t♯ 1, . . . , t♯ n) = A, and we clearly have f −1(V ) = R t♯ 1,...,t♯ n s♯ . In particular, f is continuous.
(2) We show that f is surjective. Let y ∈X♭, and let (L, L+) = (κ(y), κ(y)+). Then L is a complete non-Archimedean field (because X♭only has analytic points). By definition, L is the completion of K(x) := Frac(A♭/ supp(x)). As A♭is perfect and supp(x) is a prime ideal of A♭, the quotient A♭/ supp(x) is a perfect ring, so K(x) is also perfect. As K(x)0 is a perfect ring of definition, we can apply lemma IV.1.3.9, which implies that L is perfect.
164 V.1 Perfectoid Tate rings So, by proposition V.1.1.8, L is a perfectoid field. We get a morphism of perfectoid Huber pairs (A♭, A♭+) →(L, L+). By corollary V.1.4.6, this untilts to a morphism of perfectoid Huber pairs (A, A+) →(K, K+), and K is a perfectoid field by proposition V.1.2.9. Let x be the image by Spa(K, K+) →Spa(A, A+) of the unique closed point of Spa(K, K+).
Then |.|x is the composition of (A, A+) →(K, K+) and of the rank 1 valuation |.| giving the topology of K, so |.|x♭is the composition of (A♭, A♭+) →(L, L+) and of |.|♭. But |.|♭ is a continuous rank 1 valuation on L, so, by corollary II.2.5.11, it has to be equivalent to the continuous rank 1 valuation defining the topology of L, which means that x♭= y.
(3) Let f1, . . . , fn, f0 ∈A such that (f1, . . . , fn) = A, and let U = R f1,...,fn f0 . Again by lemma V.1.6.6, we may assume without changing U that fn = ϖN for some N ∈N. By corollary V.1.6.5 with ε some fixed number in (0, 1), there exist elements t0, . . . , tn−1, s of A♭such that, for every i ∈{0, . . . , n −1} and every x ∈X, we have |fi −t♯ i|x ≤|ϖ|1−ε x max(|fi|x, |ϖ|N+1 x ).
In particular, we get max(|fi|x, |ϖ|N x ) = max(|t♯ i|x, |ϖ|N x ), and, if |ϖN|x ≤|f0|x (for example if x ∈U) of if |ϖN|x ≤|t♯ 0|x, then |f0|x = |t♯ 0|x. It is easy to deduce from these inequalities that R t♯ 1,...,t♯ n t♯ 0 = U, so that U = f −1(R t1,...,tn t0 ).
(4) Let x, y ∈X with x ̸= y. As X is T0, we may assume that there exists a rational domain U of X such that x ∈U and y ̸∈U. By (2), there exists a rational domain V of X♭such that U = f −1(V ), and then we have f(x) ∈V and f(y) ̸∈V , and in particular f(x) ̸= f(y).
So f is injective.
(5) The statement about OX(U) and its tilt (for U ⊂X a rational domain) follows immediately from corollary V.1.5.15. (Multiplying all the equations of a rational domain by the same power of ϖ doesn’t chage the rational domain, so we may always assume that all these equations are in A+.) (6) Finally, the last statement was proved in (2).
Remark V.1.6.3. In remark 9.2.8 of , Bhatt gave a proof that f is a homeomorphism that does not use teh approximation lemma (corollary V.1.6.5). We give it here for fun. Step (1) and (2) are as in the proof of the theorem above, and then the next steps are : (3’) As in (5) of the proof above, by corollary V.1.5.15, if V ⊂X♭is a rational do-main and U = f −1(V ), then OX(U) is perfectoid and (OX♭(V ), O+ X♭(V )) is the tilt of (OX(U), O+ X(U)). As, as f is surjective, we have V = f(U).
(4’) Let f ∈A+, and let U = R f,ϖ ϖ . Choose any g ∈A♭+ such that g♯= f modulo ϖA+ (such a g always exists), and let V = R g,ϖ♭ ϖ♭ . Then we have U = f −1(V ). Indeed, if 165 V Perfectoid algebras x ∈X, then x ∈U if and only if |f|x ≤|ϖ|x; by the strong triangle inequality, this is equivalent to |g♯|x ≤|(ϖ♭)♯|x, i.e. to |g|f(x) ≤|ϖ♭|f(x), i.e. to f(x) ∈V .
(5’) Let f ∈A+. For n ∈Z≥1, we set Un = R f,ϖn ϖn . We claim that, for every n ≥1, there exists a rational subset Vn of X♭such that Un = f −1(Vn). We prove this by induction on n.
If n = 1, this is (4’). Suppose the result known for some n ≥1. Then, applying (3’) to Un shows that Vn = Spa(OX(Un)♭, O+ X(Un)♭). As Un+1 is the rational domain R fϖ−n,ϖ ϖ of Un, we can apply (4’) to get a rational domain Vn+1 of Vn such that Un+1 = f −1(Vn+1).
Finally, by corollary III.4.3.2, Vn+1 is also a rational domain in X♭, so we have proved the claim for Un+1.
(6’) Let f ∈A+ and g ∈A♭+ such that f = g♯modulo ϖA+, and let ε ∈Z[ 1 ℓ] ∩R>0. Then, if U = R ϖ1−ε f and V = R (ϖ♭)1−ε g , we have U = f −1(V ). Indeed, let x ∈X. Then x ∈U if and only if |ϖ|x ≤|ϖ|ε x|f|x, and again the strong triangle inequality implies that this is equivalent to |ϖ|x ≤|ϖ|ε x|g♯|x, i.e. to f(x) ∈V .
(7’) Let f ∈A+, n ∈Z≥1 and c ∈Z[ 1 ℓ] ∩(0, 1).
We claim that the rational domain U = R ϖnc f is the preimage by f of a quasi-compact open subset of X♭. Indeed, write U = Tn r≥1 Ur, where Ur = {x ∈X | |ϖnc−(r−1)c|x ≤|f|x ≤|ϖnc−rc|x} ⊂U ′ r = {x ∈X | ||f|x ≤|ϖnc−rc|x}.
The subsets Ur and U ′ r are rational domains of X, and, by (5’), there exists a rational domain V ′ r of X♭such that U ′ r = f −1(V ′ r). To show that Ur is the preimage of a rational domain of X♭, it suffices (thanks to (3’)) to show that it is the preimage of a rational domain Vr of V ′ r. Let g = f ϖnc−rc; then g ∈OX(U ′ r), and Ur = {x ∈U ′ r | |ϖ|c x ≤|g|x}, so the existence of such a Vr follows from (6’). Finally, U is the preimage of the quasi-compact open subset Sn r=1 Vr of X♭.
(8’) Let f, g ∈A+ and N ∈N, and let U = R f,ϖN g . We claim that U is the preimage of a quasi-compact open subset of X♭. Let U ′ = R ϖN g . By (7’), there exists a quasi-compact open subset V ′ of X♭such that U ′ = f −1(V ′). Write V ′ = S i∈I V ′ i , with I finite and the V ′ i rational domains of X♭, and let U ′ i = f −1(V ′ i ); note that the U ′ i are rational domains of X by (1). Let h = fϖN g . Then h ∈OX(U ′), so we can take its image in each OX(U ′ i). Let Ui be the rational domain R h,ϖN ϖN in U ′ i; by (5’), there exists a rational domain Vi of V ′ i such that Ui = f −1(Vi). By corollary III.4.3.2, Ui (resp. Vi) is a rational domain in X (resp. X♭). Also, we clearly have U = S i∈I Ui, so U is the preimage of the quasi-compact open subset S i∈I Vi of X♭.
(9’) Let f1, . . . , fn, g ∈A such that (f1, . . . , fn) = A, and let U = R f1,...,fn g . After multi-plying f1, . . . , fn, g by the same power of ϖ, we may assume (without changing U) that f1, . . . , fn, g ∈A+. Also, by lemma V.1.6.6, we may assume that fn is of the form ϖN.
166 V.1 Perfectoid Tate rings Then U = Tn−1 i=1 Ui, where Ui = R fi,ϖN g . By (8’), there exists quasi-compact open sub-sets V1, . . . , Vn−1 of X♭such that Ui = f −1(Vi) for i ∈{1, . . . , n−1}. Then V = Tn−1 i=1 Vi is a quasi-compact open subset of X♭, and U = f −1(V ).
(10’) Let x, y ∈X with x ̸= y. As X is T0, we may assume that there exists a rational domain U of X such that x ∈U and y ̸∈U. By (2), there exists a quasi-compact open subset V of X♭such that U = f −1(V ), and then we have f(x) ∈V and f(y) ̸∈V , and in particular f(x) ̸= f(y). So f is injective.
The main technical ingredient is the following approximation lemma.
Proposition V.1.6.4. (Lemma 6.5 of .) Let (K, K+) be a perfectoid Huber pair. We do not assume that K is a field. 1 Let (A, A+) = (K⟨X1/ℓ∞ 1 , . . . , X1/ℓn n ⟩, K+⟨X1/ℓ∞ 1 , . . . , X1/ℓn n ⟩), and let f ∈A+ be a homogeneous element of degree d ∈Z[ 1 ℓ]. For every c ∈R≥0 and every ε ∈R>0, there exists gc,ε ∈A+♭= K+♭⟨X1/ℓ∞ 1 , . . . , X1/ℓ∞ n ⟩(see corollary V.1.5.12) homogeneous of degree d such that, for every x ∈X = Spa(A, A+), |f −g♯ c,ε|x ≤|ϖ|1−ε x max(|f|x, |ϖ|c x).
Proof. Fix f and ε ∈Z[ 1 ℓ] ∩(0, 1). We also fix a ∈Z[ 1 ℓ] such that 0 < a < ε. If c ≤c′, then the result for c′ implies the result for c. So it suffices to prove the result for c = ar, with r ∈N. We show by induction on r that, for every r ∈N, there exists εr > 0 and gr ∈K+♭⟨X1/ℓ∞ 1 , . . . , X1/ℓ∞ n ⟩homogeneous of degree d such that, for every x ∈X, |f −g♯ r|x ≤|ϖ|1−ε+εr x max(|f|x, |ϖ|ar x ).
(This implies the desired inequality, because |ϖ|x < 1 for every x ∈X.) If r = 0, we take ε0 = 0 and take for g0 any element of A♭+ such that g♯ 0 = f modulo ϖA+.
Suppose that r ≥0 and that we have found εr and gr. Decreasing εr only makes the inequality more true, so we may assume that εr ∈Z[ 1 ℓ] and εr ≤ε −a. Let X♭= Spa(A♭, A♭+), and let U ♭ r ⊂X♭be the rational domain R gr,(ϖ♭)ra (ϖ♭)ra = {x ∈X♭| |gr|x ≤|ϖ♭|ra x }.
Then Ur = f −1(U ♭ r) is the rational domain R g♯ r,ϖra ϖra = R f,ϖra ϖra of X. Let h = f −g♯ r.
By the condition on gr, we have h ∈ϖra+1−ε+εrO+ X(Ur).
By corollary V.1.5.15(iii), the subrings OX(Ur)◦and A+⟨ g♯ r ϖra 1/ℓ∞ ⟩of OX(Ur) are almost equal, so ϖ−ra−1+ε−εrh is al-most an element of the second one, which means that we can find εr+1 ∈Z[ 1 ℓ] such that 0 < εr+1 < εr and that ϖ−ra−1+ε−εr+1h is in A+⟨ g♯ r ϖra 1/ℓ∞ ⟩. As h and g♯ r are homoge-neous of degree d, ϖ−ra−1+ε−εr+1h is in the ϖ-completion of L i∈Z[ 1 ℓ]∩[0,1] g♯ r ϖra i A+ d−di, where 1Lemma 6.5 of makes this assumption, but the proof seems to work in general.
167 V Perfectoid algebras A+ d−di is the set of elements of A+ that are homogeneous of degree d −di. So we can write h as a convergent sum ϖra+1−ε+εr+1 P i∈Z[ 1 ℓ]∩[0,1] g♯ r ϖra i hi, where hi ∈A+ d−di (this means that I := {i ∈Z[ 1 ℓ] ∩[0, 1] | hi ̸= 0} is finite and countable, and that hϕ(n) →0 as n →0 for every bijection ϕ : N ∼ →I). For every i, choose si ∈A♭+ d−di such that hi −s♯ i ∈ϖA+.
We may choose the si such that the sum s := P i∈Z[ 1 ℓ]∩[0,1] gr (ϖ♭)ra i si converges, and we set gr+1 = gr + (ϖ♭)ra+1−ε+εr+1s. We claim that, for every x ∈X, we have |f −g♯ r+1|x ≤|ϖ|1−ε+εr+1 x max(|f|x, |ϖ|a(r+1) x ).
Let x ∈X. First assume that |f|x > |ϖ|ar x . Then, by the induction hypothesis, we have |g♯ r|x = |fr|x > |ϖ|ra x , so it suffices to show that, for every i ∈Z[ 1 ℓ] ∩[0, 1], we have (ϖ♭)ra+1−ε+εr+1 gr (ϖ♭)ra i si ♯ x ≤|ϖ|1−ε+εr+1 x |f|x.
As |s♯ i|x ≤1, this follows from |(gi r)♯|x ≤|f|x, which holds because i ∈[0, 1]. Now we assume that |f|x ≤|ϖ|ar x , which implies that |g♯ r|x ≤|ϖ|ar x . We claim that then |f −g♯ r+1|x ≤|ϖ|a(r+1)+1−ε+εr+1 x , which clearly implies the result. As ar+1 > a(r+1)+1−ε+εr+1 (because εr+1 < εr < ε−a), it is enough to show that f −g♯ r+1 ∈ϖar+1O+ X(Ur). Note that gr+1 (ϖ♭)ar = gr (ϖ♭)ar + X i∈Z[ 1 ℓ]∩[0,1] (ϖ♭)1−ε+εr+1 gr (ϖ♭)ar i si, and that all the terms of the sum in the right hand side are in O+ X♭(U ♭ r). So we get an equality g♯ r+1 ϖar = g♯ r ϖar + X i∈Z[ 1 ℓ]∩[0,1] ϖ1−ε+εr+1 g♯ r ϖar i hi in O+ X(Ur) modulo ϖ. If we multiply this equality by ϖar, we get f −g♯ r+1 = f −g♯ r −h modulo ϖ1+raO+ X(Ur). By the choice of h, f −g♯ r −h ∈ϖ1+raO+ X(Ur), so f −g♯ r+1 ∈ϖ1+raO+ X(Ur), and this implies the desired inequality.
Corollary V.1.6.5. (See corollary 6.7(i) of .) Let (A, A+) be a perfectoid Huber pair, and let X = Spa(A, A+). Fix a pseudo-uniformizer ϖ ∈A+ of A such that ϖ = (ϖ♭)♯for ϖ♭∈A♭+.
Let c ∈R≥0, ε > 0 and f ∈A. Then there exists gc,ε ∈A♭such that, for every x ∈X, we have |f −g♯ c,ε|x ≤|ϖ|1−ε x max(|f|x, |ϖ|c x).
168 V.2 Perfectoid spaces Note that, if we take ε < 1, then the inequality of the proposition implies that, for every x ∈X, we have max(|f|x, |ϖ|N x ) = max(|g♯ c,ε|x, |ϖ|N x ).
Proof. After increasing c (and multiplying f by a power of ϖ), we may assume that c ∈N and f ∈A+. As A♭+/ϖ♭A♭+ = A+/ϖA+, we can find g0, . . . , gc ∈A♭+ and fc+1 ∈A+ such that f = g♯ 0 + ϖg♯ 1 + . . . + ϖcg♯ c + ϖc+1fc+1. It suffices to treat the case where fc+1 = 0.
Consider the continuous ring morphism ψ : B := A⟨T 1/ℓ∞ 0 , . . . , T 1/ℓ∞ c ⟩→A sending T 1/ℓm i to (g1/ℓm i )♯; this morphism sends B+ to A+. By assumption, we have f = psi(f ′), where f ′ = T0 + ϖT1 + . . . + ϖcTc. By proposition V.1.6.4, there exists g′ ∈A+♭⟨T 1/ℓ∞ 0 , . . . , T 1/ℓ∞ c ⟩ homogeneous of degree 1 such that, for every y ∈Spa(B, B+), we have |f ′ −(g′)♯|y ≤|ϖ|1−ε y max(|f ′|y, |ϖ|c y).
So, if g = ψ♭(g′), then g♯= ψ((g′)♯), and we have |f −g♯ c,ε|x ≤|ϖ|1−ε x max(|f|x, |ϖ|c x) for every x ∈Spa(A, A+).
Lemma V.1.6.6. Let (A, A+) be a Huber pair with A a Tate ring, let ϖ be a pseudo-uniformizer of A, and let f1, . . . , fn, g ∈A such that (f1, . . . , fn) = A. Then there exists N ∈N such that R f1, . . . , fn g = R f1, . . . , fn, ϖN g .
Proof. Write Pn i=1 aifi = 1, with a1, . . . , an ∈A. As A+ is open in A, there exists N ∈N such that ϖNai ∈A+ for every i ∈{1, . . . , n}. Then, if x ∈R f1,...,fn g , we have |ϖN|x = | n X i=1 ϖNaifi|x ≤max 1≤i≤n |fi|x ≤|g|x, hence x ∈R f1,...,fn,ϖN g . The other inclusion is obvious.
V.2 Perfectoid spaces Definition V.2.1. (Definition 6.15 of .) A perfectoid space is an adic space (definition III.6.5.3) that is locally isomorphic to Spa(A, A+), for (A, A+) a perfectoid Huber pair. We say that X is affinoid perfectoid if X = Spa(A, A+) for (A, A+) a perfectoid Huber pair. A morphism of perfectoid spaces is a morphism of adic spaces.
169 V Perfectoid algebras By theorem V.1.6.1, for every perfectoid Huber pair (A, A+), the space Spa(A, A+) is an affinoid perfectoid space (i.e. its structure presheaf is a sheaf). Theorem V.1.6.1 also has the following corollary.
Corollary V.2.2. (Proposition 6.17 of .) Every perfectoid space X has a tilt X♭, which is a perfectoid space over Fℓwith an isomorphism of topological spaces (.)♭: X →X♭, such that, for every affinoid perfectoid subspace U of X, if U ♭is the image of U in X♭, then the tilt of the pair (OX(U), O+ X(U)) is (OX♭(U ♭), O+ X♭(U ♭)).
Moreover, tilting induces an equivalence between the categories of perfectoid spaces over X and over X♭.
Unlike general adic spaces, perfectoid spaces admit fiber products.
Proposition V.2.0.1. (Proposition 6.18 of .) Let X →Z and Y →Z be two morphism of perfectoid spaces. Then the fiber product X ×Z Y exists in the category of adic spaces, and it is a perfectoid space.
Proof. We may assume that X, Y and Z are perfectoid affinoid, so X = Spa(A, A+), Y = Spa(B, B+) and Z = Spa(C, C+). As the maps C →A and C →B are automati-cally adic (because A, B and C are Tate rings, see proposition II.1.3.4), the completed tensor product D := \ A ⊗C B exists and is a Tate ring (see proposition II.3.2.1 and corollary II.3.1.9).
We take for D+ the completion of the integral closure of the image of A+ ⊗C+ B+ in D. Then Spa(D, D+) is the fiber product of X and Y over Z in the category of adic spaces, and it remains to prove that (D, D+) is a perfectoid pair. See the proof of proposition 6.18 of for details.
V.3 The almost purity theorem V.3.1 Statement Definition V.3.1.1. (Definition 7.1 of .) (i) A morphism (A, A+) →(B, B+) of Huber pairs is called finite ´ etale if B is a finite ´ etale A-algebra and has the corresponding canonical topology (see proposition II.4.2.2) and if B+ is the integral closure of A+ in A.
(ii) A morphism f : X →Y of adic spaces is called finite ´ etale if there is a cover of Y by open affinoid subsets V such that f −1(V ) is affinoid and the morphism (OY (V ), O+ Y (V )) →(OX(U), O+ X(U)) is finite ´ etale.
170 V.3 The almost purity theorem (iii) A morphism f : X →Y of adic spaces is called ´ etale if, for every x ∈X, there exists open neighborhoods U and V of x and f(x) and a commutative diagram U j / f|U W p V where j is an open embedding and p is finite ´ etale.
Remark V.3.1.2. The definition of an ´ etale morphism that we give here is not Huber’s definition (Huber’s definition is modelled on the definition for schemes, see definition 1.6.5 of ). How-ever, it is equivalent to it for adic spaces that are locally of finite type over a non-archimedean field (for example for adic spaces coming from rigid analytic varieties), by lemma 2.2.8 of , and it also gives a reasonable notion for adic spaces that are locally adic spectra of perfectoid pairs.
The main result of this section is the following : Theorem V.3.1.3. (Theorem 7.9 of .) Let (A, A+) be a perfectoid Huber pair, and let X = Spa(A, A+).
(i) The functor Y 7− →OY (Y ) induces an equivalence between the category of finite ´ etale maps of adic spaces Y →X and the category of finite ´ etale maps of rings A →B.
(ii) If A →B is a finite ´ etale map of rings and B+ is the integral closure of A in B, then B (with its canonical topology) is perfectoid and (B, B+) is a perfectoid Huber pair.
(iii) Tilting induces an equivalence between the categories of finite ´ etale Huber pairs over (A, A+) and over (A♭, A♭+).
This theorem will be proved in section V.3.7. It has an immediate corollary for ´ etale sites of perfectoid spaces.
Definition V.3.1.4. Let X be a perfectoid space. Then the ´ etale site X´ et of X is the category of ´ etale morphisms of perfectoid spaces Y →X, with the Grothendieck topology for which coverings are families (u : Yi →Y )i∈I such that Y = S i∈I ui(Yi).
Corollary V.3.1.5. If X is a perfectoid space, then tilting induces an isomorphism of sites X´ et ≃X♭ ´ et, and this isomorphism is functorial in X.
If (A, A+) →(B, B+) is a finite ´ etale map of perfectoid Huber pairs, we will need a way to describe the map of rings of integral elements A+ →B+. This map is not ´ etale, but we will see that it is almost ´ etale, for the correct definition of “almost”.
Definition V.3.1.6. (See sections 4.2 and 4.3 of .) Let (A, A+) be a perfectoid Huber pair, let B+ be a A+-algebra, and let M be a B+-module.
171 V Perfectoid algebras (i) We say that M is almost of finite presentation (or almost finitely presented) if, for every a ∈A00, there exists a finitely presented B+-module Ma and a morphism of B+-modules Ma →M whose kernel and cokernel are killed by a.
(ii) We say that M is almost projective if, for every B+-module N and every integer i ≥1, the B+-module Exti B+(M, N) is almost zero.
(iii) Suppose that M is a B+-algebra. We say that it is almost finite ´ etale (over B+) if it is almost of finite presentation, almost projective and if there exists e ∈(M ⊗B+ M)∗such that e2 = e, µ∗(e) = 1 and ker(µ)∗·e = 0, where µ : M ⊗B+ M →M is the multiplication map.
Remember that, by remark V.1.5.7, M∗is a (B+)∗-algebra in the situation of (iii).
Remark V.3.1.7.
(1) As in proposition V.1.5.4, we can require the conditions of definition V.3.1.6(i) only for fractional power of a well-chosen pseudo-uniformizer of A.
(2) We can define the abelian category of almost B+-modules by taking the quotient of the category of B+-modules by the Serre subcategory of almost zero B+-modules, and many of the “almost” notions have a natural interpretation in this category. However, a B+-module that is almost projective is in general not a projective object of the category of almost B+-modules; for example, B+ itself is almost projective, but it is not a projective almost B+-modules. (The category of almost modules has tensor products and internal Homs defined in the obvious way, and almost projectivity can be defined in the usual way using the internal Hom functor.) Remark V.3.1.8. The definition of almost finite ´ etale maps is motivated by the following result in ordinary commutative algebra : Let R →S be a locally free map of rings (i.e. let S be a flat R-algebra that is finitely presented as a R-module). Then R →S is ´ etale if and only if the map of rings µ : S ⊗R S →S, a ⊗b 7− →ab has a section, i.e. if and only there exists an idempotent e ∈S ⊗R S such that µ(e) = 1 and (Ker µ)e = 0.
Most of theorem V.3.1.3 will follow from the next result, which is slitghly more precise.
Theorem V.3.1.9. Let ϕ : (A, A+) →(B, B+) be a morphism of Huber pairs, with (A, A+) perfectoid. Choose a pseudo-uniformizer ϖ ∈A+ of A that is of the form ϖ = (ϖ♭)♯, with ϖ♭∈Aflat+ a pseudo-uniformizer of A♭.
(i) If ϕ is finite ´ etale, then B is also perfectoid.
(ii) Suppose that B is perfectoid. Then the following conditions are equivalent : (a) ϕ is finite ´ etale; (b) the A+-algebra B+ is almost finite ´ etale; (c) the A+/ϖA+-algebra B+/ϖB+ is almost finite ´ etale.
This theorem will be proved in section V.3.7.
172 V.3 The almost purity theorem V.3.2 The trace map and the trace pairing Let R →S be a finite locally free morphism of rings (i.e. S is a flat R-module of finite pre-sentation, or equivalently a finitely generated projective R-module). For every R-linear endo-morphism u of S, we can define its trace Tr(u) ∈R in the following way : There exists an affine covering (Spec(Ri))i∈I of Spec(R) such that S ⊗R Ri is a free Ri-module of finite rank for every i ∈I. If i ∈I, then u defines a Ri-linear endomorphism ui of S ⊗R Ri, so we can define Tr(ui) ∈Ri, and these elements glue to a global section Tr(u) of OSpec(R).
If a ∈S, left multiplication by a on S defines a R-linear endormophism on S, that we denote by ma. We get a R-linear morphism TrS/R : S →R, a 7− →Tr(ma).
Finally, the trace pairing is the R-linear morphism Tr : S ⊗R S →R, (a, b) 7− →TrS/R(ab).
Remember the following “well-known” result.
Theorem V.3.2.1. Let R →S be a finite locally free morphism of rings. Then the following conditions are equivalent : (i) The morphism R →S is ´ etale.
(ii) The trace pairing Tr : S ⊗R S →R is nondegenerate, i.e. the R-linear morphism S →HomR(S, R) that it defines by adjunction is an isomorphism (of R-modules).
(iii) There exists an idempotent e ∈S ⊗R S such that µ(e) = 1 and (Ker µ)e = 0, where µ : S ⊗R S →S is the multiplication. In other words, there exists an isomorphism of S-algebras S⊗RS ≃S×S′ (that does not preserve unit elements) such that µ : S⊗RS →S corresponds to the first projection.
We will also need the following results.
Lemma V.3.2.2. Let R →S be a finite locally free morphism of rings, and let R+ be a subring of R that is integrally closed in R and such that R = R+[ 1 ϖ], for some ϖ ∈R+. Let S+ be the integral closure of R+ in S. Then S = S+[ 1 ϖ].
Proof. Note that the rank of S as a R-module is a locally constant function on Spec(R). As Spec(R) is quasi-compact, we can write Spec(R) as a finite disjoint union of open and closed subschemes over which the rank of S is constant, and it suffices to prove the result over each of this subschemes. So we may assume that S has constant R-rank, say n.
Let a ∈S, and let f = Pn r=0(−1)rTr(∧rma)T r ∈R[T] be the characteristic polynomial of ma, where ∧rma : Vr R S →Vr R S is the rth exterior power of ma. By the Cayley-Hamilton theorem, we have f(a) = 0, so a ∈S+ if Tr(∧rma) ∈R+ for 0 ≤r ≤n. In general, as R = R+[ 1 ϖ], we may find m ∈N such that ϖmrTr(∧rma) = Tr(∧rmϖma) =∈R+ for every r ∈{0, . . . , n}, so ϖma ∈S+, and we are done.
173 V Perfectoid algebras Lemma V.3.2.3. Let R →S be a finite ´ etale map of rings, and let e ∈S ⊗R S be an idempotent as in theorem V.3.2.1(iii). Write e = Pn i=1 ai ⊗bi, with n a positive integer and ai, bi ∈S. Then, for every c ∈S, we have n X i=1 aiTrS/R(cbi) = n X i=1 biTrS/R(cai) = c.
Proof. Let µ : S ⊗R S →S be the multiplication, and let S′ = Ker µ. As (Ker µ)e = 0, we have (a ⊗1)e = (1 ⊗a)e (in S ⊗R S) for every a ∈S. We get an isomorphism of S-algebras S × S′ ∼ →S ⊗R S sending (a, b) to (a ⊗1)e + b, whose inverse sends x ∈S ⊗R S to (µ(x), x −(µ(x) ⊗1)e). Note that Pn i=1 biTrS/R(cai) = TrS⊗RS/S((c ⊗1)e), where we see S ⊗R S as a S-algebra using the map b 7− →b ⊗1. As the trace is additive on direct products of finite ´ etale S-algebras, we see that TrS⊗RS/S((c ⊗1)e) = TrS/S(c) + TrS′/S(0) = c. We prove the other equality in the same way (this time by using the map S →S ⊗R S, b 7− →1 ⊗b).
V.3.3 From almost finite ´ etale to finite ´ etale In this section, we fix a perfectoid Huber pair (A, A+), and a pseudo-uniformizer ϖ ∈A+ of A that has a compatible system (ϖ1/ℓn)n≥0 of ℓth power roots and such that ϖ divides ℓin A+.
Proposition V.3.3.1. We have a fully faithful functor from the category of almost finite ´ etale A+-algebras B0 (with morphisms taken up to almost equality) to the category of finite ´ etale (A, A+) pairs (B, B+), defined by sending B0 to the pair (B := B0[ 1 ϖ], B+), where B+ is the integral closure of A+ in B. Moreover, every pair (B, B+) in the essential image of this functor is perfectoid.
Proof. The functor is well-defined by lemma V.3.3.3. Suppose that B0 is an almost finite ´ etale A+-algebra, with image (B, B+). By lemma V.3.3.3 again, we have (B0)∗= B0, so we can almost recover B0 from (B, B+). This shows that the functor is fully faithful.
Lemma V.3.3.2. If M is an almost finitely presented A+-module, then M[ 1 ϖ] is a finitely pre-sented A-module. If moreover M is almost projective, then M[ 1 ϖ] is a projective A-module, and M is almost ϖ-torsionfree (i.e. the submodule M[ϖ∞] of ϖ-power torsion elements of M is almost zero).
Proof.
(1) We show that M[ 1 ϖ] is finitely presented as a A-module. By assumption, there exists a finitely presented A+-module N and a morphism of A+-modules N →M whose kernel and cokernel are ϖ-torsion. So we get an isomorphism N[ 1 ϖ] ≃M[ 1 ϖ], and N[ 1 ϖ] is a finitely presented A-module.
174 V.3 The almost purity theorem (2) We show that the A-module M[ 1 ϖ] is projective if M is almost projective. (See the proof of lemma 2.4.15 of .) Let ε ∈A00. As M is almost finitely presented, there exists ϕ : (A+)n →M such that ε Coker(ϕ) = 0. Let M ′ = Im(ϕ), ψ : An →M ′ be the induced surjection and j : M ′ →M be the inclusion. Then we have ε(M/M ′) = 0, so there exists a A+-linear map γ : M →M ′ such that j ◦γ : M →M is multiplication by ε. We have an exact sequence HomA+(M, (A+)n) ψ∗ →HomA+(M, M ′) →Ext1 A+(M, Ker ψ).
As M is almost projective, the third term is almost zero, so ψ∗is almost surjective. Let δ ∈A00. Then δγ ∈HomA+(M, M ′) is in the image of ψ, so there exists u : M →(A+)n such that δγ = ψ ◦u. In particular, composing by the inclusion j : M ′ →M, we see that (δε)idM = ϕ ◦u. If we take ε = ϖ1/ℓand δ = ϖ(ℓ−1)/ℓ, we see that we can find maps v : (A+)n →M and u : M →(A+)n such that v ◦u = ϖidM. But then ϖ−1v[ 1 ϖ] is a section of u[ 1 ϖ], so M[ 1 ϖ] is a direct factor of An, hence it is a projective A-module.
(3) We show that M is almost ϖ-torsionfree if M is almost projective. Let N = M[ϖ∞]. Let r ∈N. In (2), we have shown that we can find maps v : (A+)n →M and u : M →(A+)n such that v ◦u = ϖ1/ℓridM. If x ∈N, then u(x) = 0 because A+ is ϖ-torsionfree, so ϖ1/ℓrx = v(u(x)) = 0. As r was arbitrary, this shows that N is almost zero.
Lemma V.3.3.3. If B0 is an almost finite ´ etale A+-algebra, then B := B0[ 1 ϖ] is a finite ´ etale A-algebra, it is perfectoid for its canonical topology, (B0)∗= B0, and the integral closure B+ of A+ in B is a ring of integral elements of B.
Proof. We first reduce to the case where B0 is ϖ-torsionfree. Let J = B0[ϖ∞] be the ideal of ϖ-power torsion elements of B0. It suffices to show that J is almost zero. But this is the last statement of lemma V.3.3.2.
We already know that B is a finitely presented projective A-module by lemma V.3.3.2. We de-note the multiplication map on B0 by µ : B0 ⊗A+ B0 →B0. As B0 is almost finite ´ etale over A+, there exists an idempotent e ∈(B0 ⊗A+ B0)∗such that µ∗(e) = 1 and (Ker µ∗)e = 0. Note that B ⊗A B = (B0 ⊗A+ B0)[ 1 ϖ] = (B0 ⊗A+ B0)∗[ 1 ϖ], and that the multiplication ν : B ⊗A B →B is equal to µ[ 1 ϖ] and to µ∗[ 1 ϖ]. Let f be the image of e by the obvious map (B0⊗A+B0)∗→B⊗AB.
Then f is an idempotent, ν(f) = 1 and (Ker ν)f = 0. In particular, the map ν : B ⊗A B →B has a section, so it is flat. This means that the map A →B is weakly ´ etale, and we have already seen that it is of finite presentation. The fact that B is an ´ etale A-algebra now follows from [25, Lemma 0CKP].
In particular, we can put the canonical topology on B (as a A-module), which makes B into a complete topological A-algebra. Let u : M →B0 be a A+-module map with ϖ-torsion kernel and cokernel, and such that M is a finitely generated A+-module. We may assume that M is 175 V Perfectoid algebras ϖ-torsionfree. Then u induces an isomorphism M[ 1 ϖ] ∼ →B, and M with its ϖ-adic topology is an open subgroup of M[ 1 ϖ]. As ϖ−1B0 ⊂u(M) ⊂B0, this shows that B0 is open and bounded in B. Also, as B0 ⊂ϖ−1u(M), B0 also has the ϖ-adic topology. So B is a complete Tate ring.
By lemma V.3.3.4, the map a 7− →a is an almost isomorphism from B0/ϖ1/ℓB0 to B0/ϖB0.
So we can apply proposition V.1.5.10 to conclude that B is perfectoid and that B0 = (B0)∗.
It remains to show that B+ is an open subring of B (we already know that B+ ⊂B0, since B0 ⊂B0 and B0 contains the image of A+ in B). We have seen that there exists a finitely generated A+-submodule M of B such that M ⊂B0 ⊂ϖ−1M. Let b ∈B0. Then ϖbM ⊂M, so ϖb is integral over A+. This shows that ϖB0 ⊂B+, hence that B+ is open.
Lemma V.3.3.4. (Lemma 4.3.8 of .) 2 Let R →S be a weakly ´ etale map of Fp-algebras; this means that the maps R →S and S ⊗R S →S, a ⊗b 7− →ab are both flat. Then the diagram R FrobR/ R S FrobS / S (where FrobR and FrobS are the absolute Frobenius maps a 7− →ap) is a pushout square of rings. In particular, if R is perfect, so is S.
V.3.4 The positive characteristic case Proposition V.3.4.1. (Proposition 4.3.4 of .) Let (A, A+) be a perfectoid Huber pair of char-acteristic ℓ, and let ϖ ∈A+ be a pseudo-uniformizer of A. Let η : A+ →B0 be an integral map with B0 perfect. Suppose that the induced map η[ 1 ϖ] : A →B := B0[ 1 ϖ] is finite ´ etale. Then η is almost finite ´ etale.
Proof. Let J = B0[ϖ∞] be the ideal of ϖ-power torsion elements of B0. We claim that J is almost zero. Indeed, let b ∈J, and let n ∈N such that ϖnb = 0. Then, for every r ∈N, we have (ϖn/ℓrb1/ℓr)ℓr = 0, hence ϖn/ℓrb1/ℓr = 0 because B0 is perfect, hence ϖn/ℓrb = 0. By proposition V.1.5.4, this implies that J is almost zero. So replacing B0 by B0/J affects neither the hypothesis nor the conclusion, and we may assume that J = 0, i.e. that B′ 0 is ϖ-torsionfree.
Now let B′ be the integral closure of B0 in B. We show that B′ is almost equal to B0, which will allow us to replace B0 by B′. Let f ∈B′. Then the B0-module of B spanned by f N is finitely generated, so there exists r ∈N such that ϖrf n ∈B0 for every n ∈N. As B0 is perfect, this implies that ϖr/ℓnf ∈B0 for every n ∈N, so f ∈(B0)∗by proposition V.1.5.8(ii). So we 2Almostify this lemma.
176 V.3 The almost purity theorem have shown that B0 ⊂B′ ⊂(B0)∗, and the fact that B0 and B′ are almost equal follows from proposition V.1.5.8(i).
By the previous two paragraphs, we may assume that B0 is ϖ-torsionfree and integrally closed in B.
As A → B is finite ´ etale by assumption, we can find an idempotent e ∈B ⊗A B = (B0 ⊗A+ B0)[ 1 ϖ] such that µ(e) = 1 and (Ker µ)e = 0, where µ : B ⊗A B →B is the multiplication map (see remark V.3.1.8). Choose r ∈N such that ϖre ∈B0 ⊗A+ B0. As e is idempotent, we have eℓn = e for every n ∈N, hence e = e1/ℓn because B ⊗A B is perfect (by lemma V.3.3.4). As B0 ⊗A+ B0 is also perfect (by the same lemma) and injects in B ⊗A B, we see that (ϖre)1/ℓn = ϖr/ℓne ∈B0 ⊗A+ B0 for every n ∈N, so e ∈(B0 ⊗A+ B0)∗by proposition V.1.5.8(ii).
It remains to show that B0 is almost finitely presented and almost projective over A+. Let n ∈N. We just proved that ϖ1/ℓne ∈B0 ⊗A+ B0, so we can write ϖ1/ℓne = P i∈I ai ⊗bi, with I finite and ai, bi ∈B0. Consider the maps α : B →AI and β : AI →B defined by α(b) = (TrB/A(bai))i∈I and β((ci)i∈I) = P i∈I cibi. We claim that β◦α is equal to multiplication by ϖ1/ℓn. Indeed, let b ∈B. Then β(α(b)) = X i∈I biTrB/A(bai) = ϖ1/ℓnTr((b ⊗1)e), so the claim follows from lemma V.3.2.3.
Moreover, as B0 is integrally closed in B (and A+ is integrally closed in A), the map TrB/A : B →A sends B0 to A+ as B0, so α sends B0 to (A+)I. It is clear that β sends (A+)I to B0. As B0 and A+ are ϖ-torsionfree, the restrictions to α and β define maps α0 : B0 →(A+)I and β0 : (A+)I →B0 such that β0◦α0 is equal to multiplication by ϖ1/ℓn. Consider the sequence (A+)I γ0 →(A+)I β0 →B0, where γ0 = ϖ1/ℓmid −α0 ◦β0.
Then β0 ◦γ0 = 0, so we get a map Coker(γ0) →B0 whose kernel and cokernel are killed by ϖ1/ℓm.3 As Coker(γ0) is clearly finitely presented, this shows that B0 is an almost finitely presented A+-module. Let N be another A+-module, and let i ≥1 be an integer.
Then multiplication by ϖ1/ℓm on Exti A+(B0, N) factors as Exti A+(B0, N) β∗ 0 →Exti A+((A+)I, N) α∗ 0 →Exti A+(B0, N). As Exti A+((A+)I, N) = 0, this shows that multiplication by ϖ1/ℓm is 0 on Exti A+(B0, N). Hence Exti A+(B0, N) is almost zero.
Corollary V.3.4.2. (Theorem 4.3.6 of .) Let (A, A+) be a perfectoid Huber pair of character-istic ℓ, and let ϖ ∈A+ be a pseudo-uniformizer of A. Then the functor of proposition V.3.3.1 is an equivalence of categories.
In other words, we have an equivalence of categories from the category of finite ´ etale A-algebras to the category of almost finite ´ etale A+-algebras (where we identity two maps that are 3Check.
177 V Perfectoid algebras almost equal). This equivalence is given by sending a finite ´ etale map A →B to the integral closure B+ of A+ in B, and by sending an almost finite ´ etale A+-algebra B0 to the A-algebra B0[ 1 ϖ].
In particular, every finite ´ etale A-algebra is perfectoid for the canonical topology.
Proof. Let C (resp. C ′) be the category of finite ´ etale A-algebras (resp. almost finite ´ etale A+-algebras with almost equal maps identified). By lemma V.3.3.3, the formula B0 7− →B0[ 1 ϖ] define a functor from Ψ : C ′ →C , and every A-algebra in this essential image of this functor is perfectoid.
Let B be a finite ´ etale A-algebra, and let B+ be the integral closure of A+ in B. By lemma V.3.2.2, we have B = B+[ 1 ϖ]. By proposition V.3.4.1, B+ is an almost finite ´ etale A+-algebra.
So sending B to B+ defines a functor Φ : C →C ′.
It remains to show that the functors Φ and Ψ are mutually quasi-inverse. If B is a finite ´ etale A-algebra and B+ = Φ(B), then we have already seen that B = B+[ 1 ϖ] = Ψ(B+). Conversely, let B0 be an almost finite ´ etale A+-algebra, let B = B0[ 1 ϖ], and let B+ be the integral closure of A+ in B. We claim that (B0)∗= B+ ∗, which will give a functorial isomorphism between B0 and B+ in C ′. But this is proved in lemma V.3.3.3.
V.3.5 Deforming almost finite ´ etale extensions The main result of this section is the following : Theorem V.3.5.1. (Theorem 5.3.27 of .) Let (A, A+) be a perfectoid pair, and let ϖ ∈A+ be a pseudo-uniformizer of A. Then reduction modulo ϖ induces an equivalence of category from the category of almost finite ´ etale A+-algebras to the category of almost finite ´ etale A+/ϖA+-algebras.
The following definition is temporary. Eventually, we will prove that it is equivalent to defini-tion V.3.1.1(i).
Definition V.3.5.2. Let (A, A+) be a perfectoid Huber pair. A pair (B, B+) is called strongly finite ´ etale if : (i) B+ is an almost finite ´ etale A+-algebra; (ii) B+ is the integral closure of A+ in B.
Corollary V.3.5.3. Let (A, A+) be a perfectoid Huber pair. Any strongly finite ´ etale pair over (A, A+) is also perfectoid, and tilting induces an equivalence between the categories of strongly finite ´ etale pairs over (A, A+) and of finite ´ etale pairs over (A♭, A♭+).
178 V.3 The almost purity theorem Proof. Let ϖ♭∈A♭+ be a pseudo-uniformizer of A♭such that ϖ := (ϖ♭)♯is a pseudo-uniformizer of A dividing ℓin A+.
By proposition V.3.3.1, every strongly finite ´ etale pair over (A, A+) is perfectoid, and the category of strongly finite ´ etale pairs over (A, A+) is equivalent to the category of almost finite ´ etale A+-algebras; we have a similar result for (A♭, A♭+). By corollary V.3.4.2, every finite ´ etale pair over (A♭, A♭+) is strongly finite ´ etale. Finally, by theorem V.3.5.1, the category of almost ´ etale A+-algebras (resp. A♭+-algebras) is equivalent to the category of almost finite ´ etale A+/ϖA+-algebras (resp. A♭+/ϖ♭A♭+-algebras). So we get an equivalence between the category of strongly finite ´ etale pairs over (A, A+) and the category of finite ´ etale pairs over (A♭, A♭+), and this is the tilting equivalence by definition of tilting and by remark V.1.2.3.
V.3.6 The case of perfectoid fields Proposition V.3.6.1. (Proposition 3.2.10 of , proof due to Kedlaya.) Let K be a perfectoid field. If K♭is algebraically closed, then so is K.
Remember that we know that K♭is a perfectoid field by proposition V.1.2.9.
Proof. We may assume that K has characteristic 0. Let f(T) ∈K0[T] be a monic polynomial of degree d ≥1. We want to find a root of f(T) in K0. We will construct by induction a sequence (xn)n≥0 of elements of K0 such that, forr every n ∈N : (a) |f(xn)| ≤|ℓ|n; (b) if n ≥1, then |xn −xn−1| ≤|ℓ|(n−1)/d.
Condition (b) shows that (xn)n≥0 converges to some x ∈K0, and then (a) shows that f(x) = 0.
We set x0 = 0. Suppose that n ≥0 and that we have constructed x0, . . . , xn satisfying (a) and (b). We want to construct xn+1 satisfying the same properties. Write f(T + xn) = Pd i=0 biT i, with b0, . . . , bd ∈K0. We have bd = 1 by assumption. If b0 = 0, then f(xn) = 0, so we may take xm = xn for every m ≥n + 1. From now on, we assume that b0 ̸= 0. Let c = min{| b0 bj | 1 j , 1 ≤j ≤d, bj ̸= 0}.
As bd ̸= 0, we have c ≤|b0| 1 d ≤1. Let |.| be the rank 1 valuation giving the topology of K.
The value groups of |K×| and |K♭|♭are canonically isomorphic by construction of |.|♭. 4 As K♭ is algebraically closed, |K×| ≃|(K♭)×|♭is a Q-vector space, so there exists u ∈K× such that c = |u|. As c ≤1, we have u ∈K0. By definition of c, we have bi b0ui ∈K0 for every i, and there exists at least one i ≥1 such that bi b0ui ̸∈K00, i.e. such that bi b0ui is a unit in K0.
4Add a lemma ?
179 V Perfectoid algebras Choose t ∈K♭0 such that |t♯| = |ℓ|.
It is easy to see that K♭0/tK♭0 ≃K0/ℓK0.
5 Consider any lift g(T) ∈ K♭0[T] of the image of the polynomial Pd i=0 bi b0uiT i in (K0/ℓK0)[T] ≃(K♭0/tK♭0)[T]. By lemma V.3.6.2, there exists a unit y in K♭0 such that g(y) = 0.
We set xn+1 = xn + uy♯. We have to check that this xn+1 satisfies conditions (a) and (b). Firs, we have f(xn+1) = f(uy♯+ xn) = d X i=0 biui(yi)♯= b0 d X i=0 bi b0 ui(yi)♯ !
.
As y is a root of g(T), the sum between parentheses is equal to 0 modulo ℓ. So |f(xn+1)| ≤|b0||ℓ| = |f(xn)||ℓ| ≤|ℓ|n|ℓ| = |ℓ|n+1.
This proves (a). Moreover, as y is a unit (so |y♯| = 1), we have |xn+1 −xn| = |uy♯| = |u| = c ≤|b0| 1 d = |f(xn)| 1 d ≤|ℓ| n d .
This proves (b).
Lemma V.3.6.2. (Lemma 3.2.11 of .) Let K be a complete and algebraically closed non-archimedean field, and R = K0. Let f(T) ∈R[T] be a polynomial of degree e ≥1 such that the constant coefficient and at least one other coefficient of f(T) are units in R. Then f(T) has a root which is a unit of R.
Proof. Let m = K00 be the maximal ideal of R and k = R/m be its residue field. By the hy-pothesis, the image of f(T) in k[T] is a polynomial of degree ≥1 whose constant coefficient is a unit, so we can a pseudo-uniformizer ϖ of K such that the image of f(T) in R/ϖR[T] is a poly-nomial of degree ≥1 whose leading term and constant term are units in R/ϖR. We also write f : R[T] →R[T] for multiplication by f(T). Then the induced map R/ϖR[T] →R/ϖR[T] is finite free of degree ≥2, so, if R⟨T⟩is the ϖ-completion of R[T] and b f : R⟨T⟩→R⟨T⟩ is the extesion of f by continuity, the map b f is also finite free of degree ≥2. In particular, the ring S := R⟨T⟩/(f(T)) is a finite free R-algebra of dimension ≥2. As R is a henselian local ring, we can write S ≃Q i∈I Si, where I is finite and each Si is a finite free local R-algebra.
Reducing modulo m, we get k[T]/(f(T)) ≃Q i∈I Si/m, with the Si/m local. As the constant term of f(T) modulo m is nonzero, at least one of the roots of f(T) modulo m is a unit in k. So the map k[T]/(f(T)) ∼ →Q i∈I Si/m sends T to a unit in at least one of the residue fields of the Si/m; as the Si are local, the map R[T]/(f(T)) sends T to a unit in at least one of the Si, say Si0. As Si0 is a finite free R-module and K is algebraically closed, the ring Si0,red[ 1 ϖ] is isomor-phic to a nonempty product of copies of K. Projecting on one of the copies gives a morphism Si0 →K. As Si0 is integral over R and R is integrally closed in K, this map factors through a 5Add a lemma ?
180 V.3 The almost purity theorem map Si0 →R. So we have produced maps R[T]/(f(T)) →Si0 →R; the first one sends T to a unit in Si0, so the composition sends T to a unit in R. The image of T by this composition is the desired root of f(T).
Corollary V.3.6.3. (See theorems 3.2.8 and 6.2.10 of .) Let K be a perfectoid field. Then : (i) Any finite separable field extension of K is perfectoid.
(ii) If L/K is a finite separable field extension, then L/K and L♭/K♭have the same degree.
(iii) Tilting induces an equivalence between the categories of finite separable field extensions of K and of K♭.
Proof. Let Kfet (resp. K♭ fet) be the category of finite separable field extensions of K (resp.
K♭). By corollary V.3.5.3, untilting induces a fully faithful functor (.)♯: K♭ fet →Kfet that preserves degrees, and it suffices to show that this functor is essentially surjective. As (.)♯is fully faithful, it preserves automorphism groups, and so it preserves Galois extensions. By the main theorem of Galois theory, if L/K♭is a finite Galois extension, then (.)♯induces a bijection between subextensions of L/K♯and of L♯/K. So it suffices to show that every finite separable extension of K embeds in some L♯, for L/K♭a finite Galois extension.
Let C = c K♭be the completion of an algebraic closure of K♭. This is an algebraically closed perfectoid extension of K♭, and its untilt C♯over K is a perfectoid extension of K, that is also algebraically closed by proposition V.3.6.2. As C is the filtered inductive limit in the category of perfectoid K♭-algebras of all the finite separable extensions L of K♭contained in C. Let C0 ⊂C♯be the union of all the L♯, for L/K♭a finite separable extension contained in C. Then C0 is clearly algebraic over K, and it is dense in C♯. (By construction of the untilting functor.) By Krasner’s lemma (see for example section 25.2 of ), C0 is algebraically closed if and only if C♯is, so C0 is algebraically closed. This implies that every finite separable field extension of K embeds into C0, as desired.
V.3.7 Proof of theorems V.3.1.3 and V.3.1.9 V.3.7.1 Proof of theorem V.3.1.3 from theorem V.3.1.9 Suppose that theorem V.3.1.9 holds. This immediately implies (ii) and (iii) of theorem V.3.1.3.
We show (i). Let X = Spa(A, A+), and let Y →X be a finite ´ etale map of adic spaces. By definition, there is a finite cover X = Sn i=1 Spa(Ai, A+ i ) by rational domains such that, for every i, Y ×X Spa(Ai, A+ i ) ≃Spa(Bi, B+ i ) with (Ai, A+ i ) →(Bi, B+ i ) a finite ´ etale map of Huber pairs. By theorem V.3.1.9, for every i, the pair (Bi, B+ i ) is perfectoid and B+ i is almost finite 181 V Perfectoid algebras ´ etale over A+. In particular, Y is a perfectoid space, and tilting the situation gives a finite ´ etale map Y ♭→X♭:= Spa(A♭, A♭+). In other words, we may assume that A has characteristic ℓ.
Then we finish as in the proof of proposition 7.6 of : We can write (A, A+) as the completion of an inductive limit of perfections (A′, A′+)perf of strongly Noetherian Huber pairs, the finite ´ etale morphism Y →X descend to a finite ´ etale morphism to Spa of one of these (A′, A′+)perf which comes from a finite ´ etale morphism to Spa(A′, A′+) by the topological invariance of the ´ etale site, this morphism comes from a finite ´ etale A′-algebra by example 1.6.6(ii) of , and we pull this back to (A, A+) using lemma 7.3(iv) of .
V.3.7.2 Proof of theorem V.3.1.9 We already know that (i) holds in characteristic ℓby corollary V.3.4.2, and for perfectoid field by corollary V.3.6.3. In (ii), we know that (b) and (c) are equivalent by theorem V.3.5.1, and that (b) implies (a) by proposition V.3.3.1; we also know that (a) implies (b) in characteristic ℓor if A is a field by corollaries V.3.4.2 and V.3.6.3.
Let B be a finite ´ etale A-algebra, and let B+ be the integral closure of A+ in B. We still need to show that B is perfectoid and that B+ is almost finite ´ etale over A+. Consider the morphism of adic spaces f : Y := Spa(B, B+) →X := Spa(A, A+). Let x ∈X. Then the finite ´ etale map Y ×X Spa(κ(x), κ(x)+) comes from a finite ´ etale perfectoid pair over (κ(x), κ(x)+), which corresponds to a finite ´ etale perfectoid pair over (κ(x)♭, κ(x)♭+) = (κ(x♭), κ(x♭)+) (see theorem V.1.6.1).
On the other hand, (κ(x♭), κ(x♭)+) is the completion of lim − → U♭∋x♭ (OX♭(U ♭), O+ X♭(U ♭)), where we take the limit over rational domains U ♭of X♭such that x♭∈U ♭, by proposition III.6.3.1 and proposition III.6.3.7(i). As taking the category of finite ´ etale covers commutes with filtered colimits and with completions (more precisely, see corollary 10.0.5 of ), the finite ´ etale pair over (κ(x♭), κ(x♭)+) from the previous paragraph extends over a neighborhood of x♭. In other words (and using theorem V.1.6.1), we can find a rational domain V of X containing x such that the finite ´ etale OX(V )-algebra BV := B ⊗A OX(V ) is perfectoid, and that the integral closure of O+ X(V ) in BV is almost finite ´ etale over O+ X(V ).
As X is quasi-compact, this gives a finite rational convering (Vi)i∈I of X such that each Vi satisfies the two properties above. In othert words, Ui := Y ×X Vi is of the form Spa(Bi, B+ i ) with Si := B ⊗A OX(Vi) a finite ´ etale perfectoid OX(Vi)-algebra and B+ i almost finite ´ etale over O+ X(Vi). For all i, j ∈I, we have Ui ×Vj (Vi ∩Vj) ≃Uj ×Vi (Vi ∩Vj), because both sides are isomorphic to the unique finite ´ etale cover of Vi∩Vj corresponding to the finite ´ etale OX(Vi∩Vj)-algebra B ⊗A OX(Vi ∩Vj). So we can glue the Ui to get a perfectoid space Y and a map Y →X that is locally of the form Spa(B′, B′+) →Spa(A′, A′+), for (A′, A′+) →(B′, B′+) a strongly finite ´ etale map of Huber pairs. As in the proof of (i) of theorem V.3.1.3 in the previous subsection, we deduce that Y is of the form Spa(C, C+), for (A, A+) →(C, C+) strongly finite ´ etale. It remains to show that the A-algebras B and C are isomorphic. This follows from the fact 182 V.3 The almost purity theorem that they define isomorphic coherent OX-modules, which is clear from the definition of Y .
183 Index T0 space, 24 absolute value, 8 additive valuation, 8 adic (for a morphism of rings), 60 adic Point, 109 adic space, 120 adic topological ring, 55 affinoid k-algebra, 63 affinoid adic space, 120 affinoid field, 108 affinoid ring, 94 analytic point, 66 base (of a topology), 23 bounded subset (of a topological ring), 55 canonical topology, 89 Cauchy filter, 72 Cauchy sequence, 72 center (of a valuation subring), 17 characteristic group (of a valuation), 36 closed point, 24 cofinal, 47 complete, 72 completed residue field, 107 completion, 74 completion of a Huber pair, 101 constructible subset, 25 constructible topology, 26 continuous valuation, 64 continuous valuation spectrum, 64 convex subgroup, 11 convex subset (with respect to a valuation), 38 couple of definition, 55 discrete valuation, 13 discrete valuation ring, 13 domination, 7 equivalent valuation, 9 f-adic ring, 55 filter, 71 Frobenius endomorphism, 123 Gauss norm, 63, 110 generating family (of a topology), 23 generic point, 24 generization, 24 ghost components, 152 height of a totally ordered abelian group, 11 henselian local ring, 117 henselian pair, 116 horizontal generization, 38 horizontal specialization, 38 Huber pair, 94 Huber ring, 55 ideal of definition, 55 ind-constructible subset, 26 irreducible space, 24 isolated subgroup, 11 kernel of a valuation, 8 Kolmogorov space, 24 limit of a filter, 71 locally spectral space, 25 185 Index microbial valuation, 14 morphism of Huber pairs, 94 multiplicative valuation, 8 non-Archimedean field, 70 non-Archimedean topological ring, 55 Non-standard terminology., 109 open immersion, 119 open mapping theorem, 86 ordered abelian group, 7 perfect ring, 123 perfectoid field, 141 perfectoid Tate ring, 141 power-bounded element, 58 power-bounded subset, 58 primitive of degree 1, 152 pro-constructible, 26 pseudo-uniformizer, 62, 69 quasi-compact map, 23 quasi-compact space, 23 quasi-separated space, 23 rank (of a valuation), 12 rational covering, 129 rational subset, 95 Riemann-Zariski space, 17 ring of definition, 55 ring of integral elements, 94 semiperfect ring, 123 simple Laurent covering, 129 sober space, 24 specialization, 24 spectral map, 28 spectral space, 25 stably uniform Huber pair, 121 standard Laurent covering, 128 standard rational covering, 128 strict ℓ-ring, 152 strict morphism, 126, 127 strongly Noetherian, 121 subbase (of a topology), 23 support of a valuation, 8 Tate algebra, 63 Tate ring, 55 Teichm¨ uller representative, 152 tilt, 148 topologicall nilpotent subset, 58 topologically finitely generated, 63 topologically nilpotent element, 14 topologically of finite type, 123 trivial valuation, 9 uniform topological ring, 121 valuation, 9 valuation ring, 7 valuation spectrum, 19 valuation topology, 13 value group, 8 vertical generization, 35 vertical specialization, 35 Witt polynomials, 151 Witt vectors, 151 Zariski closed subset (of the adic spectrum), 101 186 Bibliography Bhargav Bhatt.
Lecture notes for a class on perfectoid spaces.
http:// www-personal.umich.edu/˜bhattb/teaching/mat679w17/lectures.
pdf, 2017.
Bhargav Bhatt, Matthew Morrow, and Peter Scholze.
Integral p-adic Hodge theory.
2019.
S. Bosch, U. G¨ untzer, and R. Remmert.
Non-Archimedean analysis, volume 261 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathemat-ical Sciences]. Springer-Verlag, Berlin, 1984. A systematic approach to rigid analytic geometry.
Siegfried Bosch. Lectures on formal and rigid geometry, volume 2105 of Lecture Notes in Mathematics. Springer, Cham, 2014.
N. Bourbaki.
´ El´ ements de math´ ematique. Fasc. XXX. Alg ebre commutative. Chapitre 6: Valuations. Actualit´ es Scientifiques et Industrielles, No. 1308. Hermann, Paris, 1964.
N. Bourbaki. ´ El´ ements de math´ ematique. Topologie g´ en´ erale. Chapitres 1 a 4. Hermann, Paris, 1971.
Nicolas Bourbaki. Espaces vectoriels topologiques. Chapitres 1 a 5. Masson, Paris, new edition, 1981. ´ El´ ements de math´ ematique. [Elements of mathematics].
Kevin Buzzard and Alain Verberkmoes.
Stably uniform affinoids are sheafy.
J. Reine Angew. Math., 740:25–39, 2018.
Brian Conrad. Number theory learning seminar 2014-2015.
stanford.edu/˜conrad/Perfseminar/, 2015.
Jean-Marc Fontaine.
Perfecto¨ ıdes, presque puret´ e et monodromie-poids (d’apres Peter Scholze). Ast´ erisque, (352):Exp. No. 1057, x, 509–534, 2013. S´ eminaire Bourbaki. Vol.
2011/2012. Expos´ es 1043–1058.
Ofer Gabber and Lorenzo Ramero. Almost ring theory, volume 1800 of Lecture Notes in Mathematics. Springer-Verlag, Berlin, 2003.
Timo Henkel. An Open Mapping Theorem for rings which have a zero sequence of units.
2014.
M. Hochster. Prime ideal structure in commutative rings. Trans. Amer. Math. Soc., 142:43– 187 Bibliography 60, 1969.
R. Huber. Continuous valuations. Math. Z., 212(3):455–477, 1993.
R. Huber.
A generalization of formal schemes and rigid analytic varieties.
Math. Z., 217(4):513–551, 1994.
Roland Huber. ´ Etale cohomology of rigid analytic varieties and adic spaces. Aspects of Mathematics, E30. Friedr. Vieweg & Sohn, Braunschweig, 1996.
Masaki Kashiwara and Pierre Schapira.
Categories and sheaves, volume 332 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathemati-cal Sciences]. Springer-Verlag, Berlin, 2006.
Kiran S. Kedlaya. On commutative nonarchimedean Banach fields. Doc. Math., 23:171– 188, 2018.
Kiran S. Kedlaya and Ruochuan Liu.
Relative p-adic Hodge theory: foundations.
Ast´ erisque, (371):239, 2015.
Falko Lorenz.
Algebra. Vol. II.
Universitext. Springer, New York, 2008.
Fields with structure, algebras and advanced topics, Translated from the German by Silvio Levy, With the collaboration of Levy.
Hideyuki Matsumura. Commutative ring theory, volume 8 of Cambridge Studies in Ad-vanced Mathematics.
Cambridge University Press, Cambridge, second edition, 1989.
Translated from the Japanese by M. Reid.
Peter Scholze. Perfectoid spaces. Publ. Math. Inst. Hautes ´ Etudes Sci., 116:245–313, 2012.
Peter Scholze. ´ etale cohomology of diamonds.
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Torsten Wedhorn.
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188 |
190111 | https://corporatefinanceinstitute.com/resources/accounting/common-size-analysis/ | Corporate Finance Institute
Home › Resources › Accounting › Common Size Analysis
Table of Contents
What is Common Size Analysis?
Formula for Common Size Analysis
Types of Common Size Analysis
Balance Sheet Common Size Analysis
Income Statement Common Size Analysis
What are the Benefits of Common Size Analysis?
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Common Size Analysis
A financial analysis tool that expresses each line item as a percentage of the base amount for a given period
Written by Jeff Schmidt
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What is Common Size Analysis?
Common size analysis, also referred to as vertical analysis, is a tool that financial managers use to analyze financial statements. It evaluates financial statements by expressing each line item as a percentage of a base amount for that period. The analysis helps to understand the impact of each item in the financial statements and its contribution to the resulting figure.
The technique can be used to analyze the three primary financial statements, i.e., balance sheet, income statement, and cash flow statement. In the balance sheet, the common base item to which other line items are expressed is total assets, while in the income statement, it is total revenues.
Summary
Common size analysis evaluates financial statements by expressing each line item as a percentage of a base amount for that period.
The formula for common size analysis is the amount of the line item divided by the amount of the base item. For example, cost of goods sold (line item) divided by revenue (base item).
Benefits of common size analysis is that it allows investors to identify large changes in a company’s financial statements, as well as the ability to compare companies of different sizes.
Formula for Common Size Analysis
Common size financial statement analysis is computed using the following formula:
Types of Common Size Analysis
Common size analysis can be conducted in two ways, i.e., vertical analysis and horizontal analysis. Vertical analysis refers to the analysis of specific line items in relation to a base item within the same financial period. For example, in the balance sheet, we can assess the proportion of inventory by dividing the inventory line using total assets as the base item.
On the other hand, horizontal analysis refers to the analysis of specific line items and comparing them to a similar line item in the previous or subsequent financial period. Although common size analysis is not as detailed as trend analysis using ratios, it does provide a simple way for financial managers to analyze financial statements.
Balance Sheet Common Size Analysis
The balance sheet common size analysis mostly uses the total assets value as the base value. A financial manager or investor can use the common size analysis to see how a firm’s capital structure compares to rivals. They can make important observations by analyzing specific line items in relation to the total assets.
For example, if the value of long-term debt in relation to the total assets value is high, it may signal that the company may become distressed.
Let’s take the example of ABC Company, with the following balance sheet:
From the table above, we calculate that cash represents 14.5% of total assets while inventory represents 12%. In the liabilities section, accounts payable is 15% of total assets, and so on.
Income Statement Common Size Analysis
The base item in the income statement is usually the total sales or total revenues. Common size analysis is used to calculate net profit margin, as well as gross and operating margins.
The ratios tell investors and finance managers how the company is doing in terms of revenues, and can be used to make predictions of future revenues and expenses. Companies can also use this tool to analyze competitors to know the proportion of revenues that goes to advertising, research and development, and other essential expenses.
We can also analyze ABC Company’s income statement:
By looking at this common size income statement, we can see that the company spent 10% of revenues on research and development and 3% on advertising.
Net income represents 10% of total revenues, and this margin can be compared to the previous year’s margin to see the company’s year-over-year performance.
What are the Benefits of Common Size Analysis?
One of the benefits of using common size analysis is that it allows investors to identify large changes in a company’s financial statements. It mainly applies when the financials are compared over a period of two or three years. Any significant movements in the financials across several years can help investors decide whether to invest in the company.
For example, large drops in the company’s profits in two or more consecutive years may indicate that the company is going through financial distress. Similarly, considerable increases in the value of assets may mean that the company is implementing an expansion or acquisition strategy, potentially making the company attractive to investors.
Common size analysis is also an excellent tool to compare companies of different sizes but in the same industry. Looking at their financial data can reveal their strategy and their largest expenses that give them a competitive edge over other comparable companies.
For example, some companies may sacrifice margins to gain a large market share, which increases revenues at the expense of profit margin. Such a strategy may allow the company to grow faster than comparable companies.
When comparing any two common size ratios, it is important to make sure that they are computed by using the same base figure. Failure to do so will render the comparison meaningless.
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190112 | http://www.grad.hr/geomteh3d/Monge/04duzina/duzina_eng.html | Orthogonal projection on a Line Segment and its True Length
True length of a line segment
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| If a line segment AB doesn't lie on a ray of projection, then its horizontal and vertical projections are line segments A'B' and A''B''. | |
| If a line segment AB is on an oblique line, then its projections have smaller length, i.e. d(A,B) > d(A',B') & d(A,B) > d(A'',B''). This fact is obtained from the properties of the right trapezoids shown in the picture on the right. | |
It is very important to determine the true length of a line segment AB given with its projections (A'B',A''B'').- We rotate the right trapezoid ABB'A' around the side A'B' for the right angle into the horizontal plane. Point A corresponds to the point Ao , point B to Bo and the following relation holds: d(A,B) = d(Ao,Bo). The length of A'Ao equals the z-coordinate of the point A, and the same holds for B, so this trapezoid A'AoB'Bo can easily be constructed in the horizontal plane. Determining the true length of a line segment-1
- If the z-coordinates of points A and B have opposite signs, after the rotation we obtain two right triangles. We can also rotate the right trapezoid ABB''A'' into the vertical plane. We denote the image of A and B with Ao and Bo. The length of A''Ao equals the y-coordinate of the point A, the same is valid for the point B. Determining the true length of a line segment-2 If the y-coordinates of the points A and B have opposite signs, the image of this trapezoid after the rotation is two right triangles. Straight line appearing in true length
- Orthogonal projection of a line segment is in true length if and only if the line segment is parallel to the image plane.
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| | | |
| Lines AB i CD are parallel to Π1, their horizontal projections are in true length. | Lines EF i GH are parallel to Π2, their vertical projections are in true length. | Line IJ is parallel to the ground line x, both its projections are in true length. |
Created by Sonja Gorjanc 3DGeomTeh - Developing project of the University of Zagreb. Translated by Helena Halas and Iva Kodrnja. |
190113 | https://www.cuemath.com/geometry/plane-definition/ | Plane Definition
In mathematics, a plane is a flat, two-dimensional surface that extends up to infinity. Planes can appear as subspaces of some multidimensional space, as in the case of one of the walls of the room, infinitely expanded, or they can enjoy an independent existence on their own, as in the setting of Euclidean geometry. The two types of planes are parallel planes and intersecting planes. Two non-intersecting planes are called parallel planes, and planes that intersect along a line are called Intersecting planes.
| Definition of a Plane
| How do you Make a Plane in Math?
| Parallel Planes
| Intersecting Planes
| Naming of Planes in Geometry
| Solved Examples on Plane
| Practice Questions on Plane
| FAQs on Plane
Definition of a Plane
In geometry, a plane is a flat surface that extends into infinity. It is also known as a two-dimensional surface. A plane has zero thickness, zero curvature, infinite width, and infinite length. It is actually difficult to imagine a plane in real life; all the flat surfaces of a cube or cuboid, flat surface of paper are all real examples of a geometric plane. We can see an example of a plane in which the position of any given point on the plane is determined using an ordered pair of numbers or coordinates. The coordinates show the correct location of the points on the plane.
The figure shown above is a flat surface extending in all directions. So, it is a plane.
Properties of Planes
A plane in math has the following properties:
How do you Make a Plane in Math?
In math, a plane can be formed by a line, a point, or a three-dimensional space. All the faces of a cuboid are planes. There is an infinite number of plane surfaces in a three-dimensional space.
Point
A point is defined as a specific or precise location on a piece of paper or a flat surface, represented by a dot. It has no width. A point has zero dimensions.
Line
A line is a combination of infinite points together. It extends in both directions. It has one dimension. The planes are difficult to draw because you have to draw the edges. But it is important to understand that the plane does not actually have edges, and it extends infinitely in all directions. The plane has two dimensions - length and width. However, since the plane is infinitely huge, its length and width cannot be estimated. To represent the idea of a plane, we can use a four-sided figure as shown below:
Therefore, we can call this figure plane QPR.
Identify Plane in a Three-Dimensional Space
In three-dimensional space, planes are all the flat surfaces on any one side of it. For example in the cuboid given below, all six faces of cuboid, those are, AEFB, BFGC, CGHD, DHEA, EHGF, and ADCB are planes. They all have only two dimensions - length and breadth.
Parallel Planes
Parallel planes are planes that never intersect. The below figure shows two planes, P and Q, that do not intersect each other. So, they are parallel planes. There are several examples of parallel planes, such as the opposite walls of the room and the floor.
Intersecting Planes
Intersecting planes are planes that are not parallel and they always intersect along a line. Two planes cannot intersect in more than one line. The below figure shows the two planes, P and Q, intersect in a single line XY. Therefore, the XY line is the common line between the P and Q planes. The two connecting walls are a real-life example of intersecting planes.
Naming of Planes in Geometry
Planes in geometry are usually referred to as a single capital (capital) letter in italics, for example, in the diagram below, the plane could be named UVW or plane P.
Important Notes
Related Articles on Plane Definition
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Solved Examples on Plane
Example 1: Sophie, a teacher, is asking her students. Are the points P, E, R, H coplanar?
Solution:
According to the definition of coplanarity, points lying in the same plane are coplanar. Points P, E, R, and H lie in the same plane. So they are coplanar. ∴ Yes, points P, E, R, and H are coplanar.
Example 2: Anna was asked to give other names for plane P. Can you help her?
Solution:
We can name the plane by its vertices. So, in the given diagram, the plane could be named plane HDF, plane HGF, and plane HGD.
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Practice Questions on Plane
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FAQs on Plane
How do you Define a Plane?
A plane is a flat two-dimensional surface. There is an infinite number of points and lines that lie on the plane. It can be extended up to infinity with all the directions. There are two dimensions of a plane- length and width.
What are the Examples of Plane Surfaces?
The surfaces which are flat are known as plane surfaces. Examples of plane surfaces are the surface of a room, the surface of a table, and the surface of a book, etc.
Is Diamond a Plane Shape?
A diamond is a 2-dimensional flat figure that has four closed and straight sides. Yes, it is a plane shape as it has two dimensions- length and width.
How Many Points do you Need for a Plane?
Any three noncollinear points make up a plane.
What is the Angle Between Two Intersecting Planes?
The angle between two intersecting planes is called the Dihedral angle.
How many Dimensions does a Plane have?
Planes are two-dimensional, but they can exist in three-dimensional space.
Is a Plane a Curved Surface?
A plane has two dimensions: length and width. All planes are flat surfaces. If it is not a flat surface, it is known as a curved surface. |
190114 | https://s2.smu.edu/~helgason/cse8394/algebra03.pdf | Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 1 Convex combinations of more than two points Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 2 We first consider the extension to three points in Rn.
Given a, b, c ∈Rn and v, w, x ∈R , y(v, w, x) = va + wb + xc , with v + w + x = 1 and v, w, x ≥0 is a (parameterized) convex combination of the given points.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 3 Let us do a little algebra, assuming that v + w ̸= 0.
y(v, w, x) = va + wb + xc = (v + w) " v v + w !
a + w v + w !
b # + xc = (1 −x) " v v + w !
a + w v + w !
b # + xc Define b′ = v v + w a + w v + w b .
Then b′ is a convex combination of a and b (Why?) and y(v, w, x) is a convex combination of b′ and c.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 4 So, assuming that the points are distinct, what is { va+wb+xc : v+w+x = 1 and 0 ≤v, w, x ≤1 } , the set of all convex combinations of a, b, c ∈Rn ?
We have that b′ = v v + w !
a + w v + w !
b (with v + w ̸= 0) and va + wb + xc = (1 −x)b′ + xc (with x = 1 −v −w) Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 5 { va + wb + xc : v + w + x = 1 and 0 ≤v, w, x ≤1 } consists of all the points on the edges and inside the triangle with corner points a, b, and c.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 6 We now consider the extension to many points in Rn.
Let A = {a1, . . . , am} be a given finite set of m points in Rn. The set A can be used to generate different polyhedral objects in Rn by combining its elements using various linear operations. The elements of A are the generators of the objects they define.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 7 One of these fundamental objects is the convex hull of the points in A (a polytope) defined by H(A) = m X i=1 λiai : m X i=1 λi = 1 and λ1, . . . , λm ≥0 The convex hull consists of all convex combinations of the generators.
In R3 one can visualize the convex hull of many points as a multi-faceted diamond.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 8 Any set is said to be convex if it contains all convex combinations of any finite set of points from that set.
H(A) is convex since it can be shown that a convex combination of convex combinations of given points is a convex combination of those points.
Example: 1 2 "1 3a1 + 2 3a2 # + 1 2 "2 3a2 + 1 3a3 # = 1 6a1 + 2 3a2 + 1 6a3 Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 9 The minimum cardinality subset F ⊂A which gen-erates H(A) is called the frame of H(A).
A frame is to a convex hull what a basis is to a linear combination.
In R3 when visualizing the convex hull of many points as a multi-faceted diamond, the corner points are the generators.
It can be shown that a point is a member of F if and only if it cannot be written as a strict convex combi-nation of two distinct points of H(A).
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 10 The fundamental tool for determining the frame F from A is the generic linear program: (LP) z = min X j∈J λj s.t.
X j∈J λjaj = ak λj ≥0 , j ∈J where A′ ⊂A and J = { j : aj ∈A′\ak }.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 11 Fundamental Results: If F ⊂A′, ak ̸= 0, and the linear program (LP) is feasible, then ak ∈F if and only if at optimality z > 1.
If F ⊂A′, ak ̸= 0, and the linear program (LP) is infeasible, then ak ∈F.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 12 Naive approaches to finding F based on iterative so-lution of problems of type (LP) have long been used.
Computationally they suffer by starting with large size sets A′ (and J) and only slowly decreasing their sizes.
Dula, Helgason, and Hickman have shown how to more efficiently compute F, in part making use of iterative solution of problems of type (LP) starting with small size sets A′ (and J) and slowly increasing their sizes.
Helgason EMIS 8394 July 8, 2004 [Convex Combinations 2] 13 References: J.H. Dul´ a and R.V. Helgason (1996), A new proce-dure for identifying the frame of the convex hull of a finite collection of points in multidimensional space, European Journal of Operational Research 92, 352-367.
J.H. Dul´ a, R.V. Helgason, and B.L. Hickman (1992), Preprocessing schemes and a solution method for the convex hull problem in multidimensional space, Com-puter Science and Operations Research: New Devel-opments in Their Interfaces, O. Balci (ed.), 59-70, Pergamon Press, U.K. |
190115 | https://madasmaths.com/archive/maths_booklets/further_topics/various/complex_numbers_part_1_exam_questions.pdf | Created by T. Madas Created by T. Madas COMPLEX NUMBERS (Exam Questions I) Created by T. Madas Created by T. Madas Question 1 () 9 3i 1 2i w −+ = − . Find the modulus and the argument of the complex number w . FP1-M , 3 2 w = , 3 arg 4 w π = − Question 2 () Solve the equation 2 2 2i 5 0 z z − − = , z ∈ . 3 1 i 2 2 z = ± + Created by T. Madas Created by T. Madas Question 3 () Find the value of x and the value of y in the following equation, given further that x∈ , y∈ . ( )( ) i 2 i 3 i x y + + = −. FP1-A , ( ) ( ) , 1, 1 x y = − Question 4 () 4i 1 i z λ λ + = + , λ ∈ . Given that z is a real number, find the possible values of λ . 2 λ = ± Created by T. Madas Created by T. Madas Question 5 () Find the values of x and y in the equation ( ) ( ) 2 2 1 i 2 i 3 10i x y + + − = + , x∈ , y ∈ . 7 x = , 1 y = Question 6 () Find the value of x and the value of y in the following equation, given further that x∈ , y∈ . ( )( ) i 3 4i 3 4i x y + + = − . FP1-E , ( ) ( ) 7 24 , , 25 25 x y = − − Created by T. Madas Created by T. Madas Question 7 () The complex number z satisfies the equation 1 18i 4 3z 2 i z − − = − , where z denotes the complex conjugate of z . Solve the equation, giving the answer in the form i x y + , where x and y are real numbers. 4 i z = − Question 8 () 3 4i z = −+ and 14 2i zw = − + . By showing clear workings, find … a) … w in the form i a b + , where a and b are real numbers. b) … the modulus and the argument of w . 2 2i w = + , 2 2 w = , arg 4 w π = Created by T. Madas Created by T. Madas Question 9 () 22 4i z = + and 6 8i z w = − . By showing clear workings, find … a) … w in the form i a b + , where a and b are real numbers . b) … the modulus and the argument of w . 1 2i w = + , 5 w = , c arg 1.11 w ≈ Question 10 () ( ) 2 7 4i 2 i 8 2 i z − = − + − + . Express z in the form i x y + , where x and y are real numbers. FP1-P , 3 7i −− Created by T. Madas Created by T. Madas Question 11 () The complex conjugate of z is denoted by z . Solve the equation 27 23i 2 3 1 i z z − + − = + , giving the answer in the form i x y + , where x and y are real numbers. 2 5i z = + Question 12 (+) Solve the following equation. 2 21 20i z = − , z ∈. Give the answers in the form i a b + , where a∈ and b∈ . FP1-K , ( ) 5 2i z = ± − Created by T. Madas Created by T. Madas Question 13 (+) The cubic equation 3 2 2 5 5 0 z z cz − + − = , c∈ , has a solution 1 2i z = − . Find in any order … a) … the other two solutions of the equations. b) … the value of c . 2 1 2i z = + , 3 1 2 z = , 12 c = Created by T. Madas Created by T. Madas Question 14 (+) The quadratic equation 2 2 1 2i 0 z z − + − = , c∈ , has a solution i z = −. Find the other solution. FP1-W , 2 2 i z = + Created by T. Madas Created by T. Madas Question 15 (+) ( ) 8 i 7 2 z z − = − , z ∈ . The complex conjugate of z is denoted by z . Determine the value of z in the above equation, giving the answer in the form i x y + , where x and y are real numbers. 2 3i z = + Question 16 (+) 3 2 26 0 z Az Bz + + + = , where A∈ , B∈ One of the roots of the above cubic equation is 1 i + . a) Find the real root of the equation. b) Determine the values of A and B . 13 z = − , 11 A = , 24 B = − Created by T. Madas Created by T. Madas Question 17 (+) The complex conjugate of z is denoted by z . Solve the equation ( ) 12 i 9 2 z z − = − , giving the answer in the form i x y + , where x and y are real numbers. 2 5i z = + Question 18 (+) The complex number z satisfies the equation ( ) 2 iz 3 3 5i z − = − , where z denotes the complex conjugate of z . Determine the value of z , giving the answer in the form i x y + , where x and y are real numbers. 1 7i z = − Created by T. Madas Created by T. Madas Question 19 (+) The cubic equation 3 2 2 4 0 z z z p − + + = , p∈ , is satisfied by 1 2i z = + . a) Find the other two roots of the equation. b) Determine the value of p . FP1-P , 3 1 2i, 2 − − , 15 p = Created by T. Madas Created by T. Madas Question 20 (+) Solve the following equation. 2 5 12i w = − , w∈ . Give the answers in the form i a b + , where a∈ and b∈ . FP1-C , ( ) 3 2i w = ± − Created by T. Madas Created by T. Madas Question 21 (+) 1 3i z = + and 2 2i w z = + . Find the exact value of the modulus of w and the exact value of the argument of w . 4 2 w = , 7 arg 12 w π = Created by T. Madas Created by T. Madas Question 22 (+) The following cubic equation is given 3 2 5 0 z az bz + + − = , where a∈ , b∈ . One of the roots of the above cubic equation is 2 i + . a) Find the other two roots. b) Determine the value of a and the value of b . 2 2 i z = − , 3 1 z = , 5 a = − , 9 b = Created by T. Madas Created by T. Madas Question 23 (+) The following cubic equation is given 3 2 6 0 z pz z q + + + = , where p∈ , q∈ . One of the three solutions of the above cubic equation is 5 i −. a) Find the other two solutions of the equation. b) Determine the value of p and the value of q . 2 3 5 i, 2 z z = + = , 8 p = − , 52 q = Created by T. Madas Created by T. Madas Question 24 (+) The complex number z is defined as ( )( ) 2 i 1 i 1 2i z = + − . It is further given that ( ) 3i 3i z P z Q z − + − = where P and Q are real constants. Find the value of P and the value of Q . SYNF-C , 3 P = , 4 Q = Created by T. Madas Created by T. Madas Question 25 () 3 i z = + and 3i w = . a) Find, in exact form where appropriate, the modulus and argument of z and the modulus and argument of w . b) Determine simplified expressions for zw and w z , giving the answers in the form i x y + , where x∈ , y ∈ . c) Find, in exact form where appropriate, the modulus and argument of zw and the modulus and argument of w z . 2, 3 z w = = , arg , arg 6 2 z w π π = = , 3 3 3i zw = −+ , 3 3 3i 4 4 w z = + , 3 6, 2 w zw z = = , ( ) 2 arg ,arg 3 3 w zw z π π = = Created by T. Madas Created by T. Madas Question 26 () Find the value of x and the value of y in the following equation, given further that x∈ , y∈ . 1 1 2 3i i 1 i x y − = − + + . FP1-I , ( ) ( ) 5 7 , , 37 37 x y = Created by T. Madas Created by T. Madas Question 27 () Find the square roots of 1 i 3 + . Give the answers in the form i a b + , where a∈ and b∈ . FP1-G , ( ) 1 6 i 2 2 ± + Created by T. Madas Created by T. Madas Question 28 () Solve the equation 13 11 3i 1 z z = − + , z ∈ , giving the answer in the form i x y + , where x and y are real numbers. 1 3i z = − Question 29 () The complex conjugate of w is denoted by w . Given further that 1 2i w = + and 2 25w z w w = − , show clearly that z is a real number, stating its value. 12 Created by T. Madas Created by T. Madas Question 30 () The following cubic equation is given 3 2 2 0 z z az b + + + = , where a∈ , b∈ . One of the roots of the above cubic equation is 1 i + . a) Find the real root of the equation. b) Find the value of a and the value of b . 4 z = − , 6 a = − , 8 b = Created by T. Madas Created by T. Madas Question 31 () The following complex numbers are given. 1 2 2i z = − , 2 3 i z = + and 3 i z a b = + where a∈ , b∈ . a) If 1 3 16 z z = , find the modulus 3 z . b) Given further that 3 2 7 arg 12 z z π = , determine the argument of 3 z . c) Find the values of a and b , and hence show 3 1 2 z z = −. FP1-Q , 3 4 2 z = , 3 3 arg 4 z π = , 4 a = − , 4 b = Created by T. Madas Created by T. Madas Question 32 () Solve the equation 4 3 2 2 14 33 26 10 0 z z z z − + − + = , z ∈ given that one of its roots is 3 i + . FP1-Q , 1 1 1 1 3 i, 3 i, i, i 2 2 2 2 z z z z = + = − = + = − Created by T. Madas Created by T. Madas Question 33 () 3 2 2 16 0 z pz qz + + + = , p∈ , q∈ . The above cubic equation has roots α , β and γ , where γ is real. It is given that ( ) 2 1 i 3 α = + . a) Find the other two roots, β and γ . b) Determine the values of p and q . ( ) 2 1 i 3 β = − , 1 2 γ = − , 7 p = − , 28 q = Created by T. Madas Created by T. Madas Question 34 () Find the value of x and the value of y in the following equation, given that , x y∈ . 1 1 1 i 1 2i x y + = + + . FP1-O , ( ) ( ) 1 , 1, 2 x y = − Created by T. Madas Created by T. Madas Question 35 () Consider the cubic equation 3 10 0 z z + + = , z ∈ . a) Verify that 1 2i + is a root of this equation. b) Find the other two roots. 1 2 1 2i, 2 z z = − = − Created by T. Madas Created by T. Madas Question 36 () The complex conjugate of z is denoted by z . Solve the equation ( ) 2 3i 2 13 4i 1 i z z + + = + + , giving the answer in the form i x y + , where x and y are real numbers. 3 i z = + Created by T. Madas Created by T. Madas Question 37 () 4 3 2 8 33 68 52 0 z z z z − + − + = , z ∈ . One of the roots of the above quartic equation is 2 3i + . Find the other roots of the equation. FP1-O , 2 3i, 2 z z = − = Question 38 () Find the values of x and y in the equation i 3 2 i 2 i 1 2i x y + = + − − , x∈ , y ∈ . 4 x = , 5 y = Created by T. Madas Created by T. Madas Question 39 () The complex conjugate of z is denoted by z . Find the two solutions of the equation ( )( ) i i 6 22i z z z − − = − , z ∈ , giving the answers in the form i x y + , where x and y are real numbers. 1 2 3i z = + , 2 28 9i 5 5 z = + Created by T. Madas Created by T. Madas Question 40 () Find the value of x and the value of y in the following equation, given further that x∈ , y∈ . 1 5i 1 i 3 2i 2 i x y − = + + − −. FP1-R , ( ) ( ) , 2,0 x y = Created by T. Madas Created by T. Madas Question 41 () Find the value of z and the value of w in the following simultaneous equations 2 1 i z w + = − 3 3i z w − = + . 1 2i, 4 i z w = −+ = −− Question 42 () It is given that 2i i z z k + = + , k ∈ and 2 2i w z = + , Im 8 w = . Determine the value of k . 4 k = Created by T. Madas Created by T. Madas Question 43 (+) Given that z and w are complex numbers prove that 2 2 4 Re Re z w z w z w + − − = , where w denotes the complex conjugate of w . FP1-M , proof Created by T. Madas Created by T. Madas Question 44 (+) Find the three solutions of the equation 2 4 4 1 0 z z + + = , z ∈ , where z denotes the complex conjugate of z . 1 1 1 , i, i 2 2 2 z = + − Question 45 (+) The complex numbers z and w are defined as 3 i z = + and 1 2i w = + . Determine the possible values of the real constant λ if 2 z w λ λ + = + . FP1-N , 0, 1 λ λ = = − Created by T. Madas Created by T. Madas Question 46 (+) The complex number z satisfies the equation 2 3 4i z = + . a) Find the possible values of … i. … z . ii. … 3 z . b) Hence, by showing detailed workings, find a solution of the equation 6 3 4 125 0 w w − + = , w∈ , ( ) 2 i z = ± + , 3 2 11i z = ± , ( ) 2 i w = ± + Created by T. Madas Created by T. Madas Question 47 (+) Solve the following quadratic equation ( ) 2 6 10 6 i 0 z z z − + + − = , z ∈ . Give the answers in the form i a b + , a∈ , b∈ . FP1-J , 1 4 i z = + , 2 2 2i z = − Created by T. Madas Created by T. Madas Question 48 (+) Solve teach of the following equations. a) 2 2i 8 0 z z + + = , z ∈ . b) 2 16 30i w + = , w∈ . 1 2 2i, 4i z z = = − , ( ) 3 5i w = ± + Created by T. Madas Created by T. Madas Question 49 (+) It is given that 2 z = and 1 2i z = + are solutions of the equation 4 3 2 3 0 z z az bz c − + + + = . where a , b and c are real constants. Determine the values of a , b and c . FP1-R , 5 a = , 1 b = −, 10 c = − Created by T. Madas Created by T. Madas Question 50 (+) The following complex numbers are given 1 i 1 i z + = − and 2 1 i w = −. a) Calculate the modulus of z and the modulus of w . b) Find the argument of z and the argument of w . In a standard Argand diagram, the points A , B and C represent the numbers z , z w + and w respectively. The origin of the Argand diagram is denoted byO . c) By considering the quadrilateral OABC and the argument of z w + , show that 3 tan 1 2 8 π = + . 1 z = , 1 w = arg 2 z π = , arg 4 w π = Created by T. Madas Created by T. Madas Question 51 (+) Solve the following quadratic equation ( ) 2 8 2 1 i 0 z z z − + + + = , z ∈ . Give the answers in the form i a b + , a∈ , b∈ . FP1-L , 1 2i z = , 2 1 4i z = − Created by T. Madas Created by T. Madas Question 52 (+) The quadratic equation ( ) 2 4 20 i 1 0 z z z A + + + + = , where A is a constant, has complex conjugate roots. If one of the roots of this quadratic is 2i z B = + , where B is a real constant, find the possible values of A . FP1-Y , 1 12i 1 4i A A = −+ = −− ∪ Created by T. Madas Created by T. Madas Question 53 (+) If 1 2i − is a root of the quartic equation 4 3 2 6 18 30 25 0 z z z z − + − + = find the other three roots. 2 3 4 1 2i, 2 i, 2 i z z z = + = − = + Question 54 () The complex conjugate of z is denoted by z . Solve the equation 2 2 z z z + = + , z ∈ . 1 z = Created by T. Madas Created by T. Madas Question 55 () It is given that cos isin z θ θ = + , 0 2 z π ≤ < . Show clearly that 2 1 i tan 1 2 z θ = − + . proof Created by T. Madas Created by T. Madas Question 56 () ( )( ) ( ) 3 4i 1 2i 1 i 1 3i q + + = + + , q∈ . a) Find the value of q . b) Hence simplify 4 arctan arctan 2 arctan3 3 + − , giving the answer in terms of π . 5 2 q = , 1 4π Created by T. Madas Created by T. Madas Question 57 () The complex conjugate of the complex number z is denoted by z . Solve the equation ( ) 2 1 2i i 2 3i 5z 1+2i z z − − + = , giving the answer in the form i x y + . 5 2i z = + Created by T. Madas Created by T. Madas Question 58 () It is given that 17 6i z = − − and 3 i w = + . Find the value of u given further that 1 3 1 10 2 u z w = + . 9 7i u = −− Created by T. Madas Created by T. Madas Question 59 () Sketch on a standard Argand diagram the locus of the points ( ) 2 1 i z = + , 3 i w = − and z w + , and use geometry to prove that tan 6 3 2 2 24 π = − + − . You must justify all the steps in this proof. FP1-L , proof Created by T. Madas Created by T. Madas Question 60 () The complex number z is given by i i a b z a b + = − , a∈ , b∈ . Show clearly that 2 2 2 2 2 1 2 z a b z a b + − = + . proof Created by T. Madas Created by T. Madas Question 61 () It is given that 1 8i 1 2i z + = − . a) Express z in the form i x y + , where x and y are real numbers. b) Find the modulus and argument of z . c) Show clearly that 2 arctan8 arctan 2 arctan 3 π + + = . 3 2i z = −+ , 13 z = , c arg 2.55 z ≈ Created by T. Madas Created by T. Madas Question 62 (+) Solve each of the following equations. a) 3 27 0 z − = . b) ( ) ( ) 2 i 2 2 w w w − − = − . ( ) 1 2 3 3, 1 3 2 z z = = −± , 1 2 2i, 1 i w w = = − Created by T. Madas Created by T. Madas Question 63 (+) ( ) ( ) 4 2 4 2 2 3i 3 2i n n z + + = + + − , n∈ . Show clearly that 0 z = for all n∈ . proof Question 64 (+) The complex conjugate of z is denoted by z . Show clearly that the equation 3 2z z z − = , is satisfied either by 0 z = or 1 z = ± . proof Created by T. Madas Created by T. Madas Question 65 (+) ( ) ( ) 5 2i 5 2i n n z = + + − , n∈ . Show clearly that z is a real number. proof Created by T. Madas Created by T. Madas Question 66 (+) The complex number z satisfies the relationship 1 1 z z + = −, 0 z ≠ . Show clearly that …. a) … 3 1 z = . b) … 8 4 1 z z + = −. FP1-W , proof Created by T. Madas Created by T. Madas Question 67 (+) ( ) ( ) 4 4 i i n n z a b b a = + + + , a∈ , b∈ , n∈ . Show that z is a real number. FP1-V , proof Created by T. Madas Created by T. Madas Question 68 (+) ( ) ( ) ( ) 3 2 4 2i 4 5i 1 3i 0 z z z − + + + − + = , z ∈ . Given that one of the solutions of the above cubic equation is 2 i z = + , find the other two solutions. FP1-U , 1 z = , 1 i z = + Created by T. Madas Created by T. Madas Question 69 () Find the solutions of the equation ( ) 4 4 16 1 w w = − , giving the answers in the form i x y + , where x∈ , y ∈ . SPX-A , 1 2 3 4 2 4 2 4 2 2, , i , i 5 5 5 5 3 z z z z = = = + = − Question 70 () Solve the quadratic equation ( ) 2 7 16 i 11 z z z − + = − , z ∈ . FP1-T , 2 3i, 5 2i z z = + = − Created by T. Madas Created by T. Madas Question 71 () ( ) ( ) 2 2 3 8i 4i 0 z z m − + − + = , z ∈ . Given that m is a real constant, find the two solutions of the above equation given further that one of these solutions is real. SPX-T , 1 2 z = , 2 4i z = + Created by T. Madas Created by T. Madas Question 72 () Solve the quadratic equation 2 4 i 4i 7 z z − + = , z ∈ . SPX-J , 2 3i, 2 i z z = −+ = + Question 73 () 4 3 2 2 2 3 4 0 z z z z − − + − = , z ∈ . By using the substitution 2 w z z = − , or otherwise, find in exact form the four solutions of the above equation. SPX-V , 1 17 1 i 3 , 2 2 z ± ± = Created by T. Madas Created by T. Madas Question 74 () Show that if n and m are natural numbers, then the equations 1 i n z = + 2 i m z = −, have no common solution for z ∈ . SPX-I , proof Created by T. Madas Created by T. Madas Question 75 () 4 3 2 20 0 z z z − + − = , z ∈ . By using the substitution 2 w z z = − , or otherwise, find in exact form the four solutions of the above equation. SPX-H , 1 21 1 i 15 , 2 2 z ± ± = Created by T. Madas Created by T. Madas Question 76 () Two distinct complex numbers 1 z and 2 z are such so that 1 2 0 z z r = = ≠ . Show clearly that 1 2 1 2 z z z z + − is purely imaginary. You may find the result 2 2 zz z r = = useful. SPX-G , proof Created by T. Madas Created by T. Madas Question 77 () The complex number z satisfies the relationship ( ) ( )( ) 5 i 4 3i 1 i n n z z + = + + , n∈ . Show that z is a real number. SPX-F , proof Created by T. Madas Created by T. Madas Question 78 () The complex numbers z and w are such so that 1 z w = = . Show clearly that 1 z w zw + + is real. SPX-D , proof Created by T. Madas Created by T. Madas Question 79 () ( ) ( ) 3 2 2 2 i 8 3i 5 i 0 z z z − − + − −+ = , z ∈ . Find the three solutions of the above equation given that one of these solutions is real. SPX-Q , 1 z = , 2 3i z = − , 2 i z = + Created by T. Madas Created by T. Madas Question 80 () Solve the quadratic equation 2 i 2 2 2 3 0 z z − − = , z ∈ . Give the answers in the form i x y + , where x and y are exact real constants. SPX-C , ( ) ( ) 1 i 3 2 , 1 i 3 2 z z = −+ − = − + Created by T. Madas Created by T. Madas Question 81 () The complex number z satisfies the equation ( ) 1 8i 1 i z z + + = + . Show clearly that 2 18 65 0 z z − + = , and hence find the possible values of z . SPX-B , 4 3i z = − , 12 5i z = + Created by T. Madas Created by T. Madas Question 82 () ( ) ( ) 3 2 2 4i 3 1 3i 14 2i 0 z z z − + − − + − = , z ∈ . Find the three solutions of the above equation given that one of these solutions is purely imaginary. SPX-E , 2i z = , 2 i z = −, 1 3i z = −+ Created by T. Madas Created by T. Madas Question 83 () It is given that = z w z w + , where z ∈ , w∈ , and 1 w > . Determine an exact simplified expression for z , in terms of w . SPX-P , 1 1 w w z w w ± = = ± − ∓ ∓ |
190116 | https://wordpandit.com/word-root-celer/ | Word Root: Celer - Wordpandit
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Word Root: Celer
Celer: The Root of Swiftness in Language and Action
Discover the origins, power, and applications of the root "celer," meaning "swift." From the speed implied in "accelerate" to the elegance of "celerity," this root reveals the significance of quickness across various fields.
Table of Contents
Introduction: The Essence of Celer
Etymology and Historical Journey
Mnemonic: Unlocking the Power of Celer
Common Celer-Related Terms
Celer Through Time
Celer in Specialized Fields
Illustrative Story: Celer in Action
Cultural Significance of the Celer Root
The Celer Family Tree
FAQs about the Celer Word Root
Test Your Knowledge: Celer Mastery Quiz
Conclusion: The Living Legacy of Celer
Introduction: The Essence of Celer
When we think of swiftness or speed, words like accelerate and celerity come to mind. These terms owe their dynamism to the root celer (pronounced sel-er), which means "swift." Originating from Latin, this root has influenced words that emphasize motion, efficiency, and agility. From the rapid pace of technology to the speed of light in science, "celer" resonates across various domains, celebrating the essence of quickness.
Etymology and Historical Journey
The root celer traces back to Latin, where it directly meant "swift" or "speedy." Ancient Romans admired celeritas (swiftness), especially in their military and communication systems. Over time, this root found its way into English, shaping words like "accelerate" and "decelerate." The Renaissance period, with its revival of classical languages, further cemented celer’s place in technical and everyday vocabulary.
Mnemonic: Unlocking the Power of Celer
To remember the meaning of "celer," imagine a cheetah—the epitome of speed—dashing across the savannah. Picture the word "celer" glowing on the cheetah’s back, symbolizing unmatched swiftness.
Mnemonic Device: "Celer means swift, like a cheetah in a drift!"
Common Celer-Related Terms
Accelerate (ak-sell-er-ate)
Definition: To increase speed or progress.
Example: "The car accelerated quickly on the open road."
Celerity (suh-ler-i-tee)
Definition: Swiftness of movement or action.
Example: "The firefighter’s celerity saved countless lives."
Decelerate (dee-sell-er-ate)
Definition: To reduce speed.
Example: "Drivers were advised to decelerate in the school zone."
Accelerant (ak-sell-er-ant)
Definition: A substance that speeds up a process, often in combustion.
Example: "Investigators identified the accelerant used to start the fire."
Accelerometer (ak-sell-er-om-uh-ter)
Definition: An instrument for measuring acceleration.
Example: "The accelerometer in the smartphone tracks movement and orientation."
Celer Through Time
The evolution of the root celer reveals how its core meaning of swiftness has adapted to different contexts:
Celeritas (Latin): Initially a military term, emphasizing rapid maneuvers and communications.
Accelerate (Modern English): Became widely associated with technological advancements, such as in vehicles and computing.
Celerity (Literary): Retains a poetic elegance, often used to describe graceful quickness in literature.
Celer in Specialized Fields
Physics and Engineering:
Term: Accelerometer
Application: Measures changes in velocity, essential in space exploration and smartphones.
Technology:
Term: Accelerate
Application: Describes the optimization of processes in computing, such as faster data transfers.
Sports:
Term: Celerity
Application: Describes athletes' agility and swift movements, critical in sprinting or fencing.
Fire Safety:
Term: Accelerant
Application: Refers to substances that increase combustion speed, used in fire investigations.
Illustrative Story: Celer in Action
Meet Leo, a competitive sprinter training for the Olympics. His coach emphasized the importance of not just raw speed but celerity—the ability to combine swiftness with grace and precision. During a critical race, Leo felt his heart accelerate as the crowd roared. Channeling his training, he surged ahead with unparalleled celerity, clinching the gold medal. His triumph demonstrated the power of celer in action—speed balanced with skill.
Cultural Significance of the Celer Root
The idea of swiftness has been celebrated across cultures. In Roman times, celeritas symbolized efficiency and power, crucial for their vast empire. Today, terms derived from "celer" echo in modern technology, such as in the acceleration of data and innovation. Literature and art often use celerity to capture the beauty of swift movements, from dancers to galloping horses.
The Celer Family Tree
Curr-/Curs- (Latin: "run")
Current: Flowing or running.
Cursory: Done quickly, often with little attention.
Tachy- (Greek: "speed")
Tachometer: Measures rotational speed.
Rapid- (Latin: "fast")
Rapid: Quick or fast movement.
FAQs About the Celer Word Root
Q: What does "celer" mean?
A: "Celer" is a Latin root that means "swift" or "speedy." It conveys the idea of quickness, whether in physical motion or the progression of events. Words like "accelerate" (to increase speed) and "celerity" (graceful swiftness) are derived from this root.
Q: How is "celerity" different from "speed"?
A: While both "celerity" and "speed" imply quickness, "celerity" often includes an element of grace, efficiency, or precision. For instance, a skilled dancer might move with celerity, highlighting not just their swiftness but their elegance and control.
Q: What is an accelerometer?
A: An accelerometer is a device that measures acceleration, which is the rate of change of velocity. It’s widely used in smartphones (e.g., to detect screen orientation), vehicles (for crash detection), and aerospace technologies (to measure g-forces during flight).
Q: What is the difference between "accelerate" and "decelerate"?
A: "Accelerate" means to increase speed or make progress faster, whereas "decelerate" means to reduce speed or slow down. For example, a car accelerates when moving from a stoplight but decelerates when approaching a red light.
Q: Why is "celerity" more common in literature than everyday speech?
A: "Celerity" is a more formal and literary term, often used to describe swiftness with an elegant or refined tone. In contrast, "speed" is a simpler, more common word used in everyday contexts. For example, an author might write about "the celerity of a hero's response" in a novel, while "speed" might describe a car's performance in a conversation.
Q: What is an accelerant, and how is it used?
A: An accelerant is a substance that increases the speed of a chemical reaction, especially combustion. In fire investigations, identifying an accelerant (like gasoline) can provide clues about the fire’s origin and whether it was intentional.
Q: Are there words related to "celer" outside of speed?
A: Yes, while most "celer"-related words pertain to swiftness or speed, some describe tools or concepts indirectly linked to the idea of motion, such as "accelerometer" (a measuring instrument) and "deceleration" (controlled reduction of movement).
Test Your Knowledge: Celer Mastery Quiz
1. What does the root "celer" mean?
Slow Swift Strong Small
Correct answer: Swift. The root "celer" comes from the Latin word for "swift." It forms the foundation of words that convey quickness or speed, such as "accelerate" and "celerity."
2. Which word describes reducing speed?
Decelerate Accelerate Celerity Accelerometer
Correct answer: Decelerate. "Decelerate" means to slow down or reduce speed. The prefix "de-" implies reversal or reduction, combined with "celer," emphasizing the act of decreasing swiftness.
3. What is "celerity" commonly used to describe?
Slow actions Graceful swiftness Loud noises Steady progress
Correct answer: Graceful swiftness. "Celerity" refers to swiftness, often with an added nuance of grace or precision. It is commonly used in literature or formal contexts to highlight elegance in rapid action.
4. What instrument measures acceleration?
Tachometer Accelerometer Barometer Thermometer
Correct answer: Accelerometer. An "accelerometer" measures the rate of change of velocity (acceleration). It’s a critical tool in various technologies, from tracking movement in smartphones to detecting crashes in vehicles.
5. Which term is associated with combustion?
Celerity Accelerant Decelerate Accelerometer
Correct answer: Accelerant. An "accelerant" speeds up the process of combustion. It is a key term in fire science and investigation, as substances like gasoline or alcohol can significantly accelerate the spread of fire.
Conclusion: The Living Legacy of Celer
The root celer highlights the timeless appeal of swiftness, whether in nature, technology, or human achievements. Its influence spans literature, science, and daily life, reminding us of the beauty and utility of speed. As the world continues to accelerate, the legacy of celer will remain integral to how we describe and celebrate movement. Embrace the power of celer to add momentum to your vocabulary and endeavors!
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190117 | https://www.youtube.com/watch?v=Je_f4RimfKI | Electromagnetic waves | Physics | Khan Academy
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Want to explore more? Check out the full electromagnetic radiation playlist here:
Electromagnetic (EM) waves are produced whenever electrons or other charged particles accelerate. The wavelength of an EM wave is the distance between consecutive points on the wave with the same phase (e.g. one crest to the next crest). The frequency of an EM wave is how many wave cycles are emitted, or pass a given point, per second. The speed of a wave is equal to its frequency times its wavelength. In vacuum, all EM waves travel at speed c = 3 x 10^8 m/s.
Sections:
00:00 - Intro
00:19 - What is an EM wave?
00:46 - How are EM waves created?
02:15 - Amplitude and phase
04:26 - Wavelength and frequency
06:00 - Wave speed
08:01 - Speed of EM waves in vacuum
09:09 - The EM spectrum
10:26 - Analog modulation
12:58 - Digital modulation
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Transcript:
Intro what's common between a Wi-Fi router our bodies and an incandescent bulb we all give out electromagnetic waves but why do we do that and why are say why are they all so different and how do we use some of them for Wireless Communications let's answer all of them let's start by asking ourselves what exactly are electromagnetic waves electromagnetic What is an EM wave? waves consist of oscillating electric and magnetic fields a changing electric field produces a changing magnetic field which produces a changing electric field and so on and so forth and that you can see in this animation the disturbance gets propagated but the beautiful thing is that this disturbance can get propagated in vacuum it does not require any medium and it carries energy that's for example how we get energy from the Sun in the form of electromagnetic waves How are EM waves created? but how do you produce these electromagnetic waves you ask well just accelerate charges accelerating charges will change electric Fields you will change magnetic fields and that's how electromagnetic waves will be generated a cool way to accelerate charge would be just to take an electron and just wiggle it up and down if you continuously wiggle an electron up and down it will continuously keep losing energy as electromagnetic waves so accelerated charges create electromagnetic waves so if you consider the Wi-Fi antenna for example you'll find that there will be electrons oscillating up and down very quickly okay and because electrons are charges oscillating up and down they will generate electromagnetic waves which we call radio waves similarly if you consider your own bodies you have electrons inside our body there are charges inside there are vibrating because just because of the temperature they're just vibrating over there and that will again generate electromagnetic waves which happen to be infrared radiation and the same thing will happen inside an incandescent bulb when the filament is heated it Heats to such a high temperature that the electrons are vibrating randomly and as a result you have electromagnetic waves being generated giving us light but of course now comes the big question why do we get radio waves over here infrared over here light rays over here what's the difference between them in all the cases it's the charges that are going up and down right then why do we get different kinds of electromagnetic waves what's the difference between them to answer that question we need to understand some of the Key properties of these electromagnetic waves so let's say this Amplitude and phase is our electromagnetic wave remember it has both electric and magnetic fields I'm not showing both of them imagine this is just electric field and it's moving so you can imagine the field is the whole wave is moving to the right for example okay and one of the Key properties of the wave is how big the wave is the height of the wave okay that's what I mean so that technically is called the amplitude of the wave so this height is what we call the amplitude so for example this would be a very small amplitude wave this would be a very large amplitude wave and of course remember when I say height I'm not there's nothing physical over here these are electric and magnetic field values right so going back to the animation it's more appropriate to think of amplitude as the maxim maximum strength of the electric or the magnetic fields another important feature of a wave is what we call its phase it's easier to understand what phase is if I were to look at just two specific points on a wave so let's cons compare only these two points on the wave and let's see what happens to them so I'll just only compare to those two points and see what happens to them as the wave moves forward I I'm going to move the wave to the right so notice they will start oscillating they'll go up and down but look they are going up and down together they are in sync with each other so we say they are oscillating in Phase with each other so you can think of phase as kind of representing where they are in their cycles of oscillation they're both these two are in Phase because they reach their valleys together they reach the zero together they reach the peak together and so on and so forth so any two points on the mountains are in Phase any two points on the valleys are in Phase with each other but what about these two points are they in face with each other let's see let's see again let's just concentrate on them and move the wave and see what happens hey no they're not what do you see you notice that when one is going up the other one is going down and so look they are not oscillating together they're not in Phase with each other in fact we say they're out of they're out of phase that's what we say but what's important is that they're not in Phase with each other okay this idea of phase will be super important later on we'll come back to it later on another important feature Wavelength and frequency of our wave is its wavelength you can think think of wavelength as the distance between any two consecutive Peaks or any two consecutive uh valleys or in fact any two consecutive points which are in Phase with each other we saw which are in Phase right so for example if you if you consider the distance between these two consecutive Peaks this length is what we'll call the wav length and it is usually represented by the symbol Lambda okay so again if you go back to our wave and this would represent short wavelengths this would represent long wavelengths okay and one last important feature of our wave is its frequency what is frequency you ask well take any point you want and then measure in 1 second how many waves go past that point number of waves passing through a point per second is what we call frequency you can think of it as number of waves passing through a point per second okay so the unit of the wavelength is meters because you're measuring distance and then what is the unit of frequency well it's number of waves per second so the frequency would be numerator is just a number so number per second would be second inverse but we often called second inverse when we use it for frequency as Hertz so for example if I say frequency is 10 Hertz it means from a given point 10 waves are passing by that point per second now here's the the cool thing about wavelength and Wave speed frequency if you know them if you know the wavelength and frequency you can figure out the speed of the wave let's see how in fact let's take some numbers directly to see how so let's take some simple numbers let's say we have a wave which has a wave length of 2 m and it has a frequency of 4 Hertz how do we figure out the speed of this wave well what we're going to do is let's imagine there's a wave that starts from here and we'll just wait for one second okay so wait for 1 second now in that 1 second I know four waves are going to pass by so let me just draw them so I'm waves going to start from here it's a peak so four waves are going to pass by in one second so here's wave number one Peak to Peak is one full wave okay wave number two wave number three and wave number four so in one second this whole thing happened in 1 second now if I figure out how much this total distance is I'm done because then that represents the total distance traveled by The Wave in 1 second and that is the speed so I need to calculate what this distance is to figure out the speed how do I do that well I know what the wavelength of this wave is right it's given to me so can you pause and think about what this total distance is going to be and as a result think about what the correlation between the velocity or the speed of the wave the frequency and the wavelength would be why why don't you pause and try all right okay so since the wavelength is 2 m and I have four waves the total distance traveled by the wave would be 2 4 it would be 8 m so in general if the wavelength is Lambda this would be Lambda and if the frequency is f there would be F number of waves over here that means the total distance CH would be f Lambda in 1 second and therefore the speed of the wave would be f Lambda now looking at this equation we Speed of EM waves in vacuum might think that hey the velocity of a wave depends on the frequency and the velocity of the wave depends on the wavelength but turns out things are a little bit more complicated it depends on which wave we are dealing with so now let's come back to electromagnetic waves turns out for electromagnetic waves in vacuum the velocity is just a constant any electromagnetic wave you take in vacuum it'll always travel at approximately 3 10 ^ 8 m/s we often call this C the speed of light speed of light in vacuum okay it is independent of the frequency of the wave it is independent of the wavelength of the wave which means for electromagnetic waves that are in vacuum since this is a constant we notice that that means if you have a high frequency wave it should have a lower wavelength if it has a high wavelength it should have a lower frequency so for electromagnetic waves in vacuum you can see that the frequency and the wavelength are inversely related to each other and now we can go back to our The EM spectrum original question why are these things giving out different kinds of electromagnetic waves well because the charges over here are oscillating at different frequencies in your Wi-Fi the charges are oscillating at lower frequencies so you get low frequency electromagnetic waves which we call radio waves inside our bodies the charges are vibrating at slightly higher frequencies and so now you get a slightly higher frequency waves which you call infrared ravs inside the light bub it's even higher frequency which we get called visible light because we can see it and just to give you a feeling for numbers okay so your radio waves that's coming from your Wi-Fi is in the order of about a billion Hertz meaning the electrons over here are oscillating about billion times per second okay and uh and when it comes to infrared it becomes 10 to the 13 for visible light becomes 10 to the 14 you can even have higher frequency waves which goes into ultraviolet rays you have xrays you have gamma rays and so on and you can also see as the frequency increases the wavelength is decreasing because we saw that the frequency and wavelength are inversely related for electromagnetic waves but that brings up the last question what really happens to these electromagnetic waves when they go and hit something for example what happens to these radio waves when you know from the Wi-Fi it goes to our phones for Analog modulation example well our phones also have a tiny antenna inside of them now in this transmitter antenna electron vibrates giving you electromagnetic waves in the receiver antenna for a phone the exact opposite happens the electromagnetic waves will fall on it and makes the electrons go up and down and that's how the energy from this oscillating electron is transferred to the electrons over here pretty cool huh but now we may be wondering well how do you transmit any information though for example the radio waves that we would be using in y5 could be about 5 GHz meaning 5 Time 5 billion Hertz that means the electrons over here are oscillating about 5 billion times per second and when the waves go and hit over here the electrons inside the antenna of our phones will also oscillate 5 billion times per second but what where is information being transferred over here how do you transform how do you transfer information that's a great question in order to transfer information we need to modulate this signal what do you mean by that say for example this is a signal that we want to send from the transmitter to receiver let's imagine radio transmitters and radio receivers you can imagine this is I don't know a signal that represents some song for example how do you transmit that well what we do is we take our radio wave and we modulate it an example of this would be you can change the amplitude of this radio wave according to the message signal and it will look somewhat like this I know it sounds it looks very complicated but what's important is look at the amplitude the amplitude nicely Mass sorry nicely matches the message signal this is called modulation in fact this is called the amplitude modulation because the amplitude of this radio wave is getting is changing in according to the message this is the message that is sent by the radio transmitter and when the when your radios receive it the electrons look even though they're vibrating with the same frequency the amplitude of the vibration will keep changing according to the message and that's how your radio receivers will detect you know what the message was this is what AM stands for amplitude modulation there are other kinds of modulation as well there's something called frequency modulation phase modulation and so on um the idea is you change some property of your radio wave in accordance to the message that you want to send but anyways since this is a continuous wave that is being modulated we call this an analog modulation your Wi-Fi does not do that your Wi-Fi does Digital modulation what we call digital modulation because it doesn't send analog signals like this your Wi-Fi sends messages in bits and zeros and months this is a digital message they are discrete see unlike the analog signal you only have two values you either get zero or one nothing in between so here we do digital modulation what would that look like well again if we were to modulate the signal using this message we might get a modular signal that would look somewhat like this again notice the amplitude is either maximum or zero okay representing your zero or one bits that you get and again when this goes and hits the receiver receiver antenna in your phone for example based on the electrons will vibrate accordingly and we'll be able to detect what this message was this is called digital modulation and all of this is an oversimplification and there's so much more to it in a nutshell this is how our communication work work today works today wireless longdistance communication satellite communication what you want to talk about all of that is by using by harnessing the power of electromagnetic waves it's just fascinating stuff |
190118 | https://www.frontiersin.org/journals/medicine/articles/10.3389/fmed.2024.1476361/full | Frontiers | The value of endocervical curettage for diagnosis of cervical precancers or worse at colposcopy of women with atypical glandular cells cytology
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ORIGINAL RESEARCH article
Front. Med., 13 December 2024
Sec. Obstetrics and Gynecology
Volume 11 - 2024 |
This article is part of the Research Topic Advancements in Diagnostic and Management Strategies for Gynecological PathologiesView all 23 articles
The value of endocervical curettage for diagnosis of cervical precancers or worse at colposcopy of women with atypical glandular cells cytology
Yusha Chen1†Fanghong Wen 2†Jiancui Chen 1Huifeng Xue 1Xiangqin Zheng3Diling Pan 4
1 Cervical Disease Diagnosis and Treatment Health Center, Fujian Maternity and Child Health Hospital College of Clinical Medical for Obstetrics & Gynecology and Pediatrics, Fujian Medical University, Fuzhou, China
2 Department of Obstetrics, Fujian Maternity and Child Health Hospital College of Clinical Medical for Obstetrics & Gynecology and Pediatrics, Fujian Medical University, Fuzhou, China
3 Department of Gynecology, Fujian Maternity and Child Health Hospital College of Clinical Medical for Obstetrics & Gynecology and Pediatrics, Fujian Medical University, Fuzhou, China
4 Department of Pathology, Fujian Maternity and Child Health Hospital College of Clinical Medical for Obstetrics & Gynecology and Pediatrics, Fujian Medical University, Fuzhou, China
Objective: This study evaluates the effectiveness of endocervical curettage (ECC) in identifying additional cervical cancer and its precursors in women with atypical glandular cells (AGC) cytology.
Methods: We conducted a retrospective analysis of medical records for women referred to colposcopy with AGC cytology between January 2019 and December 2023. The study included 433 women with AGC cytology who underwent both biopsy and ECC. Clinical characteristics such as demographics, clinical history, cytology, HPV status, colposcopic findings, and pathology were analyzed. Chi-square and Fisher's exact tests were applied to compare the characteristics of ECC-diagnosed cervical precancers or worse (HSIL+) and normal/low-grade squamous intraepithelial lesions (LSIL).
Results: The overall detection rate of HSIL+ in this population was 19.4% (86/443), with ECC alone identifying HSIL+ in 1.3% (6/443) of cases. However, ECC showed greater utility in certain subgroups. The highest additional HSIL+ detection from ECC was observed in women with HPV 16/18 infection (7.2%) and those with AGC-FN cytology (4.4%). ECC's additional yield of HSIL+ was higher in those with normal or LSIL colposcopic impressions compared to those with HSIL+ impressions. Conversely, no additional HSIL+ cases were identified by ECC alone in women under 30 years old, those with negative high-risk HPV results, or those with type 1/2 transformation zones.
Conclusion: For women with AGC cytology, ECC should be performed in patients with AGC-FN cytology, HPV 16/18 infections, type 3 transformation zones, and normal or low-grade colposcopic impressions. This approach enhances the identification of HSIL+ cases by reducing false negatives. However, for women younger than 30 years old and those with type 1/2 transformation zones, ECC offers limited benefit.
1 Introduction
Colposcopy is an important diagnostic procedure for identifying precancerous lesions of the cervix, offering a thorough examination to detect and assess abnormal areas. Nevertheless, the endocervical canal often remains difficult to visualize during colposcopy, which can result in missed significant lesions. Lesions of the cervical glandular epithelium can arise not only from the squamocolumnar junction but also from columnar cells situated further up the cervical canal, frequently involving the endocervical canal. Atypical glandular cells (AGC) are the most common cytological type of cervical glandular epithelial lesions, which include adenocarcinoma in situ (AIS) and adenocarcinoma (1).
Endocervical curettage (ECC) involves scraping cells from the endocervical canal and is used to detect abnormalities in this otherwise challenging-to-access area. The effectiveness of ECC in lowering the miss rate of colposcopy findings remains controversial. The Society of Canadian Colposcopists (2) and the American Society for Colposcopy and Cervical Pathology (ASCCP) (3) advocate for ECC in cases where atypical glandular cells are found in cytology. In contrast, the British Society for Colposcopy and Cervical Pathology (BSCCP) (4) does not support the routine use of ECC during colposcopy, even for glandular lesions.
Research has shown that ECC can effectively detect additional cervical cancer and its precursors that might be overlooked by biopsy alone in women with high-risk cytology, including atypical squamous cells, favor high-grade (ASC-H), high-grade squamous intraepithelial lesions (HSIL), and AGC (5–7). Given its rarity—typically <1% of cervical smear results (8)—AGC has not been extensively examined as a standalone condition. In China, ECC is routinely performed when AGC is identified cytologically, providing an opportunity to evaluate its practical effectiveness.
The Cervical Disease Diagnosis and Treatment Health Center at Fujian Maternity and Child Health Hospital, the largest center for cervical diseases in Fujian province, China, has a robust data collection system that includes histopathology, cytopathology, colposcopic findings, and patient details from all colposcopy exams. This retrospective study utilizes this comprehensive dataset to assess how effectively ECC identifies additional cervical cancer and its precursors in women with AGC cytology.
2 Materials and methods
2.1 Study design
Data were collected from the electronic medical records of women who underwent colposcopic examinations following abnormal cervical screening results at the Cervical Disease Diagnosis and Treatment Health Center of Fujian Maternity and Child Health Hospital. The patient selection process is detailed in Figure 1. Between January 2019 and December 2023, 19,263 colposcopies were performed at the center. Out of the 452 patients identified with AGC cytology, nine were excluded: three due to pregnancy, three following total hysterectomy, and three because of a closed cervical ostium that made endocervical curettage impossible. Therefore, the study included 433 patients with AGC cytology who had both biopsy and ECC. The time between cytology/HPV testing and colposcopy was always <3 months. Collected demographic and clinical information included age, gravidity, parity, menopausal status, cytological results, HPV status, colposcopic impressions, transformation zone type, and pathological findings. This study was approved by the Ethics Committee at Fujian Maternity and Child Health Hospital, Affiliated Hospital of Fujian Medical University (2024KY014). Informed consent was waived due to the retrospective nature of the study. The findings were reported in accordance with the Strengthening the Reporting of Observational Studies in Epidemiology (STROBE) guidelines.
Figure 1
Figure 1. Flow chart of the study. AGC, atypical glandular cells; ECC, endocervical curettage.
2.2 Cytology and HPV testing
Cytology was conducted using the liquid-based ThinPrep test. In this procedure, a cell brush is inserted into the external cervical canal to collect cells from the exocervix and endocervix. These cells are then smeared onto a slide and fixed. According to the Bethesda System (9), atypical glandular cells (AGC) can be categorized into AGC-NOS (not otherwise specified, including endocervical and endometrial) and AGC-FN (favor neoplastic).
A total of three HPV testing approaches were utilized: the hybrid capture 2 assay (Qiagen, Hilden, Germany), which detects the DNA of 13 high-risk oncogenic HPV types (including HPV types 16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, and 68); the polymerase chain reaction-reverse dot blot (PCR-RDB) HPV genotyping method (Yaneng® Limited Corporation, Shenzhen, China), capable of identifying 18 high-risk HPV types (such as HPV 16, 18, 31, 33, 35, 39, 45, 51, 52, 53, 56, 58, 59, 66, 68, 73, 82, and 83) along with 5 low-risk types (6, 11, 42, 43, and 81); and the Aptima assay (Gen-Probe Inc., San Diego, CA), which focuses on the E6/E7 mRNA of 14 high-risk HPV types (including HPV 16, 18, 31, 33, 35, 39, 45, 51, 52, 56, 58, 59, 66, and 68). HPV status is classified as HPV 16/18, non-16/18 high-risk HPV, or negative.
2.3 Colposcopy/ECC and biopsy procedures
All patients with AGC-NOS (endometrial) and negative high-risk HPV results underwent a segmented curettage procedure before colposcopy, with no abnormalities detected in the histopathology. Colposcopic examinations were conducted using a dynamic spectral imaging (DSI) colposcope (Leisegang, Germany). Up to four lesion-directed biopsies were obtained from distinct areas of the epithelium that showed acetowhitening, metaplasia, or other visible abnormalities after the application of 5% acetic acid and Lugol's iodine using Tischler biopsy forceps. If fewer than four directed biopsies were collected, additional biopsies were taken from areas of normal-appearing cervical transformation zone epithelium until a total of four biopsies were achieved. If the cervix appeared entirely normal, random biopsies were performed at the squamocolumnar junction (SCJ) in the 3, 6, 9, or 12 o'clock positions, assuming the SCJ was fully visualized. When the SCJ was not completely visible, biopsies were collected from the external cervical ostium at the corresponding 3, 6, 9, or 12 o'clock positions. Following cervical biopsy, endocervical curettage (ECC) was performed using a Kevorkian curette. Results from ECC and biopsies were categorized as normal, low-grade squamous intraepithelial lesions (LSIL), HSIL, or invasive cancer based on the Lower Anogenital Squamous Terminology system (10). The diagnosis was based on the highest grade lesion identified, with HSIL+ encompassing HSIL, adenocarcinoma in situ (AIS), and invasive cancers, while all other findings were classified as <HSIL. The pathological evaluation of cervical biopsies and ECC samples was carried out independently by two senior pathologists who were blinded to each other's results.
2.4 Statistical analysis
Statistical analysis was performed using IBM SPSS Statistics version 26.0 (IBM Corp, Armonk, NY). Statistical significance was determined using two-sided tests with a threshold of P< 0.05. Categorical variables were expressed as frequencies and percentages. For analysis, chi-square tests were used, and Fisher's exact test was applied when appropriate.
3 Results
3.1 Study sample and data features
Table 1 provides an overview of the demographic and clinical characteristics of the study participants. The mean age of the patients was 41.4 years, with a range spanning from 20 to 77 years. Among the women, 81.7% (n = 362) had experienced one to three childbirths, and a majority, 86% (n = 381), were premenopausal. Additionally, 54.2% (n = 240) had a transformation zone classified as type 3. The most common cytological result was AGC-NOS, which constituted 78.5% (n = 348) of the cases. HPV 16/18 infections were detected in 15.6% (n = 69) of the participants, whereas 22.4% (n = 99) had infections with other high-risk HPV types. Colposcopic findings were normal in 32.7% (n = 145) of cases, low-grade in 51.7% (n = 229), high-grade in 12.2% (n = 54), and cervical cancer was diagnosed in 3.4% (n = 15). Histopathological examination of biopsies revealed HSIL+ in 18.1% (n = 80) of cases, while ECC histopathology identified HSIL+ in 7.9% (n = 35). Overall, 11.7% (n = 52) of the cases were diagnosed with HSIL, 3.8% (n = 17) with AIS, 1.4% (n = 6) with SCC (cervical squamous carcinoma), and 2.5% (n = 11) with ACC (cervical adenocarcinoma).
Table 1
Table 1. Clinical features of the study cohort.
3.2 HSIL+ diagnostic yield by biopsy and ECC
The diagnostic yield for HSIL+ by biopsy and ECC is presented in Table 2. The overall concordance rate between histopathologic results of biopsy and ECC was 81.3% (360/443). In 17.4% (77/443) of cases, the biopsy revealed a higher grade of pathologic results than ECC. Conversely, ECC showed a higher grade of pathologic results in 1.3% (6/443) of cases, including 4 cases of HSIL, 1 case of AIS, and 1 case of ACC, with corresponding biopsy pathology results being normal. Biopsy alone detected 93% (80/86) of HSIL+ cases missed by ECC. Additionally, 7.7% (4/52) of HSIL cases, 5.9% (1/17) of AIS cases, and 9.0% (1/11) of ACC cases were missed by biopsy alone but were detected when biopsy was combined with ECC.
Table 2
Table 2. Comparison of biopsy and ECC histopathology diagnostic results.
3.3 Clinical characteristics associated with HSIL+ diagnostic yield by ECC
Table 3 presents the clinical features linked to HSIL+ detection via ECC. There was no significant difference between different age groups in terms of ECC-detected normal/LSIL and HSIL+ results (P = 0.257). Women with AGC-FN had a higher likelihood of HSIL+ results on ECC compared to those with AGC-NOS (28.6% vs. 8.6%, P = 0.001). The HPV16/18 positive group exhibited the highest rate of HSIL+ detection by ECC, whereas the high-risk HPV negative group was more likely to present normal or LSIL results (P< 0.001). Notably, HSIL+ was more frequently detected by ECC when colposcopy impressions were normal or LSIL compared to when colposcopy impressions were HSIL+ (P< 0.001). All HSIL+ ECC results were found in type 3 transformation zones. Although menopausal women showed a higher propensity for HSIL+ ECC results (25.7% vs. 13.0%), this finding was not statistically significant (P = 0.07).
Table 3
Table 3. Comparison of clinical characteristics between normal/LSIL and HSIL+ identified by ECC.
3.4 Stratification of HSIL+ diagnostic yield based solely on ECC
Table 4 presents the rates of HSIL+ detection via ECC and biopsy, organized by age group, cytology, HPV infection, colposcopic impression, transformation zone type, and menstrual status. The additional diagnostic yield of HSIL+ by ECC varied from 0.0% to 7.2% across different risk categories. The highest yield of HSIL+ was observed in women with HPV16/18 infection, reaching 7.2% (5/69). ECC identified an additional 4.4% (2/45) of HSIL+ cases in women with AGC-FN cytology. The most notable increase in detection rate from ECC was in patients aged 40–49, at 2.0% (3/152). For women with normal or LSIL colposcopic impressions, the HSIL+ diagnostic yield by ECC was 3.4% (6/374), whereas in the type 3 transformation zone group, ECC detected 2.5% (6/240) more HSIL+ cases. ECC also identified HSIL+ in 1.3% (5/381) of premenopausal women and 1.6% (1/62) of postmenopausal women, cases that were missed by biopsy alone. In contrast, ECC did not detect any additional HSIL+ cases in patients under 30 years, those with negative high-risk HPV infection, HSIL+ colposcopic impressions, or type 1/2 transformation zones.
Table 4
Table 4. Stratified diagnostic yield of HSIL+.
4 Discussion
This study evaluated the effectiveness of ECC in identifying HSIL+ among women referred for colposcopy due to AGC cytology. The results revealed an overall HSIL+ detection rate of 19.4% (86/443) in this population, with ECC alone detecting HSIL+ in 1.3% (6/443) of cases. This suggests that approximately 100 women would need to endure the extended discomfort of ECC procedures to identify one additional HSIL+ case not captured by biopsy alone. Nonetheless, ECC proved more beneficial in specific subgroups. The rate of HSIL+ detection by ECC alone increased to 2.0% (8/631) among women aged 40–49 years. The highest additional HSIL+ detection from ECC was observed in women infected with HPV 16/18, at 7.2%, and in those with AGC-FN cytology, where the additional detection rate was 4.4%. ECC's additional yield of HSIL+ was higher in the group with normal or LSIL colposcopic impressions compared to the HSIL+ colposcopic impression group. Conversely, no additional HSIL+ cases were missed by biopsy alone in women under 30 years of age, those with negative high-risk HPV results, or those with type 1/2 transformation zones.
AGC represent cytological abnormalities in which glandular cells show changes but do not meet the criteria for adenocarcinoma in situ or invasive adenocarcinoma of the cervix uteri. These abnormalities can range from benign changes and cervical precursor lesions of glandular or squamous origins to invasive cervical cancer and other gynecological malignancies (8). This study found that 19.4% of women with AGC had cervical intraepithelial neoplasia grade two or worse, including adenocarcinoma in situ (CIN2+/AIS+), a figure comparable to a meta-analysis (11) reporting that 19.8% of women with AGC will have high-grade squamous intraepithelial lesions (HSIL+). Although AIS and ACC often originate within the endocervical canal, this study demonstrated that all AIS cases were detectable by biopsy alone. Of the 11 ACC cases, only one with a type 3 transformation zone would have been missed without the use of ECC.
The varying detection rates of HSIL+ through ECC across different subgroups underscore the heterogeneity of AGC presentations. The additional detection of HSIL+ relies on two key factors: the risk of HSIL+ and whether the colposcopy comprehensively visualizes the transformation zone where nearly all cervical cancers develop. HPV genotype is a significant risk factor. In women with AGC cytology results, HPV 16/18 infection poses a higher risk compared to other high-risk HPV infections and HPV-negative women (12–14). Numerous studies have indicated that ECC has a higher additional detection rate for HSIL+ in patients positive for HPV 16/18 (6, 15–18). Our findings align with these studies. When AGC is identified cytologically, the cumulative incidence risk of cervical cancer is moderately lower than that of HSIL but considerably higher than that of LSIL (1), thus colposcopy is recommended for all patients regardless of HPV status (19). If colposcopy impression is normal or LSIL, HSIL+ may potentially missed by biopsy alone. Women with AGC-FN cytology have a higher risk of HSIL+, especially AIS+, compared to those with AGC-NOS (20). The type 3 transformation zone poses significant challenges for adequate visualization and sampling during colposcopy, making ECC a valuable adjunctive procedure. Conversely, the absence of missed HSIL+ cases in younger women, those with type 1/2 transformation zones, and those negative for high-risk HPV suggests that ECC may have limited utility in these subgroups.
Our findings concur with previous research indicating a relatively low yield of ECC in detecting additional HSIL+ cases in certain populations. For instance, Solomon et al. (6, 21) reported similar low detection rates of HSIL+ by ECC in women younger than 30 years old. This can be attributed to the lower incidence of cervical cancer in women younger than 30 years old (22) and the predominance of type 1/2 transformation zones in this demographic (23).
In light of the contrasting guidelines regarding ECC, our findings highlight the importance of context and patient-specific factors when considering its use. While the ASCCP recommends ECC for patients with atypical glandular cells to enhance detection rates, the hesitancy of the BSCCP to endorse routine ECC underscores the need for a more nuanced approach. Our results suggest that ECC may be particularly beneficial for women with AGC-FN cytology and specific risk factors like HPV 16/18 infections, yet show limited advantages for younger women and those with type 1/2 transformation zones. This emphasizes the necessity for tailored decision-making in clinical practice, aligning with the evolving discourse on the effectiveness of ECC. This approach could reduce unnecessary procedures in low-risk populations while ensuring high-risk cases are adequately identified and treated.
Several limitations of this study should be acknowledged. First, the retrospective design may introduce selection bias. Second, the sample size, while sufficient for detecting significant differences, may limit the generalization of our findings to broader populations. Additionally, variations in colposcopy and biopsy techniques among practitioners could influence the detection rates of HSIL+.
Future research should focus on prospective studies to validate these findings in larger and more diverse populations. Investigating the molecular and histopathological characteristics of AGC subtypes may provide deeper insights into their malignant potential and guide more precise management strategies. Additionally, exploring the cost-effectiveness of targeted ECC in high-risk subgroups could support evidence-based guidelines for colposcopy referral and follow-up.
5 Conclusion
This study aimed to improve the detection rate of HSIL+ in women with AGC while minimizing unnecessary discomfort. We identified high-risk groups warranting ECC: those with a normal/low-grade colposcopic impression, AGC-FN cytology, type 3 transformation zones, and HPV 16/18 infection. Conversely, ECC offers no benefits for women under 30 years old with type 1/2 transformation zones. These findings may reduce missed occult HSIL+ cases and contribute to the evidence base regarding the clinical use of ECC.
Data availability statement
The datasets generated and/or analysed during the current study are not publicly available due personal information protection, patient privacy regulation, and medical institutional data regulatory policies, etc., but are available from the corresponding author on reasonable request.
Ethics statement
The studies involving humans were approved by the Ethics Committee of Fujian Maternity and Child Health Hospital, Affiliated Hospital of Fujian Medical University. The studies were conducted in accordance with the local legislation and institutional requirements. The ethics committee/institutional review board waived the requirement of written informed consent for participation from the participants or the participants' legal guardians/next of kin because an exemption of informed consent was obtained due to the retrospective nature of the study.
Author contributions
YC: Funding acquisition, Writing – review & editing, Writing – original draft. FW: Writing – original draft, Methodology. JC: Writing – original draft, Project administration, Methodology, Data curation. HX: Writing – original draft, Project administration, Formal analysis, Conceptualization. XZ: Writing – review & editing, Resources, Formal analysis, Conceptualization. DP: Writing – original draft, Project administration, Funding acquisition, Formal analysis, Conceptualization.
Funding
The author(s) declare financial support was received for the research, authorship, and/or publication of this article. This study was supported by grants from Startup Fund for scientific research, the Fujian Medical University (Grant Number: 2021QH1177) and the Fujian Provincial Natural Science Foundation of China (Grant Number: 2022J011030).
Acknowledgments
The authors thank Liyu Dai for supporting in the clinical management and Dafeng Huang for encouragement.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher's note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Keywords: cervical curettage, colposcopy, biopsy, cervical lesions, atypical glandular cells
Citation: Chen Y, Wen F, Chen J, Xue H, Zheng X and Pan D (2024) The value of endocervical curettage for diagnosis of cervical precancers or worse at colposcopy of women with atypical glandular cells cytology. Front. Med. 11:1476361. doi: 10.3389/fmed.2024.1476361
Received: 05 August 2024; Accepted: 25 November 2024;
Published: 13 December 2024.
Edited by:
Cristina Secosan, Victor Babes University of Medicine and Pharmacy, Romania
Reviewed by:
Elliot M. Levine, Rosalind Franklin University of Medicine and Science, United States
Teodor Florin Georgescu, Carol Davila University of Medicine and Pharmacy, Romania
Nicolae Bacalbasa, Carol Davila University of Medicine and Pharmacy, Romania
Copyright © 2024 Chen, Wen, Chen, Xue, Zheng and Pan. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Diling Pan, pdl269@fjmu.edu.cn; Xiangqin Zheng, Zhengxq1215@163.com
†These authors have contributed equally to this work and share first authorship
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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Edited by
Cristina Secosan Victor Babes University of Medicine and Pharmacy, Romania
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Elliot M Levine Rosalind Franklin University of Medicine and Science, United States Teodor Florin Georgescu Carol Davila University of Medicine and Pharmacy, Romania Nicolae Bacalbasa Carol Davila University of Medicine and Pharmacy, Romania
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190119 | https://www.youtube.com/watch?v=acvJ_BzBEmw | Coefficient in Binomial Expansion [Hindi] | Binomial Theorem | Grade 11 | Maths | Khan Academy
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दिया है नवा का वर्ग जमा 6 एक्स की घात ती और इसकी घात है पा इसके हल के लिए हम बानो एक्सपेंशन या पास्कल त्रिभुज का उपयोग करेंगे इसके लिए मैं इन पदों में से एक ऐसा पद लेना चाहूंगा जिसके गुणांक एक्स और वा की छठवीं घात से संबंधित हो तो इस विस्तार में कोई पद एक्स की छठवीं घात और वा की छठवीं घात होगा और तब हम देखेंगे कि उस पद का गुणांक क्या होगा मैं आपको सलाह देता हूं कि इस वीडियो को रोक कर आप भी इसे अपने तरीके से हल करें मैं मानकर चलता हूं कि आपने ऐसा किया होगा और पहली नजर में आपको यह थोड़ा मुश्किल लग रहा होगा देखिए मैंने केवल यहां फिफ्थ पावर की बात कही है तो सवाल यह है कि हमें x की पावर स और y की पावर सिक्स कैसे मिलेगी इन द्विपद पदों को देखिए यहां पर y स्क का पद है और x की पावर 3 वाला पद भी है इन पावर्स पर ध्यान देंगे तो इनकी पावर फिफ्थ पावर से अधिक है अब यह सोचिए कि एक्स की पावर सिक्स और वा की पावर सिक्स वाले पद कहां से आए जरा इसे ऐसे समझते हैं कि इन पदों को गुणांक के बिना देखते हैं साथ ही हम इन्हें द्विपद पद में लेने से पहले इनके गुणज के रूप में समझे अब हम लेते हैं पहले पद को पहले पद से आरंभ करते हैं और इस घात को लेते हैं जो कि पूरे द्विपद की घात है और हम देखेंगे कि प्रत्येक पद की पावर कम होती जाएगी और ये इस प्रकार कॉपी और पेस्ट होता चला जाए इस प्रकार कॉपी और पेस्ट करेंगे हम और यह हो गया हमारा पहला पद सेकंड टर्म दूसरा पद और यह हो गया तीसरा पद चौथा पद चौथा पद फोर्थ टर्म फिफ्थ टर्म पांचवा पद और यह छठा पद हम हमेशा घातांक से एक पद ज्यादा लेते हैं अब इसके घातांक पर बात करने से पहले हम दूसरा पद भी लेते हैं तो दूसरे पद को मैं इस तरह लिखता हूं दूसरा पद गुना 6 एक्स की घात तीन इसे कॉपी पेस्ट कर लेते हैं कॉपी और यह पेस्ट यहां भी पेस्ट कर दिया इसे एक बार और पेस्ट करेंगे इसे यहां भी पेस्ट कर दिया और यह हो गया पेस्ट इसे देखने के लिए हमें इसे खिसका हो अब इस पर हम घात लगाते हैं घात लिखते हैं यह हो गई फिफ्थ पावर यह चौथी घात तीसरी घात दूसरी घात पहली घात और यह घात शून्य और नीले पदों के लिए घात होगी यह होगी शून्य घात यह पहली घात पहली घात दूसरी घात तीसरी घात चौथी घात और यह घात पांच ठीक यहां पर और इनमें प्रत्येक के सामने एक गुणांक हो सकता है जैसे इसके सामने इन सबके सामने गुणांक हो सकते हैं और उन सबको हम जोड़ते हैं लेकिन वो पद कहां है जिसकी हमने बात की थी x की पावर स और y की पावर स यदि हम यहां देखें ये है x और यहां है y का वर्ग की घात चार ये हो जाएगा y की घात आठ यहां पर है y का स्क्वायर और उसकी तीसरी तो ये y की घात छ है ये x की घात तीन है और उसका वर्ग है तो ये x की घात छ हो गया इसका अब गुणांक यानी को एफिशिएंट क्या है इसके सामने वाले पद का गुणांक क्या है यह देखना है और इसी प्रकार इस तीसरे पद को भी देखते हैं तो हम क्या पाते हैं तो गुणांक क्या होने वाला है अब हमें यह स्पष्ट है कि यह गुणांक जिसे हम यहां रखें तो हम द्विपद प्रमेय का उपयोग करके इसे ज्ञात कर सकते हैं क्या ये ये होगा मैं इस समस्या का एक हल सोचता हूं यह होगा इस गुणांक से प्राप्त गुणनफल चाहे यहां इनके गुणांक कुछ भी हो यदि हम लेते हैं थर्ड पावर 6 स्क्ड तो हम इसे हल करने की कोशिश करते हैं कि यह क्या होगा पहले देखते हैं कि यह पीले वाला भाग क्या होगा इसे ज्ञात करने के बहुत से तरीके हैं हम पास्कल त्रिभुज का प्रयोग कर सकते हैं या फिर संचय नियम का इनके प्रयोग से हम यहां का गुणांक ज्ञात कर सकते हैं यहां का गुणांक होगा 5 चु 0 और यहां का गुणांक होगा यहां का को एफिशिएंट होगा फ चूज वन हमने इसे पहले भी बहुत बार देखा है देखा कि आपने गुणांक ऊपर से चुनते हैं यह पाच यह पूरे बानो मियल का घातांक है और यहां पर इस दूसरे नंबर को चुनने के लिए हम लेते हैं दूसरे पद का घातांक तो इस प्रकार से यहां आ जाएगा 5 चुनि एक और अब देखते हैं यहां पर यह जो पीले रंग में है यहां पर होगा 5 फ चूज टू फ चूज टू अब ये 5 चूज टू किसके बराबर है यह बराबर होगा 5 फैक्टोरियल बटा 2 फटो गुना टाइम्स 5 - 2 इसे ऐसे लिखते हैं गुणा 5 माइनस 2 फैक्टोरियल और अब पा का फैक्टोरियल देखते हैं यह है 5 गुना 4 गुना 3 गुना दो यहां एक भी लिख सकते हैं उससे कोई फर्क नहीं पड़ेगा बटा दो का फैक्टोरियल यहां मैं एक लिख ही देता हूं इससे क्लियर हो जाएगा दो का फैक्टोरियल दो गुना एक और यह जो यहां पर है यह तीन का फैक्टोरियल है तो तो यहां लिखते हैं 3 2 गु 1 तो इसे हल कर लेते हैं ये तीन इससे कैंसिल हो जाएगा ये दो इस दो से कैंसिल हो गया ये एक एक से कैंसिल हो गया और ये चार डिवाइड हुआ दो से हमें मिल गया यहां दो और 5 गुना 2 होता है 10 10 ये संख्या आ जाएगी यहां ये को एफिशिएंट हमें मिला 10 अब हम इसे ज्ञात करेंगे पास्कल त्रिभुज से हम कह सकते हैं कि ये है द्विपद जब हम इसे दूसरी घात पर ले जाते हैं यह हो जाता है 12 2 1 और जब ले जाते हैं इसे तीसरी घात तब इसके गुणांक होते हैं 13 1 जब हम इसे चौथी घात तक ले जाते हैं तब यह हो जाता है 1 46 41 और जब ले जाते हैं पांचवी घात तक जिसके बारे में हम बात कर रहे हैं तब गुणांक न 5 10 फ सॉरी 10 10 5 और वन और हम जानते हैं कि जब यह तीसरा पद होगा तब ये गुणांक होगा 1 2 3 तीसरा पद तो हम जानते हैं कि यह होगा 10 चलिए हम अपने प्रश्न का उत्तर ढूंढते हैं कि x की पावर सि और y की पावर सिक्स को किससे गुणा करें अब हम इस प्रसरण को वापस लिखते हैं तो ये होगा 10 गुना ती की घात ती 3 टू द थर्ड पावर गुणा y स् पर भी पावर तीन है यानी ये है y की घात 6 y की घात 6 गुना 6 स्क्ड गु 6 का वर्ग गुना x की घात 3 का वर्ग जो होगा x की घात 3 गु 2 यानी x की घात 6 और ये बराबर होगा ये होगा 10 गुना 10 टाइम्स 27 27 गुना 36 टाइम्स 36 और यहां लिखेंगे हम x की घात 6 और y की घात 6 और यह बराबर होगा 270 गु 36 अब कैलकुलेटर की मदद लेते हैं समय बचाने के लिए जरूरी है ये 270 36 जो बराबर है 9720 तो ये होगा गुणांक को एफिशिएंट यह होगा 9720 एक्स की घात 6वा की घात 6 9720 एक्स टू द 6वा टू द 6 और यह हल हो गया |
190120 | https://www.media4math.com/library/lesson-plan-ratios-proportions-and-percents-gr-7-lesson-1-introduction-ratios-and-unit | Skip to content
Algebra >> Ratios, Proportions, and Percents >> Ratios and Rates
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Display Title Lesson Plan--Ratios, Proportions, and Percents (Gr 7)--Lesson 1--Introduction to Ratios and Unit Rates
Lesson Plan: Introduction to Ratios and Unit Rates
This lesson introduces seventh-grade students to ratios and unit rates, emphasizing their real-world applications. Through engaging activities and problem-solving exercises, students will develop proportional reasoning skills.
Lesson Objectives
Understand ratios and express them in different forms (a:b, a to b, a/b).
Calculate unit rates to compare different quantities.
Apply ratios and unit rates to real-world problems, such as speed and pricing.
Key Vocabulary
Ratio: A comparison between two quantities (e.g., 3:2).
Rate: A ratio comparing different units (e.g., miles per hour).
Unit Rate: A rate with a denominator of 1 (e.g., $2 per apple).
Teaching Examples
Calculating a Ratio: Determine the ratio of boys to girls in a class.
Using Equivalent Ratios: Scale up a recipe proportionally.
Solving a Proportion: Use equivalent ratios to find a missing value.
Subscribers to Media4Math can download this lesson plan in PDF format for easy classroom use.
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| Common Core Standards | CCSS.MATH.CONTENT.7.RP.A.1 |
| Grade Range | 6 - 8 |
| Curriculum Nodes | Algebra • Ratios, Proportions, and Percents • Ratios and Rates |
| Copyright Year | 2025 |
| Keywords | ratios, unit rates, rates | |
190121 | https://math.stackexchange.com/questions/4652805/finding-orientation-of-a-rectangle-using-the-points-sampled-on-its-surface | geometry - Finding orientation of a rectangle using the points sampled on its surface - Mathematics Stack Exchange
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Finding orientation of a rectangle using the points sampled on its surface
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If I have n n number of vectors p∈R 2 p∈R 2 on a surface of a rectangle, would it be possible to estimate the underlying rectangle orientation? The rectangle show in the figure is a virtual rectangle, we do not know the exact orientation of the virtual rectangle. Dimension of the rectangle is know, and the all the vectors are in same reference frame.
Initial thinking:
use von Mises distribution to sample rotation value centered at the mean of p p. Most of the sampled rotation are not valid.
Take two vectors p 1,p 2 p 1,p 2 randomly, and find the angle between p 1 p 1 and p 2 p 2. Continue until the distance between all the vectors are know. Although this method seems to be shedding light on the orientation, however, I do not know how to proceed further with this method.
Is there any mathematical technique that can handle this?
example rectangle with sampled points on top of it
geometry
euclidean-geometry
rotations
polygons
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asked Mar 5, 2023 at 21:55
goldfinchgoldfinch
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I think the first thing to do would be to find the largest distance between any two of the points. If that distance is ≤≤ the short side of the rectangle, then it can be in any orientation. But if that distance is close to the length of a diagonal, as in your example, then the rectangle has to be close to the two orientations with those points in diagonally opposite corners.Paul Sinclair –Paul Sinclair 2023-03-06 19:00:33 +00:00 Commented Mar 6, 2023 at 19:00
One other step would be to find the convex hull of the points. Only the points on the hull tell you anything useful about the orientation. If the hull points are in a rectangle, the interior points have to be as well, so there is nothing to learn from them. This will significantly cut down on the points you have to examine to arrive at an orientation. In your example, it looks like about 7 points are on the hull (estimated by eyeball, it might be only 5 or 6). The rest can be ignored.Paul Sinclair –Paul Sinclair 2023-03-06 20:01:17 +00:00 Commented Mar 6, 2023 at 20:01
Thank you. I will try your suggestions, and update the results here.goldfinch –goldfinch 2023-03-06 20:44:45 +00:00 Commented Mar 6, 2023 at 20:44
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190122 | https://www.doubtnut.com/qna/56435607 | The figure shows a plot of terminal voltage 'V' verus the current 'I' of a given cell. Calculate from the graph (a) emf of the cell and (b) internal resistance of the cell. The figure shows a plot of terminal voltage 'V' verus the current 'I' of a given cell. Calculate from the graph (a) emf of the cell and (b) internal resistance of the cell.
More from this Exercise
(a) For emf of the cell, current I = 0 Then T.P.D. = 6V (Maximum) called emf of the cell so emf = 6V (b) From V=E−IR V=4V I=1 Amp 4=6−1.r⇒r=6−4=2Ω = Internal Resistance of the cell.
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The emf of a cell having internal resistance is
A cell of emf 'ε' and internal resistance 'r' is connected across a variable resistor 'R'. Plot a graph showing variation of terminal voltage 'V' of the cell versus the current 'l'. Using the plot, show how the emf of the cell and its internal resistance can be determined.
Knowledge Check
For a cell, the graph between the potential difference (V) across the terminals of the cells and the current I drawn from the cell is as shown in figure. Calculate the e.m.f. and the internal resistance of the cell. ocr_image
for a cell, the graph between the p.d. (V) across the terminals of the cell ad te current I drawn from the cell is shown in the fig. the emf and the internal resistance of the cell is E and r respectively. ocr_image
For a cell, a graph is plotted between the potential difference V across the terminals of the cell and the current I drawn the cell. The emf and the internal resistance of the cell are E and r, respectively. Then ocr_image
A cell of emf ε and internal resistance r is connected across a variable resistor R. Plot a graph showing variation of terminal voltage V of the cell versus the current I. Using the plot, show how the emf of the cell and its internal resistance can be determined.
A cell of emf 2V produces 0.3A current when connected to a resistor of 5Ω. Calculate the terminal voltage and internal resistance of the cell.
The potential difference of a cell in an open circuit is 6 V, which falls to 4 V when a current of 2 A is drawn from the cell. Calculate the emf and the internal resistance of the cell.
Potential differences across the terminals of a cell were measured (in volt) against different currents (in ampere) flowing through the cell. A graph was drawn which was a straight line ABC as shown in figure
Determine from graph (i) emf of the cell (ii) maximum current obtained from the cell and (iii) internal resistance of the cell.
For a cell the graph between the p.d. (v) across the terminals of the cells and the current (I) drawn from the cell as shown. The emf and internal resistance is
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"Física universitaria es una colección de tres volúmenes que cumple con los requisitos de alcance y secuencia para los cursos de física de dos y tres semestres con base en el cálculo. El volumen 1 abarca la mecánica, el sonido, las oscilaciones y las ondas. El volumen 2 abarca la termodinámica, la electricidad y el magnetismo, y el volumen 3 abarca la óptica y la física moderna. Este libro de texto hace hincapié en la conexión entre la teoría y la práctica, por lo que los conceptos de la física se tornan interesantes y accesibles para los estudiantes. Al mismo tiempo, se mantiene el rigor matemático propio de la materia. Los ejemplos frecuentes y categóricos se centran en cómo abordar un problema, cómo trabajar con las ecuaciones y cómo comprobar y generalizar el resultado.
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Copyright: 2016
"University Physics is a three-volume collection that meets the scope and sequence requirements for two- and three-semester calculus-based physics courses. Volume 1 covers mechanics, sound, oscillations, and waves. Volume 2 covers thermodynamics, electricity, and magnetism, and Volume 3 covers optics and modern physics. This textbook emphasizes connections between theory and application, making physics concepts interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. Frequent, strong examples focus on how to approach a problem, how to work with the equations, and how to check and generalize the result.
This is the official print version of this OpenStax textbook. OpenStax makes full-color hardcover and B&W paperback print copies available for students who prefer a hardcopy textbook to go with the free digital version of this OpenStax title. The textbook content is exactly the same as the OpenStax digital book. This textbook is available for free download at the www.openstax.org website, but as many students prefer to study with hardcopy printed books, we offer affordable OpenStax textbooks for sale through this storefront and on Amazon as well as through campus bookstores."
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University Physics Vol 2
Author(s):
Edition: 1
Copyright: 2016
"University Physics is a three-volume collection that meets the scope and sequence requirements for two- and three-semester calculus-based physics courses. Volume 1 covers mechanics, sound, oscillations, and waves. Volume 2 covers thermodynamics, electricity, and magnetism, and Volume 3 covers optics and modern physics. This textbook emphasizes connections between theory and application, making physics concepts interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. Frequent, strong examples focus on how to approach a problem, how to work with the equations, and how to check and generalize the result."
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University Physics Vol 1
Author(s):
Edition:
Copyright: 2016
"University Physics is a three-volume collection that meets the scope and sequence requirements for two- and three-semester calculus-based physics courses. Volume 1 covers mechanics, sound, oscillations, and waves. Volume 2 covers thermodynamics, electricity, and magnetism, and Volume 3 covers optics and modern physics. This textbook emphasizes connections between theory and application, making physics concepts interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. Frequent, strong examples focus on how to approach a problem, how to work with the equations, and how to check and generalize the result."
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Physics for High School
Author(s):
Edition: 1
Copyright: 2020
"Physics for High School by OpenStax was developed under the guidance and support of experienced high school teachers and subject matter experts. Beginning with an introduction to physics and scientific processes and followed by chapters focused on motion, mechanics, thermodynamics, waves, and light, this book incorporates a variety of tools to engage and inspire students. Hands-on labs, worked examples, and highlights of how physics is applicable everywhere in the natural world are embedded throughout the book, and each chapter incorporates a variety of assessment types such as practice problems, performance tasks, and traditional multiple choice items. Additional instructor resources are included as well, including direct instruction presentations and a solutions manual."
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College Physics for AP
Author(s):
Edition: 2
Copyright: 2022
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College Physics 2e
Author(s):
Edition: 2
Copyright: 2022
"College Physics 2e introduces topics conceptually and progresses through clear explanations in the context of career-oriented, practical applications, and meets the scope and sequence of an algebra-based physics course. The narrative of physics and scientific discovery has been even further expanded to focus on including more diverse contributors to the field. Building on the success of the first edition the authors have increased focus on interdisciplinary connections, including enhancements to biological and medical applications. The problem solving approach has been revised to further unify conceptual, analytical, and calculation skills within the learning process, the authors have integrated a wide array of strategies and supports throughout the text for students."
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Organic Chemistry
Author(s):
Edition: 10
Copyright: 2023
"John McMurry's Organic Chemistry is renowned as the most clearly written book available for organic chemistry. In John McMurry's words, 'I wrote this book because I love writing. I get great pleasure and satisfaction from taking a complicated subject, turning it around until I see it clearly from a new angle, and then explaining it in simple words.'"
In Organic Chemistry: A Tenth Edition from OpenStax, McMurry continues this tradition while updating scientific discoveries, highlighting new applications, scrutinizing every piece of art, and providing example problems to assist students.
Organic Chemistry: A Tenth Edition continues to meet the scope and sequence of a two-semester organic chemistry course that follows a functional group approach.
John McMurry decided to publish Organic Chemistry: A Tenth Edition under an open license as a tribute to his son, Peter McMurry, who passed away from cystic fibrosis in December 2019.
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Chemistry: Atoms First Pak
Author(s):
Edition: 2
Copyright: 2019
Chemistry: Atoms First 2e is a peer-reviewed, openly licensed introductory textbook produced through a collaborative publishing partnership between OpenStax and the University of Connecticut and UConn Undergraduate Student Government Association. This text is an atoms-first adaptation of OpenStax Chemistry 2e. The intention of “atoms-first” involves a few basic principles: first, it introduces atomic and molecular structure much earlier than the traditional approach, and it threads these themes through subsequent chapters. This approach may be chosen as a way to delay the introduction of material such as stoichiometry that students traditionally find abstract and difficult, thereby allowing students time to acclimate their study skills to chemistry. Additionally, it gives students a basis for understanding the application of quantitative principles to the chemistry that underlies the entire course. It also aims to center the study of chemistry on the atomic foundation that many will expand upon in a later course covering organic chemistry, easing that transition when the time arrives.
The second edition has been revised to incorporate clearer, more current, and more dynamic explanations, while maintaining the same organization as the first edition. Substantial improvements have been made in the figures, illustrations, and example exercises that support the text narrative.
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Chemistry 2e
Author(s):
Edition: 2
Copyright: 2019
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Microbiology
Author(s):
Edition: 1
Copyright: 2016
"Microbiology covers the scope and sequence requirements for a single-semester microbiology course for non-majors. The book presents the core concepts of microbiology with a focus on applications for careers in allied health. The pedagogical features of the text make the material interesting and accessible while maintaining the career-application focus and scientific rigor inherent in the subject matter. Microbiology’s art program enhances students’ understanding of concepts through clear and effective illustrations, diagrams, and photographs.
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Concepts of Biology
Author(s):
Edition: 1
Copyright: 2013
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Biology
Author(s):
Edition: 2
Copyright: 2018
"Biology 2e is designed to cover the scope and sequence requirements of a typical two-semester biology course for science majors. The text provides comprehensive coverage of foundational research and core biology concepts through an evolutionary lens. Biology includes rich features that engage students in scientific inquiry, highlight careers in the biological sciences, and offer everyday applications. The book also includes various types of practice and homework questions that help students understand—and apply—key concepts.
The 2nd edition has been revised to incorporate clearer, more current, and more dynamic explanations, while maintaining the same organization as the first edition. Art and illustrations have been substantially improved, and the textbook features additional assessments and related resources."
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Biology for AP
Author(s):
Edition: 1
Copyright: 2018
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Anatomy and Physiology
Author(s):
Edition: 2
Copyright: 2022
This is the official print version of this OpenStax textbook. OpenStax makes full-color hardcover and B&W paperback print copies available for students who prefer a hardcopy textbook to go with the free digital version of this OpenStax title. The textbook content is exactly the same as the OpenStax digital book. This textbook is available for free download at the OpenStax dot org website, but as many students prefer to study with hardcopy books, we offer affordable OpenStax textbooks for sale through Amazon as well as most campus bookstores.
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Nutrition for Nurses
Author(s):
Edition: 1
Copyright: 2024
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Statistics for High School
Author(s):
Edition: 1
Copyright: 2020
"Statistics for High School by OpenStax was developed under the guidance and support of experienced high school teachers and subject matter experts. Statistics offers instruction in grade-level appropriate concepts and skills in a logical, engaging progression that begins with sampling and data and covers topics such as probability, random variables, the normal distribution, and hypothesis testing. This content was developed with students in mind, incorporating statistics labs, worked exercises, and additional opportunities for assessment that incorporate real-world statistical applications. For instructors, resources are available to support the implementation of the Statistics textbook, including a Getting Started Guide, direct instruction presentations, and a solutions manual.
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Introductory Statistics 2e
Author(s):
Edition: 2
Copyright: 2023
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Precalculus
Author(s):
Edition: 2
Copyright: 2021
"Precalculus 2e provides a comprehensive exploration of mathematical principles and meets scope and sequence requirements for a typical precalculus course. The text proceeds from functions through trigonometry and ends with an introduction to calculus. The modular approach and the richness of content ensure that the book addresses the needs of a variety of courses. Precalculus 2e offers a wealth of examples with detailed, conceptual explanations, building a strong foundation in the material before asking students to apply what they’ve learned.
The Precalculus 2e revision focused on improving relevance and representation as well as mathematical clarity and accuracy. Introductory narratives, examples, and problems were reviewed and revised using a diversity, equity, and inclusion framework. Many contexts, scenarios, and images have been changed to become even more relevant to students’ lives and interests. To maintain our commitment to accuracy and precision, examples, exercises, and solutions were reviewed by multiple faculty experts. All improvement suggestions and errata updates from the first edition were considered and unified across the different formats of the text."
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Prealgebra 2e
Author(s):
Edition: 2
Copyright: 2020
"Prealgebra 2e is designed to meet scope and sequence requirements for a one-semester prealgebra or basic math course. The book’s organization makes it easy to adapt to a variety of course syllabi. The text introduces the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. Each topic builds upon previously developed material to demonstrate the cohesiveness and structure of mathematics.
The second edition contains detailed updates and accuracy revisions to address comments and suggestions from users. Dozens of faculty experts worked through the text, exercises and problems, graphics, and solutions to identify areas needing improvement. Though the authors made significant changes and enhancements, exercise and problem numbers remain nearly the same in order to ensure a smooth transition for faculty."
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Intermediate Algebra
Author(s):
Edition: 2
Copyright: 2020
"Intermediate Algebra 2e is designed to meet the scope and sequence requirements of a one-semester intermediate algebra course. The book’s organization makes it easy to adapt to a variety of course syllabi. The text expands on the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. The material is presented as a sequence of clear steps, building on concepts presented in prealgebra and elementary algebra courses.
The second edition contains detailed updates and accuracy revisions to address comments and suggestions from users. Dozens of faculty experts worked through the text, exercises and problems, graphics, and solutions to identify areas needing improvement. Though the authors made significant changes and enhancements, exercise and problem numbers remain nearly the same in order to ensure a smooth transition for faculty."
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Contemporary Mathematics
Author(s):
Edition: 1
Copyright: 2023
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Elementary Algebra
Author(s):
Edition: 2
Copyright: 2020
"Elementary Algebra 2e is designed to meet scope and sequence requirements for a one-semester elementary algebra course. The book's organization makes it easy to adapt to a variety of course syllabi. The text expands on the fundamental concepts of algebra while addressing the needs of students with diverse backgrounds and learning styles. Each topic builds upon previously developed material to demonstrate the cohesiveness and structure of mathematics."
The second edition contains detailed updates and accuracy revisions to address comments and suggestions from users. Dozens of faculty experts worked through the text, exercises and problems, graphics, and solutions to identify areas needing improvement. Though the authors made significant changes and enhancements, exercise and problem numbers remain nearly the same in order to ensure a smooth transition for faculty."
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College Algebra 2e
Author(s):
Edition: 2
Copyright: 2021
"College Algebra 2e provides a comprehensive exploration of algebraic principles and meets scope and sequence requirements for a typical introductory algebra course. The modular approach and richness of content ensure that the book addresses the needs of a variety of courses. College Algebra 2e offers a wealth of examples with detailed, conceptual explanations, building a strong foundation in the material before asking students to apply what they’ve learned.
The College Algebra 2e revision focused on improving relevance and representation as well as mathematical clarity and accuracy. Introductory narratives, examples, and problems were reviewed and revised using a diversity, equity, and inclusion framework. Many contexts, scenarios, and images have been changed to become even more relevant to students’ lives and interests. To maintain our commitment to accuracy and precision, examples, exercises, and solutions were reviewed by multiple faculty experts. All improvement suggestions and errata updates from the first edition were considered and unified across the different formats of the text."
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Calculus Volume 2
Author(s):
Edition: 1
Copyright: 2016
"Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 2 covers integration, differential equations, sequences and series, and parametric equations and polar coordinates."
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Calculus Volume 3
Author(s):
Edition: 1
Copyright: 2016
"Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 3 covers parametric equations and polar coordinates, vectors, functions of several variables, multiple integration, and second-order differential equations."
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Calculus Volume 1
Author(s):
Edition: 1
Copyright: 2016
"Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 1 covers functions, limits, derivatives, and integration."
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Algebra and Trigonomtry
Author(s):
Edition: 2
Copyright: 2021
"Algebra and Trigonometry provides a comprehensive exploration of mathematical principles and meets scope and sequence requirements for a typical introductory algebra and trigonometry course. The modular approach and the richness of content ensure that the book addresses the needs of a variety of courses. Algebra and Trigonometry 2e offers a wealth of examples with detailed, conceptual explanations, building a strong foundation in the material before asking students to apply what they’ve learned."
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Writing Guide with Handbook
Author(s):
Edition: 1
Copyright: 2021
"Writing Guide with Handbook aligns to the goals, topics, and objectives of many first-year writing and composition courses. It is organized according to relevant genres, and focuses on the writing process, effective writing practices or strategies—including graphic organizers, writing frames, and word banks to support visual learning—and conventions of usage and style. The text includes an editing and documentation handbook, which provides information on grammar and mechanics, common usage errors, and citation styles.
Writing Guide with Handbook breaks down barriers in the field of composition by offering an inviting and inclusive approach to students of all intersectional identities. To meet this goal, the text creates a reciprocal relationship between everyday rhetoric and the evolving world of academia. Writing Guide with Handbook builds on students’ life experiences and their participation in rhetorical communities within the familiar contexts of personal interaction and social media. The text seeks to extend these existing skills by showing students how to construct a variety of compelling compositions in a variety of formats, situations, and contexts.
The authors conceived and developed Writing Guide with Handbook in 2020; its content and learning experiences reflect the instructional, societal, and individual challenges students have faced. The authors invite students and instructors to practice invitational discussions even as they engage in verbal and written argument. Instructors will be empowered to emphasize meaning and voice and to teach empathy as a rhetorical strategy. Students will be empowered to negotiate their identities and their cultures through language as they join us in writing, discovering, learning, and creating."
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Preparing for College Success
Author(s):
Edition: 1
Copyright: 2023
"OpenStax’s Preparing for College Success is a comprehensive resource that provides secondary students critical information to prepare for their “to and through” college journey.
The book addresses the evolving challenges and opportunities of today’s diverse students. College match, preparing for transition, time management and study skills, building relationships, financial literacy, and career planning are all reflected throughout the material.
Preparing for College Success also includes an array of student-centered features including scenarios and advice from current college students that can support robust classroom conversation or individual learning."
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College Success
Author(s):
Edition: 1
Copyright: 2020
"OpenStax College Success is a comprehensive and contemporary resource that serves First Year Experience, Student Success, and College Transition courses. Developed with the support of hundreds of faculty and coordinators, the book addresses the evolving challenges and opportunities of today’s diverse students. Engagement, self-analysis, personal responsibility, and student support are reflected throughout the material. College Success also includes an array of student surveys and opinion polls, and OpenStax will regularly provide the results to adopting faculty."
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Principles of Accounting, Volume 1: Financial Accounting
Author(s):
Edition: 1
Copyright: 2019
"Principles of Accounting is designed to meet the scope and sequence requirements of a two-semester accounting course that covers the fundamentals of financial and managerial accounting. Due to the comprehensive nature of the material, we are offering the book in two volumes. This book is specifically designed to appeal to both accounting and non-accounting majors, exposing students to the core concepts of accounting in familiar ways to build a strong foundation that can be applied across business fields. Each chapter opens with a relatable real-life scenario for today’s college student. Thoughtfully designed examples are presented throughout each chapter, allowing students to build on emerging accounting knowledge. Concepts are further reinforced through applicable connections to more detailed business processes. Students are immersed in the “why” as well as the “how” aspects of accounting in order to reinforce concepts and promote comprehension over rote memorization."
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Principles of Accounting, Volume 2: Managerial Accounting
Author(s):
Edition: 1
Copyright: 2019
"Principles of Accounting is designed to meet the scope and sequence requirements of a two-semester accounting course that covers the fundamentals of financial and managerial accounting. Due to the comprehensive nature of the material, we are offering the book in two volumes. This book is specifically designed to appeal to both accounting and non-accounting majors, exposing students to the core concepts of accounting in familiar ways to build a strong foundation that can be applied across business fields. Each chapter opens with a relatable real-life scenario for today’s college student. Thoughtfully designed examples are presented throughout each chapter, allowing students to build on emerging accounting knowledge. Concepts are further reinforced through applicable connections to more detailed business processes. Students are immersed in the “why” as well as the “how” aspects of accounting in order to reinforce concepts and promote comprehension over rote memorization."
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Principles of Finance
Author(s):
Edition: 1
Copyright: 2022
"Designed to meet the scope and sequence of your course, Principles of Finance provides a strong foundation in financial applications using an innovative use-case approach to explore their role in business decision-making. An array of financial calculator and downloadable Microsoft Excel data exercises also engage students in experiential learning throughout. With flexible integration of technical instruction and data, this title prepares students for current practice and continual evolution."
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Business Ethics
Author(s):
Edition: 1
Copyright: 2018
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Business Law Essentials I
Author(s):
Edition: 1
Copyright: 2019
"Business Law I Essentials is a brief introductory textbook designed to meet the scope and sequence requirements of courses on Business Law or the Legal Environment of Business. The concepts are presented in a streamlined manner, and cover the key concepts necessary to establish a strong foundation in the subject. The textbook follows a traditional approach to the study of business law. Each chapter contains learning objectives, explanatory narrative and concepts, references for further reading, and end-of-chapter questions.
Business Law I Essentials may need to be supplemented with additional content, cases, or related materials, and is offered as a foundational resource that focuses on the baseline concepts, issues, and approaches."
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Introductory Business Statistics
Author(s):
Edition: 1
Copyright: 2017
"Introductory Business Statistics is designed to meet the scope and sequence requirements of the one-semester statistics course for business, economics, and related majors. Core statistical concepts and skills have been augmented with practical business examples, scenarios, and exercises. The result is a meaningful understanding of the discipline, which will serve students in their business careers and real-world experiences."
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Principles of Economics 3e
Author(s):
Edition: 3
Copyright: 2022
"Principles of Economics 3e covers the scope and sequence of most introductory economics courses. The third edition takes a balanced approach to the theory and application of economics concepts. The text uses conversational language and ample illustrations to explore economic theories, and provides a wide array of examples using both fictional and real-world scenarios.
The third edition has been carefully and thoroughly updated to reflect current data and understanding, as well as to provide a deeper background in diverse contributors and their impacts on economic thought and analysis. For example, the third edition highlights the research and views of a broader group of economists. Brief references and deeply explored socio-political examples have been updated to showcase the critical – and sometimes unnoticed – ties between economic developments and topics relevant to students."
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Principles of Microeconomics
Author(s):
Edition: 3
Copyright: 2022
"Principles of Microeconomics 3e covers the scope and sequence of most one semester introductory microeconomics courses. The third edition takes a balanced approach to the theory and application of microeconomics concepts. The text uses conversational language and ample illustrations to explore economic theories, and provides a wide array of examples using both fictional and real-world applications.
The third edition has been carefully and thoroughly updated to reflect recent developments, as well as to provide a deeper background in diverse contributors and their impacts on economic thought and analysis. For example, the third edition highlights the research and views of a broader group of economists."
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Principles of Macroeconomics 3e
Author(s):
Edition: 3
Copyright: 2022
"Principles of Macroeconomics 3e covers the scope and sequence of most one semester introductory macroeconomics courses. The third edition takes a balanced approach to the theory and application of macroeconomics concepts. The text uses conversational language and ample illustrations to explore economic theories, and provides a wide array of examples using both fictional and real-world scenarios.
The third edition has been carefully and thoroughly updated to reflect current data and understanding, as well as to provide a deeper background in diverse contributors and their impacts on economic thought and analysis. For example, the third edition highlights the research and views of a broader group of economists. Brief references and deeply explored socio-political examples have also been updated to showcase the critical – and sometimes unnoticed – ties between economic developments and topics relevant to students."
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Introduction to Business
Author(s):
Edition: 1
Copyright: 2018
"Designed to meet the scope and sequence of your course, OpenStax Introduction to Anthropology is a four-field text integrating diverse voices, engaging field activities, and meaningful themes like Indigenous experiences and social inequality to engage students and enrich learning. The text showcases the historical context of the discipline, with a strong focus on anthropology as a living and evolving field. There is significant discussion of recent efforts to make the field more diverse—in its practitioners, in the questions it asks, and in the applications of anthropological research to address contemporary challenges. In addressing social inequality, the text drives readers to consider the rise and impact of social inequalities based on forms of identity and difference (such as gender, ethnicity, race, and class) as well as oppression and discrimination. The contributors to and dangers of socioeconomic inequality are fully addressed, and the role of inequality in social dysfunction, disruption, and change is noted."
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Entrepreneurship
Author(s):
Edition: 1
Copyright: 2020
"This textbook is intended for use in introductory Entrepreneurship classes at the undergraduate level. Due to the wide range of audiences and course approaches, the book is designed to be as flexible as possible. Theoretical and practical aspects are presented in a balanced manner, and specific components such as the business plan are provided in multiple formats. Entrepreneurship aims to drive students toward active participation in entrepreneurial roles, and exposes them to a wide range of companies and scenarios."
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Organizational Behavior
Author(s):
Edition: 1
Copyright: 2019
"This OpenStax resource aligns to introductory courses in Organizational Behavior. The text presents the theory, concepts, and applications with particular emphasis on the impact that individuals and groups can have on organizational performance and culture. An array of recurring features engages students in entrepreneurial thinking, managing change, using tools/technology, and responsible management."
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Principles of Management
Author(s):
Edition: 1
Copyright: 2019
"Principles of Management is designed to meet the scope and sequence requirements of the introductory course on management. This is a traditional approach to management using the leading, planning, organizing, and controlling approach. Management is a broad business discipline, and the Principles of Management course covers many management areas such as human resource management and strategic management, as well as behavioral areas such as motivation. No one individual can be an expert in all areas of management, so an additional benefit of this text is that specialists in a variety of areas have authored individual chapters."
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Principles of Marketing
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Edition: 1
Copyright: 2023
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College Success Concise
Author(s):
Edition: 1
Copyright: 2023
"OpenStax College Success Concise serves First Year Experience, Student Success, and College Transition courses, and can also be used as a supplementary resource in courses across the curriculum. With the input of hundreds of instructors and academic success experts, the authors carefully prioritized the most critical topics to align to briefer courses.
The offering covers material such as college culture, time management, mindset, study skills, test preparation, financial literacy, health, and planning for the future.
While much of the material is very similar to the original College Success book, this version was holistically edited and updated. Users will see additions such as a new section on group work and greatly expanded coverage of stress management and wellbeing."
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Introduction to Philosophy
Author(s):
Edition: 1
Copyright: 2022
"Designed to meet the scope and sequence of your course, Introduction to Philosophy surveys logic, metaphysics, epistemology, theories of value, and history of philosophy thematically. To provide a strong foundation in global philosophical discourse, diverse primary sources and examples are central to the design, and the text emphasizes engaged reading, critical thinking, research, and analytical skill-building through guided activities."
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U.S. History
Author(s):
Edition: 1
Copyright: 2014
"U.S. History is designed to meet the scope and sequence requirements of most introductory courses. The text provides a balanced approach to U.S. history, considering the people, events, and ideas that have shaped the United States from both the top down (politics, economics, diplomacy) and bottom up (eyewitness accounts, lived experience). U.S. History covers key forces that form the American experience, with particular attention to issues of race, class, and gender.
This is the official print version of this OpenStax textbook. OpenStax makes full-color hardcover and B&W paperback print copies available for students who prefer a hardcopy textbook to go with the free digital version of this OpenStax title. The textbook content is exactly the same as the OpenStax digital book. This textbook is available for free download at the www.openstax.org website, but as many students prefer to study with hardcopy printed books, we offer affordable OpenStax textbooks for sale through this storefront and on Amazon as well as through campus bookstores."
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World History Vol 1
Author(s):
Edition: 1
Copyright: 2023
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World History Vol 2
Author(s):
Edition: 1
Copyright: 2022
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Introduction to Intellectual Property
Author(s):
Edition: 1
Copyright: 2021
"Introduction to Intellectual Property provides a clear, effective introduction to patents, copyright, trademarks, and trade secrets. The text may be used by students and instructors in formal courses, as well as those applying intellectual property considerations to entrepreneurship, marketing, law, computer science, engineering, design, or other fields. The luminaries involved with this project represent the forefront of knowledge and experience, and the material offers considerable examples and scenarios, as well as exercises and references.
Introduction to Intellectual Property was originally developed by the Michelson 20MM Foundation, released under the title The Intangible Advantage.
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Astronomy 2e
Author(s):
Edition: 2
Copyright: 2022
"Designed to meet the scope and sequence of your course, Astronomy 2e is written in clear non-technical language, with the occasional touch of humor and a wide range of clarifying illustrations. It has many analogies drawn from everyday life to help non-science majors appreciate, on their own terms, what our modern exploration of the universe is revealing. The book can be used for either a one-semester or two-semester introductory course.
The second edition has been updated according to new exploration and discoveries. The second edition also includes a significant amount of new art and images."
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190124 | https://contentexplorer.smarterbalanced.org/target/m-g4-c1-ta | Grade 4 - Claim 1 - Target A | Smarter Content Explorer
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Mathematics
Target A
Use the four operations with whole numbers to solve problems.
Sample Item
Grade 4
Test
Grade 4
Claim 1
Concepts and Procedures
Students can explain and apply mathematical concepts and carry out mathematical procedures with precision and fluency.
Grade
Grade 4
Content Domain
Operations and Algebraic Thinking
Standards
OA-1 OA-2 OA-3
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Major
Overview
Standards
Clarifications
Range ALDs
Evidence
Item Guidelines
Task Models
Standards
OA-1
Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as... + More
#### OA-2
Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive... + More
#### OA-3
Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter... + More
Clarifications
Tasks for this target will require students to use the four operations to solve straightforward, one-step or multi-step contextual word problems, including problems where the remainder must be interpreted.
Range Achievement Level Descriptors
+Level 1
+Level 2
+Level 3
+Level 4
Evidence Required
1
The student solves contextual problems involving multiplicative comparisons, e.g., by using drawings and equations with a... + More
#### 2
The student solves straightforward, contextual problems using the four operations.
Item Guidelines
Depth of Knowledge
M-DOK1
Recall includes the recall of information such as fact, definition, term, or a simple procedure, as well as performing a simple algorithm or applying a formula. That is, in mathematics a one-step, well-defined, and straight algorithmic procedure should be... + More
#### M-DOK2
Skill/Concept includes the engagement of some mental processing beyond a habitual response. A Level 2 assessment item requires students to make some decisions as to how to approach the problem or activity, whereas Level 1 requires students to demonstrate a... + More
Allowable Item Types
Equation/Numeric
Allowable Stimulus Materials
multiplication equations, verbal statements of multiplicative comparison, contextual problems involving multiplicative comparison, one-step contextual word problems, measurements limited to: kilometers (km), meters (m), centimeters (cm), kilograms (kg), grams (g), pounds (lb), ounces (oz), liters...
More
Key/Construct Relevant Vocabulary
Remainder, sum, difference, quotient, product, equation, times as much, times as many, equation
Allowable Tools
None
Target-Specific Attributes
Numbers used in this target must be whole numbers. In describing a multiplicative comparison, the language “times as much” or “times as many” is preferable to “times more than.”
Accessibility
Item writers should consider the following Language and Visual Element/Design guidelines when developing items. Language Key Considerations: Use simple, clear, and easy-to-understand language needed to assess the construct or aid in the understanding of the...
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Development Notes
Interpreting multiplication equations as multiplicative comparisons and representing verbal statements of multiplicative comparisons as multiplication equations (4.OA.1) will be assessed in Claim 4. Items asking students to solve a word problem by using an equation...
More
Task Models
Select a Task Model below
1a
1b
1c
2
Task Model 1a
Item Types
Equation/Numeric
#### Depth of Knowledge
M-DOK1
#### Standards
OA-2
Target Evidence Statement
The student solves contextual problems involving multiplicative comparisons, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
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Prompt Features: The student is prompted to solve a contextual problem involving multiplicative comparison. Stimulus Guidelines: Numbers should fit in the parameters of up to 4-digit by 1-digit, or 2-digit by 2-digit multiplication problems, and up...
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Stimulus
The student is presented with a contextual problem involving multiplicative comparison with an unknown product.
Example 1
Example Stem: A cat has 4 times as many toys as a puppy. The puppy has 12 toys. How many toys does the cat have?
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Rubric: (1 point) The student solves for an unknown and enters the correct number (e.g., 48).
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How to determine if a segment passes through the positive X axis?
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A segment is defined by two endpoints (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2).
The positive X axis for simplicity can be defined as the segment going from (0,0)(0,0) to (10 9,0)(10 9,0) (I'm working on a program where the maximum X X value is 10 6 10 6 therefore this is good enough). Although a formula that doesn't rely on endpoints (for the axis) is also acceptable (and probably even better).
What I currently have is an algorithm that detects segment intersections, so I use the segment from (0,0)(0,0) to (10 9,0)(10 9,0) explained above. The algorithm I'm using is a O(1)O(1) check that uses cross product for getting the orientation of points (code below).
But I'm wondering if there's a more elegant way to do this. I just want to refactor the code since this is the only intersection I need to do, and it looks a bit ugly to have all this code just for this. I suspect even a more elegant way would need some cross product stuff, but any idea is appreciated!
C++ code of my current algorithm.
```
struct Segment {
Point p, q;
int orientation(Point p, Point q, Point r) {
ll val = (q.y - p.y) (r.x - q.x) - (q.x - p.x) (r.y - q.y);
if (val == 0) return 0;
return (val > 0) ? 1 : 2;
}
bool intersect(const Segment& s) {
Point p1 = p;
Point q1 = q;
Point p2 = s.p;
Point q2 = s.q;
int o1 = orientation(p1, q1, p2);
int o2 = orientation(p1, q1, q2);
int o3 = orientation(p2, q2, p1);
int o4 = orientation(p2, q2, q1);
return (o1 != o2 && o3 != o4);
}
};
```
geometry
computational-geometry
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edited Dec 5, 2021 at 23:43
Chris VilchesChris Vilches
asked Dec 5, 2021 at 23:28
Chris VilchesChris Vilches
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1 If(y1y2>0) // the segment does NOT cross the x-axis Daniel Mathias –Daniel Mathias 2021-12-06 00:06:55 +00:00 Commented Dec 6, 2021 at 0:06
@DanielMathias Thanks, but my question involves only the positive part of the x axis, i.e. (x,0)(x,0) where x x is [0, ∞). Your formula works for the entire x axis (negative included).Chris Vilches –Chris Vilches 2021-12-06 00:58:49 +00:00 Commented Dec 6, 2021 at 0:58
My comment was only a hint, and not a full solution. If the segment does cross the x-axis, then you need to find where it crosses, which I believe is x 1+y 1 y 1−y 2(x 2−x 1)x 1+y 1 y 1−y 2(x 2−x 1)Daniel Mathias –Daniel Mathias 2021-12-06 01:05:50 +00:00 Commented Dec 6, 2021 at 1:05
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If you trace the intersect function step by step, and if you exclude the cases where y 1=0 y 1=0 and where y 2=0,y 2=0, you might notice that two of the calls to orientation are merely testing whether y 1>0 y 1>0 and whether y 2>0.y 2>0. The return statement then checks whether those two results are different, that is, whether y 1 y 1 and y 2 y 2 are on opposite sides of the x x axis.
In the cases where y 1=0 y 1=0 and where y 2=0,y 2=0, the two calls to orientation will return two different values (indicating an intersection with the x x axis) except, bizarrely, in the case where y 1=y 2=0 y 1=y 2=0 (in which case the algorithm will tell you that the segment does not intersect the x x axis).
One of the two remaining calls to orientation -- the one that compares the point (10 9,0)(10 9,0) with the points (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) -- is always going to return the same value as every other time, unless you're so unlucky that someone gives you points like (x 1,y 1)=(10 9+1,1)(x 1,y 1)=(10 9+1,1) and (x 2,y 2)=(10 9+1,−1).(x 2,y 2)=(10 9+1,−1). So ideally you would skip this call to orientation and merely check whether the one remaining call -- which compares (0,0)(0,0) with (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) -- returns the result that puts (0,0)(0,0) on the desired side of the line through (x 1,y 1)(x 1,y 1) and (x 2,y 2).(x 2,y 2).
This last call to orientation will use (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) as the first two arguments (in one order or the other) and (0,0)(0,0) as the third argument, so it will set val to either
(y 1−y 2)x 1−(x 1−x 2)y 1(y 1−y 2)x 1−(x 1−x 2)y 1
or the exact opposite of that quantity (changing the sign), and it will reliably always set val to the same sign for every example where the line through (x 1,y 1)(x 1,y 1) and (x 2,y 2)(x 2,y 2) intersects the positive x x axis. Which sign that is just depends on the order in which it receives the first two arguments.
Since all we need to do is get a consistent sign, we can assume we will compute (y 1−y 2)x 1−(x 1−x 2)y 1(y 1−y 2)x 1−(x 1−x 2)y 1 rather than the opposite. To find out which sign is the one that says "intersects positive x x axis," we just need to work out one example. Let's suppose (x 1,y 1)=(1,1)(x 1,y 1)=(1,1) and (x 2,y 2)=(1,−1).(x 2,y 2)=(1,−1). This is a segment that intersects the positive x x axis, and
(y 1−y 2)x 1−(x 1−x 2)y 1=2×1−0×1=2.(y 1−y 2)x 1−(x 1−x 2)y 1=2×1−0×1=2.
One more detail to figure out. Consider the segment defined by (x 1,y 1)=(0,1)(x 1,y 1)=(0,1) and (x 2,y 2)=(0,−1).(x 2,y 2)=(0,−1). This segment passes through the x x axis at (0,0).(0,0). Do you consider that to be an intersection with "the positive x x axis" or not? Note that the algorithm you are currently using will tell you that this segment does intersect the positive x x axis; I am merely unsure whether that is the answer you wanted or a defect in the implementation.
So what I would do is to write a function that returned the result true if both of the following two conditions are true:
First condition: either y 1=0,y 1=0, or y 2=0,y 2=0, or y 1 y 1 and y 2 y 2 have opposite signs.
Second condition: (y 1−y 2)x 1−(x 1−x 2)y 1>0(y 1−y 2)x 1−(x 1−x 2)y 1>0 (unless you think passing through (0,0)(0,0) should count as an intersection with "the positive x x axis," in which case use ≥≥ instead of >>).
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answered Dec 6, 2021 at 1:06
David KDavid K
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Thanks for the explanation! The problem I'm solving has the property that no segments pass through the origin (0,0)(0,0) so that case is not necessary to consider.Chris Vilches –Chris Vilches 2021-12-06 01:31:32 +00:00 Commented Dec 6, 2021 at 1:31
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Here is a simple code segment to implement my suggestion.
The first line returns false if the segment is entirely above or below the x axis. Otherwise, the second line returns true if, and only if, the x-intercept is non-negative, as you desire.
{
if(p.yq.y>0)return 0;
return p.x+(q.x-p.x)p.y/(p.y-q.y)>=0;
}
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edited Dec 6, 2021 at 2:11
answered Dec 6, 2021 at 1:53
Daniel MathiasDaniel Mathias
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Worked first try, thanks a lot! Passed all tests (over 200 cases) so I can tell it's OK.Chris Vilches –Chris Vilches 2021-12-06 01:58:40 +00:00 Commented Dec 6, 2021 at 1:58
@ChrisVilches Simple is good :o)Daniel Mathias –Daniel Mathias 2021-12-06 01:59:40 +00:00 Commented Dec 6, 2021 at 1:59
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Segment {(x 1,y 1),(x 2,y 2)}{(x 1,y 1),(x 2,y 2)} crosses X X axis when
y 1⋅y 2<0∧x=∈(x 1,x 2)∧x>0 y 1⋅y 2<0∧x=∈(x 1,x 2)∧x>0
where
x=(x 2⋅y 1−x 1⋅y 2)y 1−y 2 x=(x 2⋅y 1−x 1⋅y 2)y 1−y 2
In C++ this may look like
bool intersect(double x1, double y1, double x2, double y2) {
if ( y1 y2 ) > .0
return false;
double x = (x2 y1 - x1 y2);
return x > 0 && x > min(x1, x2) && x < max(x1,x2);
}
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answered Dec 6, 2021 at 2:14
Yuri GinsburgYuri Ginsburg
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I just found another way to do this.
bool intersects_positive_x_axis() {
if (p.cross(q) > 0) return q < p;
return p < q;
}
Where p p and q q are the two endpoints of the segment, c r o s s c r o s s is the cross product operator, and << is the comparison by bearing (angle). Since I'm doing a polar sort with a radial sweep algorithm, I already had this implemented.
If p×q>0 p×q>0 then the segment endpoints are in counter clockwise order, but if the order by bearing is not the same, then it means the segment must be crossing the positive side of the x-axis. This is because points in the first quadrant have the lowest bearing value, and points in the fourth quadrant have the highest.
This code ignores the intersection with the point (0,0)(0,0), which has not been tested.
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answered Dec 7, 2021 at 15:56
Chris VilchesChris Vilches
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--- --- --- --- --- --- --- --- --- --- ---
| A proof that the square root of 2 is irrational Here you can read a step-by-step proof with simple explanations for the fact that the square root of 2 is an irrational number. It is the most common proof for this fact and is by contradiction. How do we know that square root of 2 is an irrational number? In other words, how do we know that √2 doesn't have a pattern in its decimal sequence? Maybe the pattern is very well hidden and is really long, billions of digits? Here is where mathematical proof comes in. The proof that √2 is indeed irrational is usually found in college level math texts, but it isn't that difficult to follow. It does not rely on computers at all, but instead is a "proof by contradiction": if √2 WERE a rational number, we'd get a contradiction. I encourage all high school students to study this proof since it illustrates so well a typical proof in mathematics and is not hard to follow. A proof that the square root of 2 is irrational Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero. We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further. From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something. From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me! Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction. If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get: | | | | --- | 2 | = | (2k)2/b2 | | 2 | = | 4k2/b2 | | 2b2 | = | 4k2 | | b2 | = | 2k2 | This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!! WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational. --- See also What is a mathematical proof? 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Hospital infection
Hospital infection
Published: 14 October 2018
Infection used to be one of the biggest killers in hospitals before modern hygiene standards were introduced. But the risk of infection hasn't gone away.
Hospital infection has been an issue for as long as there have been hospitals. Putting so many weak and vulnerable people in close proximity, many of them arriving with contagious diseases is bound to encourage the spread of infection.
What is an infection?
An infection happens when infectious agents (commonly called germs) enters the body and begin to reproduce and spread, releasing toxins into the blood and disrupting bodily functions. The infectious agents are simple micro-organisms such as bacteria, viruses, fungi, and prions.
How an infection spreads depends on the type of micro-organism. Infection can be transmitted in a variety of ways, including:
skin contact
bodily fluids
contact with faeces
airborne droplets or particles
touching an object that an infected person has also touched.
Some infections, such as cold viruses, are mild and little more than a nuisance, but others are severe and life-threatening. The body’s immune system usually protects the body against infections, but if the infection spreads too rapidly or cannot be controlled, the immune system may be overwhelmed and the infection may become dangerous.
The common symptoms of infection are triggered by the immune systems response to an infecting micro-organism and include fever, feeling cold or shivery, localised redness, swelling or discharge of pus, nasal congestion, coughing, diarrhoea or vomiting. These are all signs that the body is trying to rid itself of the infection.
Hospital infections
In the past, even when an operation or treatment was successful, the patient often died from infection. Infections in hospital wards, particularly after surgery, were so common that the phenomenon was known as hospitalism or ward fever.
Inexplicably, an infection might sweep through an entire ward, killing the most vulnerable. No wonder surgery was a last resort for most patients. Victorian surgeon James Y. Simpson, the discoverer of chloroform, claimed that:
a man laid on the operating table in one or our surgical hospitals is exposed to more chance of death than was the English soldier on the field of Waterloo.
As early as 1847, the Hungarian doctor Ignaz Semmelweis had realised that infection rates were far worse in maternity wards attended by doctors than in wards where midwives only delivered the babies. Doctors and medical students handled a wider variety of patients as well as dead bodies, whereas midwives only came into contact with generally healthy pregnant women.
Semmelweis suspected that it was the doctors who were bringing the infections into the ward. He introduced a simple antiseptic hand-washing procedure for medical staff entering his wards and found the infection rates fell.
The infections were being transferred from patient to patient by unwitting doctors, who rarely washed their own hands or instruments between procedures. But because no one knew what caused infections, Semmelweis couldn't prove that hand-washing prevented the spread of infection and his findings were not widely adopted. Tragically Semmelweis himself died from blood poisoning caused by a cut on his finger.
The germ theory of disease
The French scientist Louis Pasteur speculated that the spread of microorganisms (called germs) in the body could explain infectious disease. This became known as the germ theory of disease. Although he never tested the theory, Pasteur suggested that germ-killing chemicals might be able to limit the spread of infections.
In the UK, surgeon Joseph Lister had been investigating surgical infection and was inspired by Pasteur’s research. Lister develop the antisepsis system in order to prevent infections from entering a wound during and after surgery.
Lister's system killed infectious agents using carbolic acid as a disinfectant for washing the hands, cleaning surgical instruments and on post-surgical dressings.
Lister published a paper about his system in the Lancet, but because Pasteur’s germ theory was still untested many surgeons were sceptical. Without the belief that germs were real and dangerous, many surgeons found the antiseptic system unnecessarily complicated and excessive.
Some surgeons were opposed because they thought that Lister was proposing carbolic acid as a treatment for infection rather than as one possible chemical barrier or antiseptic for preventing infection entering the body. They objected that carbolic acid was too toxic for wound dressings because, in strong solutions it irritated and burned the wound and surrounding skin, and an overdose could cause excessive vomiting.
The importance of hospital sanitation
Some of Lister’s critics argued that good hygiene and cleanliness were all that was needed to control infection, and that Lister’s elaborate antisepsis system was simply another form of sanitation.
Lister responded by producing statistical evidence from his wards at Glasgow Infirmary to demonstrate the effectiveness of the antisepsis system, but his data was limited and failed to convince many of his critics. Most surgeons preferred to wait until there was stronger evidence for the germ theory and antisepsis before committing themselves.
But one person who did convince hospital administrators about the importance of preventing hospital infection was Florence Nightingale. She used statistical evidence based on the data she gathered at military hospitals hospitals during the Crimean War to demonstrate that infections, not war wounds, were the biggest killers of soldiers in army hospitals.
When Nightingale returned to Britain, she collaborated with the epidemiologist William Farr in the preparation of hospital statistics for her book Notes on Hospitals (1859).
The compelling statistical evidence she provided convinced many hospitals that improved sanitation was sufficient to reduce infection-related deaths and diseases in hospitals.
Like many doctors at the time, Nightingale believed in the traditional view that miasmas (bad air) and not germs spread infection. Her campaign was based on empirical evidence rather than scientific proof.
Nevertheless, her campaign to clean up hospitals improved standards of care as well as hygiene and led to fewer deaths from infection. She also set up a training school in London, where nurses were instructed the principles of good hygiene.
Lister didn’t think that sanitation was important if proper antiseptic procedures were in place. His sanitary procedure in the operating room simply consisted of removing his coat, rolling up his sleeves and pinning a clean but unsterilised towel over his waistcoat and trousers.
Such was his faith in germ theory and antisepsis that cleanliness seemed irrelevant to him. In an address to the British Medical Association he declared:
My patients have the dirtiest wounds and sores in the world’ but, he explained, ‘aesthetically they are dirty, though surgically they are clean.
In fact both good hygiene and antiseptic procedures are needed to prevent and control hospial infection. Both the sanitarians and the Listerians had yet to be convinced by each other's evidence.
Modern infection control
In 1877 the German scientist Robert Koch provided convincing scientific evidence that germs were the cause of infectious diseases such as anthrax and typhoid. His laboratory work on bacterial infection also showed that dry heat and steam sterilisation were as effective as chemical agents in controlling and preventing the growth of bacteria.
Lister's antisepsis, Nightingale's hospital hygiene and Koch's bateriology combined in a new approach to infection control in hospitals, which was called asepsis. In modern hospitals asepsis is the norm, combining steam sterilisation of instruments and clothing in the operating theatre with disinfectant cleaning of surfaces and antiseptic cleaning and dressings for wounds.
The discovery of antibiotics in the 1940s made a dramatic difference to the control of infections in the body. For the first time, there was an effective drug treatment and cure for infections.
Gradually, hospitals became complacent about hygiene standards. Why would you put so much effort into preventing infections when you could just focus on treating the infected patient?
Unfortunately, bacteria and viruses evolve at astonishing rates, and some strains of 'superbugs' have become resistant to antibiotics. Overuse of antibiotics has, once again, made hospital infection potentially life-threatening.
Antibiotic resistant 'superbugs' such as MRSA exist inside and outside hospitals so modern health care services have to be vigilant about infections in the hospital and those brought in by visitors. Hospital infection was almost thought of as a thing of the past, but it has not gone away, and hygiene and aseptic controls are as important as ever.
Suggestions for further research
Surgical Asepsis and Principles of Sterile Technique, (website)
Slide presentation: Asepsis and Antisepsis in the Operating Room, slideshare.net (website)
‘Asepsis and Bacteriology’ by Thomas Schlich (ejournal)
‘The History of Germ Theory’ by Jemima Hodkinson (ejournal)
‘Antiseptic Surgery’, William Watson Cheyne, 1882 (ebook)
‘On the Effects of the Antiseptic System of Treatment’, Joseph Lister, 1870 (ebook)
'One False Move', 1963 public information film (video)
Find out more about infection
### Joseph Lister’s antisepsis system
Not so long ago even the smallest procedure could be deadly if infection entered the body. Antisepsis gave us a way to make surgery safe.
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Hospitals
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Read more stories about the history of hospitals and the practice of hospital medicine.
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LIving with a superbug
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In 2016, 54-year-old Craig was infected with Mycobacterium Abscessus, a rare water-borne bacteria, during a routine operation. |
190128 | https://assets.publishing.service.gov.uk/media/6140b7008fa8f503ba3dc8d1/Maths_guidance_KS_1_and_2.pdf | Mathematics guidance: key stages 1 and 2 Non-statutory guidance for the national curriculum in England June 2020 2 Contents Acknowledgements 3 Summary 4 Introduction 5 Ready-to-progress criteria: year 1 to year 6 9 Year 1 guidance 16 Year 2 guidance 48 Year 3 guidance 82 Year 4 guidance 143 Year 5 guidance 208 Year 6 guidance 278 Appendix: number facts fluency overview 332 3 Acknowledgements We would like to thank and acknowledge the following people involved in the production of this publication: Dr Debbie Morgan (Director for Primary Mathematics, NCETM); Clare Christie (Maths Consultant and Lead Practitioner, Ashley Down Primary School); Professor Anne Watson (Emeritus Professor in Mathematics Education, University of Oxford); Dr Yvette Sturdy (Development Editor, DiGiV8 Ltd.); Jalita Jacobson (Editor and Illustrator); Steve Evans (Illustrator); Dr Xinfeng Huang, (Associate Professor, Shanghai Normal University); Elizabeth Lambert (Assistant Director for Primary, NCETM); Emma Patman (Assistant Director for Primary, NCETM); Carol Knights (Director for Secondary, NCETM); Kris Dong (Deputy Head of Primary, National Nord Anglia Chinese International School Shanghai); Ellie Swidryk (Mastery Specialist Teacher, Shrubland Street Primary School); Thomas Hassall (Mastery Specialist Teacher, St Andrew’s Junior School); David Byron (Mastery Specialist Teacher, Cathedral Primary School, Bristol); Thomas Henesey (Mastery Specialist Teacher, Kingsholm C of E Primary School); Ione Haroun (Mastery Specialist Teacher, St White's Primary School); Farzana Khan (Mastery Specialist Teacher, Astwood Bank Primary School); Joanne Lane (Mastery Specialist Teacher, Stourport Primary Academy); Cat Potter (Head of Mathematics, Trinity Academy Bristol). 4 Summary This publication provides non-statutory guidance from the Department for Education. It has been produced to help teachers and schools make effective use of the National Curriculum to develop primary school pupils’ mastery of mathematics. Who is this publication for? This guidance is for: local authorities school leaders, school staff and governing bodies in all maintained schools, academies and free schools 5 Introduction Aims of the publication This publication aims to: bring greater coherence to the national curriculum by exposing core concepts in the national curriculum and demonstrating progression from year 1 to year 6 summarise the most important knowledge and understanding within each year group and important connections between these mathematical topics What is included in the publication? This publication identifies the most important conceptual knowledge and understanding that pupils need as they progress from year 1 to year 6. These important concepts are referred to as ready-to-progress criteria and provide a coherent, linked framework to support pupils’ mastery of the primary mathematics curriculum. The ready-to-progress criteria for all year groups are provided at the end of the introduction (Ready-to-progress criteria), and each criterion is explained within the corresponding year-group chapter. Please note that the publication does not address the whole of the primary curriculum, but only the areas that have been identified as a priority. It is still a statutory requirement that the whole of the curriculum is taught. However, by meeting the ready-to-progress criteria, pupils will be able to more easily access many of the elements of the curriculum that are not covered by this guidance. The year-group chapters Each chapter begins with a table that shows how each ready-to-progress criterion for that year group links to pupils’ prior knowledge and future applications. Each year-group chapter then provides: teaching guidance for each ready-to-progress criterion, including core mathematical representations, language structures and discussion of connections to other criteria example assessment questions for each ready-to-progress criterion guidance on the development of calculation and fluency Representations of the mathematics A core set of representations have been selected to expose important mathematical structures and ideas, and make them accessible to pupils. Consistent use of the same representations across year groups help to connect prior learning to new learning. The 6 example below demonstrates the use of tens frames and counters extended from year 1, where each counter represents 1 and a filled frame represents 10, to year 4 where each counter represents 100 a filled frame represents 1,000. Figure 1: using a tens frame and counters Figure 2: using a tens frame and counters Language structures The development and use of precise and accurate language in mathematics is important, so the guidance includes ‘Language focus’ features. These provide suggested sentence structures for pupils to use to capture, connect and apply important mathematical ideas. Once pupils have learnt to use a core sentence structure, they should be able to adapt and reason with it to apply their understanding in new contexts. Language focus “8 plus 6 is equal to 14, so 8 hundreds plus 6 hundreds is equal to 14 hundreds.” “14 hundreds is equal to 1,400.” Making Connections ‘Making connections’ features discuss important connections between ready-to-progress criteria within a year group. The example below describes how division with remainders is connected to multiplication and fractions criteria. Making connections Pupils must have automatic recall of multiplication facts and related division facts, and be able to recognise multiples (4NF–1 ) before they can solve division problems with remainders. For example, to calculate , pupils need to be able to identify the largest multiple of 7 that is less than 55 (in this case 49). They must then recall how many sevens there are in 49, and calculate the remainder. Converting improper fractions to mixed numbers (4F–2 ) relies on solving division problems with remainders. For example, converting to a mixed number depends on the calculation . 7 Assessment Example assessment questions are provided for each ready to progress criterion. These questions demonstrate the depth and breadth of understanding that pupils need to be ready to progress to the next year group. Calculation and fluency Each chapter ends with a section on the development of calculation methods and fluency. Pupils should be able to choose and use efficient calculation methods for addition, subtraction, multiplication and division. They must also have automatic recall of a core set of multiplicative and additive facts to enable them to focus on learning new concepts. Appendix: number facts fluency overview sets out when the multiplication tables and core additive facts should be taught, and in what order. How to use this publication This publication can support long-term, medium-term and short-term planning, and assessment. At the long-term planning stage, this guidance can be used to ensure that the most important elements that underpin the curriculum are covered at the right time, and to ensure that there is continuity and consistency for pupils as they progress from one year group to the next. At the medium-term planning stage, teachers can use the guidance to inform decisions on how much teaching time to set aside for the different parts of the curriculum. Teaching time can be weighted towards the ready-to-progress criteria. The ready-to-progress tables at the start of each year group and the ‘Making connections’ features support medium-term planning by demonstrating how to make connections between mathematical ideas and develop understanding based on logical progression. At the short-term planning stage, the guidance can be used to inform teaching strategy, and the representations and ‘Language focus’ features can be used to make concepts more accessible to pupils. The publication can also be used to support transition conversations between teachers of adjacent year groups, so that class teachers understand what pupils have been taught in the previous year group, how they have been taught it, and how effectively pupils have understood and remembered it. 8 Ready-to-progress criteria and the curriculum The ready-to-progress criteria in this document are organised into 6 strands, each of which has its own code for ease of identification. These are listed below. Measurement and Statistics are integrated as applications of number criteria, and elements of measurement that relate to shape are included in the Geometry strand. Ready-to-progress criteria strands Code Number and place value NPV Number facts NF Addition and subtraction AS Multiplication and division MD Fractions F Geometry G Special educational needs and disability (SEND) Pupils should have access to a broad and balanced curriculum. The National Curriculum Inclusion Statement states that teachers should set high expectations for every pupil, whatever their prior attainment. Teachers should use appropriate assessment to set targets which are deliberately ambitious. Potential areas of difficulty should be identified and addressed at the outset. Lessons should be planned to address potential areas of difficulty and to remove barriers to pupil achievement. In many cases, such planning will mean that pupils with SEN and disabilities will be able to study the full national curriculum. The guidance in this document will support planning for all SEND pupils by highlighting the most important concepts within the national curriculum so that teaching and targeted support can be weighted towards these.
9 Ready-to-progress criteria: year 1 to year 6 The table below is a summary of the ready-to-progress criteria for all year groups. Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 NPV 1NPV–1 Count within 100, forwards and backwards, starting with any number. 3NPV–1 Know that 10 tens are equivalent to 1 hundred, and that 100 is 10 times the size of 10; apply this to identify and work out how many 10s there are in other three-digit multiples of 10. 4NPV–1 Know that 10 hundreds are equivalent to 1 thousand, and that 1,000 is 10 times the size of 100; apply this to identify and work out how many 100s there are in other four-digit multiples of 100. 5NPV–1 Know that 10 tenths are equivalent to 1 one, and that 1 is 10 times the size of 0.1. Know that 100 hundredths are equivalent to 1 one, and that 1 is 100 times the size of 0.01. Know that 10 hundredths are equivalent to 1 tenth, and that 0.1 is 10 times the size of 0.01. 6NPV–1 Understand the relationship between powers of 10 from 1 hundredth to 10 million, and use this to make a given number 10, 100, 1,000, 1 tenth, 1 hundredth or 1 thousandth times the size (multiply and divide by 10, 100 and 1,000). 2NPV–1 Recognise the place value of each digit in two-digit numbers, and compose and decompose two-digit numbers using standard and non-standard partitioning. 3NPV–2 Recognise the place value of each digit in three-digit numbers, and compose and decompose three-digit numbers using standard and non-standard partitioning. 4NPV–2 Recognise the place value of each digit in four-digit numbers, and compose and decompose four-digit numbers using standard and non-standard partitioning. 5NPV–2 Recognise the place value of each digit in numbers with up to 2 decimal places, and compose and decompose numbers with up to 2 decimal places using standard and non-standard partitioning. 6NPV–2 Recognise the place value of each digit in numbers up to 10 million, including decimal fractions, and compose and decompose numbers up to 10 million using standard and non-standard partitioning. 1NPV–2 Reason about the location of numbers to 20 within the linear number system, including comparing using < > and = 2NPV–2 Reason about the location of any two-digit number in the linear number system, including identifying the previous and next multiple of 10. 3NPV–3 Reason about the location of any three-digit number in the linear number system, including identifying the previous and next multiple of 100 and 10. 4NPV–3 Reason about the location of any four-digit number in the linear number system, including identifying the previous and next multiple of 1,000 and 100, and rounding to the nearest of each. 5NPV–3 Reason about the location of any number with up to 2 decimals places in the linear number system, including identifying the previous and next multiple of 1 and 0.1 and rounding to the nearest of each. 6NPV–3 Reason about the location of any number up to 10 million, including decimal fractions, in the linear number system, and round numbers, as appropriate, including in contexts. 10 Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 NPV 3NPV–4 Divide 100 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 100 with 2, 4, 5 and 10 equal parts. 4NPV–4 Divide 1,000 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 1,000 with 2, 4, 5 and 10 equal parts. 5NPV–4 Divide 1 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in units of 1 with 2, 4, 5 and 10 equal parts. 6NPV–4 Divide powers of 10, from 1 hundredth to 10 million, into 2, 4, 5 and 10 equal parts, and read scales/number lines with labelled intervals divided into 2, 4, 5 and 10 equal parts. 5NPV–5 Convert between units of measure, including using common decimals and fractions. NF 1NF–1 Develop fluency in addition and subtraction facts within 10. 2NF–1 Secure fluency in addition and subtraction facts within 10, through continued practice. 3NF–1 Secure fluency in addition and subtraction facts that bridge 10, through continued practice. 1NF–2 Count forwards and backwards in multiples of 2, 5 and 10, up to 10 multiples, beginning with any multiple, and count forwards and backwards through the odd numbers. 3NF–2 Recall multiplication facts, and corresponding division facts, in the 10, 5, 2, 4 and 8 multiplication tables, and recognise products in these multiplication tables as multiples of the corresponding number. 4NF–1 Recall multiplication and division facts up to , and recognise products in multiplication tables as multiples of the corresponding number. 5NF–1 Secure fluency in multiplication table facts, and corresponding division facts, through continued practice. 4NF–2 Solve division problems, with two-digit dividends and one-digit divisors, that involve remainders, and interpret remainders appropriately according to the context. 3NF–3 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10). 4NF–3 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 100) 5NF–2 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 1 tenth or 1 hundredth). 11 Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 AS 1AS–1 Compose numbers to 10 from 2 parts, and partition numbers to 10 into parts, including recognising odd and even numbers. 2AS–1 Add and subtract across 10. 3AS–1 Calculate complements to 100. 6AS/MD–1 Understand that 2 numbers can be related additively or multiplicatively, and quantify additive and multiplicative relationships (multiplicative relationships restricted to multiplication by a whole number). 1AS–2 Read, write and interpret equations containing addition ( ), subtraction ( ) and equals ( ) symbols, and relate additive expressions and equations to real-life contexts. 2AS–2 Recognise the subtraction structure of ‘difference’ and answer questions of the form, “How many more…?”. 3AS–2 Add and subtract up to three-digit numbers using columnar methods. 6AS/MD–2 Use a given additive or multiplicative calculation to derive or complete a related calculation, using arithmetic properties, inverse relationships, and place-value understanding. 2AS–3 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract only ones or only tens to/from a two-digit number. 3AS–3 Manipulate the additive relationship: Understand the inverse relationship between addition and subtraction, and how both relate to the part–part–whole structure. Understand and use the commutative property of addition, and understand the related property for subtraction. 6AS/MD–3 Solve problems involving ratio relationships. 2AS–4 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract any 2 two-digit numbers. 6AS/MD–4 Solve problems with 2 unknowns. 12 13 Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 MD 2MD–1 Recognise repeated addition contexts, representing them with multiplication equations and calculating the product, within the 2, 5 and 10 multiplication tables. 3MD–1 Apply known multiplication and division facts to solve contextual problems with different structures, including quotitive and partitive division. 4MD–1 Multiply and divide whole numbers by 10 and 100 (keeping to whole number quotients); understand this as equivalent to making a number 10 or 100 times the size. 5MD–1 Multiply and divide numbers by 10 and 100; understand this as equivalent to making a number 10 or 100 times the size, or 1 tenth or 1 hundredth times the size. For year 6, MD ready-to-progress criteria are combined with AS ready-to-progress criteria (please see above). 2MD–2 Relate grouping problems where the number of groups is unknown to multiplication equations with a missing factor, and to division equations (quotitive division). 4MD–2 Manipulate multiplication and division equations, and understand and apply the commutative property of multiplication. 5MD–2 Find factors and multiples of positive whole numbers, including common factors and common multiples, and express a given number as a product of 2 or 3 factors. 4MD–3 Understand and apply the distributive property of multiplication. 5MD–3 Multiply any whole number with up to 4 digits by any one-digit number using a formal written method. 5MD–4 Divide a number with up to 4 digits by a one-digit number using a formal written method, and interpret remainders appropriately for the context. 14 Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 F 3F–1 Interpret and write proper fractions to represent 1 or several parts of a whole that is divided into equal parts. 6F–1 Recognise when fractions can be simplified, and use common factors to simplify fractions. 3F–2 Find unit fractions of quantities using known division facts (multiplication tables fluency). 5F–1 Find non-unit fractions of quantities. 6F–2 Express fractions in a common denomination and use this to compare fractions that are similar in value. 3F–3 Reason about the location of any fraction within 1 in the linear number system. 4F–1 Reason about the location of mixed numbers in the linear number system. 6F–3 Compare fractions with different denominators, including fractions greater than 1, using reasoning, and choose between reasoning and common denomination as a comparison strategy. 4F–2 Convert mixed numbers to improper fractions and vice versa. 5F–2 Find equivalent fractions and understand that they have the same value and the same position in the linear number system. 3F–4 Add and subtract fractions with the same denominator, within 1. 4F–3 Add and subtract improper and mixed fractions with the same denominator, including bridging whole numbers. 5F–3 Recall decimal fraction equivalents for , , and , and for multiples of these proper fractions. G 1G–1 Recognise common 2D and 3D shapes presented in different orientations, and know that rectangles, triangles, cuboids and pyramids are not always similar to one another. 2G–1 Use precise language to describe the properties of 2D and 3D shapes, and compare shapes by reasoning about similarities and differences in properties. 3G–1 Recognise right angles as a property of shape or a description of a turn, and identify right angles in 2D shapes presented in different orientations. 5G–1 Compare angles, estimate and measure angles in degrees (°) and draw angles of a given size. 15 Strand Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 G 5G–2 Compare areas and calculate the area of rectangles (including squares) using standard units. 1G–2 Compose 2D and 3D shapes from smaller shapes to match an example, including manipulating shapes to place them in particular orientations. 3G–2 Draw polygons by joining marked points, and identify parallel and perpendicular sides. 4G–1 Draw polygons, specified by coordinates in the first quadrant, and translate within the first quadrant. 6G–1 Draw, compose, and decompose shapes according to given properties, including dimensions, angles and area, and solve related problems. 4G–2 Identify regular polygons, including equilateral triangles and squares, as those in which the side-lengths are equal and the angles are equal. Find the perimeter of regular and irregular polygons. 4G–3 Identify line symmetry in 2D shapes presented in different orientations. Reflect shapes in a line of symmetry and complete a symmetric figure or pattern with respect to a specified line of symmetry. 16 Year 1 guidance Ready-to-progress criteria Previous experience Year 1 ready-to-progress criteria Future applications Begin to develop a sense of the number system by verbally counting forward to and beyond 20, pausing at each multiple of 10. 1NPV–1 Count within 100, forwards and backwards, starting with any number. Count through the number system. Place value within 100. Compare and order numbers. Add and subtract within 100. Play games that involve moving along a numbered track, and understand that larger numbers are further along the track. 1NPV–2 Reason about the location of numbers to 20 within the linear number system, including comparing using < > and = Reason about the location of larger numbers within the linear number system. Compare and order numbers. Read scales. Begin to experience partitioning and combining numbers within 10. 1NF–1 Develop fluency in addition and subtraction facts within 10. Add and subtract across 10. All future additive calculation. Add within a column during columnar addition when the column sums to less than 10 (no regrouping). Subtract within a column during columnar subtraction when the minuend of the column is larger than the subtrahend (no exchanging). Distribute items fairly, for example, put 3 marbles in each bag. Recognise when items are distributed unfairly. 1NF–2 Count forwards and backwards in multiples of 2, 5 and 10, up to 10 multiples, beginning with any multiple, and count forwards and backwards through the odd numbers. Recall the 2, 5 and 10 multiplication tables. Carry out repeated addition and multiplication of 2, 5, and 10, and divide by 2, 5 and 10. Identify multiples of 2, 5 and 10. Unitise in tens. Identify odd and even numbers. 17 Previous experience Year 1 ready-to-progress criteria Future applications Understand the cardinal value of number words, for example understanding that ‘four’ relates to 4 objects. Subitise for up to to 5 items. Automatically show a given number using fingers. 1AS–1 Compose numbers to 10 from 2 parts, and partition numbers to 10 into parts, including recognising odd and even numbers. Add and subtract within 10. Devise and record number stories, using pictures, numbers and symbols (such as arrows). 1AS–2 Read, write and interpret equations containing addition ( ), subtraction ( ) and equals ( ) symbols, and relate additive expressions and equations to real-life contexts. Represent composition and decomposition of numbers using equations. See, explore and discuss models of common 2D and 3D shapes with varied dimensions and presented in different orientations (for example, triangles not always presented on their base). 1G–1 Recognise common 2D and 3D shapes presented in different orientations, and know that rectangles, triangles, cuboids and pyramids are not always similar to one another. Describe properties of shape. Categorise shapes. Identify similar shapes. Select, rotate and manipulate shapes for a particular purpose, for example: rotating a cylinder so it can be used to build a tower rotating a puzzle piece to fit in its place 1G–2 Compose 2D and 3D shapes from smaller shapes to match an example, including manipulating shapes to place them in particular orientations. Find the area or volume of a compound shape by decomposing into constituent shapes. Rotate, translate and reflect 2D shapes. Identify congruent shapes. 18 1NPV–1 Count forwards and backwards within 100 Count within 100, forwards and backwards, starting with any number. 1NPV–1 Teaching guidance Counting to and across 100, forwards and backwards, is a skill that will need to be practised regularly throughout year 1. Counting provides a good opportunity to link number names to numerals, and to the position of numbers in the linear number system. Practice should include: reciting number names, without the support of visual representations, to allow pupils to focus on and develop fluency in the verbal patterns counting with the support of visual representations and gestural patterns, for example pupils can point to numerals on a 100 square or number line, or tap out the numbers on a Gattegno chart starting the counting sequence with numbers other than 1 or 100 Figure 3: 100 square Figure 4: 0 to 100 number line 19 Figure 5: Gattegno chart showing multiples of 1, 10, 100 and 1,000 It is important to draw pupils’ attention to structures in the number system, for example, linking 1, 2, 3, 4, 5… to 31, 32, 33, 34, 35…, and this can be supported by the use of the visual representations above. Because number names in English do not always reflect the structure of the numbers, pupils should also practise using dual counting, first counting with number names, and then repeating the count with words based on the number structures. Language focus “…seven, eight, nine, ten, eleven, twelve, thirteen… twenty, twenty-one, twenty-two…” “…seven, eight, nine, one-ten, one-ten-one, one-ten-two, one-ten-three… two-tens, two-tens-one, two-tens two…” When counting backwards, pupils often find it challenging to identify which number they should say after they have said a multiple of 10. A partially marked number line can be used for support. Figure 6: 0 to 100 number line to support backwards counting Counting backwards from 20 down to 10 requires additional focused practice, due to the irregularity of these number names. By the end of year 1, pupils must be able to count forwards and backwards, within 100, without visual aids. 20 Making connections Being able to count fluently, both forwards and backwards, is necessary for pupils to be able to reason about the location of numbers in the linear number system (1NPV–2) Sequencing in ones will extend to sequencing in multiples of 2, 5 and 10 (1NF–2). 1NPV–1 Example assessment questions Fill in the missing numbers. 8 9 11 12 37 38 40 42 43 63 62 60 58 57 Assessment guidance: To assess criterion 1NPV–1, teachers must listen to each pupil count. This can be done through specifically planned tasks, or by carefully watching and listening to an individual pupil during daily counting as part of class routines. 1NPV–2 Numbers to 20 in the linear number system Reason about the location of numbers to 20 within the linear number system, including comparing using < > and = 1NPV–2 Teaching guidance Pupils should be introduced to the number line as a representation of the order and relative size of numbers. Pupils should: begin to develop the ability to mentally visualise a number line, with consecutive whole numbers equally spaced along it draw number lines, with consecutive whole numbers equally spaced along them identify or place numbers up to 20 on marked and unmarked number lines Pupils should use efficient strategies and appropriate reasoning, including working backwards from a multiple of 10, to identify or place numbers on marked number lines. 21 Figure 7: identifying 5, 12 and 19 on a marked 0 to 20 number line Language focus “a is 5 because it is halfway between 0 and 10.” “b is 12 because it is 2 more than 10.” “c is 19 because it is one less than 20.” Pupils should also be able to estimate the value or position of numbers on unmarked number lines, using appropriate proportional reasoning, rather than counting on from a start point or back from an end point. Pupils should learn to estimate the value/position of a number relative to both ends of the number line, beginning with a 0 to 10 number line, then moving on to 0 to 20 and 10 to 20 number lines. When pupils are asked to mark all numbers on, for example, an unmarked 0 to 10 number line, they typically start at 1 and run out of space as they approach 10, and so bunch the larger numbers together. Pupils should learn to look at the full length of the number line and mark on the midpoint first. They should be able to reason about the location of numbers relative to both ends, and the midpoint, of a number line, for example, “16 is about here because it is just over halfway between 10 and 20.” Figure 8: placing 16 on an unmarked 0 to 20 number line There is no need for pupils to be completely accurate in their estimation of value or position of numbers. Rather they should make reasonable judgements that demonstrate they are developing proportional thinking. Pupils should use their knowledge of the position of the numbers 0 to 10 on a number line to help them to estimate the value or position of the numbers 10 to 20. Figure 9: placing 16 on an unmarked 10 to 20 number line 22 Making connections Being able to count fluently (1NPV–1), both forwards and backwards, is necessary for pupils to be able to reason about the location of numbers in the linear number system. 1NPV–2 Example assessment questions Label these numbers on the number line. 9 15 3 12 Estimate the value of the missing numbers. Mahmood is using 10cm paper strips to measure things in the classroom. a. How long do you think the eraser is? 23 b. How long do you think the pencil is? Images drawn to scale Mia measures 2 different leaves with a ruler. How long is each leaf? Images drawn to scale Assessment guidance: The example questions above can be set as a written task. However, teachers will need to watch pupils closely to assess wheher pupils are developing efficient strategies and appropriate proportional reasoning skills. Teachers should assess pupils in small groups. 24 1NF–1 Fluently add and subtract within 10 Develop fluency in addition and subtraction facts within 10. 1NF–1 Teaching guidance It is very important for pupils to be able to add and subtract within 10, fluently, by the end of year 1. This should be taught and practised until pupils move beyond counting forwards or backwards in ones, to more efficient strategies and eventually to automatic recall of these number facts. This is necessary before pupils move on to additive calculation with larger numbers. The 66 addition facts within 10 are shown on the grid below. The number of addition facts to be learnt is reduced when commutativity is applied and pupils recognise that 3 + 2, for example, is the same as 2 + 3. Pupils must also have automatic recall of the corresponding subtraction facts, for example 5 – 3 and 5 – 2. + 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Pupils should learn to compose and partition numbers within 10 (1AS–1) before moving on to formal addition and subtraction. Although 1NF–1 (this criteiron) and 1AS–2 are presented as separate ready-to-progress criteria, they depend upon one another and should be developed, in tandem, throughout year 1. 25 As part of their work on 1AS–1, pupils are likely to already have memorised some number bonds within 10 (for example, number bonds to 5, to 10 and some doubles facts). However, at this stage, most pupils won’t remember all of their number facts by rote learning, so they should also be taught to derive additive facts within 10 from previously memorised facts or knowledge. Examples of appropriate strategies include the following. Example strategy 1: Example strategy 2: Language focus “I know that double 3 is equal to 6, so 4 plus 3 is equal to 7.” Figure 10: tens frames with counters showing derivation of a ‘near-double’ addition calculation Language focus “If I subtract 2 from an even number I get the previous even number, so 6 minus 2 is equal to 4.” Figure 11: tens frames with counters showing that subtracting 2 from an even number gives the previous even number Pupils need extensive practice to meet this criterion. You can find out more about fluency for addition and subtraction within 10 here in the calculation and fluency section: 1NF–1 Making connections Understanding how numbers within 10 can be composed and partitioned (1AS–1) underpins fluency in addition and subtraction facts within 10. Pupils need to be able carry out these calculations when they are presented as equations, and when they are presented as contextual word problems as described in 1AS–2. The 1NF–1 Example assessment questions below provide contextual problems only. 26 1NF–1 Example assessment questions I cycled 4km to get to my friend’s house, and then cycled another 3km with my friend. How far have I cycled? There are 9 children. 6 of them have scooters and the rest do not. How many of the children do not have scooters? Sarah had £6. Then she spent £3. How much money does she have left? I have 1 metre of red ribbon. I have 5 metres of blue ribbon. How many metres of ribbon do I have altogether? 1NF–2 Count forwards and backwards in multiples of 2, 5 and 10 Count forwards and backwards in multiples of 2, 5 and 10, up to 10 multiples, beginning with any multiple, and count forwards and backwards through the odd numbers. 1NF–2 Teaching guidance Pupils must be able to count in multiples of 2, 5 and 10 by the end of year 1 so that they are ready to progress to multiplication involving groups of 2, 5 and 10 in year 2. As with counting in ones, within 100 (1NPV–1 ), this is a skill that will need to be practised throughout year 1. As with 1NPV–1, forwards and backwards counting practice should include: reciting just the number names (for example, “ten, twenty, thirty…”), without the support of visual representations counting with the support of visual representations and gestural patterns, for example pupils can point to numerals on a number line or 100 square, or tap out the numbers on a Gattegno chart starting the forwards counting sequence with numbers other than 2, 5 or 10 The 100 square and Gattegno chart are provided in criterion 1NPV–1 in figures 1 and 3 respectively. Figure 12: number line to support counting in multiples of 2 Figure 13: number line to support counting in multiples of 10 27 Figure 14: number line to support counting in multiples of 5 When pupils are confident with the counting sequences, they should learn to enumerate objects arranged in groups of 2, 5 or 10 by skip counting, so they begin to connect counting in multiples with finding the total quantity for repeated groups. Pupils should first identify how many objects are in each repeated group, and then skip count in this number. They should leave year 1 understanding that when objects are grouped equally, it is more efficient to skip count than to count in ones. Recognising that a group of 5, for example, can be treated as a single unit is called unitising, and is the basis of multiplicative reasoning. Figure 15: counting in tens – 9 packs of 10 pencils Language focus “The pencils are in groups of 10, so we will count in tens.” Pupils should also practise counting in two ways: counting the total number of objects using skip counting, or counting the number of repeated groups. This will prepare pupils for multiplication and division in year 2. 28 Language focus “Ten, twenty, thirty…” “1 group of 10, 2 groups of 10, 3 groups of 10…” In time, shortened to: “1 ten, 2 tens, 3 tens…” Once pupils have learnt to recognise 2p, 5p and 10p coins, they should be expected to apply their knowledge of counting in multiples of 2, 5 and 10 to find the value of a set of like coins, and to find how many of a particular denomination is required to pay for a given item – see 1NF–2 (questions 6 and 7). Pupils should also learn to recite the odd number sequence, both forwards and backwards. This can initially be supported by a number line with odd numbers highlighted. Figure 16: number line to support counting through the odd number sequence Pupils need extensive practice to meet this criterion. You can find out more about fluency for counting in multiples of 2, 5 and 10 here in the calculation and fluency section: 1NF–2 Making connections The patterns and structure in the number system which pupils learn from counting in ones (1NPV–1) will prepare pupils for learning to count in multiples of 2, 5 and 10. In 1AS–1 pupils learn to identify odd and even numbers within 10. Counting in multiples of 2 (even numbers), and through the odd number sequence, demonstrates to pupils that the odd and even number patterns continue through the number system. 29 1NF–2 Example assessment questions Task: Provide each pupil with an even number of counters up to 20, then ask pupils to: a. put the counters into groups of 2 b. count in multiples of 2 find out how many counters there are These sticks are grouped into bundles of 10. How many sticks are there altogether? How many wheels are there altogether? Count in groups of 2. There are 5 hedgehogs in each group. How many hedgehogs are there altogether? There are 5 dots on each token. How many dots are there altogether? How much money is in each purse? 30 Task: Provide each pupil with 2p, 5p and 10p coins (real or otherwise), then ask pupils to show how to pay for: a. the drum with 2p coins b. the boat with 5p coins c. the dinosaur with 10p coins Assessment guidance: To assess whether pupils can recite the number sequences, teachers must listen to each pupil count. This can be done through specifically planned tasks, or by carefully watching and listening to an individual pupil during daily counting as part of class routines. The example questions and tasks above can be used to assess whether pupils can enumerate objects in groups of 2, 5 or 10. However, simply providing the correct answers to the example questions does not demonstrate that a pupil has met this part of the criterion – teachers should assess pupils in small groups to ensure that they are counting in multiples of 2, 5 or 10 rather than in ones. 1AS–1 Compose and partition numbers to 10 Compose numbers to 10 from 2 parts, and partition numbers to 10 into parts, including recognising odd and even numbers. 1AS–1 Teaching guidance Learning to ‘see’ a whole number and the parts within it at the same time is an important stage in the development of pupils’ understanding of number. Composing numbers (putting parts together to make a whole) underpins addition, and decomposing a number into parts (partitioning) underpins subtraction. Exploring different ways that a number can be partitioned and put back together again helps pupils to understand that addition and subtraction are inverse operations. Pupils should be presented with varied cardinal (quantity) representations, both concrete and pictorial, which support identification of the ‘numbers within a number’. The examples below provide different ways of showing that 8 can be composed from 2 numbers. The representations draw attention to the parts within the whole. 31 Figure 17: 8 represented as 3 fingers and 5 fingers Figure 18: 8 represented as 6 and 2 with base 10 number boards Figure 19: 8 represented as two 4-value dice Figure 20: 8 represented as 2 rows of 4 Figure 21: 8 represented as tally marks: 5 and 3 Figure 22: 8 represented on a bead string: 7 and 1 Pupils should learn to interpret and sketch partitioning diagrams to represent the ways numbers can be partitioned or combined. At this stage, these should be used alongside quantity images to support development of the understanding of quantity. Pupils should be able to relate the numerals in the partitioning diagrams to the quantities in images, and use the language of parts and wholes to describe the relationship between the numbers. Figure 23: using a partitioning diagram to represent 6 flags, consisting of 4 spotty flags and 2 stripy flags Language focus “There are 6 flags. 4 are spotty and 2 are stripy.” “6 is the whole. 4 is a part. 2 is a part.” 32 Pupils should also experience working with manipulatives and practise partitioning a whole number of items into parts, then putting the parts back together. They should understand that the total quantity is conserved. Pupils should repeatedly partition and recombine the whole, in different ways. Pupils should learn how to work systematically to partition each of the numbers to 10 into 2 parts. They should recognise that there is a finite number of ways that a given number can be partitioned. Pupils should pay attention to the patterns observed when working systematically, for example: in each step below, the value of one part increases by 1 and the value of the other part decreases by 1, while the whole remains the same number pairs are repeated, but with the values reversed, for example when 6 is the whole, the parts can be 2 and 4, or 4 and 2 (pupils must be able to identify what is the same and what is different between these two options; this lays the foundations for understanding the commutative property of addition) Pupils must be able to describe and understand these patterns. Figure 24: working systematically to partition 6 Once pupils have learnt to write addition and subtraction equations, they should use these to express the different ways that numbers can be composed and decomposed (see 1AS–2), for example: 6 2 4 6 4 2 6 4 2 6 2 4 33 Pupils should learn to recognise odd and even numbers, up to 10, based on whether they can be composed of groups of 2 or not. Base 10 number boards, or tens frames with counters shown arranged in twos, can be used to expose the structure of odd and even numbers. Figure 25: odd and even numbers up to 10 Making connections Composing and decomposing numbers within 10 can be expressed with addition and subtraction equations (1AS–2), and is the basis of fluency in addition and subtraction facts within 10 (1NF–1). Here, pupils learn to identify odd and even numbers, while in 1NF–2 they develop fluency in the odd and even number sequences through practising skip counting. 34 1AS–1 Example assessment questions Mother duck is in the water with her 6 ducklings. There are 2 ponds. How many ducklings could be in each pond? Fill in the missing numbers. a. b. 7 7 6 7 5 7 4 7 3 7 2 7 1 7 0 7 35 Task: Provide each pupil with a tens frame and counters in 2 colours, then ask pupils to use the manipulatives to answer questions such as the following. “I am holding 9 counters altogether. How many counters are there in my closed hand?” Underline the numbers that are in the wrong sorting circle. Write the missing numbers in these odd and even sequences. 1 3 9 2 6 10 Assessment guidance: The focus of this criterion is understanding that numbers can be composed from, and partitioned into, smaller numbers. Pupils are assessed separately on their fluency in number facts within 10, in criterion 1NF–1. Therefore manipulatives such as counters and tens frames, or counters and partitioning diagram templates, should be made available to pupils during assessment of this criterion so that the questions are not dependent on pupils’ emerging number facts fluency. Note that Example assessment question 2 relies on pupils having learnt to write and interpret addition and subtractions equations. This question should only be used to assess understanding of composition and partitioning after pupils have met criterion 1AS–2. 36 1AS–2 Read, write and interpret additive equations Read, write and interpret equations containing addition ( ), subtraction ( ) and equals ( ) symbols, and relate additive expressions and equations to real-life contexts. 1AS–2 Teaching guidance Pupils must learn to use the mathematical symbols , and . Expressions or equations involving these symbols should be introduced as a way to represent numerical situations and mathematical stories. An expression such as 3 5 should not be interpreted as asking “What is 3 5 ?” but, rather, as a way to represent the additive structures discussed below, either within a real-life context or within an abstract numerical situation. It is important that pupils do not think of the equals symbol as meaning ‘and the answer is’. They should instead understand that the expressions on each side of an equals symbol have the same value. All examples used to teach this criterion should use quantities within 10, and be supported by manipulatives or images, to ensure that pupils are able to focus on the mathematical structures and to avoid the cognitive load of having to work out the solutions. For each of the 4 additive structures described below (aggregation, partitioning, augmentation and reduction), pupils should learn to link expressions (for example, 5 2 and 6 – 2) to contexts before they learn to link equations (for example, 5 2 7 and 6 – 2 4 ) to contexts. For each case, pupils’ understanding should be built up in steps: Pupils should first learn to describe the context using precise language (see the language focus boxes below). Pupils should then learn to write the associated expression or equation. Pupils should then use precise language to describe what each number in the expression or equation represents. Pupils need to be able to write and interpret expressions and equations to represent aggregation (combining 2 parts to make 1 whole) and partitioning (separating 1 whole into 2 parts). 37 Figure 26: addition as aggregation Language focus “There are 5 flowers in one bunch. There are 2 flowers in the other bunch. There are 7 flowers altogether.” “We can write this as 5 plus 2 is equal to 7.” “The 5 represents the number of flowers in 1 bunch.“ “The 2 represents the number of flowers in the other bunch.” “The 7 represents the total number of flowers.” Pupils must understand that, in partitioning situations, the subtraction symbol represents a splitting up or differentiating of the whole. The problem “There are 6 children altogether. 2 children are wearing coats. How many are not wearing coats?” is represented by 6 – 2 4 . Here, the subtraction symbol represents the separation of the 2 children wearing coats, and so, the number of children not wearing coats is exposed. 38 Figure 27: subtraction as partitioning Language focus “There are 6 children altogether. 2 children are wearing coats. 4 children are not wearing coats.” “We can write this as 6 minus 2 is equal to 4.” “The 6 represents the total number of children.” “The 2 represents the number of children that are wearing coats.” “The 4 represents the number of children that are not wearing coats.” Pupils must also be able to write and interpret expressions and equations to represent augmentation (increasing a quantity by adding more) and reduction (decreasing a quantity by taking some away). Note that ‘take away’ should only be used to describe the subtraction operation in reduction contexts. Figure 28: addition as augmentation 39 Language focus “First 4 children were sitting on the bus. Then 3 more children got on the bus. Now 7 children are sitting on the bus.” “We can write this as 4 plus 3 is equal to 7.” “The 4 represents the number of children that were on the bus at the start.” “The 3 represents the number of children that got on the bus.” “The 7 represents the number of children that are on the bus now.” Figure 29: subtraction as reduction Language focus “First there were 4 children in the bumper car. Then 1 child got out. Now there are 3 children in the bumper car.” “We can write this as 4 minus 1 is equal to 3.” “The 4 represents the number of children that were in the car at the start.” “The 1 represents the number of children that got out of the car.” “The 3 represents the number of children that are in the car now.” In the course of learning to read, write and interpret addition and subtractions equations, pupils should also learn that equations can be written in different ways, including: 40 varying the position of the equals symbol (for example, 5 – 2 3 and 3 5 – 2 ) for addition, the addends can be written in either order and the sum remains the same (commutativity) Figure 30: aggregation or partitioning context: 5 cakes altogether, 2 with cherries and 3 without 2 3 5 5 2 3 3 2 5 5 3 2 5 – 3 2 2 5 – 3 5 – 2 3 3 5 – 2 Pupils must also learn to relate addition and subtraction contexts and equations to mathematical diagrams such as bar models, number lines, tens frames with counters, and partitioning diagrams. Figure 31: bar model and subtraction equation ( – 7 2 5 ) Figure 32: number line and addition equation (2 3 5 ) Figure 33: tens frame with counters and addition equation (3 1 4 ) Figure 34: cherry partitioning model and subtraction equation ( – 7 3 4 ) Making connections Once pupils have completed this criterion, they should represent the composition and partitioning of numbers to 10 (1AS–1) using addition and subtraction equations. This criterion and 1AS–1 provide the conceptual prerequisites for pupils to develop fluency in addition and subtraction within 10 (1NF–1). 41 1AS–2 Example assessment questions Write an equation to represent this story. First I had 6 balloons. Then 2 floated away. Now I have 4 balloons. Write an equation to represent this story. There are 2 apples. There are 3 oranges. Altogether there are 5 pieces of fruit. Which equation matches the picture? Can you explain your choice? 3 + 3 = 6 8 = 4 + 3 4 = 3 + 1 4 + 3 = 7 Holly looks at this picture. She writes 4 –1 3 . Explain how Holly’s equation represents the picture. Write an equation to represent this picture. Explain how your equation matches the picture. 42 Assessment guidance: For pupils to meet this criterion, they need to demonstrate mastery of the structures. Correct calculation of the solutions to calculations is not required (this is assessed in 1NF–1). Where a question requires pupils to explain their reasoning, this should be done verbally. 1G–1 Recognise common 2D and 3D shapes Recognise common 2D and 3D shapes presented in different orientations, and know that rectangles, triangles, cuboids and pyramids are not always similar to one another. 1G–1 Teaching guidance Pupils need a lot of experience in exploring and discussing common 2D and 3D shapes. In the process, they should learn to recognise and name, at a minimum: rectangles (including squares), circles, and triangles cuboids (including cubes), cylinders, spheres and pyramids Pupils need to be able to recognise common shapes when they are presented in a variety of orientations and sizes and relative proportions, including large shapes outside the classroom (such as a rectangle marked on the playground or a circle on a netball court). Pupils should be able to describe, using informal language (for example, “long and thin”), the differences between non-similar examples of the same shapes, and recognise that these are still examples of the given shape. Figure 35: non-similar cylinders Pupils should practise distinguishing a given named shape type from plausible distractors. These activities should involve exploring shapes (for example, shapes cut from card) rather than only looking at pictures. 43 Figure 36: distinguishing triangles from plausible distractors Language focus Shape a: “This is not a triangle because it has 4 sides.” Shape b or e: “This is a triangle because it has 3 straight sides.” Shape c or d: “This is not a triangle because it has 6 sides.” Shape f: “This is not a triangle because some sides are curved.” Making connections Categorising examples and non-examples (for example, determining which shapes are triangles and which are not) is an important mathematical skill. Pupils should be developing this skill here and in other contexts, such as in 1AS–1, where they categorise numbers according to whether they are even or not even. 1G–1 Example assessment questions Task: Lay out a selection of 3D shapes, then ask “Can you find 3 different cuboids?” Task: Lay out a selection of 2D shapes that include triangles and plausible distractors, and other 2D shapes, then instruct pupils to choose 3 shapes as follows: a triangle a shape that reminds you of a triangle but is not one a shape which is nothing like a triangle Task: Lay out a selection of shapes, hold up one of them, and ask: “I wonder whether this shape is a rectangle. What do you think?” “Can you find a shape which is a rectangle?” (if the original shape is not a rectangle) 44 Task: Lay out a selection of shapes, hold up a cylinder, and instruct pupils to find another shape which “ is a bit like this one”. Ask pupils to explain their reasoning. Assessment guidance: Practical work, carried out in small groups, provides the most reliable method of assessing whether pupils have met this criterion. Teachers should assess pupils based, not just on their answers, but on the reasoning they use to reach their answers, for example, in question 4, a pupil may choose a cone because “it has a circle too”. When selecting shapes, careful attention should be paid to providing plausible distractors to allow assessment of reasoning. Pupils may use informal language, especially when discussing plausible distractors, for example, the shape presented in question 3 could be a parallelogram, which pupils could describe as being “a bit like a rectangle, but squashed.” Ask pupils to explain why they chose that one. 1G–2 Compose 2D and 3D shapes from smaller shapes Compose 2D and 3D shapes from smaller shapes to match an example, including manipulating shapes to place them in particular orientations. 1G–2 Teaching guidance The ability to compose and decompose shapes, and see shapes within shape, is a skill which runs through to key stage 3 and key stage 4, and beyond. In year 1, it is vital that pupils work practically, exploring shapes (for example, shapes cut from card, pattern blocks and tangrams) and putting them together to make new shapes. Pupils must be able to arrange 2D shapes to match an example compound shape. To begin with, the constituent shapes in a given example image should be the same size and colour as the actual shapes that pupils are using. This allows pupils to begin by laying the pieces over the example image, rotating individual pieces to match the exemplars. By the end of year 1, though, pupils should be able to copy a pattern block picture, and make a good attempt at copying a tangram picture, without overlaying the pieces on the example. Figure 37: example pattern block picture Figure 38: example tangram picture 45 Tangrams are more challenging to complete than pattern block pictures because: they contain several different-sized triangles, which pupils must distinguish from one another to complete the task placement of the parallelogram may require pupils to turn it over rather than just rotate it Pupils must also be able to arrange 3D shapes to match an example compound shape, for example joining a given number of multilink blocks to match an example. As a first step pupils can each make their own shape from a given number of blocks, and then compare the different shapes that have been made. Comparing compound shapes, and identifying the ones that are the same, will require pupils to rotate the shapes in various directions, and provides an opportunity to develop spatial language including: left, right, top, middle, bottom, on top of, below, in front of, behind and between. Figure 39: example compound 3D shapes composed of cubes Pupils must also learn to copy compound shapes composed of other 3D shapes, including cuboids, cylinders, spheres and pyramids. Figure 40: example compound shape composed of 4 different 3D shapes Making connections In 1AS–1, pupils learn to compose and decompose numbers to 10. Here children are using composition and decomposition in the context of shapes, recognising that shapes can be combined to form a larger shape and decomposed to return to the original shapes. 46 1G–2 Example assessment questions Task: Give each pupil some multilink cubes. Present pupils with a shape composed of multilink cubes and ask them to copy it. Task: Give each pupils some pattern block pieces. Present pupils with a pattern block picture and ask them to copy it. Task: Give each pupil some building blocks. Build a tower from different-shaped building blocks, then ask pupils to copy it . Task: Give each pupil a set of identical 2D shapes (for example, equilateral triangles of equal size). Make a pattern from the set of identical shapes and ask pupils to copy it. Assessment guidance: Practical work, carried out in small groups, provides the most reliable method of assessing whether pupils have met this criterion. Teachers should carefully watch pupils to assess their ability to rotate shapes to match those within the patterns, pictures and arrangements, and to place shapes relative to other shapes. Calculation and fluency 1NF–1 Fluently add and subtract within 10 Develop fluency in addition and subtraction facts within 10. The main addition and subtraction calculation focus in year 1 is developing fluency in additive facts within 10, as outlined in the 1NF–1 Teaching guidance Fluency in these facts allows pupils to more easily master addition and subtraction with 2-digit numbers in year 2, and underpins all future additive calculation. Pupils should practise carrying out addition and subtraction calculations, and working with equations in different forms, such as those below, until they achieve automaticity. Pupils should begin to recognise the inverse relationship between addition and subtraction, and use this to calculate. For example, if a pupil knows 6 4 10 , then they should be able to reason that 10 – 4 6 and 10 – 6 4 . 47 Pupils should also be expected to solve contextual addition and subtraction calculations with the 4 structures described in 1AS–2 (aggregation, partitioning, augmentation and reduction), for calculation within 10. 5 2 6 4 1 8 3 4 5 8 1 7 6 2 10 5 8 7 7 2 6 3 9 5 3 4 9 7 3 5 2 10 Pupils will need extensive practice, throughout the year, to achieve the fluency required to meet this criterion. 1NF–2 Count forwards and backwards in multiples of 2, 5 and 10 Count forwards and backwards in multiples of 2, 5 and 10, beginning with any multiple, and count forwards and backwards through the odd numbers. Pupils must be fluent in counting in multiples of 2, 5 and 10 by the end of year 1. Although this is the basis of multiplication and division by 2, 5, and 10, pupils do not need to be introduced to the words ‘multiplication’ and ‘division’ or to the multiplication and division symbols ( and ) in year 1, and are not expected to solve calculations presented as written equations. However, through skip counting (using practical resources, images such as number lines, or their fingers) pupils should begin to solve contextual multiplication and quotitive division problems, involving groups of 2, 5 or 10, for example: “I have four 5p coins. How much money do I have altogether?” “There are 10 apples in each bag. How many bags do I need to have 60 apples?” Pupils will need extensive practice, throughout the year, to achieve the fluency required to meet this criterion. 48 Year 2 guidance Ready-to-progress criteria Year 1 conceptual prerequesites Year 2 ready-to-progress criteria Future applications Know that 10 ones are equivalent to 1 ten. Know that multiples of 10 are made up from a number of tens, for example, 50 is 5 tens. 2NPV–1 Recognise the place value of each digit in two-digit numbers, and compose and decompose two-digit numbers using standard and non-standard partitioning. Compare and order numbers. Add and subtract using mental and formal written methods. Place the numbers 1 to 9 on a marked, but unlabelled, 0 to 10 number line. Estimate the position of the numbers 1 to 9 on an unmarked 0 to 10 number line. Count forwards and backwards to and from 100. 2NPV–2 Reason about the location of any two-digit number in the linear number system, including identifying the previous and next multiple of 10. Compare and order numbers. Round whole numbers. Subtract ones from a multiple of 10, for example: Develop fluency in addition and subtraction facts within 10. 2NF–1 Secure fluency in addition and subtraction facts within 10, through continued practice. All future additive calculation. Add within a column during columnar addition when the column sums to less than 10 (no regrouping). Subtract within a column during columnar subtraction when the minuend of the column is larger than the subtrahend (no exchanging). 49 Year 1 conceptual prerequesites Year 2 ready-to-progress criteria Future applications Learn and use number bonds to 10, for example: Partition numbers within 10, for example: 2AS–1 Add and subtract across 10, for example: Add and subtract within 100: add and subtract any 2 two-digit numbers, where the ones sum to 10 or more, for example: Use knowledge of unitising to add and subtract across other boundaries, for example: Add within a column during columnar addition when the column sums to more than 10 (regrouping), for example, for: Subtract within a column during columnar subtraction when the minuend of the column is smaller than the subtrahend (exchanging), for example, for: Solve missing addend problems within 10, for example: 2AS–2 Recognise the subtraction structure of ‘difference’ and answer questions of the form, “How many more…?”. Solve contextual subtraction problems for all three subtraction structures (reduction, partitioning and difference) and combining with other operations. Add and subtract within 10, for example: Know that a multiple of 10 is made up from a number of tens, for example, 50 is 5 tens. 2AS–3 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract only ones or only tens to/from a two-digit number. Add and subtract using mental and formal written methods. 50 Year 1 conceptual prerequesites Year 2 ready-to-progress criteria Future applications Add and subtract within 10. Know that a multiple of 10 is made up from a number of tens, for example, 50 is 5 tens. 2AS–4 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract any 2 two-digit numbers. Add and subtract numbers greater than 100, recognising unitising, for example: so Count in multiples of 2, 5 and 10. 2MD–1 Recognise repeated addition contexts, representing them with multiplication equations and calculating the product, within the 2, 5 and 10 multiplication tables. Use multiplication to represent repeated addition contexts for other group sizes. Memorise multiplication tables. Count in multiples of 2, 5 and 10 to find how many groups of 2, 5 or 10 there are in a particular quantity, set in everyday contexts. 2MD–2 Relate grouping problems where the number of groups is unknown to multiplication equations with a missing factor, and to division equations (quotitive division). Division with other divisors. Recognise common 2D and 3D shapes presented in different orientations. 2G–1 Use precise language to describe the properties of 2D and 3D shapes, and compare shapes by reasoning about similarities and differences in properties. Identify similar shapes. Describe and compare angles. Draw polygons by joining marked points Identify parallel and perpendicular sides. Identify regular polygons Find the perimeter of regular and irregular polygons. Compare areas and calculate the area of rectangles (including squares) using standard units. Compare areas and calculate the area of rectangles (including squares) using standard units. 51 2NPV–1 Place value in two-digit numbers Recognise the place value of each digit in two-digit numbers, and compose and decompose two-digit numbers using standard and non-standard partitioning. 2NPV–1 Teaching guidance Pupils need to be able to connect the way two-digit numbers are written in numerals to their value. They should demonstrate their reasoning using full sentences. Language focus “This is the number 42. The 4 shows we have 4 groups of ten. The 2 shows we have 2 extra ones.” Pupils should recognise that 42, for example, can be composed either of 42 ones, or of 4 tens and 2 ones. They should be able to group objects into tens, with some left over ones, to count efficiently and to demonstrate an understanding of the number. Pupils need to be capable of identifying the total quantity in different representations of groups of ten and additional ones. Within these representations the relative positions of the tens and the ones should be varied. Figure 41: varied representations of two-digit numbers as groups of ten and additional ones Pupils need to be able to partition two-digit numbers into tens and ones parts, and represent this using diagrams, and addition and subtraction equations. Figure 42: partitioning 28 into 20 and 8 20 8 28 28 20 8 8 20 28 28 8 20 28 20 8 8 28 20 28 8 20 20 28 8 52 It is also important for pupils to be able to think flexibly about number, learning to: partition into a multiple of ten and another two-digit number, in different ways (for example, 68 can be partitioned into 50 and 18, into 40 and 28, and so on) partition into a two-digit number and a one-digit number, in different ways (for example, 68 can be partitioned into 67 and 1, 66 and 2, and so on) Making connections Learning about place value should include connections with addition and subtraction in the form of partitioning two-digit numbers according to tens and ones, and writing associated additive equations. Pupils should also partition two-digit numbers in ways other than according to place value to prepare them to solve addition and subtraction calculations involving two-digit numbers. 2NPV–1 Example assessment questions Daisy has used 10cm rods and 1cm cubes to measure the length of this toy boat. How long is the boat? What is the total value of these coins? Monika watches a cartoon for 20 minutes and a news programme for 5 minutes. How long does she watch television for? Fill in the missing numbers. 47 7 8 60 Jed collects 38 conkers and gives 8 of them to Dylan. How many conkers does Jed have left? 53 2NPV–2 Two-digit numbers in the linear number system Reason about the location of any two-digit number in the linear number system, including identifying the previous and next multiple of 10. 2NPV–2 Teaching guidance Pupils need to be able to identify or place two-digit numbers on marked number lines. They should use efficient strategies and appropriate reasoning, including working backwards from a multiple of 10. Figure 43: identifying 36 and 79 on a marked 0 to 100 number line Language focus “a is 36 because it is one more than the midpoint of 35.” “b is 79 because it is one less than 80.” Pupils should also be able to estimate the value or position of two-digit numbers on unmarked numbers lines, using appropriate proportional reasoning, rather than counting on from a start point or back from an end point. For example, here pupils should reason: “60 is about here on the number line because it’s just over half way”. Figure 44: placing 60 on an unmarked 0 to 100 number line To prepare for future work on rounding, pupils should also learn to identify which 2 multiples of 10 a given two-digit number is between. 54 2NPV–2 Example assessment questions Look at lines A, B and C. Estimate how long they are by comparing them to the 100cm lines? The table shows the results of a survey which asked pupils to choose their favourite sport. Which sports were chosen by between 20 and 30 pupils? Favourite sport Number of pupils netball 24 basketball 19 tennis 12 football 32 hockey 6 swimming 28 gymnastics 15 Sophie thinks of a number. She says, “My number is between 40 and 50. It has 7 in the ones place.” What is Sophie’s number? Estimate the position of 60 on this number line: 55 The bar chart shows the number of pupils in each year-group in a school. How many pupils are in year 1? Fill in the missing numbers. 2NF–1 Fluently add and subtract within 10 Secure fluency in addition and subtraction facts within 10, through continued practice. 2NF–1 Teaching guidance In year 1, pupils should have learnt to add and subtract fluently within 10 (1NF–1). However, pupils may not still be fluent by the beginning of year 2, so this fluency should now be secured and maintained. Pupils should practise additive calculation within 10 until they have automatic recall of the additive facts. Fluency in these facts is required for pupils to succeed with addition and subtraction across 10 (2AS–1) and for additive calculation with larger numbers (2AS–3 and 2AS–4). The 66 addition facts within 10 are shown on the grid below. The number of addition facts to be learnt is reduced when commutativity is applied and pupils recognise that 3 + 2, for example, is the same as 2 + 3. Pupils must also have automatic recall of the corresponding subtraction facts, for example 5 – 3 and 5 – 2. 56 + 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Making connections Fluency in these addition and subtraction facts is required for addition and subtraction across 10 (2AS–1) and for additive calculation with larger numbers (2AS–3 and 2AS–4). 2NF–1 Assessment guidance Assessment guidance: For pupils to have met criterion 2NF–1, they need to be able to add and subtract within 10 without counting forwards or backwards in ones on their fingers, on a number line or in their heads. Pupils need to be able to automatically recall the facts. Teachers should assess pupils in small groups – simply providing the correct answers to calculations in a written test does not demonstrate that a pupil has met the criterion. 57 2AS–1 Add and subtract across 10 Add and subtract across 10, for example: 8 5 13 13 – 5 8 2AS–1 Teaching guidance Pupils need to have a strategy for confidently and fluently carrying out calculations such as: 7 5 12 15 – 9 6 For both addition and subtraction across 10, tens frames and partitioning diagrams can be used to support pupils as they learn about these strategies. First, pupils should learn to add three one-digit numbers by making 10, for example, 7 3 2 10 2 . They can then relate this to addition of two numbers across 10, by partitioning one of the addends, for example 7 5 7 3 2. Figure 45: tens frames with counters, and a partitioning diagram, showing 7 5 12 Pupils can subtract across 10 by using: the ‘subtracting through 10’ strategy (partitioning the subtrahend) – part of the subtrahend is subtracted to reach 10, then the rest of the subtrahend is subtracted from 10 58 or the ‘subtracting from 10’ strategy (partitioning the minuend) – the subtrahend is subtracted from 10, then the difference between the minuend and 10 is added Figure 46: using the ‘subtracting through 10’ strategy to calculate 15 minus 9 Figure 47: using the ‘subtracting from 10’ strategy to calculate 15 minus 9 You can find out more about fluency and recording for these calculations here in the calculation and fluency section: 2AS–1 Making connections This criterion depends on criterion 2NPV–1 where, as part of composing numbers according to place value, pupils learnt to: add a ten and some ones to make numbers between 11 and 19, for example: subtract the ones from a number between 11 and 19 to make 10, for example: 59 2AS–1 Example assessment questions Amisha spends £5 on a book and £8 on a T-shirt. How much does she spend altogether? I have a 15cm length of ribbon. I cut off 6cm. How much ribbon is left? I have 17 pencils. 9 have been sharpened. How many have not been sharpened? A garden fence was 8m long. Then the gardener added 7 more metres of fencing. How long is the garden fence now? Assessment guidance: For pupils to have met criterion 2AS–1, they need to be able to add and subtract across 10 without counting forwards or backwards in ones on their fingers, on a number line or in their heads. Teachers should assess pupils in small groups – simply providing the correct answers to the example questions above does not demonstrate that a pupil has met the criterion. The full set of addition and subtraction facts which children need to be fluent in is shown in the appendix. 2AS–2 Solve comparative addition and difference problems Recognise the subtraction structure of ‘difference’ and answer questions of the form, “How many more…?”. 2AS–2 Teaching guidance Pupils need to be able to solve problems with missing addends using known number facts or calculation strategies, for example: 19 25 Pupils need to be able to recognise problems about difference, and relate them to subtraction. For example, they should understand that they need to calculate 3 5 or 5 3 to solve the problem: There are 5 red cars and 3 blue cars. What is the difference between the number of red cars and blue cars? 60 Figure 48: finding the difference – bar model and number line Pupils should be able to recognise contextual problems involving finding a difference, phrased as ‘find the difference’, ‘how many more’ and ‘how many fewer’, such as those shown in the 2AS–2 below. Pupils may solve these problems by relating them to either a missing addend equation or to subtraction, applying known facts and strategies. 61 2AS–2 Example assessment questions The bar chart shows how many points some pupils scored in a quiz. a. How many more points did John score than Sara? b. How many fewer points did Harry score than Saskia? c. What is the difference between Saskia’s score and Paul’s score? I have £19 and want to buy a game which costs £25. How much more money do I need? Felicity has 34 marbles and Dan has 30 marbles. What is the difference between the number of marbles they have? It takes me 20 minutes to walk to school. So far I have been walking for 12 minutes. How much longer do I have to walk for? Liam is 90cm tall. Karim is 80cm tall. How much taller is Liam than Karim? 62 2AS–3 Add and subtract within 100 – part 1 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract only ones or only tens to/from a two-digit number. 2AS–3 Teaching guidance Pupils should be able to apply known one-digit additive facts to: adding and subtracting 2 multiples of ten (for example, 40 30 70 and 70 – 30 40) adding and subtracting ones to/from a two-digit number (for example, 64 3 67 and 67 – 3 64) adding and subtracting multiples of ten to/from a two-digit number (for example, 45 30 75 and 75 – 30 45) the special case of subtracting ones from a multiple of ten, by using complements of 10 (for example 30 – 3 27) Tens frames, Dienes and partitioning diagrams can be used to support pupils as they learn how to relate these calculations to one-digit calculations. Figure 49: Dienes and equations to support adding a multiple of 10 to a two-digit number 63 Figure 50: tens frames with counters, and number lines, to support subtracting ones from a multiple of 10 Throughout, pupils should use spoken language to demonstrate their reasoning. Language focus “4 plus 3 is equal to 7. So 4 tens and plus 3 tens is equal to 7 tens.” “10 minus 3 is equal to 7. So 30 minus 3 is equal to 27.” Pupils should also be able to apply strategies for addition or subtraction across 10, from above, to addition or subtraction bridging a multiple of 10. You can find out more about fluency and recording for all of these calculation types here in the calculation and fluency section: 2AS–3 Making connections Learning to subtract ones from a multiple of 10 should be connected to pupils’ understanding of the location of two-digit numbers in the linear number system, for example pupils understand that 27 is 3 ‘before’ 30, so . 64 2AS–3 Example assessment questions A bouncy ball costs 60p. Circle the coins which you could use to pay for it. Is there more than one answer? Sophie’s book has 50 pages. So far she has read 9 pages. How many more pages does Sophie have left to read? 65 What is the total cost of: a. the bedtime stories book and the train set? b. the doll’s house and the plane? c. the scooter and the teddy? d. the boat, the train set and the drum? Oak class raise £68 for their class fund. They spend £40 on new paints. How much money do they have left? Assessment guidance: Simply providing the correct answers to the example questions above does not demonstrate that pupils have met criterion 2AS–3. For pupils to meet the criterion, they must be able to use known addition and subtraction facts within 10 to solve the calculations efficiently. Before using the questions above, assess pupils in small groups by watching how they solve calculations such as: 40 30 90 – 40 31 5 48 – 6 20 59 72– 30 46 4 80 3 Language focus “I know that 2 plus 5 is equal to 7, so 20 plus 50 is equal to 70. There are 9 ones as well, so 20 plus 59 is equal to 79.” Pupils who are using extensive written notes (for example, sketching ‘jumping back’ in tens on a number line as a way to solve 90 – 40) have not yet met criterion 2AS–3. 66 2AS–4 Add and subtract within 100 – part 2 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract any 2 two-digit numbers. 2AS–4 Teaching guidance As for 2AS–3, Dienes and partitioning diagrams can be used to support pupils as they learn about strategies for carrying out these calculations. To add 2 two-digit numbers, pupils need to combine one-digit addition facts with their understanding of two-digit place value. Pupils should first learn to add 2 multiples of ten and 2 ones before moving on to the addition of 2 two-digit numbers, for example: 40 20 5 3 60 8 68 40 5 20 3 60 8 68 45 23 60 8 68 Figure 51: Dienes and an equation to support adding 2 two-digit numbers Language focus “First I partition both numbers. Then I add the tens. Then I add the ones. Then I combine all of the tens and all of the ones.” Pupils can then learn to be more efficient, by partitioning just one addend, for example: 45 23 45 20 3 65 3 When pupils learn to subtract one two-digit number from another, the progression is similar to that for addition. Pupils can first learn to subtract a multiple of ten and some ones from a two-digit number, and then connect this to the subtraction of one two-digit number from another, for example: 67 45 – 20 – 3 25 – 3 22 45 – 23 45 – 20 – 3 25 – 3 22 or 45 – 3 – 20 42 – 20 22 45 – 23 45 – 3 – 20 42 – 20 22 There is an important difference compared to the addition strategy: pupils should not partition both two-digit numbers for subtraction as this can lead to errors, or calculations involving negative numbers, when bridging a multiple of 10, for example: 63 –17 60 –10 7 – 3 63 –17 60 –10 3 – 7 You can find out more about fluency and recording for addition and subtraction of two-digit numbers here in the calculation and fluency section: 2AS–4 Making connections Pupils should also be able to apply strategies for addition or subtraction across 10 (2AS–1) to calculations such as and . 68 2AS–4 Example assessment questions a. Daisy spends £32 in the shop. Circle the 2 items she buys. b. What is the total cost of the bicycle and construction set? c. Jalal pays for the bicycle using a £50 note. How much change does he get? d. Yu Yan wants to buy the construction set. She has saved £15. How much more money does Yu Yan need to save? 69 2MD–1 Multiplication as repeated addition Recognise repeated addition contexts, representing them with multiplication equations and calculating the product, within the 2, 5 and 10 multiplication tables. 2MD–1 Teaching guidance Pupils must first be able to recognise equal groups. To better understand and identify equal groups, pupils should initially explore both equal and unequal groups. Pupils should then learn to describe equal groups with words. Figure 52: recognising equal groups – 3 groups of 5 eggs Language focus “There are 3 equal groups of eggs.” “There are 5 eggs in each group.” “There are 3 groups of 5.” Based on their existing additive knowledge, pupils should be able to represent equal-group contexts with repeated addition expressions, for example 5 5 5. They should then learn to write multiplication expressions to represent the same contexts, for example 3 5 . Pupils must be able to explain how each term in a multiplication expression links to the context it represents. Pupils must also be able to understand equivalence between a repeated addition expression and a multiplication expression: 5 5 5 3 5. Pupils should then learn to calculate the total number of items (the product), for contexts based on the 2, 5 and 10 multiplication tables, initially by skip counting. They should be able to write complete multiplication equations, for example , and explain how each term links to the context. 3 5 15 70 Language focus “The 3 represents the number of groups.” “The 5 represents the number of eggs in each group.” “The 15 represents the total number of eggs.” Pupils should be able to relate multiplication to situations where the total number of items cannot be seen, for example by representing 3 5 with three 5-value counters. Figure 53: three 5-value counters You can find out more about fluency and recording for the 2, 5 and 10 multiplication tables here in the calculation and fluency section: 2MD–1. Making connections Pupils must be able to write and solve addition problems with 3 or more addends before they can connect repeated addition to multiplication. 2MD–1 Example assessment questions Write these addition expressions as multiplication expressions. The first one has been completed for you. 5 5 5 5 5 5 5 2 2 2 2 2 __ 2 2 2 __ 10 10 10 __ There are 7 year-groups in Winterdale School. Each year-group has 2 classes. How many classes are in the school? Sally buys 3 cinema tickets. Each ticket costs £5. How much does Sally spend? Write the multiplication expression and calculate the cost. 71 There are 10 children sitting at each table in a dining hall. There are 8 tables. How many children are there? The pictogram shows how many socks each child has. How many socks does Asif have? Write a story to go with this equation. 6 10 60 Complete the calculations. 7 5 10 4 9 2 72 2MD–2 Grouping problems: missing factors and division Relate grouping problems where the number of groups is unknown to multiplication equations with a missing factor, and to division equations (quotitive division). 2MD–2 Teaching guidance Pupils need to be able to represent problems where the total quantity and group size is known, using multiplication equations with missing factors. For example, “There are 15 biscuits. If I put them into bags of 5, how many bags will I need?’’ can be represented by the following equation: 5 15 Pupils can use skip counting or their emerging 2, 5 and 10 multiplication table fluency to calculate the missing factor. Figure 54: 3 bags of 5 biscuits alongside three 5-value counters Pupils should then learn that unknown-factor problems can also be represented with division equations (quotitive division), for example, 15 5 .They should be able to use skip counting or their multiplication-table fluency to find the quotient: 15 5 3. Pupils should be able to describe how each term in the division equation links to the context and describe the division equation in terms of ‘division into groups’. Language focus “The 15 represents the total number of biscuits.” “The 5 represents the number of biscuits in each bag.” “The 3 represents the number of bags.” “15 divided into groups of 5 is equal to 3.” 73 Pupils also need to be able to solve division calculations that are not set in contexts. They should recognise that they need to skip count in the divisor (2, 5 or 10), or use the associated multiplication fact, to find the quotient. For example, to calculate 60 10, they can skip count in tens (counting the required number of tens) or apply the fact that 6 10 60 . You can find out more about fluency and recording for division by 2, 5 or 10 here in the calculation and fluency section: 2MD–2 2MD–2 Example assessment questions Miss Robinson asked Harry to get 60 apples from the kitchen. The apples come in bags of 10. How many bags does Harry need to get? Diego has some 5p coins. He has 40p altogether. How many 5p coins does Diego have? The pictogram shows how many socks each child has. Lena has 8 socks. How would this be represented on the pictogram? Draw it. There are 5 balloons in a pack. I need 15 balloons for my party. How many bags should I buy? Fill in the missing numbers. 5 30 50 10 2 14 74 2G–1 Describe and compare 2D and 3D shapes Use precise language to describe the properties of 2D and 3D shapes, and compare shapes by reasoning about similarities and differences in properties. 2G–1 Teaching guidance Building on 1G–1, pupils should continue to explore and discuss common 2D and 3D shapes, now extending to include quadrilaterals and other polygons, and cuboids, prisms and cones. Pupils must now learn to use precise language to describe 2D shapes, including the terms ’sides’ and ‘vertex’/’vertices’. They should learn to identify the sides of a given 2D shape and to identify a vertex as a point where two sides meet. Pupils must learn that a polygon is a 2D shape which has only straight sides and then learn to identify a given polygon by counting the number sides (or vertices). Pupils should practise running their finger along each side as they count the sides (or practise touching each vertex as they count the vertices). Later, pupils may mark off the sides or vertices on an image as they count. It is important that they learn to count the sides/vertices accurately, counting each once and only once. Pupils must know that it is the number of sides/vertices that determines the type of polygon, rather than whether the given shape looks like their mental image of a particular polygon. For example, although pupils may informally describe the shape below as “like a square with 2 corners cut off”, they should be able to recognise and explain that, because it has 6 straight sides, it is a hexagon. Figure 55: an irregular hexagon Language focus “This shape is a hexagon because it has exactly 6 straight sides.” When discussing 3D shapes, pupils should be able to correctly use the terms ‘edges’, ‘vertex’/‘vertices’ and ‘faces’. Pupils need to be able to accurately count the number of edges, vertices and faces for simple 3D shapes, such as a triangular-based pyramid or a cuboid, using sticky paper (if necessary) to keep track of the edges/vertices/faces as they count. Pupils should be able to identify the 2D shapes that make up the faces of 3D 75 shapes, including identifying pyramids according to the shape of their base (‘square-based’ and ‘triangle-based’). Pupils should gain experience describing and comparing standard and non-standard exemplars of polygons. They should explore shapes (for example, shapes cut from card) rather than only looking at pictures. Examples of irregular polygons should not be restricted to those where every side-length is different and every internal angle is a different size. Examples should include polygons in which: some side-lengths are equal some internal angles are equal there are a variety of sizes of internal angles (acute, right angled, obtuse and/or reflex) there are pairs of parallel and perpendicular sides Figure 56: a variety of different pentagons (regular and irregular) Language focus “These shapes are all pentagons because they all have exactly 5 straight sides.” Pupils don’t need to be able to identify or name angle types or parallel/perpendicular sides in year 2, but it is important that they gain visual experience of them as preparation for identifying and naming them in key stage 2. In a similar way, pupils should gain experience describing and comparing a wide variety of 3D shapes including cuboids (and cubes), prisms, cones, pyramids, spheres and cylinders. Pupils should explore shapes practically as well as looking at pictures. Pupils should also explore and discuss regular polyhedrons such as octahedrons and dodecahedrons, although year 2 pupils do not need to remember the names of these shapes. As well as discussing sides and vertices (as a precursor to evaluating perimeter and angle), pupils should begin to use informal language to discuss and compare the space inside 2D shapes (as a precursor to evaluating area). Pupils should be able to reason about the shape and size of the space inside a 2D shape, relative to other 2D shapes, using language such as ‘long and thin’, ‘short and wide’, ‘larger and ‘smaller’. 76 2G–1 Example assessment questions How many sides does this shape have? What is the name of this shape? Sketch a hexagon. Try to think of a hexagon that will look different to those drawn by other pupils. Task: Lay out a selection of 3D shapes, then instruct pupils to find a shape that has: a. fewer than 5 edges b. more than 5 faces c. exactly 1 vertex d. all faces the same shape e. no flat faces f. no straight edges g. both a square face and a triangular face a. Circle all of the octagons. b. Explain why the shapes you have not circled are not octagons. Task: Present pupils with a cylinder and a cone (the 3D shapes rather than pictures), then instruct pupils to: a. describe something that is the same about the 2 shapes b. describe something that is different about the 2 shapes Task: Lay out a selection of 3D shapes, then ask pupils to identify all of the shapes that have a square face. 77 Here are 4 rectangles. Which do you think is the largest rectangle? Which do you think is the smallest rectangle? If each rectangle was a slice of your favourite food, which one would you choose to have? Which of the shapes B to H are exactly the same shape as shape A, but just a different size? Assessment guidance: Practical work, carried out in small groups, should form part of the assessment of this criterion. For question 3, pupils must be able to name common 3D shapes, but do not need to be able to name every shape they might select to fulfil the required criteria. For questions 7 and 8, assessment should be discussion based – pupils need not necessarily provide a ‘correct’ answer, but should demonstrate an emerging sense of the size and shape of the space within 2D shapes. For question 8, for example, some pupils may say that shape F is the same shape as A but just stretched, and some pupils may say that F is a different shape to A because F is “long and thin” and A is “shorter and fatter”: both of these answers would show that pupils are developing awareness of the overall shape rather than just attending to the number of sides and vertices. 78 Calculation and fluency 2AS–1 Add and subtract across 10 Add and subtract across 10, for example: 8 5 13 13 – 5 8 At first, pupils will use manipulatives, such as tens frames, to understand the strategies for adding and subtracting across 10. However, they should not be using the manipulatives as a tool for finding answers. Pupils should be able to carry out these calculations mentally, using their fluency in complements to 10 and partitioning. Pupils are fluent in these calculations when they no longer rely on extensive written methods, such as equation sequences or partitioning diagrams. Pupils do not need to memorise all additive facts for adding and subtracting across 10, but they need to be able to recall appropriate doubles (double 6, 7, 8 and 9) and corresponding halves (half of 12, 14, 16 and 18), and use these known facts for calculations such as 6 6 12 and 18 – 9 9 . Year 2 pupils will need lots of practice to be able to add and subtract across 10 with sufficient fluency to make progress with the year 3 curriculum. They should also continue to practise adding and subtracting within 10. 2AS–3 Add and subtract within 100 – part 1 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract only ones or only tens to/from a two-digit number For pupils to become fluent with the strategies for these two-digit additive calculations, as well as having automatic recall of one-digit additive facts, they must also be conceptually fluent with the connections between one-digit facts and two-digit calculations. This conceptual fluency is based on: being able to unitise (for example, understanding 40 50 as 4 units of ten + 5 units of ten) an understanding of place-value Pupils should be able to solve these calculations mentally and be able to demonstrate their reasoning either verbally or with manipulatives or drawings. Note that this is different from using manipulatives or drawings to calculate an answer, which pupils should not need to do. 79 2AS–4 Add and subtract within 100 – part 2 Add and subtract within 100 by applying related one-digit addition and subtraction facts: add and subtract any 2 two-digit numbers These calculations involve more steps than those in 2AS–3. To avoid overload of working memory, pupils should learn how to record the steps using informal written notation or equation sequences, as shown below. This is particularly important for calculations where addition of the ones involves bridging a multiple of 10, as these require a further calculation step. Figure 57: adding 26 and 37 by partitioning both addends Figure 58: adding 26 and 37 by partitioning one addend Figure 59: subtracting 17 from 63 by subtracting the tens first Figure 60: subtracting 17 from 63 by subtracting the ones first Pupils do not need to learn formal written methods for addition and subtraction in year 2, but column addition and column subtraction could be used as an alternative way to record two-digit calculations at this stage. For calculations such as 26 37, pupils can begin to think about the 2 quantities arranged in columns under place-value headings of tens and ones. They can use counters or draw dots for support: Figure 61: adding 2 two-digit numbers using 10s and 1s columns 80 2MD–1 Multiplication as repeated addition Recognise repeated addition contexts, representing them with multiplication equations and calculating the product, within the 2, 5 and 10 multiplication tables. Pupils must be able to carry out calculations connected to the 2, 5 and 10 multiplication tables, for example: 4 5 Pupils should practise skip counting in multiples of 2, 5 and 10, up to 10 groups of each, until they are fluent. When carrying out a multiplication calculation by skip counting, they may keep track of the number of twos, fives or tens using their fingers or by tallying. Pupils may also recite, using the language of the multiplication tables to keep track (1 times 5 is 5, 2 times 5 is 10…). They can also use or draw 2-, 5- or 10-value counters to support them in solving multiplicative problems. Pupils who are sufficiently fluent in year 2 multiplicative calculations are not reliant on drawing arrays or using number lines as tools to calculate. Pupils should have sufficient conceptual understanding to recognise these as models of multiplication and division, and explain how they link to calculation statements. However they should not need to use them as methods for carrying out calculations. Pupils need to be able to represent 4 fives (or 5, 4 times) as both 4 5 and 5 4 . They should be able to use commutativity to solve, for example, 2 sevens, using their knowledge of 7 twos. 2MD–2 Grouping problems: missing factors and division Relate grouping problems where the number of groups is unknown to multiplication equations with a missing factor, and to division equations (quotitive division). Pupils need to be able to solve missing-factor and division problems connected to the 2, 5 and 10 multiplication tables, for example: 5 20 20 5 Pupils should solve division (and missing-factor) problems, such as these, by connecting division to their emerging fluency in skip counting and known multiplication facts. Pupils should not be solving statements such as 20 5 by sharing 20 between 5 using manipulatives or by drawing dots. Pupils should also not rely on drawing arrays or number lines as tools for calculation. 81 As for 2MD–1, pupils can keep track of the number of twos, fives or tens using their fingers or by tallying. They may also recite, using the language of the multiplication tables, or draw 2-, 5- or 10-value counters. Eventually pupils should be fluent in isolated multiplication facts (for example, 4 fives are 20) and use these to solve missing-factor multiplication problems and division problems. 82 Year 3 guidance Ready-to-progress criteria Year 2 conceptual prerequisite Year 3 ready-to-progress criteria Future applications Know that 10 ones are equivalent to 1 ten, and that 40 (for example) can be composed from 40 ones or 4 tens. Know how many tens there are in multiples of 10 up to 100. 3NPV–1 Know that 10 tens are equivalent to 1 hundred, and that 100 is 10 times the size of 10; apply this to identify and work out how many 10s there are in other three-digit multiples of 10. Solve multiplication problems that that involve a scaling structure, such as ‘ten times as long’. Recognise the place value of each digit in two-digit numbers, and compose and decompose two-digit numbers using standard and non-standard partitioning. 3NPV–2 Recognise the place value of each digit in three-digit numbers, and compose and decompose three-digit numbers using standard and non-standard partitioning. Compare and order numbers. Add and subtract using mental and formal written methods. Reason about the location of any two-digit number in the linear number system, including identifying the previous and next multiple of 10. 3NPV–3 Reason about the location of any three-digit number in the linear number system, including identifying the previous and next multiple of 100 and 10. Compare and order numbers. Estimate and approximate to the nearest multiple of 1,000, 100 or 10. Count in multiples of 2, 5 and 10. 3NPV–4 Divide 100 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 100 with 2, 4, 5 and 10 equal parts. Read scales on graphs and measuring instruments. 83 Year 2 conceptual prerequisite Year 3 ready-to-progress criteria Future applications Add and subtract across 10, for example: 3NF–1 Secure fluency in addition and subtraction facts that bridge 10, through continued practice. Add and subtract mentally where digits sum to more than 10, for example: Add and subtract across other powers of 10, without written methods, for example: Add within a column during columnar addition when the column sums to more than 10 (regrouping), for example, for: Subtract within a column during columnar subtraction when the minuend of the column is smaller than the subtrahend (exchanging), for example, for: Calculate products within the 2, 5 and 10 multiplication tables. 3NF–2 Recall multiplication facts, and corresponding division facts, in the 10, 5, 2, 4 and 8 multiplication tables, and recognise products in these multiplication tables as multiples of the corresponding number. Use multiplication facts during application of formal written layout. Use division facts during short division and long division. Automatically recall addition and subtraction facts within 10, and across 10. Unitise in tens: understand that 10 can be thought of as a single unit of 1 ten. 3NF–3 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10), for example: Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 100), for example: and so and so 84 Year 2 conceptual prerequisite Year 3 ready-to-progress criteria Future applications Automatically recall number bonds to 9 and to 10. Know that 10 ones are equivalent to 1 ten, and 10 tens are equivalent to 1 hundred. 3AS–1 Calculate complements to 100, for example: Calculate complements to other numbers, particularly powers of 10. Calculate how much change is due when paying for an item. Automatically recall addition and subtraction facts within 10 and across 10. Recognise the place value of each digit in two- and three-digit numbers. Know that 10 ones are equivalent to 1 ten, and 10 tens are equivalent to 1 hundred. 3AS–2 Add and subtract up to three-digit numbers using columnar methods. Add and subtract other numbers, including four-digits and above, and decimals, using columnar methods. Have experience with the commutative property of addition, for example, have recognised that and have the same sum. Be able to write an equation in different ways, for example, and Write equations to represent addition and subtraction contexts. 3AS–3 Manipulate the additive relationship: Understand the inverse relationship between addition and subtraction, and how both relate to the part–part–whole structure. Understand and use the commutative property of addition, and understand the related property for subtraction. All future additive reasoning. Recognise repeated addition contexts and represent them with multiplication equations. Relate grouping problems where the number of groups is unknown to multiplication equations with a missing factor, and to division equations (quotitive division). 3MD–1 Apply known multiplication and division facts to solve contextual problems with different structures, including quotitive and partitive division. 85 Year 2 conceptual prerequisite Year 3 ready-to-progress criteria Future applications 3F–1 Interpret and write proper fractions to represent 1 or several parts of a whole that is divided into equal parts. Use unit fractions as the basis to understand non-unit fractions, improper fractions and mixed numbers, for example: is 2 one-fifths is 6 one-fifths, so 3F–2 Find unit fractions of quantities using known division facts (multiplication tables fluency). Apply knowledge of unit fractions to non-unit fractions. Reason about the location of whole numbers in the linear number system. 3F–3 Reason about the location of any fraction within 1 in the linear number system. Compare and order fractions. Automatically recall addition and subtraction facts within 10. Unitise in tens: understand that 10 can be thought of as a single unit of 1 ten, and that these units can be added and subtracted. 3F–4 Add and subtract fractions with the same denominator, within 1. Add and subtract improper and mixed fractions with the same denominator, including bridging whole numbers. Recognise standard and non-standard examples of 2D shapes presented in different orientations. Identify similar shapes. 3G–1 Recognise right angles as a property of shape or a description of a turn, and identify right angles in 2D shapes presented in different orientations. Compare angles. Estimate and measure angles in degrees. Compose 2D shapes from smaller shapes to match an exemplar, rotating and turning over shapes to place them in specific orientations. 3G–2 Draw polygons by joining marked points, and identify parallel and perpendicular sides. Find the area or volume of a compound shape by decomposing into constituent shapes. Find the perimeter of regular and irregular polygons. 86 3NPV–1 Equivalence of 10 hundreds and 1 thousand Know that 10 tens are equivalent to 1 hundred, and that 100 is 10 times the size of 10; apply this to identify and work out how many 10s there are in other three-digit multiples of 10. 3NPV–1 Teaching guidance Pupils need to experience: what 100 items looks like making a unit of 1 hundred out of 10 units of 10, for example using 10 bundles of 10 straws to make 100, or using ten 10-value place-value counters Figure 62: ten 10-value place-value counters in a tens frame Language focus “10 tens is equal to 1 hundred.” Pupils must then be able to work out how many tens there are in other three-digit multiples of 10. Figure 63: eighteen 10-value place-value counters in 2 tens frames Language focus “18 tens is equal to 10 tens and 8 more tens.” “10 tens is equal to 100.” “So 18 tens is equal to 100 and 8 more tens, which is 180.” 87 The reasoning here can be described as grouping or repeated addition – pupils group or add 10 tens to make 100, then add another group of 8 tens. Pupils need to be able to apply this reasoning to measures contexts, as shown in the 3NPV–1 below. It is important for pupils to understand that there are tens within this new unit of 100, in different contexts. Pupils should be able to explain that numbers such as 180 and 300 are multiples of 10, because they are each equal to a whole number of tens. They should be able to identify multiples of 10 based on the fact that they have a zero in the ones place. As well as understanding 100 and other three-digit multiples of 10 in terms of grouping and repeated addition, pupils should begin to describe 100 as 10 times the size of 10. Figure 64: place-value chart illustrating the scaling relationship between ones, tens and hundreds Language focus “100 is 10 times the size of 10.” Making connections Learning to identify the number of tens in three-digit multiples of 10 should be connected to pupils understanding of multiplication and the grouping structure of division (2MD–1). Pupils should, for example, be able to represent 180 as 18 tens using the multiplication equations or , and be able to write the corresponding division equations . In 3MD–1 they will learn that can represent the structure of 180 divided into groups of 10 (quotitive division), as here, and that it can also represent 180 shared into 10 equal shares of 18 each (partitive division). 88 3NPV–1 Example assessment questions How many 10cm lengths can a 310cm length of ribbon be cut into? The school office sells 52 poppies for 10p each. How much money have they collected altogether? I take 10ml of medicine every day. How many days will a 250ml bottle last? Marek is 2 years old, and has a mass of 10kg. His father’s mass is 10 times as much. What is the mass of Marek’s father? Janey saves up £100. This is 10 times as much money as her brother has. How much money does her brother have? Circle the numbers that are multiples of 10. Explain your answer. 640 300 105 510 330 409 100 864 3NPV–2 Place value in three-digit numbers Recognise the place value of each digit in three-digit numbers, and compose and decompose three-digit numbers using standard and non-standard partitioning. 3NPV–2 Teaching guidance Pupils should be able to identify the place value of each digit in a three-digit number. They must be able to combine units of ones, tens and hundreds to compose three-digit numbers, and partition three-digit numbers into these units. Pupils need to experience variation in the order of presentation of the units, so that they understand that 40 300 2 is equal to 342, not 432. Figure 65: two representations of the place-value composition of 342 Pupils also need to solve problems relating to subtraction of any single place-value part from the whole number, for example: 342 – 300 342 – 302 89 As well as being able to partition numbers in the ‘standard’ way (into individual place-value units), pupils must also be able to partition numbers in ‘non-standard’ ways, and carry out related addition and subtraction calculations, for example: Figure 66: partitioning 830 into 430 and 400 Figure 67: partitioning 654 into 620 and 34 You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 3NPV–2 and 3NF–3 3NPV–2 Example assessment questions What number is represented by these counters? What number is represented by this expression? 1 10 10 100 100 10 10 Fill in the missing numbers to complete these partitioning diagrams. 90 Fill in the missing numbers. 600 70 1 3 500 40 461 60 1 20 3 823 953 50 3 846 40 800 203 90 290 3 628 20 628 8 Fill in the missing symbols (<, > or =). 100 60 5 105 60 300 40 2 300 24 783 80 783 3 839 9 30 839 39 There are 365 days in a year. If it rains on 65 days of the year, on how many days does it not rain? A bamboo plant was 4m tall. Then it grew by another 83cm. How tall is the bamboo plant now? Express your answer in centimetres. In the school library there are 25 books on the trolley and 250 books on the shelves. How many books are there altogether? Francesco had 165 marbles. Then he gave 45 marbles to his friend. How many marbles does Francesco have now? The tree outside Cecily’s house in 308cm tall. How much further would it have to grow to reach the bottom of Cecily’s bedroom window, at 3m 68cm? 91 3NPV–3 Three-digit numbers in the linear number system Reason about the location of any three-digit number in the linear number system, including identifying the previous and next multiple of 100 and 10. 3NPV–3 Teaching guidance Pupils need to be able to identify or place three-digit numbers on marked number lines with a variety of scales. Pupils should also be able to estimate the value or position of three-digit numbers on unmarked numbers lines, using appropriate proportional reasoning. Pupils should apply this skill to taking approximate readings of scales in measures and statistics contexts, as shown in the 3NPV–3 below. For more detail on identifying, placing and estimating positions of numbers on number lines, see year 2, 2NPV–2. Pupils must also be able to identify which pair of multiples of 100 or 10 a given three-digit number is between. To begin with, pupils can use a number line for support. In this example, for the number 681, pupils must identify the previous and next multiples of 100 and 10. Figure 68: using a number line to identify the next and previous multiple of 100 Figure 69: using a number line to identify the next and previous multiple of 10 Language focus “The previous multiple of 100 is 600. The next multiple of 100 is 700.” “The previous multiple of 10 is 680. The next multiple of 10 is 690.” Pupils need to be able to identify previous and next multiples of 100 or 10 without the support of a number line. 92 Figure 70: identifying the nearest multiple of 100 Finally, pupils should also be able to count forwards and backwards from any three-digit number in steps of 1 or 10. Pay particular attention to counting over ‘boundaries’, for example: 210, 200, 190 385, 395, 405 Making connections Here, pupils must apply their knowledge that 10 tens is equal to 1 hundred (see 3NPV–1) to understand that each interval of 100 on a number line or scale is made up of 10 intervals of 10. This also links to 3NPV–4 , in which pupils need to be able to read scales divided into 2, 4, 5 and 10 equal parts. 3NPV–3 Example assessment questions Fill in the missing numbers. 900 700 600 400 200 370 390 420 440 Estimate to fill in the missing numbers. Estimate and mark the position of these numbers on the number line. 600 200 480 840 762 195 93 Fill in the missing numbers. 100 100 less more 800 10 10 less more 390 100 100 less more 100 10 10 less more 800 previous next multiple multiple of 100 of 100 630 previous next multiple multiple of 100 of 100 347 previous next multiple multiple of 10 of 10 492 previous next multiple multiple of 10 of 10 347 94 Look at lines A, B and C. Can you estimate how long they are by comparing them to the 1,000cm lines? Estimate the mass, in grams, shown on this weighing scale. 95 3NPV–4 Reading scales with 2, 4, 5 or 10 intervals Divide 100 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 100 with 2, 4, 5 and 10 equal parts. 3NPV–4 Teaching guidance By the end of year 3, pupils must be able to divide 100 into 2, 4, 5 or 10 equal parts. This is important because these are the intervals commonly found on measuring instruments and graph scales. Figure 71: Bar models showing 100 partitioned into 2, 4, 5 and 10 equal parts Pupils should practise counting in multiples of 10, 20, 25, and 50 from 0, or from any multiple of these numbers, both forwards and backwards. This is an important step in becoming fluent with these number patterns. Pupils will have been practising counting in multiples of 1, 2 and 5 since year 1, and this supports counting in units of 10, 20 and 50. However, counting in units of 25 is not based on any multiples with which pupils are already familiar, so they typically find this the most challenging. Language focus “Twenty-five, fifty, seventy-five, one hundred” needs to be a fluent spoken language pattern, which pupils can continue over 100. 96 Pupils should be able to apply this skip counting beyond 100, to solve contextual multiplication and division measures problems, as shown in the 3NPV–4 below (questions 5 and 7). Pupils should also be able to write and solve multiplication and division equations related to multiples of 10, 20, 25 and 50 up to 100. Pupils need to be able to solve addition and subtraction problems based on partitioning 100 into multiples of 10, 20 and 50 based on known number bonds to 10. Pupils should also have automatic recall of the fact that 25 and 75 are bonds to 100. They should be able to automatically answer a question such as “I have 1m of ribbon and cut off 25cm. How much is left in centimetres?” Making connections Dividing 100 into 10 equal parts is also assessed as part of 3NPV–1. Reading scales builds on number line knowledge from 3NPV–3 . Conversely, experience of working with scales with 2, 4, 5 or 10 divisions in this criterion improves pupils’ estimating skills when working with unmarked number lines and scales as described in 3NPV–3. 3NPV–4 Example assessment questions Fill in the missing numbers. 0 25 100 0 10 20 40 60 80 90 100 60 40 0 97 What were Jenny and Asif’s scores? Miss Scot weighs herself. How much does she weigh, in kilograms? How many centimetres long is the ribbon? How many 25p cupcakes can I buy for £5? How many 50cm lengths of wood can I cut from a 3m plank? We raise £100 at the school fair and divide the money equally between 5 charities. How much does each charity get? Fill in the missing numbers. 100 4 20 100 100 50 25 100 Stan counts from 0 in multiples of 25. Circle the numbers he will say. 100 25 240 155 400 275 505 350 98 3NF–1 Fluently add and subtract within and across 10 Secure fluency in addition and subtraction facts that bridge 10, through continued practice. 3NF–1 Teaching guidance Before pupils begin work on columnar addition and subtraction (3AS–1), it is essential that pupils have automatic recall of addition and subtraction facts within and across 10. These facts are required for calculation within the columns in columnar addition and subtraction. All mental calculation also depend on these facts. Identifying core number facts: columnar addition Identifying core number facts: columnar subtraction Figure 72: columnar addition of 465 and 429 Figure 73: columnar subtraction of 286 from 749 Within-column calculations: Within-column calculations: Pupils should already have automatic recall of addition and subtraction facts within 10, from year 1 (1NF–1). In year 2 (2AS–1), pupils learnt strategies for addition and subtraction across 10. However, year 3 pupils are likely to need further practice, and reminders of the strategies, to develop sufficient fluency. Pupils should practise until they achieve automaticity in the mental application of these strategies. Without this practice many pupils are likely to still be reliant on counting on their fingers to solve within-column calculations in columnar addition and subtraction. The full set of addition calculations that pupils need for columnar addition are shown on the next page. The number of facts to be learnt is reduced when commutativity is applied and pupils recognise that 7 5 , for example, is the same as 5 7 . Automaticity in subtraction facts should also be developed through the application of the relationship between addition and subtraction, for example, pupils should recognise that if 7 5 12 then12– 5 7 . 99 + 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Pupils need extensive practice to meet this criterion. You can find out more about fluency for addition and subtraction within and across 10 here in the calculation and fluency section: 3NF–1 Making connections Fluency in these addition and subtraction facts is required for within-column calculation in columnar addition and subtraction ( 3AS–2). 3NF–1 Example assessment questions Mr Kahn drove 8km to get to his friend’s house, and then drove another 3km with his friend to get to the gym. How far did Mr Kahn drive? There are 12 children. 5 of them can ride a bicycle and the rest cannot. How many of the children cannot ride a bicycle? Maja had £17. Then she spent £9. How much money does she have left? 100 I have 6 metres of red ribbon and 6 metres of blue ribbon. How many metres of ribbon do I have altogether? Hazeem is growing a sunflower and a bean plant. So far, his sunflower plant is 14cm tall and his bean plant is 8cm tall. How much taller is the sunflower plant than the bean plant? Assessment guidance: For pupils to have met criterion 3NF–1, they need to be able to add and subtract within and across 10 without counting forwards or backwards in ones on their fingers, on a number line or in their heads. Pupils need to be able to automatically recall the facts within 10, and be able to mentally apply strategies for calculation across 10, with accuracy and speed. Teachers should assess pupils in small groups – simply providing the correct answers to the example questions above does not demonstrate that a pupil has met the criterion. The full set of addition and subtraction facts which children need to be fluent in is shown in the appendix. 3NF–2 Recall of multiplication tables Recall multiplication facts, and corresponding division facts, in the 10, 5, 2, 4 and 8 multiplication tables, and recognise products in these multiplication tables as multiples of the corresponding number. 3NF–2 Teaching guidance The national curriculum requires pupils to recall multiplication table facts up to 12 12 , and this is assessed in the year 4 multiplication tables check. In year 3, the focus should be on learning facts in the 10, 5, 2, 4 and 8 multiplication tables. Language focus When pupils commit multiplication table facts to memory, they do so using a verbal sound pattern to associate the 3 relevant numbers, for example, “six fours are twenty-four”. It is important to provide opportunities for pupils to verbalise each multiplication fact as part of the process of developing fluency. While pupils are learning the individual multiplication tables, they should also learn that: 101 the factors can be written in either order and the product remains the same (for example, we can write 3 4 12 or 4 3 12 to represent the third fact in the 4 multiplication table) the products within each multiplication table are multiples of the corresponding number, and be able to recognise multiples (for example, pupils should recognise that 64 is a multiple of 8, but that 68 is not) adjacent multiples in, for example, the 8 multiplication table, have a difference of 8 Figure 74: number line and array showing that adjacent multiples of 8 (32 and 40) have a difference of 8 As pupils develop automatic recall of multiplication facts, they should learn how to use these to derive division facts. They must be able to use these ‘known division facts’ to solve division calculations, instead of using the skip counting method they learnt in year 2 (2MD–2). Pupils should learn to make the connection between multiplication and division facts as they learn each multiplication table, rather than afterwards. Language focus “4 times 5 is 20, so 20 divided by 5 is 4.” It is useful for pupils to learn the 10 and 5 multiplication tables one after the other, and then the 2, 4 and 8 multiplication tables one after the other. The connections and patterns will help pupils to develop fluency and understanding. You can find out more about developing automatic recall of multiplication tables here in the calculation and fluency section: 3NF–2 102 Making connections Alongside developing fluency in multiplication tables facts and corresponding division facts, pupils must learn to apply them to solve contextual problems with different multiplication and division structures (3MD–1). In 3F–2 pupils use known division facts to find unit fractions of quantities, for example , so . 3NF–2 Example assessment questions A spider has 8 legs. If there are 5 spiders, how many legs are there altogether? A book costs £5. How much do 6 books cost? 18 socks are put into pairs. How many pairs are there? Felicity wants to buy a scooter for £60. If she pays with £10 notes, how many notes does she need? Circle the numbers that are multiples of 4. 14 24 40 34 16 32 25 Assessment guidance: The multiplication tables check in year 4 will assess pupils’ fluency in all multiplication tables. At this stage, teachers should assess fluency in facts within the 10, 5, 2, 4 and 8 multiplication tables. Once pupils can automatically recall multiplication facts, and have covered criterion 3MD–1, they should be able to apply their knowledge to contextual questions like those shown here. Teachers should ensure that pupils answer these questions using automatic recall of the appropriate multiplication facts – for question 1, for example, if a pupil counts up in multiples of 8, or draws 5 spiders and counts the legs in ones, the pupil has not met this criterion. 103 3NF–3 Scaling number facts by 10 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10), for example: 8 6 14 and 14 6 8 so 80 60 140 and 140 60 80 3 4 12 and 12 4 3 so 30 4 120 and 120 4 30 3NF–3 Teaching guidance During year 3, pupils develop automaticity in addition and subtraction facts within 20 (3NF–1), and learn to recall multiplication table facts and related division facts for the 10, 5, 2, 4 and 8 multiplication tables. To be ready to progress to year 4, pupils must also be able to combine these facts with unitising in tens, including: scaling known additive facts within 10, for example, 90 60 30 scaling known additive facts that bridge 10, for example, 80 60 140 scaling known multiplication tables facts, for example, 30 4 120 scaling division facts derived from multiplication tables, for example, 120 4 30 For calculations such as 80 60 140 , pupils can begin by using tens frames and counters as they did for calculation across 10 (2AS–1), but now using 10-value counters. Figure 75: tens frames with 10-value counters showing 80 + 60 = 140 104 8 + 6 14 80 + 60 140 14 – 6 8 140 – 60 80 14 – 8 6 140 – 80 60 You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 3NPV–2 and 3NF–3 Similarly, pupils can use 10-value counters to understand how a known multiplicative fact, such as 3 5 15 , relates to a scaled calculation, such as 3 50 150 . Pupils should be able reason in terms of unitising in tens. Figure 76: 3-by-5 array of 10-value place-value counters 3 5 15 3 50 150 Language focus “3 times 5 is equal to 15.” “3 times 5 tens is equal to 15 tens.” “15 tens is equal to 150.” 3 5 15 30 5 150 Language focus “3 times 5 is equal to 15.” “3 tens times 5 is equal to 15 tens.” “15 tens is equal to 150.” Multiplication calculations in this criterion should all be related to the 5, 10, 2, 4 and 8 multiplication tables. It is important for pupils to understand all of the calculations in this criterion in terms of working with units of 10. 105 Making connections This criterion builds on: additive fluency within and across 10 (3NF–1) 3NF–2 , where pupils develop fluency in multiplication and division facts 3NPV–1, where pupils need to be able to work out how many tens there are in any three-digit multiple of 10 Meeting this criterion also requires pupils to be able to fluently multiply whole numbers by 10 (3NF–2 ). 3NF–3 Example assessment questions A garden table costs £80 and 2 garden chairs each cost £60. How much do the 2 chairs and the table cost altogether? 130 people are expected at a concert. So far 70 people have arrived. How many more people are due to arrive? A family ticket for a safari park is £40. 3 families go together. How much do the 3 family tickets cost altogether? Fill in the missing numbers. 30 110 7 60 106 3AS–1 Calculate complements to 100 Calculate complements to 100, for example: 3AS–1 Teaching guidance Calculating complements to 100 is an important skill, because it is a prerequisite for calculating how much change is due when paying for an item. When pupils calculate complements (the amount needed to complete a total), a common error is to end up with a total that is too large: When calculating complements to 100, pupils typically make an extra ‘unit’ of 10, making 110 instead of 100. When finding change from a whole number of pounds, pupils typically make an extra £1, for example, they incorrectly calculate the change due from £5 for a cost of £3.40 as £2.60 It is important for pupils to spend time specifically learning about calculating complements, including the risk of creating ‘extra units’. This should begin in year 3, with calculating complements to 100. Pupils should compare correct calculations with the corresponding common incorrect calculations for complements to 100. They should be able to discuss the pairs of calculations and understand the source of the error in the incorrect calculations. Incorrect complement to 100 Correct complement to 100 Figure 77: partitioning diagram and calculation showing incorrect complement to 100: 62 and 48 Figure 78: partitioning diagram and calculation showing correct complement to 100: 62 and 38 A shaded 100 grid can be used to show why there are only 9 full tens in the correct complements to 100. The 10th ten is composed of the ones digits. 107 Figure 79: a 100 grid shaded in 2 colours to represent 62 and 38 as a complement to 100 Once pupils understand why given complements to 100 are correct or not, they should learn to work in steps to calculate complements themselves: First make 10 ones. Then work out the number of additional tens needed. Pupils must understand that the tens digits should bond to 9, not to 10. Check that the 2 numbers sum to 100. Figure 80: calculating a complement to 100 Language focus “First we make 10 ones. The ones digits add up to 1 ten, so we need 9 more tens.” Making connections Pupils will need to calculate the majority of two-digit complements to 100 as described above. However, pupils should memorise the pair 75 and 25 (see 3NPV–4 ). 108 3AS–1 Example assessment questions Which of these are correct complements to 100 and which have an extra 10? Tick the correct column. Explain your answers. Correct bond to 100 Incorrect bond to 100 (extra 10) Explanation 28 + 72 61 + 49 55 + 45 43 + 67 84 + 16 39 + 71 Fill in the missing numbers. 65 100 100 29 100 42 100 83 A dressmaker had 1m of ribbon. Then she used 22cm of it. How many centimetres of ribbon does she have left? A toy shop sells ping-pong balls for 65p each. If I use a £1 coin to pay for a ping-pong ball, how much change will I get, in pence? Mr Jones has 100 stickers. 47 of them are gold and the rest are silver. How many are silver? 109 3AS–2 Columnar addition and subtraction Add and subtract up to three-digit numbers using columnar methods. 3AS–2 Teaching guidance Pupils must learn to add and subtract using the formal written methods of columnar addition and columnar subtraction. Pupils should master columnar addition, including calculations involving regrouping (some columns sum to 10 or more), before learning columnar subtraction. However, guidance here is combined due to the similarities between the two algorithms. Beginning with calculations that do not involve regrouping (no columns sum to 10 or more) or exchange (no columns have a minuend smaller than the subtrahend), pupils should: learn to lay out columnar calculations with like digits correctly aligned learn to work from right to left, adding or subtracting the least significant digits first Teachers should initially use place-value equipment, such as Dienes, to model the algorithms and help pupils make connections to what they already know about addition and subtraction.
110 Figure 81: columnar addition with no regrouping: calculation and Dienes representation Figure 82: columnar subtraction with no exchange: calculation and Dienes representation Pupils should use unitising language to describe within-column calculations. Language focus “3 ones plus 5 ones is equal to 8 ones.” “4 tens plus 2 tens is equal to 6 tens.” “5 ones minus 3 ones is equal to 2 ones.” “6 tens minus 2 tens is equal to 4 tens.” Pupils must also learn to carry out columnar addition calculations that involve regrouping, and columnar subtraction calculations that involve exchange. Regrouping and exchange build on pupils’ understanding that 10 ones is equivalent to 1 ten, and that 10 tens is equivalent to 1 hundred. Dienes can be used to model the calculations, and to draw attention to the regrouping/exchange. 111 Dienes (or any other place-value apparatus) should be used, only initially, to support pupils understanding of the structure of the algorithms, and should not be used as a tool for finding the answer. Once pupils understand the algorithms, they should use known facts to perform the calculation in each column (3NF–1). For calculations with more than 2 addends, pupils should add the digits within a column in the most efficient order. Figure 83: making efficient choices for within-column calculations when adding 3 addends Pupils must learn that, although columnar methods can be used for any additive calculation, they are not always the most appropriate choice. For example, 164 + 36 can be calculated by recognising that 64 and 36 is a complement to 100, while 120 + 130 maybe be calculated by unitising in tens (12 tens + 13 tens = 25 tens) or by recognising that 20 + 30 = 50. Throughout, pupils should continue to recognise the inverse relationship between addition and subtraction. Pupils may represent calculations using partitioning diagrams or bar models, and should learn to check their answers using the inverse operation. Figure 84: using addition to check a subtraction calculation You can find out more about fluency for these calculations here in the calculation and fluency section: 3AS–2 Making connections The within-column calculations in columnar addition and subtraction use the facts practised to fluency in 3NF–1. Any additive calculation can be carried out using columnar methods. However other methods may sometimes be more efficient, such as those taught in 3NPV–3 and 3NF–3 . 112 3AS–2 Example assessment questions Solve these calculations using columnar addition or columnar subtraction. a. 89 – 23 b. 127 43 49 c. 402 130 78 d. 462– 256 e. 345–72 f. 407 –129 Year 3 want to buy some sports equipment which costs £472. So far they have raised £158. How much more money do they need to raise? Cheryl has £135. She spends £53 on some new trainers. How much money does she have left? There are 172 non-fiction books in the school library and 356 fiction books. How many books are there in the library altogether? Fill in the missing numbers. Mahsa carries out the following columnar addition calculation. Write a columnar subtraction calculation that she could do to check that her calculation is correct. Complete the following calculations. Choose carefully which method to use. 175 25 63 89 42 50 250 300 776– 200 523– 247 400–35 113 3AS–3 Manipulate the additive relationship Manipulate the additive relationship: Understand the inverse relationship between addition and subtraction, and how both relate to the part–part–whole structure. Understand and use the commutative property of addition, and understand the related property for subtraction. 3AS–3 Teaching guidance Pupils will begin year 3 with an understanding of some of the individual concepts covered in this criterion, and will already be familiar with using partitioning diagrams and bar models. However, pupils need to leave year 3 with a coherent understanding of the additive relationship, and how addition and subtraction equations relate to the various additive structures. Pupils must understand that the simplest addition and subtraction equations describe the relationship between 3 numbers, where one is a sum of the other two. They should understand that both addition and subtraction equations can be used to describe the same additive relationship. They should practise writing the full set of 8 equations that are represented by a given partitioning diagram or bar model. Figure 85: partitioning diagrams showing the additive relationship between 25, 12 and 37 25 12 37 37 –12 25 12 25 37 37 – 25 12 37 25 12 25 37 –12 37 12 25 12 37 – 25 Pupils should learn and use the correct names for the terms in addition and subtraction equations. addend addend sum minuend– subtrahend difference Pupils understanding should go beyond the fact that addition and subtraction are inverse operations. They need understand how the terms in addition and subtraction equations are related to each other, and to the parts and whole within an additive relationship, and use this understanding to manipulate equations. 114 Figure 86: connecting the terms in addition and subtraction equations to the part-part-whole structure With experience of the commutative property of addition, pupils can now learn that, because of the relationship between addition and subtraction, the commutative property has a related property for subtraction. Language focus “If we swap the values of the subtrahend and difference, the minuend remains the same.” Both of the following equations are therefore correct: 37– 25 12 37–12 25 Pupils should use their understanding of the additive relationship and how it is related to parts and a whole, the inverse relationship between addition and subtraction, and the commutative property, to manipulate equations. They must recognise that if 2 of the 3 numbers in a given additive relationship are known, the unknown number can always be determined: addition is used to find an unknown whole, while subtraction is used to find an unknown part, irrespective of how the problem is presented. For example, 34 ?
56 has an unknown part so is solved using subtraction, even though the problem is written as addition. Pupils need to be able to solve: missing-addend problems (addend ?
sum ) missing-subtrahend problems (minuend ?
difference ) missing-minuend problems (? subtrahend difference ) Pupils have been solving missing-number problems since year 1. However, with smaller numbers, they were able to rely on missing-number facts (known number bonds) or counting on (for example, counting on 3 fingers to get from 12 to 15 for 12 ?
15 ). Now that pupils are using numbers with up to three digits, they will need to rearrange missing-number equations to solve them, using formal written methods where necessary. Teachers should not assume that pupils will be able to do this automatically: pupils will need to spend time learning the properties of the additive relationship, and practising rearranging equations. They should be able to identify the type of each problem in terms 115 of whether a part or the whole is unknown, and can sketch partitioning diagrams or bar models to help them do this. Missing-addend problems Type of problem: missing part Rewrite the addition equation as a subtraction equation, for example: Language focus “There is a missing part. To find the missing part, we subtract the other part from the whole.” Missing-subtrahend problems Type of problem: missing part Rewrite the subtraction equation by swapping the subtrahend and the difference, for example: Language focus “There is a missing part. To find the missing part, we subtract the other part from the whole.” Missing-minuend problems Type of problem: missing whole Rewrite the subtraction equation as an addition equation, for example: Language focus “There is a missing whole. To find the missing whole, we add the 2 parts.” 116 Making connections Once pupils have identified the calculation required to solve a missing-number problem, they need to be able to perform that calculation. Pupils must be fluent in identifying and applying appropriate addition and subtraction strategies. 3AS–3 Example assessment questions Fill in the missing numbers. 364 857 785 180 1 45 721 250cm 65cm 117 3MD–1 Multiplication and division structures Apply known multiplication and division facts to solve contextual problems with different structures, including quotitive and partitive division. 3MD–1 Teaching guidance At this stage, pupils will be developing fluency in the 5, 10, 2, 4 and 8 multiplication tables (3NF–2 ), so should be able to solve multiplication problems about groups of 5, 10, 2, 4 or 8. Pupils have already begun to learn that if the factors are swapped, the product remains the same (3NF–2). Language focus “factor times factor is equal to product” “The order of the factors does not affect the product.” Pupils should also learn that the commutative property allows them to use their known facts to solve problems about 5, 10, 2, 4 or 8 equal groups (for example, 2 groups of 7). An array can be used to illustrate how the commutative property relates to different grouping interpretations – the example below shows that 7 groups of 2 and 2 groups of 7 both correspond to the same total quantity (14). Figure 87: using an array to show that 7 groups of 2 and 2 groups of 7 both correspond to the same total quantity This means that pupils can use their knowledge that 7 2 14 to solve a problem about 2 groups of 7, even though they have not yet learned the 7 multiplication table. Pupils should already be solving division calculations using known division facts corresponding to the 5, 10, 2, 4 and 8 multiplication tables (3NF–2). They must also be able to use these known facts to solve both quotitive (grouping) and partitive (sharing) contextual division problems. The same array that was used to illustrate the commutative 118 property of multiplication can also be used to show how known division facts can be applied to the two different division structures. Quotitive division Partitive division I need 14 ping-pong balls. There are 2 ping-pong balls in a pack. How many packs do I need? Figure 88: using an array and bar model to show that 14 divided into groups of 2 is equal to 7 Language focus “7 times 2 is 14, so 14 divided by 2 is 7.” “14 divided into groups of 2 is equal to 7.” I need 7 packs of ping-pong balls. £14 is shared between 2 children. How much money does each child get? Figure 89: using an array and bar model to show that 14 shared between 2 is equal to 7 Language focus “7 times 2 is 14, so 14 divided by 2 is 7.” “£14 shared between 2 is equal to £7 each.” Each child gets £7. At this stage, pupils only need to be able to apply division facts corresponding to division by 5, 10, 2, 4 and 8 to solve division problems with the two different contexts. In year 4, pupils will learn, for example, that if they know that 2 7 14 , then they know both 14 2 7 and 14 7 2 . When pupils are solving contextual problems, dividing into groups of 5, 10, 2, 4 or 8 (quotitive division) or sharing into 5, 10, 2, 4 or 8 parts (partitive division), they should calculate by recalling a known multiplication fact rather than by skip counting, as described in 3NF–2 and illustrated above. 119 Making connections Using the commutative law of multiplication reduces the number of multiplication tables facts that pupils need to memorise, both in year 3 (3NF–2 ) and beyond. Being able to calculate the size of a part in partitive contexts using known division facts is a prerequisite for finding unit fractions of quantities (3F–2). 3MD–1 Example assessment questions Circle the expressions that match the picture. 2 6 6 6 6 2 2 6 6 2 6 6 If one sweet costs 3p, how much do 8 sweets cost? I need to buy 32 metres of fencing to go around my garden. The fencing is sold in 8-metre lengths. How many 8-metre lengths do I need to buy? There are 24 strawberries in a tub. I share them equally between the 4 people in my family. How many does each person get? A gardener has 5 plant pots. She plants 6 seeds in each pot. How many seeds does she plant altogether? 120 3F–1 Use and understand fraction notation Interpret and write proper fractions to represent 1 or several parts of a whole that is divided into equal parts. 3F–1 Teaching guidance Pupils should learn that when a whole is divided into equal parts, fraction notation can be used to describe the size of each equal part relative to the whole. Because it is the size of a part relative to the whole which determines the value of a fraction, it is important that pupils talk about, and identify, both the whole and the part from the start of their work on fractions. They should not begin, for example, by talking about ‘1 out of 3 parts’ without reference to a whole. Pupils should begin by working with concrete resources and diagrams. First they should learn to identify the whole and the number of equal parts, then to describe one particular equal part relative to the whole. Figure 90: a circle divided into 3 equal parts, with one part shaded Language focus “The whole is divided into 3 equal parts. 1 of these parts is shaded.” Pupils must be able to use this precise language to describe a unit fraction of a: shape/area (as in the above example) measure (for example, a length of ribbon or a beaker of water) set (for example, a group of sheep where all are white except one, which is black) Pupils should then learn to interpret and write unit fractions, relating to these contexts, using mathematical notation. They should continue to describe the whole, the number of parts and the particular part, and relate this to the written fraction. 121 SayWrite“The whole has been divided…”The fraction bar: –“…into 3 equal parts.”The denominator: 3“1 of these parts is shaded.”The numerator: 1 Language focus “The whole is divided into 3 equal parts. Each part is one-third of the whole.” A clear understanding of unit fractions is the foundation for all future fractions concepts. Pupils should spend sufficient time working with unit fractions to achieve mastery before moving on to non-unit fractions. Pupils should learn that a non-unit fraction is made up of a quantity of unit fractions. They should practise using unitising language to describe, for example, 5 eighths as 5 one-eighths (here, we are unitising in eighths). Language focus “The whole is divided into 8 equal parts and 5 of those parts are shaded. of the shape is shaded. is 5 one-eighths.” Pupils should also experience examples where all parts of the shape are shaded (or all parts of the measure or set are highlighted) and the numerator is equal to the denominator. They should understand, for example that 5 5 represents all 5 equal parts, and is equivalent to the whole. Teaching should draw attention to the fact that in order to identify a fraction, the parts need to be equal. Comparing situations where the parts are equal and those where they are not is a useful activity (see 3F–1, questions 2 and 4). 122 Making connections Showing, describing and representing a unit fraction of a shape, measure or set involves dividing it into a number of equal parts. The theme of dividing a quantity into a given number of equal parts runs through many topics, including: partitive division (3MD–1) finding a unit fraction of a value using known division facts (3F–2). 3F–1 Example assessment questions What fraction of each diagram is shaded? Does each diagram show the given fraction? Explain your answers. What fraction of each diagram is shaded/highlighted? 123 Tick or cross each diagram to show whether 3 5 is shaded. Explain your answers. a. Shade 1 10 of this set. b. Shade 3 4 of this shape. c. Circle 4 5 of the flowers. d. Colour 1 3 of the line. 124 3F–2 Find unit fractions of quantities Find unit fractions of quantities using known division facts (multiplication tables fluency). 3F–2 Teaching guidance In 3F–1 pupils learnt how fractions act as operators on a whole. Now they must learn to evaluate the outcome of that operation, for unit fractions of quantities, and connect this to what they already know about dividing quantities into equal parts using known division facts, for example: 1 10 30 10 3 so of 30 3 1 4 36 4 9 so of 36 9 In year 3, pupils learn the 5, 10, 2, 4 and 8 multiplication tables, so examples in this criterion should be restricted to finding 1 5 , 1 10 , 1 2 , 1 4 or 1 8 of quantities. This allows pupils to focus on the underlying concepts instead of on calculation. Pupils should begin by working with concrete resources or diagrams representing sets. As in 3F–1, they should identify the whole, the number of equal parts, and the size of each part relative to the whole written as a unit fraction. They should then extend their description to quantify the number of items in a part, and connect this to the unit-fraction operator. Figure 91: 12 oranges divided into 4 equal parts Language focus “The whole is 12 oranges. The whole is divided into 4 equal parts.” “Each part is of the whole. of 12 oranges is 3 oranges.” 125 Once pupils can confidently and accurately describe situations where the value of a part is visible, they should learn to calculate the value of a part when it cannot be seen. Bar models are a useful representation here. Pupils should calculate the size of the parts using known division facts, initially with no reference to unit fractions. This is partitive division (3MD–1). Figure 92: using a bar model to represent 15 divided into 5 equal parts Then pupils should make the link between partitive division and finding a fraction of a quantity. They should understand that the situations are the same because both involve dividing a whole into a given number of equal parts. Pupils must understand that, therefore, division facts can be used to find a unit fraction of a quantity. Figure 93: using a bar model to represent one-fifth of 15 126 Language focus “To find of 15, we divide 15 into 5 equal parts.” “15 divided by 5 is equal to 3, so of 15 is equal to 3.” Making connections This criterion builds directly on 3F–1, where pupils learnt to associate fraction notation with dividing a shape, measure or set into a number of equal parts. The focus of this criterion is understanding that finding a unit fraction of a quantity is the same structure as partitive division (3MD–1). In 3NF–1 pupils develop fluency in the 5, 10, 2, 4 and 8 multiplication tables and associated division facts. These division facts are applied here to find , , , or of quantities. 3F–2 Example assessment questions Rohan saved £32. He spends 1 4 of his money on a toy. How much does he spend? Find: a. 1 5 of 35 b. 1 10 of 40 c. 1 8 of 24 The school caretaker buys 50 litres of paint. She uses 1 5 of it to paint the year 3 classroom. How many litres of paint is this? There are 16 apples in a fruit bowl. Some children eat 1 4 of the apples. How many are left? 127 3F–3 Fractions within 1 in the linear number system Reason about the location of any fraction within 1 in the linear number system. 3F–3 Teaching guidance So far (in 3F–1 and 3F–2) pupils will probably have only experienced fractions as operators, for example, 1 3 of this shape, 3 5 of this line, or 1 4 of this quantity. Pupils must also develop an understanding of fractions as numbers, each of which has a place in the linear number system (for example, the number 1 4 in contrast to the operator 1 4 of something). Pupils will already have learnt to place whole numbers on number lines, and now they must learn that other numbers (fractions) lie between these whole numbers, beginning in year 3 with fractions within 1. Pupils should learn to count, forwards and backwards, in multiples of unit fractions, with the support of number lines. Figure 94: number line to support counting to 1 in multiples of one quarter Figure 95: number line to support counting to 1 in multiples of one fifth Pupils should practise dual counting. Language focus “One fifth, two fifths, three fifths…” “1 one-fifth, 2 one-fifths, 3 one-fifths…” This reinforces the understanding that non-unit fractions are repeated additions of unit fractions. 128 Pupils also need to understand that a fraction with the numerator equal to the denominator is equivalent to 1. Practice should involve counting, for example, both “… three-fifths, four-fifths, five-fifths” and “… three-fifths, four-fifths, one”. Pupils should then learn to label marked number lines, within 1. Identifying points between labelled intervals is an important skill in graphing and measures. A common mistake that pupils make is to count the number of marks between labelled intervals, rather than the number of parts, for example, on the number line below they may count 3 marks and incorrectly deduce that the number line is marked in thirds. Figure 96: labelling a 0 to 1 number line marked in quarters Language focus “Each whole-number interval is divided into 4 equal parts, so we count in quarters.” Pupils must also be able to estimate the value or position of fractions on 0 to 1 number lines that do not have fractional marks. Figure 97: estimating the position of a fraction on a 0 to 1 number line Pupils also need to be able to reason, for example, that 1 4 is nearer to 0 than 1 3 is, because 1 4 is smaller than 1 3 ; they should consider the number of parts the 0 to 1 interval is divided into, and understand that the greater the denominator, the more parts there are, and therefore 1 4 < 1 3 . 129 Figure 98: estimating the relative positions of one quarter and one third on a 0 to 1 number line Making connections Having a visual image of fractions in the linear number system helps pupils to add and subtract fractions with the same denominator, for example (3F–4). It also supports comparison of fractions, and reading scales. 3F–3 Example assessment questions Label the points on this number line. How tall is this plant? Give your answer as a fraction of a metre. a. Which is larger, 6 8 or 3 8 ? Explain your answer. b. Which is larger, 1 4 or 1 3 ? Explain your answer. 130 Gemma and Kasper look at this number line. Gemma says the arrow is pointing to the number 3 4 . Kasper says the arrow is pointing to the number 3 5 . Who is correct? Explain your answer. Add the missing labels to the measuring jug. 131 3F–4 Add and subtract fractions within 1 Add and subtract fractions with the same denominator, within 1. 3F–4 Teaching guidance To add and subtract fractions, pupils must already understand that non-unit fractions are repeated additions of unit fractions, for example, three-eighths is 3 one-eighths. In other words, pupils must have begun to unitise with unit fractions in the same way that they learnt to unitise, for example, in tens (30 is 3 tens). Addition and subtraction of fractions with the same denominator then follows logically: just as pupils learnt that 3 tens plus 2 tens is 5 tens, they can reason that 3 one-eighths plus 2 one-eighths is equal to 5 one-eighths. Figure 99: adding and subtracting fractions with the same denominator: pie model, bar model and number line Language focus “3 one-eighths plus 2 one-eighths is equal to 5 one-eighths.” “5 one-eighths minus 2 one-eighths is equal to 3 one-eighths.” 132 Pupils should be able to understand and use the following generalisations. Language focus “When adding fractions with the same denominators, just add the numerators.” “When subtracting fractions with the same denominators, just subtract the numerators.” Making connections In 3F–1 and 3F–3, pupils learnt that non-unit fractions are made up of multiples of a unit fraction. This criterion is based on pupils being able to unitise with unit fractions. Pupils already learnt to unitise in tens and combine this understanding with known addition and subtraction facts in 3NPV–1 and 3NF–3 . 3F–4 Example assessment questions Complete the calculations. 5 1 9 9 6 2 8 8 5 3 12 12 9 6 11 11 5 7 14 14 9 10 0 Diego writes: 3 5 8 12 12 12 Mark writes: 3 5 8 12 12 24 Who is correct? Explain the mistake that has been made. 133 Decide whether each calculation is correct or not. Explain your answers. Correct( ) or incorrect ( )?Explanation Sofia had a jug containing 7 10 of a litre of juice. She drank 4 10 of a litre. How much does she have left? 134 3G–1 Recognise right angles Recognise right angles as a property of shape or a description of a turn, and identify right angles in 2D shapes presented in different orientations. 3G–1 Teaching guidance Pupils must have learnt about fractions before beginning work on this criterion. In particular they should recognise one-, two- and three-quarters of a circle. Pupils must be able to describe and represent quarter, half and three-quarter turns (clockwise and anti-clockwise). Pupils should begin by making quarter turns with their bodies, following instructions such as “Stand, and make a quarter turn clockwise. Walk in a straight line. Stop. Make a quarter turn anticlockwise.” They should be able to relate these movements to the quarter turn of a clock hand. Pupils should learn that the angle relative to the starting orientation, created by a quarter turn (in either direction), is called a right angle – programmable robots and geo-strips are useful tools for illustrating this. Pupils should then learn to follow instructions involving 1 2 turn and 3 4 turns, clockwise and anticlockwise. They should recognise that the result of making a 1 2 turn clockwise is the same as the result of making a 1 2 turn anticlockwise. Pupils should also understand 1 2 and 3 4 turns as repeated 1 4 turns, and therefore as repeated turns through a right angle. Pupils should recognise that a right angle is the ‘amount of turn’ between 2 lines, and is independent of the length of those lines. They should be presented with a right angle created by 2 long lines (such as two metre sticks) and a right angle created by 2 short lines (such as 2 geo-strips), and understand that both are right angles. It is important that pupils know that it is incorrect to describe the right-angle made from longer lines as a ‘bigger right angle’, or that made from the shorter lines as a ‘smaller right angle’. Pupils should practise identifying right angles in their environment, for example, the corner of their desk, the panes of a window, or the hands on a clock at 3pm or 9pm. They should learn to use various tools to confirm that angles are right angles, for example a card circle with a quarter circle cut out, or a piece of paper of any size that has been folded in half and half again to create a right angle. Pupils must then learn to identify right angles in polygons. This should involve both handling shapes (for example, cut from cardboard) and working with images of shapes. When pupils are handling shapes, they should practise rotating the shapes to check each angle against a right-angle checker. Images of shapes should be presented in a variety of orientations, so that pupils’ ability to identify right angles is not dependent on the lines being horizontal and vertical. Pupils must also learn to interpret and use the standard convention for marking right angles (as illustrated below). 135 Figure 100: irregular pentagon with 3 right angles marked using standard convention Whether working with cut-out shapes or images, pupils should be able to state whether a given angle is greater than or smaller than a right angle, using the angle-checker. Pupils should recognise that: the only polygon in which every vertex can be a right angle is a quadrilateral quadrilaterals that have 4 right angles are rectangles irrespective of the length of their sides a quadrilateral that has all side-lengths equal and every vertex a right angle is a regular rectangle that can also be called a square Making connections In 3G–2, children will learn that 2 lines are at right angles are termed ‘perpendicular’. Composing and drawing shapes in 3G–2 provides another context in which to identify right angles. 136 3G–1 Example assessment questions Here is a map of a treasure island. a. Follow the instructions and say where you end up. Each time, start at the camp, facing north. i. Go forwards 3 squares. Make a quarter turn clockwise. Go forwards 2 squares. Make a quarter turn anticlockwise. Go forwards 2 squares. Where are you? ii. Make a three-quarter turn clockwise. Go forward 3 squares. Make a quarter turn anticlockwise. Go forward 1 square. Where are you? b. Start at the camp, facing North. Write some instructions, like the ones above, to get to the treasure. Draw an irregular hexagon with one right angle on this grid. 137 Mark all of the right angles in these shapes. Use a right-angle checker to help you. 3G–2 Draw polygons and identify parallel and perpendicular sides Draw polygons by joining marked points, and identify parallel and perpendicular sides. 3G–2 Teaching guidance Pupils must learn to draw polygons by joining marked points, precisely, using a ruler. Pupils should be able to mark vertices themselves on a grid (square or isometric), as well as join already-marked points. Pupils must be able to identify a pair of parallel or perpendicular lines, as well as horizontal and vertical lines. They should be able to explain why a pair of lines are parallel or perpendicular. Language focus “These 2 lines are parallel because they are always the same distance apart. They will never meet no matter how far we extend them.” “These 2 lines are perpendicular because they are at right angles to each other.” Pupils should be able to select or create shapes according to parameters that include these terms, such as joining 2 isosceles triangles to make a parallelogram. 138 Pupils may use standard notation to mark parallel sides. Making connections In 3G–1, children learnt to identify right angles. In this criterion they should identify right angles in shapes they have drawn or made, and know that a right angle is made at the point where two perpendicular lines meet. 3G–2 Example assessment questions Task: Provide each pupil with 2 trapezium pieces from a pattern block set. Then ask them to make 3 different shapes by joining the pieces and discuss the properties of each shape they make. Here are 5 vertices of a regular hexagon. Mark the sixth vertex and join the points to draw the hexagon. Here are 2 sides of a square. Complete the square. 139 Look at these 5 quadrilaterals. Mark all the pairs of parallel sides. Hint: you can extend sides to help you. Mark the missing vertex of this quadrilateral so that 2 of the sides are perpendicular. Calculation and fluency Number, place value and number facts: 3NPV–2 and 3NF–3 3NPV–2: Recognise the place value of each digit in three-digit numbers, and compose and decompose three-digit numbers using standard and non-standard partitioning. 3NF–3: Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10), for example: 8 6 14 and 14 6 8 so 80 60 140 and 140 60 80 3 4 12 and 12 4 3 so 30 4 120 and 120 4 30 Representations such as place-value counters and partitioning diagrams (3NPV–2), and tens-frames with place-value counters (3F–3), can be used initially to help pupils understand calculation strategies and make connections between known facts and related calculations. However, pupils should not rely on such representations for calculating. For the calculations in 3NF–3, for example, pupils should instead be able to calculate by verbalising the relationship. 140 Language focus “8 plus 6 is equal to 14, so 8 tens plus 6 tens is equal to 14 tens.” “14 tens is equal to 140.” Pupils should be developing fluency in both formal written and mental methods for addition and subtraction. Mental methods can include jottings to keep track of calculation, or language structures as exemplified above. Pupils should select the most efficient method to calculate depending on the numbers involved. 3NF–1 Fluently add and subtract within and across 10 Secure fluency in addition and subtraction facts that bridge 10, through continued practice. Pupils who are fluent in addition and subtraction facts within and across 10 have the best chance of mastering columnar addition and columnar subtraction. Teachers should make sure that fluency in addition and subtraction facts is given the same prominence as fluency in multiplication tables. Pupils should continue to practise calculating with additive facts within 10. Pupils may initially use manipulatives, such as tens frames and counters, to apply the strategies for adding and subtracting across 10 described in year 2 (2AS–1). However, they should not be using the manipulatives as a tool for finding answers, and by the end of year 3 pupils should be able to carry out these calculations mentally, using their fluency in complements to 10 and partitioning. Pupils do not need to memorise all additive facts for adding and subtracting across 10, but need to be able to recall appropriate doubles (double 6, 7, 8 and 9) and corresponding halves (half of 12, 14, 16 and 18), and use these known facts for calculations such as 6 6 12 and 18 – 9 9 . 141 3AS–2 Columnar addition and subtraction Add and subtract up to three-digit numbers using columnar methods. Pupils must be able to add 2 or more numbers using columnar addition, including calculations whose addends have different numbers of digits. Figure 101: columnar addition for calculations involving three-digit numbers For calculations with more than 2 addends, pupils should add the digits within a column in the most efficient order. For the third example above, efficient choices could include: beginning by making 10 in the ones column making double 8 in the tens column Pupils must be able to subtract 1 three-digit number from another using columnar subtraction. They should be able to apply the columnar method to calculations where the subtrahend has fewer digits than the minuend, and they must be able to exchange through 0. Figure 102: columnar subtraction for calculations involving three-digit numbers Pupils should make sensible decisions about how and when to use columnar subtraction. For example, when the minuend and subtrahend are very close together pupils may mentally find the difference, avoiding the need for column subtraction. For example, for 402–398 , pupils can see that 398 is 2 away from 400, and then there is 2 more to get to 402, so the difference is 4. This is more efficient than the corresponding columnar subtraction calculation which requires exchange through the zero. 142 3NF–2 Recall of multiplication tables Recall multiplication facts, and corresponding division facts, in the 10, 5, 2, 4 and 8 multiplication tables, and recognise products in these multiplication tables as multiples of the corresponding number. Pupils who are fluent in these multiplication table facts can solve the following types of problem by automatic recall of the relevant fact rather than by skip counting or reciting the relevant multiplication table: identifying products 8 4 3 5 10 10 solving missing-factor problems 5 45 6 48 22 2 using relevant multiplication table facts to solve division problems 35 5 40 8 Pupils should also be fluent in interpreting contextual multiplication and division problems, identifying the appropriate calculation and solving it using automatic recall of the relevant fact. This is discussed, and example questions are given, in 3MD–1. As pupils become fluent with the multiplication table facts, they should also develop fluency in related calculations as described in 3NF–3 (scaling number facts by 10). 143 Year 4 guidance Ready-to-progress criteria Year 3 conceptual prerequesite Year 4 ready-to-progress criteria Future applications Know that 10 tens are equivalent to 1 hundred, and that 100 is 10 times the size of 10. 4NPV–1 Know that 10 hundreds are equivalent to 1 thousand, and that 1,000 is 10 times the size of 100; apply this to identify and work out how many 100s there are in other four-digit multiples of 100. Solve multiplication problems that that involve a scaling structure, such as ‘10 times as long’. Recognise the place value of each digit in three-digit numbers, and compose and decompose three-digit numbers using standard and non-standard partitioning. 4NPV–2 Recognise the place value of each digit in four-digit numbers, and compose and decompose four-digit numbers using standard and non-standard partitioning. Compare and order numbers. Add and subtract using mental and formal written methods. Reason about the location of any three-digit number in the linear number system, including identifying the previous and next multiple of 10 and 100. 4NPV–3 Reason about the location of any four-digit number in the linear number system, including identifying the previous and next multiple of 1,000 and 100, and rounding to the nearest of each. Compare and order numbers. Estimate and approximate to the nearest multiple of 1,000, 100 or 10. Divide 100 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 100 with 2, 4, 5 and 10 equal parts. 4NPV–4 Divide 1,000 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 1,000 with 2, 4, 5 and 10 equal parts. Read scales on graphs and measuring instruments. Recall multiplication and division facts in the 5 and 10, and 2, 4 and 8 multiplication tables, and recognise products in these multiplication tables as multiples of the corresponding number. 4NF–1 Recall multiplication and division facts up to , and recognise products in multiplication tables as multiples of the corresponding number. Use multiplication facts during application of formal written methods. Use division facts during application of formal written methods. 144 Year 3 conceptual prerequesite Year 4 ready-to-progress criteria Future applications Use known division facts to solve division problems. Calculate small differences, for example: 4NF–2 Solve division problems, with two-digit dividends and one-digit divisors, that involve remainders, for example: and interpret remainders appropriately according to the context. Correctly represent and interpret remainders when using short and long division. Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10), for example: 4NF–3 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 100), for example: and so and so Apply place-value knowledge to known additive and multiplicative number facts, extending to a whole number of larger powers of ten and powers of ten smaller than one, for example: Multiply two-digit numbers by 10, and divide three-digit multiples of 10 by 10. 4MD–1 Multiply and divide whole numbers by 10 and 100 (keeping to whole number quotients); understand this as equivalent to making a number 10 or 100 times the size. Convert between different metric units of measure. Apply multiplication and division by 10 and 100 to calculations involving decimals, for example: Understand the inverse relationship between multiplication and division. Write and use multiplication table facts with the factors presented in either order. 4MD–2 Manipulate multiplication and division equations, and understand and apply the commutative property of multiplication. Recognise and apply the structures of multiplication and division to a variety of contexts. 4MD–3 Understand and apply the distributive property of multiplication. Recognise when to use and apply the distributive property of multiplication in a variety of contexts. 145 Year 3 conceptual prerequesite Year 4 ready-to-progress criteria Future applications Reason about the location of fractions less than 1 in the linear number system. 4F–1 Reason about the location of mixed numbers in the linear number system. Compare and order fractions. Identify unit and non-unit fractions. 4F–2 Convert mixed numbers to improper fractions and vice versa. Compare and order fractions. Add and subtract fractions where calculation bridges whole numbers. Add and subtract fractions with the same denominator, within 1 whole, for example: 4F–3 Add and subtract improper and mixed fractions with the same denominator, including bridging whole numbers, for example: Draw polygons by joining marked points. 4G–1 Draw polygons, specified by coordinates in the first quadrant, and translate within the first quadrant. Draw polygons, specified by coordinates in the 4 quadrants. Measure lines in centimetres and metres. Add more than 2 addends. Recall multiplication table facts. 4G–2 Identify regular polygons, including equilateral triangles and squares, as those in which the side-lengths are equal and the angles are equal. Find the perimeter of regular and irregular polygons. Draw, compose and decompose shapes according to given properties, dimensions, angles or area. 4G–3 Identify line symmetry in 2D shapes presented in different orientations. Reflect shapes in a line of symmetry and complete a symmetric figure or pattern with respect to a specified line of symmetry. Draw polygons, specified by coordinates in the 4 quadrants: draw shapes following translation or reflection in the axes. 146 4NPV–1 Equivalence of 10 hundreds and 1 thousand Know that 10 hundreds are equivalent to 1 thousand, and that 1,000 is 10 times the size of 100; apply this to identify and work out how many 100s there are in other four-digit multiples of 100. 4NPV–1 Teaching guidance 4NPV–1 follows on from what children learnt in year 3 about the relationship between the units of 10 and 100 (see 3NPV–1). Pupils need to experience: what 1,000 items looks like making a unit of 1 thousand out of 10 units of 100, for example using 10 bundles of 100 straws to make 1,000, or using ten 100-value place-value counters Figure 103: ten 100-value place-value counters in a tens frame Language focus “10 hundreds is equal to 1 thousand.” Pupils must then be able to work out how many hundreds there are in other four-digit multiples of 100. Figure 104: eighteen 100-value place-value counters in 2 tens frames Language focus “18 hundreds is equal to 10 hundreds and 8 more hundreds.” “10 hundreds is equal to 1,000.” “So 18 hundreds is equal to 1,000 and 8 more hundreds, which is 1,800.” 147 The reasoning here can be described as grouping or repeated addition – pupils group or add 10 hundreds to make 1,000, then add another group of 8 hundreds. Pupils need to be able to apply this reasoning to measures contexts, as shown in the 4NPV–1 below. It is important for pupils to understand that there are hundreds within this new unit of a thousand, in different contexts. Pupils should be able to explain that numbers such as 1,800 and 3,000 are multiples of 100, because they are each equal to a whole number of hundreds. They should be able to identify multiples of 100 based on the fact that they have zeros in both the tens and ones places. As well as understanding 1,000 and other four-digit multiples of 100 in terms of grouping and repeated addition, pupils should be able to describe these numbers in terms of scaling by 10. Figure 105: place-value chart illustrating the scaling relationship between ones, tens, hundreds and thousands Language focus “1000 is 10 times the size of 100.” “1,800 is 10 times the size of 180.” 148 Making connections Learning to identify the number of hundreds in four-digit multiples of 100 should be connected to pupils’ understanding of multiplication and the grouping structure of division (2MD–1). Pupils should, for example, be able to represent 1,800 as 18 hundreds using the multiplication equations or , and be able to write the corresponding division equations and . Criterion 4MD–1 requires pupils to interpret the multiplication equations in terms of the scaling structure of multiplication, for example 1,800 is 100 times the size of 18. 4NPV–1 Example assessment questions How many 100g servings of rice are there in a 2,500g bag? One large desk costs a school £100. How much will 14 large desks cost? My school field is 100m long. How many times do I have to run its length to run 3km? My cup contains 100 ml of fizzy drink. The bottle contains 10 times as much. How many millilitres are there in the bottle? A rhino mother weighs about 1,000kg. She weighs about 10 times as much as her baby. What is the approximate weight of the baby rhino? Circle the lengths that could be made using 1 metre (100cm) sticks. 3,100cm 8,000cm 1,005cm 6,600cm 7,090cm 1,000cm 149 4NPV–2 Place value in four-digit numbers Recognise the place value of each digit in four-digit numbers, and compose and decompose four-digit numbers using standard and non-standard partitioning. 4NPV–2 Teaching guidance Pupils should be able to identify the place value of each digit in a four-digit number. They must be able to combine units of ones, tens, hundreds and thousands to compose four-digit numbers, and partition four-digit numbers into these units. Pupils need to experience variation in the order of presentation of the units, so that they understand that 40 + 300 + 2 + 5000 is equal to 5,342, not 4,325. Figure 106: 2 representations of the place-value composition of 5,342 Pupils also need to solve problems relating to subtraction of any single place-value part from the whole number, for example: 5,342 – 300 5,342 – 5,302 As well as being able to partition numbers in the ‘standard’ way (into individual place-value units), pupils must also be able to partition numbers in ‘non-standard’ ways, and carry out related addition and subtraction calculations, for example: Figure 107: partitioning 7,830 into 7,430 and 400 Figure 108: partitioning 5,050 into 2,000 and 3,050 150 4NPV–2 Example assessment questions Complete the calculations. 90 7 6,000 400 900 70 600 4 9 7,000 60 400 9,000 700 6 40 4,382 4,000 4,382 300 4,382 80 4,382 2 8,451 5,000 300 5,614 9,575 50 0 6,140 5,000 4 2,000 1 ,430 50 A football stadium can hold 6,430 people. So far 4,000 tickets have been sold for a match. How many tickets are left? On a field trip, the children need to walk 4,200m. So far they have walked 3km. How much further do they have to walk? Mr. Davis has 2 cats. One cat weighs 4,200g. The other cat weighs 3,050g. Their basket weighs 2kg. How much does the basket weigh with both cats inside it? 4NPV–3 Four-digit numbers in the linear number system Reason about the location of any four-digit number in the linear number system, including identifying the previous and next multiple of 1,000 and 100, and rounding to the nearest of each. 4NPV–3 Teaching guidance Pupils need to be able to identify or place four-digit numbers on marked number lines with a variety of scales. Pupils should also be able to estimate the value or position of four-digit numbers on unmarked numbers lines, using appropriate proportional reasoning. Pupils should apply this skill to taking approximate readings of scales in measures and statistics contexts, as shown in the Example assessment questions below. For more detail on identifying, placing and estimating positions of numbers on number lines, see year 2, 2NPV–2. 151 Pupils must also be able to identify which pair of multiples of 1000 or 100 a given four-digit number is between. To begin with, pupils can use a number line for support. In this example, for the number 8,681, pupils must identify the previous and next multiples of 1,000 and 100. Figure 109: using a number line to identify the previous and next multiple of 1,000 Figure 110: using a number line to identify the previous and next multiple of 100 Language focus “The previous multiple of 1,000 is 8,000. The next multiple of 1,000 is 9,000.” “The previous multiple of 100 is 8,600. The next multiple of 100 is 8,700.” Pupils need to be able to identify previous and next multiples of 1000 or 100 without the support of a number line. Pupils should then learn to round a given four-digit number to the nearest multiple of 1,000 by identifying the nearest of the pair of multiples of 1,000 that the number is between. Similarly, pupils should learn to round to the nearest multiple of 100. They should understand that they need to examine the digit in the place to the right of the multiple they are rounding to, for example when rounding to the nearest multiple of 1,000 pupils must examine the digit in the hundreds place. Again, pupils can initially use number lines for support, but should be able to round without that support by the end of year 4. Figure 111: identifying the nearest multiple of 1,000 with a number line for support 152 Language focus “The closest multiple of 1,000 is 9,000.” “8,681 rounded to the nearest thousand is 9,000.” Finally, pupils should also be able to count forwards and backwards from any four-digit number in steps of 1, 10 or 100. Pay particular attention to counting over ‘boundaries’, for example: 2,100, 2,000, 1,900 2,385, 2,395, 2,405 Making connections Here, pupils must apply their knowledge that 10 hundreds is equal to 1 thousand (see 4NPV–1 ) to understand that each interval of 1,000 on a number line or scale is made up of 10 intervals of 100. This also links to 4NPV–4 in which pupils need to be able to read scales divided into 2, 4, 5 and 10 equal parts. However, for the current criterion pupils are not expected to make precise placements, but instead approximate using proportional reasoning. 4NPV–3 Example assessment questions a. Which 2 numbers round to 5,600 when rounded to the nearest hundred? b. Round each number to the nearest thousand. c. Estimate the value of each number. 153 a. Estimate how much liquid is in the beaker. b. Estimate how much liquid needs to be added to make 1 litre. Estimate and mark the position of 600g on this scale. 154 The bar chart shows the number of red and blue cars that passed a school in one day. a. Estimate the number of red and blue cars that passed the school on this day. b. Estimate the number of blue cars that passed the school on this day. c. Add the following data for other coloured cars to the bar chart. Yellow Black White Number of cars 100 1,050 1,995 155 Fill in the missing numbers. 600 700 900 1 ,100 1 ,300 5,001 5,002 5,003 3,650 3,950 4,250 4,350 1 ,075 1 ,085 1 ,095 4NPV–4 Reading scales with 2, 4, 5 or 10 intervals Divide 1,000 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 1,000 with 2, 4, 5 and 10 equal parts. 4NPV–4 Teaching guidance By the end of year 4, pupils must be able to divide 1,000 into 2, 4, 5 or 10 equal parts. This is important because these are the intervals commonly found on measuring instruments and graph scales. Figure 112: bar models showing 1,000 partitioned into 2, 4, 5 and 10 equal parts Pupils should practise counting in multiples of 100, 200, 250, and 500 from 0, or from any multiple of these numbers, both forwards and backwards. This is an important step in becoming fluent with these number patterns. Pupils will have been practising counting in multiples of 1, 2 and 5 since year 1, and this supports counting in units of 100, 200 and 156 500. Pupils typically find counting in multiples of 250 the most challenging, because they only started to encounter this pattern in year 3, when counting in multiples of 25. Language focus “Twenty-five, fifty, seventy-five, one hundred” needs to be a fluent spoken language pattern. Fluency in this language pattern provides the basis to count in multiples of 250. Pupils should be able to apply this skip counting, beyond 1,000, to solve contextual multiplication and division measures problems, as shown in the 4NPV–4 below (questions 5 and 7). Pupils should also be able to write and solve multiplication and division equations related to multiples of 100, 200, 250 and 500 up to 1,000. Pupils need to be able to solve addition and subtraction problems based on partitioning 1,000 into multiples of 100, 200 and 500 based on known number bonds to 10. Pupils should also have automatic recall of the fact that 250 and 750 are bonds to 1,000. They should be able to immediately answer a question such as “I have 1 litre of water and pour out 250ml. How much is left?” Making connections 4MD–2 requires pupils to manipulate multiplication and division equations. They should therefore be able to write and manipulate multiplication and division equations related to the composition of 1,000 as discussed here. Dividing 1,000 into 10 equal parts is also assessed as part of 4NPV–1. Reading scales builds on number line knowledge from 4NPV–3 . Conversely, experience of working with scales with 2, 4, 5 or 10 divisions in this criterion improves pupils’ estimating skills when working with unmarked number lines and scales as described in 4NPV–3. 157 4NPV–4 Example assessment questions Fill in the missing numbers. 3,000 4,000 4,500 5,000 2,400 3,000 3,200 1 ,500 2,000 2,750 What is the reading on each of these scales? The beaker contains 1 litre of water. If I pour out 600ml, how much is left? Mark the new water level on the picture. A motorway repair team can build 250m of motorway barrier in 1 day. In 5 working days, how many metres of motorway barrier can can they build? How many 500ml bottles can I fill from a 3 litre container of water? 158 The pictogram shows how many cans a class recycled in 2020. How many cans did the class recycle in 2020? 1kg of strawberries is shared equally between 5 people. How many grams of strawberries do they each get? I have already swum 750m. How much further do I need to go to swim 2km? Fill in the missing parts, and write as many different equations as you can think of to describe the structure. 159 The bar charts show the number of red and blue cars that passed 3 different schools on a given day. How many red and blue cars passed each school? School A School B School C Fill in the missing numbers. 1 ,000 4 200 1 ,000 1 ,000 500 250 1 ,000 160 4NF–1 Recall of multiplication tables Recall multiplication and division facts up to 12 12 , and recognise products in multiplication tables as multiples of the corresponding number. 4NF–1 Teaching guidance The national curriculum requires pupils to recall multiplication table facts up to 12 12 , and this is assessed in the multiplication tables check. For pupils who do not have automatic recall of all of the facts by the time of the check, fluency in facts up to 9 9 should be prioritised in the remaining part of year 4. The facts to 9 9 are particularly important for progression to year 5, because they are required for formal written multiplication and division. The 36 multiplication facts that are required for formal written multiplication are as follows. During application of formal written multiplication, pupils may also need to multiply a one-digit number by 1. Multiplication of the numbers 1 to 9 by 1 are not listed here because these calculations do not need to be recalled in the same way. While pupils are learning the individual multiplication tables, they should also learn that: the factors can be written in either order and the product remains the same (for example, we can write 3 4 12 or 4 3 12 to represent the third fact in the 4 multiplication table) the products within each multiplication table are multiples of the corresponding number, and be able to recognise multiples (for example, pupils should recognise, 64 is a multiple of 8, but that 68 is not) adjacent multiples in, for example, the 8 multiplication table, have a difference of 8 161 Figure 113: number line and array showing that adjacent multiples of 8 (32 and 40) have a difference of 8 Language focus When pupils commit multiplication table facts to memory, they do so using a verbal sound pattern to associate the 3 relevant numbers, for example, “nine sevens are sixty-three”. It is important to provide opportunities for pupils to verbalise each multiplication fact as part of the process of developing fluency. It is useful for pupils to learn the multiplication tables in the following order/groups: 10 then 5 multiplication tables 2, 4 and 8 multiplication tables one after the other 3, 6, and 9 multiplication tables one after the other 7 multiplication table 11 and 12 muliplication tables The connections and patterns will help pupils to develop fluency and understanding. Pupils must also be able to apply their automatic recall of multiplication table facts to solve division problems, for example, solving 28 7 , by recalling that 28 4 7 . You can find out more about developing automatic recall of multiplication tables here in the calculation and fluency section: 4NF–1 162 Making connections Solving division problems with remainders (4NF–2 ), relies on automatic recall of multiplication facts. Criterion 4MD–2 involves linking individual multiplication facts to related multiplication and division facts. Once pupils have automatic recall of the multiplication table facts, they then have access to a whole set of related facts. For example, if pupils know that , they also know that , and . Converting mixed numbers to improper fractions (4F–2 ), also relies on automatic recall of multiplication facts. For example, converting to an improper fraction involves calculating 3 times 6 sixths plus 1 more sixth, so requires knowledge of the multiplication fact . Efficiently calculating the perimeter of a regular polygon (4G–2 ), or finding the side-length of a regular polygon, given the perimeter, depends on recall of multiplication and division facts. 4NF–1 Example assessment questions A regular hexagon has sides of 7cm. What is its perimeter? A regular octagon has a perimeter of 72cm. What is the length of each of the sides? It takes Latoya 8 minutes to walk to school. It takes Tatsuo 3 times as long. How long does it take Tatsuo to walk to school? An egg box contains 6 eggs. I need 54 eggs. How many boxes should I buy? 8 children spend a day washing cars and earn £40 altogether. If they share the money equally how much do they each get? Circle the numbers that are multiples of 3. 16 18 23 9 24 163 Assessment guidance: The multiplication tables check will assess pupils’ fluency. Once pupils can automatically recall multiplication facts, they should be able to apply their knowledge to questions like those shown here. 4NF–2 Division problems with remainders Solve division problems, with two-digit dividends and one-digit divisors, that involve remainders, for example: 74 9 8 r 2 and interpret remainders appropriately according to the context. 4NF–2 Teaching guidance Pupils should recognise that a remainder arises when there is something ‘left over’ in a division calculation. Pupils should recognise and understand why remainders only occur when the dividend is not a multiple of the divisor. This can be achieved by discussing the patterns seen when the dividend is incrementally increased by 1 while the divisor is kept the same. Total number of apples (dividend) Number of apples in each tray (divisor) Number of trays (quotient) Number of apples left over (remainder) Equation 12 4 3 0 13 4 3 1 14 4 3 2 15 4 3 3 16 4 4 0 17 4 4 1 18 4 4 2 19 4 4 3 20 4 5 0 164 A common mistake made by pupils is not making the maximum number of groups possible, for example: 17 4 3 r 5 (incorrect) The table above can be used to help pupils recognise and understand that the remainder is always smaller than the divisor. Note that when pupils use the short division algorithm in year 5, if they ‘carry over’ remainders that are larger than the divisor, the algorithm will not work. Language focus “If the dividend is a multiple of the divisor there is no remainder.” “If the dividend is not a multiple of the divisor, there is a remainder.” “The remainder is always less than the divisor.” Once pupils can correctly perform division calculations that involve remainders, they need to recognise that, when solving contextual division problems, the answer to the division calculation must be interpreted carefully to determine how to make sense of the remainder. The answer to the calculation is not always the answer to the contextual problem. Consider the following context: Four scouts can fit in each tent. How many tents will be needed for thirty scouts? Figure 114: pictorial representation and counters: with 30 scouts and 4 per tent, 7 tents are insufficient 30 4 7 r 2 Pupils may simply say that “7 remainder 2 tents are needed”, but this does not answer the question. Pupils should identify what each number in the equation represents to help them correctly interpret the result of the calculation in context. 165 Language focus “The 30 represents the total number of scouts.” “The 4 represents the number of scouts in each tent.” “The 7 represents the number of full tents.” “The 2 represents the number of scouts left over.” “We need another tent for the 2 left-over scouts. 8 tents are needed.” Figure 115: pictorial representation and counters: with 30 scouts and 4 per tent, 7 tents are insufficient Making connections Pupils must have automatic recall of multiplication facts and related division facts, and be able to recognise multiples (4NF–1 ) before they can solve division problems with remainders. For example, to calculate , pupils need to be able to identify the largest multiple of 7 that is less than 55 (in this case 49). They must then recall how many sevens there are in 49, and calculate the remainder. Converting improper fractions to mixed numbers (4F–2 ) relies on solving division problems with remainders. For example, converting to a mixed number depends on the calculation . 4NF–2 Example assessment questions Which of these division calculations have the answer of 3 r 2? 23 7 17 5 32 6 7 2 14 4 30 8 I have 60 metres of bunting for the school fair. What length of bunting will be left over if I cut it into lengths of 8 metres? 166 It takes 7 minutes to make a pom-pom. How many complete pom-poms can Malik make in 30 minutes? 23 apples are shared equally between 4 children. How many whole apples does each child get? Ruby writes: 37 5 6 r 7 Explain what mistake Ruby has made, and write the correct answer. Decide whether each calculation has a remainder or not. Explain how you can do this without doing each calculation? Has a remainder?(Yes or No) 4NF–3 Scaling number facts by 100 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 100), for example: 8 6 14 and 14 6 8 so 800 600 1 ,400 and 1 ,400 600 800 3 4 12 and 12 4 3 so 300 4 1 ,200 and 1 ,200 4 300 4NF–3 Teaching guidance Pupils should begin year 4 with automatic recall of addition and subtraction facts within 20 (3NF–1). By the end of year 4, pupils should be able to recall all multiplication table facts and related division facts. To be ready to progress to year 5, pupils must also be able to combine these facts with unitising in hundreds, including: scaling known additive facts within 10, for example, 900 600 300 scaling known additive facts that bridge 10, for example, 800 600 1 ,400 167 scaling known multiplication tables facts, for example, 300 4 1 ,200 scaling division facts derived from multiplication tables, for example, 1 ,200 4 300 For calculations such as 800 + 600 = 1,400, pupils can begin by using tens frames and counters as they did for calculation across 10 (2AS–1) and across 100 (3NF–3 ), but now using 100-value counters. Figure 116: tens frames with 100-value counters showing 800 + 600 = 1,400 8 + 6 14 800 + 600 1,400 14 – 6 8 1 ,400 – 600 800 14 – 8 6 1 ,400 – 800 600 Similarly, pupils can use 100-value counters to understand how a known multiplicative fact, such as 3 5 15 , relates to a scaled calculation, such as 300 5 1 ,500 . Pupils should be able reason in terms of unitising in hundreds, or in terms of scaling a factor by 100. 168 Figure 117: 3-by-5 array of 100-value place-value counters 3 5 15 3 500 1 ,500 Language focus “3 times 5 is equal to 15.” “3 times 5 hundreds is equal to 15 hundreds.” “15 hundreds is equal to 1,500.” 3 5 15 300 5 1 ,500 Language focus “3 times 5 is equal to 15.” “3 hundreds times 5 is equal to 15 hundreds.” “15 hundreds is equal to 1,500.” Language focus “If I multiply one factor by 100, I must multiply the product by 100.” Pupils must be able to make similar connections for known division facts. 15 3 5 1 ,500 300 5 1 ,500 3 500 Language focus “If I multiply the dividend by 100 and the divisor by 100, the quotient remains the same.” “If I multiply the dividend by 100 and keep the divisor the same, I must multiply the quotient by 100.” 169 It is important for pupils to understand all of the calculations in this criterion in terms of working with units of 100, or scaling by 100. You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 4NPV–2 and 4NF–3 Making connections This criterion builds on pupils’ additive fluency and also on: 4NPV–1 , where pupils need to be able to work out how many hundreds there are in any four-digit multiple of 100 4NF–1 , where pupils develop fluency in multiplication and division facts Meeting this criterion also requires pupils to be able to fluently multiply whole numbers by 100 (4MD–1). 4NF–3 Example assessment questions I need 1kg of flour to make some bread. I have 800g. How many more grams of flour do I need? A builder can buy bricks in pallets of 600. How many pallets should she buy if she needs 1,800 bricks? Dexter ran round a 400m running track 6 times. How far did he run? I mix 700ml of orange juice and 600ml of lemonade to make a fruit drink for a party. What volume of fruit drink have I made in total? A farmer had 1,200m of fencing to put up round his fields. He put up the same amount of fencing each day, and it took him 6 days to put up all the fencing. How many metres of fencing did he put up each day? Fill in the missing numbers. 300 1 ,100 4,200 600 170 4MD–1 Multiplying and dividing by 10 and 100 Multiply and divide whole numbers by 10 and 100 (keeping to whole number quotients); understand this as equivalent to making a number 10 or 100 times the size. 4MD–1 Teaching guidance As well as being able to calculate multiplication and division by 10 and 100, pupils need to start to think about multiplication as scaling, so that they can conceive of dividing by 10 and 100 when there are decimal answers in year 5 (5NF–2). If pupils only understand multiplying by 10 or 100 in terms of the repeated addition/grouping structure of multiplication (for example, 23 100 is 100 groups of 23), they will struggle to conceptualise 23 100 . However, if pupils understand 23 100 as ‘23, made one-hundred times the size’, the inverse of this, 23 100 , can be thought of as ‘23 made one-hundredth times the size’. To meet criteria 4MD–1, pupils should be able to use and understand the language of 10 or 100 times the size, and understand division as the inverse action; in year 5, once pupils have learnt about tenths and hundredths they should apply the language to division by 10 and 100 (1 tenth or 1 hundredth times the size). Language focus “23, made 100 times the size, is 2,300.” “23 multiplied by 100 is equal to 2,300.” “First we had 23 ones. Now we have 23 hundreds.” “1,450 is 10 times the size of 145.” “1,450 divided by 10 is equal to 145.” “First we had 145 tens. Now we have 145 ones.” Pupils know that 1,000 is 10 times the size of 100 (4NPV–1 ), and that 100 is 10 times the size of 10 (3NPV–1). Pupils should now extend this ‘10 times the size’ relationship to other numbers, beginning with those with 1 significant figure. The Gattegno chart can be used to help pupils see, for example, that 80, made 10 times the size is 800: pupils can move their finger or a counter up from 80 to 800. They should connect this action to multiplication by 10, and be able to solve/write the corresponding multiplication calculation (80 10 800 ). Similarly, because 80 is 10 times the size of 8, they can solve 80 10 8 , moving their finger or a counter down from 80 to 8. 171 Figure 118: using the Gattegno chart to multiply and divide by 10 Pupils may also work with place-value charts. Figure 119: using a place-value chart to multiply and divide by 100 Repeated association of the written form (for example, 10 ) and the verbal form (“ten times the size”) will help pupils become fluent with the links. Pupils should also be able to relate multiplying by 10 or by 100 with the idea of multiplying a quantity of items – here they can use the verbal form “ten times as many”, for example, “200 pencils is 10 times as many as 20 pencils.” Both the Gattegno chart and the place-value chart also help pupils to see that multiplying by 100 is equivalent to multiplying by 10, and then multiplying by 10 again (and that dividing by 100 is equivalent to dividing by 10 and dividing by 10 again). The same representations can be used to extend to numbers with more than one significant digit. 13 10 130 130 10 13 130 10 1 ,300 1 ,300 10 130 13 100 1 ,300 1 ,300 100 13 172 Making connections In 3NPV–1 and 4NPV–1 , respectively, children learnt that 100 is 10 times the size of 10, and 1,000 is 10 times the size of 100. Here they applied this idea to scaling other numbers. Until 4MD–1, pupils understood, for example, to represent 23 groups of 100. Here pupils must relate the same equation to a completely different structure – the scaling of 23 by 100 (making 23 one hundred times the size). Pupils learn more about how one equation can represent different multiplicative structures in 4MD–2. 4MD–1 Example assessment questions Fill in the missing numbers. 100 11 100 100 3,500 100 10 14 10 10 250 10 Bethany has 15 marbles. Nasir has 100 times as many. How many marbles does Nasir have? Sumaya’s walk from her home to school is 130m. Millie’s walk is 10 times as far. How far does Millie walk to get to school? Fill in the missing numbers. 100 600 1 ,500 10 100 8 1 ,200 10 173 4MD–2 Manipulating the multiplicative relationship Manipulate multiplication and division equations, and understand and apply the commutative property of multiplication. 4MD–2 Teaching guidance Pupils will begin year 4 with an understanding of some of the individual concepts covered in this criterion, but they need to leave year 4 with a coherent understanding of multiplicative relationships, and how multiplication and division equations relate to the various multiplicative structures. Pupils need to be able to apply the commutativite property of multiplication in 2 different ways. The first can be summarised as ‘1 interpretation, 2 equations’. Here pupils must understand that 2 different equations can correspond to one context, for example, 2 groups of 3 is equal to 6 can be represented by 2 3 6 and by 3 2 6 . Spoken language can support this understanding. Language focus “2 groups of 3 is equal to 6.” “3, two times is equal to 6.” “2 groups of 3 is equal to 3, two times.” The second way that pupils must understand commutativity, can be summarised as ‘one equation, two interpretations’. Here, pupils must understand that a single equation, such as 2 7 14 , can be interpreted in two ways. Language focus “2 groups of 7 is equal to 14.” “7 groups of 2 is equal to 14.” “2 groups of 7 is equal to 7 groups of 2.” Pupils should understand that both interpretations correspond to the same total quantity (product). 174 Language focus “factor times factor is equal to product” “The order of the factors does not alter the product.” An array is an effective way to illustrate this. Figure 120: using an array to show that 7 groups of 2 and 2 groups of 7 both correspond to the same total quantity 175 Pupils must be able to describe what each number in the equation represents for the 2 different interpretations, in context. 7 2 14 2 7 14 Interpretation 1 Interpretation 2 Figure 121: 7 groups of 2 – 7 nests of 2 eggs and seven 2-value counters Language focus “The 2 represents number of eggs in each nest/group”. “The 7 represents the number of nests/groups.” “The 14 represents the total number of eggs/product.” Figure 122: 2 groups of 7 – 2 nests of 7 eggs and two 7-value counters Language focus “The 2 represents the number of nests/groups.” “The 7 represents the number of eggs in each nest/group.” “The 14 represents the total number of eggs/product.” Pupils should be able to bring together these ideas to understand that either of a pair of multiplication equations can have two different interpretations. Figure 123: schematic diagram summarising the commutative property of multiplication and the different grouping interpretations Pupils must understand that, because division is the inverse of multiplication, any multiplication equation can be rearranged to give division equations. The value of the product in the multiplication equation becomes the value of the dividend in the corresponding division equations. 2 7 14 7 2 14 14 2 7 14 7 2 176 This means that the commutative property of multiplication has a related property for division. Language focus “If we swap the values of the divisor and quotient, the dividend remains the same.” As with multiplication, any division equation can be interpreted in two different ways, and these correspond to quotitive and partitive division. 14 7 2 Partitive division Quotitive division Figure 121: 7 groups of 2 – 7 nests of 2 eggs and seven 2-value counters Language focus “14 shared between 7 is equal to 2.” “The 14 represents the total number of eggs.” “The 7 represents the number of nests/shares.” “The 2 represents the number of eggs in each nest/share.” Figure 122: 2 groups of 7 – 2 nests of 7 eggs and two 7-value counters Language focus “14 divided into groups of 7 is equal to 2. “The 14 represents the total number of eggs.” “The 7 represents the number of eggs in each nest/group.” “The 2 represents the number of nests/groups.” Pupils need to be able to fluently move between the different equations in a set, and understand how one known fact (such as 7 2 14 ) allows them to solve 4 different calculations each with two possible interpretations. You can find out more about fluency in manipulating multiplication and division equations here in the calculation and fluency section: 4MD–2 177 Making connections Being able to move between the grouping and sharing structures of division supports calculation. For example, irrespective of the calculation context, pupils may find it more helpful to think of the sharing (partitive) structure when calculating (1,600 shared or partitioned into 2 equal shares/parts is 800), rather than thinking about the grouping structure (800 twos in 1,600). Conversely, pupils may find it more helpful to think of in terms of the grouping (quotitive) structure (how many groups of 25 there are in 200) rather than thinking about the sharing structure (200 shared or partitioned into 25 equal shares). 4MD–2 Example assessment questions Using pictures of vases of flowers, draw two pictures which can be represented by the equation 5 4 20 . Write as many multiplication and division equations as you can to represent each picture. a. b. Write a story that could be represented by this equation 3 7 21 . Using pictures of apples in bowls, draw 2 pictures which can be represented by the equation 18 3 6 . Use 15 16 240 to write 3 other related multiplication and division equations. 45kg of animal feed is shared between some horses. They each get 5kg. How many horses were there? 1m 40cm of ribbon was cut into equal pieces. Each piece is 14cm long. How many pieces of ribbon are there? Fill in the missing numbers. 20 5 3,000 250 100 5,400 178 4MD–3 The distributive property of multiplication Understand and apply the distributive property of multiplication. 4MD–3 Teaching guidance The first place that pupils will have encountered the distributive law is within the multiplication tables themselves. Pupils have seen, for example, that adjacent multiples in the 6 times table have a difference of 6. Number lines and arrays can be used to illustrate this. Figure 124: number line and array showing that adjacent multiples of 6 (24 and 30) have a difference of 6 Pupils should be able to represent such relationships using mixed operation equations, for example: 5 6 4 6 6 or 5 6 4 6 1 6 4 6 5 6 6 or 4 6 5 6 1 6 Pupils should learn that multiplication takes precedence over addition. They should then extend this understanding beyond the multiplication tables, for example, if they are given the equation 20 6 120 , they should be able to determine that 21 6 126 , or vice versa. Pupils also need to be able to apply the distributive property to non-adjacent multiples. The array chart used to show the connection between 4 6 and 5 6 can be adapted to show the connection between 5 6 , 3 6 and 2 6 . 179 Figure 125: Array showing the relationship between , and As for adjacent multiples, pupils should be able to use mixed operation equations to represent the relationships: 5 6 3 6 2 6 3 6 5 6 2 6 2 6 5 6 3 6 Again, pupils should understand that multiplication takes precedence: the multiplications are calculated first, and then the products are added or subtracted. Pupils can use language patterns to support their reasoning. Language focus “5 is equal to 3 plus 2, so 5 times 6 is equal to 3 times 6 plus 2 times 6.” This illustrates the distributive property of multiplication: ( ) a b c a b a c and ( ) a b c a b a c Note that the examples of adjacent multiples above are simply a special case of this in which b or c is equal to 1. Pupils should then use the distributive property and known multiplication table facts to multiply 2-digit numbers (above 12) by one-digit numbers. 14 3 , for example, can be calculated by relating it to 10 3 and 4 3 : 180 Figure 126: array showing the relationship between , and 14 3 10 3 4 3 30 12 42 Pupils should recognise that factors can be partitioned in ways other than into ‘10 and a bit’. For example 14 3 could also be related to double 7 3 . Figure 127: array showing that is double 14 3 7 3 7 3 21 21 42 Pupils should be expected to extend their understanding of the distributive law to support division. For example, they should be able to connect a calculation such as 65 5 to known multiplication and division facts: “we have 3 more fives than 10 fives” or “we have 1 more 5 than 12 fives”. Making connections Pupils need to be fluent in multiplication tables to (4NF–1 ) and be able to multiply by 10 (4MD–1) to be able to efficiently apply the distributive property. Mastery of this criterion supports fluency in the 11 and 12 multiplication tables, since multiplication by 11 or 12 can be derived from multiplication by 10 and by 1 or 2, using the distributive property. The formal written methods of multiplication also depend upon the distributive property of multiplication. 181 4MD–3 Example assessment questions I had 6 flowers, each with 8 petals. If I get one more flower, how many petals do I have altogether? Jordan buys 10 packs of soft drinks for a party. Each pack contains 6 cans. One pack is lemonade and the rest are cola. How many cans of cola are there? I have a 65cm length of string. How many 5cm lengths can I make from it? Fill in the missing symbols (<, > or =). 4 6 5 6 6 3 6 3 4 6 3 6 6 4 6 6 4 4 6 5 6 4 6 6 4 5 6 Fill in the missing numbers. 4 7 3 7 16 4 10 4 4 16 4 8 4 4 A box of chocolates costs £7. How much do 16 boxes cost? Felicity is putting flowers into bunches of 5. So far she has made 4 bunches. She has 15 more flowers. a. How many bunches will she make altogether? b. How many flowers does she have altogether? 3 72 216 Explain how you can use this fact to calculate: a. 3 73 b. 4 72 182 4F–1 Mixed numbers in the linear number system Reason about the location of mixed numbers in the linear number system. 4F–1 Teaching guidance Sometimes pupils get quite far in their maths education only thinking of a fraction as a part of a whole, rather than as a number in its own right. In year 3, pupils learnt about the location in the linear number system of fractions between 0 and 1 (3F–2). For pupils to be able to add and subtract fractions across 1, or those greater than 1 (4F–3 ), they need to understand how mixed numbers fit into the linear number system. Pupils should develop fluency counting in multiples of unit fractions, using number lines for support. Counting draws attention to the equivalence of, for example, 4 4 1 and 2, or 5 5 2 and 3. Pupils should practise counting both forwards and backwards. Figure 128: number line to support counting in multiples of one quarter Figure 129: number line to support counting in multiples of one fifth Pupils should then learn to label marked number lines, extending beyond 1. A common mistake that pupils make is to count the number of marks between labelled intervals, rather than the number of parts. For example, on the number line below they may count 3 marks and incorrectly deduce that the number line is marked in thirds. Figure 130: labelling a number line marked in quarters 183 Language focus “Each interval is divided into 4 equal parts, so we count in quarters.” Pupils should also be able to estimate the value or position of mixed numbers on number lines which do not have fractional marks. Pupils must understand that it is not the absolute size of the numerator and denominator which determine the value of the fractional part of the mixed number, but the relationship between the numerator and denominator (whether the numerator is a large or small part of the denominator). Pupils need to be able to reason, for example, that 7 8 1 is close to 2, but that 7 30 1 is closer to 1. Figure 131: placing a mixed number on a number line with no fractional marks Pupils must also be able to identify the previous and next whole number, and will then be able to round to the nearest whole number, which further supports estimation and approximation. Language focus “ is between 1 and 2.” “The previous whole number is 1.” “The next whole number is 2.” Making connections Having a visual image of mixed numbers in the linear number system helps pupils to add and subtract fractions with the same denominator (including mixed numbers), for example (4F–3 ). It also supports comparison of mixed numbers, and reading scales. 184 4F–1 Example assessment questions Add labels to each mark on the number lines. What are the values of a, b, c and d? Estimate the position of the following numbers on the number line. 2 9 2 2 3 3 7 3 1 5 1 How much water is in the beaker? Write your answer as a mixed number. Circle the larger number in each of these pairs. Explain your reasoning. 3 9 3 8 9 3 1 3 4 1 8 4 1 3 2 2 3 1 185 4F–2 Convert between mixed numbers and improper fractions Convert mixed numbers to improper fractions and vice versa. 4F–2 Teaching guidance It should be noted that this criterion covers content that is in year 5 of the national curriculum. It has been included here, in year 4, to improve coherence. Pupils have already learnt that fractions where the numerator is equal to the denominator have a value of 1. This knowledge should be extended to other integers. For example, if we know that 5 5 1 , we can repeat groups of 5 5 to see that 10 5 2 and 15 5 3 and so on. Figure 132: number line showing integers expressed as fractions Language focus “When the numerator is a multiple of the denominator, the fraction is equivalent to a whole number.” Pupils can then learn to express mixed numbers as improper fractions. Figure 133: number line showing mixed number–improper fraction equivalence To convert from mixed numbers to improper fractions, pupils should use their multiplication tables facts to find the improper-fraction equivalent to the integer part of the mixed number, and then add on the remaining fractional part. Figure 134: bar model showing mixed number–improper fraction equivalence 186 Language focus “There are 2 groups of five-fifths, which is 10 fifths and 3 more fifths. This is 13 fifths.” Pupils should then learn to convert in the other direction – from improper fractions to mixed numbers. Pupils need to understand the improper fraction in terms of a multiple of a unit fraction. For example, they should be able to see 21 8 as 21 one-eighths. Pupils can initially use unit-fraction counters (here, 1 8 value counters) to represent this. Conversion to a mixed number then builds on division with remainders (4NF–2 ). In this example, since 8 one-eighths is equal to 1 whole, pupils must find how many wholes can be made from the 21 eighths, and how many eighths are left over. Figure 135: using unit fraction counters to support conversion from an improper fraction to a mixed number Language focus “We have 21 eighths. 8 eighths is equal to 1 (whole).” “21 eighths is equal to 2 groups of 8 eighths, and 5 more eighths. This is 2 and 5 eighths.” Once pupils have an understanding of the conversion process, they should not need to use counters or other similar manipulatives or drawings. 187 Making connections For pupils to succeed with this criterion, they must first be fluent in multiplication facts (4NF–1 ) and division with remainders (4NF–2 ). Converting, for example, to an improper fraction involves calculating (4NF–1). The reverse process, (converting to a mixed number) involves calculating (4NF–2). Converting between mixed numbers and improper fractions also supports efficient addition and subtraction of fractions with the same denominator (4F–3 ). Being able to move easily between the two will allow pupils to choose the most efficient calculation approach. 4F–2 Example assessment questions Which of these fractions are equivalent to a whole number? Explain how you know. 48 6 48 7 48 8 48 9 48 10 Express the following mixed numbers as improper fractions. 1 8 4 4 9 6 11 12 3 2 3 8 Express the following improper fractions as mixed numbers. 17 2 13 6 28 10 41 7 Sarah wants to convert 17 4 to a mixed number. She writes: 17 5 4 4 3 Explain what mistake Sarah has made, and write the correct answer. The school kitchen has 17 packs of butter. Each pack weighs 1 4 kg. How many kilograms of butter do they have altogether? Express your answer as a mixed number. I have a 1 2 6 m length of string. How many 1 2 m lengths can I cut? 188 4F–3 Add and subtract improper and mixed fractions (same denominator) Add and subtract improper and mixed fractions with the same denominator, including bridging whole numbers, for example: 7 4 11 5 5 5 7 2 5 8 8 8 3 3 2 4 1 5 5 5 7 8 1 4 2 5 5 5 8 7 4F–3 Teaching guidance To meet this criterion, pupils must be able to carry out the following types of calculation (with a common denominator): add and subtract 2 improper fractions add and subtract a proper fraction to/from a mixed number add and subtract one mixed number to/from another In year 3, pupils used their understanding that a non-unit fraction is a multiple of a unit fraction, for example, 2 5 is 2 lots of 1 5 .This allowed them to reason that 2 2 4 5 5 5 , because 2 lots of 1 5 plus 2 lots of 1 5 is equal to 4 lots of 1 5 , or 4 5 . Pupils should now apply this strategy to addition and subtraction involving improper fractions. Language focus “7 one-fifths plus 4 one-fifths is equal to 11 one-fifths.” Figure 136: bar model showing addition of improper fractions with the same denominator 189 The generalisation that pupils learnt for adding and subtracting fractions with the same denominator within 1 whole therefore also extends to improper fractions. Language focus “When adding fractions with the same denominators, just add the numerators.” “When subtracting fractions with the same denominators, just subtract the numerators.” When adding and subtracting mixed numbers, pupils can convert them to improper fractions and apply this generalisation. This is sometimes a good approach. However this isn’t always necessary, or the most efficient choice, and pupils should learn to use their knowledge of the composition of mixed numbers to add and subtract. Figure 137: number line and pie model showing subtraction of a fraction from a mixed number (same denominator) Pupils must also be able to add and subtract across a whole number, for example: 2 4 1 5 5 5 7 8 1 4 2 5 5 5 8 7 Addition strategies include: adding to reach the whole number, then adding the remaining fraction Figure 138: number line showing strategy for adding across a whole number 190 adding the fractional parts first to give an improper fraction, which is then converted to a mixed number and added Figure 139: pie model showing strategy for adding across a whole number Before pupils attempt to subtract across a whole number (for example, 1 4 5 5 8 ), they should first learn to subtract a proper fraction from a whole number (for example, 3 5 8 ). Pupils can find this challenging, so number lines and bar or pie models can be useful. Once pupils can do this, they should be able to subtract across the whole number by using: a ‘subtracting through the whole number’ strategy (partitioning the subtrahend) – part of the subtrahend is subtracted to reach the whole number, then the rest of the subtrahend is subtracted from the whole number or a ‘subtracting from the whole number’ strategy (partitioning the minuend) – the subtrahend is subtracted from the whole number part of the minuend, then the remaining fractional part of the minuend is added Subtracting through the whole number 1 4 1 1 3 5 5 5 5 5 3 5 2 5 8 8 8 7 Subtracting from the whole number 1 4 4 1 5 5 5 5 1 1 5 5 2 5 8 8 7 7 Finally, pupils need to be able add one mixed number to another, and subtract one mixed number from another. 2 3 5 9 9 9 4 1 5 5 3 2 9 9 9 5 1 4 Pupils should initially solve problems where addition or subtraction of the fractional parts does not bridge a whole number. Both pie models, as above, and partitioning diagrams 191 can be used to support addition, by partitioning both of the mixed numbers into their whole number and fractional parts. Pupils should not extend the method for addition – partitioning both mixed fractions – to subtraction of one mixed number from another. Pupils should instead use a similar strategy to that used for subtraction of one two-digit number from another (2AS–2), only partitioning the subtrahend. Figure 140: pie model showing strategy for subtracting across a whole number This is important when subtraction of the fractional parts bridges a whole number, to avoid pupils writing a calculation that involves negative fractions. Making connections In 4F–2 , pupils learnt to convert between improper fractions and mixed numbers. One strategy for adding and subtracting mixed numbers is to convert them to improper fractions before adding/subtracting and then to convert the answer back again. Pupils also need to draw on these conversions when addition of mixed numbers results in an improper fraction. For example, pupils should be expected to convert an answer such as to . In 4F–1, pupils learnt to position mixed numbers on a number line. This supports bridging through a whole number, for example understanding that . Some of the strategies covered in this criterion are analogous to those that pupils have used for whole numbers. The 2 strategies presented for adding and subtracting across a whole number mirror those for adding and subtracting across 10, which pupils learnt in year 2 (2NF–1). The strategies presented for adding and subtracting two mixed numbers mirror those for adding and subtracting 2 two-digit numbers (2AS–2) – for addition, partition both addends; and for subtraction, partition only the subtrahend. If pupils can see these connections, they will be able to calculate more confidently with fractions, seeing them as just another type of number. 192 4F–3 Example assessment questions It is a 3 4 2 km cycle ride to my friend’s house, and a further 3 4 kmride to the park. How far do I have to cycle altogether? I have 5m of rope. I cut off 4 10 m. How much rope is left? Fill in the missing numbers. 4 4 4 7 7 7 1 4 6 7 7 7 2 2 3 The table below shows the number of hours Josie read each day during a school week. For how long did Josie read altogether? MonTuesWedThursFrihours1hourhourshourshours A tailor has 7 10 3 m of ribbon. She uses 9 10 1 m to complete a dress. How much ribbon is left? 4G–1 Draw polygons specified by coordinates or by translation Draw polygons, specified by coordinates in the first quadrant, and translate within the first quadrant. 4G–1 Teaching guidance Pupils should already be adept at placing markings at specific points, and joining these accurately with a ruler to draw a polygon (3G–1). Pupils can begin by describing translations of polygons drawn on squared paper, by counting how many units to the left/right and up/down the polygon has been translated. 193 Figure 141: translation of a polygon on a square grid Language focus “The polygon has been translated 4 squares to the right and 3 squares down.” Pupils should then learn to translate polygons on squared paper according to instructions that describe how many units to move the polygon to the left/right and up/down. Pupils can translate each point of the polygon individually, for example, translating each point right 4 and down 3 to mark the new points, and then joining them. Alternatively, pupils can translate and mark one point, then mark the other points of the polygon relative to the translated point. In year 4, pupils must start to use coordinate geometry, beginning with the first quadrant. Initially, pupils can work with axes with no number labels, marking specified points as a translation from the origin, described as above. For example, “Start at the origin and mark a point along 5, and up 4.” Figure 142: marking a point relative to the origin (O) on a grid with no number labels 194 Finally, pupils should learn to use coordinate notation with number labels on the axes. They must be able to mark the position of points specified by coordinates, and write coordinates for already marked points. Figure 143: marking a point on a grid with number labels When pupils first start to mark points, they should still start at the origin, moving along and then up as specified by the coordinates. If they do not do this, they are likely to place a point such as (5, 4) by just looking for a 5 and a 4, and possibly end up placing the point at (4, 5). Language focus “First count along the x-axis, then count along the y-axis.” Pupils should then be able to draw polygons by marking and joining specified coordinates. Making connections In 4NPV–3 , 4NPV–4 and 4F–1, and in previous year groups, pupils learnt to place or identify specified points (numbers) on a number line or scale. In this criterion children learn to place or identify specified points with reference to 2 number lines. 195 4G–1 Example assessment questions Translate the quadrilateral so that point A moves to point B. A kite has been translated from position A to position B. Describe the translation. Mark the points, and join them to make a square. (3,1) (2,4) (5,5) (6,2) 196 This triangle is translated so that point A moves to (4, 3). Draw the shape in its new position. Mark the following points, and join them to make a polygon. (5, 0) (3, 1) (5, 2) (7, 1) a. What is the name of the polygon that you have drawn? b. Translate the polygon you have just drawn left 2 and up 3. What are the coordinates of the vertices of this new polygon? 197 4G–2 Perimeter: regular and irregular polygons Identify regular polygons, including equilateral triangles and squares, as those in which the side-lengths are equal and the angles are equal. Find the perimeter of regular and irregular polygons. 4G–2 Teaching guidance Pupils must be able to identify regular polygons, and reason why a given polygon is regular. Language focus “This is a regular polygon, because all of the sides are the same length, and all of the interior angles are equal.” Pupils often define a regular polygon as having equal side-lengths and neglect to mention the angles – it is important that pupils consider both sides and angles when assessing and describing whether a polygon is regular. Pupils should examine and discuss a wide range of irregular shapes, including examples with equal angles, but unequal side-lengths (shape d below), examples with equal angles, but unequal interior angles and unequal side-lengths (shape c below), and examples with equal side-lengths, but unequal angles (shape e below). Figure 144: a regular octagon and 4 irregular octagons Pupils can make different polygons with equal length geo-strips to explore regular shapes and irregular shapes with equal side-lengths but unequal angles. Pupils should compare angles using informal language, and begin to discuss whether angles are smaller than a right angle (acute), larger than a right angle but smaller than a ‘straight line’ (180°) (obtuse), or larger than a ‘straight line’ (180°) (reflex) in preparation for measuring angles with a protractor in year 5. 198 Pupils should also learn that equilateral triangles are regular triangles, and that squares are regular quadrilaterals. Pupils need to understand perimeter as the total distance around the outside of a shape, and be able to measure or calculate the perimeter of shapes with straight sides. Pupils should be able to measure side-lengths in centimetres or metres, as appropriate, using skills developed from year 1 onwards. They should use an appropriate strategy to find the perimeter of a given polygon, according to the property of the shape. Shape type Strategy for calculating the perimeter rectilinear shapes on centimetre-square-grids count the number of centimetre lines around the shape Figure 145: incorrect and correct methods for finding the perimeter of a rectilinear shape on a square grid Drawn to scale. polygons with equal side-lengths use multiplication: polygons with unequal side-lengths use addition: perimeter = sum of the side-lengths 199 Shape type Strategy for calculating the perimeter rectangles use doubling and addition: or Figure 146: 2 strategies for calculating the perimeter of a rectangle Drawn to scale. As well as working with ‘small’ shapes in the classroom, pupils should gain experience working with larger shapes, such as calculating the perimeter of a rectilinear area drawn on the playground in metres. Making connections Pupils must be fluent in multiplication table facts (4NF–1 ) to efficiently calculate the perimeter of polygons with equal side-lengths. They must also be able to apply appropriate strategies for adding more than 2 numbers to calculate the perimeter of irregular polygons. 200 4G–2 Example assessment questions Taro uses some 8cm sticks to make these shapes. Name each shape and find its perimeter. Drawn to scale. What is the perimeter of this shape? Drawn to actual size. Sarah draws a rhombus with a perimeter of 36cm. What is the length of each side? Here is a plan of a school playground. How many metres of fencing is needed to put a fence around the perimeter? Drawn to scale. 201 Name each shape and say whether it is regular or irregular. Explain your reasons. 4G–3 Identify line symmetry in 2D shapes Identify line symmetry in 2D shapes presented in different orientations. Reflect shapes in a line of symmetry and complete a symmetric figure or pattern with respect to a specified line of symmetry. 4G–3 Teaching guidance Identifying line symmetry requires pupils to be able to decompose shapes: where line symmetry exists within a shape, the shape can be split into two parts which are a reflection of one another. Pupils should learn to compose a symmetrical shape from two congruent shapes. Figure 147: composing shapes from two identical isosceles triangles Pupils should then learn to identify line symmetry in given symmetrical shapes. They can begin by folding or cutting paper shapes, or using a mirror, but should eventually be able to do this by mapping corresponding points in relation to the proposed line of symmetry. They should be able to explain why a particular shape is symmetrical, or why a given line is a line of symmetry. 202 Language focus “This is a line of symmetry because it splits the shape into two equal parts which are mirror images.” Pupils need to be able to identify line symmetry regardless of the orientation a shape is presented in, including cases where the line of symmetry is neither a horizontal nor a vertical line. For the second part of this criterion, pupils must be able to reflect shapes in a line of symmetry, both where the line dissects the original shape (see 4G–3 Example assessment questions ) and where it does not, and complete symmetrical patterns. Figure 148: reflecting a shape in a line of symmetry Figure 149: a symmetrical pattern 4G–3 Example assessment questions Draw one or more lines of reflection symmetry in each of these irregular hexagons. 203 Reflect the three shapes in the mirror line. Complete the shape by reflecting it in the mirror line. Name the polygon that you have completed. 204 Draw a line of symmetry on each shape. Are you able to draw more than one line of symmetry on any of the shapes? Complete the symmetrical pattern. Calculation and fluency Number, place value and number facts: 4NPV–2 and 4NF–3 4NPV–2 Recognise the place value of each digit in four-digit numbers, and compose and decompose four-digit numbers using standard and non-standard partitioning. 4NF–3 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 100), for example: 8 6 14 and 14 6 8 so 800 600 1 ,400 and 1 ,400 600 800 3 4 12 and 12 4 3 so 300 4 1 ,200 and 1 ,200 4 300 Representations such as place-value counters and partitioning diagrams (4NPV–2 ) and tens-frames with place-value counters (4NF–3), can be used initially to help pupils understand calculation strategies and make connections between known facts and related calculations. However, pupils should not rely on such representations for calculating. For the calculations in 4NF–3, for example, pupils should instead be able to calculate by verbalising the relationship. 205 Language focus “8 plus 6 is equal to 14, so 8 hundreds plus 6 hundreds is equal to 14 hundreds.” “14 hundred is equal to 1,400.” Pupils should be developing fluency in both formal written and mental methods for addition and subtraction. Mental methods can include jottings to keep track of calculation, or language structures as exemplified above. Pupils should select the most efficient method to calculate depending on the numbers involved. Addition and subtraction: extending 3AS–3 Pupils should also extend columnar addition and subtraction methods to four-digit numbers. Pupils must be able to add 2 or more numbers using columnar addition, including calculations whose addends have different numbers of digits. Figure 150: columnar addition for calculations involving four-digit numbers For calculations with more than 2 addends, pupils should add the digits within a column in the most efficient order. For the third example above, efficient choices could include: beginning by making 10 in the ones column making double-6 in the hundreds column Pupils must be able to subtract one four-digit number from another using columnar subtraction. They should be able to apply the columnar method to calculations where the subtrahend has fewer digits than the minuend, and must be able to exchange through 0. 206 Figure 151: columnar subtraction for calculations involving four-digit numbers Pupils should make sensible decisions about how and when to use columnar subtraction. For example, when the minuend is a multiple of 1,000, they may transform to an equivalent calculation before using column subtraction, avoiding the need to exchange through zeroes. Figure 152: transforming a columnar subtraction calculation to an equivalent calculation 4NF–1 Recall of multiplication tables Recall multiplication and division facts up to 12 12 , and recognise products in multiplication tables as multiples of the corresponding number. Recall of all multiplication table facts should be the main multiplication calculation focus in year 4. Pupils who leave year 4 fluent in these facts have the best chance of mastering short multiplication in year 5. Pupils who are fluent in multiplication table facts can solve the following types of problem by automatic recall of the relevant fact rather than by skip counting or reciting the relevant multiplication table: 8 9 3 12 6 6 (identify products) 5 45 8 48 121 11 (solve missing-factor problems) 35 7 63 9 (use relevant multiplication table facts to solve division problems) 207 Pupils should also be fluent in interpreting contextual multiplication and division problems, identifying the appropriate calculation and solving it using automatic recall of the relevant fact. Examples are given in 4NF–1 Example assessment questions. As pupils become fluent with the multiplication table facts, they should also develop fluency in related calculations as described in 4NF–3 (scaling number facts by 100). Pupils should also develop fluency in multiplying and dividing by 10 and 100 (4MD–1). 4MD–2 Manipulating the multiplicative relationship Manipulate multiplication and division equations, and understand and apply the commutative property of multiplication. Pupils who are fluent in manipulating multiplicative expressions can solve the following types of problem: 4 7 6 5 9 9 (apply understanding of the inverse relationship between multiplication and division to solve missing-dividend problems) 72 8 35 5 81 9 (solve missing-divisor problems) To solve missing-divisor problems, pupils can use their understanding that the divisor and the quotient can be swapped, so that 72 8 , for example, is solved using 72 8 . Alternatively they can use their understanding of the relationship between multiplication and division, and solve the related missing-factor problem (here 8 72 ). In either case, pupils will then need to apply their multiplication table fluency (here ‘9 eights are 72’) to finally identify the missing divisor. 208 Year 5 guidance Ready-to-progress criteria Year 4 conceptual prerequesite Year 5 ready-to-progress criteria Future applications Know that 10 hundreds are equivalent to 1 thousand, and that 1,000 is 10 times the size of 100; apply this to identify and work out how many 100s there are in other four-digit multiples of 100. 5NPV–1 Know that 10 tenths are equivalent to 1 one, and that 1 is 10 times the size of 0.1. Know that 100 hundredths are equivalent to 1 one, and that 1 is 100 times the size of 0.01. Know that 10 hundredths are equivalent to 1 tenth, and that 0.1 is 10 times the size of 0.01. Solve multiplication problems that have the scaling structure, such as ‘ten times as long’. Understand that per cent relates to ‘number of parts per hundred’, and write percentages as a fraction with denominator 100, and as a decimal fraction. Recognise the place value of each digit in four-digit numbers, and compose and decompose four-digit numbers using standard and non-standard partitioning. 5NPV–2 Recognise the place value of each digit in numbers with up to 2 decimal places, and compose and decompose numbers with up to 2 decimal places using standard and non-standard partitioning. Compare and order numbers, including those with up to 2 decimal places. Add and subtract using mental and formal written methods. Reason about the location of any four-digit number in the linear number system, including identifying the previous and next multiple of 1,000 and 100, and rounding to the nearest of each. 5NPV–3 Reason about the location of any number with up to 2 decimals places in the linear number system, including identifying the previous and next multiple of 1 and 0.1 and rounding to the nearest of each. Compare and order numbers, including those with up to 2 decimal places. Estimate and approximate to the nearest 1 or 0.1. Divide 1,000 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in multiples of 1,000 with 2, 4, 5 and 10 equal parts. 5NPV–4 Divide 1 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in units of 1 with 2, 4, 5 and 10 equal parts. Read scales on graphs and measuring instruments. 209 Year 4 conceptual prerequesite Year 5 ready-to-progress criteria Future applications Divide 100 and 1,000 into 2, 4, 5 and 10 equal parts. Find unit fractions of quantities using known division facts (multiplication tables fluency). 5NPV–5 Convert between units of measure, including using common decimals and fractions. Read scales on measuring instruments, and on graphs related to measures contexts. Solve measures problems involving different units by converting to a common unit. Recall multiplication and division facts up to . Solve division problems, with two-digit dividends and one-digit divisors, that involve remainders, for example: 5NF–1 Secure fluency in multiplication table facts, and corresponding division facts, through continued practice. Use multiplication facts during application of formal written layout. Use division facts during short division and long division. Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10 or 100), for example: 5NF–2 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 1 tenth or 1 hundredth), for example: Recognise number relationships within the context of place value to develop fluency and efficiency in calculation. Multiply and divide whole numbers by 10 and 100 (keeping to whole number quotients); understand this as equivalent to scaling a number by 10 or 100. 5MD–1 Multiply and divide numbers by 10 and 100; understand this as equivalent to making a number 10 or 100 times the size, or 1 tenth or 1 hundredth times the size. Convert between different metric units of measure. 210 Year 4 conceptual prerequesite Year 5 ready-to-progress criteria Future applications Recall multiplication and division facts up to , and recognise products in multiplication tables as multiples of the corresponding number. Recognise multiples of 10, 100 and 1,000. Apply place-value knowledge to known additive and multiplicative number facts. Multiply and divide whole numbers by 10 and 100 (keeping to whole number quotients). 5MD–2 Find factors and multiples of positive whole numbers, including common factors and common multiples, and express a given number as a product of 2 or 3 factors. Solve contextual division problems. Simplify fractions. Express fractions in the same denomination. Recall multiplication facts up to . Manipulate multiplication and division equations. 5MD–3 Multiply any whole number with up to 4 digits by any one-digit number using a formal written method. Solve contextual and non-contextual multiplication problems using a formal written method. Recall multiplication and division facts up to . Manipulate multiplication and division equations. Solve division problems, with two-digit dividends and one-digit divisors, that involve remainders, for example: and interpret remainders appropriately according to the context. 5MD–4 Divide a number with up to 4 digits by a one-digit number using a formal written method, and interpret remainders appropriately for the context. Solve contextual and non-contextual division problems using a formal written method. Recall multiplication and division facts up to . Find unit fractions of quantities using known division facts (multiplication-tables fluency). Unitise using unit fractions (for example, understand that there are 3 one-fifths in three-fifths). 5F–1 Find non-unit fractions of quantities. Solve multiplication problems that have the scaling structure. 211 Year 4 conceptual prerequesite Year 5 ready-to-progress criteria Future applications Recall multiplication and division facts up to . Reason about the location of fractions in the linear number system. 5F–2 Find equivalent fractions and understand that they have the same value and the same position in the linear number system. Compare and order fractions. Use common factors to simplify fractions. Use common multiples to express fractions in the same denomination. Add and subtract fractions with different denominators and mixed numbers, using the concept of equivalent fractions. Divide powers of 10 into 2, 4, 5 and 10 equal parts. 5F–3 Recall decimal fraction equivalents for , , and , and for multiples of these proper fractions. Read scales on graphs and measuring instruments. Know percentage equivalents of common fractions. Recognise right angles as a property of shape or a description of a turn, and identify right angles in 2D shapes presented in different orientations. Identify whether the interior angles of a polygon are equal or not. 5G–1 Compare angles, estimate and measure angles in degrees (°) and draw angles of a given size. Solve problems involving missing angles. Compose polygons from smaller shapes. Recall multiplication facts up to . 5G–2 Compare areas and calculate the area of rectangles (including squares) using standard units. Calculate the area of compound rectilinear shapes and other 2D shapes, including triangles and parallelograms, using standard units. Use the relationship between side-length and perimeter, and between side-length and area to calculate unknown values. 212 5NPV–1 Tenths and hundredths Know that 10 tenths are equivalent to 1 one, and that 1 is 10 times the size of 0.1. Know that 100 hundredths are equivalent to 1 one, and that 1 is 100 times the size of 0.01. Know that 10 hundredths are equivalent to 1 tenth, and that 0.1 is 10 times the size of 0.01. 5NPV–1 Teaching guidance This criterion follows on from what pupils learnt in years 3 and 4 about the relationship between adjacent place-value positions. The value of a given digit is made 10 times the size if it is moved 1 position to left, and is made one tenth times the size if it is moved 1 position to the right. Pupils should learn, therefore, that we can extend the place-value chart to include positions to the right of the ones place. Figure 153: place-value chart illustrating the scaling relationship between adjacent positions, including tenths and hundredths Pupils should understand that: a ‘1’ in the tenths column has a value of one tenth, and is one tenth the size of 1 a ‘1’ in the hundredths column has a value of one hundredth, and is one hundredth the size of 1 Pupils should learn that the decimal point is used between the ones digit and the tenths digit, so that we can write decimal numbers without using a place-value chart. They should learn that one tenth is written as 0.1 and one hundredth is written as 0.01. 213 Pupils must be able to describe the relationships between 1, 0.1 and 0.01. Language focus “1 is 10 times the size of one-tenth.” “One-tenth is 10 times the size of one-hundredth.” “1 is 100 times the size of one-hundredth.” As well as understanding one-tenth and one-hundredth as scaling 1, pupils must understand them in terms of regrouping and exchanging. Dienes can be used to illustrate the relative size of 1, one-tenth and one-hundredth, with a ‘flat’ now representing 1. Pupils should experience how 10 hundredths can be regrouped into one-tenth, and how both 10 tenths and 100 hundredths can be regrouped into 1 one. Conversely they should be able to exchange 1 one for 10 tenths or for 100 hundredths and 1 tenth for 10 hundredths. Figure 154: using Dienes to represent 1, one-tenth and one-hundredth Figure 155: using Dienes to represent the equivalence of 1 tenth and 10 hundredths 214 Pupils must describe the equivalence between the different quantities using unitising language (unitising in ones, tenths and hundredths). Language focus “10 tenths is equal to 1 one.” “10 hundredths is equal to 1 tenth.” “100 hundredths is equal to 1 one.” Once pupils understand the relative size of these new units, they should learn to use place-value counters to represent the equivalence. Pupils must then be able to work out how many tenths there are in other multiples of 0.1 and how many hundredths there are in other multiples of 0.01. Initially pupils should work with values that involve only the tenths place (for example 0.4) or only the hundredths place (for example 0.04). Once they have learnt to write numbers with tenths and hundredths (5NPV–2), they should be able to reason, for example, that: 18 hundredths is equal to 1 tenth and 8 hundredths, and is written as 0.18 18 tenths is equal to 1 one and 8 tenths, and is written as 1.8 Figure 156: eighteen 0.01-value place-value counters in 2 tens frames Language focus “18 hundredths is equal to 10 hundredths and 8 more hundredths.” “10 hundredths is equal to 1 tenth.” “So 18 hundredths is equal to 1 tenth and 8 more hundredths, which is 0.18.” Pupils need to be able to apply this reasoning to measures contexts, as shown in the Example assessment questions below. 215 This learning should also be connected to pupils’ understanding of fractions – they should understand that one-tenth can be written as both 0.1 and 1 10 and that one hundredth can be written as both 0.01 and 1 100 . Pupils should be able to write, for example, 18 hundredths as both 0.18 and 18 100 . Making connections Pupils need to be able to write numbers in decimal notation (5NPV–2) to be able to make the connection between, for example 18 tenths and 1.8. Meeting 5NPV–1 also supports 5NF–2 (applying place value to known number facts) because, if pupils can unitise in tenths and hundredths, and, for example understand that 12 tenths = 1.2 and 12 hundredths = 0.12, then they can reason that: (5 tenths plus 7 tenths is equal to 12 tenths, which is 1.2) (3 hundredths times 4 is equal to 12 hundredths, which is 0.12) 5NPV–1 Example assessment questions An apple weighs about 0.1kg. Approximately how many apples are there in a 1.8kg bag? I have a 0.35m length of wooden rod. How many 0.01m lengths can I cut it into? Mrs Jasper is juicing oranges. Each orange makes about 0.1 litres of juice. If Mrs Jasper juices 22 oranges, approximately how many litres of orange juice will she get? Circle all of the numbers that are equal to a whole number of tenths. 0.2 4.8 1 0.01 10 0.83 Fill in the missing numbers. 0.01 1 0.1 1 1 0.01 0.
Fill in the missing numbers. tenths 3.9 hundredths 0.22 hundredths 8 216 Match the numbers on the left with the equivalent fractions on the right. 0.20 0.02 0.12 0.21 5NPV–2 Place value in decimal fractions Recognise the place value of each digit in numbers with up to 2 decimal places, and compose and decompose numbers with up to 2 decimal places using standard and non-standard partitioning. 5NPV–2 Teaching guidance Pupils must be able to read, write and interpret decimal fractions with up to 2 decimal places. Pupils should work first with decimal fractions with one significant digit (for example, 0.3 and 0.03). The Gattegno chart is a useful tool here. Figure 157: Gattegno chart showing thousands, hundreds, tens, ones, tenths and hundredths The number 300 is spoken as “three hundred” rather than as “three-zero-zero”, and this helps pupils to identify the value of the 3 in 300. However, decimal fractions are usually spoken as digits, for example, 0.03 is spoken as “zero-point-zero-three” (or “nought-point-nought-three”) rather as “three hundredths”. As such, pupils need to practise speaking decimal fractions in both ways and learn to convert from one to the other. 217 Language focus “Three hundredths is zero-point-zero-three.” Pupils must then learn to work with decimal fractions with 2 significant digits (for example, 0.36). For any given decimal fraction of this type, pupils must be able to connect the spoken words (zero-point-three-six), the value in decimal notation (0.36), describing the number of tenths and hundredths (3 tenths and 6 hundredths) and visual representations (such as place-value counters and the Gattegno chart). Figure 158: 4 different represenations of 0.36 Pupils should be able to identify the place value of each digit in numbers with up to 2 decimal places. They must be able to combine units of hundredths, tenths, ones, tens, hundreds and thousands to compose numbers, and partition numbers into these units. Pupils need to experience variation in the order of presentation of the units, so that they understand that 0.4 3 0.02 50 is equal to 53.42, not 43.25. Figure 159: 2 representations of the place-value composition of 53.42 218 Pupils also need to solve problems relating to subtraction of any single place-value part from the whole number, for example: 53.42 – 3 53.42 – 53.02 As well as being able to partition numbers in the ‘standard’ way (into individual place-value units), pupils must also be able to partition numbers in ‘non-standard’ ways and carry out related addition and subtraction calculations, for example: Figure 160: partitioning 7.83 into 7.43 and 0.4 Figure 161: partitioning 0.25 into 0.22 and 0.03 You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 5NPV–2 and 5NF–2 5NPV–2 Example assessment questions Complete the calculations. 4 0.07 0.2 0.4 0.02 70 20 0.07 4 0.4 20 700 Circle the numbers that add together to give a total of 0.14 0.04 0.12 0.1 0.2 219 Fill in the missing numbers. 3.87 0.8 25.14 – 0.04 19.7 – 9 99.99 90 84.51 50 0.3 5.61 95.75 – 0.5 6.14 5 0.04 2 1.43 0.05 I have 3.7kg of modelling clay. If we use 2kg, how much will be left? I will use 0.65 litres of milk for one recipe, and 0.23 litres of milk for another. How much milk will I use altogether? Ilaria jumped 3.19m in a long jump competition. Emma jumped 3.12m. How much further did Ilaria jump than Emma? Maya cycled 7.3km to get to her friend’s house, and then cycled a further 0.6km to the park. How far did Maya cycle altogether? 5NPV–3 Decimal fractions in the linear number system Reason about the location of any number with up to 2 decimals places in the linear number system, including identifying the previous and next multiple of 1 and 0.1 and rounding to the nearest of each. 5NPV–3 Teaching guidance Pupils need to become familiar with the relative positons, on a number line, of numbers with 1 and 2 decimal places. They will need to see number lines with both tenths and intermediate hundredths values marked, and learn, for example, that 0.5 is the same as 0.50 and 3 is the same as 3.0 or 3.00. Pupils should recognise the magnitude and position of a given decimal fraction, irrespective of the precision it is given to, for example, 5 tenths lies between 0.45 and 0.55 on the number line below, whether it is represented as 0.5 or 0.50. Figure 162: 0 to 1 number line marked and labelled in intervals of 5 hundredths Pupils need to be able to identify or place decimal fractions on number lines marked in tenths and/or hundredths. They should use efficient strategies and appropriate reasoning, including identifying the midpoints or working backwards from a whole number or a multiple of one tenth. 220 Figure 163: identifying 0.14 and 0.41 on a 0 to 0.5 number line marked with intervals of hundredths Language focus “a is 0.14 because it is 1 hundredth less than the midpoint of 0.1 and 0.2, which is 0.15.” “b is 0.41 because it is 1 hundredth more than 0.4.” Pupils need to be able to estimate the value or position of decimal fractions on unmarked or partially marked numbers lines, using appropriate proportional reasoning, rather than counting on from a start point or back from an end point. For example, here pupils should reason: “8.6 is about here on the number line because it’s just over half way”. Figure 164: placing 8.6 on an unmarked 8 to 9 number line Here, pupils should reason: “8.75 is about here on the number line because it’s the midpoint of 8.7 and 8.8.” Figure 165: placing 8.75 on an 8 to 9 number line marked only in tenths Pupils must also be able to identify which whole numbers, or which pair of multiples of 0.1, a given decimal fraction is between. To begin with, pupils can use a number line for support. In this example, for the number 8.61, pupils must identify the previous and next whole number, and the previous and next multiple of 0.1. Figure 166: using a number line to identify the previous and next whole number 221 Figure 167: using a number line to identify the previous and next multiple of 0.1 Language focus “The previous whole number is 8. The next whole number is 9.” “The previous multiple of 0.1 is 8.6. The next multiple of 0.1 is 8.7.” By the end of year 5 pupils need to be able to complete this type of task without the support of a number line. Pupils should then learn to round a given decimal fraction to the nearest whole number by identifying the nearest of the pair of whole numbers that the decimal fraction is between. Similarly, pupils should learn to round to the nearest multiple of 0.1. They should understand that they need to examine the digit in the place to the right of the unit they are rounding to, for example when rounding to the nearest whole number, pupils must examine the digit in the tenths place. Again, pupils can initially use number lines for support, but should be able to round without that support by the end of year 5. Figure 168: identifying the nearest whole number with a number line for support Language focus “The closest whole number is 9.” “8.61 rounded to the nearest whole number is 9.” 222 Finally, pupils should also be able to count forwards and backwards from any decimal fraction in steps of 1, 0.1 or 0.01. Pay particular attention to counting over ‘boundaries’, for example: 2.1, 2.0, 1.9 2.85, 2.95, 3.05 Making connections Here, pupils must apply their knowledge that 10 tenths is equal to 1 one (see 5NPV–1) to understand that each interval of 1 on a number line or scale is made up of 10 intervals of 0.1. Similarly, they must use their knowledge that 10 hundredths is equal to 1 tenth to understand that each interval of 0.1 on a number line or scale is made up of 10 intervals of 0.01. This also links to 5NPV–4, in which pupils need to be able to read scales divided into 2, 4, 5 and 10 equal parts. 5NPV–3 Example assessment questions Place each of these numbers on the number line. 0.6 0.16 0.91 0.09 0.69 The table shows how far some children jumped in a long-jump competition. NameDistance jumped (m)Jamal3.04Reyna3.40Faisal2.85Ilaria3.19Charlie3.09Kagendo2.90 a. Who jumped the furthest and won the competition? b. Who came third in the competition? 223 c. How much further did Kagendo jump then Faisal? d. How much further did Ilaria jump than Charlie? Fill in the missing symbols (<, > or =). 0.3 0.5 0.03 0.05 0.50 0.5 9 9.00 0.2 0.15 0.11 0.09 1.01 1.1 3 2.99 140 1.40 Here is a weighing scale. Estimate the mass in kilograms that the arrow is pointing to. Estimate and mark the position of 0.7 litres on this beaker. 224 Fill in the missing numbers. 5.01 5.02 5.03 3.65 3.95 4.25 4.35 27.9 27.8 27.7 A farmer weighed each of 6 new-born lambs. Round the mass of each lamb to the nearest whole kilogram. Rounded to nearest whole kilogram 5.19kg 6.7kg 4.08kg 6.1kg 6.45kg 4.91kg I need 4.25 metres of ribbon. a. How much is this to the nearest tenth of a metre? b. How much is this to the nearest metre? c. If ribbon is sold only in whole metres, how many metres do I need to buy? 225 5NPV–4 Reading scales with 2, 4, 5 or 10 intervals Divide 1 into 2, 4, 5 and 10 equal parts, and read scales/number lines marked in units of 1 with 2, 4, 5 and 10 equal parts. 5NPV–4 Teaching guidance By the end of year 5, pupils must be able to divide 1 into 2, 4, 5 or 10 equal parts. This is important because these are the intervals commonly found on measuring instruments and graph scales. Figure 169: bar models showing 1 partitioned into 2, 4, 5 and 10 equal parts Pupils should practise counting in multiples of 0.1, 0.2, 0.25 and 0.5 from 0, or from any multiple of these numbers, both forwards and backwards. This is an important step in becoming fluent with these number patterns. Language focus “Twenty-five, fifty, seventy-five, one hundred” needs to be a fluent spoken language pattern. Fluency in this language pattern provides the basis to count in multiples of 0.25. Pupils should be able to apply this skip counting, beyond 1, to solve contextual multiplication and division measures problems, as shown in 5NPV–4 Example assessment questions below (questions 8 to 10). Pupils should also be able to write, solve and manipulate multiplication and division equations related to multiples of 0.1, 0.2, 0.25 and 0.5 up to 1, and connect this to their knowledge of fractions, and decimal-fraction equivalents (5F–3). 226 Pupils need to be able to solve addition and subtraction problems based on partitioning 1 into multiples of 0.1, 0.2 and 0.5 based on known number bonds to 10. Pupils should also have automatic recall of the fact that 0.25 and 0.75 are bonds to 1. They should be able to immediately answer a question such as “I have 1 litre of water and pour out 0.25 litres. How much is left?” Making connections Dividing 1 into 10 equal parts is also assessed as part of 5NPV–1. Reading scales also builds on number-line knowledge from 5NPV–3. Conversely, experience of working with scales with 2, 4, 5 or 10 divisions in this criterion improves pupils’ estimating skills when working with unmarked number lines and scales as described in 5NPV–3. In 5F–3 pupils need to be fluent in common fraction–decimal equivalents, for example, . This criterion provides the foundations for 5F–3. 5NPV–4 Example assessment questions Fill in the missing parts, and write as many different equations as you can think of to represent the bar model. Fill in the missing numbers. 7.5 7 6 4.4 4.6 5.2 2.5 3 3.75 227 5 children have been growing sunflowers. The bar chart shows how tall each child’s sunfower has grown. How tall is each flower?’ The bar chart below shows long-jump distances for 6 children. a. How far did the winning child jump? b. What was the difference between the two longest jumps? Complete the labelling of these scales. 228 What is the reading on each of these scales, in kilograms? Here is a 1 litre beaker with some liquid in. How much more liquid, in litres, do I need to add to the beaker to make 1 litre? A motorway repair team can build 0.2km of motorway barrier in 1 day. In 6 working days, how many kilometres of motorway barrier can they build? How many 0.25 litre servings of orange juice are there in a 2 litre carton? 0.25m of ribbon costs £1. How much does 2m of ribbon cost? Fill in the missing numbers. 1 0.2 5 m 1m 1 5 1 0.8 4 m 1m 1 5 1 1 1 0.2 0.2 5 0.2m 4 m 229 Here is a part of a number line divided into 4 equal parts. In which section (a, b, c or d) does each of these numbers belong? Explain your answers. 4.3 4.03 4.09 4.76 4.41 4.69 5NPV–5 Convert between units of measure Convert between units of measure, including using common decimals and fractions. 5NPV–5 Teaching guidance Pupils should first memorise the following unit conversions: 1km = 1,000m 1m = 100cm 1cm = 10mm 1 litre = 1,000ml 1kg = 1,000g £1 = 100p It is essential that enough time is given to this foundational stage before moving on. Practical experience of these conversions will help pupils to avoid common errors in recalling the correct power of 10 for a given conversion. For example, they can walk 1km while counting the number of metres using a trundle measuring wheel. Once pupils can confidently recall these conversions, they should apply them to whole number conversions, from larger to smaller units and vice versa, for example, £4 = 400p and 8,000g = 8kg. Pupils must then learn to convert from and to fraction and decimal-fraction quantities of larger units, within 1, for example 0.25km = 250m. They should be able to carry out conversions that correspond to some of the common 2, 4, 5 and 10 part measures intervals, as exemplified below for kilometre–metre conversions. 230 Distance in km expressed as a fraction Distance in km expressed as a decimal fraction Distance in metres 0.2km 200m 0.25km 250m 0.5km 500m 0.75m 750m 0.1km 100m all other multiples of , for example, 0.7km 700m For finding 3 4 of a unit, pupils should have sufficient fluency in the association between 3 4 and 0.75, 75 and 750 that they should not need to first work out 1 4 and multiply by 3. For all conversions, pupils should begin by stating the single unit conversion rate as a step to the fraction or decimal-fraction conversion. Language focus “1m is 100cm.” “So is 75cm.” Pupils should learn to derive other common conversions over 1. To convert, for example, 3,700 millilitres to litres, they should not need to think about dividing by 1,000 and moving the digits 3 places. Instead they should be able to use single unit conversion rates and their understanding of place value. 231 Language focus “1,000ml is 1 litre.” “So 3,000ml is 3 litres, and 3,700ml is 3.7 litres.” For pounds and pence, and metres and centimetres, pupils should also be able to carry out conversions that correspond to 100 parts, for example, 52p = £0.52, and 43cm = 0.43m. Language focus “100p is £1.” “So 50p is £0.50, and 52p is £0.52.” Pupils can use ratio tables for support. 1m 100cm 75cm 1,000ml 1 litre 3,700ml 3.7 litres 100p £1 52p £0.52 Pupils must learn to solve measures problems involving different units by converting to a common unit. Making connections To succeed with this criterion, pupils must be fluent in the division of 1,000, 100 and 1 into 2, 4, 5 and 10 equal parts (4NPV–4 , 3NPV–4 and 5NPV–4 respectively). They must also be able to recall common fraction-decimal equivalents (5F–3). The fraction conversions in this criterion are special cases of finding fractions of quantities (5F–1). 232 5NPV–5 Example assessment questions Fill in the missing numbers to complete these conversions between units. 3 1 4 2 1 4 1.8 litres ml km m 5 cm mm £8.12 p 4 kg g 3.4m cm 21mm cm 2,250ml litres 650cm m 8,300m km 165p £ 750g kg Put these volumes in order from smallest to largest. 0.75 litres 1.1 litres 0.3 litres 1 5 litre 900ml 1 2 1 litres Put these lengths in order from smallest to largest. 0.45m 10mm 208cm 1 2 2 m 80cm 0.9m 1 2 cm Maya needs to post 3 parcels. The mass of each parcel is shown below. How much do the parcels weigh altogether, in kilograms? Parcel Mass of parcel A 3.2kg B 4,500g0 C Finn has a 1 2 7 m length of wood. How many 3 4 m length pieces can he cut from it? 233 I need 1 4 1 kg of flour for a recipe. I pour some flour into the weighing scales. How much more flour do I need for the recipe? Fill in the values in the empty circles so that each row and column of 3 circles adds to 5km. 234 5NF–1 Secure fluency in multiplication and division facts Secure fluency in multiplication table facts, and corresponding division facts, through continued practice. 5NF–1 Teaching guidance Before pupils begin work on formal multiplication and division (5MD–3 and 5MD–4), it is essential that pupils have automatic recall of multiplication and division facts within the multiplication tables. These facts are required for calculation within the ‘columns’ during application of formal written methods. All mental multiplicative calculation also depends on these facts. Identifying core number facts: short multiplication Identifying core number facts: short division Figure 170: short multiplication of 342 by 7 Figure 171: short division of 4,952 by 8 Within-column calculations: Within-column calculations: Pupils should already have automatic recall of multiplication table facts and corresponding division facts, from year 3 (5, 10, 2, 4 and 8 multiplication tables, 3NF–2 ) and year 4 (all multiplication tables up to and including 12, 4NF–1). Pupils’ fluency in multiplication facts is assessed in the summer term of year 4 in the statutory multiplication tables check, and this will identify some pupils who need additional practice. However, even pupils who were fluent in the multiplication tables at the time of the multiplication tables check will benefit from further practice to maintain and secure fluency. Pupils must also be able to fluently derive related division facts, including division facts with remainders before they begin to learn formal written methods for multiplication and division (5MD–3 and 5MD–4). The multiplication facts to 9 9 , and related division facts, are particularly important as these are the facts required for formal written multiplication and division. The 36 multiplication facts that are required for formal written multiplication are as follows. 235 You can find out more about multiplicative fluency here in the calculation and fluency section: 5NF–1 Making connections Fluency in these multiplicative facts is required for: mental calculation, when combined with place-value knowledge, for example, if pupils know that , then they can calculate and (5NF–2) identifying factors and multiples (5MD–1) within-column calculation in short multiplication (5MD–3) and short division (5MD–4) calculating fractions of quantities (5F–1) finding equivalent fractions (5F–2) calculating area (5G–2) 5NF–1 Assessment guidance Assessment for this criterion should focus on whether pupils have fluency in multiplication facts and division facts. Pupils can be assessed through a time-limited written check. 236 5NF–2 Scaling number facts by 0.1 or 0.01 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 1 tenth or 1 hundredth), for example: 5NF–2 Teaching guidance Pupils must be able to combine known additive and multiplicative facts with unitising in tenths and hundredths, including: scaling known additive facts within 10, for example, 0.09 – 0.06 0.03 scaling known additive facts that bridge 10, for example, 0.8 0.6 1.4 scaling known multiplication tables facts, for example 0.03 4 0.12 scaling division facts derived from multiplication tables, for example, 0.12 4 0.03 scaling calculation of complements to 100, for example 0.62 0.38 1 For calculations such as 0.8 0.6 1.4 , pupils can begin by using tens frames and counters as they did for calculation across 10 (2AS–1), calculation across 100 (3NF–3 ) and calculation across 1,000 (4NF–3), but now using 0.1-value counters (or 0.01 value counters for calculations such as 0.08 0.06 0.14 ). 237 Figure 172: tens frames with 0.1-value counters showing .
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0 8 0 6 1 4 8 6 14 14 – 6 8 14 – 8 6 0.8 0.6 1.4 1.4 – 0.6 0.8 1.4 – 0.8 0.6 Language focus “8 plus 6 is equal to 14, so 8 tenths plus 6 tenths is equal to 14 tenths.” “14 tenths is equal to 1 one and 4 tenths.” Pupils must also be able to scale additive calculations related to complements to 100 (3AS–1), for example: 62 38 100 so 0.62 0.38 1 238 Figure 173: a 100 grid shaded in 2 colours to represent 0.62 and 0.38 as a complement to 1 Pupils can initially use 0.1- or 0.01-value counters to understand how a known multiplicative fact, such as 3 5 15 , relates to scaled calculations, such as 3 0.5 1.5 or 3 0.05 0.15 . Pupils should be able reason in terms of unitising in tenths or hundreds, or in terms of scaling a factor by one-tenth or one-hundredth. Figure 174: 3-by-5 array of 0.01-value place-value counters 3 5 15 3 0.05 0.15 Language focus “3 times 5 is equal to 15.” “3 times 5 hundredths is equal to 15 hundredths.” “15 hundredths is equal to 0.15.” 3 5 15 0.03 5 0.15 Language focus “3 times 5 is equal to 15.” “3 hundredths times 5 is equal to 15 hundredths.” “15 hundredths is equal to 0.15.” 239 Language focus “If I make one factor one-hundredth times the size, I must make the product one-hundredth times the size.” Pupils must be able to make similar connections for known division facts, for example, for scaling by one-hundredth: 15 3 5 0.15 0.03 5 0.15 3 0.05 Language focus “If I make the dividend one-hundredth times the size and the divisor one-hundredth times the size, the quotient remains the same.” “If I make the dividend one-hundredth times the size and keep the divisor the same, I must make the quotient one-hundredth times the size.” It is important for pupils to understand all of the calculations in this criterion in terms of working with units of 0.1 or 0.01, or scaling by one-tenth or one-hundredth for multiplicative calculations. You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 5NPV–2 and 5NF–2 Making connections This criterion builds on: known addition and subtraction facts 5NF–1, where pupils secure fluency in multiplication and division facts 5NPV–1, where pupils need to be able to work out how many tenths or hundredths there are in given numbers Meeting this criterion also requires pupils to be able to fluently divide whole numbers by 10 or 100 (5MD–1). 240 5NF–2 Example assessment questions Circle the numbers that sum to 0.13 0.1 0.5 0.05 0.8 0.08 0.3 Are these calculations correct? Mark each correct calculation with a tick () and each incorrect calculation with a cross (). Explain your answers. 0.05 0.05 0.010 0.04 0.06 0.1 0.13 0.7 0.2 0.61 0.49 1 0.73 0.27 1 0.4 0.5 0.45 I live 0.4km away from school. Every day I walk to school in the morning and home again in the afternoon. a. How far do I walk each day? b. How far do I walk in 5 days? Some children are making bunting for the school fair. If each child makes 0.4m of bunting, and there are 12 children, how many metres of bunting do they make altogether? A chef needs 2.4kg of potatoes for a recipe. If one potato weighs about 0.3kg, approximately how many potatoes does the chef need? A bottle contains 0.7 litres of fruit drink. Maria need 5 litres of drink for a party. How many bottles does she need to buy? I need 0.5kg of brown flour and 0.7kg of white flour for a recipe. What is the total mass of flour that I need? What is the total volume of liquid in these measuring beakers, in litres? 241 5MD–1 Multiplying and dividing by 10 and 100 Multiply and divide numbers by 10 and 100; understand this as equivalent to making a number 10 or 100 times the size, or 1 tenth or 1 hundredth times the size. 5MD–1 Teaching guidance In years 3 and 4, pupils considered multiplication and division by 10 and 100 both in terms of scaling (for example, 2,300 is 100 times the size of 23) and in terms of grouping or unitising (for example, 2,300 is 23 groups of 100). To meet criterion 5MD–1, pupils should be able to use and understand the language of 10 or 100 times the size to describe multiplication of numbers, including decimal fractions, by 10 or 100. They should understand division as the inverse action, and should be able to use and understand the language of one-tenth or one-hundredth times the size to describe division of numbers by 10 or 100, including to calculations that give decimal fraction quotients. Pupils already know the following relationships between powers of ten, and can describe them using scaling language (“…times the size”). Figure 175: multiplicative relationships between powers of 10: 10 times the size and one-tenth times the size Figure 176: multiplicative relationships between powers of 10: 100 times the size and one-hundredth times the size Pupils should extend the ‘ten times the size’/‘one-tenth times the size’ relationship to multiplicative calculations that ‘cross’ 1, beginning with those with 1 significant figure. The Gattegno chart can be used to help pupils see, for example, that 8, made one-tenth times the size is 0.8; pupils can move their finger or a counter down from 8 to 0.8. Pupils must connect this action to division by 10, and be able to solve/write the corresponding division calculation (8 10 0.8 ). Similarly, because 8 is 10 times the size of 0.8, they can solve 0.8 10 , moving their finger or a counter up from 0.8 to 8. 242 Figure 177: using the Gattegno chart to multiply and divide by 10, crossing 1 Language focus “8, made one-tenth of the size, is 0.8.” “8 divided by 10 is equal to 0.8.” “First we had 8 ones. Now we have 8 tenths.” “0.8, made 10 times the size, is 8.” “0.8 multiplied by 10 is equal to 8.” “First we had 8 tenths. Now we have 8 ones.” Pupils may also work with place-value charts. Figure 178: using a place-value chart to multiply and divide by 10, crossing 1 Similarly, pupils should be able to: divide ones by 100 and carry out inverse multiplications, for example, 8 100 0.08 and 0.08 100 8 243 divide tenths by 10 and carry out inverse multiplications, for example, 0.8 10 0.08 and 0.08 10 0.8 This understanding should then be extended to multiplicative calculations that ‘cross’ 1 and involve numbers with more than one significant digit, for example: 13 100 0.13 0.13 100 13 26.5 10 2.65 2.65 10 26.5 4,710 100 47.1 47.1 100 4,710 Throughout this criterion, repeated association of the written form (for example, 10 ) and the verbal form (for example, “one tenth times the size”) will help pupils become fluent with the links. Pupils should also be able to use appropriate language to describe the relationships in different contexts, including measures. Language focus “8cm is 10 times the length of 0.8cm.” “0.25kg is one-hundredth times the mass of 25kg.” “150 pencils is 10 times as many as 15 pencils.” Both the Gattegno chart and place-value charts can be used for support throughout this criterion, but by the end of year 5 pupils must be able to calculate without them. These representations can also help pupils to see that multiplying by 100 is equivalent to multiplying by 10, and then multiplying by 10 again (and that dividing by 100 is equivalent to dividing by 10 and dividing by 10 again). Making connections In 5NPV–1, pupils learnt to describe the relationship between different powers of 10 in terms of scaling. Here they applied this idea to scale other numbers by 100, 10, one-tenth and one-hundredth. This criterion also supports scaling known additive and multiplicative number facts by 1 tenth or 1 hundredth. For example, the known fact can be used to solve : one factor (3) has been scaled by one tenth, so the product (15) must be scaled by one tenth. 244 5MD–1 Example assessment questions Fill in the missing numbers. 10 4.03 10 10 0.2 10 100 21.7 100 100 5,806 100 Ruby ran 2.3km. Her mum ran 10 times this distance. How far did Ruby’s mum run? A zookeeper weighs an adult gorilla and its baby. The adult gorilla has a mass of 149.3kg. The baby gorilla has a mass one-tenth times that of the adult gorilla. How much does the baby gorilla weigh, in kilograms? The length of a new-born crocodile is about 0.25m. The length of an adult female crocodile is about 2.5m. Approximately how many times as long as a new-born crocodile is an adult female crocodile? Fill in the missing numbers. 100 5 273 100 10 6 42 10 100 0.79 1.35 100 10 0.75 16.2 10 Use the following to complete the equations: 10 100 10 100 Use each term only once. 543 5.43 0.12 1.2 51.5 5,150 40.3 4.03 245 5MD–2 Find factors and multiples Find factors and multiples of positive whole numbers, including common factors and common multiples, and express a given number as a product of 2 or 3 factors. 5MD–2 Teaching guidance Pupils should already know and be able to use the words ‘multiple’ and ‘factor’ in the context of the multiplication tables. They should know, for example, that the products within the 6 multiplication table are all multiples of 6, and should be familiar with the generalisation: factor factor product In year 5, pupils should learn the definitions of the terms ‘multiple’ and ‘factor’, and understand the inverse relationship between them. Language focus “A multiple of a given number is the product of the given number and any whole number.” “A factor of a given number is a whole number that the given number can be divided by without giving a remainder.” “21 is a multiple of 3. 3 is a factor of 21.” Pupils must be able to identify factors and multiples within the multiplication tables, and should learn to work systematically to identify all of the factors of a given number. They should be able to express products in the multiplication tables as products of 3 factors, where relevant, for example, 48 2 3 8 . Pupils already know how to scale known multiplication table facts by 10 or 100 (3NF–3 and 4NF–3), and must now learn to apply this to identify factors and multiples of larger numbers, as exemplified below. 7 3 21 7 300 2,100 700 3 2,100 Language focus “21 is a multiple of 3, so… 2,100 is a multiple of 300” 2,100 is a multiple of 3” 246 Pupils should learn to express multiples of 10 or 100 as products of 3 factors, for example: 7 3 21 so 7 3 10 210 Pupils should learn that these factors can be written in any order (commutative property of multiplication) and that any pair of the factors can be multiplied together first (associative property of multiplication). Applying commutativity Applying associativity (example) Pupils should be able to recognise whether any given number is a multiple of 2, 5, or 10 by attending to the final digit and, conversely, recognise 2, 5, or 10 as factors. Pupils should also be able to recognise multiples and factors linked to their experience of dividing powers of 10 into 2, 4 or 5 equal parts, by attending to the appropriate digit(s), for example: 175 is a multiple of 25 25 is a factor of 175 (attending to the final 2 digits) 8,500 is a multiple of 500 500 is factor of 8,500 (attending to the final 3 digits) 380 is a multiple of 20 20 is a factor of 380 (attending to the final 2 digits) Pupils should learn to identify factors and multiples for situations other than those described above by using short division or divisibility rules. For example, to determine whether 392 is a multiple of 8 (or whether 8 is a factor of 392) pupils can use the divisibility rule for 8 or use short division to determine whether 392 8 results in a quotient without a remainder. Pupils must learn how to find common factors and common multiples of small numbers in preparation for simplifying fractions and finding common denominators. They must also learn to recognise and use squared numbers and use the correct notation (for example, 2 3 9 ), and learn to establish whether a given number (up to 100) is prime. 247 Making connections Pupils must be fluent in their multiplication tables to meet this criterion (5NF–1), and must also be able to scale multiplication facts by 10 or 100. Short division (5MD–4) can be used to identify factors when other strategies are not applicable. 5MD–2 Example assessment questions Write all of the numbers from 1 to 30 in the correct places on this Venn diagram. Circle any number that is a multiple of both 3 and 7. 42 43 47 49 Find a common factor of 48 and 64 that is greater than 6. How many common multiples of 4 and 6 are there that are less than 40? Circle any number that is a factor of both 24 and 36. 2 4 6 8 10 12 a. Find a multiple of 30 that is between 200 and 300. b. Find a multiple of 40 that is between 300 and 400. c. Find a multiple of 50 that is between 400 and 500. Show that 3 is a factor of 231. 248 Fill in the table with examples of 2-, 3- and 4-digit numbers that are multiples of 9, 25 and 50. 2-digit number 3-digit number 4-digit number Multiples of 9 Multiples of 25 Multiples of 50 Give two 2-digit factors of 270. Find 3 numbers which are multiples of 25 but not multiples of 50. Fill in the missing numbers. 6 32 6 4 6 5 4 5 480 8 10 72 2 6 5 105 7 140 5MD–3 Multiply using a formal written method Multiply any whole number with up to 4 digits by any one-digit number using a formal written method. 5MD–3 Teaching guidance Pupils should learn to multiply multi-digit numbers by one-digit numbers using the formal written method of short multiplication. They should begin with examples that do not involve regrouping, such as the two-digit one-digit calculation shown below, and learn that, like columnar addition and subtraction, the algorithm begins with the least significant digit (on the right). When pupils first learn about short multiplication, place-value equipment (such as place-value counters) should be used to model the algorithm and help pupils relate it to what they already know about multiplication. Pupils should understand that short multiplication is based on the distributive property of multiplication which they learnt about in year 4 (4MD–3 ). Figure 179: place-value counters showing 34 2 249 Informal written method Expanded multiplication algorithm Short multiplication Initially, pupils should use unitising language to help them understand and apply short multiplication. Language focus “2 times 4 ones is equal to 8 ones: write 8 in the ones column.” “2 times 3 tens = 6 tens: write 6 in the tens column.” Pupils may also use place-value headings while they learn to use the formal written method, as illustrated above. However, by the end of year 5, they must be able to use the short multiplication algorithm without using unitising language and place-value headings. Once pupils have mastered the basic principles of short multiplication without regrouping, they must learn to use the algorithm where regrouping is required, for multiplication of numbers with up to 4 digits by one-digit numbers. Pupils can again use unitising language, now for support with regrouping, until they are able to apply the algorithm fluently. Figure 180: multiplying 367 by 4 using short multiplication 4 6 tens 24 tens 2 hundreds 4 s tens plus 2 more tens 2 hundreds 6 4 3 hundreds 4 12 hundreds 7 ones 2 1 thousand hundreds plus 2 more 1 thousand 28 ones 2 tens 8 on 4 hundred d e n s r te u n h d e s s 250 Pupils must be able to use short multiplication to solve contextual multiplication problems with: the grouping structure (see 5MD–3 5MD-3, questions 3 to 6) the scaling structure (see 5MD–3 Example assessment questions, questions 7 and 8) Pupils should also be able to use short multiplication to solve missing-dividend problems (for example, ? 2,854 3 ) Pupils must learn that, although short multiplication can be used to multiply any number by a one-digit number, it is not always the most appropriate choice. For example, 201 4 can be calculated mentally by applying the distributive property of multiplication (200 4 800 , plus 4 more). You can find out more about recording and fluency for these calculations here in the calculation and fluency section: 5MD–3 Making connections Pupils must be fluent in multiplication facts within the multiplication tables (5NF–1) before they begin this criterion. Once pupils have learnt short division (5MD–4) they should be able to use short multiplication to check their short division calculations, and vice versa. Pupils should be able to use short multiplication, where appropriate, when calculating a non-unit-fraction of a quantity (5F–1). 5MD–3 Example assessment questions Fill in the missing numbers. 278 6 7 1 ,297 2,854 3 6 372 251 Draw a line to match each multiplication expression with the correct addition expression. 120 + 18 80+4 120 + 24 Josh cycles 255 metres in 1 minute. If he keeps cycling at the same speed, how far will he cycle in 8 minutes? A factory packs biscuits into boxes of 9. The factory produces 1,350 packets of biscuits in a day. How many biscuits is that? Ellen has 1 large bag of 96 marbles, and 4 smaller bags each containing 76 marbles. How many marbles does she have altogether? There are 6 eggs in a box. If a farmer needs to deliver 1,275 boxes of eggs to a supermarket, how many eggs does she need? Aryan’s grandmother lives 235 kilometres away from Aryan. His aunt lives 3 times that distance away from Aryan. How far away does Aryan’s aunt live from him? How far is this to the nearest 100 kilometres? Felicity can make 5 hairbands in 1 hour. A factory can make 235 times as many. How many hairbands can the factory make in 1 hour? Fill in the missing numbers. Liyana writes: 9,565 7 1 ,365 Use short multiplication to check whether Liyana’s equation is correct. Assessment guidance: Pupils need to be able to identify when multiplication is the appropriate operation to use to solve a given problem. Assessment of whether a pupil has mastered multiplication sufficiently to progress to year 6 should also include questions which require other operations to solve. 252 5MD–4 Divide using a formal written method Divide a number with up to 4 digits by a one-digit number using a formal written method, and interpret remainders appropriately for the context. 5MD–4 Teaching guidance Pupils should learn to divide multi-digit numbers by one-digit numbers using the formal written method of short division. They should begin with examples that do not involve exchange, such as the two-digit one-digit calculation shown below. Pupils should learn that division is the only operation for which the formal algorithm begins with the most significant digit (on the left). When pupils first learn about short division, place-value equipment (such as place-value counters) should be used to model the algorithm and help pupils relate it to what they already know about division. They should understand that short division is based on the distributive property of multiplication which they learnt about in year 4 (4MD–3 ). Short division with place-value counters Short division Figure 181: dividing 84 by 4 using short division with place-value counters Figure 182: dividing 84 by 4 using short division with place-value headings Initially, pupils should use unitising language to help them understand and apply short division. Language focus “8 tens divided by 4 is equal to 2 tens: write 2 in the tens column.” “4 ones divided by 4 is equal to 1 one: write 1 in the ones column.” 253 Pupils may also use place-value headings while they learn to use the formal written method, as illustrated above. However, by the end of year 5, pupils must be able to use the short division algorithm without using unitising language and place-value headings. Once pupils have mastered the basic principles of short division without exchange, they must learn to use the algorithm where exchange is required, for division of numbers with up to 4 digits by one-digit numbers. Pupils can again use unitising language, now for support with exchange, until they are able to apply the algorithm fluently. Figure 183: dividing 612 by 4 using short division 2 hundreds 20 tens plus s 1 more ten 21tens 21 tens 4 5 6 tens remainder 1 d e 1t 1 en hundreds 4 hundred r ma s inder 2 hu 10 one plus 2 more ones 12 ones 12 ones te 4 n red n 3 one s Pupils must be able to use short division to solve contextual division problems with: the quotitive structure (see 5MD–4 Example assessment questions, questions 5 and 8) the partitive structure (see 5MD–4 Example assessment questions, questions 2, 6 and 7) Pupils should also be able to use short division to find unit fractions of quantities, and to solve missing-factor problems (for example, ?
5 1 ,325 ) and missing-divisor problems (for example, 952 ?
7 ). Pupils must be able to carry out short division calculations that involve a remainder and, for contextual problems, interpret the remainder appropriately as they learnt to do in year 4 (4NF–2 ). Pupils must learn that, although short division can be used to divide any number by a one-digit number, it is not always the most appropriate choice. For example, 804 4 , can be calculated mentally by partitioning, dividing and adding the partial quotients (804 4 200 , plus 1 more). You can find out more about recording and fluency for these calculations here in the calculation and fluency section: 5MD–4 254 Making connections Pupils must be fluent in division facts corresponding to the multiplication tables (5NF–1) before they begin this criterion. Pupils should be able to use short multiplication (5MD–3) to check their short division calculations, and vice versa. Pupils should be able to use short division, where appropriate, to find a unit fraction of a quantity, and as the first step in finding a non-unit fraction of a quantity (5F–1). 5MD–4 Example assessment questions Fill in the missing numbers. 4,728 8 952 7 5 1 ,325 176 4 I have 1 2 1 litres of juice which I need to share equally between 6 glasses. How many millilitres of juice should I pour into each glass? A school fair raises £5,164. The school keeps 1 4 of the money for new playground equipment and gives the rest to charity. How much money does the school keep? Fryderyk has saved 4 times as much money as his sister Gabriel. If Fryderyk has saved £348, how much has Gabriel saved? A farmer has 3,150 eggs to pack into boxes of 6. How many boxes does she need? Sharif wants to walk a long distance, for charity, over 6 weekends. The total distance Sharif wants to walk is 293km. Approximately how far should he walk each weekend? Maria makes 1,531g of cake mix. She puts 250g into a small cake tin and wants to share the rest equally between 3 large cake tins. How many grams of cake mix should she put in each large cake tin? 174 children are going on a trip. 4 children can fit into each room in the hostel. How many rooms are needed? Fill in the missing numbers. David writes: 785 9 7,065 Use short division to check whether David’s calculation is correct. 255 Assessment guidance: Pupils need to be able to identify when division is the appropriate operation to use to solve a given problem. Assessment of whether a pupil has mastered division sufficiently to progress to year 6 should also include questions which require other operations to solve. 5F–1 Find non-unit fractions of quantities Find non-unit fractions of quantities. 5F–1 Teaching guidance Pupils should already be able to find unit fractions of quantities using known division facts corresponding to multiplication table facts (3F–2). Language focus “To find of 15, we divide 15 into 5 equal parts.” “15 divided by 5 is equal to 3, so of 15 is equal to 3.” By the end of year 5 pupils must be able to find unit and non-unit fractions of quantities, including for situations that go beyond known multiplication and division facts. Pupils already understand the connection between a unit fraction of a quantity and dividing that quantity by the denominator. Now they should learn to reason about finding a non-unit fraction of a quantity, using division (to find the unit fraction) then multiplication (to find multiples of the unit fraction), and link this to their understanding of parts and wholes. Initially, calculations should depend upon known multiplication and division facts, so that pupils can focus on reasoning. 256 Figure 184: bar model to support finding three-fifths of 40 40 5 8 1 So, of 40 8 5 8 3 24 3 So, of 40 24 5 Language focus “Three-fifths is equal to 3 one-fifths.” “To find 3 one-fifths of 40, first find one-fifth of 40 by dividing by 5, and then multiply by 3.” Once pupils can carry out these calculations fluently, and explain their reasoning, they should extend their understanding to calculate unit and non-unit fractions of quantities for calculations that go beyond known multiplication table facts. For example, they should be able to: apply place-value understanding to known number facts to find 3 7 of 210 use short division followed by short multiplication to find 4 9 of 3,411 Pupils should also be able to construct their own bar models to solve more complex problems related to fractions of quantities. For example: Miss Reeves has some tangerines to give out during break-time. She has given out 5 6 of the tangerines, and has 30 left. How many tangerines did Miss Reeves have to begin with? Figure 185: using a bar model to solve more complex problems related to fractions of quantities 257 Making connections Pupils must be fluent in multiplication facts within the multiplication tables, and corresponding division facts (5NF–1). They must be able to confidently scale these facts by 10 or 100 (3NF–3 and 4NF–3) to find, for example, of 210. Pupils also need to be able to calculate using short multiplication (5MD–3) and short division (5MD–4) to be able to find, for example, of 3,411. 5F–1 Example assessment questions Find: 3 8 of 32 2 9 of 45 3 5 of 30 2 7 of 630 4 9 of 315 2 5 of 3,500 5 8 of 2,720 Stan bought 15 litres of paint and used 2 3 of it decorating his house. How much paint has he used? My granny lives 120km from us. We are driving to see her and are 5 6 of the way there. How far have we driven so far? I am 3 4 of the way through my holiday. I have 3 days of holiday left. How many days have I already been on holiday for? A school is trying to raise £7,500 for charity. They have raised 5 6 of the total so far. How much have they raised? 4 5 of the runners in a race have finished the race so far. If 92 people have finished, how many runners were in the race altogether? There are 315 cows on a farm. 3 5 of the cows are having calves this year. How many cows are not having calves? 258 5F–2 Find equivalent fractions Find equivalent fractions and understand that they have the same value and the same position in the linear number system. 5F–2 Teaching guidance Pupils must understand that more than one fraction can describe the same portion of a quantity, shape or measure. They should begin with an example where one of the fractions is a unit fraction, and the connection to the equivalent fraction uses known multiplication table facts. 1 3 4 12 Figure 186: circle divided into 4 equal parts with 1 part shaded Language focus “The whole is divided into 4 equal parts and 1 of those parts is shaded.” Figure 187: circle divided into 12 equal parts with 3 parts shaded Language focus “The whole is divided into 12 equal parts and 3 of those parts is shaded.” Figure 188: diagram showing that of 12 cakes is equal to 3 cakes Figure 189: diagram showing that of 12 cakes is equal to 3 cakes 259 Pupils should learn that 2 different fractions describing the same portion of the whole share the same position on a number line, have the same numerical value and are called equivalent fractions. Figure 190: number line showing that 1 4 and 3 12 are equivalent Pupils need to understand that equivalent fractions, such as 1 4 and 3 12 , have the same numerical value because the numerator and denominator within each fraction have the same proportional relationship. In each case the numerator is 1 4 of the denominator (and the denominator is 4 times the numerator). Language focus “ and are equivalent because 1 is the same portion of 4 as 3 is of 12.” Attending to the relationship between the numerator and denominator will prepare pupils for comparing fractions with different denominators in year 6 (6F–3). Pupils should also be able to identify the multiplicative relationship between the pair of numerators, and understand that it is the same as that between the pair of denominators. Pupils should learn to find equivalent fractions of unit fractions by using one of these multiplicative relationships (the ‘vertical’ relationship between the numerator and denominator, or the ‘horizontal’ relationship between the pairs of numerators and denominators). Figure 191: diagram showing the multiplicative relationships between the numerators and denominators in 1 4 and 3 12 260 In a similar way, pupils must then learn to find equivalent fractions of non-unit fractions, for example, 3 6 5 10 or 3 8 12 32 . Making connections Pupils must be fluent in multiplication facts within the multiplication tables, and corresponding division facts (5NF–1). Being able to find unit and non-unit fractions of a quantity (5F–1) helps pupils to see that equivalent fractions have the same value. 5F–2 Example assessment questions Find different ways to write the fraction of each shape or quantity that is shaded or highlighted. Draw lines to match the unit fractions on the left with their equivalent fractions on the right. 261 Mark each fraction on the number line. 9 24 36 48 12 16 10 40 9 72 Hint: convert each fraction to an equivalent fraction with a denominator of 8. Use the numbers 3, 24, 8 and 1 to complete this chain of equivalent fractions. 2 6 Fill in the missing digits. 12 4 8 3 5 40 3 21 63 20 30 15 Fill in the missing number. Sally and Tahira each have a 1m ribbon. Sally cuts her ribbon into 5 equal parts and uses 1 of them to make a hair tie. Tahira cuts her ribbon into 10 equal parts and uses 3 of them to make a bracelet. Have Sally and Tahira used the same amount of ribbon? Explain your answer. 262 5F–3 Recall decimal equivalents for common fractions Recall decimal fraction equivalents for , , and , and for multiples of these proper fractions. 5F–3 Teaching guidance Pupils know that both proper fractions and decimals fractions can be used to represent values between whole numbers. They now need to learn that the same value can be represented by both a decimal fraction and a proper fraction, and be able to recall common equivalents, beginning with unit fractions. Unit fraction Decimal fraction 0.5 0.25 0.2 0.1 Pupils should also be able to explain the equivalencies. A shaded hundred grid is a useful representation here. Figure 192: hundred grid divided into 4 equal parts: 1 4 is equal to 25 hundredths 1 25 4 100 0.25 Pupils should then extend their understanding and automatic recall to multiples of these unit fractions/decimal fractions, up to 1. 263 Figure 193: 0 to 1 number lines illustrating common proper fraction – decimal fraction equivalents Pupils must be able to use these common equivalents with little effort, applying them to solve comparison and measures problems such as those shown in the example assessment questions. For a given problem, posed using a mixture of decimal fractions and proper fractions, pupils should be able to make a sensible decision about whether to carry out the calculation using decimal fractions or proper fractions. Finally, pupils need to extend this knowledge beyond the 0 to 1 interval. They should know for example that 3.2km and 1 5 3 km are 2 different ways of writing the same distance. Making connections This criterion builds on 5NPV–4, where pupils learnt to divide 1 into 2, 4, 5 or 10 equal parts and to read scales marked in multiples of multiples of 0.1, 0.2, 0.25 or 0.5. Criterion 5NPV–5 requires pupils to convert between units of measure, including using the common decimal fraction and proper fraction equivalents in this criterion. 264 5F–3 Example assessment questions Fill in the missing symbols (<, > or =). 1 10 0.75 1 4 0.4 1 5 0.5 3 4 0.75 4 5 0.8 1 2 0.2 Write these measurements as mixed numbers. 1.2km 5.75m 25.5kg Write these measurements as decimals. 1 4 1 litres 1 2 10 cm 4 5 4 m My brother weighs 27.3kg. I weigh 1 2 27 kg. How much more than my brother do I weigh? Year 6 set off on a 3 4 2 km woodland walk. By lunch, they had walked 1.75km. How much further do they need to walk? Here are two parcels: What is the total combined weight of the parcels, in kilograms? Put each set of numbers in order from smallest to greatest. a. 1.4 1 4 4 4.1 4.4 b. 1 5 3 3.5 3 5 1 1.3 265 5G–1 Compare, estimate, measure and draw angles Compare angles, estimate and measure angles in degrees (°) and draw angles of a given size. 5G–1 Teaching guidance In year 3, pupils learnt to identify right angles and to determine whether a given angle is larger or smaller than a right angle (3G–1). Pupils should now compare angles, including the internal angles of polygons, and be able to identify the largest and smallest angles when there is a clear visual difference. Pupils should be able to use the terms acute, obtuse and reflex when describing and comparing angles, and use conventional markings (arcs) to indicate angles. Language focus “An acute angle is smaller than a right angle.” “An obtuse angle is larger than a right angle but less than the angle on a straight line.” “A reflex angle is larger than the angle on a straight line, but less than the angle for a full turn.” Figure 194: irregular pentagon with 3 acute internal angles, 1 obtuse internal angle and 1 reflex internal angle Language focus “D is the smallest angle. It is an acute angle.” “C is the largest angle. It is a reflex angle.” 266 Pupils must learn that we can measure the size of angles just as we can measure the length of sides. They should learn that the unit used is called degrees and indicated by the ° symbol. Pupils should know that there are 360° in a full turn, 90° in a quarter turn or right angle, and 180° in a half turn or on a straight line. Pupils must know that the position of the arc indicating an angle does not affect the size of the angle, which is determined by the amount of turn between the two lines. Similarly, they should know that the length of the lines does not affect the size of the angle between them. Figure 195: the position of the arc indicating an angle does not affect the size of the angle Figure 196: the length of the lines does not affect the size of the angle between them Before pupils learn to use protractors, they should learn to estimate and approximate common angles, and angles that are close to them, including 90°, 180°, other multiples of 10°, and 45°. They should use sets of ‘standard angle’ measuring tools (for example, cut out from card) for support in approximating, and to check estimates. Figure 197: a selection of ‘standard angle’ measuring tools 267 Once pupils can make reasonable estimates of angle size, they must learn to make accurate measurements, using a protractor, for angles up to 180°. It is good practice to make an estimate before taking an accurate measurement, and pupils should use learn to use their estimates for support in reading the correct value off the protractor. Pupils should also, now, be able to use the more formal definitions of acute, obtuse and reflex. Language focus “An acute angle is less than 90°.” “An obtuse angle is greater than 90° but less than 180°.” “A reflex angle is greater than 180° but less than 360°.” 5G–1 Example assessment questions Do not use a protractor for questions 1, 2 and 3. Here is an irregular pentagon. a. Which is the largest angle in this pentagon? b. Which is the smallest angle? c. Which angle is 100°? 268 Here are 6 angles. a. Which is the largest angle? b. Which is the smallest angle? c. Which angle is 45°? This pentagon has a line of symmetry. Estimate the size of each angle. Measure and label each of the angles in these shapes using a protractor. a. Draw an angle of 68°. b. Draw an angle of 103°. 269 5G–2 Compare and calculate areas Compare areas and calculate the area of rectangles (including squares) using standard units. 5G–2 Teaching guidance Pupils need to know that the area of a shape is the space within a shape. When there is a clear visual difference, pupils should be able to compare the area of shapes without making a quantitative evaluation of each area. For example, pupils can see that the circle has a larger area than the decagon. Figure 198: a decagon and a circle with a clear visual difference in area Pupils should learn that, when there is not a clear visual difference between areas, a common unit can be used to quantify the areas and enable comparison. They should understand that any unit can be used, but that the square centimetre (cm2) is the standard unit of measure for area that they will use most frequently. Pupils should gain a sense of the size of a square centimetre, and the notation used, before they begin to quantify other areas using this unit. Figure 199: a square centimetre Pupils need to be able to find the area of shapes drawn on square-centimetre grids by counting squares, including shapes for which some of the area is made up of half-squares. They should understand that different shapes can have the same area. 270 Figure 200: a rectangle, square and triangle with equal areas Drawn to actual size. Pupils should then learn that the area of a rectangle can be calculated by multiplying the length by the width. They should learn why this is the case by examining rectangles drawn on square-centimetre grids, and understand that the factors can be written in either order: the area of the rectangle below is equal to 4 rows of 5 square centimetres, or 5 columns of 4 square centimetres. This should build on pupils understanding of the grouping structure of multiplication and array representations. Figure 201: the area of a rectangle can be calculated by multiplying the length by the width Drawn to actual size. Language focus “To find the area of a rectangle, multiply the length by the width.” Pupils should learn that the area of larger shapes and spaces, such as the floor or ceiling of the classroom, or the playground, is expressed in square metres (m2). Pupils should experience working with large spaces directly, as well as drawings representing them. 271 Making connections Pupils must be able to multiply two numbers together in order to calculate the area of a rectangle, including: known multiplication facts within the multiplication tables (5NF–1) (for example, to calculate the area of a 9cm by 4cm rectangle) scaling known multiplication facts by 10 or 100 (3NF–3 , 4NF–3 and 5NF–2) (for example, to calculate the area of a rectangle or a rectangle) other mental or written methods (for example, to calculate the area of a rectangle) 5G–2 Example assessment questions For each pair of shapes, tick the shape with the larger shaded area. 272 Find the area of these shapes drawn on a square-centimetre grid. Drawn to actual size. Here are three shapes on a triangular grid. Put the shapes in order from smallest to largest according to their area. 273 a. Draw a rectangle with an area of 12cm2 on this square-centimetre grid. b. Draw a hexagon with an area of 12cm2 on this square-centimetre grid. Drawn to actual size. Find the area of each of these rectangles. Leila is putting some tiles on the wall behind her kitchen sink. Each tile is square, with sides equal to 10cm. Here is the area she has tiled so far. If Leila adds one more row of tiles on top of these ones, what is the total area she will have tiled? 274 Each half of a volleyball court is a 9m 9m square. What is the total area of a volleyball court? Drawn to scale. Estimate the area of your classroom floor. Calculation and fluency Number, place value and number facts: 5NPV–2 and 5NF–2 5NPV–2 Recognise the place value of each digit in numbers with up to 2 decimal places, and compose and decompose numbers with up to 2 decimal places using standard and non-standard partitioning. 5NF–2 Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 1 tenth or 1 hundredth), for example: Representations such as place-value counters and partitioning diagrams (5NPV–2) and tens-frames with place-value counters (5NF–2) can be used initially to help pupils understand calculation strategies and make connections between known facts and related calculations. However, pupils should not rely on such representations for calculating. For the calculations in 5NF–2, for example, pupils should instead be able to calculate by verbalising the relationship. 275 Language focus “8 plus 6 is equal to 14, so 8 tenths plus 6 tenths is equal to 14 tenths.” “14 tenths is equal to 1 one and 4 tenths.” Pupils should maintain fluency in both formal written and mental methods for addition and subtraction. Mental methods can include jottings to keep track of calculation, or language structures as exemplified above. Pupils should select the most efficient method to calculate depending on the numbers involved. Addition and subtraction: extending 3AS–3 Pupils should also extend columnar addition and subtraction methods to numbers with up to 2 decimal places. Pupils must be able to add 2 or more numbers using columnar addition, including calculations whose addends have different numbers of digits. Figure 202: columnar addition for calculations involving numbers with up to 2 decimal places For calculations with more than 2 addends, pupils should add the digits within a column in the most efficient order. For the third example above, efficient choices could include: beginning by making 10 in the tenths column making double-6 in the ones column Pupils must be able to subtract one number from another using columnar subtraction, including numbers with up to 2 decimal places. They should be able to apply the columnar method to calculations presented as, for example, 21.8 – 9.29 or 58 14.69 , where the subtrahend has more decimal places than the minuend. Pupils must also be able to exchange through 0. 276 Figure 203: columnar subtraction for calculations involving numbers with up to 2 decimal places Pupils should make sensible decisions about how and when to use columnar methods. For example, when subtracting a decimal fraction from a whole numbers, pupils may be able to use their knowledge of complements, avoiding the need to exchange through zeroes. For example, to calculate 8 – 4.85 pupils should be able to work out that the decimal complement to 5 from 4.85 is 0.15, and that the total difference is therefore 3.15. 5NF–1 Secure fluency in multiplication and division facts Secure fluency in multiplication table facts, and corresponding division facts, through continued practice. Pupils who have automatic recall of multiplication table facts and corresponding division facts have the best chance of mastering formal written methods. The facts up to 9 9 are required for calculation within the ‘columns’ during application of formal written methods, and all mental multiplicative calculation also depends on these facts. Pupils will need regular practice of multiplication tables and associated division facts (including calculating division facts with remainders) to maintain the fluency they achieved by the end of year 4. Pupils should also maintain fluency in related calculations including: scaling known multiplicative facts by 10 or 100 (3NF–3 and 4NF–3) multiplying and dividing by 10 and 100 for calculations that involve whole numbers only (4MD–1) They should develop fluency in: scaling multiplicative facts by one-tenth or one-hundredth (5NF–2) multiplying and dividing by 10 and 100, for calculations that bridge 1 (5MD–1) 277 5MD–3 Multiply using a formal written method Multiply any whole number with up to 4 digits by any one-digit number using a formal written method. Pupils must be able to multiply whole numbers with up to 4 digits by one-digit numbers using short multiplication. Figure 204: short multiplication for multiplication of 2-, 3- and 4-digit numbers by one-digit numbers Pupils should be fluent in interpreting contextual problems to decide when multiplication is the appropriate operation to use, including as part of multi-step problems. Pupils should use short multiplication when appropriate to solve these calculations. Examples are given in 5MD–3. 5MD–4 Divide using a formal written method Divide a number with up to 4 digits by a one-digit number using a formal written method, and interpret remainders appropriately for the context. Pupils must be able to divide numbers with up to 4 digits by one-digit numbers using short division, including calculations that involve remainders. Pupils do not need to be able to express remainders arising from short division, using proper fractions or decimal fractions. Figure 205: short division for division of 2-, 3- and 4-digit numbers by one-digit numbers Pupils should be fluent in interpreting contextual problems to decide when division is the appropriate operation to use, including as part of multi-step problems. Pupils should use short division when appropriate to solve these calculations. For contextual problems, pupils must be able to interpret remainders appropriately as they learnt to do in year 4 (4NF–2 ). Examples are given in 5MD–4 Example assessment questions. 278 Year 6 guidance Ready-to-progress criteria Year 5 conceptual prerequesite Year 6 ready-to-progress criteria Key stage 3 applications Understand the relationship between powers of 10 from 1 hundredth to 1,000 in terms of grouping and exchange (for example, 1 is equal to 10 tenths) and in terms of scaling (for example, 1 is ten times the size of 1 tenth). 6NPV–1 Understand the relationship between powers of 10 from 1 hundredth to 10 million, and use this to make a given number 10, 100, 1,000, 1 tenth, 1 hundredth or 1 thousandth times the size (multiply and divide by 10, 100 and 1,000). Understand and use place value for decimals, measures, and integers of any size. Interpret and compare numbers in standard form , where n is a positive or negative integer or zero. Recognise the place value of each digit in numbers with units from thousands to hundredths and compose and decompose these numbers using standard and non-standard partitioning. 6NPV–2 Recognise the place value of each digit in numbers up to 10 million, including decimal fractions, and compose and decompose numbers up to 10 million using standard and non-standard partitioning. Understand and use place value for decimals, measures, and integers of any size. Order positive and negative integers, decimals, and fractions. Use a calculator and other technologies to calculate results accurately and then interpret them appropriately. Reason about the location of numbers between 0.01 and 9,999 in the linear number system. Round whole numbers to the nearest multiple of 1,000, 100 or 10, as appropriate. Round decimal fractions to the nearest whole number or nearest multiple of 0.01 6NPV–3 Reason about the location of any number up to 10 million, including decimal fractions, in the linear number system, and round numbers, as appropriate, including in contexts. Order positive and negative integers, decimals, and fractions; use the number line as a model for ordering of the real numbers; use the symbols =, ≠, <, >, ≤, ≥ Round numbers and measures to an appropriate degree of accuracy (for example, to a number of decimal places or significant figures). Use approximation through rounding to estimate answers and calculate possible resulting errors expressed using inequality notation a < x ≤ b 279 Year 5 conceptual prerequesite Year 6 ready-to-progress criteria Key stage 3 applications Divide 1000, 100 and 1 into 2, 4, 5 and 10 equal parts, and read scales/number lines with 2, 4, 5 and 10 equal parts. 6NPV–4 Divide powers of 10, from 1 hundredth to 10 million, into 2, 4, 5 and 10 equal parts, and read scales/number lines with labelled intervals divided into 2, 4, 5 and 10 equal parts. Use standard units of mass, length, time, money, and other measures, including with decimal quantities. Construct and interpret appropriate tables, charts, and diagrams. Be fluent in all key stage 2 additive and multiplicative number facts (see Appendix: number facts fluency overview ) and calculation. Manipulate additive equations, including applying understanding of the inverse relationship between addition and subtraction, and the commutative property of addition. Manipulate multiplicative equations, including applying understanding of the inverse relationship between multiplication and division, and the commutative property of multiplication. 6AS/MD–1 Understand that 2 numbers can be related additively or multiplicatively, and quantify additive and multiplicative relationships (multiplicative relationships restricted to multiplication by a whole number). Understand that a multiplicative relationship between 2 quantities can be expressed as a ratio or a fraction. Express 1 quantity as a fraction of another, where the fraction is less than 1 and greater than 1. Interpret mathematical relationships both algebraically and geometrically. Interpret when the structure of a numerical problem requires additive, multiplicative or proportional reasoning. Make a given number (up to 9,999, including decimal fractions) 10, 100, 1 tenth or 1 hundredth times the size (multiply and divide by 10 and 100). Apply place-value knowledge to known additive and multiplicative number facts (scaling facts by 10, 100, 1 tenth or 1 hundredth). Manipulate additive equations. Manipulate multiplicative equations. 6AS/MD–1 Use a given additive or multiplicative calculation to derive or complete a related calculation, using arithmetic properties, inverse relationships, and place-value understanding. Recognise and use relationships between operations including inverse operations. Use algebra to generalise the structure of arithmetic, including to formulate mathematical relationships. Understand and use standard mathematical formulae; rearrange formulae to change the subject. 280 Year 5 conceptual prerequesite Year 6 ready-to-progress criteria Key stage 3 applications Recall multiplication and division facts up to . Apply place-value knowledge to known additive and multiplicative number facts. 6AS/MD–3 Solve problems involving ratio relationships. Use ratio notation, including reduction to simplest form. Divide a given quantity into 2 parts in a given part:part or part:whole ratio; express the division of a quantity into 2 parts as a ratio. Be fluent in all key stage 2 additive and multiplicative number facts and calculation. Manipulate additive equations. Manipulate multiplicative equations. Find a fraction of a quantity. 6AS/MD–4 Solve problems with 2 unknowns. Reduce a given linear equation in two variables to the standard form y = mx + c; calculate and interpret gradients and intercepts of graphs of such linear equations numerically, graphically and algebraically. Use linear and quadratic graphs to estimate values of y for given values of x and vice versa and to find approximate solutions of simultaneous linear equations. Recall multiplication and division facts up to . Find factors and multiples of positive whole numbers, including common factors and common multiples. Find equivalent fractions and understand that they have the same value and the same position in the linear number system. 6F–1 Recognise when fractions can be simplified, and use common factors to simplify fractions. Use the concepts and vocabulary of prime numbers, factors (or divisors), multiples, common factors, common multiples, highest common factor, lowest common multiple, prime factorisation, including using product notation and the unique factorisation property. Simplify and manipulate algebraic expressions by taking out common factors. 281 Year 5 conceptual prerequesite Year 6 ready-to-progress criteria Key stage 3 applications Recall multiplication and division facts up to . Find factors and multiples of positive whole numbers. Find equivalent fractions. Reason about the location of fractions and mixed numbers in the linear number system. 6F–2 Express fractions in a common denomination and use this to compare fractions that are similar in value. Order positive and negative integers, decimals and fractions. Use the 4 operations, including formal written methods, applied to integers, decimals, proper and improper fractions, and mixed numbers, all both positive and negative. Use and interpret algebraic notation, including: a/b in place of coefficients written as fractions rather than as decimals. Reason about the location of fractions and mixed numbers in the linear number system. Find equivalent fractions. 6F–3 Compare fractions with different denominators, including fractions greater than 1, using reasoning, and choose between reasoning and common denomination as a comparison strategy. Order positive and negative integers, decimals, and fractions; use the number line as a model for ordering of the real numbers; use the symbols =, ≠, <, >, ≤, ≥ Find the perimeter of regular and irregular polygons. Compare angles, estimate and measure angles in degrees (°) and draw angles of a given size. Compare areas and calculate the area of rectangles (including squares) using standard units. 6G–1 Draw, compose, and decompose shapes according to given properties, including dimensions, angles and area, and solve related problems. Draw shapes and solve more complex geometry problems (see Mathematics programmes of study: key stage 3 - Geometry and measures). 282 6NPV–1 Powers of 10 Understand the relationship between powers of 10 from 1 hundredth to 10 million, and use this to make a given number 10, 100, 1,000, 1 tenth, 1 hundredth or 1 thousandth times the size (multiply and divide by 10, 100 and 1,000). 6NPV–1 Teaching guidance An understanding of the relationship between the powers of 10 prepares pupils for working with much larger or smaller numbers at key stage 3, when they will learn to read and write numbers in standard form (for example, 8 600,000,000 6 10 ). Pupils need to know that what they learnt in year 3 and year 4 about the relationship between 10, 100 and 1,000 (see 3NPV–1 and 4NPV–1 ), and in year 5 about the relationship between 1, 0.1 and 0.01 (5NPV–1) extends through the number system. By the end of year 6, pupils should have a cohesive understanding of the whole place-value system, from decimal fractions through to 7-digit numbers. Pupils need to be able to read and write numbers from 1 hundredth to 10 million, written in digits, beginning with the powers or 10, as shown below, and should understand the relationships between these powers of 10. one hundredth one tenth one one hundred one thousand ten thousand one hundred thousand one million ten million 0 . 0 1 0 . 1 1 1 0 ten 1 0 0 1 , 0 0 0 1 0 , 0 0 0 1 0 0 , 0 0 0 1 , 0 0 0 , 0 0 0 1 0 , 0 0 0 , 0 0 0 Pupils should know that each power of 10 is equal to 1 group of 10 of the next smallest power of 10, for example 1 million is equal to 10 hundred thousands. Figure 206: ten 100,000-value place-value counters in a tens frame 283 Language focus “10 hundred-thousands is equal to 1 million.” Pupils should also understand this relationship in terms of scaling by 10 or one-tenth. Language focus “1,000,000 is 10 times the size of 100,000.” “100,000 is one-tenth times the size of 1,000,000.” Pupils must also understand the relationships between non-adjacent powers of 10 up to a scaling by 1,000 or 1 thousandth (or grouping of up to 1,000 of a given power). Language focus “10 thousands is equal to 10,000.” “10,000 is 10 times the size of 1,000.” “1,000 is one-tenth times the size of 10,000.” Pupils must also be able to write multiples of these powers of 10, including when there are more than 10 of given power of 10, for example, 18 hundred thousands is written as 1,800,000. Pupils should be able to restate the quantity in the appropriate power of 10, for example 18 hundred thousands is equal to 1 million 8 hundred thousand. Once pupils understand the relationships between powers of ten, they should extend this to other numbers in the Gattegno chart. They must be able to identify the number that is 10, 100, 1,000, 1 tenth, 1 hundredth or 1 thousandth times the size of a given number, and associate this with multiplying or dividing by 10, 100 and 1,000. This will prepare pupils for multiplying by decimals in key stage 3, when they will learn, for example, that dividing by 100 is equivalent to multiplying by 0.01. 284 Figure 207: using the Gattegno chart to multiply and divide by 100 Language focus “50,000 is 100 times the size of 500.” “500 multiplied by 100 is equal to 50,000.” “500 is one-hundredth times the size of 50,000.” “50,000 divided by 100 is equal to 500.” Pupils should recognise the inverse relationship between, for example making a number 100 times the size, and returning to the original number by making it one-hundredth times the size. This understanding should then be extended to multiplicative calculations that involve numbers with more than one significant digit, extending what pupils learnt in 5MD–1 about multiplying and dividing by 10 and 100. 1 ,659 100 165,900 165,900 100 1 ,659 21 ,156 10 211 ,560 211 ,560 10 21 ,156 47.1 1 ,000 47,100 47,100 1 ,000 47.1 Pupils can use the Gattegno chart for support throughout this criterion, but by the end of year 6 they must be able to calculate without it. You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 6NPV–1 and 6NPV–2 285 Making connections Writing multiples of powers of 10 depends on 6NPV–2. In 6AS/MD–2 pupils use their understanding of place-value and scaling number facts to manipulate equations. 6NPV–1 Example assessment questions Complete the sentences. a. 500 made 1,000 times the size is . b. 0.7 made 100 times the size is . c. 800,000 made 10 times the size is . d. 4,000,000 made one-thousandth times the size is . e. 9,000 made one-hundredth times the size is . f. 3 made one-tenth times the size is . The distance from London to Bristol is about 170km. The distance from London to Sydney, Australia is about 100 times as far. Approximately how far is it from London to Sydney? A newborn elephant weighs about 150kg. A newborn kitten weighs about 150g. How many times the mass of a newborn kitten is a newborn elephant? Walid has a place-value chart and three counters. He has represented the number 1,110,000. a. Find 2 different numbers that Walid could make so that 1 number is one-hundredth times the size of the other number. b. Find 2 different numbers that Walid could make so that 1 number is 1,000 times the size of the other number. Fill in the missing numbers. 10 4.3 10 10 27,158 10 286 100 729 100 100 5,806 100 1 ,000 14.3 1 ,000 1 ,000 2,670,000 1 ,000 Use the following to complete the equations: 10 100 1 ,000 10 100 1 ,000 Use each term only once. 543 5.43 3,169 3,169,000 515 5,150 276,104 27,610.4 35,000 35 427 42,700 6NPV–2 Place value in numbers up to 10,000,000 Recognise the place value of each digit in numbers up to 10 million, including decimal fractions, and compose and decompose numbers up to 10 million using standard and non-standard partitioning. 6NPV–2 Teaching guidance Pupils must be able to read and write numbers up to 10,000,000, including decimal fractions. Pupils should be able to use a separator (such as a comma) every third digit from the decimal separator to help read and write numbers. Pupils must be able to copy numbers from calculator displays, inserting thousands separators and decimal points correctly. This will prepare them for secondary school, where pupils will be expected to know how to use calculators. Pupils need to be able to identify the place value of each digit in a number. 287 Language focus “In 67,000.4… the 6 represents 6 ten-thousands; the value of the 6 is 60,000 the 7 represents 7 thousands; the value of the 7 is 7,000 the 4 represents 4 tenths; the value of the 4 is 0.4” Pupils must be able to combine units from millions to hundredths to compose numbers, and partition numbers into these units, and solve related addition and subtraction calculations. Pupils need to experience variation in the order of presentation of the units, so that they understand, for example, that 5,034,000.2 is equal to 4,000 + 30,000 + 0.2 + 5,000,000. Pupils should be able to represent a given number in different ways, including using place-value counters and Gattegno charts, and write numbers shown using these representations. Pupils should then have sufficient understanding of the composition of large numbers to compare and order them by size. Pupils also need to be able to solve problems relating to subtraction of any single place-value part from a number, for example: 381 ,920 – 900 381 ,920 – 380,920 As well as being able to partition numbers in the ‘standard’ way (into individual place-value units), pupils must also be able to partition numbers in ‘non-standard’ ways, and carry out related addition and subtraction calculations, for example: 518.32 30 548.32 381 ,920 – 60,000 321 ,920 Pupils can initially use place-value counters for support with this type of partitioning and calculation, but by the end of year 6 must be able to partition and calculate without them. You can find out more about fluency and recording for these calculations here in the calculation and fluency section: Number, place value and number facts: 6NPV–1 and 6NPV–2 288 6NPV–2 Example assessment questions What is the value of the digit 5 in each of these numbers? a. 720,541 b. 5,876,023 c. 1,587,900 d. 651,920 e. 905,389 f. 2,120,806.50 g. 8,002,345 h. 701,003.15 Write a seven-digit number that includes the digit 8 once, where the digit has a value of: a. 8 million b. 8 thousand c. 8 hundred d. 80 thousand Fill in the missing symbols (< or >). 7,142,294 7,124,294 99,000 600,000 6,090,100 690,100 1 ,300,610 140,017 589,940 1 ,010,222 Put these numbers in order from smallest to largest. 8,102,304 8,021,403 843,021 8,043,021 289 6NPV–3 Numbers up to 10 million in the linear number system Reason about the location of any number up to 10 million, including decimal fractions, in the linear number system, and round numbers, as appropriate, including in contexts. 6NPV–3 Teaching guidance Pupils have already learnt about the location of whole numbers with up to 4 digits in the linear number system (1NPV–2, 2NPV–2, 3NPV–3 and 4NPV–3 ) and about the location of decimal fractions with up to 2 decimal places between whole numbers in the linear number system (5NPV–3). Pupils must now extend their understanding to larger numbers. Pupils need to be able to identify or place numbers with up to 7 digits on marked number lines with a variety of scales, for example placing 12,500 on a 12,000 to 13,000 number line, and on a 10,000 to 20,000 number line. Figure 208: placing 12,500 on a 12,000 to 13,000 number line marked, but not labelled, in multiples of 100 Figure 209: placing 12,500 on a 10,000 to 20,000 number line marked, but not labelled, in multiples of 1,000 Pupils need to be able to estimate the value or position of numbers on unmarked or partially marked numbers lines, using appropriate proportional reasoning. Figure 210: estimating the position of 65,000 on an unmarked 50,000 to 100,00 number line 290 In the example below, pupils should reason: “a must be about 875,000 because it is about halfway between the midpoint of the number line, which is 850,000, and 900,000.” Figure 211: identifying 875,000 on a 800,00 to 900,000 number line marked only with a midpoint Pupils should understand that, to estimate the position of a number with more significant digits on a large-value number line, they must attend to the leading digits and can ignore values in the smaller place-value positions. For example, when estimating the position of 5,192,012 on a 5,100,000 to 5,200,000 number line they only need to attend to the first 4 digits. Pupils must also be able to round numbers in preparation for key stage 3, when they will learn to round numbers to a given number of significant figures or decimal places. They have already learnt to round numbers with up to 4 digits to the nearest multiple of 1,000, 100 and 10, and to round decimal fractions to the nearest whole number or multiple of 0.1. Now pupils should extend this to larger numbers. They must also learn that numbers are rounded for the purpose of eliminating an unnecessary level of detail. They must understand that rounding is a method of approximating, and that rounded numbers can be used to give estimated values including estimated answers to calculations. Pupils should only be asked to round numbers to a useful and appropriate level: for example, rounding 7-digit numbers to the nearest 1 million or 100,000, and 6-digit numbers to the nearest 100,000 or 10,000. Pupils may use a number line for support, but by the end of year 6, they need to be able to round numbers without a number line. As with previous year groups (3NPV–3 , 4NPV–3 and 5NPV–3), pupils should first learn to identify the previous and next given multiple of a power of 10, before identifying the closest of these values. In the examples below, for 5,192,012, pupils must be able to identify the previous and next multiples of 1 million and 100,000, and round to the nearest of each. Figure 212: using a number line to identify the previous and next multiple of 1 million 291 Figure 213: using a number line to identify the previous and next multiple of 100,000 Language focus “The previous multiple of 1 million is 5 million. The next multiple of 1 million is 6 million.” “The previous multiple of 100,000 is 5,100,000. The next multiple of 100,000 is 5,200,000.” Figure 214: identifying the nearest multiple of 1 million and the nearest multiple of 100,000 Language focus “The closest multiple of 1 million is 5 million.” “5,192,012 rounded to the nearest million is 5 million.” “The closest multiple of 100,000 is 5,200,000.” “5,192,012 rounded to the nearest 100,000 is 5,200,000.” Pupils should explore the different reasons for rounding numbers in a variety of contexts, such as the use of approximate values in headlines, and using rounded values for 292 estimates. They should discuss why a headline, for example, might use a rounded value, and when precise figures are needed. Finally, pupils should also be able to count forwards and backwards, and complete number sequences, in steps of powers of 10 (1, 10, 100, 1,000, 10,000 and 100,000). Pay particular attention to counting over ‘boundaries’, for example: 2,100,000 2,000,000 1,900,000 378,500 379,500 380,500 Making connections Here, pupils must apply their knowledge from 6NPV–1, that each place value unit is made up of 10 of the unit to its right, to understand how each interval on a number line or scale is made up of 10 equal parts. This also links to 6NPV–4, in which pupils need to be able to read scales divided into 2, 4, 5 and 10 equal parts. 6NPV–3 Example assessment questions Show roughly where each of these numbers is located on the number line below. 2,783,450 7,000,500 5,250,000 8,192,092 99,000 Estimate the values of a, b, c and d. For each number: write the previous and next multiple of 1 million circle the previous or next multiple of 1 million which is closest to the number 293 Fill in the missing numbers. 6,361 ,040 6,371 ,040 6,381 ,040 6,391 ,040 6,401 ,040 6,411 ,040 2,004,567 2,003,567 2,002,567 2,001 ,567 2,000,567 1 ,999,567 7,730,004 7,930,004 8,030,004 8,230,004 8,430,004 9,149,301 9,129,301 9,119,301 9,089,301 What might the missing number be in this web page? A swimming pool holds approximately 82,000 litres of water. The capacity of the swimming pool has been rounded to the nearest multiple of 1,000. Fill in the missing numbers to complete the sentences. a. The minimum amount of water that the pool could hold is . b. The maximum amount of water that the pool could hold is . 294 6NPV–4 Reading scales with 2, 4, 5 or 10 intervals Divide powers of 10, from 1 hundredth to 10 million, into 2, 4, 5 and 10 equal parts, and read scales/number lines with labelled intervals divided into 2, 4, 5 and 10 equal parts. 6NPV–4 Teaching guidance It is important for pupils to be able to divide powers of 10 into 2, 4, 5 or 10 equal parts because these are the intervals commonly found on measuring instruments and graph scales. Pupils have already learnt to divide 1, 100 and 1,000 in this way (5NPV–4, 3NPV–4 and 4NPV–4 respectively), and must now extend this to larger powers of 10. Pupils should be able to make connections between powers of 10, for example, describing similarities and differences between the values of the parts when 1 million, 1,000 and 1 are divided into 4 equal parts. Figure 215: bar models showing 1 million, 1,000 and 1 partitioned into 4 equal parts Pupils should be able to skip count in these intervals forwards and backwards from any starting number (for example, counting forward from 800,000 in steps of 20,000, or counting backwards from 5 in steps of 0.25). This builds on counting in steps of 10, 20, 25 and 50 in year 3 (3NPV–4 ), in steps of 100, 200, 250 and 500 in year 4 (4NPV–4), and in steps of 0.1, 0.2, 0.25 and 0.5 in year 5 (5NPV–4). Pupils should practise reading measurement and graphing scales with labelled power-of-10 intervals divided into 2, 4, 5 and 10 equal parts. Pupils need to be able to write and solve addition, subtraction, multiplication and division equations related to powers of 10 divided into 2, 4, 5 and 10 equal parts, as exemplified for 1 million and 4 equal parts below. Pupils should be able to connect finding equal parts of a power of 10 to finding 1 2 , 1 4 , 1 5 or 1 10 of the value. 295 750,000 250,000 1 ,000,000 1 ,000,000 – 250,000 750,000 1 ,000,000 – 750,000 250,000 1 ,000,000 4 250,000 1 ,000,000 250,000 4 4 250,000 1 ,000,000 250,000 4 1 ,000,000 1 4 of 1 ,000,000 250,000 Making connections Dividing powers of 10 into 10 equal parts is also assessed as part of 6NPV–1. Reading scales also builds on number-line knowledge from 6NPV–3. Conversely, experience of working with scales with 2, 4, 5 or 10 divisions in this criterion improves pupils’ estimating skills when working with unmarked number lines and scales as described in 6NPV–3. 296 6NPV–4 Example assessment questions If 1 10 of a 1kg bag of flour is used, how much is left? In 2005, the population of Birmingham was about 1 million. At that time, about 1 5 of the population was over 60 years old. Approximately how many over-60s lived in Birmingham in 2005? A builder ordered 1,000kg of sand. She has about 300kg left. What fraction of the total amount is left? Fill in the missing parts. Fill in the missing numbers. 297 The bar chart shows the approximate populations of 3 different towns. What are the populations? What mass does each scale show? 298 Some children are trying to raise £200,000 for charity. The diagram shows how much they have raised so far. a. How much money have they raised? b. How much more money do they need to raise to meet their target? 6AS/MD–1 Quantify additive and multiplicative relationships Understand that 2 numbers can be related additively or multiplicatively, and quantify additive and multiplicative relationships (multiplicative relationships restricted to multiplication by a whole number). 6AS/MD–1 Teaching guidance Throughout key stage 2, pupils have learnt about and used 2 types of mathematical relationship between numbers: additive relationships and multiplicative relationships. In year 6, pupils should learn to represent the relationship between 2 given numbers additively or multiplicatively, as well as use such a representation to calculate a missing number, including in measures and statistics contexts. Consider the following: Holly has cycled 20km. Lola has cycled 60km. We can describe the relationship between the distances either additively (Lola has cycled 40km further than Holly; Holly has cycled 40km fewer than Lola) or multiplicatively 299 (Lola has cycled 3 times the distance that Holly has cycled). The relationship between the numbers 20 and 60 can be summarised as follows. Figure 216: additive relationship between 20 and 60 Figure 217: multiplicative relationship between 20 and 60 Language focus “The relationship between 2 numbers can be expressed additively or multiplicatively.” As pupils progress into key stage 3, the ability to relate, recognise and use multiplicative relationships is essential. A pupil who can think multiplicatively would, for example, calculate the cost of 1.2m of ribbon at 75p per metre as 1.2 75p , whereas a pupil who was still thinking only in terms of additive relationships would use the approach of finding the cost of 0.2m (15p) and adding it to the cost of 1m (75p). During key stage 3, pupils will regularly use calculators to solve problems with this type of structure, and the multiplicative approach is more efficient because it involves fewer steps. Given any 2 numbers (related by a whole-number multiplier), pupils must be able to identify the additive relationship (in the example above, 40 and 40 ) and the multiplicative relationship (in the example above 3 and 3 ). Though multiplicative relationships should be restricted to whole-number multipliers, pupils should be able to connect division by the whole number to scaling by a unit fraction: in the example above, this corresponds to understanding that because 60 3 20 , 20 is one-third times the size of 60. When given a sequence of numbers, pupils should be able to identify whether the terms are all related additively or multiplicatively, identify the specific difference or multiplier and use this to continue a sequence either forwards or backwards. Pupils will need to use formal written methods to calculate larger numbers in sequences. 300 Figure 218: completing a sequence where the difference between adjacent terms is 7.5 Figure 219: completing a sequence where each term is 5 times the previous Making connections In 6AS/MD–4 pupils solve problems with 2 unknowns, where the relationship between the unknowns may be additive, multiplicative or both, for example: find 2 numbers, where one is 3 times the size of the other, and the difference between them is 40. 301 6AS/MD–1 Example assessment questions Fill in the missing numbers. 300 1 ,200 75 3 0.1 10 300 1 ,200 75 3 0.1 10 Write an expression in each box to show the relationship between numbers 25 and 75. Is there more than one way to answer this question? Explain. The examples below show the first 2 numbers in a sequence. Find 2 different ways to continue each sequence, using addition for the first and multiplication for the second. a. 4 16 or 4 16 b. 2 200 or 2 200 c. 0.01 10 or 0.01 10 Complete these sequences. 0.5 5 9.5 27.5 32 0.5 0.75 1 25 125 625 0.2 6 180 302 6AS/MD–2 Derive related calculations Use a given additive or multiplicative calculation to derive or complete a related calculation, using arithmetic properties, inverse relationships, and place-value understanding. 6AS/MD–2 Teaching guidance In previous year groups in key stage 2 pupils have learnt about and used the commutative and associative properties of addition (3AS–3), and the commutative, associative and distributive properties of multiplication (4MD–2 and 4MD–3 ). Pupils have also implicitly used the compensation property of addition, for example, when partitioning two-digit numbers in different ways in year 2: 70 + 2 = 72 60 + 12 = 72 In year 6, pupils should learn the compensation property of addition. Language focus “If one addend is increased and the other is decreased by the same amount, the sum stays the same.” Pupils should be able to use the compensation property of addition to complete equations such as 25 35 27.5 ?
, and to help them solve calculations such as 27.5 + 32.5. Similarly, pupils may have implicitly used the compensation property of multiplication, for example, when recognising connections between multiplication table facts: 5 8 10 4 In year 6, pupils should learn the compensation property of multiplication. Language focus “If I multiply one factor by a number, I must divide the other factor by the same number for the product to stay the same.” 303 Pupils should be able to use the compensation property of multiplication to complete equations such as 0.3 320 3 ?
, and to help them solve calculations such as 0.3 320 . Pupils have extensive experience about the effect on the product of scaling one factor from 3NF–3 , 4F–3 and 5NF–2, where they learnt to scale known number facts by 10, 100, one-tenth and one-hundredth. Now they can generalise. Language focus “If I multiply one factor by a number, and keep the other factor the same, I must multiply the product by the same number.” Pupils should practise combining their knowledge of arithmetic properties and relationships to solve problems such as the examples here and in the Example assessment questions below. Example problem 1 Question: Explain how you would use the first equation to complete the second equation: 2,448 34 72 72 24,480 Explanation: Use the inverse relationship between multiplication and division to restate the equation: 72 34 2,448 Apply understanding of place-value: the product can be made 10 times the size by making one of the factors 10 times the size. 72 340 24,480 Example problem 2 Question: Explain how you would use the first equation to complete the second equation: 921 349 572 92.1 44.9 Explanation: Apply understanding of place value, making the sum and addends 1 tenth times the size. 92.1 34.9 57.2 Apply the compensation property of addition to solve the equation: add 10 to the first addend and subtract 10 from the second addend. 92.1 44.9 47.2 Pupils should learn to write a series of written equations to justify their solutions. Being able to work fluently with related equations in this way will prepare pupils for manipulating algebraic equations in key stage 3 and writing proofs. 304 Pupils can already apply place-value understanding to known multiplication facts to scale one factor, for example, 3 4 12 , so 3 40 120 . Now they should extend this to scaling both factors, for example, 3 4 12 , so 30 40 1 ,200 . Making connections In this criterion, pupils use their understanding from 6NPV–1 of scaling numbers by 10, 100 and 1,000. 6AS/MD–2 Example assessment questions Fill in the missing numbers. 327 278 330 25 48 50 327 + 515 = 842 Use this calculation to complete the following equations. 61.5 84.2 8,420 – 3,270 85,200 – 52,500 21 ,760 256 85 Use this calculation to complete the following equations. 256 8.5 2,560 85 2,156 85 3,128 23 136 Use the division calculation so solve the following calculation. Explain your answer. 24 136 Fill in the missing number. 25 60 60 120 305 6AS/MD–3 Solve problems involving ratio relationships Solve problems involving ratio relationships. 6ASMD–3 Teaching guidance Pupils already have the arithmetic skills to solve problems involving ratio. They should now learn to describe 1-to-many (and many-to-1) correspondence structures. Language focus “For every 1 cup of rice you cook, you need 2 cups of water.” “For every 10 children on the school trip, there must be 1 adult.” Pupils should learn to complete ratio tables, given a 1-to-many or many-to-1 relationship. cups of rice 1 2 3 4 5 6 cups of water 2 4 6 8 10 12 number of children 10 20 30 40 50 60 number of adults 1 2 3 4 5 6 Pupils must recognise that proportionality is preserved in these contexts, for example, there is always twice the volume of water needed compared to the volume or rice, regardless of how much rice there is. This will prepare pupils for key stage 3, when they will learn to describe correspondence structures using ratio notation and to express ratios in their simplest forms. Pupils should be able to recognise a 1-to-many or many-to-1 structure, without it being explicitly given and use the relationship to solve problems. For example, here pupils should recognise that, in both examples, for every 1 red bead there are 3 blue beads (or for every 3 blue beads there is 1 red bead), irrespective of the arrangement of the beads. 306 Figure 220: bead strings, each with the structure ‘for every 1 red bead, there are 3 blue beads’ For examples like this, pupils should also be able to include the total quantity in a table. number of red beads 1 2 3 4 number of blue beads 3 6 9 12 total number of beads 4 8 12 16 Pupils should also be able to answer questions such as: if there were 5 red beads, how many blue beads would there be? if there were 21 blue beads, how many beads would there be altogether? if there were 40 beads altogether, how many red beads and how many blue beads would there be? Pupils must also learn to describe and solve problems related to many-to-many structures. Language focus “For every 2 yellow beads there are 3 green beads”. Pupils may initially use manipulatives, such as cubes or beads, for support, but by the end of year 6, they must be able to complete many-to-many correspondence tables and solve related problems without manipulatives. number of yellow beads 2 4 6 8 number of green beads 3 6 9 12 total number of beads 5 10 15 20 Pupils should also begin to prepare for using the unitary method at key stage 3, when it is required for unit conversions, percentage calculations and other multiplicative problems. For example, if they are given a smoothie recipe for 2 people (20 strawberries, 1 banana and 150ml milk), they should be able to adjust the recipe by multiplying or dividing by a 307 whole number, for example, dividing the quantities by 2 to find the amounts for 1 person, or multiplying the quantities by 3 to find the amounts for 6 people. At key stage 3, pupils would then, for example, be able to use the unitary method to adjust the recipe for 5 people, via calculating the amounts for 1 person. Making connections To recognise a one-to-many or many-to-one structure, pupils need to be able to identify multiplicative relationships between given numbers (6AS/MD–2). 6AS/MD–3 Example assessment questions For every 1 litre of petrol, Miss Smith’s car can travel about 7km. a. How many kilometres can Miss Smith’s car travel on 6 litres of petrol? b. Miss Smith lives about 28km from school. How many litres of petrol does she use to get to school? For every 3m of fence I need 4 fence panels. The fence will be 15m long. How many fence panels will I need? I am decorating a cake with fruit. I use 2 raspberries for every 3 strawberries. Altogether I put 30 berries on the cake. a. How many raspberries did I use? b. How many strawberries did I use? For every 500g of excess baggage I take on an aeroplane, I must pay £7.50. I have 3.5kg of excess baggage. How much must I pay? 308 Lily and Ralph are eating grapes. The diagram represents the relationship between the number of grapes that the children eat. Fill in the missing numbers. Number of grapes that Lily eats Number of grapes that Ralph eats 1 20 3 Giya is planting flowers in her garden. For every 5 red flowers she plants, she plants 3 yellow flowers. If Giya plants 18 yellow flowers, how many red flowers does she plant? I am making a necklace. So far, it has 4 black beads and 1 white bead. How many more white beads would I need to add so that there are 4 white beads for every 1 black bead? 6AS/MD–4 Solve problems with 2 unknowns Solve problems with 2 unknowns. 6AS/MD–4 Teaching guidance Pupils need to be able to solve problems with 2 unknowns where: there are an infinite number of solutions there is more than 1 solution there is only 1 solution Pupils may have seen equations with 2 unknowns before, for example, when recognising connections between multiplication table facts: 5 10 309 In year 6, pupils must recognise that an equation like this has many (an infinite number) of solutions. They should learn to provide example solutions by choosing a value for one unknown and then calculating the other unknown. Pupils should be able to solve similar problems where there is more than one solution, but not an infinite number, for example: Danny has some 50p coins and some 20p coins. He has £1.70 altogether. How many of each type of coin might he have? In these cases, pupils may choose a value for the first unknown and be unable to solve the equation for the other unknown (pupils may first set the number of 50p pieces at 2, giving £1, only to find that it is impossible to make up the remaining 70p from 20p coins). Pupils should then try a different value until they find a solution. For a bound problem with only a few solutions, like the coin example, pupils should be able to find all possible solutions by working systematically using a table like that shown below. They should be able to reason about the maximum value in each column. Figure 221: finding the 2 solutions to the coin problem: one 50p coin and six 20p coins, or three 50p coins and one 20p coin Pupils must also learn to solve problems with 2 unknowns that have only 1 solution. Common problems of this type involve 2 pieces of information being given about the relationship between the 2 unknowns – 1 piece of additive information and either another piece of additive information or a piece of multiplicative information. Pupils should learn to draw models to help them solve this type of problem. 310 Example problem 1 Question: The sum of 2 numbers is 25, and the difference between them is 7. What are the 2 numbers? Solution: Figure 222: using a bar model to solve a problem with 2 unknowns – example 1 a = 9 + 7 = 16 b = 9 The numbers are 16 and 9. Example problem 2 Question: The sum of 2 numbers is 48. One number is one-fifth times the size of the other number. What are the 2 numbers? Solution: Figure 223: using a bar model to solve a problem with 2 unknowns – example 2 a = 8 b = 5 8 40 The numbers are 8 and 40. Pupils should also be able to use bar modelling to solve more complex problems with 2 unknowns and 1 solution, such as: 4 pears and 5 lemons cost £3.35. 4 pears and 2 lemons cost £2.30. What is the cost of 1 lemon? Figure 224: using a bar model to solve a problem with 2 unknowns – example 3 311 Solving problems with 2 unknowns and 1 solution will prepare pupils for solving simultaneous equations in key stage 3. Pupils should practise solving a range of problems with 2 unknowns, including contextual measures and geometry problems. Making connections Within this criterion, pupils must be able to use their understanding of how 2 numbers can be related additively or multiplicatively (6AS/MD–1). In 6G–1 pupils solve geometry problems with 2 unknowns, for example, finding the unknown length and unknown width of a rectangle with a perimeter of 14cm. 6AS/MD–4 Example assessment questions A baker is packing 60 cakes into boxes. A small box can hold 8 cakes and a large box can hold 12 cakes. a. How many different ways can he pack the cakes? b. How can he pack the cakes with the fewest number of boxes? 1 eraser and 5 pencils cost a total of £3.35. 5 erasers and 5 pencils cost a total of £4.75. a. How much does 1 eraser cost? b. How much does 1 pencil cost? An adult ticket for the zoo costs £2 more than a child ticket. I spend a total of £33 buying 3 adult and 2 child tickets. a. How much does an adult ticket cost? b. How much does a child ticket cost? The balances show the combined masses of some large bags of dog food and some small bags of dog food. How much does each bag-size cost? A rectangle with side-lengths a and b has a perimeter of 30cm. a is a 2-digit whole number and b is a 1-digit whole number. What are the possible values of a and b? 312 The diagram shows the total cost of the items in each row and column. Fill in the 2 missing costs. 6F–1 Simplify fractions Recognise when fractions can be simplified, and use common factors to simplify fractions. 6F–1 Teaching guidance In year 5, pupils learnt to find equivalent fractions (5F–2). Now pupils must build on this and learn to recognise when fractions are not in their simplest form. They should use their understanding of common factors (5MD–2) to simplify fractions. Pupils should learn that when the numerator and denominator of a fraction have no common factors (other than 1) then the fraction is in its simplest form. Pupils should learn that a fraction can be simplified by dividing both the numerator and denominator by a common factor. They must realise that simplifying a fraction does not change its value, and the simplified fraction has the same position in the linear number system as the original fraction. Pupils should begin with fractions where the numerator and denominator have only one common factor other than 1, for example 6 15 . 313 Figure 225: simplifying 6 15 by dividing the numerator and denominator by the common factor of 3 Language focus “A fraction can be simplified when the numerator and denominator have a common factor other than 1.” Pupils should then learn to simplify fractions where the numerator and denominator share several common factors, for example 4 12 . Pupils should understand that they should divide the numerator and denominator by the highest common factor to express a fraction in its simplest form, but that the simplification can also be performed in more than 1 step. Figure 226: simplifying 4 12 by dividing the numerator and denominator by the highest common factor Figure 227: simplifying 4 12 in 2 steps Language focus “To convert a fraction to its simplest form, divide both the numerator and the denominator by their highest common factor.” Pupils should learn to always check their answer when simplifying a fraction to confirm that it is in its simplest form and the only remaining common factor is 1. 314 Pupils should be able to simplify fractions: where the numerator is a factor of the denominator (and therefore also the highest common factor), for example, 3 9 or 7 28 , resulting in a simplified fraction that is a unit fraction where the numerator is not a factor of the denominator, for example, 4 14 or 15 20 , resulting in a simplified fraction that is a non-unit fraction In year 4 pupils learnt to convert between mixed numbers and improper fractions (4F–2 ) and to add and subtract fractions to give a sum greater than 1 (4F–3 ). This criterion on simplifying fractions provides an opportunity for pupils to continue to practise these skills as they learn how to simplify fractions with a value greater than 1. Pupils should consider calculations such as 9 11 12 12 and understand that the resulting improper fraction, 20 12 , can be simplified either directly, or by first converting to a mixed number and then simplifying the fractional part. Figure 228: simplifying 20 12 to 5 3 , then converting to a mixed number Figure 229: converting 20 12 to 5 3 , then simplifying 6F–1 Example assessment questions Sort these fractions according to whether they are expressed in their simplest form or not. 3 15 2 5 4 20 25 36 1 6 7 21 18 30 9 17 5 15 11 20 23 30 Fraction in its simplest form Fraction not in its simplest form 315 Solve these calculations, giving each answer in the simplest form. 2 4 9 9 3 1 7 7 4 2 15 15 5 5 2 12 12 12 2 7 4 13 13 13 4 4 5 5 7 5 3 10 10 10 8 8 1 9 9 9 7 9 10 10 3 2 13 e i g hts + 11 eights 7 1 s i xth minus 1 2 sixths 17 t h irds minus 5 thirds Ahmed says, “To simplify a fraction, you just halve the numerator and halve the denominator.” Is Ahmed’s statement always true, sometimes true or never true? Explain your answer. Put these numbers in order from smallest to largest by simplifying them to unit fractions. 3 18 5 20 4 8 2 18 4 12 6 60 How much water is in this beaker? Write your answer as a fraction of a litre in its simplest form. 316 6F–2 Express fractions in a common denomination Express fractions in a common denomination and use this to compare fractions that are similar in value. 6F–2 Teaching guidance Pupils should already be able to identify multiples of numbers, and common multiples of numbers within the multiplication tables (4NF–1 and 5MD–2). To be ready to progress to key stage 3, given 2 fractions pupils must be able to express them with the same denominator. Pupils should first work with pairs of fractions where one denominator is a multiple of the other, for example, 1 5 and 4 15 . They should learn that the denominator that is the multiple (here 15) can be used as a common denominator. Pupils should then be able to apply what they already know about writing equivalent fractions (5F–2) to express the fractions in a common denomination. Language focus “We need to compare the denominators of and .” "15 is a multiple of 5.” “We can use 15 as the common denominator.” “We need to express both fractions in fifteenths.” Figure 230: expressing 1 5 in fifteenths Pupils must then learn to work with pairs of fractions where one denominator is not a multiple of the other, for example, 1 3 and 3 8 . Pupils should recognise when 1 denominator is not a multiple of the other (here, 3 is not a factor of 8 and 8 is not a multiple of 3) and learn to identify a new denominator that is a common multiple of both denominators. Again, pupils should then be able to apply what they already know about 317 writing equivalent fractions to express the fractions in a common denomination. Language focus “We need to compare the denominators of and .” “8 is not a multiple of 3.” “24 is a multiple of both 3 and 8.” “We can use 24 as the common denominator.” “We need to express both fractions in twenty-fourths.” Figure 231: expressing 1 5 and 3 8 in twenty-fourths At key stage 3, pupils will learn to find the lowest common multiple of any 2 numbers. At key stage 2, being able to find a common multiple of the denominators by multiplying the 2 denominators is sufficient. Language focus “If one denominator is not a multiple of the other, we can multiply the two denominators to find a common denominator.” This does not always result in the lowest common denominator (for example 1 6 and 2 9 can be expressed with a common denominator of 18 rather than 54), and if pupils can identify a lower common denominator then they should use it. Pupils may also recognise when the resulting pair of common denomination fractions can be simplified. Pupils should also be able to find a common denominator for more than 2 fractions, such as 1 3 , 4 15 and 2 5 , when 1 of the denominators is a multiple of the other denominators (in this case 15). 318 Once pupils know how to express fractions with a common denominator, they can use this to compare and order fractions. Language focus “If the denominators are the same, then the larger the numerator, the larger the fraction.” Making connections In 6F-1 pupils learnt to simplify fractions, and applied this to compare and order fractions that can be simplified to unit fractions (see 6F-1 Example assessment questions , question 4). In this criterion, pupils learnt to express fractions in a common denomination, and use this to compare and order fractions. Pupils should be able to identify which method is appropriate for a given pair or set of fractions. Pupils learn more about comparing fractions, and choosing an appropriate method in 6F-3 . 6F–2 Example assessment questions Fill in the missing symbols (<, > or =). You will need to simplify some of the fractions and express each pair with a common denominator. 5 2 7 3 6 3 10 5 7 3 9 4 5 6 7 8 2 7 3 10 2 3 6 9 3 el evenths missing symbol 1 third 1 f ifth missing symbol 2 elevenths Express each set of fractions with a common denominator. Then put them in order from smallest to largest. a. 4 20 1 4 3 10 2 5 b. 2 9 1 3 1 6 4 18 319 Ahmed has a beaker containing 7 10 of a litre of water. Imran has a beaker containing 3 5 of a litre of water. Express the fractions with a common denominator to work out whose beaker contains the most water. Ben and Felicity are both trying to raise the same amount of money for charity. So far, Ben has raise 3 4 of the amount, while Felicity has raised 5 7 of the amount. Express the fractions with a common denominator to work out who is closest to meeting their target. 6F–3 Compare fractions with different denominators Compare fractions with different denominators, including fractions greater than 1, using reasoning, and choose between reasoning and common denomination as a comparison strategy. 6F–3 Teaching guidance In 6F–2 pupils learnt to compare any 2 fractions by expressing them with a common denominator. However fractions can often be compared by reasoning, without the need to express them with a common denominator. Pupils can already compare unit fractions. Language focus “If the numerators are both 1, then the larger the denominator, the smaller the fraction.” Pupils should now extend this to compare other fractions with the same numerator, for example, because 1 5 is greater than 1 6 we know that 2 5 is greater than 2 6 . Figure 232: bar models to compare 2 5 and 2 6 320 Language focus “If the numerators are the same, then the larger the denominator, the smaller the fraction.” Pupils should be able to use reasoning in other ways when comparing fractions: For each fraction they should be able to visualise where it is positioned on a number line, for example, thinking about whether it is greater than or less than 1 2 or whether it is close to 0 or 1. Pupils should be able to reason about the relationship between the numerator and the denominator of each fraction, asking themselves ‘Is this fraction a large or small part of the whole?’. They should be able to reason, for example, 5 6 is greater than 7 11 because 5 is a larger part of 6 than 7 is of 11. For fractions that are a large part of the whole, pupils should be able to reason about how close each fraction is to the whole. For example, 7 8 is 1 8 less than the whole, while 6 7 is 1 7 less than the whole. Since 1 8 is less than than 1 7 , 7 8 must be larger than 6 7 . For a given pair or set of fractions, pupils must learn to assess whether it is more appropriate to compare them using reasoning or to express them in a common denomination. Making connections When given a pair or set of fractions to compare, pupils may need to convert some of the fractions to their simplest form (6F–1). They then need to assess whether it is more appropriate to compare them using reasoning (this criterion) or to express them in a common denomination (6F–2). 321 6F–3 Example assessment questions Which number(s) could go in the missing-number box to make this statement true? 1 1 1 4 10 Without using a common denominator, put each set of fractions in order from smallest to largest.a. 10 8 7 8 5 8 3 8 8 8 4 8 2 8 b. 1 6 1 5 1 8 1 7 1 10 1 9 c. 3 3 3 8 3 11 3 100 3 5 3 halves Sabijah and Will are in a running race. Sabijah has run 9 10 of the race. Will has run 8 9 of the race. Who is further ahead? Explain your reasoning. Fill in the missing symbols (<, > or =). 5 4 6 7 8 7 9 11 Think of a number that can go in each box so that the fractions are arranged in order from smallest to largest. 1 1 4 3 3 5 7 322 6G–1 Draw, compose and decompose shapes Draw, compose, and decompose shapes according to given properties, including dimensions, angles and area, and solve related problems. 6G–1 Teaching guidance Through key stage 2, pupils have learnt to measure perimeters, angles and areas of shapes, and have learnt to draw polygons by joining marked points (3G–2) and draw angles of a given size (5G–1). By the end of year 6, pupils must be able to draw, compose and decompose shapes defined by specific measurements. Composing and decomposing shapes prepares pupils for solving geometry problems at key stage 3, for example, finding the area of a trapezium by decomposing it to a rectangle and 2 triangles. Pupils should be able to draw a named shape to meet a given measurement criterion, for example: drawing a rectangle, on squared-centimetre paper, with a perimeter of 14cm (Example 1 below) drawing a pentagon, on squared-centimetre paper, with an area of 10cm2 (Example 2 below) drawing a triangle at actual size, based on a sketch with labelled lengths and angles (see 6G–1, question 5 ) Example 1 Task: draw a rectangle with a perimeter of 14cm. Example solution: Figure 233: a 5cm by 2cm rectangle on a squared-centimetre grid Drawn to actual size. Example 2 Task: draw a pentagon with an area of 10cm2. Example solution: Figure 234: an irregular pentagon with an area of 10cm2 Drawn to actual size. 323 Examples like these involve more than 1 unknown and have more than 1 solution. Pupils should learn to choose a value for 1 of the variables and work out other unknowns from this. For example, to draw a rectangle with a perimeter of 14cm, the width could be chosen to be 1cm and the length then calculated to be 6cm, or the width could be chosen to be 2cm and the length then calculated to be 5cm. There are 6 possible whole-number solutions to this problem (counting the same rectangle in a different orientation as a separate solution), and pupils may provide any one of them to complete the task. Pupils should be able to reason about dimensions or areas given for part of a shape to determine the values for other parts of a shape or for a compound shape. Example 3 Problem: find the perimeter of the large rectangle on the right. Figure 235: problem involving a compound shape made from 3 identical rectangles Drawn to scale, not actual size Solution: perimeter of large rectangle = 35cm Figure 236 solving a problem involving a compound shape made from 3 identical rectangles Drawn to scale, not actual size 324 Further examples are provided below in 6G–1, questions 4 and 8. Making connections In 6AS/MD-4 pupils learnt to solve problems with 2 unknowns. Drawing a shape to match given properties can correspond to a problem with 2 unknowns (Example 1 above) or more (Example 2 above). 6G–1 Example assessment questions Lois has started drawing a shape on this squared-centimetre grid. Complete her shape so that it has an area of 14cm2. Drawn to actual size 325 Here is a rhombus on a triangular grid. Draw a different shape with the same area on the grid. Draw a hexagon on this squared-centimetre grid. Include one side of length 4cm and one side of length 3cm. Here is a square made from 4 smaller squares. The area of the large square is 64cm2. What is the length of 1 side of each small square? 326 Here is a sketch of a triangle. It is not drawn to scale. Draw the full-size triangle accurately. Use an angle measurer (protractor) and a ruler. Not drawn to scale Here is a picture of a pentagon made from a regular hexagon and an equilateral triangle. The perimeter of the triangle is 24cm. What is the perimeter of the pentagon? Drawn to scale, not actual size 327 Calculation and fluency Number, place value and number facts: 6NPV–1 and 6NPV–2 6NPV–1 Understand the relationship between powers of 10 from 1 hundredth to 10 million, and use this to make a given number 10, 100, 1,000, 1 tenth, 1 hundredth or 1 thousandth times the size (multiply and divide by 10, 100 and 1,000). 6NPV–2 Recognise the place value of each digit in numbers up to 10 million, including decimal fractions, and compose and decompose numbers up to 10 million using standard and non-standard partitioning. Pupils should develop fluency in multiplying numbers by 10, 100 and 1,000 to give products with up to 7 digits, and dividing up to 7-digit numbers by 10, 100 and 1,000. Pupils should be able to carry out calculations based on their understanding of place-value as well as non-standard partitioning, for example: 4,000 30,000 0.2 5,000,000 381 ,920 – 900 518.32 30 381 ,920 – 60,000 Pupils should also be able to apply their place-value knowledge for larger numbers to known additive and multiplicative number facts, including scaling both factors of a multiplication calculation, for example: 8 6 14 800,000 600,000 1 ,400,000 3 4 12 3 40,000 120,000 300 400 120,000 Representations such as place-value counters, partitioning diagrams and Gattegno charts can be used initially to help pupils understand calculation strategies and make connections between known facts and related calculations. However, pupils should not rely on such representations for calculating. Pupils should maintain fluency in both formal written and mental methods for calculation. Mental methods can include jottings to keep track of calculations. Pupils should select the most efficient method to calculate depending on the numbers involved. Pupils should learn to check their calculations with a calculator so that they know how to use one. This will help pupils when they progress to key stage 3. 328 Addition and subtraction: formal written methods Pupils should continue to practise adding whole numbers with up to 4 digits, and numbers with up to 2 decimal places, using columnar addition. This should include calculations with more than 2 addends, and calculations with addends that have different numbers of digits. Figure 237: range of columnar addition calculations For calculations with more than 2 addends, pupils should add the digits within a column in the most efficient order. For the second example above, efficient choices could include: beginning by making 10 in the ones column making double-6 in the hundreds column Pupils should continue to practise using columnar subtraction for numbers with up to 4 digits, and numbers with up to 2 decimal places. This should include calculations where the minuend and subtrahend have a different numbers of digits or decimal places, and those involving exchange through 0. Figure 238: range of columnar subtraction calculations Pupils should make sensible decisions about how and when to use columnar methods. For example, when subtracting a decimal fraction from a whole number, pupils may be able to use their knowledge of complements, avoiding the need to exchange through zeroes. For example, to calculate 8 – 4.85 pupils should be able to work out that the decimal complement to 5 from 4.85 is 0.15, and that the total difference is therefore 3.15. 329 Pupils should learn to check their columnar addition and subtraction calculations with a calculator so that they know how to use one. This will help pupils when they progress to key stage 3. Multiplication: extending 5MD–3 In year 5, pupils learnt to multiply any whole number with up to 4 digits by any 1-digit number using short multiplication (5MD–3). They should continue to practise this in year 6. Pupils should also learn to use short multiplication to multiply decimal numbers by 1-digit numbers, and use this to solve contextual measures problems, including those involving money. Figure 239: range of short multiplication calculations Pupils should be able to multiply a whole number with up to 4 digits by a 2-digit whole number by applying the distributive property of multiplication (4MD–3 ). This results in multiplication by a multiple of 10 (which they can carry out by writing the multiple of 10 as a product of 2 factors (5MD–3) and multiplication by a one-digit number. 124 26 124 20 124 6 124 2 10 124 6 2,480 744 3,224 Pupils should be able to represent this using the formal written method of long multiplication, and explain the connection to the partial products resulting from application of the distributive law. 330 Figure 240: long multiplication calculation Pupils should be fluent in interpreting contextual problems to decide when multiplication is the appropriate operation to use, including as part of multi-step problems. Pupils should use short or long multiplication as appropriate to solve these calculations. Pupils should learn to check their short and long multiplication calculations with a calculator so that they know how to use one. This will help pupils when they progress to key stage 3. Division: extending 5MD–4 In year 5, pupils learnt to divide any whole number with up to 4 digits by a 1-digit number using short division, including with remainders (5MD–4). They should continue to practise this in year 6. Pupils should also learn to use short division to express remainders as a decimal fraction. Figure 241: range of short division calculations For contextual problems, pupils must be able to interpret remainders appropriately as they learnt to do in year 4 (4NF–2 ). This should be extended to making an appropriate decision about how to represent the remainder. Consider the question “4 friends equally share the cost of a £109 meal. How much does each of them pay?” Pupils should recognise that an answer of £27 remainder 1 is not helpful in this context, and that they need to express the answer as a decimal fraction (£27.25) to provide a sufficient answer to the question. Pupils should also be able to divide any whole number with up to 4 digits by a 2-digit number, recording using either short or long division. Pupils are likely to need to write out multiples of the divisor to carry out these calculations and can do this efficiently using a 331 ratio table – they can write out all multiples up to 10 (working in the most efficient order) or write out multiples as needed. 1 17 2 34 3 51 4 68 5 85 6 7 8 136 Figure 242: long division calculation ( , 8 211 17 ) Pupils should be fluent in interpreting contextual problems to decide when division is the appropriate operation to use, including as part of multi-step problems. Pupils should use short or long division as appropriate to solve these calculations. Pupils should learn to check their short and long division calculations with a calculator so that they know how to use one. This will help pupils when they progress to key stage 3. 332 Appendix: number facts fluency overview Addition and subtraction facts The full set of addition calculations that pupils need to be able to solve with automaticity are shown in the table below. Pupils must also be able to solve the corresponding subtraction calculations with automaticity. + 0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 Pupils must be fluent in these facts by the end of year 2, and should continue with regular practice through year 3 to secure and maintain fluency. It is essential that pupils have automatic recall of these facts before they learn the formal written methods of columnar addition and subtraction. The Factual fluency progression table at the end of this appendix summarises the order in which pupil should learn these additive number facts. 333 Multiplication and division facts The full set of multiplication calculations that pupils need to be able to solve by automatic recall are shown in the table below. Pupils must also have automatic recall of the corresponding division facts. Pupils must be fluent in these facts by the end of year 4, and this is assessed in the multiplication tables check. Pupils should continue with regular practice through year 5 to secure and maintain fluency. The 36 most important facts are highlighted in the table. Fluency in these facts should be prioritised because, when coupled with an understanding of commutativity and fluency in the formal written method for multiplication, they enable pupils to multiply any pair of numbers. The Factual fluency progression table at the end of this appendix summarises the order in which pupil should learn these multiplicative number facts. Pupils should learn the multiplication tables in the ‘families’ described in the progression table – making connections between the multiplication tables in each family will enable pupils to develop automatic recall more easily, and provide a deeper understanding of multiplication and division. 334 Factual fluency progression Year 1 Year 2 Year 3 Year 4 Year 5 Additive factual fluency Addition and subtraction within 10. Addition and subtraction across 10. Secure and maintain fluency in addition and subtraction within and across 10, through continued practice. Multiplicative factual fluency Recall the 10 and 5 multiplication tables, and corresponding division facts. Recall the 3, 6 and 9 multiplication tables, and corresponding division facts. Secure and maintain fluency in all multiplication tables, and corresponding division facts, through continued practice. Recall the 2, 4 and 8 multiplication tables, and corresponding division facts. Recall the 7 multiplication table, and corresponding division facts. Recall the 11 and 12 multiplication tables, and corresponding division facts. 335 © Crown copyright 2020 This publication (not including logos) is licensed under the terms of the Open Government Licence v3.0 except where otherwise stated. Where we have identified any third party copyright information you will need to obtain permission from the copyright holders concerned. To view this licence: visit www.nationalarchives.gov.uk/doc/open-government-licence/version/3 email psi@nationalarchives.gsi.gov.uk write to Information Policy Team, The National Archives, Kew, London, TW9 4DU About this publication: enquiries www.education.gov.uk/contactus download www.gov.uk/government/publications Reference: DfE-00100-2020 Follow us on Twitter: @educationgovuk Like us on Facebook: facebook.com/educationgovuk |
190129 | https://www.ck12.org/flexi/cbse-math/inverse-trigonometric-functions/what-are-the-properties-of-inverse-trigonometric-functions/ | What are the properties of inverse trigonometric functions? - Characteristics & Properties | CK-12 Foundation
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Inverse Trigonometric Functions
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What are the properties of inverse trigonometric functions?
Flexi Says:
The inverse trigonometric functions have several important properties. Here are some of them:
Domain and Range:
The domain and range of inverse trigonometric functions are opposite to those of trigonometric functions. For example, the domain of sin−1x is −1≤x≤1 and its range is −π 2≤y≤π 2.
Odd/Even Property:
The functions sin−1x, tan−1x, and csc−1x are odd functions, i.e., f(−x)=−f(x). The functions cos−1x, sec−1x, and cot−1x are neither odd nor even.
Additive Property:
The inverse trigonometric functions do not have the additive property like the trigonometric functions. For example,sin−1(x+y) is not equal to sin−1x+sin−1y.
Principal Value:
Every inverse trigonometric function has a principal value. For example, the principal value of sin−1x lies in the interval −π 2≤y≤π 2.
Composition:
The composition of a trigonometric function and its inverse gives the original value. For example,sin(sin−1x)=x and sin−1(sinx)=x for −π 2≤x≤π 2.
Periodicity:
Unlike trigonometric functions, inverse trigonometric functions are not periodic.
Analogy / Example
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What is the typical shape of the graph for y = arctan x?What is the domain and range of inverse cosec(x)?Identities of inverse trigonometric functions with negative angles.
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190130 | https://www.purplemath.com/modules/systlin6.htm | Published Time: Thu, 13 Mar 2025 18:47:06 GMT
Systems of Linear Equations: Gaussian Elimination | Purplemath
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Systems of Linear Equations: Gaussian Elimination
DefinitionsGraphingSpecial CasesSubstitutionElimination/AdditionGaussian EliminationMore Examples
Purplemath
Solving three-variable, three-equation linear systems is more difficult, at least initially, than solving the two-variable, two-equation systems, because the computations involved are more involved; there are just so many opportunities for careless mistakes to creep in. (I speak from painful experience.) So, when moving on from two-variable linear systems to more complicated situations, you will need to be very neat in your working, and you should plan to use lots of scratch paper. Lots and lots of scratch paper.
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(The methodology for solving these larger systems of equations is an extension of the two-variable solving-by-addition method, so make sure you know this method well and can consistently use it correctly.)
Though the method of solution is based on addition/elimination, trying to do actual the actual addition tends quickly to become confusing, so there is a systematized method for solving three-or-more-variables linear systems. This method is called "Gaussian elimination".
(This solution method is named after Carl Friedrich Gauss, though Europeans had actually gotten this method from Isaac Newton a couple centuries earlier, who had come up with it independently about fifteen hundred years after the Chinese had developed it.)
Let's start simple, and work our way up to messier examples.
Solve the following system of equations.
5x + 4 y – z = 0
10 y – 3 z = 11
z = 3
It's fairly easy to see how to get started in solving this system. The third equation in the system, the one at the bottom of the stack above, is a simple one-variable equality; namely, that z=3. The second equation, the middle one in the stack above, has only the variables y and z. The simplest next step for me, then, it to substitute the z-value from the third equation back into the second equation. Then I'll solve the resulting one-variable equation for the value of y; in other words, I'll use that second equation to back-solve for y.
10 y – 3(3) = 11
10 y – 9 = 11
10 y = 20
y = 2
This gives me the second value in this three-variable system. The only variable left is x. The first equation in the system, the one at the top of the stack above, contains this variable. I already have the values of y and z, so I'll plug their value in for their variables in the first equation, and back-solve for x.
5x + 4(2) – (3) = 0
5 x + 8 – 3 = 0
5 x + 5 = 0
5 x = –5
x = –1
This is the third value for the solution to this system of equations. Keeping in mind that the variables will be listed in this three-coordinate point alphabetically, my answer is:
(x, y, z) = (−1, 2, 3)
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The reason this system was easy to solve is that the system was "upper triangular"; this refers to the equations having the form of a triangle in the upper corner, because the first row contained terms with all three variables, the second row contained only terms with the second and third variable, and the third row contained a term only with the third variable. Outlining this region creates that upper-corner triangle, as shown below.
(This formation is also called "echelon" form (pronounced "ESH-eh-lahn") or "row-echelon" form, from the military definition of "echelon" being where each row (of soldiers or tanks or whatever) is shorter than the one immediately behind it. In this case, each row (of variable terms) in the matrix of equations is shorter than the row above it.)
The point is that, in this format, the system is simple to solve. And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the "row operations" (that is, the multiplying through, the dividing through, and the adding down) that we saw when we learned the "addition/elimination method" of solving systems of equations on the previous page.
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Solve the following system of equations using Gaussian elimination.
–3 x + 2 y – 6 z = 6
5 x + 7 y – 5 z = 6
x + 4 y – 2 z = 8
No equation is solved for a variable, so I'll have to do the multiplication-and-addition thing to simplify this system. In order to keep track of my work, I'll write down each step as I go. But I'll do my computations on scratch paper. Here is how I did it:
The first thing to do is to get rid of the leading x-terms in two of the rows. For now, I'll just look at which rows will be easy to clear out; I can switch the rows' order later, if needed, to put the system into upper triangular form.
There is no rule that says I have to use the x-term from the first row, and, in this case, I think it will be simpler to use the x-term from the third row, since its coefficient is simply "1". So I'll multiply the third row by 3, and add it to the first row. I first do my actual computations on scratch paper:
...and then I write down the results:
(When we were solving two-variable systems on the previous page, we could multiply a row, rewriting the system off to the side, and then add down. There is no space for this in a three-variable system, which is why we need the scratch paper.)
Please take careful note: I didn't actually do anything to the third row — I had used the third row, but I hadn't actually changed the third row — so I merely copied it over, unchanged, into the new matrix of equations. I neither changed nor used the second row, so I copied it over, unchanged, to the new matrix as well. The only row that was actually changed was the first row; in the new matrix, you can see that I wrote in the results of my scratch-work.
Do not confuse "using" with "changing".
(To make my steps clear for the grader (and for me, when I review later for the test), I have annotated the arrow between the two steps to show that I'd multiplied the third row by three, and then had added the result to the first row, in order to get a new first row.)
I generally prefer to work with smaller numbers, and I notice that all the coefficients in the (new) first row are even. So, to get smaller coefficients, I'll multiply the first row by one-half (or, which is the same thing, divide through by 2):
Now that's I've removed the x-term from R 1, I turn my attention to R 2. I could divide this row by −5 and then add this to the third row (leaving the in-use second row unchanged), but this will give me loads of fractions in the third row. I'd like to avoid fractions as long as possible. So, instead, I'll multiply the third row by −5 and add the result to the second row (leaving the in-use third row unchanged). I do this work on scratch paper:
I haven't done anything with the first row, so I'll copy it to the new system unchanged. I worked with the third row (in my scratch-work), but I only worked on the second row, so the second row will be updated in the new system, but the third row will be copied over unchanged.
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Okay, now the x-column is cleared out except for the leading term in the third row. So, to work further toward triangular form, I next have to work on the y-column.
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Note: Since the third equation has an x-term, I cannot use it on either of the other two equations any more — or I'll undo my progress in clearing those rows of x-terms. I can work on the equation (using other rows to clear the y-term), but not with it (that is, I cannot apply it to the other two rows).
Looking at the first and second rows, I see that I could divide the first row by 7 and the second row by 13, and then the y-terms will cancel. But this will give me loads of fractions. I could multiply to the least common multiple, which is 91, but then the numbers will start getting kinda big. Neither of these options is "wrong"; it's just that I'm lazy, so I'm going to see if there might be a better option.
If I multiply the first row by 2 and add the result to the second row, this will give me a leading 1 in the second row. I won't have gotten rid of the leading y-term in the second row, but I will have converted it (without bogging down in fractions) to a form that is simpler to deal with. (You should keep an eye out for this sort of simplification.) First I do the scratch work:
I update the second row, leaving the first (that I'd worked with but not on) and third (with which I'd done nothing) unchanged.
Now I can use the second row to clear out the y-term in the first row. I'll multiply the second row by −7 and add the result to the first row. First I do my scratch work:
I worked with R 2, but only on R 1; I did nothing with R 3. So I leave R 2 and R 3 unchanged, updating the system with the new R 1.
I can now see what the value of z is but, just to be thorough, I'll divide through on the first row by 43. Then I'll rearrange the rows to put them in upper-triangular form:
Now I can start the process of back-solving:
y – 7(1) = –4
y – 7 = –4
y = 3
That's a second variable done. Now I'll do the final back-solving:
x + 4(3) – 2(1) = 8
x + 12 – 2 = 8
x + 10 = 8
x = –2
I now have values for each of the three variables. I found these values in, as it happened to turn out, reverse-alphabetical order. I'll need to remember to write my solution with the variables in the correct, alphabetical order.
(x, y, z) = (−2, 3, 1)
Note: There is nothing sacred about the steps I used in solving the above system; there was nothing special about how I solved this system. You could work in a different order or simplify or combine different rows in different ways, and still come up with the correct answer. These systems are sufficiently complicated that there is unlikely ever to be only one right way of computing the answer. So don't stress over "how did she know to do that next?", because there is no rule. I didn't "know"; I just did whatever I happened to notice first, or whatever happened to struck my fancy; I did whatever seemed simplest or whatever first sprang to mind.
Don't worry if you would have used completely different steps from a classmate or your instructor. As long as each step along your way is correct, then you will come up with the same answer.
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In the above example, I could have gone further in my computations and been more thorough-going in my row operations, clearing out all the y-terms other than that in the second row and all the z-terms other than that in the first row. This would then have been the entire process, from beginning to end:
When a system is reduced in this way, I can just read off the values of x, y, and z directly from the system; I don't have to bother with the back-substitution. This more-complete method of solving is called "Gauss-Jordan elimination" (the "Jordan" part being named after Wilhelm Jordan). When the equations are reduced to this point, where you can simply read off the solution, the system is said to be in "reduced" row-echelon form.
Many textbooks only go as far as Gaussian elimination, but I've always found it easier to continue on and do Gauss-Jordan. And no instructor should ever count off for your doing extra steps or being "too complete" in your work. Unless otherwise specified, do what you like.
Note that I did two row operations at once in that last step before switching the rows. As long as I'm not working with and working on the same row in the same step, this is okay. In that last step, I was working with the first row, using it to work on the second and third rows.
And yes, solving systems of linear equations (by hand) is often, even usually, this long and involved. Like I said before: lots and lots of scratch paper.
The next page has more worked examples.
URL:
You can use the Mathway widget below to practice solving systems of linear equations with three or more variables (or skip the widget and continue on to the next page). Try the entered exercise, or type in your own exercise. Click the button and select "Solve using Gaussian Elimination" to compare your answer to Mathway's.
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(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)
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Isocitric acid is optically active but citric acid is not. Identify the probable structure of citric acid. а. ОН HOOCH2C . соон COOH b. НО СООН ноосн,ссоон С. OH ноосH-C. COOH Соон d. A and
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Question: Isocitric acid is optically active but citric acid is not. Identify the probable structure of citric acid. а. ОН HOOCH2C . соон COOH b. НО СООН ноосн,ссоон С. OH ноосH-C. COOH Соон d. A and
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Here’s the best way to solve it.Solution Share Share Share done loading Copy link Isocitric acid is a TCA intermediate and an isomer of citric acid. It is a chiral …
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Isocitric acid is optically active but citric acid is not. Identify the probable structure of citric acid. а. ОН HOOCH2C . соон COOH b. НО СООН ноосн,ссоон С. OH ноосH-C. COOH Соон d. A and
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190132 | https://www.desmos.com/calculator/yl60qp93x9 | Cubic function | Desmos
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190133 | https://www.youtube.com/watch?v=2fW8QUjNl0I | Calculus 1 Using the IVT to show that a zero exists on an interval for a function
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but if you guys can just pay attention to what I am doing over here um Again by using the intermediate value theorem it's basically saying if a function is continuous so here's our function is that continuous for all we know yes absolutely right then it's saying on a closed interval do we see you guys see we have a closed interval right so what they're saying is there is a number any number that's in between our two intervals right so any number that we can kind of pick inside of there we know it's going to be in our um is going to be a value that's inside the function now why the intermediate value is very important is because well in between that closed interval if I have a positive and a negative number the only way to go from positive to negative if you have an F ofx that becomes negative and an fx becomes positive that means somehow whatever this function looks like we had to cross the x- axxis right that means that that F of C has to equal zero and those are your critical numbers right those are like our zeros of a function you know and of anything to do this to show that F of C exists what we need to do is evaluate the function on both intervals when we do that we get eight when we evaluate for f of three we get that becomes 9 9 = 1 so you guys see we started off with a positive number on our interval and we ended with a negative number correct what that tells us then is there has to be a zero that exists by the intermediate value theorem so on your on your test what you would basically say is um f of C so you'd write verbatim FX is continuous on the closed interval 0 to three and negative one or zero has to be less than or equal to or is greater than equal to negative 1 which is greater than equal to 8 then there exists C where zero has to be less than or equal to C Which is less than or equal to three where F of C equals z by the intermate value there and when you have a problem like that they are going to want to see you writing that out do not worry |
190134 | https://math.stackexchange.com/questions/4775465/angle-between-tangent-circles-in-constant-diameter-cam | geometry - Angle between tangent circles in constant diameter cam - Mathematics Stack Exchange
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Angle between tangent circles in constant diameter cam
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I am playing around with eccentric cams in a 3D printing project I'm working on and I've got this sketch for a constant diameter cam but I want to better understand its properties in order to optimize my mechanism. Specifically, I am trying to find a function that describes the dwell period of the cam based on the ratio of the radiuses of the two concentric circles that make up the shape.
In my drawing there are two concentric circles A and B. Then there are two more circles C and D who are tangent to B, and whose center points are coincident with the intersection of A and the other. From these circles we have an angle ijk which describes the dwell period of the cam. Just from playing with it and by eye I can see that as the ratio of the concentric radiuses Br/Ar approaches 1 ijk approaches 180 degrees, and as Br/Ar approaches 0 ijk approaches 60 degrees and the cam becomes a reuleaux triangle. It's been a long time since I took a geometry class however and I can't seem to figure out how to derive the actual function that would define ijk as a function of Br/Ar.
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asked Sep 25, 2023 at 19:46
JoeBruzekJoeBruzek
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1 Just note that i k=r A+r B i k=r A+r B to get: sin∠i j k 2=1 2(1+r A r B).sin∠i j k 2=1 2(1+r A r B). Intelligenti pauca –Intelligenti pauca 2023-09-25 20:08:59 +00:00 Commented Sep 25, 2023 at 20:08
Incredible! I see that it is correct by looking at values, but how do you know that i k=r A+r B i k=r A+r B JoeBruzek –JoeBruzek 2023-09-25 20:45:20 +00:00 Commented Sep 25, 2023 at 20:45
2 i k i k is the radius of circles C and D.Intelligenti pauca –Intelligenti pauca 2023-09-25 20:47:57 +00:00 Commented Sep 25, 2023 at 20:47
1 Aha, yes of course. Thank you!JoeBruzek –JoeBruzek 2023-09-25 21:00:14 +00:00 Commented Sep 25, 2023 at 21:00
Cam calculations require two parts. One is the cam lobe shape and the other (missing) is the follower radius. Also, you need to specify if the follower is translating type, or swinging type.John Alexiou –John Alexiou 2023-09-25 22:08:11 +00:00 Commented Sep 25, 2023 at 22:08
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190135 | https://math.stackexchange.com/questions/2649250/compute-the-dihedral-angle-of-a-regular-pyramid | Stack Exchange Network
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Compute the dihedral angle of a regular pyramid
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$\begingroup$
Given a regular pyramid, defined as a right pyramid with a base which is a regular polygon, with the vertex above the centroid of the base, I would like to compute the dihedral angle between adjacent faces, as well as the angle between a lateral face and the base, in terms of the number of sides and the slant edge length. Can someone walk through the calculation?
Using Google I found this helpful article about computing dihedral angles by Greg Egan.
We have isosceles triangle $CAD$ above the plane of triangle $CBD$, making a dihedral angle of $\epsilon$. Triangle $CBD$ is the orthogonal projection of $CAD$, so also isosceles. The apex angle between adjacent edges in triangle $CAD$ is $\alpha$. The angle of triangle $CBD$ is $\beta.$ Edge $AC$ makes a slant angle of $\gamma$ with the projected edge $BC$.
There is a plane perpendicular to slant edge $AC$ through the opposite vertex $D$, which intersects edge $AC$ at point $F$, and $BC$ at point $G$. Angle $GFD$ is $\delta.$
If triangle $CAD$ is the side of a regular pyramid, and $CBD$ its base, then $2\delta$ will be the dihedral angle between adjacent faces, and $\epsilon$ will be the dihedral angle between slant face and base.
I can see that plane $DFG$ is perpendicular to slant edge $AC$, hence why triangles $AFD$ and $AFG$ are right triangles. But why are $DGC$ and $FGD$ right angles?
How do I arrive at $$\sin\delta=\cos(\beta/2)/\cos(\alpha/2)$$?
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edited Feb 14, 2018 at 8:50
John DanielsJohn Daniels
asked Feb 13, 2018 at 18:06
John DanielsJohn Daniels
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$\begingroup$ Im not able to access the page that youve cited just now, but Id guess that $F$ and $G$ are defined to be the feet of the altitudes to $D$, creating right angles by definition. $\endgroup$
amd
– amd
2018-02-13 22:28:45 +00:00
Commented Feb 13, 2018 at 22:28
$\begingroup$ @amd the plane GFD is defined to be perpendicular to the line AC of intersection of two planes CAD and CBD. I'm sure you're right that implies DG is an altitude of BDC, but for some reason I'm blanking. Actually I guess what I want to see is that DG is perpendicular to plane BAC. So DG is perpendicular to both BC and FG. $\endgroup$
John Daniels
– John Daniels
2018-02-14 00:12:21 +00:00
Commented Feb 14, 2018 at 0:12
$\begingroup$ I don't think you have formulated the problem with enough information. For example, consider a square pyramid, and let $s = 1.$ Now you have the number of sides and $s,$ but if the square is a square of side $1/4$ you get very different dihedral angles than if the square has side $1.$ The formula you cite at the end is derived for regular polyhedra, where $\alpha$ is determined by the type of polyhedron and therefore the edge of the regular polygon, $2s \sin\frac\alpha2,$ is also determined. If you meant the polygon to have side $1,$ that needs to be part of the problem statement. $\endgroup$
David K
– David K
2018-02-14 02:30:34 +00:00
Commented Feb 14, 2018 at 2:30
$\begingroup$ @DavidK I think the formula $2s\sin(\alpha/2)$ for the side opposite $\alpha$ should apply to any isosceles triangle. I don't think I have assumed equilateral triangular faces or regular polyhedra anywhere in the equations quoted. If the dihedral angle depends on the ratio of the slant edge to the base edge, I will be happy to accept that. $\endgroup$
John Daniels
– John Daniels
2018-02-14 02:58:15 +00:00
Commented Feb 14, 2018 at 2:58
$\begingroup$ The issue is not whether the side has length $2s \sin\frac\alpha2$ -- of course it does! The issue is that in your formulation, we cannot find the value of of $2s \sin\frac\alpha2$, whereas in the page you linked to, that value can easily be found, because that page most definitely _does assume it is dealing with a regular polyhedron. On the other hand, if the polygon's side is $1,$ then $s$ is the ratio of the slant edge to the base edge, and we're both happy. $\endgroup$
David K
– David K
2018-02-14 04:19:15 +00:00
Commented Feb 14, 2018 at 4:19
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A vectorial approach would be quite lean and effective.
Given two faces of the pyramid, sharing the common edge $V P_n$, and containing the contiguous base points $P_{n-1}$ and $P_{n+1}$, the dihedral angle between these two faces would be the angle made by the two vectors ($t_m, t_p$), normal to the common edge and lying on the respective face.
Clearly, that will be also the angle made by the normal vectors to the faces, provided that one is taken in the inward, and the other in the outward direction.
That is, by the right-hand rule, $$ {\bf n}_{\,m} = \mathop {P_{\,n} P_{\,n - 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to \quad \quad {\bf n}_{\,p} = \mathop {P_{\,n} P_{\,n + 1} }\limits^ \to \; \times \;\mathop {P_{\,n} V}\limits^ \to $$
Then the dihedral angle $\alpha$ will be simply computed from the dot product $$ \cos \alpha = {{{\bf n}_{\,m} \; \cdot \;{\bf n}_{\,p} } \over {\left| {{\bf n}_{\,m} } \right|\;\;\left| {{\bf n}_{\,p} } \right|}} $$
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edited Jul 24, 2020 at 8:21
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answered Feb 14, 2018 at 18:27
G CabG Cab
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$\begingroup$ One thing that's awkward about my solution is that I have to talk about perpendicularity of lines that do not actually intersect. I'm sure there's a right way to talk about those. Vectors solve this problem. $\endgroup$
ziggurism
– ziggurism
2018-02-14 21:35:41 +00:00
Commented Feb 14, 2018 at 21:35
$\begingroup$ @ziggurism: to my knowledge, vectorial approach is the most effective: You can easily calculate all angles, lengths and surfaces of interest. $\endgroup$
G Cab
– G Cab
2018-02-14 21:39:44 +00:00
Commented Feb 14, 2018 at 21:39
$\begingroup$ Yes of course that's true, but if a geometry problem is presented in a synthetic way (Euclidean geometry style), I like to solve it that way. $\endgroup$
ziggurism
– ziggurism
2018-02-14 21:42:24 +00:00
Commented Feb 14, 2018 at 21:42
$\begingroup$ @ziggurism: ok, I only wish you do not want to use also ..the greek numeral system. $\endgroup$
G Cab
– G Cab
2018-02-14 23:20:41 +00:00
Commented Feb 14, 2018 at 23:20
$\begingroup$ Let's compromise: roman numerals? $\endgroup$
ziggurism
– ziggurism
2018-02-15 02:43:55 +00:00
Commented Feb 15, 2018 at 2:43
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The short answer is, $DG$ lies in the plane $BCD$ perpendicular to $AB$, and is therefore (parallel to a line) perpendicular to $AB$. $DG$ also lies in the plane $DFG$ perpendicular to $AC$, and therefore is (parallel to a line) perpendicular to $AC$. Being perpendicular to both $AB$ and $AC$, $DG$ is perpendicular to every line in the plane $ABC$ that it intersects, including both $FG$ and $BC$.
Let's walk through the derivation of all three formulas in detail. Following the linked article, we see:
First, from the slant face isosceles triangle $\triangle CAD$, using the right triangle with altitude $AE,$ and $AC=s,$ we have $EC=s\sin(\alpha/2).$
Altitude $AB$ is perpendicular to plane $CAD,$ so triangle $ABC$ is a right triangle, and we have $BC=AC\cos\gamma=s\cos\gamma.$ From the base isosceles triangle $\triangle CBD$ we have $EC=s\cos\gamma\sin(\beta/2).$
Equating the two expressions gives $$\cos\gamma = \frac{\sin\left(\frac\alpha2\right)}{\sin\left(\frac\beta2\right)}\tag{1}\label{1}.$$
2. Next, why are $FG$ and $BC$ (that is, plane $ABC$) perpendicular to $DG$?
The trick is to realize that the perpendicular to a plane through the vertex of a triangle is orthogonal to all lines in that plane, including the two edges that meet it, as well as (a parallel line to) the opposite edge.
By construction, plane $DFG$ is perpendicular to $AC$. Therefore $AC$ is perpendicular to any line in $DFG$, ergo $AC$ is perpendicular to $FG$ and $FD$.
Additionally, line $DG$ lies in plane $BCD,$ so it (or a line parallel to it) is perpendicular to $AB$, as $AB$ is perpendicular to plane $BCD.$ Line $DG$ is also in plane $DFG,$ so perpendicular to $AC$. Therefore it is perpendicular to plane $ABC.$ And therefore also to $FG.$ Angles $\angle DGF$ and $\angle DGC$ are right angles (line $BC$ is also in plane $ABC$ and so perpendicular to $DG$).
Therefore looking at right triangle $\triangle FGD$ we have $GD=s\sin\alpha\sin\delta.$
And looking at right triangle $\triangle BGD$ we see $GD=s\cos\gamma\sin\beta.$
Equating gives $$\sin\delta=\frac{\cos\gamma\sin\beta}{\sin\alpha} \stackrel{\eqref{1}}= \frac{\sin\left(\frac\alpha2\right)\cdot2\sin\left(\frac\beta2\right)\cos\left(\frac\beta2\right)}{\sin\left(\frac\beta2\right)\cdot2\sin\left(\frac\alpha2\right)\cos\left(\frac\alpha2\right)}=\frac{\cos\left(\frac\beta2\right)}{\cos\left(\frac\alpha2\right)}\tag{2}\label{2}.$$
3. Looking at right triangle $\triangle ABE$ we have $BE=AE\cos\epsilon.$
From triangle $\triangle AEC$ we have $AE=s\cos\left(\frac\alpha2\right).$
And from triangle $\triangle BEC$ we have $BE=s\cos\left(\gamma\right)\cos\left(\frac\beta2\right).$
Thus $$\cos\epsilon=\frac{\sin\left(\frac\alpha2\right)\cos\left(\frac\beta2\right)}{\cos\left(\frac\alpha2\right)\sin\left(\frac\beta2\right)}=\frac{\tan\left(\frac\alpha2\right)}{\tan\left(\frac\beta2\right)}\tag{3}\label{3}.$$
So given the vertex angle $\alpha$ and the projection of that angle into the base $\beta$, we have the edge angle $\gamma$, the dihedral angle $\delta$ between $ABC$ and $ACD$, and the dihedral angle $\epsilon$ between $ACD$ and $BCD$.
Applying this to the regular pyramid, where $CAD$ is a lateral face and $\triangle ABC$ is a vertical (perpendicular to the base) cross section through an edge between adjacent faces, then the dihedral angle between adjacent side faces is $2\delta.$
In terms of the number of sides of the pyramid $n$, since there are $n$ angles of measure $\beta$ around a vertex in the plane, we have $$\beta=\frac{2\pi}{n}.$$ And in terms of the slant edge length $AC=s$, and base side length $CD=\ell$, we have $$\ell=2s\sin\left(\frac\alpha2\right).$$ So in terms of $n,\ell,$ and $s,$ we write $$\begin{align}\cos\gamma&=\frac{\ell}{2s\sin\left(\frac{\pi}{n}\right)},\\sin\delta&=\frac{2s\cos\left(\frac{\pi}{n}\right)}{\sqrt{4s^2-\ell^2}},\\cos\epsilon&=\frac{\ell}{\tan\left(\frac{\pi}{n}\right)\sqrt{4s^2-\ell^2}}.\end{align}$$
For example, putting $n=4$, $s=\ell$ gives us the regular square pyramid with $\gamma=\frac{\pi}{4},$ $\sin\delta=\sqrt{\frac{2}{3}},$ and $\cos\epsilon=\frac{1}{\sqrt{3}}.$ Which matches the angles of the regular octahedron in the table of Platonic solid dihedral angles.
Or with $n=3$, $s=\ell$ we get $\cos\gamma=1/\sqrt{3}$ and $\cos(2\delta)=\cos\epsilon=1/3,$ again agreeing with known geometry of the regular tetrahedron.
For a non-Platonic example, a regular hexagonal pyramid with height $h=\sqrt{3}\ell$ and slant edge length $s=2\ell$ will have $\cos(2\delta)=-3/5$, the angle of a $3$-$4$-$5$ triangle.
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edited Feb 14, 2018 at 17:57
answered Feb 14, 2018 at 15:26
ziggurismziggurism
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$\begingroup$ Very helpful answer, but I would just say there is a complication where $\beta > 90^\circ$ for then $G$ no longer lies on line segment $BC$ but is located out to the left of $B$, eg in the case of the tetrahedron where $\beta = 120^\circ$. $\endgroup$
Ross Ure Anderson
– Ross Ure Anderson
2023-01-04 17:17:48 +00:00
Commented Jan 4, 2023 at 17:17
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An alternate approach to obtaining the formula for $\delta$ is to take a unit normal $\underline{\mathbf{u}}$ for a given slope and then rotate this by the angle $\beta$ about the $z$-axis, ie about the vector $\underline{\mathbf{k}}$, to produce a unit normal $\underline{\mathbf{v}}$ for the adjacent slope, as shown in Fig. 1. This approach avoids the complication which arises when $\beta > 90^\circ$, for then $G$ is no longer on the line segment $BC$, but extends out to the left beyond $B$ (eg. in the case of the tetrahedron we have $\alpha = 60^\circ$ and $\beta = 120^\circ$, giving dihedral angle $2\delta = \arccos (1/3)$).
Since $\underline{\mathbf{u}}$ and $\underline{\mathbf{v}}$ are both outward pointing, and the dihedral angle between two planes equals the angle between their normals (or its complement), we have : $$ \underline{\mathbf{u}} \cdot \underline{\mathbf{v}} = \cos(180 - 2\delta) = -\cos 2\delta. $$
From Fig. 1 we have : $$ \underline{\mathbf{u}} = \sin \epsilon \; \underline{\mathbf{i}} + \cos \epsilon \; \underline{\mathbf{k}}. $$
Rotating $\underline{\mathbf{u}}$ about the $z$-axis by angle $\beta$ leaves the $\underline{\mathbf{k}}$ component unchanged and changes the component $\sin \epsilon \; \underline{\mathbf{i}}$ to $\sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \; \underline{\mathbf{j}}$ so that : $$ \underline{\mathbf{v}} = \sin \epsilon \cos \beta \; \underline{\mathbf{i}} + \sin \epsilon \sin \beta \;\underline{\mathbf{j}} + \cos \epsilon \; \underline{\mathbf{k}} $$
Thus : \begin{eqnarray} 2 \sin^2 \textstyle{\frac{1}{2}} \delta -1 = -\cos 2\delta & = & \underline{\mathbf{u}} \cdot \underline{\mathbf{v}} \nonumber \ & = & \sin^2 \epsilon \cos \beta + \cos^2 \epsilon \label{eq:Rodrigues} \tag{1} \end{eqnarray}
We require an expression that involves $\alpha$ and $\beta$ only. Denote the altitude of the regular pyramid by $h$, and the radius of its regular polygonal base by $r$. Then : \begin{equation} \sin \epsilon = \frac{h}{s \cos \frac{1}{2} \alpha}, \hspace{1em} \mbox{and} \hspace{1em} \cos \epsilon = \frac{r \cos \frac{1}{2} \beta}{s \cos \frac{1}{2} \alpha} \label{eq:angle-epsilon} \tag{2} \end{equation}
with : \begin{equation} \frac{h}{s} = \sin \gamma \hspace{1em} \mbox{and} \hspace{1em} \frac{r}{s} = \cos \gamma. \label{eq:angle-gamma} \tag{3} \end{equation}
As in the article we can readily establish by computing the length of EC in two ways that : \begin{equation} \cos \gamma = \frac{\sin \frac{1}{2} \alpha}{\sin \frac{1}{2} \beta}. \label{eq:angle-gamma-formula} \tag{4} \end{equation}
Thus putting equations (\ref{eq:Rodrigues}) - (\ref{eq:angle-gamma-formula}) together we now have : \begin{eqnarray} 2 \sin^2 \textstyle{\frac{1}{2}} \delta -1 & = & \frac{\sin^2 \gamma}{\cos^2 \frac{1}{2} \alpha} \cdot \cos \beta + \cos^2 \gamma \cdot \frac{\cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \ & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \left( 1 - \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot (\cos^2 \frac{1}{2}\beta - \sin^2 \frac{1}{2}\beta) + \left( \frac{\sin^2 \frac{1}{2} \alpha}{\sin^2 \frac{1}{2} \beta} \right) \cdot \cos^2 \frac{1}{2} \beta \right] \ & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + \sin^2 \frac{1}{2} \alpha \right] \ \Rightarrow 2 \sin^2 \delta & = & \frac{1}{\cos^2 \frac{1}{2} \alpha} \left[ \cos^2 \frac{1}{2} \beta - \sin^2 \frac{1}{2} \beta + 1 \right] \ & = & \frac{2 \cos^2 \frac{1}{2} \beta}{\cos^2 \frac{1}{2} \alpha} \ \Rightarrow \mbox{since } \alpha, \beta, \gamma \in (0, 180^\circ), \hspace{1.5em} \sin \delta & = & \frac{\cos \frac{1}{2} \beta}{\cos \frac{1}{2} \alpha}. \end{eqnarray}
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answered Dec 30, 2022 at 14:49
Ross Ure AndersonRoss Ure Anderson
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Statistical Guess
Notes on College & High school level Statistics
Sample Proportion
Sample Proportion
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The full name is Sampling Distribution of the Sample Proportion, which's denoted by p-hat.
Sampling Distribution of the Sample Proportion
p-hat
Refer to youtube: The Sampling Distribution of the Sample Proportion
Refer to article: The Sample Proportion
Refer to article: Sampling Distribution of the Sample Proportion, p-hat
Sample Proportion is the proportion of success in a sample.
Sample Proportion
Sample Proportion(p-hat) is a random variable,
specifically a Binomial Random Variable.
Binomial Random Variable
So let X denotes the number of success in the sample, which is the Binomial Random Variable with parameter n and _p_,
Mean and Variance of Sample Proportions
Mean and Variance of Sample Proportions
Recall that the _binomial random variable X_:
Hence, we derived the Mean & Variance of Sample Proportion p-hat from X:
p-hat
Why is that?
That’s why we say:
p-hat is an unbiased estimator for p of population.
p-hat
p
And for Standard Deviation of Sample Proportion:
Example
Solve:
Probabilities with Sample Proportions
Probabilities with Sample Proportions
It’s just to find out the probability area in the Normal Distribution.
All you need is the Mean, Standard Deviation and the point you’re to measure.
Example
Solve:
0.63
0.019
0.60
0.06
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Notes on College & High school level Statistics
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190137 | https://pmc.ncbi.nlm.nih.gov/articles/PMC10171266/ | Left upper lobe atelectasis - PMC
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J Bras Pneumol
. 2023 May 4;49(2):e20230064. doi: 10.36416/1806-3756/e20230064
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Left upper lobe atelectasis
Atelectasia do lobo superior esquerdo
Edson Marchiori
Edson Marchiori
Universidade Federal do Rio de Janeiro, Rio de Janeiro (RJ) Brasil.
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1, Bruno Hochhegger
Bruno Hochhegger
Universidade Federal de Ciências da Saúde de Porto Alegre, Porto Alegre (RS) Brasil.
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2, Gláucia Zanetti
Gláucia Zanetti
Universidade Federal do Rio de Janeiro, Rio de Janeiro (RJ) Brasil.
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Universidade Federal do Rio de Janeiro, Rio de Janeiro (RJ) Brasil.
Universidade Federal de Ciências da Saúde de Porto Alegre, Porto Alegre (RS) Brasil.
Collection date 2023.
© 2023 Sociedade Brasileira de Pneumologia e Tisiologia
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License which permits unrestricted non-commercial use, distribution, and reproduction in any medium provided the original work is properly cited.
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PMCID: PMC10171266 PMID: 37194819
A 68-year-old man presented with complaints of persistent, severe cough and weight loss in the last six months. Chest X-rays showed opacity in the upper third of the left lung and volume loss (Figure 1).
Figure 1. Posteroanterior and lateral chest X-rays (in A and in B, respectively) showing opacity and left upper lobe volume loss. Note an upward and forward shift of the major fissure and a mass in the left hilar region, preventing complete displacement of the fissure, which assumed an S shape (the Golden S sign).
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The aforementioned chest X-ray findings are indicative of left upper lobe atelectasis. Atelectasis appears as parenchymal opacification and volume loss caused by loss of lung aeration. Atelectasis can be lobar, segmental, or subsegmental, or it can involve an entire lung, with varying imaging appearances. It can occur through different mechanisms, including passive retraction of the lung parenchyma, scar tissue formation, compression of lung tissue, surfactant deficiency, and bronchial obstruction. Obstructive atelectasis occurs when airway obstruction inhibits regional lung ventilation partially or completely. Perfusion to the area is maintained; however, gas uptake into the blood continues. Eventually, all of the gas in that segment will be absorbed and, without return of ventilation, the airway will collapse. The causes of bronchial obstruction are varied, the most common being bronchial tumors in adults and foreign bodies in children. Children are especially susceptible to resorption atelectasis in the presence of an aspirated foreign body because they have poorly developed collateral pathways for ventilation.1-3
The radiological signs of pulmonary atelectasis can be divided into direct signs and indirect signs. Displacement of fissures is the most important direct sign of pulmonary atelectasis. The indirect signs are basically related to the loss of lung volume and include increased lung opacity, elevation of the diaphragm, mediastinal shift, and compensatory hyperinflation of the remaining lung parenchyma. Total atelectasis caused by bronchial obstruction appears as an opaque hemithorax, with displacement of the mediastinum toward the affected side. Left upper lobe atelectasis appears as a superior and anterior displacement of the major fissure. Anterior displacement of the major fissure is more easily seen on lateral chest X-rays. Segmental atelectasis of the left upper lobe with preservation of the lingula, as seen in our patient, can result in findings similar to those of atelectasis of the right upper lobe. Although the Golden S sign was initially used in order to describe signs of right upper lobe atelectasis, it can be applicable to atelectasis involving any lung lobe. The Golden S sign represents bowing or displacement of a fissure caused by a mass (generally bronchial carcinoma) that prevents complete displacement of the fissure.1-3
On the basis of the aforementioned findings, a diagnosis of left upper lobe atelectasis was made. A bronchoscopy was then performed, showing a tumor obstructing the left upper lobe bronchus, the lingular segments being spared.
REFERENCES
1.Webb WR, Muller NL, Naidich DP. High-resolution CT of the lung. 4. Philadelphia: Lippincott Williams & Wilkins; 2008. [Google Scholar]
2.Müller NL, Silva CI. Imaging of the Chest. Philadelphia: Saunders-Elsevier; 2008. [Google Scholar]
3.Marchiori E, Hochhegger B, Zanetti G. Opaque hemithorax. J Bras Pneumol. 2017;43(3):161–161. doi: 10.1590/S1806-37562017000000024. [DOI] [PMC free article] [PubMed] [Google Scholar]
Articles from Jornal Brasileiro de Pneumologia are provided here courtesy of Sociedade Brasileira de Pneumologia e Tisiologia (Brazilian Thoracic Society)
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190138 | https://ocw.mit.edu/courses/6-0002-introduction-to-computational-thinking-and-data-science-fall-2016/2b524c63c8236d001970e454a3707bf9_soZv_KKax3E.pdf | MITOCW | watch?v=soZv_KKax3E The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation, or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu.
PROFESSOR: Good afternoon, everybody. Welcome to Lecture 8. So we're now more than halfway through the lectures.
All right, the topic of today is sampling. I want to start by reminding you about this whole business of inferential statistics. We make references about populations by examining one or more random samples drawn from that population.
We used Monte Carlo simulation over the last two lectures. And the key idea there, as we saw in trying to find the value of pi, was that we can generate lots of random samples, and then use them to compute confidence intervals. And then we use the empirical rule to say, all right, we really have good reason to believe that 95% of the time we run this simulation, our answer will be between here and here.
Well, that's all well and good when we're doing simulations. But what happens when you to actually sample something real? For example, you run an experiment, and you get some data points. And it's too hard to do it over and over again.
Think about political polls. Here was an interesting poll. How were these created? Not by simulation. They didn't run 1,000 polls and then compute the confidence interval. They ran one poll-- of 835 people, in this case. And yet they claim to have a confidence interval. That's what that margin of error is. Obviously they needed that large confidence interval.
So how is this done? Backing up for a minute, let's talk about how sampling is done when you are not running a simulation. You want to do what's called probability sampling, in which each member of the population has a non-zero probability of being included in a sample.
There are, roughly speaking, two kinds. We'll spend, really, all of our time on something called simple random sampling. And the key idea here is that each member of the population has an equal probability of being chosen in the sample so there's no bias.
Now, that's not always appropriate. I do want to take a minute to talk about why. So suppose Now, that's not always appropriate. I do want to take a minute to talk about why. So suppose we wanted to survey MIT students to find out what fraction of them are nerds-- which, by the way, I consider a compliment. So suppose we wanted to consider a random sample of 100 students. We could walk around campus and choose 100 people at random. And if 12% of them were nerds, we would say 12% of the MIT undergraduates are nerds-- if 98%, et cetera.
Well, the problem with that is, let's look at the majors by school. This is actually the majors at MIT by school. And you can see that they're not exactly evenly distributed. And so if you went around and just sampled 100 students at random, there'd be a reasonably high probability that they would all be from engineering and science. And that might give you a misleading notion of the fraction of MIT students that were nerds, or it might not.
In such situations we do something called stratified sampling, where we partition the population into subgroups, and then take a simple random sample from each subgroup. And we do that proportional to the size of the subgroups. So we would certainly want to take more students from engineering than from architecture. But we probably want to make sure we got somebody from architecture in our sample.
This, by the way, is the way most political polls are done. They're stratified. They say, we want to get so many rural people, so many city people, so many minorities-- things like that. And in fact, that's probably where the election recent polls were all messed up. They did a very, retrospectively at least, a bad job of stratifying.
So we use stratified sampling when there are small groups, subgroups, that we want to make sure are represented. And we want to represent them proportional to their size in the population. This can also be used to reduce the needed size of the sample. If we wanted to make sure we got some architecture students in our sample, we'd need to get more than 100 people to start with. But if we stratify, we can take fewer samples. It works well when you do it properly. But it can be tricky to do it properly. And we are going to stick to simple random samples here.
All right, let's look at an example. So this is a map of temperatures in the United States. And so our running example today will be sampling to get information about the average temperatures. And of course, as you can see, they're highly variable. And we live in one of the cooler areas.
The data we're going to use is real data-- and it's in the zip file that I put up for the class-- from the US Centers for Environmental Information. And it's got the daily high and low temperatures for 21 different American cities, every day from 1961 through 2015. So it's an interesting data set-- a total of about 422,000 examples in the dataset. So a fairly good sized dataset. It's fun to play with.
All right, so we're sort of in the part of the course where the next series of lectures, including today, is going to be about data science, how to analyze data. I always like to start by actually looking at the data-- not looking at all 421,000 samples, but giving a plot to sort of give me a sense of what the data looks like. I'm not going to walk you through the code that does this plot. I do want to point out that there are two things in it that we may not have seen before.
Simply enough, I'm going to use numpy.std to get standard deviations instead of my own code for it. And random.sample to take simple random samples from the population.
random.sample takes two arguments. The first is some sort of a sequence of values. And the second is an integer telling you how many samples you want. And it returns a list containing sample size, randomly chosen distinct elements.
Distinct elements is important, because there are two ways that people do sampling. You can do sampling without replacement, which is what's done here. You take a sample, and then it's out of the population. So you won't draw it the next time. Or you can do sampling with replacement, which allows you to draw the same sample multiple times-- the same example multiple times.
We'll see later in the term that there are good reasons that we sometimes prefer sampling with replacement. But usually we're doing sampling without replacement. And that's what we'll do here. So we won't get Boston on April 3rd multiple times-- or, not the same year, at least.
All right. So here's the histogram the code produces. You can run it yourself now, if you want, or you can run it later. And here's what it looks like. The daily high temperatures, the mean is 16.3 degrees Celsius. I sort of vaguely know what that feels like. And as you can see, it's kind of an interesting distribution. It's not normal. But it's not that far, right? We have a little tail of these cold temperatures on the left. And it is what it is. It's not a normal distribution. And we'll later see that doesn't really matter.
OK, so this gives me a sense. The next thing I'll get is some statistics. So we know the mean is 16.3 and the standard deviation is approximately 9.4 degrees. So if you look at it, you can believe that.
Well, here's a histogram of one random sample of size 100. Looks pretty different, as you might expect. Its standard deviation is 10.4, its mean 17.7. So even though the figures look a little different, in fact, the means and standard deviations are pretty similar. If we look at the population mean and the sample mean-- and I'll try and be careful to use those terms-- they're not the same. But they're in the same ballpark. And the same is true of the two standard deviations.
Well, that raises the question, did we get lucky or is something we should expect? If we draw 100 random examples, should we expect them to correspond to the population as a whole?
And the answer is sometimes yeah and sometimes no. And that's one of the issues I want to explore today.
So one way to see whether it's a happy accident is to try it 1,000 times. We can draw 1,000 samples of size 100 and plot the results. Again, I'm not going to go over the code. There's something in that code, as well, that we haven't seen before. And that's the ax.vline plotting command. V for vertical. It just, in this case, will draw a red line-- because I've said the color is r-- at population mean on the x-axis. So just a vertical line. So that'll just show us where the mean is. If we wanted to draw a horizontal line, we'd use ax.hline. Just showing you a couple of useful functions.
When we try it 1,000 times, here's what it looks like. So here we see what we had originally, same picture I showed you before. And here's what we get when we look at the means of 100 samples. So this plot on the left looks a lot more like it's a normal distribution than the one on the right. Should that surprise us, or is there a reason we should have expected that to happen?
Well, what's the answer? Someone tell me why we should have expected it. It's because of the central limit theorem, right? That's exactly what the central limit theorem promised us would happen. And, sure enough, it's pretty close to normal. So that's a good thing.
And now if we look at it, we can see that the mean of the sample means is 16.3, and the standard deviation of the sample means is 0.94. So if we go back to what we saw here, we see that, actually, when we run it 1,000 times and look at the means, we get very close to what we had initially. So, indeed, it's not a happy accident. It's something we can in general expect.
All right, what's the 95% confidence interval here? Well, it's going to be 16.28 plus or minus 1.96 times 0.94, the standard deviation of the sample means. And so it tells us that the confidence interval is, the mean high temperature, is somewhere between 14.5 and 18.1.
Well, that's actually a pretty big range, right? It's sort of enough to where you wear a sweater or where you don't wear a sweater. So the good news is it includes the population mean.
That's nice. But the bad news is it's pretty wide.
Suppose we wanted it tighter bound. I said, all right, sure enough, the central limit theorem is going to tell me the mean of the means is going to give me a good estimate of the actual population mean. But I want it tighter bound. What can I do?
Well, let's think about a couple of things we could try. Well, one thing we could think about is drawing more samples. Suppose instead of 1,000 samples, I'd taken 2,000 or 3,000 samples.
We can ask the question, would that have given me a smaller standard deviation? For those of you who have not looked ahead, what do you think? Who thinks it will give you a smaller standard deviation? Who thinks it won't? And the rest of you have either looked ahead or refused to think. I prefer to believe you looked ahead.
Well, we can run the experiment. You can go to the code. And you'll see that there is a constant of 1,000, which you can easily change to 2,000. And lo and behold, the standard deviation barely budges. It got a little bit bigger, as it happens, but that's kind of an accident. It just, more or less, doesn't change. And it won't change if I go to 3,000 or 4,000 or 5,000. It'll wiggle around. But it won't help much. What we can see is doing that more often is not going to help.
Suppose we take larger samples? Is that going to help? Who thinks that will help? And who thinks it won't? OK. Well, we can again run the experiment. I did run the experiment. I changed the sample size from 100 to 200. And, again, you can run this if you want. And if you run it, you'll get a result-- maybe not exactly this, but something very similar-- that, indeed, as I increase the size of the sample rather than the number of the samples, the standard deviation drops fairly dramatically, in this case from 0.94 0.66. So that's a good thing.
I now want to digress a little bit before we come back to this and look at how you can visualize this-- Because this is a technique you'll want to use as you write papers and things like that-- is how do we visualize the variability of the data? And it's usually done with something called an error bar. You've all seen these things here. And this is one I took from the literature. This is plotting pulse rate against how much exercise you do or how frequently you exercise.
And what you can see here is there's definitely a downward trend suggesting that the more you exercise, the lower your average resting pulse. That's probably worth knowing. And these error bars give us the 95% confidence intervals for different subpopulations.
And what we can see here is that some of them overlap. So, yes, once a fortnight-- two weeks for those of you who don't speak British-- it does get a little bit smaller than rarely or never. But the confidence interval is very big. And so maybe we really shouldn't feel very comfortable that it would actually help.
The thing we can say is that if the confidence intervals don't overlap, we can conclude that the means are actually statistically significantly different, in this case at the 95% level. So here we see that the more than weekly does not overlap with the rarely or never. And from that, we can conclude that this is actually, statistically true-- that if you exercise more than weekly, your pulse is likely to be lower than if you don't.
If confidence intervals do overlap, you cannot conclude that there is no statistically significant difference. There might be, and you can use other tests to find out whether there are. When they don't overlap, it's a good thing. We can conclude something strong. When they do overlap, we need to investigate further.
All right, let's look at the error bars for our temperatures. And again, we can plot those using something called pylab.errorbar. Lab So what it takes is two values, the usual x-axis and y-axis, and then it takes another list of the same length, or sequence of the same length, which is the y errors. And here I'm just going to say 1.96 times the standard deviations. Where these variables come from you can tell by looking at the code. And then I can say the format, I want an o to show the mean, and then a label. Fmt stands for format.
errorbar has different keyword arguments than plot. You'll find that you look at different ways like histograms and bar plots, scatterplots-- they all have different available keyword arguments. So you have to look up each individually. But other than this, everything in the code should look very familiar to you.
And when I run the code, I get this. And so what I've plotted here is the mean against the sample size with errorbars. And 100 trials, in this case. So what you can see is that, as the sample size gets bigger, the errorbars get smaller. The estimates of the mean don't necessarily get any better.
In fact, we can look here, and this is actually a worse estimate, relative to the true mean, than the previous two estimates. But we can have more confidence in it. The same thing we saw on Monday when we looked at estimating pi, dropping more needles didn't necessarily give us a more accurate estimate. But it gave us more confidence in our estimate. And the same thing is happening here. And we can see that, steadily, we can get more and more confidence.
So larger samples seem to be better. That's a good thing. Going from a sample size of 50 to a sample size of 600 reduced the confidence interval, as you can see, from a fairly large confidence interval here, ran from just below 14 to almost 19, as opposed to 15 and a half or so to 17. I said confidence interval here. I should not have. I should have said standard deviations. That's an error on the slides.
OK, what's the catch? Well, we're now looking at 100 samples, each of size 600. So we've looked at a total of 600,000 examples. What has this bought us? Absolutely nothing. The entire population only contained about 422,000 samples. We might as well have looked at the whole thing, rather than take a few of them. So it's like, you might as well hold an election rather than ask 800 people a million times who they're going to vote for. Sure, it's good. But it gave us nothing.
Suppose we did it only once. Suppose we took only one sample, as we see in political polls.
What can we can conclude from that? And the answer is actually kind of surprising, how much we can conclude, in a real mathematical sense, from one sample. And, again, this is thanks to our old friend, the central limit theorem.
So if you recall the theorem, it had three parts. Up till now, we've exploited the first two. We've used the fact that the means will be normally distributed so that we could use the empirical rule to get confidence intervals, and the fact that the mean of the sample means would be close to the mean of the population.
Now I want to use the third piece of it, which is that the variance of the sample means will be close to the variance of the population divided by the sample size. And we're going to use that to compute something called the standard error-- formerly the standard error of the mean.
People often just call it the standard error. And I will be, alas, inconsistent. I sometimes call it one, sometimes the other.
It's an incredibly simple formula. It says the standard error is going to be equal to sigma, where sigma is the population standard deviation divided by the square root of n, which is going to be the size of the sample. And then there's just this very small function that implements it. So we can compute this thing called the standard error of the mean in a very straightforward way.
We can compute it. But does it work? What do I mean by work? I mean, what's the relationship of the standard error to the standard deviation? Because, remember, that was our goal, was to understand the standard deviation so we could use the empirical rule.
Well, let's test the standard error of the mean. So here's a slightly longer piece of code. I'm going to look at a bunch of different sample sizes, from 25 to 600, 50 trials each. So getHighs is just a function that returns the temperatures. I'm going to get the standard deviation of the whole population, then the standard error of the means and the sample standard deviations, both. And then I'm just going to go through and run it. So for size and sample size, I'm going to append the standard error of the mean. And remember, that uses the population standard deviation and the size of the sample. So I'll compute all the SEMs. And then I'm going to compute all the actual standard deviations, as well. And then we'll produce a bunch of plots--or a plot, actually.
All right, so let's see what that plot looks like. Pretty striking. So we see the blue solid line is the standard deviation of the 50 means. And the red dotted line is the standard error of the mean.
So we can see, quite strikingly here, that they really track each other very well. And this is saying that I can anticipate what the standard deviation would be by computing the standard error.
Which is really useful, because now I have one sample. I computed standard error. And I get something very similar to what I get of the standard deviation if I took 50 samples and looked at the standard deviation of those 50 samples. All right, so not obvious that this would be true, right? That I could use this simple formula, and the two things would track each other so well.
And it's not a coincidence, by the way, that as I get out here near the end, they're really lying on top of each other. As the sample size gets much larger, they really will coincide.
So one, does everyone understand the difference between the standard deviation and the standard error? No. OK. So how do we compute a standard deviation? To do that, we have to look at many samples-- in this case 50-- and we compute how much variation there is in those 50 samples.
For the standard error, we look at one sample, and we compute this thing called the standard error. And we argue that we get the same number, more or less, that we would have gotten had we taken 50 samples or 100 samples and computed the standard deviation. So I can avoid taking all 50 samples if my only reason for doing it was to get the standard deviation. I can take one sample instead and use the standard error of the mean. So going back to my temperature-- instead of having to look at lots of samples, I only have to look at one. And I can get a confidence interval. That make sense? OK.
There's a catch. Notice that the formula for the standard error includes the standard deviation of the population-- the standard deviation of the sample. Well, that's kind of a bummer.
Because how can I get the standard deviation of the population without looking at the whole population? And if we're going to look at the whole population, then what's the point of sampling in the first place?
So we have a catch, that we've got something that's a really good approximation, but it uses a value we don't know. So what should we do about that? Well, what would be, really, the only obvious thing to try? What's our best guess at the standard deviation of the population if we have only one sample to look at? What would you use? Somebody? I know I forgot to bring the candy today, so no one wants to answer any questions.
AUDIENCE: The standard deviation of the sample?
PROFESSOR: The standard deviation of the sample. It's all I got. So let's ask the question, how good is that?
Shockingly good. So I looked at our example here for the temperatures. And I'm plotting the sample standard deviation versus the population standard deviation for different sample sizes, ranging from 0 to 600 by one, I think.
So what you can see here is when the sample size is small, I'm pretty far off. I'm off by 14% here. And I think that's 25. But when the sample sizes is larger, say 600, I'm off by about 2%.
So what we see, at least for this data set of temperatures-- if the sample size is large enough, the sample standard deviation is a pretty good approximation of the population standard deviation.
Well. Now we should ask the question, what good is this? Well, as I said, once the sample reaches a reasonable size-- and we see here, reasonable is probably somewhere around 500-- it becomes a good approximation. But is it true only for this example? The fact that it happened to work for high temperatures in the US doesn't mean that it will always be true.
happened to work for high temperatures in the US doesn't mean that it will always be true.
So there are at least two things we should consider to asking the question, when will this be true, when won't it be true. One is, does the distribution of the population matter? So here we saw, in our very first plot, the distribution of the high temperatures. And it was kind of symmetric around a point-- not perfectly. But not everything looks that way, right?
So we should say, well, suppose we have a different distribution. Would that change this conclusion? And the other thing we should ask is, well, suppose we had a different sized population. Suppose instead of 400,000 temperatures I had 20 million temperatures. Would I need more than 600 samples for the two things to be about the same?
Well, let's explore both of those questions. First, let's look at the distributions. And we'll look at three common distributions-- a uniform distribution, a normal distribution, and an exponential distribution. And we'll look at each of them for, what is this, 100,000 points.
So we know we can generate a uniform distribution by calling random.random. Gives me a uniform distribution of real numbers between 0 and 1. We know that we can generate our normal distribution by calling random.gauss. In this case, I'm looking at it between the mean of 0 and a standard deviation of 1. But as we saw in the last lecture, the shape will be the same, independent of these values.
And, finally, an exponential distribution, which we get by calling random.expovariate. Very And this number, 0.5, is something called lambda, which has to do with how quickly the exponential either decays or goes up, depending upon which direction. And I'm not going to give you the formula for it at the moment. But we'll look at the pictures. And we'll plot each of these discrete approximations to these distributions.
So here's what they look like. Quite different, right? We've looked at uniform and we've looked at Gaussian before. And here we see an exponential, which basically decays and will asymptote towards zero, never quite getting there. But as you can see, it is certainly not very symmetric around the mean.
All right, so let's see what happens. If we run the experiment on these three distributions, each of 100,000 point examples, and look at different sample sizes, we actually see that the difference between the standard deviation and the sample standard deviation of the population standard deviation is not the same.
We see, down here-- this looks kind of like what we saw before. But the exponential one is really quite different. You know, its worst case is up here at 25. The normal is about 14. So that's not too surprising, since our temperatures were kind of normally distributed when we looked at it. And the uniform is, initially, much better an approximation.
And the reason for this has to do with a fundamental difference in these distributions, something called skew. Skew is a measure of the asymmetry of a probability distribution. And what we can see here is that skew actually matters. The more skew you have, the more samples you're going to need to get a good approximation. So if the population is very skewed, very asymmetric in the distribution, you need a lot of samples to figure out what's going on. If it's very uniform, as in, for example, the uniform population, you need many fewer samples. OK, so that's an important thing. When we go about deciding how many samples we need, we need to have some estimate of the skew in our population.
All right, how about size? Does size matter? Shockingly-- at least it was to me the first time I looked at this-- the answer is no. If we look at this-- and I'm looking just for the uniform distribution, but we'll see the same thing for all three-- it more or less doesn't matter. Quite amazing, right?
If you have a bigger population, you don't need more samples. And it's really almost counterintuitive to think that you don't need any more samples to find out what's going to happen if you have a million people or 100 million people. And that's why, when we look at, say, political polls, they're amazingly small. They poll 1,000 people and claim they're representative of Massachusetts.
This is good news. So to estimate the mean of a population, given a single sample, we choose a sample size based upon some estimate of skew in the population. This is important, because if we get that wrong, we might choose a sample size that is too small. And in some sense, you always want to choose the smallest sample size you can that will give you an accurate answer, because it's more economical to have small samples than big samples.
And I've been talking about polls, but the same is true in an experiment. How many pieces of data do you need to collect when you run an experiment in a lab. And how much will depend, again, on the skew of the data. And that will help you decide.
When you know the size, you choose a random sample from the population. Then you compute the mean and the standard deviation of that sample. And then use the standard deviation of that sample to estimate the standard error. And I want to emphasize that what you're getting here is an estimate of the standard error, not the standard error itself, which would require you to know the population standard deviation. But if you've chosen the sample size to be appropriate, this will turn out to be a good estimate.
And then once we've done that, we use the estimated standard error to generate confidence intervals around the sample mean. And we're done. Now this works great when we choose independent random samples. And, as we've seen before, that if you don't choose independent samples, it doesn't work so well. And, again, this is an issue where if you assume that, in an election, each state is independent of every other state, and you'll get the wrong answer, because they're not.
All right, let's go back to our temperature example and pose a simple question. Are 200 samples enough? I don't know why I chose 200. I did. So we'll do an experiment here. This is similar to an experiment we saw on Monday.
So I'm starting with the number of mistakes I make. For t in a range number of trials, sample will be random.sample of the temperatures in the sample size. This is a key step. The first time I did this, I messed it up. And instead of doing this very simple thing, I did a more complicated thing of just choosing some point in my list of temperatures and taking the next 200 temperatures. Why did that give me the wrong answer? Because it's organized by city. So if I happen to choose the first day of Phoenix, all 200 temperatures were Phoenix-- which is not a very good approximation of the temperature in the country as a whole.
But this will work. I'm using random.sample. I'll then get the sample mean. Then I'll compute my estimate of the standard error by taking that as seen here. And then if the absolute value of the population minus the sample mean is more than 1.96 standard errors, I'm going to say I messed up. It's outside. And then at the end, I'm going to look at the fraction outside the 95% confidence intervals.
And what do I hope it should print? What would be the perfect answer when I run this? What fraction should lie outside that? It's a pretty simple calculation. Five, right? Because if they all were inside, then I'm being too conservative in my interval, right? I want 5% of the tests to fall outside the 95% confidence interval.
If I wanted fewer, then I would look at three standard deviations. Instead of 1.96, then I would expect less than 1% to fall outside. So this is something we have to always keep in mind when expect less than 1% to fall outside. So this is something we have to always keep in mind when we do this kind of thing. If your answer is too good, you've messed up. Shouldn't be too bad, but it shouldn't be too good, either. That's what probabilities are all about. If you called every election correctly, then your math is wrong.
Well, when we run this, we get this lovely answer, that the fraction outside the 95% confidence interval is 0.0511. That's exactly-- well, close to what you want. It's almost exactly 5%. And if I run it multiple times, I get slightly different numbers. But they're all in that range, showing that, here, in fact, it really does work.
So that's what I want to say, and it's really important, this notion of the standard error. When I talk to other departments about what we should cover in 60002, about the only thing everybody agrees on was we should talk about standard error. So now I hope I have made everyone happy. And we will talk about fitting curves to experimental data starting next week.
All right, thanks a lot. |
190139 | https://www.iitianacademy.com/cie-as-a-level-physics-9702-topic-7-waves-unit-7-3-doppler-effect-for-sound-waves-study-notes/ | CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes
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Home / CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes – New Syllabus
CIE AS/A Level Physics 7.3 Doppler effect for sound waves Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
understand that when a source of sound waves moves relative to a stationary observer, the observed frequency is different from the source frequency (understanding of the Doppler effect for a stationary source and a moving observer is not required)
use the expression f o=f s×V V±V s for the observed frequency when a source of sound waves moves
relative to a stationary observer
AS/A Level Physics Study Notes- All Topics
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7.3.1 understand that when a source of sound waves moves relative to a stationary observer, the observed frequency is different from the source frequency (understanding of the Doppler effect for a stationary source and a moving observer is not required)
Doppler effect is the change in frequency of a wave received by an observer moving relative to the source of waves.
You may have noticed a change in pitch of the note heard when an emergency vehicle passes you while sounding its siren. The pitch is higher as the vehicle approaches you, and lower as it moves away (recedes). This is an example of the Doppler effect; you can hear the same thing if a train passes at speed while sounding its whistle.
Above Figure shows that the frequency of waves emitted by a stationary source is same in all directions. Thus, the observer at rest perceives frequency f.
When the source starts moving towards the observer, shorter wavelength waves are moving towards the observer and therefore, high frequency is observed.
Similarly, when the source starts to move away from the observer, wavelength becomes longer and the received frequency if less than the actual frequency emitted from the source.
In nutshell, waves get compressed when source moves towards the observer and expand when the source starts moving away from the observer.
Doppler Effect in light:
A similar kind of phenomena is observed in electromagnetic or light waves.
Red shift: When the observer is moving away from the source, the frequency of the waves is reduced. Thus, a shift towards red side of the spectrum is observed ( due to increase in wavelength).
Blue shift: When the observer is moving towards the source, the frequency of the waves is increased. Thus, a shift towards blue side of the spectrum is observed ( due to decrease in wavelength).
Applications of Doppler effect:
Radar Guns: It detects the change in the speed of the vehicle from the change in frequency of the reflected radio waves.
Measuring speed of galaxies in astronomy: The Doppler shift of light from stars and galaxies give information about the speed and direction of their movement.
Sound navigation and Ranging (SONAR): It is used by ships to detect underwater objects (like submarines) by detecting the change in frequency.
7.3.2 use the expression f o=f s×V V±V s for the observed frequency when a source of sound waves moves relative to a stationary observer
The frequency and wavelength observed by an observer will change according to the speed v s at which the source is moving relative to the stationary observer.
Considering the Figure given in the previous section, the wavelength of sound emitted by a stationary source is given as λ o = (\frac{v}{f s}), where v is the wave speed and f s is the source frequency.
When the source starts moving with a velocity vs, the wavelength of the signal received by the observer changes to:
Moving towards observer:λ=f s v v−v s(shorter wavelength)
Moving away from observer:λ=f s v v+v s(longer wavelength)
Therefore, the frequency received by the observer will be given as:
f o=v λ
Moving towards observer:f o=f s v v–v s
Moving away from observer:f o=f s v v+v s
Trick to remember: Source moving closer: Frequency received will be high (small denominator, v – v s)
Source moving away: Frequency received will be low (large denominator, v + v s)
When both source and observor are moving:
The general formula of Doppler effect is given as:
f o=f s v±v o v±v s
Here, v o is the speed of the observor.
Case I: Source is moving towards the observer at rest.
f o=f s v v−v s
Case II: Source is moving away from the observer at rest.
f o=f s v v+v s
Case III: Observor moving away from a stationary source.
f o=f s v−v o v
Case IV: Observor moving towards a stationary source.
f o=f s v+v o v
Question
A train with a whistle that emits a note of frequency 800 Hz is approaching a stationary observer at a speed of 60 m s−1.Calculate the frequency of the note heard by the observer.
Answer/ExplanationSpeed of sound in air =330 m s−1
Step 1 Select the appropriate form of the Doppler equation. Here, the source is approaching the observer so we choose the minus sign:
f o=f s v v−v s
Step 2 Substitute values from the question and solve:
f o= 800×330 330−60 = 978 Hz
So, the observer hears a note whose pitch is raised significantly, because the train is travelling at a speed that is a significant fraction of the speed of sound.
Question
Two cars A and B are moving toward each other at a speed of 140 m/s. If the frequency of the horn emitted by A is 800 Hz, then what is the apparent frequency of the horn heard by the passenger sitting in car B?
Answer/Explanation
The velocity of sound in air is 360 m/s.
f o=f s v+v o v−v s
f o=800(360+140 360−140)Hz
= 1818 Hz.
12.5 The Doppler effect for sound waves
You may have noticed a change in pitch of the note heard when an emergency vehicle passes you while sounding its siren. The pitch is higher as the vehicle approaches you, and lower as it moves away (recedes). This is an example of the Doppler effect; you can hear the same thing if a train passes at speed while sounding its whistle.
Figure 12.11 shows why this change in frequency is observed. It shows a source of sound emitting waves with a constant frequency f S, together with two observers A and B.
If the source is stationary (Figure 12.11a), waves arrive at A and B at the same rate, and so both observers hear sounds of the same frequency f s.
If the source is moving towards A and away from B (Figure 12.11b), the situation is different. From the diagram, you can see that the waves are squashed together in the direction of A and spread apart in the direction of B.
Observer A will observe, or detect, waves whose wavelength is shortened. More wavelengths per second arrive at A, and so A observes a sound of higher frequency than f s. Similarly, the waves arriving at B have been stretched out and B will observe a frequency lower than f s.
Figure 12.11: Sound waves (green lines) emitted at constant frequency by a a stationary source, and b a source moving with speed v s. The separation between adjacent green lines is equal to one wavelength.
An equation for observed frequency
There are two different speeds involved in this situation. The source is moving with speed v s. The sound waves travel through the air with speed v, which is unaffected by the speed of the source. (Remember, the speed of a wave depends only on the medium it is travelling through.)
The frequency and wavelength observed by an observer will change according to the speed v s at which the source is moving relative to the stationary observer. Figure 12.12 shows how we can calculate the observed wavelength λ 0 and the observed frequency f 0.
The wave sections shown in Figure 12.12 represent the f s wavelengths emitted by the source in 1 s. Provided the source is stationary (Figure 12.12a), the length of this section is equal to the wave speed v. The wavelength observed by the observer is simply:
λ 0=v f s
The situation is different when the source is moving away (receding) from the observer (Figure 12.12b).
In 1 s, the source moves a distance v s. Now the section of f s wavelengths will have a length equal to v+v s.
Figure 12.12: Sound waves, emitted at constant frequency by a a stationary source, and b a source moving with speed v S away from the observer (that is, the person hearing the sound).
The observed wavelength is now given by:
λ 0=(v+v s)f s
The observed frequency is given by:
f 0=v λ 0=f s×v(v+v s)
where f 0 is the observed frequency, f S is the frequency of the source, v is the speed of the wave and v S is the speed of the source relative to the observer.
This shows us how to calculate the observed frequency when the source is moving away from the observer. If the source is moving towards the observer, the section of f s wavelengths will be compressed into a shorter length equal to v−v S, and the observed frequency will be given by:
f 0=v λ 0=f s×v(v−v s)
We can combine these two equations to give a single equation for the Doppler shift in frequency due to a moving source:
observed frequency, f 0=f s×v(v±v s)
where the plus sign applies to a receding source and the minus sign to an approaching source. Note these important points:
The frequency fs of the source is not affected by the movement of the source.
The speed v of the waves as they travel through the air (or other medium) is also unaffected by the movement of the source.
Note that a Doppler effect can also be heard when an observer is moving relative to a stationary source, and also when both source and observer are moving. There is more about the Doppler effect and light in Chapter 31.
WORKED EXAMPLE
3 A train with a whistle that emits a note of frequency 800 Hz is approaching a stationary observer at a speed of 60 m s−1.
Calculate the frequency of the note heard by the observer.
speed of sound in air =330 m s−1
Step 1 Select the appropriate form of the Doppler equation. Here the source is approaching the observer so we choose the minus sign:
f 0=f s×v(v−v s)
Step 2 Substitute values from the question and solve:
f 0=800×330(330−60)=800×330 270=978 Hz≈980 Hz
So, the observer hears a note whose pitch is raised significantly, because the train is travelling at a speed that is a significant fraction of the speed of sound.
Question
10 A plane’s engine emits a note of constant frequency 120 Hz. It is flying away from a stationary observer at a speed of 80 m s−1. Calculate:
a the observed wavelength of the sound received by the observer
b its observed frequency.
(Speed of sound in air =330 m s−1.)
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190140 | https://pubmed.ncbi.nlm.nih.gov/2555167/ | The extreme mutator effect of Escherichia coli mutD5 results from saturation of mismatch repair by excessive DNA replication errors - PubMed
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. 1989 Nov;8(11):3511-6.
doi: 10.1002/j.1460-2075.1989.tb08516.x.
The extreme mutator effect of Escherichia coli mutD5 results from saturation of mismatch repair by excessive DNA replication errors
R M Schaaper1,M Radman
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1 Laboratory of Molecular Genetics, National Institute of Environmental Health Sciences, Research Triangle Park, NC 27709.
PMID: 2555167
PMCID: PMC401508
DOI: 10.1002/j.1460-2075.1989.tb08516.x
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The extreme mutator effect of Escherichia coli mutD5 results from saturation of mismatch repair by excessive DNA replication errors
R M Schaaper et al. EMBO J.1989 Nov.
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. 1989 Nov;8(11):3511-6.
doi: 10.1002/j.1460-2075.1989.tb08516.x.
Authors
R M Schaaper1,M Radman
Affiliation
1 Laboratory of Molecular Genetics, National Institute of Environmental Health Sciences, Research Triangle Park, NC 27709.
PMID: 2555167
PMCID: PMC401508
DOI: 10.1002/j.1460-2075.1989.tb08516.x
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Escherichia coli mutator mutD5 is the most potent mutator known. The mutD5 mutation resides in the dnaQ gene encoding the proofreading exonuclease of DNA polymerase III holoenzyme. It has recently been shown that the extreme mutability of this strain results, in addition to a proofreading defect, from a defect in mutH, L, S-encoded postreplicational DNA mismatch repair. The following measurements of the mismatch-repair capacity of mutD5 cells demonstrate that this mismatch-repair defect is not structural, but transient. mutD5 cells in early log phase are as deficient in mismatch repair as mutL cells, but they become as proficient as wild-type cells in late log phase. Second, arrest of chromosomal replication in a mutD5-dnaA(Ts) strain at a nonpermissive temperature restores mismatch repair, even from the early log phase of growth. Third, transformation of mutD5 strains with multicopy plasmids expressing the mutH or mutL gene restores mismatch repair, even in rapidly growing cells. These observations suggest that the mismatch-repair deficiency of mutD strains results from a saturation of the mutHLS-mismatch-repair system by an excess of primary DNA replication errors due to the proofreading defect.
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References
Mol Gen Genet. 1984;195(3):418-23 - PubMed
J Bacteriol. 1985 Sep;163(3):1007-15 - PubMed
J Mol Biol. 1983 Jun 5;166(4):557-80 - PubMed
J Bacteriol. 1974 Feb;117(2):477-87 - PubMed
Cell. 1988 Jun 17;53(6):837-40 - PubMed
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二次函数
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2023年科普中国×百度百科科学100词合作建设
基本定义
一般地，把形如y=ax²+bx+c（a≠0）（a、b、c是常数）的函数叫做二次函数，其中a称为二次项系数，b为一次项系数，c为常数项。x为自变量，y为因变量。等号右边自变量的最高次数是2。
顶点坐标
交点式为 y=a（x-x1）（x-x2）（仅限于与x轴有交点的抛物线），
与x轴的交点坐标是A（X1，0）和B（x2,0）。
注意：“变量”不同于“未知数”，不能说“二次函数是指未知数的最高次数为二次的多项式函数”。“未知数”只是一个数（具体值未知，但是只取一个值），“变量”可在一定范围内任意取值。在方程中适用“未知数”的概念（函数方程、微分方程中是未知函数，但不论是未知数还是未知函数，一般都表示一个数或函数——也会遇到特殊情况），但是函数中的字母表示的是变量，意义已经有所不同。从函数的定义也可看出二者的差别。
历史
大约在公元前480年，古巴比伦人和中国人已经使用配方法求得了二次方程的正根，但是并没有提出通用的求解方法。公元前300年左右，欧几里得提出了一种更抽象的几何方法求解二次方程。
7世纪印度的婆罗摩笈多是第一位懂得使用代数方程的人，它同时容许有正负数的根。
11世纪阿拉伯的花拉子密 独立地发展了一套公式以求方程的正数解。亚伯拉罕·巴希亚（亦以拉丁文名字萨瓦索达著称）在他的著作Liber embadorum中，首次将完整的一元二次方程解法传入欧洲。
据说施里德哈勒是最早给出二次方程的普适解法的数学家之一。但这一点在他的时代存在着争议。这个求解规则是：在方程的两边同时乘以二次项未知数的系数的四倍；在方程的两边同时加上一次项未知数的系数的平方；然后在方程的两边同时开二次方（引自婆什迦罗第二）
函数性质
1.二次函数的图像是抛物线，但抛物线不一定是二次函数。开口向上或者向下的抛物线才是二次函数。抛物线是轴对称图形。对称轴为直线 。对称轴与抛物线唯一的交点为抛物线的顶点P。特别地，当b=0时，抛物线的对称轴是y轴（即直线x=0）。
2.抛物线有一个顶点P，坐标为P 。当 时，P在y轴上；当 时，P在x轴上。
3.二次项系数a决定抛物线的开口方向和大小。当a>0时，抛物线开口向上；当a<0时，抛物线开口向下；|a|越小，则抛物线的开口越大；|a|越大，则抛物线的开口越小
4.一次项系数b和二次项系数a共同决定对称轴的位置。当a与b同号时（即ab>0），对称轴在y轴左侧；当a与b异号时（即ab<0）（可巧记为：左同右异），对称轴在y轴右侧。
5.常数项c决定抛物线与y轴交点。抛物线与y轴交于(0, c）
6.抛物线与x轴交点个数： 时，抛物线与x轴有2个交点。 时，抛物线与x轴有1个交点。当 时，抛物线与x轴没有交点。
7.当 时，函数在 处取得最小值 ；在 上是减函数，在 上是增函数；抛物线的开口向上；函数的值域是 。
当 时，函数在 处取得最大值 ；在 上是增函数，在 上是减函数；抛物线的开口向下；函数的值域是 。
当 时，抛物线的对称轴是y轴，这时，函数是偶函数，解析式变形为y=ax²+c(a≠0)。
8.定义域：R
9.值域：当a>0时，值域是 ；当a<0时，值域是 。6
奇偶性：当b=0时，此函数是偶函数；当b不等于0时，此函数是非奇非偶函数。
周期性：无
解析式：
①一般式：
⑴a≠0
⑵若a>0，则抛物线开口朝上；若a<0，则抛物线开口朝下。
⑶顶点： ；
⑷
若Δ>0，则函数图像与x轴交于两点：
和 ；
若Δ=0，则函数图像与x轴交于一点：
若Δ
②顶点式： 此时顶点为（h,k)
时，对应顶点为 ，其中, ；
③交点式：
函数图像与x轴交于 和 两点。
表达式
顶点式
y=a(x-h)²+k(a≠0,a、h、k为常数),顶点坐标为（h,k)，对称轴为直线x=h，顶点的位置特征和图像的开口方向与函数y=ax²的图像相同，当x=h时，y最大（小）值=k.有时题目会指出让你用配方法把一般式化成顶点式。
例：已知二次函数y的顶点(1,2)和另一任意点(3,10)，求y的解析式。
解：设y=a(x-1)²+2，把(3,10)代入上式，解得y=2(x-1)²+2。
注意：与点在平面直角坐标系中的平移不同，二次函数平移后的顶点式中，h>0时，h越大，图像的对称轴离y轴越远，且在x轴正方向上，不能因h前是负号就简单地认为是向左平移。
具体可分为下面几种情况：
当h>0时，y=a(x-h)²的图像可由抛物线y=ax²向右平行移动h个单位得到；
当h>0时，y=a(x+h)²的图像可由抛物线y=ax²向左平行移动h个单位得到；
当h>0,k>0时，将抛物线y=ax²向右平行移动h个单位，再向上移动k个单位，就可以得到y=a(x-h)²+k的图像；
当h>0,k>0时，将抛物线y=ax²向左平行移动h个单位，再向下移动k个单位，就可以得到y=a(x+h)²-k的图像；
当h＜0,k>0时，将抛物线y=ax²向左平行移动|h|个单位，再向上移动k个单位可得到y=a(x-h)²+k的图像；
当h＜0,k＜0时，将抛物线y=ax²向左平行移动|h|个单位，再向下移动|k|个单位可得到y=a(x-h)²+k的图像。4
交点式
[仅限于与x轴即y=0有交点时的抛物线，即b2-4ac≥0] .
已知抛物线与x轴即y=0有交点A(x1, 0)和B(x2, 0),我们可设 ,然后把第三点代入x、y中便可求出a。
由一般式变为交点式的步骤： (韦达定理)
重要概念：a，b，c为常数，a≠0，且a决定函数的开口方向。a>0时，开口方向向上；a绝对值可以决定开口大小。a的绝对值越大开口就越小，a的绝对值越小开口就越大。
f(x)=f[x0]+fx0,x1+fx0,x1,x2(x-x1)+...fx0,...xn...(x-xn-1)+Rn(x)由此可引导出交点式的系数 (y为截距) 二次函数表达式的右边通常为二次三项式。
欧拉交点式：
若ax²+bx+c=0有两个实根x1,x2，则 此抛物线的对称轴为直线 。
三点式
方法1：
已知二次函数上三个点，(x1, y1)、(x2, y2)、(x3, y3)。把三个点分别代入函数解析式y=a(x-h)²+k(a≠0,a、h、k为常数)，有：
得出一个三元一次方程组，就能解出a、b、c的值。
方法2：
已知二次函数上三个点，(x1, y1)、(x2, y2)、(x3, y3)
利用拉格朗日插值法，可以求出该二次函数的解析式为：
与X轴交点的情况:
当 时，函数图像与x轴有两个交点，分别是(x1, 0)和(x2, 0)。
当 时，函数图像与x轴只有一个切点，即 。
当 时，抛物线与x轴没有公共交点。x的取值范围是虚数（ ）
函数图像
基本图像
在平面直角坐标系中作出二次函数y=ax2+bx+c的图像，可以看出，在没有特定定义域的二次函数图像是一条永无止境的抛物线。 如果所画图形准确无误，那么二次函数图像将是由
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Expression for mean free path: We know from postulates of a kinetic theory that the molecules of a gas are in random motion and they collide with each other. Between two successive collisions, a molecule moves along a straight path with uniform velocity. This path is called mean free path.
Consider a system of molecules each with diameter d. Let n be the number of molecules per unit volume.
Assume that only one molecule is in motion and all others are at rest as shown in the Figure.
Mean free path
If a molecule moves with average speed v in a time t, the distance travelled is vt. In this time t, consider the molecule to move in an imaginary cylinder of volume πd2vt. It collides with any molecule whose center is within this cylinder. Therefore, the number of collisions is equal to the number of molecules in the volume of the imaginary cylinder. It is equal to πd2vtn. The total path length divided by the number of collisions in time t is the mean free path.
Mean free path, λ = "distance travelled"/"Number of collisions"
λ = "vt"/("n"π"d"^2"vt") = 1/("n"π"d"^2) ........(1)
Though we have assumed that only one molecule is moving at a time and other molecules are at rest, in actual practice all the molecules are in random motion. So the average relative speed of one molecule with respect to other molecules has to be taken into account. After some detailed calculations the correct expression for mean free path
∴ λ = 1/(sqrt2"n"π"d"^2) .........(2)
Equation (2) implies that the mean free path is inversely proportional to number density. When the number density increases the molecular collisions increases so it decreases the distance travelled by the molecule before collisions.
Case 1: Rearranging the equation (2) using ‘m’ (mass of the molecule)
∴ λ = "m"/(sqrt2π"d"^2"mn")
But mn = mass per unit volume = ρ (density of the gas)
∴ λ = "m"/(sqrt2π"d"^2ρ) ..........(3)
Also we know that PV = NkT
P = "N"/"V""kT" = nkT ⇒ n = "P"/"kT"
Substituting n = "P"/"kT" in equation (2), we get
λ = "kT"/(sqrt2π"d"^2"P") ........(4)
The equation (4) implies the following:
(i) Mean free path increases with increasing temperature. As the temperature increases, the average speed of each molecule will increase. It is the reason why the smell of hot sizzling food reaches several meters away from the smell of cold food.
(ii) Mean free path increases with decreasing pressure of the gas and diameter of the gas molecules.
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Mean Free Path
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Q III. 7.Q III. 6.Q III. 8.
Chapter 9: Kinetic Theory of Gases - Evaluation [Page 185]
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Samacheer Kalvi Physics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Kinetic Theory of GasesEvaluation | Q III. 7. | Page 185
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190143 | https://engineering.stackexchange.com/questions/63139/how-to-calculate-the-forces-on-a-steel-cable | structural engineering - How to calculate the forces on a steel cable - Engineering Stack Exchange
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How to calculate the forces on a steel cable
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A steel cable is mounted to two rigid supports, 12 meters apart. One support is 5 meters higher than the other, so the cable is inclined about 22.6 degrees from horizontal. Let's call that angle θ θ and say the cable is tensioned with a force of F T F T.
Let's assume the weight of the cable is negligible compared to the rest of the forces involved. So far, the forces on each of the two cable supports should be F x=F T c o s(θ)/2 F x=F T c o s(θ)/2 and F y=F T s i n(θ)/2 F y=F T s i n(θ)/2, with only the directions changing, correct?
If a load is fixed to an arbitrary point along the cable, let's say 1/3 of the way (4 meters horizontal distance from the leftmost support), how can I calculate the way the load force transfers onto the two support points?
Going further, in order to estimate the expected amount of cable sag for a given load, what parameters do I need to know for the cable I'm using, and how would this be calculated?
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asked May 16 at 22:33
jamesjames
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1 The forces on each support are incorrect. The pre tension F T F T is developed in the entire cable.Eric –Eric 2025-05-16 23:44:46 +00:00 Commented May 16 at 23:44
1 Furthermore, the last two questions are really apart of the same question. The deflection becomes the key to determine the tensions,. The force distribution is not independent of the deflection. It doesn't seem trivial to me as a physics problem( I'm inclined to use spring models over a continuous beam model ). I'll try to think about it, but someone probably has experience with it.Eric –Eric 2025-05-17 00:09:01 +00:00 Commented May 17 at 0:09
1 I want to add that also the deflection is dependent on the pre-tension (as far as I can see). It seems quite coupled to me.Eric –Eric 2025-05-17 00:18:24 +00:00 Commented May 17 at 0:18
2 The parameters you need are E and A (but beware that EA for a cable is not the same as for the material itself). The forces are solved using a force triangle at the point of application of the load, and will be iterative, that is as the cable deflects the geometry will change and hence the force triangle will change. If you don't understand this you need a textbook, not a textbox.Greg Locock –Greg Locock 2025-05-17 00:44:14 +00:00 Commented May 17 at 0:44
Hi Greg, thanks for your comment. I am asking here to get a starting point for understanding and further investigation. If this site were to be only for professional engineers, there wouldn't be much point, would there? Anyway, if you have a textbook to recommend I will happily take a look at it.james –james 2025-05-17 15:37:37 +00:00 Commented May 17 at 15:37
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This is a nontrivial question.
Your assumption for T x=1/2 F t c o s α T x=1/2 F t c o s α is not right. Both supports have F x=F t c o s(α)F x=F t c o s(α)
Steps for a Calculation (Elastic Cable):
Define Knowns:
Geometry: L=12 L=12 m, H=5 H=5 m, L 1=4 L 1=4 m.
Load: W W.
Cable properties: E E, A A.
Initial condition: Initial tension F T F T (which gives an initial strain and thus an initial unstretched length S u n s t r e t c h e d=S c h o r d/(1+F T/(E A))S u n s t r e t c h e d=S c h o r d/(1+F T/(E A)) where S c h o r d=13 S c h o r d=13 m) $.
The Unknown: The vertical position of the load point, y P y P (or the sag s=(L 1 tan θ)−y P s=(L 1 tanθ)−y P).
L segments:
Geometric lengths of segments as a function of y P y P: S 1(y P)=L 2 1+y 2 P−−−−−−−√S 1(y P)=L 1 2+y P 2 S 2(y P)=(L−L 1)2+(H−y P)2−−−−−−−−−−−−−−−−−−√S 2(y P)=(L−L 1)2+(H−y P)2
Equilibrium equations at the load point P (solving for T 1 T 1 and T 2 T 2 in terms of W W and y P y P): T 1(y P)L 1 S 1(y P)=T 2(y P)L−L 1 S 2(y P)T 1(y P)L 1 S 1(y P)=T 2(y P)L−L 1 S 2(y P)T 1(y P)y P S 1(y P)+T 2(y P)H−y P S 2(y P)=W T 1(y P)y P S 1(y P)+T 2(y P)H−y P S 2(y P)=W (Solve these for T 1(y P)T 1(y P) and T 2(y P)T 2(y P)).
Elastic compatibility equation (relating sagged lengths, tensions, and unstretched length): S u n s t r e t c h e d=S 1(y P)1+T 1(y P)E A+S 2(y P)1+T 2(y P)E A S u n s t r e t c h e d=S 1(y P)1+T 1(y P)E A+S 2(y P)1+T 2(y P)E A Or, if working with elongations: S 1(y P)+S 2(y P)=S u n s t r e t c h e d(1+some effective average tension E A)S 1(y P)+S 2(y P)=S u n s t r e t c h e d(1+some effective average tension E A) -- this is less direct. The compatibility equation S 1(y P)(1−T 1(y P)E A)+S 2(y P)(1−T 2(y P)E A)=S u n s t r e t c h e d S 1(y P)(1−T 1(y P)E A)+S 2(y P)(1−T 2(y P)E A)=S u n s t r e t c h e d (if S 1,S 2 S 1,S 2 are current lengths and S u n s t r e t c h e d S u n s t r e t c h e d is the total) is also an option, or more accurately: S u n s t r e t c h e d,s e g m e n t 1+S u n s t r e t c h e d,s e g m e n t 2=S u n s t r e t c h e d,t o t a l S u n s t r e t c h e d,s e g m e n t 1+S u n s t r e t c h e d,s e g m e n t 2=S u n s t r e t c h e d,t o t a l S 1(y P)=S u n s t r e t c h e d,s e g m e n t 1(1+T 1(y P)/E A)S 1(y P)=S u n s t r e t c h e d,s e g m e n t 1(1+T 1(y P)/E A)S 2(y P)=S u n s t r e t c h e d,s e g m e n t 2(1+T 2(y P)/E A)S 2(y P)=S u n s t r e t c h e d,s e g m e n t 2(1+T 2(y P)/E A)
The partitioning of S u n s t r e t c h e d,t o t a l S u n s t r e t c h e d,t o t a l into S u n s t r e t c h e d,s e g m e n t 1 S u n s t r e t c h e d,s e g m e n t 1 and S u n s t r e t c h e d,s e g m e n t 2 S u n s t r e t c h e d,s e g m e n t 2 is also initially unknown and depends on the final configuration.
The most straightforward compatibility equation is usually the current total stretched length S 1(y P)+S 2(y P),S 1(y P)+S 2(y P),is the result of stretching the original total unstretched length S u n s t r e t c h e d S u n s t r e t c h e d by the work done by the tensions.
It is often written as:The unstretched length S u n s t r e t c h e d S u n s t r e t c h e d must equal the sum of the "relaxed" lengths of the two segments:
S u n s t r e t c h e d=S 1(y P)−T 1(y P)S 1(y P)E A+S 2(y P)−T 2(y P)S 2(y P)E A S u n s t r e t c h e d=S 1(y P)−T 1(y P)S 1(y P)E A+S 2(y P)−T 2(y P)S 2(y P)E A
This simplifies to:S u n s t r e t c h e d=S 1(y P)(1−T 1(y P)E A)+S 2(y P)(1−T 2(y P)E A)S u n s t r e t c h e d=S 1(y P)(1−T 1(y P)E A)+S 2(y P)(1−T 2(y P)E A)
Correction to the above compatibility equation based on standard definition (L=L 0(1+T/E A)L=L 0(1+T/E A)):
The initial unstretched length is S 0 S 0. The current length S i S i of a segment under tension T i T i is S i=S 0,i(1+T i/(E A))S i=S 0,i(1+T i/(E A)). So S 0,i=S i/(1+T i/(E A))S 0,i=S i/(1+T i/(E A)).
Therefore, the total initial unstretched length S u n s t r e t c h e d S u n s t r e t c h e d is: S u n s t r e t c h e d=S 1(y P)1+T 1(y P)E A+S 2(y P)1+T 2(y P)E A S u n s t r e t c h e d=S 1(y P)1+T 1(y P)E A+S 2(y P)1+T 2(y P)E A This is the equation you need to solve for y P y P.
Solve for y P y P: This equation is non-linear and typically requires numerical methods.
Calculate Sag: s=(L 1 tan θ)−y P=(4⋅5/12)−y P=(5/3)−y P s=(L 1 tanθ)−y P=(4⋅5/12)−y P=(5/3)−y P.
Calculate Final Tensions and Support Reactions: Once y P y P is known, find T 1 T 1 and T 2 T 2 from the equilibrium equations, and then the support reactions.
This is a non-trivial problem, usually solved by structural analysis programs.
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answered May 17 at 2:52
kamrankamran
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I very much appreciate your time and effort in explaining this. I think the cable elongation can be considered negligible, which helps to simplify things a bit. I will have to revisit and study this more later. For now though, a simple question. I think of F T F T as the force that I would apply to tension the cable. Let's say I apply 1 kN of force from one end of the cable using a tensioner/turnbuckle. Would the force I applied not be evenly divided between the two support points, so 500 N per support point? I am asking to correct any misunderstandings I might have here.james –james 2025-05-17 15:45:40 +00:00 Commented May 17 at 15:45
1 @james, no, imagine the cable is is being tensioned by attaching it at one end to a spring. If you draw the Freebody diagram for each end you find out the tension in the cable is equal to reaction at each support. Same as the game of tug of war: the tension in the rope is equal to force at each end. BTW if we assume there is no elongation, we can immediately get the reaction treating the cable as a beam.kamran –kamran 2025-05-17 19:45:01 +00:00 Commented May 17 at 19:45
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190144 | https://math.libretexts.org/Courses/Northeast_Wisconsin_Technical_College/College_Algebra_(NWTC)/05%3A_Exponential_and_Logarithmic_Functions/5.03%3A_Exponential_Equations_and_Inequalities | 5.3: Exponential Equations and Inequalities - Mathematics LibreTexts
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5: Exponential and Logarithmic Functions
College Algebra (NWTC)
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Tue, 09 May 2023 23:46:51 GMT
5.3: Exponential Equations and Inequalities
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Joshua Halpern
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5. 5: Exponential and Logarithmic Functions
6. 5.3: Exponential Equations and Inequalities
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5.3: Exponential Equations and Inequalities
Last updated May 9, 2023
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5.2: Properties of Logarithms
5.4: Logarithmic Equations and Inequalities
Page ID 127040
Carl Stitz & Jeff Zeager
Lakeland Community College & Lorain County Community College
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Example 6.3.1.
2. Example 6.3.2
3. Example 6.3.3
4. Example 6.3.4
5. 6.3.1. Exercises
6. 6.3.2. Answers
7. Reference
In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2 x=128. After a moment’s calculation, we find 128=2 7, so we have 2 x=2 7. The one-to-one property of exponential functions, detailed in Theorem 6.4, tells us that 2 x=2 7 if and only if x=7. This means that not only is x=7 a solution to 2 x=2 7, it is the only solution. Now suppose we change the problem ever so slightly to 2 x=129. We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to write 129=2 log 2(129). We’d then have 2 x=2 log 2(129), which means our solution is x=log 2(129). This makes sense because, after all, the definition of log 2(129) is ‘the exponent we put on 2 to get 129.’ Indeed we could have obtained this solution directly by rewriting the equation 2 x=129 in its logarithmic form log 2(129)=x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log 2(129)=ln(129)ln(2). Another way to arrive at this answer is as follows
2 x=129 ln(2 x)=ln(129)Take the natural log of both sides.xln(2)=ln(129)Power Rule x=ln(129)ln(2)
‘Taking the natural log’ of both sides is akin to squaring both sides: since f(x)=ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable x. We summarize below the two common ways to solve exponential equations, motivated by our examples.
Steps for Solving an Equation involving Exponential Functions
Isolate the exponential function.
If convenient, express both sides with a common base and equate the exponents.
Otherwise, take the natural log of both sides of the equation and use the Power Rule.
Example 6.3.1.
Solve the following equations. Check your answer graphically using a calculator.
2 3x=16 1−x
2000=1000⋅3−0.1t
9⋅3 x=7 2x
75=100 1+3e−2t
25 x=5 x+6
e x−e−x 2=5
Solution.
Since 16 is a power of 2, we can rewrite 2 3x=16 1−x as 2 3x=(2 4)1−x. Using properties of exponents, we get 2 3x=2 4(1−x). Using the one-to-one property of exponential functions, we get 3x=4(1−x) which gives x=4 7. To check graphically, we set f(x)=2 3x and g(x)=16 1−x and see that they intersect at x=4 7≈0.5714.
We begin solving 2000=1000⋅3−0.1t by dividing both sides by 1000 to isolate the exponential which yields 3−0.1t=2. Since it is inconvenient to write 2 as a power of 3, we use the natural log to get ln(3−0.1t)=ln(2). Using the Power Rule, we get −0.1tln(3)=ln(2), so we divide both sides by −0.1ln(3) to get t=−ln(2)0.1ln(3)=−10ln(2)ln(3). On the calculator, we graph f(x)=2000 and g(x)=1000⋅3−0.1x and find that they intersect at x=−10ln(2)ln(3)≈−6.3093.
We first note that we can rewrite the equation 9⋅3 x=7 2x as 3 2⋅3 x=7 2x to obtain 3 x+2=7 2x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use the natural log: ln(3 x+2)=ln(7 2x). The power rule gives (x+2)ln(3)=2xln(7). Even though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just constants. The equation (x+2)ln(3)=2xln(7) is actually a linear equation and as such we gather all of the terms with x on one side, and the constants on the other. We then divide both sides by the coefficient of x, which we obtain by factoring.
(x+2)ln(3)=2xln(7)xln(3)+2ln(3)=2xln(7)2ln(3)=2xln(7)−xln(3)2ln(3)=x(2ln(7)−ln(3))Factor.x=2ln(3)2ln(7)−ln(3)
Graphing f(x)=9⋅3 x and g(x)=7 2x on the calculator, we see that these two graphs intersect at x=2ln(3)2ln(7)−ln(3)≈0.7866.
Our objective in solving 75=100 1+3e−2t is to first isolate the exponential. To that end, we clear denominators and get 75(1+3e−2t)=100. From this we get 75+225e−2t=100, which leads to 225e−2t=25, and finally, e−2t=1 9. Taking the natural log of both sides gives ln(e−2t)=ln(1 9). Since natural log is log base e, ln(e−2t)=−2t. We can also use the Power Rule to write ln(1 9)=−ln(9). Putting these two steps together, we simplify ln(e−2t)=ln(1 9) to −2t=−ln(9). We arrive at our solution, t=ln(9)2 which simplifies to t=ln(3). (Can you explain why?) The calculator confirms the graphs of f(x)=75 and g(x)=100 1+3e−2x intersect at x=ln(3)≈1.099.
We start solving 25 x=5 x+6 by rewriting 25=5 2 so that we have (5 2)x=5 x+6, or 5 2x=5 x+6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u=5 x, we have u 2=(5 x)2=5 2x so the equation 5 2x=5 x+6 becomes u 2=u+6. Solving this as u 2−u−6=0 gives u=−2 or u=3. Since u=5 x, we have 5 x=−2 or 5 x=3. Since 5 x=−2 has no real solution, (Why not?) we focus on 5 x=3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln(5 x)=ln(3) so that xln(5)=ln(3) or x=ln(3)ln(5). On the calculator, we see the graphs of f(x)=25 x and g(x)=5 x+6 intersect at x=ln(3)ln(5)≈0.6826.
At first, it’s unclear how to proceed with e x−e−x 2=5, besides clearing the denominator to obtain e x−e−x=10. Of course, if we rewrite e−x=1 e x, we see we have another denominator lurking in the problem: e x−1 e x=10. Clearing this denominator gives us e 2x−1=10e x, and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a ‘quadratic in disguise.’ If we let u=e x, then u 2=e 2x so the equation e 2x−1=10e x can be viewed as u 2−1=10u. Solving u 2−10u−1=0, we obtain by the quadratic formula u=5±26. From this, we have e x=5±26. Since 5−26<0, we get no real solution to e x=5−26, but for e x=5+26, we take natural logs to obtain x=ln(5+26). If we graph f(x)=e x−e−x 2 and g(x)=5, we see that the graphs intersect at x=ln(5+26)≈2.312
The authors would be remiss not to mention that Example 6.3.1 still holds great educational value. Much can be learned about logarithms and exponentials by verifying the solutions obtained in Example 6.3.1 analytically. For example, to verify our solution to 2000=1000⋅3−0.1t, we substitute t=−10ln(2)ln(3) and obtain
2000=?1000⋅3−0.1(−10ln(2)ln(3))2000=?1000⋅3 ln(2)ln(3)2000=?1000⋅3 log 3(2)Change of Base 2000=?1000⋅2 Inverse Property 2000=✓2000
The other solutions can be verified by using a combination of log and inverse properties. Some fall out quite quickly, while others are more involved. We leave them to the reader.
Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1 applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using sign diagrams as demonstrated below.
Example 6.3.2
Solve the following inequalities. Check your answer graphically using a calculator.
2 x 2−3x−16≥0
e x e x−4≤3
xe 2x<4x
Solution.
Since we already have 0 on one side of the inequality, we set r(x)=2 x 2−3x−16. The domain of r is all real numbers, so in order to construct our sign diagram, we need to find the zeros of r. Setting r(x)=0 gives 2 x 2−3x−16=0 or 2 x 2−3x=16. Since 16=2 4 we have 2 x 2−3x=2 4, so by the one-to-one property of exponential functions, x 2−3x=4. Solving x 2−3x−4=0 gives x=4 and x=−1. From the sign diagram, we see r(x)≥0 on (−∞,−1]∪[4,∞), which corresponds to where the graph of y=r(x)=2 x 2−3x−16, is on or above the x-axis.
The first step we need to take to solve e x e x−4≤3 is to get 0 on one side of the inequality. To that end, we subtract 3 from both sides and get a common denominator
e x e x−4≤3 e x e x−4−3≤0 e x e x−4−3(e x−4)e x−4≤0 Common denomintors.12−2e x e x−4≤0
We set r(x)=12−2e x e x−4 and we note that r is undefined when its denominator e x−4=0, or when e x=4. Solving this gives x=ln(4), so the domain of r is (−∞,ln(4))∪(ln(4),∞). To find the zeros of r, we solve r(x)=0 and obtain 12−2e x=0. Solving for e x, we find e x=6, or x=ln(6). When we build our sign diagram, finding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4 which means ln(3)<ln(4)<ln(5)<ln(6)<ln(7). While the prospect of determining the sign of r(ln(3)) may be very unsettling, remember that e ln(3)=3, so r(ln(3))=12−2e ln(3)e ln(3)−4=12−2(3)3−4=−6 We determine the signs of r(ln(5)) and r(ln(7)) similarly.5 From the sign diagram, we find our answer to be (−∞,ln(4))∪[ln(6),∞). Using the calculator, we see the graph of f(x)=e x e x−4 is below the graph of g(x)=3 on (−∞,ln(4))∪(ln(6),∞), and they intersect at x=ln(6)≈1.792.
As before, we start solving xe 2x<4x by getting 0 on one side of the inequality, xe 2x−4x<0. We set r(x)=xe 2x−4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x)=0 produces xe 2x−4x=0. We factor to get x(e 2x−4)=0 which gives x=0 or e 2x−4=0. To solve the latter, we isolate the exponential and take logs to get 2x=ln(4), or x=ln(4)2=ln(2). (Can you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1)=0, we choose ln(1 2), ln(3 2) and ln(3). Evaluating,6 we get
r(ln(1 2))=ln(1 2)e 2ln(1 2)−4ln(1 2)=ln(1 2)e ln(1 2)2−4ln(1 2)Power Rule=ln(1 2)e ln(1 4)−4ln(1 2)=1 4ln(1 2)−4ln(1 2)=−15 4ln(1 2)
Since 1 2<1, ln(1 2)<0 and we get r(ln(1 2)) is (+), so r(x)<0 on (0,ln(2)). The calculator confirms that the graph of f(x)=xe 2x is below the graph of g(x)=4x on these intervals.7
Example 6.3.3
Recall from Example 6.1.2 that the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T(t)=70+90e−0.1t. When will the coffee be warmer than 100∘F?
Solution
We need to find when T(t)>100, or in other words, we need to solve the inequality 70+90e−0.1t>100. Getting 0 on one side of the inequality, we have 90e−0.1t−30>0, and we set r(t)=90e−0.1t−30. The domain of r is artificially restricted due to the context of the problem to [0,∞), so we proceed to find the zeros of r. Solving 90e−0.1t−30=0 results in e−0.1t=1 3 so that t=−10ln(1 3) which, after a quick application of the Power Rule leaves us with t=10ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1<3, 0=ln(1)0. So we choose t=0 as a test value in [0,10ln(3)). Since 3<4, 10ln(3)<10ln(4), so the latter is our choice of a test value for the interval (10ln(3),∞). Our sign diagram is below, and next to it is our graph of y=T(t) from Example 6.1.2 with the horizontal line y=100.
In order to interpret what this means in the context of the real world, we need a reasonable approximation of the number 10ln(3)≈10.986. This means it takes approximately 11 minutes for the coffee to cool to 100∘F. Until then, the coffee is warmer than that.8
We close this section by finding the inverse of a function which is a composition of a rational function with an exponential function.
Example 6.3.4
The function f(x)=5e x e x+1 is one-to-one. Find a formula for f−1(x) and check your answer graphically using your calculator.
Solution
We start by writing y=f(x), and interchange the roles of x and y. To solve for y, we first clear denominators and then isolate the exponential function.
y=5e x e x+1[12pt]x=5e y e y+1 Switch x and y[12pt]x(e y+1)=5e y xe y+x=5e y x=5e y−xe y x=e y(5−x)e y=x 5−x ln(e y)=ln(x 5−x)y=ln(x 5−x)
We claim f−1(x)=ln(x 5−x). To verify this analytically, we would need to verify the compositions (f−1∘f)(x)=x for all x in the domain of f and that (f∘f−1)(x)=x for all x in the domain of f−1. We leave this to the reader. To verify our solution graphically, we graph y=f(x)=5e x e x+1 and y=g(x)=ln(x 5−x) on the same set of axes and observe the symmetry about the line y=x. Note the domain of f is the range of g and vice-versa.
6.3.1. Exercises
In Exercises 1 - 33, solve the equation analytically.
2 4x=8[expeqnfirst]
3(x−1)=27
5 2x−1=125
4 2x=1 2
8 x=1 128
2(x 3−x)=1
3 7x=81 4−2x
9⋅3 7x=(1 9)2x
3 2x=5
5−x=2
5 x=−2
3(x−1)=29
(1.005)12x=3
e−5730k=1 2
2000e 0.1t=4000
500(1−e 2x)=250
70+90e−0.1t=75
30−6e−0.1x=20
100e x e x+2=50
5000 1+2e−3t=2500
150 1+29e−0.8t=75
25(4 5)x=10
e 2x=2e x
7e 2x=28e−6x
3(x−1)=2 x
3(x−1)=(1 2)(x+5)
7 3+7x=3 4−2x
e 2x−3e x−10=0
e 2x=e x+6
4 x+2 x=12
e x−3e−x=2
e x+15e−x=8
3 x+25⋅3−x=10[expeqnlast]
In Exercises 34 - 39, solve the inequality analytically.
e x>53[expineqfirst]
1000(1.005)12t≥3000
2(x 3−x)<1
25(4 5)x≥10
150 1+29e−0.8t≤130
150 1+29 e−0.8 t 70+90e−0.1t≤75[expineqlast]
In Exercises 40 - 45, use your calculator to help you solve the equation or inequality.
2 x=x 2[calcexpineqfirst]
e x=ln(x)+5
e x=x+1
e−x−xe−x≥0
3(x−1)<2 x
e x<x 3−x[calcexpineqlast]
[onetoonelogexercise] Since f(x)=ln(x) is a strictly increasing function, if 0<a<b then ln(a)6 without a sign diagram. Use this technique to solve the inequalities in Exercises 34 - 39. (NOTE: Isolate the exponential function first!)
[hyperbolicsine] Compute the inverse of f(x)=e x−e−x 2. State the domain and range of both f and f−1.
In Example 6.3.4, we found that the inverse of f(x)=5e x e x+1 was f−1(x)=ln(x 5−x) but we left a few loose ends for you to tie up.
Show that (f−1∘f)(x)=x for all x in the domain of f and that (f∘f−1)(x)=x for all x in the domain of f−1.
Find the range of f by finding the domain of f−1.
Let g(x)=5x x+1 and h(x)=e x. Show that f=g∘h and that (g∘h)−1=h−1∘g−1. (We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a specific example of the property.)
With the help of your classmates, solve the inequality e x>x n for a variety of natural numbers n. What might you conjecture about the “speed” at which f(x)=e x grows versus any polynomial?
6.3.2. Answers
x=3 4
x=4
x=2
x=−1 4
x=−7 3
x=−1,0,1
x=16 15
x=−2 11
x=ln(5)2ln(3)
x=−ln(2)ln(5)
No solution.
x=ln(29)+ln(3)ln(3)
x=ln(3)12ln(1.005)
k=ln(1 2)−5730=ln(2)5730
t=ln(2)0.1=10ln(2)
x=1 2ln(1 2)=−1 2ln(2)
t=ln(1 18)−0.1=10ln(18)
x=−10ln(5 3)=10ln(3 5)
x=ln(2)
t=1 3ln(2)
t=ln(1 29)−0.8=5 4ln(29)
x=ln(2 5)ln(4 5)=ln(2)−ln(5)ln(4)−ln(5)
x=ln(2)
x=−1 8ln(1 4)=1 4ln(2)
x=ln(3)ln(3)−ln(2)
x=ln(3)+5ln(1 2)ln(3)−ln(1 2)=ln(3)−5ln(2)ln(3)+ln(2)
x=4ln(3)−3ln(7)7ln(7)+2ln(3)
x=ln(5)
x=ln(3)
x=ln(3)ln(2)
x=ln(3)
x=ln(3), ln(5)
x=ln(5)ln(3)
(ln(53),∞)
[ln(3)12ln(1.005),∞)
(−∞,−1)∪(0,1)
(−∞,ln(2 5)ln(4 5)]=(−∞,ln(2)−ln(5)ln(4)−ln(5)]
(−∞,ln(2 377)−0.8]=(−∞,5 4ln(377 2)]
[ln(1 18)−0.1,∞)=[10ln(18),∞)
x≈−0.76666,x=2,x=4
x≈0.01866,x≈1.7115
x=0
(−∞,1]
≈(−∞,2.7095)
≈(2.3217,4.3717)
x>1 3(ln(6)+1)
f−1=ln(x+x 2+1). Both f and f−1 have domain (−∞,∞) and range (−∞,∞).
Reference
1 You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most ‘mathy’ of the logarithms.)
2 This is also the ‘if’ part of the statement log b(u)=log b(w) if and only if u=w in Theorem 6.4.
3 Please resist the temptation to divide both sides by ‘ln’ instead of ln(2). Just like it wouldn’t make sense to divide both sides by the square root symbol ‘′ when solving x2=5, it makes no sense to divide by ‘ln’.
4 This is because the base of ln(x) is e>1. If the base b were in the interval 0<b<1, then log b(x) would decreasing.
5 We could, of course, use the calculator, but what fun would that be?
6 A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method.
7 Note: ln(2)≈0.693.
8 Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book
This page titled 5.3: Exponential Equations and Inequalities is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Carl Stitz & Jeff Zeager via source content that was edited to the style and standards of the LibreTexts platform.
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190145 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8773483/ | A Rapid Method for Postmortem Vitreous Chemistry—Deadside Analysis - PMC
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A Rapid Method for Postmortem Vitreous Chemistry—Deadside Analysis
Brita Zilg
Brita Zilg
1 Forensic Research Laboratory, Department of Oncology-Pathology, Karolinska Institute, 171 77 Stockholm, Sweden; brita.zilg@ki.se (B.Z.); kanar.alkass@ki.se (K.A.)
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2 Department of Forensic Genetics and Forensic Toxicology, National Board of Forensic Medicine, 587 58 Linkoping, Sweden; robert.kronstrand@rmv.se
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3 Division of Clinical Chemistry and Pharmacology, Department of Biomedical and Clinical Science, Faculty of Medicine and Health Science, Linköping University, 581 85 Linkoping, Sweden; soren.berg@liu.se
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1 Forensic Research Laboratory, Department of Oncology-Pathology, Karolinska Institute, 171 77 Stockholm, Sweden; brita.zilg@ki.se (B.Z.); kanar.alkass@ki.se (K.A.)
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1 Forensic Research Laboratory, Department of Oncology-Pathology, Karolinska Institute, 171 77 Stockholm, Sweden; brita.zilg@ki.se (B.Z.); kanar.alkass@ki.se (K.A.)
2 Department of Forensic Genetics and Forensic Toxicology, National Board of Forensic Medicine, 587 58 Linkoping, Sweden; robert.kronstrand@rmv.se
3 Division of Clinical Chemistry and Pharmacology, Department of Biomedical and Clinical Science, Faculty of Medicine and Health Science, Linköping University, 581 85 Linkoping, Sweden; soren.berg@liu.se
Correspondence: henrik.druid@ki.se; Tel.: +46-70-602-7141
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Nina Heldring: Academic Editor
Brita Zilg: Academic Editor
Received 2021 Dec 1; Accepted 2021 Dec 23; Collection date 2022 Jan.
© 2021 by the authors.
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PMCID: PMC8773483 PMID: 35053180
Abstract
Vitreous fluid is commonly collected for toxicological analysis during forensic postmortem investigations. Vitreous fluid is also often analyzed for potassium, sodium, chloride and glucose for estimation of time since death, and for the evaluation of electrolyte imbalances and hyperglycemia, respectively. Obtaining such results in the early phase of a death investigation is desirable both in regard to assisting the police and in the decision-making prior to the autopsy. We analyzed vitreous fluid with blood gas instruments to evaluate/examine the possible impact of different sampling and pre-analytical treatment. We found that samples from the right and left eye, the center of the eye as well as whole vitreous samples gave similar results. We also found imprecision to be very low and that centrifugation and dilution were not necessary when analyzing vitreous samples with blood gas instruments. Similar results were obtained when analyzing the same samples with a regular multi-analysis instrument, but we found that such instruments could require dilution of samples with high viscosity, and that such dilution might impact measurement accuracy. In conclusion, using a blood gas instrument, the analysis of postmortem vitreous fluid for electrolytes and glucose without sample pretreatment produces rapid and reliable results.
Keywords: vitreous, postmortem, glucose, electrolytes, forensic medicine
1. Introduction
Autopsy has long been considered the gold standard in reaching a diagnosis when a person has died of uncertain causes. While this is still true regarding a large number of illnesses that are visible macroscopically or microscopically, there are many serious medical conditions that may escape detection. The major drawback is that autopsy diagnostics are traditionally based on morphology. Although computer tomography and magnetic resonance imaging has been introduced in the routine casework at many forensic medicine facilities in the last few decades, these radiological methods can also only provide structural information. Forensic toxicology is the only exception from the tradition of morphological diagnostics, which allows for the detection and quantification of alcohol and drugs by means of chemical analysis. With the help of postmortem reference concentrations, the pathologist may be able to diagnose, or rule out, an intoxication as the cause of death . Toxicology was most likely introduced because intoxication does not usually cause any visible morphological signs/traces at autopsy.
Chemical analyses of postmortem samples today are not limited to toxicology; analysis of endogenous biomolecules in postmortem samples can also be used to identify pathologies. For instance, an increase in glial fibrillary acidic protein or neurofilament light protein in cerebrospinal fluid or serum can indicate brain injury and increased troponin T may be used as an indicator of myocardial infarction. Negative results can be equally important in the evaluation of the possible causes of death in the early stages of the investigation. Pioneering work in this field was performed by Coe [4,5] and over several decades, postmortem biochemistry has become increasingly appreciated as an asset in forensic pathology casework . Most research in this field has explored the changes in various analytes in serum, pericardial fluid, cerebrospinal fluid, and vitreous fluid. The problem with postmortem serum is that it usually shows substantial hemolysis, which may interfere with, or even preclude, some analyses. In contrast, vitreous fluid is a transparent fluid with a very low amount of cells, at least during the early postmortem interval. It is also easily accessible before autopsy by a puncture at the lateral aspect of the eyeball. Hyperglycemia can be clearly identified by analyzing vitreous fluid glucose levels [7,8], whilst vitreous potassium levels can provide an estimate of the postmortem interval [9,10,11,12] and levels of sodium and chloride can reveal dehydration or water intoxication [13,14].
In hospitals, rapid analyses often form the basis for treatment measures in an emergency setting, whereas it is generally believed that urgent actions are not needed when a person dies. However, many biochemical changes start to occur during the agonal stages of death and the early postmortem interval. Glucose and oxygen are consumed until the cells die, and many cell functions, including membrane pumps which require energy, will fail as the ATP levels drop, at which stage an uncontrolled diffusion of ions and molecules flow in both directions. Therefore, postmortem biochemistry is easier to interpret if samples are collected as soon as possible after death. Early sampling and analysis are also important for the postmortem workup since results obtained in the early stages can form the basis for important decisions regarding how to proceed with the case examination and any additional sampling and analysis which may be warranted, just like in the emergency room. Even the police may want certain information urgently, such as an estimate of the time of death, and this can be reported quickly based on an analysis of vitreous potassium. We have used blood gas instruments to analyze postmortem vitreous fluid for 20 years and have reported our experience of using such analyses for the diagnosis of hyperglycemia, sodium and chloride imbalances and for the estimation of the postmortem interval by analysis of vitreous potassium [8,11,14]. There are also numerous other publications on these subjects, however the authors have usually sent samples to a clinical chemistry laboratory for analysis. The interpretation of vitreous chemistry results has been a matter of discussion; in particular, concerns have been raised regarding the possible influence of factors such as centrifugation, dilution and storage of the samples [15,16,17,18]. The possible impact of differences in concentrations between the eyes has also been an issue . Even though it has been observed in forensic toxicology that the pre-analytical factors are much more important than the errors introduced during sample extraction and analysis —which is most likely true for most biochemical analysis—it is important to understand the imprecision of point-of-care instruments, such as blood gas instruments. Hence, in this study, we aimed to investigate the reliability of blood gas instruments with regard to the results for potassium, sodium, chloride and glucose, in comparison with results obtained by analysis at external laboratories. We also studied the possible impact of sampling technique and the pretreatment/handling of the samples before analysis.
2. Materials and Methods
2.1. Sample Collection
Our standard procedure for collecting vitreous fluid is to puncture the lateral aspect of the eye and withdraw a 0.25 mL sample from the center of the vitreous compartment, using a 16-gauge needle into a 1 mL or 10 mL syringe. The tip of the needle is then typically visible through the pupil. For the current study, all of the vitreous fluid was collected, unless otherwise specified. The vitreous fluid is viscous, and becomes more liquefied with age . However, even in younger subjects, it is possible to collect virtually all of the vitreous fluid if needed by using the procedure described above. For some of the analyses, the vitreous fluid was injected directly from the syringe by which the sample was collected. For test tube samples, the fluid was aspirated by the instrument.
2.2. Analytical Principles of Blood Gas Instruments
The blood gas instruments used for this study were ABL700, ABL835 and ABL90 flex (all from Radiometer Copenhagen, Brønshøj, Denmark). Each of these instruments is able to determine levels of Na+, Cl−, K+, Ca++, glucose, lactate, pH, pCO 2, pO 2, total Hb and specific forms of Hb (i.e., COHb, MetHb, O 2 Hb, HHb, HbF and SHb). For this project, the concentrations of potassium, sodium, chloride and glucose were measured. The analysis of these parameters was performed by sensor units with selective permeable membranes. The sensor units employ two different measuring principles, potentiometry for electrolytes and amperometry for glucose and lactate. For electrolyte measurements, the sensing element is an ion-selective sensor with a membrane with a specific ion carrier (potassium and chloride) or an ion-selective ceramic pin (sodium). The concentration of glucose was measured using an amperometric method. The sensor has a silver cathode and a platinum anode in contact with an electrolyte solution.
2.3. Precision Test of the Blood Gas Instrument
Prefabricated test solutions (Radiometer, Copenhagen) of glucose 4.4, 11.1 and 15.7 mmol/L and potassium 4.3 mmol/L were used, and additionally, solutions containing 10 and 30 mmol/L of potassium were prepared. The samples were measured using the ABL90 flex blood gas instrument (Radiometer Copenhagen). Samples of glucose and potassium with three different concentrations were aliquoted and analyzed eight times on one day (within-series imprecision) and one time on eight different days (between-series imprecision).
2.4. Comparison Different Instruments
Whole vitreous samples were collected from 50 consecutive postmortem cases. The samples were centrifuged and then divided into three aliquots. One sample was sent to an external lab (Aleris Medilab, Stockholm, Sweden) and was analyzed with a Beckman Coulter AU5800 instrument. There the samples were diluted 1:2 with water. Samples which had a potassium concentration exceeding the measuring range were diluted 1:4. Another aliquot was sent to a different external lab and analyzed with an inductively coupled plasma-mass spectrometry (ICP-MS) instrument at ALS Scandinavia, Luleå, Sweden (at the time of the study, it was named ALS Global). The third aliquot was analyzed undiluted with two blood gas instruments, ABL700 and ABL90 flex (Radiometer, Copenhagen).
2.5. Influence of Cells and Centrifugation
To investigate if cells present in whole vitreous samples affect the measurements of the electrolyte concentrations, 22 whole vitreous samples were studied. Each sample was vortexed for one minute. Then, 0.2 mL was aspirated and measured without further treatment using an ABL 835 blood gas instrument, and 0.015 mL was taken for cell counting. The remaining fluid was centrifuged for 10 min at 1900× g. The supernatant was collected until 0.2 mL of fluid remained in the tube. The pellet was vortexed again for one minute and then sonicated (Bransonic 12, Branson Ultrasonic Corporation, Danbury, CT, USA) for two hours. The samples from the supernatant and the pellet were also measured using the blood gas instrument. For the staining of cells, 0.015 mL of the sample was briefly vortexed with 0.015 mL of Trypan Blue (0.4%, Sigma Aldrich, Stockholm, Sweden, filtered and diluted 1:10). After 10 min of incubation, the fluid was filled into a Bürker chamber. The cells of several squares (0.1 mm 3 each) were counted. The cell count was calculated using the following formula: counted cells × dilution factor × 10,000/number of C-squares = cells/mL. From this collection of samples, two groups of 6–8 samples with cell numbers >100,000 or <20,000 cells/mL were analyzed with ABL835.
2.6. Comparison of Undiluted and Diluted Samples
Whole vitreous fluid from 29 postmortem cases was sampled with a 16-gauge needle and a 1 mL syringe. A 0.2 mL portion was injected directly into the ABL700 and 0.2 mL was diluted with 0.2 mL of DDH 2 O water and vortexed. The samples were analyzed with an ABL700 blood gas instrument.
2.7. Impact of Hyaluronidase and Hyaluronic Acid
Vitreous fluid was collected from eight postmortem cases. Each sample was vortexed and 1 mL was aliquoted to each of three Eppendorf tubes. The content in one tube was untreated. To the second tube, 100 μL of distilled water was added and 100 μL of hyaluronidase was added to the third tube (Sigma, 50 mg/mL). All samples were well vortexed and incubated for 1 h at room temperature before analysis. To mimic the impact of a high concentration of hyaluronic acid, sodium hyaluronate (Hyalgan, Takeda Pharma, Stockholm, Sweden, 10 mg/mL) was added to duplicates of a prepared NaCl solution at increasing volumes.
2.8. Differences between Whole Vitreous and Sample from the Centre of the Vitreous
Vitreous samples from 27 postmortem cases were collected with a 16-gauge needle and a 10 mL syringe. The whole vitreous was withdrawn from the right eye. To this end, the needle usually had to be moved around to different positions within the vitreous compartment to enable complete sampling, and more than 2 mL could be obtained in most cases. Only 0.25 mL was aspirated from the left eye from the center of the vitreous; the tip of the needle was typically visible through the pupil. Both samples were measured immediately with an ABL 835 blood gas instrument using the 0.095 mL program. In addition to electrolytes and glucose, lactate concentrations were also registered.
2.9. Comparison of Samples Collected from the Left and Right Eye
In 16 postmortem cases, 0.25 mL of vitreous fluid was sampled from the center of each eye with a 16-gauge needle and a 1 mL syringe. Fluid was taken from the right and left eye in separate syringes and analyzed immediately with an ABL700 blood gas instrument.
2.10. Effect of Spiking of Vitreous Samples
In 33 postmortem cases, analysis was directly performed on an ABL700 blood gas instrument, and 0.3 mL of the remaining vitreous sample was transferred to an Eppendorf tube. A solution containing K, Na, Cl and glucose in the concentration of 70, 350, 420 and 160 mmol/L was made, and 30 μL of this solution was added to the aliquots of the vitreous samples. The results were compared with the calculated increase in concentrations of the solutes.
2.11. Effect of Delayed Analysis
Vitreous fluid from 13 cases was sampled with a 16-gauge needle and a 10 mL syringe. The samples were analyzed with an ABL90 blood gas instrument directly after sampling, without vortexing. The remaining portion of the samples were then stored in a refrigerator and analyzed again one or several days later, after being vortexed for 1 min. This test aimed to reflect the situation in practice; ideally the samples are analyzed immediately after collection, but occasionally, the analysis cannot be done until a few days later for various reasons.
2.12. Statistical Methods
The Mann–Whitney U-test was used for comparison between groups. ANOVA was used for comparisons between multiple groups. The association between variables was analyzed with linear regression. A p< 0.05 was considered statistically significant. All statistical analyses were performed with SPSS v. 25 (SPSS Inc., Chicago, IL, USA).
3. Results
3.1. Imprecision of the Blood Gas Instrument
Table 1 shows the within-series and between-series imprecision. The CV% for the within-series imprecision was very low, and was maximally 3.24% between-series for the analysis of 10.0 mmol/L potassium. Please note that the calibration interval of the instrument was 1–25 mmol/L for potassium, and high precision was still obtained for the 30 mmol/L solution. For potassium, almost all within-series measurements showed the same concentration for each test solution.
Table 1.
Within-series and between-series imprecision of ABL90 flex instrument.
| Within-Series Imprecision | Between-Series Imprecision |
:---: |
| | Glucose | Potassium | | Glucose | Potassium |
| | Solution 1 4.4 mmol/L | Solution 2 11.1 mmol/L | Solution 3 15.7 mmol/L | Solution 2 4.3 mmol/L | 10 mmol/L | 30 mmol/L | | Solution 1 4.4 mmol/L | Solution 2 11.1 mmol/L | Solution 3 15.7 mmol/L | Solution 2 4.3 mmol/L | 10 mmol/L | 30 mmol/L |
| Mean | 4.26 | 10.55 | 15.13 | 4.40 | 10.00 | 29.96 | Mean | 4.26 | 10.61 | 15.23 | 4.41 | 9.75 | 29.49 |
| SD | 0.07 | 0.09 | 0.15 | 0.00 | 0.00 | 0.07 | SD | 0.07 | 0.23 | 0.31 | 0.04 | 0.32 | 0.55 |
| CV% | 1.75 | 0.88 | 0.98 | 0.00 | 0.00 | 0.25 | CV% | 1.75 | 2.16 | 2.01 | 0.80 | 3.24 | 1.87 |
| Bias% | −3.85 | −4.81 | −3.86 | 2.33 | 0.00 | −0.12 | Bias% | −3.85 | −4.25 | −3.23 | 2.62 | −2.50 | −1.71 |
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3.2. Comparison between Different Analytical Instruments
In Figure 1, Figure 2, Figure 3 and Figure 4 the results obtained with the ABL90 flex and other instruments are shown. Almost identical results were obtained with the ABL 90 flex and ABL700 (Figure 1). Analysis with a Beckman Coulter AU5800 showed significantly (p< 0.001) higher sodium concentrations (Figure 2), but very similar potassium and chloride concentrations. ICP-MS analysis showed consistently lower sodium concentrations compared with the ABL 700 (p = 0.002), ABL90 flex (p< 0.001) and Beckman Coulter AU5800 (p< 0.001) and also higher chloride concentrations than the ABL90 flex (p< 0.019) and Beckman Coulter AU5800 (p = 0.007) (Figure 3 and Figure 4). Up to about 20 mmol/L, the ICP-MS showed similar potassium concentrations as the ABL 90 flex and Beckman Coulter AU5800, but showed an upward deviation at higher potassium levels.
Figure 1.
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Comparison results using ABL90 flex and ABL700.
Figure 2.
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Comparison of ABL90 flex and Beckman Coulter AU5800.
Figure 3.
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Comparison of ABL90 flex and ICP-MS.
Figure 4.
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Comparison of Beckman Coulter AU5800 and ICP-MS.
3.3. Influence of Cells/Centrifugation
Two groups of samples with >100,000 and <20,000 cells/mL were analyzed; the median was 1,097,500 and 5000 cells/mL for the group with a high and low cell-count number, respectively. Table 2 shows the means and the standard deviations of the electrolytes for each group of samples. There were no significant differences in electrolyte concentrations between untreated samples, and supernatant and pellets of the centrifuged samples in either group (Mann–Whitney U-test). The higher levels of potassium and lower levels of sodium and chloride in the group with the high cell number reflect the longer postmortem interval in this group, which is why more desquamated cells were found.
Table 2.
Impact of cells and centrifugation. Means and standard deviations of the concentrations of potassium, sodium and chloride.
| | Cell Number (Cells/mL) | Untreated (mmol/L) | Supernatant (mmol/L) | Pellet (mmol/L) |
:---: :---:
| K+ | >100,000 (n = 6) | 19.8 ± 3.8 | 19.9 ± 3.7 | 19.8 ± 3.4 |
| | <20,000 (n = 7) | 16.9 ± 3.5 | 17.0 ± 3.5 | 17.2 ± 3.6 |
| Na+ | >100,000 (n = 8) | 120.8 ± 13.1 | 122 ± 12.9 | 123 ± 12.7 |
| | <20,000 (n = 7) | 129.1 ± 8.7 | 130 ± 9.0 | 132 ± 9.1 |
| Cl− | >100,000 (n = 8) | 104 ± 10.9 | 104 ± 11.2 | 109 ± 13.2 |
| | <20,000 (n = 7) | 111 ± 10.8 | 111 ± 11.1 | 114 ± 11.1 |
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3.4. Comparison of Undiluted and Diluted Samples
Figure 5 shows undiluted vitreous samples compared with samples that were diluted 1:2 with distilled water. Dilution seems to have no impact on potassium or glucose, whereas sodium and chloride show slightly lower values than expected when diluted. Please note that the diluted concentrations for sodium and chloride are below the calibration range of the ABL700 instrument.
Figure 5.
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Comparison of undiluted and diluted samples, n = 29.
3.5. Hyaluronan, Hyaluronidas
Almost identical results were obtained for potassium, sodium and chloride when adding hyaluronidase or distilled water (Table 3a). When corrected for the dilution factor, these results were very similar to the analyzed results of the untreated sample. The addition of sodium hyaluronate to water solutions of these electrolytes caused a gradual increase in the concentrations of sodium and chloride, which was proportional to the content in the added hyaluronate (dissolved in 145 mmol/L NaCl) (Table 3b), whereas potassium showed the same values as the samples with the addition of distilled water.
Table 3.
Effect of addition of (a) hyaluronidase to postmortem vitreous samples and (b) sodium hyaluronate to water solutions of electrolytes.
(a)
Case #Sample + DDH 2 OConcentration RatiosCase #Sample + HyaluronidaseConcentration Ratios
K+Na+Cl−K+ RatioNa+ RatioCl− RatioK+Na+Cl−K+ RatioNa+ RatioCl− Ratio
1 11 130 107 0.92 0.92 0.93 1 10.9 129 106 0.92 0.91 0.92
2 11.6 130 96 0.94 0.93 0.92 2 11.6 129 96 0.94 0.92 0.92
3 7.8 127 110 0.93 0.92 0.92 3 7.8 127 110 0.93 0.92 0.92
4 13 122 107 0.92 0.91 0.91 4 13.1 122 107 0.93 0.91 0.91
5 10.6 119 106 0.92 0.92 0.93 5 10.5 118 105 0.91 0.91 0.92
6 11.9 126 111 0.92 0.91 0.92 6 11.9 126 111 0.92 0.91 0.92
7 16.3 122 105 0.93 0.92 0.92 7 16.3 122 106 0.93 0.92 0.93
8 21.1 95 77 0.92 0.91 0.92 8 21 95 77 0.92 0.91 0.92
(b)
1 mL solution +K+Na+Cl−Na+ CorrCl− Corr
10 µL DDH 2 O 9.7 110 113 111 114
10 µL DDH 2 O 9.7 110 113 111 114
20 µL DDH 2 O 9.6 109 112 111 114
20 µL DDH 2 O 9.6 108 112 110 114
50 µL DDH 2 O 9.4 105 109 110 114
50 µL DDH 2 O 9.4 105 109 110 114
100 µL DDH 2 O 9.0 101 104 111 114
100 µL DDH 2 O 8.9 100 104 110 114
10 µL hyaluronate 9.7 111 115 112 116
10 µL hyaluronate 9.7 111 115 112 116
20 µL hyaluronate 9.7 112 115 114 117
20 µL hyaluronate 9.7 112 115 114 117
50 µL hyaluronate 9.4 113 116 119 122
50 µL hyaluronate 9.4 113 116 119 122
100 µL hyaluronate 9.1 115 117 127 129
100 µL hyaluronate 9.0 115 117 127 129
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3.6. Differences between Whole Vitreous Fluid and Vitreous from the Center of the Eye
The analytical results based on the 0.25 mL sample collected from the left eye and the whole vitreous sample collected from the right eye are shown in Figure 6. There was a close correlation (r 2 = 0.97–1.00, slope 1.03–1.09) for potassium, chloride and glucose, whereas sodium levels were somewhat more dispersed (r 2 = 0.86, slope 0.86). No significant differences between the central sample and the whole vitreous sample were seen with regards to electrolyte concentrations, glucose or lactate (ANOVA and Mann–Whitney U-test, p> 0.05). Since it is known that lactate increases in vitreous fluid with PMI [8,20], and that postmortem diffusion might cause an uneven distribution, we included lactate levels in this comparison but found no difference in concentrations between the central sample and whole vitreous sample.
Figure 6.
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Comparison of whole and central vitreous.
3.7. Comparison of Samples Taken from the Left and Right Eye
The results of whole vitreous samples collected from the right and left eye correlated well (Figure 7). Linear regression showed r 2 of 0.95–0.99 with a slope coefficient of 0.93–1.02. There were no significant differences in the concentration of any of the solutes between the two groups of samples (Mann–Whitney U-test, p> 0.05).
Figure 7.
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Comparison of samples taken from center of the right and left eye, n = 16.
3.8. Spike Test
In Figure 8, the effects of spiking postmortem vitreous samples with electrolytes and glucose are shown. At higher concentrations, the measured sodium and chloride concentrations showed a trend towards higher values than the calculated concentration. However, as can be seen in the Bland–Altman plots, there was only little overall bias between the measured and calculated values for each of the analytes.
Figure 8.
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Comparison of actual concentration and calculated concentration (vitreous + spike).
3.9. Comparison of Vortexed and Unvortexed Samples
Figure 9 shows the results for vitreous samples that were analyzed immediately after sampling, unvortexed, and for samples that were analyzed one or several days later after being vortexed. Neither the delay in analysis nor vortexing had any effect on the measurements of glucose or any of the electrolytes.
Figure 9.
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Comparison of samples not vortexed (day 1) and vortexed (day 2), n = 13.
4. Discussion
Given that it is well understood that a postmortem investigation benefits from relevant medical information in the early stages in supporting decision-making during, or even before an autopsy, it is remarkable that rapid biochemical analysis is not applied more often in routine forensic casework. Such “deadside” analysis can be regarded as being as important as bedside analysis in the clinical setting. Already in 1966, Reh reported a rapid method for the detection of diabetic coma by analyzing glucose in cerebrospinal fluid with an enzymatic method . Zugibe (1966) similarly analyzed sodium and potassium in samples from cases of acute myocardial infarction using flame photometry and found that the sodium/potassium ratio was increased . Despite such possible use of in-house, simple analysis of electrolytes and glucose, forensic medical facilities tend to rely on the services of local clinical chemistry laboratories. This is most likely based on the assumption that results from such laboratories are more reliable, given that their methodology is used extensively to analyze samples from living patients.
In this study, we investigated the feasibility of analyzing postmortem vitreous fluid with a blood gas instrument and we conducted several tests and comparisons to validate this strategy as a proof of concept. We showed that the imprecision of the ABL 90 flex instrument was very low for glucose and potassium (Table 1). Indeed, the CV% for potassium was no higher than 3.24, and this was for only one of the concentrations in the between-series study. Vitreous potassium is used extensively for estimating the postmortem interval (PMI), particularly when the true interval is so long that the classical methods used in the early phase no longer work. If we assume the measuring error to be 4%, i.e., a value of 12.5 mmol/L is obtained rather than 12.0, then the estimated PMI using the equations of Zilg et al. , Madea et al. , Sturner and Gantner and Bortolotti et al. would be 40.9, 34.8, 50.2 and 58.8 h, respectively, instead of 37.8, 32.2, 46.6 and 55.9 h, respectively. This example shows that the difference in results of these and many other equations surpass the analytical error. At very long postmortem intervals, the effect becomes even more pronounced.
We also showed that two blood gas instruments, the ABL90 flex (a simple instrument with cassette system) and a regular instrument, the ABL700, produced indistinguishable results in a large set of postmortem vitreous samples (Figure 1). When comparing the results of the ABL90 flex and Beckman Coulter AU5800, the latter instrument reported higher sodium values, but similar potassium and chloride results. The ICP-MS instrument used by ALS Global also provided similar potassium values, but generally lower potassium and higher chloride values. ICP-MS and atomic absorption spectrophotometry is supposed to provide the “true” numbers of atoms , but this is not necessarily what we want to know. Rather, we want to know whether a test result can be compared with established reference levels for a certain method. If there is an understanding of what the deviations mean, such a method should be feasible for clinical practice as well as in postmortem casework. By and large, both the blood gas instruments and the Beckman Coulter provided similar results, and we have previously shown the feasibility of the blood gas instrument readings of electrolytes and glucose in postmortem casework [8,11,14]. The difference between the ABL90 flex and Beckman Coulter AU5800 regarding sodium was statistically significant, but since the difference is not large, neither numerically or in terms of percentage, the interpretation of dehydration (or overhydration) will not be affected.
At longer PMIs, the vitreous fluid volume is reduced due to evaporation, and the fluid gradually shows an increasingly grey-brown tinge due to increased amounts of cells and cell debris, prompting other researchers in this field to centrifuge the samples to obtain a more transparent fluid for analysis. Thus, the question is if the cells present in the fluid retain a higher content of potassium, and lower content of sodium and chloride than the extracellular fluid. We therefore analyzed untreated vitreous fluid samples, and compared these with the supernatant and pellet after centrifugation. We found that there was no difference in concentrations in these three aliquots (Table 2), implying that the ion concentrations had equilibrated before the cells had detached.
Whenever high concentrations are found, and exceed the calibration range for a method, the standard procedure is to dilute the sample appropriately to obtain a concentration that matches the measuring range. Hence, we investigated the effect of a relevant dilution of the samples and found that the results were not much affected (Figure 5). While the potassium levels were not affected, there was a reduction in both the sodium and chloride levels. The most likely explanation for this is that dilution causes a reduction in the concentration levels, which are below the calibration range for sodium and chloride, but within the calibration range for potassium.
Concerns have been raised as to the possible impact of the viscosity of vitreous fluid on the accuracy of the analytical results. Blana et al. reported that heat and hyaluronidase treatment of vitreous samples resulted in a slight increase and decrease, respectively, of measured electrolyte concentrations. We also investigated the possible impact of hyaluronidase treatment on electrolyte results, and found no differences compared to the results of untreated samples. Further, we studied the possible effect of the addition of hyaluronidase to water solutions of sodium and chloride, but found no influence on the analytical results, even at hyaluronidase concentrations that exceed those that are present in the vitreous fluid of phakic eyes .
Regarding the collection of vitreous fluid, we collected a small amount, approximately 0.25 mL, from each eye, which was pooled in the same syringe in order to still retain sufficient fluid for toxicological analysis. However, most other investigators have used whole vitreous samples for analysis. Given that the diffusion of ions and other solutes from the retinal cells after death may produce different concentrations in the central and peripheral compartments of the vitreous, we compared the results of analysis of samples collected from the central vitreous with the results of analysis of the whole vitreous. Figure 7 shows that almost the same concentrations of electrolytes was found in both samples.
Furthermore, we tested whether spiking authentical postmortem samples with electrolytes and glucose would produce results that correlated with mathematical calculations. Figure 8 shows that the results did not deviate much, implying that the matrix effects of hyaluronic acid and other molecules prevalent in the vitreous do not have any substantial impact on the distribution of either electrolytes or glucose.
Finally, we performed a test that should resemble the situation in practice, i.e., the possible impact of delay of analysis and sedimentation of particles in the solution. Samples that were stored in a refrigerator for one day were vortexed before analysis. We did not find any change in the concentrations of the analytes using such a practice (Figure 9). This means that though it is desirable to obtain a result as soon as possible, similar results will be obtained if the analysis is performed a day after sampling.
Today, most instruments use ion-selective electrodes for determining the concentrations of potassium, sodium and chloride. It is therefore not surprising that we observed similar results with blood gas instruments and a multi-analysis instrument that is typically used in clinical chemistry laboratories. The small difference in concentrations and any imprecision is most likely due to different lining conditions, implying that samples with higher viscosity had to be diluted to avoid clogging, and hence the concentrations measured fell short of the optimal measuring range of the analytes. Given such concerns regarding the internal conditions of laboratory instruments, it is tempting to consider simple hand-held instruments, such as an i-STAT® analyzer. Monzon et al. (2018) evaluated the feasibility of such an analyzer in regard to electrolytes and glucose in vitreous and found low imprecision. However, the calibration range for potassium was reported to be 2.0–9.0 mmol, implying that many samples would have to be diluted before analysis .
5. Conclusions
We have shown that analysis of electrolytes and glucose with blood gas instruments in postmortem samples, provides rapid and reliable results without dilution, enzymatic treatment or centrifugation. We also found no difference in the concentrations of the electrolytes or glucose in samples from the right and left eye, and the results in samples from the central part of the vitreous were the same compared to the results of whole vitreous samples. We therefore conclude that a small, pooled sample from the center of each eye is feasible for these chemical analyses, which leaves a sufficient amount of vitreous fluid for toxicological analysis.
Acknowledgments
The authors would like to acknowledge the assistance of the staff at the Swedish National Board of Forensic Medicine.
Author Contributions
H.D.: designed and supervised all parts of the study. H.D., B.Z.: writing—original draft preparation. K.A.: methodology and sampling. R.K.: validation. S.B.: data curation and statistical analysis. All authors read, contributed to, and approved the final manuscript. All authors have read and agreed to the published version of the manuscript.
Funding
Part of this work was supported by the Swedish National Board of Forensic Medicine.
Institutional Review Board Statement
All samples were collected from humans undergoing a forensic medicine examination upon the request of the police, and analysis of these samples was a part of the medico-legal investigation. All such casework is exempt from ethical review board evaluation. The European law regulating personal integrity, GDPR, excludes deceased subjects. All information was anonymized before data analysis. Ethical approval for these studies was granted by the Regional Ethics Review Board in Stockholm, No 2008/231-31/3 and 2017/1235-32.
Informed Consent Statement
Not applicable.
Data Availability Statement
All relevant data are provided in the article.
Conflicts of Interest
The authors declare no conflict of interest.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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All relevant data are provided in the article.
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Abstract
1. Introduction
2. Materials and Methods
3. Results
4. Discussion
5. Conclusions
Acknowledgments
Author Contributions
Funding
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Data Availability Statement
Conflicts of Interest
Footnotes
References
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190148 | https://www.cuemath.com/geometry/angle-addition-postulate/ | LearnPracticeDownload
Angle Addition Postulate
The angle addition postulate in geometry states that if we place two or more angles side by side such that they share a common vertex and a common arm between each pair of angles, then the sum of those angles will be equal to the total sum of the resulting angle. For example, if ∠AOB and ∠BOC are adjacent angles on a common vertex O sharing OB as the common arm, then according to the angle addition postulate, we have ∠AOB + ∠BOC = ∠AOC.
| | |
--- |
| 1. | Angle Addition Postulate Definition |
| 2. | Angle Addition Postulate Formula |
| 3. | FAQs on Angle Addition Postulate |
Angle Addition Postulate Definition
The definition of angle addition postulate states that "If a ray is drawn from point O to point P which lies in the interior region of ∠MON, then ∠MOP + ∠NOP = ∠MON". This postulate can be applied to any pair of adjacent angles in math. In other words, the angle addition postulate can be defined as 'the sum of two angles joined together through a common arm and a common vertex is equal to the sum of the resulting angle formed'.
Angle Addition Postulate Formula
If an angle AOC is given where O is the vertex joining rays OA and OC, and there lies a point B in the interior of ∠AOC, then the angle addition postulate formula is given as ∠AOB+∠BOC = ∠AOC. If ∠AOC is divided into more than two angles such as ∠AOB, ∠BOD, and ∠DOC, then also we can apply the formula of angle addition postulate as ∠AOB+∠BOD+∠DOC = ∠AOC.
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Angle Addition Postulate Examples
Example 1: In the figure given below, if ∠POS is a right angle, ∠2 = 30°, and ∠3 = 40°. Find the value of ∠1.
Solution: It is given that ∠POS is a right angle. It means that ∠POS = 90°. Now, by using the angle addition postulate formula, we can write ∠1 + ∠2 + ∠3 = 90°. Given, ∠2 = 30° and ∠3 = 40°. Substituting these values in the above equation, we get,
∠1 + 30° + 40° = 90°
∠1 + 70° = 90°
∠1 = 90° - 70°
∠1 = 20°
Therefore, the value of ∠1 is 20°.
2. Example 2: In the given figure, XYZ is a straight line. Find the value of x using the angle addition postulate.
Solution: It is given that XYZ is a straight line. It means that ∠XYO and ∠OYZ form a linear pair of angles.
⇒ ∠XYO + ∠OYZ = 180° (using angle addition postulate and linear pair of angles property)
⇒ (3x + 5) + (2x - 5) = 180°
⇒ 5x = 180°
⇒ x = 36
Therefore, the value of x is 36.
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Practice Questions on Angle Addition Postulate
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FAQs on Angle Addition Postulate
What is Angle Addition Postulate in Geometry?
The angle addition postulate in geometry is a mathematical axiom which states that if there is a ray drawn from O to Q which is any point inside the region of angle POR, then the sum of angles ∠POQ and ∠QOR is equal to ∠POR. It can be represented in the form of a mathematical equation as ∠POQ + ∠QOR = ∠POR.
What is the Angle Addition Postulate Formula?
The formula of angle addition postulate in math is used to express the sum of two adjacent angles. If there are two angles (∠AOB and ∠BOC) joined together sharing a common arm OB and a common vertex O, then the angle addition postulate formula is ∠AOB + ∠BOC = ∠AOC.
How to Find x in Angle Addition Postulate?
If there is any missing angle 'x' when two or more angles are joined together, then we can subtract the sum of remaining angles from the total sum to find the value of x. For example, if two angles ∠PQR and ∠RQS are joined together such that ∠RQS = 40°, ∠PQR = x, and ∠PQS = 70°, then the value of x will be (70 - 40)° = 30°.
How to Use Angle Addition Postulate?
The angle addition postulate can be used to find the sum of two or more adjacent angles and to find the missing values of angles. It establishes a relation between the measurement of angles joined together.
How do you Find the Angle Addition Postulate?
The angle addition postulate is a mathematical fact that can be considered true without any proof. It tells us that the sum of two or more angles joined together is equal to the sum of the larger angle formed.
How is the Angle Addition Postulate Used in Real Life?
In real life, the angle addition postulate is used in construction (bridges, buildings, etc), architecture, designing, etc.
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1.9
Avogadros Law
Written by:Divya Ladha
Fact-checked by:The CK-12 Editorial Team
Last Modified: Jul 25, 2025
Avogadro’s Law: Understanding Gas Volume and Particle Relationship
Imagine you are inflating a balloon. As you blow more air into it, the balloon expands. Have you ever wondered why this happens? The answer lies in a fundamental principle of chemistry called Avogadro’s Law. This law explains the relationship between the volume of a gas and the number of gas particles present.
When a balloon is inflated, air (a mixture of gases) is added, increasing the number of gas molecules (n) inside. As a result, the volume (V) of the balloon increases proportionally, causing it to expand. Conversely, if air is released, the number of gas molecules decreases, leading to a decrease in volume, and the balloon deflates. This principle is crucial in designing hot air balloons, airbags, and respiratory devices, where gas volume control is essential.
Avogadro’s Law
Avogadro’s Law states that "Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules." This means that the volume of a gas is directly proportional to the number of gas particles (moles) present, provided temperature and pressure remain constant.
Mathematically, Avogadro’s Law is expressed as:
@$\begin{align}V \propto n\end{align}@$
or
@$\begin{align}\frac{V_1}{n_1} = \frac{V_2}{n_2}\end{align}@$
where:
V = Volume of the gas
n = Number of moles of gas
This equation shows that if the number of moles of a gas increases, its volume will also increase, and vice versa, as long as temperature and pressure remain constant.
Understanding Avogadro’s Law
In 1811, Amedeo Avogadro hypothesized that the fundamental unit of matter is not always a single atom. He suggested that gases are composed of molecules, some of which are diatomic (e.g. O2, N2, H2). This idea helped explain Gay-Lussac’s Law of Combining Volumes and laid the foundation for the mole concept and Avogadro’s number.
Avogadro’s Number
Avogadro’s Number (NA) is a fundamental constant that represents the number of particles (atoms, ions, or molecules) in one mole of a substance:
NA = 6.022 × 1023 particles per mole
This means that one mole of any gas occupies the same volume at standard temperature and pressure (STP). The molar volume of a gas at STP (273.15 K or 0°C, 1 bar) is 22.4 L/mol.
This concept is crucial in stoichiometric calculations involving gaseous reactions.
PLIX: Avogadro's Law
Case Study: Applying Avogadro's Law in Aerosol Drug Formulation for Accurate Dosage Control
A pharmaceutical company is developing a new process to produce an aerosol spray used for asthma treatment. In their research lab, they fill a 44.8 L container with a gas at STP (Standard Temperature and Pressure). They want to determine how many moles of the active compound in gaseous form are present. Later, to test dosage consistency, they compare the volume of 2.5 moles of the same gas at STP and observe it occupies 56 L. The team uses Avogadro's Law to ensure dosage accuracy and validate that the volume of the gas is directly proportional to the number of moles when temperature and pressure are constant.
Examples on Avogadro’s Law
Example 1:
A gas occupies 5.0 L at 1.5 moles. If the number of moles is increased to 3.0 moles, what will be the new volume at constant temperature and pressure?
Using Avogadro’s Law,
@$\begin{align}\frac{V_1}{n_1} = \frac{V_2}{n_2}\end{align}@$
@$\begin{align}\frac{5.0}{1.5} = \frac{V_2}{3.0}\end{align}@$
@$\begin{align}V_2 = \frac{(5.0 \times 3.0)}{1.5} = 10.0 \text{ L}\end{align}@$
Thus, the new volume is 10.0 L.
Example 2:
A gas at STP occupies 44.8 L. How many moles of gas are present?
At STP, 1 mole of any gas occupies 22.4 L.
@$\begin{align}n = \frac{V}{22.4}\end{align}@$
@$\begin{align}n = \frac{44.8}{22.4} = 2.0 \text{ moles}\end{align}@$
Thus, the gas sample contains 2.0 moles.
Example 3:
An LPG cylinder contains 11.2 L of propane gas at STP. How many propane molecules are present?
Find the moles:
@$\begin{align}n = \frac{V}{22.4} = \frac{11.2}{22.4} = 0.5 \text{ moles}\end{align}@$
Find the number of molecules:
@$\begin{align}\text{Molecules} = n \times N_A = 0.5 \times 6.022 \times 10^{23}= 3.011 \times 10^{23} \text{ molecules}\end{align}@$
Thus, the LPG cylinder contains 3.011×1023 molecules of propane.
| |
| Summary of Avogadro's Law |
| Avogadro’s Law states that equal volumes of all gases contain the same number of molecules at constant temperature and pressure. Mathematical expression of Avogadro’s Law is V ∝ n Avogadro’s Law explains why gases behave similarly under standard conditions. Avogadro’s Law led to the concept of molar volume and Avogadro’s number (6.022 × 10²³ molecules per mole), which is a fundamental constant in chemistry. |
Review Questions on Avogadro’s Law
What is the volume occupied by 0.25 moles of a gas at STP?
If 8.96 L of hydrogen gas is present at STP, how many hydrogen molecules does it contain?
A 5.0 L sample of nitrogen gas contains 0.2 moles. How much volume will 0.5 moles occupy under the same conditions?
Explain how Avogadro’s Law is applied in respiration.
What is the significance of Avogadro’s Number?
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190151 | https://ics.uci.edu/~eppstein/pubs/BerEpp-IJCGA-92.pdf | International Journal of Computational Geometry & Applications, Vol. 2, No. 3 (1992) 241–255 c ⃝World Scientific Publishing Company POLYNOMIAL-SIZE NONOBTUSE TRIANGULATION OF POLYGONS MARSHALL BERN Xerox Palo Alto Research Center, 3333 Coyote Hill Road Palo Alto, California 94304, U.S.A.
and DAVID EPPSTEIN Department of Information and Computer Science, University of California Irvine, California 92717, U.S.A.
Received 11 May 1991 Revised 6 June 1992 ABSTRACT We describe methods for triangulating polygonal regions of the plane so that no triangle has a large angle. Our main result is that a polygon with n sides can be trian-gulated with O(n2) nonobtuse triangles. We also show that any triangulation (without Steiner points) of a simple polygon has a refinement with O(n4) nonobtuse triangles.
Finally we show that a triangulation whose dual is a path has a refinement with only O(n2) nonobtuse triangles.
Keywords: Computational geometry, mesh generation, triangulation, angle condition.
1. Introduction One of the classical motivations for problems in computational geometry has been automatic mesh generation for finite element methods. In particular, mesh generation has motivated a number of triangulation algorithms, such as finding a triangulation that minimizes the maximum angle.1 A triangulation algorithm takes a geometric input, typically a point set or polygonal region, and produces an output that is a triangulation of the input. For a point set, this usually means partitioning the region bounded by its convex hull into triangles, such that the vertices of the triangles are exactly the input vertices. Triangulating a polygonal region usually means partitioning the region into triangles such that the vertices of the triangles are exactly the vertices of the region’s boundary. In both cases, the output must be a simplicial complex, that is, triangles intersect only at shared vertices or edges.
Mesh generation applications, however, invariably place further requirements on the triangulation, stemming from numerical analysis. Very flat and very sharp angles are undesirable.2,3 To satisfy angle bounds, it may be necessary to augment 1 the input by adding new vertices, called Steiner points. Because the complexity of the finite element computation depends on the size of the mesh, the number of Steiner points should be kept small. Until recently, computational geometers neglected the more realistic Steiner versions of triangulation problems, and concen-trated on problems that do not allow extra vertices.
An upper bound of 90◦has special importance in mesh generation. A triangu-lation of a point set with maximum angle 90◦is necessarily the Delaunay triangu-lation. In finite element methods for solving certain partial differential equations, a nonobtuse mesh leads to a matrix with nice numerical propertices, such as all off-diagonal elements being nonpositive.4 For a more geometric motivation, notice that each (closed) triangle in a triangulation contains the center of its circumscrib-ing circle exactly when all angles measure at most 90◦. The “perpendicular planar dual” of such a triangulation can be formed by simply joining perpendicular bisec-tors of edges. Practitioners often use a mesh and its dual—sometimes with bent edges in the dual—to discretize vector fields.4 In general, a straight-line embedding of the planar dual of a triangular mesh can be formed by placing dual vertices at the meeting point of angle bisectors5; however, a perpendicular planar dual is more accurate and simplifies the calculation of flow or forces across element boundaries.
1.1. Summary of Results In this paper we consider the problem of triangulating a polygonal region so that all angles measure at most 90◦. We give the first algorithm that produces such a nonobtuse triangulation with size (number of triangles) bounded by a polynomial in n, the number of sides of the input polygon. The only previous provably-correct algorithms, due to Baker et al.,4 have no size guarantees. Indeed, the size of the triangulations produced by the algorithms of Baker et al. depend on the geometry of the input, and hence can be arbitrarily large, even when n is fixed.
Our main result is that an arbitrary polygon can be triangulated with O(n2) nonobtuse triangles. The polygon need not be simple; it may contain polygonal holes. We also consider the problem of refining a given triangulation of a polygon (without Steiner points) into a nonobtuse triangulation. We achieve O(n4) triangles for arbitrary simple polygons, and O(n2) for the special case of a polygon with a “path triangulation”, that is, a triangulation whose dual is a path.
All our algorithms can be made to run in time O(n log n + k), where k is the size of the output.
Our angle bound is the best possible for polynomial-size triangulations. Our work with John R. Gilbert6 shows that any smaller fixed bound on the largest angle would require the number of triangles to depend not only on the size of the input, but also on its aspect ratio. Consider for example an input that is an extremely long, thin rectangle R. A triangle with a maximum angle of, say, 89◦must have longest side no longer than a constant multiple of R’s smaller dimension, and hence a large number of these triangles would be needed to cover R.
2 1.2. Related Work The original computational geometry result motivated by mesh generation is that the Delaunay triangulation of a point set maximizes the minimum angle.7,8,9 (It is a curiously similar fact that, for points in convex position, the farthest-point Delaunay triangulation minimizes the minimum angle.10) For polygons the mini-mum angle is maximized by the constrained Delaunay triangulation.11,12 It is also known how to compute the triangulations that minimize the maximum angle,1 maximize the minimum triangle height,13 and minimize the maximum triangle “eccentricity”13 for both point sets and polygons.
Theoretical work on Steiner triangulation problems is less common. In an ear-lier paper,6 we developed algorithms based on quadtrees for a number of Steiner triangulation problems. In a certain strong sense, we solved the problems of Steiner-triangulating point sets and polygonal regions with no small angles (when the min-imum angle bound is sufficiently small, for example no more than 18.4◦). For each input our algorithms produce an output that has size within a constant factor of the optimal size. The number of triangles required necessarily depends not only on the size of the input, but also on its aspect ratio. We also showed how to triangulate point sets with no large angles. If the angle bound is less than 90◦, the solution is essentially identical to that for no small angles; but if the only requirement is that there be no obtuse angles, the point set can be triangulated with only linearly many Steiner points.
Gerver14 showed how to compute a nonpolynomial-size triangulation of a poly-gon with no angles larger than 72◦, assuming all interior angles of the input measure at least 36◦. As mentioned above, Baker et al.4 gave a nonpolynomial algorithm for nonobtuse triangulation, thereby showing that every polygon has such a trian-gulation. They gave a simple algorithm for nonobtuse triangulation of a polygon whose vertices have integer coordinates. This algorithm imposes a unit grid on the polygon, so it typically uses an exponential (in the size of the encoding of the input) number of triangles. They went on to give a somewhat more complicated algorithm for the general problem. This second algorithm also avoids angles smaller than 14◦, and is thus inherently nonpolynomial as well. Our earlier result that point sets can be triangulated with O(n) nonobtuse triangles led us to suspect that polygons too could be triangulated with a polynomial number of nonobtuse triangles. Subse-quent to the work reported here, Eppstein devised an algorithm that triangulates a convex polygon with O(n1.85) nonobtuse triangles (included in Ref. 18).
2. Obtuse with subdivided legs In this section, we develop the main subroutine of our nonobtuse triangulation algorithm. In the next section, we shall present the full algorithm. Throughout this paper, we use hypotenuse to mean the longest side of a triangle and legs to mean the other two sides. A subdivision point is a vertex of a polygon at which the angle measures 180◦. A nonobtuse (nonacute) angle is one measuring at most 3 (respectively, at least) 90◦. The measure of angle ̸ abc is denoted |̸ abc|.
Suppose we are given a triangle ace with hypotenuse ae as its horizontal base, with |̸ ace| ≥90◦, and with subdivided legs.
In this section we show how to triangulate this input into nonobtuse triangles, without adding any new subdivision points to the legs. Our overall stategy is to replace the legs by ones sloped slightly closer to the horizontal, while keeping the base fixed. The region between the old legs and the new legs is triangulated in such a way as to reduce the total number of subdivision points by one. The remaining subdivision points propagate towards the polygon boundary.
Figure 1. Base cases.
We have two different reduction methods, called apex merger and side merger.
Roughly speaking, an apex merger combines the first subdivision point on one leg with apex c; a side merger combines two closely-spaced subdivision points. The reduction methods given here have been simplified somewhat from the reduction methods given in the conference version of this paper.15 Our first two lemmas and Figure 1 give the basis of an induction. Lemmas 3 and 4 then give “coroutines” for the inductive step; each of these lemmas calls the other on a triangle with at most n −1 subdivision points.
Lemma 1.
If ace is a right triangle with one subdivision point on a leg, ace can be triangulated into three right triangles, with one subdivision point on base ae.
Proof.
Simply add the edge from the subdivision point to the opposite vertex of ace; then drop the altitude from the subdivision point perpendicular to base ae. 2 Lemma 2.
If triangle ace is obtuse with no subdivision points, then it can be triangulated into two nonobtuse triangles, adding one subdivision to base ae.
Proof.
Simply drop the altitude from c to ae. 2 Lemma 3.
Assume nonacute triangle ace has n ≥1 subdivision points, all on one leg. Using O(n) nonobtuse triangles, either we can reduce the untriangulated portion of ace to a nonacute triangle ac′e with n −1 subdivision points, or we can completely triangulate ace adding at most n + 1 subdivision points to base ae.
Proof.
Without loss of generality, assume ac is the subdivided leg, and let b be the subdivision point closest to c. If |̸ ace| = 90◦, we simply add edge be and reduce to the case of an obtuse triangle with n −1 subdivision points.
So assume |̸ ace| > 90◦, and consider extending a perpendicular to ac at b and a perpendicular to ce at c. Assume these meet at point c′ inside the (closed) triangle ace. Now imagine adding edges that project the subdivision points along ac onto 4 c’ c’ b e c a e c b a Figure 2. (a) Apex merger.
(b) Side merger.
ac′, alternately projecting perpendicularly away from ac and perpendicularly onto ac′. The first subdivision point b projects away (to point c’), the next one down from b projects onto, and so forth. See Figure 2(a). Now if ̸ bac′ is sufficiently small (relative to the spacing of subdivision points), then none of these projection edges cross each other, and triangle abc′ will be divided into some number of quadrilaterals and one right triangle.
Each quadrilateral has the nice property that it can be divided into two right triangles with a diagonal. This move, in which we reduce the number of subdivision points by combining the apex and an adjacent subdivision point using the meeting point of perpendiculars, is called an apex merger.
So assume that the perpendiculars to ac at b and to ce at c do not meet at a point c′ with sufficiently small ̸ bac′. Instead define c′ to be the last point (i.e., furthest from b) in triangle ace along the perpendicular to ac at b, such that the projection edges from ac do not cross. (Two line segments cross if they intersect at a point that is interior to each segment.) Notice that |̸ c′ce| > 90◦since the apex merger did not succeed.
One possibility is that c′ lies on base ae. (This case occurs whenever n ≤2 and the apex merger fails.) We add the projection edges from ac to ac′, along with c′c, and drop an altitude from c to c′e. Quadrilaterals are triangulated as before into two right triangles. Now there is no untriangulated portion of ace, and the number of Steiner points on base ae is at most n + 1. Look at Figure 2(b) and imagine c′ lying on base ae.
The other possibility is that c′ lies interior to ace.
Now by our choice of c′ some adjacent pair of projection edges meet along ac′, as shown in Figure 2(b).
Along with the projection edges and quadrilateral diagonals, we add cc′, c′e, and an altitude from c to c′e. Triangle ac′e has at most n−1 subdivision points, although one of them has now moved to the right leg (necessitating Lemma 4 below). This reduction step is called a side merger. 2 Notice that the algorithm given in the proof of Lemma 3 either reduces the number of subdivision points by one without adding anything to base ae, or it completely triangulates ace.
The same will be true of our method for the case of subdivisions on both legs.
Therefore, we assert that the algorithm given by Lemmas 1–4 adds at most n + 1 points to the base, and at most n if ace is right.
Because each merger uses O(n) triangles and removes one subdivision point, the total number of triangles will be O(n(n −n′ + 2)), where n′ is the number of subdivisions added to the base. We use these bounds recursively in the proof of the next lemma.
5 c’ c’ e d c b a e d c b a Figure 3. In both apex (a) and side (b) mergers, c′de is triangulated recursively.
Lemma 4.
Assume triangle ace has n ≥2 subdivision points, with at least one on each leg. Using O(n(n −n′ + 2)) triangles, either we can reduce the untriangulated portion of ace to a nonacute triangle ac′e with n′ ≤n −1 subdivision points, or we can completely triangulate ace adding n′ ≤n + 1 subdivisions to the base.
Proof.
Let b and d be the closest subdivision points to c on ac and ce, respectively.
Let nr represent the number of subdivision points on the right leg ce.
We first try the apex merger of c with b. Assume that the perpendiculars to ac at b and to ce at c meet at point c′ in ace, such that ̸ cac′ is small enough that no projection edges cross, as shown in Figure 3(a). In this case, we add edges c′d and c′e. Triangle c′de has nr −1 subdivisions, all one leg, so we can triangulate it recursively, outputting at most nr subdivisions on base c′e. Thus we have reduced the problem to triangulating ac′e, an obtuse triangle with one fewer subdivisions.
So assume the apex merger of c and b is not possible, and let c′ be the last point in triangle ace along the perpendicular to ac at b such that no pair of adjacent projection edges from ab onto ac′ cross.
Since the apex merger failed, ̸ c′cd is obtuse.
Now if c′ lies on base ae, then we add edges bc′, c′c, and c′d; drop an altitude from c to c′d; and triangulate the quadrilaterals along ab. These steps add n −nr subdivision points to base ae and reduce the problem to triangulating c′de, an obtuse triangle with nr subdivisions (on two legs). Triangle c′de is triangulated recursively, adding at most nr + 1 subdivision points to ae, for a total of at most n + 1. Because each merger removes a subdivision point at the cost of O(n) new triangles, the total number of triangles used will be O(n(n −n′ + 2)), where n′ is the number of subdivision points added to ae.
The remaining possibility is that c′ lies interior to ace. In this case, we can perform a side merger. We add edges c′c, c′d, c′e, and an altitude from c onto c′d, as shown in Figure 3(b). We then triangulate c′de recursively. After this step, triangle ac′e has at most nr +1 subdivisions on leg c′e and n−nr −2 on leg ab′ (one fewer due to the merger and another fewer because c′ is the apex and not counted as a subdivision point), so the total number of subdivision points is at most n −1.
As throughout, the number of triangles used in this step is O(n) per subdivision point removed. 2 Theorem 1.
An obtuse or right triangle with n subdivisions on its legs can be triangulated with O(n2) nonobtuse triangles, without adding any new subdivisions to the legs. 2 6 3. Arbitrary Polygons We now show how the method of the previous section can be applied to an arbitrary polygon, possibly with holes.
Our overall strategy is to partition the polygon into some simple shapes, by cutting it with horizontal and vertical lines.
Each such shape can be either triangulated directly or divided into obtuse triangles with subdivided legs, that can then be further triangulated using the methods developed to prove Theorem 1.
Draw a vertical line segment through each vertex of the polygon, extending to the boundaries of the polygon. These lines divide the polygons into slabs, which we define as quadrilaterals with two vertical sides, possibly with subdivision points on the vertical sides. In degenerate cases, two vertices may lie on the same vertical line; this causes no problems.
Each vertex of a slab will be either an original input vertex, or a point where a vertical line touches the polygon boundary. Draw a horizontal line segment through each slab vertex, and extend the line segment to the last possible vertical segment.
In other words, each endpoint of a horizontal segment should lie either on a vertical segment, or on the vertex inducing the horizontal, and each horizontal segment should be as long as possible with this property. A polygon so divided is shown in Figure 4(a).
Lemma 5.
The above steps partition the polygon into regions of four types: (1) rectangles with unsubdivided sides; (2) right triangles with hypotenuse on the boundary of the polygon and vertical leg possibly subdivided; (3) obtuse trian-gles with two sides on the boundary of the polygon, and one leg vertical and pos-sibly subdivided; (4) slabs with two sides on the boundary of the polygon, and two possibly-subdivided vertical sides, that cannot be simultaneously crossed by a horizontal line.
Proof.
Any other shape must have a vertex from which a line segment can be extended vertically or horizontally. The construction only creates verticals and horizontals, so any other line segments must be on the polygon boundary. 2 Theorem 2.
Any n-vertex polygon (with holes) can be triangulated with O(n2) nonobtuse triangles.
Proof.
We apply the procedure above to subdivide the polygon into the shapes listed in Lemma 5. Triangulating the rectangles is trivial. The right and obtuse triangles can be triangulated by the method of the previous section. The slabs can be divided by a diagonal into two obtuse triangles, that can then be triangulated by the method of the previous section. See Figure 4(b).
We create O(n) horizontal and vertical line segments, so there are O(n2) rectan-gles. Nonrectangular faces each contain a portion of the polygon boundary between vertical line segments, so there are O(n) such faces. There are O(n) subdivisions in all these regions, arising from the horizontal line segments. Therefore the method of the previous section, when applied to the nonrectangular faces, creates O(n2) triangles. 2 7 Figure 4. (a) Polygon cut by horizontals and verticals. (b) Triangulation.
4. Refining a Given Triangulation In this section we consider the problem of refining a given triangulation (without interior Steiner points) of a simple polygon into a nonobtuse triangulation. The edges of the input triangulation must appear in the output, possibly subdivided.
This problem has application to computing a mesh for a domain with interfaces, perhaps modeling an object made of more than one material.
The refinement problem is a special case of—and perhaps a first step towards solving—the more general problem of refining an arbitrary straight-line planar graph into nonobtuse triangles. The more general problem has applications to computing a mesh on a polyhedral surface, and to learning polygons from examples given by a “helpful teacher”. Salzberg et al.16 have shown that, if the inside and outside of a polygon (up to the convex hull) can be simultaneously triangulated without obtuse angles, then a teacher who picks examples as advantageously as possible can teach a polygon to a nearest-neighbor classifier, using a number of examples proportional to the size of the triangulation.
Our strategy for this problem is to again cut the input into cases that we can handle: rectangles and obtuse triangles with subdivided legs. The cutting proce-dure, however, grows more complicated. We process triangles in a preorder traversal of the tree that is the planar dual of the initial triangulation. Each triangle after the first one then has one “inbound” edge (the one shared with a triangle earlier in the order) and two “outbound” edges. Ideally each triangle is obtuse with its hypotenuse an outbound edge. Our preprocessing step dices up “backwards” obtuse triangles, propagating O(n2) subdivision points up the tree, in order to achieve this ideal situation. The ideal situation can be triangulated with quadratic increase in complexity.
Lemma 6.
If ace is a nonobtuse triangle with subdivisions on only one side, then ace can be triangulated into nonobtuse triangles, adding new subdivisions only to the other two sides.
Proof.
Let ac be the side with subdivisions. Consider the perpendicular projection 8 e’ e c a Figure 5. A nonobtuse triangle with subdivisions on only one side.
e′ of e onto ac. Add edges between e and the subdivision points adjacent to e′ (but do not add e′ or edges to e′), as shown in Figure 5.
This divides ace into one nonobtuse triangle without subdivisions and either one or two nonacute triangles that can be triangulated by the method of Section 2. 2 Theorem 3.
Any triangulation of a polygon (without holes) with n vertices can be refined into a nonobtuse triangulation with O(n4) triangles.
Proof.
Let T be the tree that has a vertex for each triangle of the input and an edge between each pair of triangles that share a side. Root T at any vertex corresponding to a triangle with an exterior side, that is, one lying along the boundary of the polygon. We now label triangle sides as inbound or outbound. Label the side that a triangle shares with its parent in T inbound; label the other two sides (whether exterior or not) outbound.
At this point, if each triangle is nonobtuse or is obtuse with its hypotenuse labeled outbound, then we can refine the triangulation as follows. We split the root triangle into two right triangles by dropping an altitude. Each subsequent triangle in the tree starts with some number of subdivision points on its inbound side, and we triangulate using either Lemma 6 or the method of Section 2, adding new subdivisions only to outbound sides.
So assume that there is at least one backwards obtuse triangle with its hy-potenuse labeled inbound. See Figure 6. We first process triangles in “upward” order given by a postorder traversal of T. A backwards obtuse triangle that has no backwards descendants is handled by simply dropping the altitude onto its hy-potenuse; this introduces a subdivision point to an outbound side of the parent triangle.
We now carry this through inductively, by showing how to partition a triangle, that has subdivisions on its outbound sides, into rectangles and right triangles with subdivided legs and hypotenuse that is part of an outbound side. The partitioning introduces new subdivision points to the inbound side, which is an outbound side of the parent; these points will be further propagated up the tree. We may also introduce new subdivision points to the outbound sides; these points are resolved later.
Assume the triangle is ace and its inbound side is horizontal base ae. If ̸ cae 9 Root c e a Figure 6. Triangle ace is backwards obtuse.
and ̸ cea are both acute, then we partition ace by sending vertical lines from c and from all subdivision points onto ae. We then draw horizontal lines from each subdivision point up to the last possible vertical, as shown in Figure 7(a). This partitioning introduces new subdivisions only to the inbound side (and to some internal verticals). If one of ̸ cae and ̸ cea is right, then we partition in the same way. This time, however, the partitioning introduces new subdivisions to the vertical side of ace, which may be an inbound side of a child triangle; these points will be resolved in the downward pass of the algorithm.
So assume that one of ̸ cae and ̸ cea is obtuse, say ̸ cea. We draw horizontal lines from each subdivision point on ac and ce across to the other side. Now from vertex e, from each subdivision point, and from each new vertex (where a horizontal hits ac or ce), extend a vertical line, up or down, until it contacts the last possible horizontal, base ae counting as a horizontal. See Figure 7(b). As in the case of the right angle, this partitioning introduces new subdivisions to the outbound sides as well as to the inbound side. The subdivision points on the outbound sides (inbound sides of children) are resolved in the next—downward—pass of the algorithm.
The downward pass triangulates in a preorder traversal of T.
Each original triangle is either unchanged (which implies that it is not backwards obtuse) or it has been partitioned as above into rectangles and small triangles with hypotenuse part of an original outbound side. Unchanged nonobtuse triangles are triangulated e c a e c a Figure 7. Within obtuse triangles, the upward pass leaves subdivisions only on legs.
10 using Lemma 6. Unchanged obtuse triangles are triangulated using the algorithm of Section 2. Finally a partitioned triangle is handled as follows. Subdivisions added to sides of rectangles are passed straight through; this adds yet more subdivision points to the legs of the small triangles and slices rectangles into smaller rectangles that can be triangulated at the end. Then the small triangles with hypotenuses along outbound edges are triangulated in any order using the algorithm of Section 2.
The total number of vertices (subdivision points and apexes of obtuse triangles) to propagate downwards is O(n2). This bound follows from the fact that, in the upwards pass of the algorithm, each apex of a backwards obtuse angle adds only O(1) new subdivision points per original triangle. The downwards pass of the algo-rithm is quadratic as before, so overall we have obtained a nonobtuse triangulation with O(n4) triangles. 2 In the case that the dual graph of the triangulation is simply a path, we can improve the size bound to O(n2) by exploiting the fact that each triangle has an exterior side. Incidentally, this condition defines an interesting class of polygons: call a polygon a path polygon if it has a Steiner triangulation whose dual is a path. Equivalently, a path polygon is one that can be swept by an extensible line segment with endpoints on its boundary, without retracing any area. Path polygons simultaneously generalize monotone polygons and spiral polygons (polygons with only one chain of reflex vertices); they can be recognized in time O(n2) by dynamic programming.
Theorem 4.
Assume we are given a triangulation of a polygon with n vertices, such that the planar dual of the triangulation is a path. Then this triangulation can be refined into a nonobtuse triangulation with O(n2) triangles.
Proof.
Assume we are at a generic step of the upwards pass, in original triangle ace. Assume outbound side ac is subdivided, ce is exterior, and horizontal base ae is inbound. We distinguish three cases. If |̸ cae| ≥90◦, then the procedure of Section 2 will triangulate ace in the downwards pass of the algorithm, adding new subdivisions only to exterior side ce; thus, the upwards pass of the algorithm need not partition ace. In the second case, |̸ ace| ≥90◦, and we drop the altitude from c to ae, forming subdivision point c′ on ae. Triangle c′ce has an exterior hypotenuse, so it requires no further subdivision. The other triangle, acc′, has reduced to the third case, in which both angles alongside the subdivided side ac are acute.
In the third case, the opposite vertex e projects perpendicularly onto the sub-divided side ac. Add edges from e to the subdivision points on either side of its perpendicular projection e′, as in Lemma 6. This splits ace into either two or three triangles, depending on whether there exist subdivisions on each side of e′. An ob-tuse triangle with hypotenuse ce can be handled without further subdivision, even if—as will be the case in lower levels of the recursion—side ce is not exterior but only outbound. A nonobtuse triangle without subdivisions—such as a triangle in the middle—can be handled as is by the downwards pass, since Lemma 6 will apply.
The last triangle (leftmost in Figure 8) has at least one fewer subdivision point, so we partition it recursively. (The recursive call starts by dropping an altitude to ae, as the second case necessarily applies.) 11 c e a Figure 8. The upward pass on a triangle in a path polygon.
Notice that the upward pass only leaves behind subdivision points that do not propagate beyond their original triangle in the downward pass. In other words, each subdivision on ac either drops an altitude to ae or is left in a triangle that the downward pass processes towards exterior side ce. Thus the downward pass results in O(n2) complexity overall. 2 5. Conclusions and Open Problems We have shown how to triangulate arbitrary polygons using a polynomial number of right and acute triangles. This result demonstrates a strong separation between the complexities of two desirable properties of finite element meshes. Forbidding angles larger than 90◦incurs only polynomial cost in size, but forbidding angles smaller than a fixed bound incurs cost dependent upon the geometry of the domain.
There are still a number of open problems in “mesh generation theory”.
First of all, we would like to extend the results of Section 4. The general version of the refinement problem is to refine an arbitrary straight-line planar graph. In particular, this would require an algorithm for triangulated polygons with holes, and further, a way to handle line-segment holes so that Steiner points on each side of a line-segment hole match up. As mentioned above, an interesting special case of refining planar graphs is simultaneously triangulating the inside and outside of a polygon. Edelsbrunner and Tan17 have recently given the first polynomial-size solution to a problem related to refining planar graphs: finding a “conforming Delaunay triangulation”, that is, a set of Steiner points such that the Delaunay triangulation of all vertices contains the input graph.
Open Problem 1.
Can every straight-line planar graph be refined into a polynomial-size nonobtuse triangulation?
Second, the question of lower bounds is interesting, especially in light of Epp-stein’s recent subquadratic algorithm for convex polygons.18 Open Problem 2.
Can any nontrivial lower bound be shown for the size of a nonobtuse triangulation?
Paterson19 observed that for the problem of refining a given triangulation, the complexity must be quadratic, even for convex polygons and a largest angle bound arbitrarily close to 180◦. His example is a convex polygon with an “accordion” triangulation, shown in Figure 9. There are Θ(n) long, skinny, nearly horizontal 12 Figure 9. Lower bound example for refinement.
triangles; the polygon also has Θ(n) vertices evenly spaced along the top of this stack of triangles. Because each vertex at the top must have a “downwards” edge, there must be a path downward from each vertex through the stack of triangles.
Separate paths cannot merge because of the angle bound.
Thus each path has complexity Ω(n), and the total triangulation has complexity Ω(n2). Salzberg et al.
gave a similar lower-bound example for their learning problem.16 Third, it would be interesting to give an algorithm that uses only acute angles, so that all circumcenters lie in the interiors of their elements. By carefully warping rectangles and right triangles, it is possible to turn our linear-size nonobtuse trian-gulations of point sets into acute triangulations (included in the journal version of our previous paper6). For polygon input, however, some new ideas are needed, as it is not possible to divide an obtuse triangle into acute triangles, adding subdivision points only to the base.
Finally, there remains the question of extending these results to higher dimen-sions.
We believe the correct analog of a nonobtuse triangle is a simplex that contains its circumcenter; this generalization most closely follows the mesh gen-eration application.
Perhaps the size bound for a nonobtuse triangulation of a d-dimensional polyhedron will turn out to be O(nd), matching a lower-bound ex-ample for the inside-outside triangulation problem.16 Acknowledgements We would like to thank Warren Smith for drawing our attention to this problem, Anna Lubiw for the term “path polygon”, and David Dobkin, John Gilbert, and Mike Paterson for some helpful discussions.
References 1. H. Edelsbrunner, T.S. Tan, and R. Waupotitsch. A polynomial time algorithm for the minmax angle triangulation. 6th ACM Symp. Comp. Geom., 1990, 44–52. Final version to appear in SIAM J. Stat. Sci. Comput.
2. I. Babuˇ ska and A. Aziz. On the angle condition in the finite element method. SIAM.
J. Numer. Analysis 13 (1976), 214–227.
3. I. Fried. Condition of finite element matrices generated from nonuniform meshes.
AIAA J. 10 (1972), 219–221.
13 4. B.S. Baker, E. Grosse, and C.S. Rafferty.
Nonobtuse triangulation of polygons.
Discrete and Comp. Geom. 3 (1988), 147–168.
5. M. Bern and J.R. Gilbert. Drawing the planar dual. To appear in Inform. Proc.
Letters, 1992.
6. M. Bern, D. Eppstein, and J.R. Gilbert. Provably good mesh generation. 31st IEEE Symp. Found. Comp. Sci., 1990, 231–241. Final version to appear in appear in J.
Comput. Syst. Science.
7. D.T. Lee and B.J. Schachter. Two algorithms for constructing a Delaunay triangu-lation. Int. J. of Computer and Information Sciences 9 (1980), 219–242.
8. D. Mount and A. Saalfeld. Globally-equiangular triangulations of co-circular points in O(n log n) time. 4th ACM Symp. Comp. Geom., 1988, 143–152.
9. R. Sibson. Locally equiangular triangulations. Computer J., 21 (1978), 243–245.
10. D. Eppstein. The farthest point Delaunay triangulation minimizes angles. Comp.
Geom. Theory and Applic. 1 (1992), 43–48.
11. D.T. Lee and A.K. Lin.
Generalized Delaunay triangulation for planar graphs.
Discrete and Comp. Geom. 1 (1986), 201–217.
12. L.P. Chew. Constrained Delaunay triangulations. Algorithmica 4 (1989), 97–108.
13. M. Bern, D. Eppstein, H. Edelsbrunner, S. Mitchell, and T.S. Tan. Edge insertion for optimal triangulations. 1st Latin American Symp. on Theoretical Informatics, 1992, 46–60.
14. J.L. Gerver. The dissection of a polygon into nearly equilateral triangles. Geom.
Dedicata 16 (1984), 93–106.
15. M. Bern and D. Eppstein. Polynomial-size nonobtuse triangulation of polygons. 7th ACM Symp. Comp. Geom., 1991, 342–250.
16. S. Salzberg, A. Delcher, D. Heath, and S. Kasif. Learning with a helpful teacher.
12th Int. Joint Conf. on Art. Intelligence, Sydney, Australia, 1991.
17. H. Edelsbrunner and T.S. Tan. An upper bound for conforming Delaunay triangu-lations. Tech. Report UIUCDCS-R-92-1739, U. of Illinois at Urbana-Champaign.
Also to appear in 8th ACM Symp. Comp. Geom., 1992.
18. M. Bern, D. Dobkin, and D. Eppstein. Triangulating polygons without large angles.
To appear in 8th ACM Symp. Comp. Geom., 1992.
19. M.S. Paterson. Personal communication, 1990.
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Infertility
Summarytoggle arrow icon
Infertility is the inability to achieve pregnancy after 12 months of regular unprotected sexual intercourse in female individuals < 35 years of age and after 6 months in those ≥ 35 years of age, or the need for medically assisted reproduction to achieve pregnancy. Causes of female infertility include diminished ovarian reserve, ovulatory dysfunction, and tubal, pelvic, uterine, and cervical disorders. Causes of male infertility include male hypogonadism and sperm transport disorders. Diagnosis involves assessing both partners to determine the underlying causes and typically includes hormone tests to assess ovulatory function, imaging to evaluate for uterine abnormalities and tubal patency, and semen analysis. Management includes treatment of the underlying cause and often medically assisted reproduction, including ovulation induction, intrauterine insemination, and assisted reproductive technology (e.g., in vitro fertilization).
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Definitionstoggle arrow icon
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Epidemiologytoggle arrow icon
Infertility affects approximately 1 in 6 people in their lifetime.
Epidemiological data refers to the US, unless otherwise specified.
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Etiologytoggle arrow icon
Female infertility
Male infertility
Sperm transport disorders
The cause of infertility remains unexplained in up to a third of affected couples.
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Diagnosistoggle arrow icon
General principles
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Female individualstoggle arrow icon
Initial assessment
Approach
Ovulation tests
Options include:
Imaging
Further assessment
Further assessment is typically performed by reproductive endocrinology and may include the following.
Although ovarian reserve testing can predict responsiveness to ovarian stimulation, it does not predict reproductive potential better than the patient's age and should not be performed to evaluate reproductive potential in female individuals without infertility.
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Male individualstoggle arrow icon
Initial assessment
Further assessment
Further assessment is typically performed by urology and may include the following.
Antisperm antibodies form when the blood testis barrier is disrupted, usually due to infections or trauma of the male genital tract, and can lead to immobilization and agglutination of sperm or have a spermatoxic effect.
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Managementtoggle arrow icon
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Ovulation induction
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190153 | https://math.stackexchange.com/questions/520959/help-in-question-related-to-locus-of-pair-of-tangent-to-a-circle | coordinate systems - Help in question related to locus of pair of tangent to a circle? - Mathematics Stack Exchange
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Help in question related to locus of pair of tangent to a circle? [closed]
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This the question in my text-book:
The tangent to x 2+y 2=a 2 x 2+y 2=a 2 having inclination α α and β β intersect at P P. If cot α cotα + cot β=0 cotβ=0, then the locus of P P is...
I really don't know how to approach this question so any help will be appreciable.
circles
coordinate-systems
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edited Mar 20, 2016 at 10:56
Kamil Jarosz
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DeiknymiDeiknymi
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cot α+cot β=0⟹sin α cos β+sin β cos α=0 cotα+cotβ=0⟹sinα cosβ+sinβ cosα=0 What else can be done with this?abiessu –abiessu 2013-10-10 03:03:19 +00:00 Commented Oct 10, 2013 at 3:03
thank's but i did that much myself but what after that Deiknymi –Deiknymi 2013-10-10 03:05:30 +00:00 Commented Oct 10, 2013 at 3:05
Ah, perhaps that would be good to add to the question so that we don't waste your time telling you things you already know... I presume that you have already applied the "sine of sum of angles" formula then?abiessu –abiessu 2013-10-10 03:18:56 +00:00 Commented Oct 10, 2013 at 3:18
ah! no i didn't do it can you elaborate it a little bit Deiknymi –Deiknymi 2013-10-10 03:25:00 +00:00 Commented Oct 10, 2013 at 3:25
Is this the answer:The family of all circles with radius > a?rnjai –rnjai 2013-10-10 03:40:03 +00:00 Commented Oct 10, 2013 at 3:40
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HINT:
As cot α+cot β=0 cotα+cotβ=0 we have tan α=−tan β=tan(−β)tanα=−tanβ=tan(−β)
⟹α=n π−β⟹α=n π−β where n n is any integer
So, here α=−β α=−β or π−β π−β
From the Article 148,150 148,150 of The elements of coordinate geometry by Loney,
the equation of the tangent at (x 1,y 1)(x 1,y 1) of the Circle x 2+y 2=a 2 x 2+y 2=a 2 is x x 1+y y 1=a 2 x x 1+y y 1=a 2
Now, the Parametric Equation of the circle is x=a cos θ,y=a sin θ x=a cosθ,y=a sinθ
So, the equation of the tangent becomes x cos θ+y sin θ=a⟺y=−x cot θ+a csc θ x cosθ+y sinθ=a⟺y=−x cotθ+a cscθ whose gradient is −cot θ=cot(−θ)−cotθ=cot(−θ) or cot(π−θ)cot(π−θ)
Replace θ θ with α,β α,β to find two simultaneous equation in sin β,cos β sinβ,cosβ
Eliminate β β using sin 2 β+cos 2 β=1 sin 2β+cos 2β=1
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answered Oct 10, 2013 at 4:33
lab bhattacharjeelab bhattacharjee
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Starting with cot α+cot β=0 cotα+cotβ=0, we have the following:
sin α cos β+sin β cos α=0 sinα cosβ+sinβ cosα=0
⟹sin(α+β)=0⟹α+β=0+n π⟹sin(α+β)=0⟹α+β=0+n π
Then we have that α=n π−β α=n π−β. From this, we can conclude that one of the x,y x,y coordinates of P P is 0 0. Specifically, the angle of incidence is mirrored about the 0+n π 0+n π angle, and the circle is centered at the origin (as specified by the equation of it), and thus the tangent lines are mirrored about the x x or y y axis as well.
For the non-zero coordinate, consider the triangle whose hypotenuse is that coordinate, and which uses a a as the length of one of its legs. Note that the third leg of this triangle is the segment along one of the tangent lines described in the question stretching from the tangent point with the circle to the axis where it meets in the angle α α. The angle α α is complementary to the angle adjacent to the a a leg in this triangle, so we would use a csc α a cscα as the value of the length of the hypotenuse, which means that our point P P occurs at
P=(±a sin α,0)P=(±a sinα,0)
or
P=(0,±a sin α)P=(0,±a sinα)
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edited Oct 16, 2013 at 14:18
answered Oct 10, 2013 at 3:41
abiessuabiessu
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Can you explain why the x coordinate of P is 0?Betty Mock –Betty Mock 2013-10-12 18:57:47 +00:00 Commented Oct 12, 2013 at 18:57
@BettyMock: thank you for challenging my derivation as well, I was incorrect. My answer has the corrected logic.abiessu –abiessu 2013-10-13 14:21:07 +00:00 Commented Oct 13, 2013 at 14:21
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We use the following notations:-
The center of the circle is O.
P is at (X, Y).
The points of contacts are A and B.
Some of the answers confirmed that the tangents cut each other at right angles. We continue from that point on and finish the rest (which is then just a one line proof.)
From the figure, OAPB is a square. Thus,
X 2+Y 2=O P 2=A B 2=2 a 2 X 2+Y 2=O P 2=A B 2=2 a 2
And hence the locus of P is x 2+y 2=2 a 2 x 2+y 2=2 a 2
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answered Oct 10, 2013 at 16:36
MickMick
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Furthermore, the 4 corner points [namely, (a, a), (a, -a), (-a, a) and (-a, -a)] should be excluded from the locus because the given condition restricts the inclinations of the tangents from being 0.Mick –Mick 2013-10-10 16:58:38 +00:00 Commented Oct 10, 2013 at 16:58
Actually, I see no reason why the tangent lines should meet at a right angle. Why do you think that they do?abiessu –abiessu 2013-10-11 11:29:16 +00:00 Commented Oct 11, 2013 at 11:29
I agreed that I did not check the original question and just assumed the correctness of some of the analyses. As pointed out, these analyses were not necessarily correct. However, as I have already pointed out in my work, I just took it from there. Based on the orthogonal assumption (if allowed), I did provide a neat finishing touch.Mick –Mick 2013-10-11 16:18:14 +00:00 Commented Oct 11, 2013 at 16:18
That is a very neat locus for the one scenario. :-)abiessu –abiessu 2013-10-11 17:10:00 +00:00 Commented Oct 11, 2013 at 17:10
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The answer which was checked and upvoted was wrong. Hope this one is better.
Let T be the intersection point of the 2 lines.
If cot α α +cot β β = 0 the same relationship holds for the tan which is 1/cot. So tan α α = -tan β β and α=−β±n π α=−β±n π. Let us assume that α=−β α=−β and that line 1 intersects the x-axis at point (b,0). Line 2 is the same line with negative slope, so it has to intersect the x-axis at the point (-b,0).
So the triangle with vertices (-b,0),(b,0) and T is isoceles. If we knew what b is, we would be done.
What we do know is that line 1 is tangent to the circle at a point p = a(c o s ϕ,s i n ϕ)a(c o s ϕ,s i n ϕ) where ϕ=π 2−θ ϕ=π 2−θ. This is because the radius from (0,0) to a(c o s ϕ,s i n ϕ)a(c o s ϕ,s i n ϕ) is perpendicular to line 1; so the triangle with vertices (0,0),(0,b) and a(c o s ϕ,s i n ϕ)a(c o s ϕ,s i n ϕ) is a right triangle. That doesn't leave any choice for ϕ ϕ.
The distance from p to the y-axis is then a c o s ϕ a c o s ϕ. The height h of the isoceles triangle with vertices a(c o s ϕ,s i n ϕ)a(c o s ϕ,s i n ϕ),(−a c o s ϕ,a s i n ϕ)(−a c o s ϕ,a s i n ϕ), and T is h = a(c o s ϕ)(s i n θ)a(c o s ϕ)(s i n θ) (since θ θ is the base angle for that triangle).
So the height of the entire triangle with vertices at (-b,0), (b,0) and T is a s i n ϕ+a(c o s ϕ)(s i n θ)a s i n ϕ+a(c o s ϕ)(s i n θ) which you can simplify down to a s i n θ(1+s i n θ)a s i n θ(1+s i n θ).
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edited Oct 12, 2013 at 22:45
answered Oct 10, 2013 at 3:33
Betty MockBetty Mock
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1 tan α⋅tan β=−1 tanα⋅tanβ=−1 is the condition for orthogonality , right?lab bhattacharjee –lab bhattacharjee 2013-10-10 03:51:56 +00:00 Commented Oct 10, 2013 at 3:51
I don't think this analysis is correct, please consider proving more clearly how the two tangent lines are perpendicular to each other.abiessu –abiessu 2013-10-11 11:31:50 +00:00 Commented Oct 11, 2013 at 11:31
1 Yes, I agree with lab. α α and β β being the inclinations of the two tangents ⟹tan α⋅tan β=−1⟹tanα⋅tanβ=−1 if the two tangents were indeed perpendicular, since the tangent (function) of the inclination gives the slope of the line. So yes, I also agree with abiessu that this analysis is incorrect.Jose Arnaldo Bebita Dris –Jose Arnaldo Bebita Dris 2013-10-11 11:49:33 +00:00 Commented Oct 11, 2013 at 11:49
You guys are right -- I didn't show the lines were orthogonal, and this answer is not correct.Betty Mock –Betty Mock 2013-10-12 18:53:10 +00:00 Commented Oct 12, 2013 at 18:53
I'm confused where ϕ,θ ϕ,θ come from, and how P P could possibly be a cos ϕ≤a a cosϕ≤a in distance from the y y-axis. The answer is (±a sin α,0)(±a sinα,0), not any quantity in direct relationship with a linear combination of a cos α,a sin α a cosα,a sinα.abiessu –abiessu 2013-10-13 14:29:21 +00:00 Commented Oct 13, 2013 at 14:29
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I think it is simply the outside part of the horizontal line through the center (as P must be outside).
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answered Oct 13, 2013 at 15:29
w wichiramalaw wichiramala
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What does this mean? Can you write down what the coordinates of P P are?abiessu –abiessu 2013-10-13 16:35:39 +00:00 Commented Oct 13, 2013 at 16:35
This implies t a n α+t a n β=0 t a n α+t a n β=0. Hence the 2 slopes are opposite. Thus P is on X-axis, except inside the circle. The points on Y-axis outside the circle is also valid.w wichiramala –w wichiramala 2013-10-16 09:29:28 +00:00 Commented Oct 16, 2013 at 9:29
Good point, I updated my answer to account for this as well.abiessu –abiessu 2013-10-16 14:20:14 +00:00 Commented Oct 16, 2013 at 14:20
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We know that the equation of any tangent to the circle x 2+y 2=a 2 x 2+y 2=a 2 is y−m x=a 1+m 2−−−−−−√y−m x=a 1+m 2.
In this question let the tangent passes through a genearl point P(h,k)P(h,k) so this means
k−m h=a 1+m 2−−−−−−√k−m h=a 1+m 2
Squaring both sides we get-
(k−m h)2=(a 1+m 2−−−−−−√)2(k−m h)2=(a 1+m 2)2
m 2(h 2−a 2)−2 m k h+k 2−a 2=0 m 2(h 2−a 2)−2 m k h+k 2−a 2=0
Which is a quadratic in m m as two tangents can be drawn passing through the given point P P let there slopes be m 1=tan α,m 2=tan β m 1=tanα,m 2=tanβ then-
m 1+m 2=2 k h h 2−a 2 m 1+m 2=2 k h h 2−a 2
Since cot α+cot β=0 cotα+cotβ=0 also means tan α+tan β=0 tanα+tanβ=0 so m 1+m 2=0 m 1+m 2=0 thus which means k h=0 k h=0 replacing h h by x x and k k by y y we get the locus of P P as x y=0 x y=0
Remark-abiessu has given a much general/better answer and I have posted this answer because the answer given is just x y=0 x y=0(incidentally I also have the same book with the same question as has the OP) so this method will also work.If we just want to match the answers .Still This should not be considered a general answer.
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edited Mar 20, 2016 at 11:08
answered Mar 20, 2016 at 10:53
FreelancerFreelancer
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190154 | https://diamhomes.ewi.tudelft.nl/~pgroeneboom/mono.pdf | A monotonicity property of the power function of multivariate tests Piet Groeneboom and Donald R. Truax Delft University of Technology and University of Oregon Abstract Let S = Pn k=1 XkX′ k, where the Xk are independent observations from a 2-dimensional normal N(µk, Σ) distribution, and let Λ = Pn k=1 µkµ′ kΣ−1 be a diagonal matrix of the form λI, where λ ≥0 and I is the identity matrix. It is shown that the density φ of the vector ˜ ℓ= (ℓ1, ℓ2) of characteristic roots of S can be written as G(λ, ℓ1, ℓ2)φ0(˜ ℓ), where G satisfies the FKG condition on R3 +. This implies that the power function of tests with monotone acceptance region in ℓ1 and ℓ2, i.e. a region of the form {g(ℓ1, ℓ2) ≤c}, where g is nondecreasing in each argument, is nondecreasing in λ. It is also shown that the density φ of (ℓ1, ℓ2) does not allow a decomposition φ(ℓ1, ℓ2) = G(λ, ℓ1, ℓ2)φ0(˜ ℓ), with G satisfying the FKG condition, if Λ = diag(λ, 0) and λ > 0, implying that this approach to proving monotonicity of the power function fails in general.
Key words and phrases: monotonicity of power functions, noncentral Wishart matrix, char-acteristic roots, orthogonal groups, Euler angles, correlation inequalities, hypergeometric functions of matrix arguments, FKG inequality, pairwise total positive of order two.
1 1 Introduction Let X be a normally distributed random p × n matrix with expectation EX = µ and in-dependent columns with common covariance matrix Σ. Here and in the sequel we assume n ≥p. Let ˜ ℓdenote the vector of characteristic roots of XX′ and let ˜ λ denote the vector of characteristic roots of the noncentrality matrix µµ′Σ−1. It is shown in Perlman and Olkin (1980) that any test of the hypothesis µ = 0 versus µ ̸= 0 with acceptance region {g(˜ ℓ) ≤c}, where g is nondecreasing in each argument, is unbiased. Furthermore they make the con-jecture that the power function of such a test is nondecreasing in each component λi of the vector of noncentrality parameters ˜ λ and suggest that this result could be proved by showing that the density of φ of ˜ ℓcan be written φ(˜ ℓ) = G(˜ λ˜ ℓ)φ0(˜ ℓ), where G is pairwise TP2 (totally positive of order 2) in the pairs (ℓi, ℓj), i ̸= j, and (λi, ℓj), 1 ≤i, j ≤p (loc. cit. Proposition 2.6 (ii) and Remark 3.2).
We show in this note that the suggested TP2 property does not hold in general (see section 4), but that the following partial result of this type does hold: if the dimension of the observations equals 2 and ˜ λ = (λ, λ), then the density φ of ˜ ℓcan be written φ(˜ ℓ) = G(λ, ˜ ℓ)φ0(˜ ℓ), where G satisfies the FKG condition on R3 + (we use the notation R+ = {x ∈ R : x ≥0}). This means G(λ1, ˜ ℓ)G(λ2, ˜ ℓ) ≤G(λ1 ∧λ2, ˜ ℓ1 ∧˜ ℓ2)G(λ1 ∨λ2, ˜ ℓ1 ∨˜ ℓ2), (1.1) for (λi, ˜ ℓi) ∈R3 +, i = 1, 2. Here we use the conventions x ∧y = min(x, y), x ∨y = max(x, y), if x, y ∈R and x∧y = (x1∧y1, . . . , xn∧yn), x∨y = (x1∨y1, . . . , xn∨yn), if x = (x1, . . . , xn) ∈Rn and y = (y1, . . . , yn) ∈Rn. since in our case the function G is strictly positive on R3 +, proving that G satisfies the FGK condition on R3 + is equivalent to proving that G is pairwise TP2 on R3 + (cf. Perlman and Olkin (1980), Remark 2.3). This means that the power function is monotone “on the diagonal” in the 2-dimensional case. We believe that this property holds generally (i.e. also for dimensions higher than 2), but were not able to adapt our method of proof to the higher dimensional case.
The key lemmas in our approach are given in Section 2. They give integral inequalities for diagonal elements of an orthogonal matrix under densities of an exponential type with respect to Haar measure on the orthogonal group. These lemmas are similar in spirit to correlation inequalities for spin configurations in Kelly and Sherman (1968).
The results in Section 3 follow easily from the Lemmas in Section 2 by using the integral representation of the hypergeometric function 0F1 ¡ 1 2n; 1 4Λ, L ¢ , where Λ = diag(λ1, λ2), L = diag(ℓ1, ℓ2), which is given in James (1961). If Λ = λI, with λ ≥0, this integral reduces to an integral over the orthogonal group O(n) (instead of a repeated integral involving the orthogonal groups O(2) and O(n)). The density φ(˜ ℓ) of the characteristic roots ℓ1 and ℓ2 of XX′ can then be written φ(˜ ℓ) = G(λ, ˜ ℓ)φ0(˜ ℓ), where G(λ, ˜ ℓ) = 0F1 ¡ 1 2n; 1 4λI, L ¢ exp(−λ) and φ0 is the density under the null hypothesis µ = 0. The TP2 properties of the function G follow from the corresponding properties of the hypergeometric function 0F1 ¡ 1 2n; λI, L ¢ . The 2 monotonicity result for the power function follows from this by using the FKG inequality due to Fortuin, Ginibre and Kasteleyn (1971). For an exposition on the FKG inequality and its uses we refer to Kemperman (1977) and Perlman and Olkin (1980).
2 Some preparatory lemmas Lemma 2.1 Let a1 ≥a2 ≥0 and let H be an n × n orthogonal matrix, where n ≥2. Then the diagonal elements h11 and h22 have a non-negative covariance under the density f(h11, h22) = exp ( 2 X i=1 aihii ) Á Z O(n) exp ( 2 X i=1 aihii ) dH (2.1) with respect to Haar measure on O(n), where dH denotes Haar measure on O(n).
Proof. First consider the special orthogonal group SO(n) of orthogonal matrices with deter-minant equal to one. Any H ∈SO(n) can be written as a product Hn−1 . . . H1 of rotations H1, . . . , Hn−1, where Hk = H(1)(θ1k) . . . H(k)(θkk) (2.2) and H(i)(θik) is a rotation by the angle θik in the (xi, xi+1)-plane, oriented such that the rotation from the i-th unit vector ei to the (i + 1)th unit vector ei+1 is positive. The range of the angles θik is as follows: ½ 0 ≤θik < 2π, i = 1, 0 ≤θik < π, i > 1.
(2.3) These parameters are called Euler angles, see e.g. Vilenkin (1968), chapter IX. In terms of these parameters, Haar measure on SO(n) is given by dH = cn n−1 Y k=1 k Y j=1 sinj−1 θjkdθjk (2.4) where cn = n Y k=1 Γ(k/2)/(2πk/2), (2.5) see Vilenkin (1968), p. 439. By induction it is seen that hn1 = n−1 Y k=1 sin θkk, h1n = (−1)n−1 n−1 Y k=1 sin θk,n−1.
(2.6) Note that the distribution of (h11, h22) under Haar measure on the orthogonal group is the same as the distribution of (ϵ1hn1, ϵ2h1n), where ϵ1 and ϵ2 are independent random variables with the same distribution P{ϵi = 1} = P{ϵi = −1} = 1 2 and (hn1, h1n) is distributed according to Haar measure on SO(n), independent of (ϵ1, ϵ2). Thus, taking the expectation 3 with respect to (ϵ1, ϵ2), we get Z O(n) h11h22f(h11, h22) dH = c1E ( ϵ1ϵ2 Z 2π 0 dθ11 Z 2π 0 dθ1,n−1 Z π 0 dθ22 Z π 0 dθ2,n−1 . . .
Z π 0 n=1 Y k=1 (sin θkk sin θk,n−1) (sin θn−1,n−1)n−2 · n−2 Y k=1 ³ sink−1 θkk sink−1 θk,n−1 ´ ·f à ϵ1a1 n−1 Y k=1 sin θkk, ϵ2a2 n−1 Y k=1 sin θk,n−1 !
dθn−1,n−1 ) = c2 Z π/2 0 dθ11 Z π/2 0 dθ1,n−1 Z π/2 0 dθ22 Z π/2 0 dθ2,n−1 . . .
Z π/2 0 n−1 Y k=1 (sin θkk sin θk,n−1) sinh à a1 n−1 Y k=1 sin θkk !
· sinh à a2 n−1 Y k=1 sin θk,n−1 n−2 Y k=1 sink−1 θkk sink−1 θk,n−1 !
· sinn−2 θn−1,n−1 dθn−1,n−1.
Note that for n = 2 there is only one parameter θ11, for n = 3 there are three parameters θ11, θ22, θ33, θ13, θ23, etc. The constants c1 and c2 are defined by c1 = ½Z 2π 0 dθ11 Z 2π 0 dθ1,n−1 Z π 0 dθ22 Z π 0 dθ2,n−1 . . .
Z π 0 n−2 Y k=1 ³ sink−1 θkk sink−1 θk,n−1 ´ sinn−2 θn−1,n−1 dθn−1,n−1 )−1 and c2 = (Z π/2 0 dθ11 Z π/2 0 dθ1,n−1 . . .
Z π/2 0 π/2 Y 0 cosh à a1 n−1 Y k=1 sin θkk !
cosh à a2 n−1 Y k=1 sin θk,n−1 !
· Ãn−2 Y k=1 sink−1 θkk sink−1 θk,n−1 !
sinn−2 θn−1,n−1 dθn−1,n−1 )−1 Now let S = [0, π/2]2n−3 and define the density q on S by q(θ11, . . . , θn−1,n−1, θ1,n, . . . , θn−2,n−1) 4 = c2 cosh à a1 n−1 Y k=1 sin θkk !
cosh à a2 n−1 Y k=1 sin θk,n−1 !
(2.7) · (n−2 Y k=1 sink−1 θkk sink−1 θk,n−1 ) sinn−2 θn−1,n−1.
Let ˜ θ = (θ11, . . . , θn−1,n−1, θ1,n−1, . . . , θn−2,n−1), and g1(˜ θ) = Ãn−1 Y k=1 sin θkk !
tanh à a1 n−1 Y k=1 sin θkk !
, (2.8) g2(˜ θ) = Ãn−1 Y k=1 sin θk,n−1 !
tanh à a2 n−1 Y k=1 sin θk,n−1 !
.
(2.9) Then Z O(n) h11h22f(h11, h22)dH = Z π/2 0 dθ11 . . .
Z π/2 0 Ãn−1 Y k=1 sin θkk sin θk,n−1 !
· tanh à a1 n−1 Y k=1 sin θkk !
tanh à a2 n−1 Y k=1 sin θk,n−1 !
q(˜ θ) dθn−1,n−1 (2.10) = E {g1(θ)g2(θ)} where the expectation is taken with respect to the density q on S.
The density q is pairwise TP2, since = ∂2 ∂θij∂θkl log q(˜ θ) ≥0 for any pair of different components θij and θkl of ˜ θ, and since q > 0 on S. Thus, again by the fact that q > 0 on S, it follows that q satisfies the FKG condition on S (cf. Perlman and Olkin (1980), Remark 2.3). Since g1 and g2 are both nondecreasing in each argument on S, the FKG inequality implies E{g1(˜ θ)g2(˜ θ)} ≥Eg1(˜ θ)Eg2(˜ θ) (2.11) (see e.g. Perlman and Olkin (1980), Remark 2.5). By computations similar to those used in computing R O(n) h11h22f(h11, h22) dH it is seen that Z O(n) h11f(h11, h22) dH = Eg1(˜ θ) (2.12) Z O(n) h22f(h11, h22) dH = Eg2(˜ θ) (2.13) The result now follows from (2.10) to (2.13).
2 5 Lemma 2.2 Under the same conditions as in Lemma 2.1, the diagonal elements h11 and h22 of H satisfy Z O(n) hiif(h11, h22) dH ≥0, i = 1, 2, (2.14) where f is given by (2.1).
Proof. Using the notation of the proof of Lemma 2.1 we have Z O(n) h11f(h11, h22) dH = Eg1(˜ θ) = Z S Ãn−1 Y k=1 sin θkk !
tanh à a1 n−1 Y k=1 sin θkk !
q(˜ θ) d˜ θ, (2.15) where S = [0, π/2]2n−3; see (2.7), (2.8) and (2.12). The expression on the right-hand side of (2.15) is clearly non-negative (and strictly positive if a1 > 0). The proof for h22 is completely similar.
2 3 Total positivity and monotonicity Theorem 3.1 Let L = diag(ℓ1, ℓ2) and Λ = diag(λ, λ), where ℓi ≥0, i = 1, 2,, and λ > 0.
Then the hypergeometric function 0F1(1 2n; 1 4Λ, L) is TP2 in (ℓ1, ℓ2) and in (ℓj, λ), j = 1, 2, for each n ≥2.
Proof. We use the following integral representation 0F1 ¡ 1 2n; 1 4Λ, L ¢ = Z O(2) Z O(n) exp n trD′ λH1DℓH′ 2 o dH1dH2, (3.1) where H1 ∈O(2), H2 ∈O(n) and dH1 and dH2 denote Haar measure on O(2) and O(n), respectively; Dℓis a 2 × n matrix defined by (Dℓ)ij = ℓ1/2 i δij and Dλ is a 2 × n matrix defined by (Dλ)ij = λ1/2 i δij where δij is Kronecker’s delta (see e.g. James (1961)). When Λ = diag(λ, λ) we obtain the following integral representation 0F1 ¡ 1 2n; 1 4Λ, L ¢ = Z O(n) exp ½ λ1/2 2 X j=1 ℓ1/2 j hjj ¾ dH (3.2) since in this case Z O(n) exp © trD′ λH1DℓH′ 2 ª dH2 = Z O(n) exp ½ λ1/2 n X i=1 2 X j=1 h(1) ij h(2) ij ℓ1/2 j ¾ dH2 (3.3) = Z O(n) exp ½ λ1/2 2 X j=1 ℓ1/2 j hjj ¾ dH 6 where H1 = (h(1) ij ) and H2 = (h(2) ij ). The last equality in (3.3) holds, since 2 X i=1 2 X j=1 h(1) ij h(2) ij ℓ1/2 j = tr © H1A(L)H′ 2 ª , (3.4) where A(L) is the n × n matrix defined by A(L)ii = ℓ1/2 i , i = 1, 2, and A(L)ij = 0 for other values of (i, j), and where H1 is the n×n orthogonal matrix defined by (H1)ij = h(1) ij , 1 ≤i, j ≤2, (H1)ii = 1, i > 2.
Here we use that the function Ψ : A 7→ Z O(n) exp {trAH} dH, A an n × n matrix, is invariant under transformations A 7→H1A, H1 ∈O(n).
Let F = 0F1 ¡ 1 2n; 1 4Λ, L ¢ . Then ∂2 ∂ℓ1∂ℓ2 log F = 1 4λ(ℓ1ℓ2)−1 2 Z O(n) h11h22 exp ½ λ 1 2 2 X j=1 ℓ1/2 j hjj ¾ dH/F (3.5) −1 2λ(ℓ1ℓ2)−1 2 Z O(n) h11 exp ½ λ1/2 2 X j=1 ℓ1/2 j hjj ¾ dH/F · Z O(n) h22 exp ½ λ1/2 2 X j=1 ℓ1/2 j hjj ¾ dH/F and ∂2 ∂λ∂ℓi log F = 1 4(λℓi)−1 2 Z O(n) hii exp ½ λ 1 2 2 X j=1 ℓ 1 2 j hjj ¾ dH/F +1 4ℓ −1 2 i Z O(n) hii 2 X j=1 ℓ 1 2 j hjj exp ½ λ 1 2 2 X j=1 ℓ 1 2 j hij ¾ dH/F (3.6) −1 4ℓ −1 2 i Z O(n) 2 X j=1 ℓ 1 2 j hjj exp ½ λ 1 2 j 2 X j=1 ℓ 1 2 j hjj ¾ dH/F · Z O(n) hii exp ½ λ 1 2 2 X j=1 ℓ 1 2 j hjj ¾ dH/F 7 By Lemmas 2.1 and 2.2 it follows that 3.5 and 3.6 are nonnegative. Hence F is pairwise TP2 in (ℓ1, ℓ2) and (ℓj, λ), j = 1, 2.
2 The following corollary shows that the power function is monotone “on the diagonal”.
Corollary 3.1 Let ˜ ℓ= (ℓ1, ℓ2) be distributed according to the density φλ(˜ ℓ) = exp(−λ) 0F1 ¡ 1 2n; 1 4Λ, L ¢ φ0(˜ ℓ), (3.7) where Λ = diag(λ, λ), L = diag(ℓ1, ℓ2), φ0(˜ ℓ) = ( k(ℓ1 −ℓ2)(ℓ1ℓ2) 1 2 (n−3) exp © −1 2(ℓ1 + ℓ2) ª , ℓ1 ≥ℓ2 ≥0 0, otherwise (3.8) and k > 0 is a constant such that φ0 is a probability density. Then the function λ → Z R2 g(˜ ℓ)φλ(˜ ℓ)d˜ ℓ, λ ≥0, is nondecreasing for each g which is nondecreasing in the components ℓ1 and ℓ2 of ˜ ℓ.
Proof. Define G(λ, ℓ1, ℓ2) = exp(−λ) 0F1 ¡ 1 2n; λI, L ¢ .
(3.9) Then G > 0 on the rectangle R3 +. Since ∂2 ∂ℓ1∂ℓ2 log G(λ, ℓ1, ℓ2) ≥0 and ∂2 ∂ℓj∂λ log G(λ, ℓ1, ℓ2) ≥0 for each (λ, ℓ1, ℓ2) ∈R3 +, it follows that G is pairwise TP2 on R3 +. Since G > 0 on R3 +, this implies that G satisfies the FKG condition on R3 + (cf. Perlman and Olkin (1980), Remark 2.3). The result now follows from Proposition 2.6 (ii) and Remark 2.7 in Perlman and Olkin (1980).
2 4 A Counterexample We show that the approach to proving monotonicity of the power function by showing that 0F1(1 2n; 1 4, L) is pairwise TP2 (which worked “on the diagonal” in Section 3), fails in general.
Take n = 2, Λ = diag(λ, 0), λ > 0, L = (ℓ1, ℓ2), ℓi ≥0, i = 1, 2. Then by the same line of argument as used in Lemma 2.1 we have ∂2 ∂ℓ1∂ℓ2 0 F1 ¡ 1 2n; 1 4, L ¢ = ∂2 ∂ℓ1ℓ2 Z O(2) Z O(2) exp ½ tr 1 2 H1L 1 2 H′ 2 ¾ dH1dH2 = 1 4λ(ℓ1ℓ2)−1 2 Z O(2) Z O(2) h(1) 11 h(2) 11 h(1) 12 h(2) 12 exp ½ λ 1 2 2 X j=1 h(1) 1j h(2) 1j ℓ 1 2 j ¾ dH1dH2 = 1 π2 λ(ℓ1ℓ2)−1 2 Z π/2 0 dθ1 Z π/2 0 cos θ1 cos θ2 sin θ1 sin θ2 · sinh ³ λ 1 2 ℓ 1 2 1 cos θ1 cos θ2 ´ sinh ³ λ 1 2 ℓ 1 2 2 sin θ1 sin θ2 ´ dθ2, 8 where H1 = (h(1) ij ) and H2 = (h(2) ij ). Define the density q on [0, π/2]2 by q(θ1, θ2) = k. cosh ³ λ 1 2 ℓ 1 2 1 cos θ1 cos θ2 ´ cosh ³ λ 1 2 ℓ 1 2 2 sin θ1 sin θ1 ´ , (4.1) where k > 0 is chosen such that q is a probability and define g1(θ1, θ2) = −cos θ1 cos θ2 tanh ³ λ 1 2 ℓ 1 2 1 cos θ1 cos θ2 ´ g2(θ1, θ2) = sin θ1 sin θ2 tanh ³ λ 1 2 ℓ 1 2 2 sin θ1 sin θ2 ´ .
(4.2) The density q clearly satisfies the FKG condition on S and hence, since g1 and g2 are both increasing in θ1 and θ2 on S, we have by the FKG inequality Eg1(θ1, θ2)g2(θ1, θ2) ≥Eg1(θ1, θ2)Eg2(θ1, θ2), (4.3) where the expectation is taken with respect to the density q on S. Moreover, the inequality in 4.3 is strict (cf. Perlman and Olkin (1980), Proposition 2.4 (ii)). Let F = 0F1(1, Λ, L).
Then ∂2 ∂ℓ1∂ℓ2 log F = µ ∂2 ∂ℓ1∂ℓ2 F ¶ /F −∂F ∂ℓ1 ∂F ∂ℓ2 /F 2 = 1 4λ(ℓ1ℓ2)−1 2 (−Eg1g2 + Eg1Eg2) < 0, (4.4) implying that F is not TP2 in the pair (ℓ1, ℓ2).
However, it is shown by a completely different method in Perlman and Olkin (1980) that any test of the type described in Section 1 has a power function which is increasing in λ, if Λ = diag(λ, 0).
Acknowledgement. We have benefited from helpful remarks by Tom Koornwinder on an earlier draft of this manuscript. We also want to thank Michael Keane and Michael Perlman who encouraged us to publish this manuscript that was originally written in 1983, while the first author was visiting MSRI in Berkeley.
5 References 1. Fortuin, C.M., J. Ginibre, & P.W. Kasteleyn, (1971), Correlation inequalities on some partially ordered sets. Commun.Math.Phys. 22, 89-103.
2. James, A.T. (1955), A generating function for averages over the orthogonal group.
Proc.Roy.Soc. London Ser. A 229, 367-375.
3. James, A.T. (1961), The distribution of noncentral means with known covariance, Ann.
Math. Statist. 32, 874-882.
4. Kelly, D.G. & S. Sherman, (1968), General Griffiths inequalities on correlations in Ising ferromagnets. J. Mathematical Phys. 9, 466-484.
9 5. Kemperman, J.H.B. (1977), On the FKG inequality for measures on a partially ordered space. Indag. Math 39, 313-331.
6. Perlman, M.D. & I.Olkin (1980), Unbiasedness of invariant tests for MANOVA and other multivariate problems. Ann. Statst. 8, 1326-1341.
7. Vilenkin, N.J. (1968), Special functions and the theory of group representations, Trans-lations of Mathematical Monographs, 22, American Math. Soc., Providence, Rhode Island, U.S.A.
10 |
190155 | https://fiveable.me/key-terms/formal-logic-i/inverse | Inverse - (Formal Logic I) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Formal Logic I
Inverse
👁️🗨️formal logic i review
key term - Inverse
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Definition
In logic, the inverse of a conditional statement is formed by negating both the hypothesis and the conclusion of that statement. It’s represented in the form 'If not P, then not Q' for a conditional statement 'If P, then Q'. Understanding the inverse is crucial in exploring logical implications, as it helps in evaluating the truth values and relationships between propositions.
5 Must Know Facts For Your Next Test
The inverse of a conditional statement does not necessarily share the same truth value as the original statement; in fact, they can be false even if the original statement is true.
When you have 'If P, then Q', the inverse would be 'If not P, then not Q', allowing for analysis of different logical scenarios.
Understanding inverses is essential for testing the validity of arguments in logical proofs and reasoning.
Inverses can be particularly useful when discussing fallacies, as recognizing an invalid inverse can help clarify misconceptions about logical relationships.
The truth table for a conditional statement helps illustrate how inverses relate to other forms like the contrapositive and converse.
Review Questions
How does the inverse differ from the original conditional statement in terms of truth values?
The inverse can differ significantly in truth values compared to the original conditional statement. While a conditional statement 'If P, then Q' may be true, its inverse 'If not P, then not Q' could be false. This highlights that just because one condition holds true doesn't guarantee that its negated version will also hold true. Understanding these differences is essential for accurately assessing logical implications.
Why is it important to understand the inverse when working with material conditionals?
Understanding the inverse is important when working with material conditionals because it allows one to critically analyze logical statements and their implications. By examining both a conditional statement and its inverse, one can uncover potential flaws in reasoning or assumptions. This understanding also facilitates better evaluations of logical arguments by highlighting how different forms relate to each other.
Evaluate how recognizing an inverse can impact your ability to identify logical fallacies in arguments.
Recognizing an inverse can greatly enhance your ability to identify logical fallacies within arguments by revealing incorrect assumptions about causal relationships. If someone claims that a false inverse leads to a true conclusion, this can indicate a misunderstanding of logical principles. Being aware of how inverses operate helps you dismantle flawed reasoning and encourages clearer thought processes when analyzing arguments.
Related terms
Contrapositive:
The contrapositive of a conditional statement flips and negates both the hypothesis and conclusion, expressed as 'If not Q, then not P'.
Material Conditional:
A logical connective that expresses a relationship between two propositions, where the conditional statement is considered false only when the first proposition is true and the second is false.
Logical Implication:
A relationship between statements where one statement logically follows from another, often used to evaluate the validity of arguments.
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190156 | https://dictionary.cambridge.org/us/dictionary/english/possess | Cambridge Dictionary +Plus
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English (US)
Meaning of possess in English
possess verb [T] (OWN)
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C1
to have or own something, or to have a particular quality:
I don't possess a single DVD (= I don't have even one DVD).
In the past the root of this plant was thought to possess magical powers.
More examplesFewer examples
We're trying to bring out the artistic talents that many people possess without realizing it.
She had already sold everything of value that she possessed.
Ruth possessed great writing skills.
He was charged with possessing a fake passport.
"I'm arresting you on suspicion of illegally possessing drugs, " said the police officer.
SMART Vocabulary: related words and phrases
Having and owning - general words
acquire
acquisition
alluvion
attach
attach something to something phrasal verb
be endowed with something idiom
bore
fully
holder
interest someone in something phrasal verb
lay
lord/master/mistress/king/queen of all you survey idiom
make something (all) your own idiom
reside in something/someone phrasal verb
revert
revert to something phrasal verb
shared ownership
stake
stake something out phrasal verb
survey
See more results »
possess verb [T] (CONTROL)
(of a wish or an idea) to take control over a person's mind, making that person behave in a very strange way:
[ + to infinitive ] Whatever possessed him to wear that appalling jacket!
SMART Vocabulary: related words and phrases
Controlling and being in charge
agentive
aggrandize
assert your authority
assume
assumption
get the better of someone idiom
get your hooks into someone/something idiom
get/fall into the wrong hands idiom
govern
guiding principle
paternalist
paternalistic
paternalistically
peremptorily
peremptory
slow
tame
well in hand
what someone says, goes idiom
wrangler
See more results »
(Definition of possess from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)
possess | Intermediate English
possess verb [T] (OWN)
Add to word list
Add to word list
to have or own something, or to have a particular quality:
Those states are the countries that possess nuclear weapons.
She possesses the unusual talent of knowing when to say nothing.
possess verb [T] (CONTROL)
(of a desire or an idea) to take control over a person’s mind, making that person behave in a strange way:
I don’t know what possessed me to start yelling like that.
possessor
noun [ C usually sing ] us
/pəˈzes·ər/
I am now the proud possessor of a driver’s license!
(Definition of possess from the Cambridge Academic Content Dictionary © Cambridge University Press)
possess | Business English
possess
verb [ T ]
uk
/pəˈzes/ us
Add to word list
Add to word list
to have or own something:
More than five million German households possess exercise equipment.
These bonds possess many favorable investment attributes.
to have a particular quality:
possess knowledge/qualities/skills Sales managers must possess strong leadership qualities.
(Definition of possess from the Cambridge Business English Dictionary © Cambridge University Press)
Examples of possess
possess
We also possess the equivalent data of the instructors, have access to all coursework, grades, and every student alumni's information.
From International Business Times
This fundamental right to acquire, possess, and sell property is the backbone of opportunity and the most practical means to pursue human happiness.
From Heritage.org
He is a caring husband and father, a thoughtful speaker, and possessed of an inspirational biography.
From The Atlantic
The competitive advantages that seed-producing plants possess has led to their dominance in most contemporary natural habitats.
From Phys.Org
She was also clairvoyant, possessing the ability to "perceive" events before they happened.
From Voice of America
These activists possess clear public interest demands, utilize creative approaches, and evoke positive media reaction.
From Foreign Policy
The area possesses seven unique qualities that combine to make it the place where creatively enthusiastic people want to be.
From Huffington Post
They didn't need to fuss about illusion, because they knew they possessed the real thing.
From Hollywood Reporter
That is, these things, these coins possess so little value that stores let you take one or leave one.
From NPR
Nor does she possess a whit of those other characters' lightheartedness.
From San Francisco Chronicle
The skills those technicians and specialists possess are required to produce a modern tank.
From CNN
But how many can be said to possess both legs and teeth?
From Plain Dealer
For example, a medical student may be brilliant but typically possesses no direct experience when treating a patient.
From Huffington Post
They possess an incredible willingness to do the work that needs to be done.
From Huffington Post
People possess a tolerance threshold for various tasks in the home.
From Phys.Org
These examples are from corpora and from sources on the web. Any opinions in the examples do not represent the opinion of the Cambridge Dictionary editors or of Cambridge University Press or its licensors.
What is the pronunciation of possess?
Translations of possess
in Chinese (Traditional)
擁有, 具有, 控制…
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in Chinese (Simplified)
拥有, 具有, 控制…
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in Spanish
poseer, tener…
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in Portuguese
possuir…
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(व्यक्तिकडे किंवा वस्तु मध्ये) असणे…
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～を所有する, 持つ, 所有（しょゆう）する…
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sahip olmak, ...ın/in sahibi olmak…
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posséder…
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posseir, tenir…
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bezitten…
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எதையாவது வைத்திருப்பது அல்லது சொந்தமாக வைத்திருப்பது அல்லது ஒரு குறிப்பிட்ட தரத்தை வைத்திருப்பது…
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(कोई चीज़) प्राप्त करना या या अधिकार में रखना, कोई विशेष गुण होना…
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ધરાવવું, પાસે હોવું…
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eje…
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äga, ha, besitta…
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memiliki…
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besitzen…
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eie, ha, besitte…
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володіти…
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обладать, владеть…
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ఏదైనా కలిగి ఉండు లేదా దానిని స్వంతం చేసుకొను లేదా ఒక ప్రత్యేక గుణం కలిగి ఉండు…
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يَمْلُك…
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অধিকারী হওয়া বা নির্দিষ্ট থাকা…
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mít, vlastnit…
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memiliki…
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ครอบครอง…
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sở hữu, có…
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posiadać…
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소유하다…
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avere, possedere…
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cut someone some slack
to not judge someone as severely as you usually would because they are having problems at the present time
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possess (OWN)
possess (CONTROL)
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possess (OWN)
possess (CONTROL)
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190157 | https://chem.libretexts.org/Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/02%3A_General_Chemistry_Review/2.04%3A_Using_logarithms_-_Log_vs._Ln | 2.4: Using logarithms - Log vs. Ln - Chemistry LibreTexts
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2: General Chemistry Review
CHEM 2141: Survey of Physical Chemistry
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Fri, 21 Feb 2025 20:05:44 GMT
2.4: Using logarithms - Log vs. Ln
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Contents
1. Home
2. Campus Bookshelves
3. University of North Carolina Charlotte
4. CHEM 2141: Survey of Physical Chemistry
5. 2: General Chemistry Review
6. 2.4: Using logarithms - Log vs. Ln
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CHEM 2141: Survey of Physical Chemistry
Front Matter
1: Introduction
2: General Chemistry Review
3: The First Law of Thermodynamics
4: Entropy and The Second and 3rd Law of Thermodynamics
5: Helmholtz and Gibbs Energies
6: Chemical Kinetics I
7: Chemical Kinetics II
8: Optional- Special topics
9: Quantum Basics
10: Molecular Spectroscopy
Back Matter
2.4: Using logarithms - Log vs. Ln
Last updated Feb 21, 2025
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3: The First Law of Thermodynamics
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Table of contents
1. Contributors and Attributions
A common question exists regarding the use of logarithm base 10 (log or log 10) vs. logarithm base e (ln). The logarithm base e is called thenatural logarithmsince it arises from the integral:
ln(a)=∫1 a d x x
Of course, one can convert from ln to logwith a constant multiplier.
ln(10 loga)=log(a)ln(10)≈2.3025log(a)
but 10 loga=a so
lna≈2.4025loga
The analysis of the reaction order and rate constant using the method of initial rates is performed using the log 10 function. This could have been done using the ln function just as well. The initial rate is given by
r o=k′[A]0 a
The analysis can proceed by taking the logarithm base 10 of each side of the equation
logr o=logk′+alog[A]0
or the ln of each side of the equation
lnr o=lnk′+aln[A]0
as long as one is consistent.
Once can think of the log or the ln as a way to 'linearize data' that has some kind of power law dependence. The only difference between these two functions is a scaling factor (ln10≈2.3025) in the slope. Obviously, if you multiply both sides of the equation by the same number the relative values of the constants remains the same on both sides.
Contributors and Attributions
Stefan Franzen (North Carolina State University)
2.4: Using logarithms - Log vs. Ln is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.
10: Using logarithms - Log vs. Ln is licensed CC BY-NC-SA 4.0.
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Before we dive into understanding the slope of the secant line formula, let's remember what slope and secant mean. Slope is like how steep a hill is when you walk up it. A secant line on a curve is a straight line that touches two points on the curve. If those two points get closer and closer together, the secant line turns into the tangent line, which shows how steep the curve is at that exact point.
In this maths formula article, we will learn the Slope of the Secant Line Formula along with some solved examples.
What is the Slope of Secant Line?
The slope of a secant line refers to the average rate of change between two points on a curve. It represents how the function's values change over an interval between these two points. Mathematically, the slope of a secant line is calculated by dividing the change in the function's output ((\Delta y)) by the change in the input ((\Delta x)) between the two points. This is often symbolized as (\frac{\Delta y}{\Delta x}).
In essence, the slope of a secant line provides an approximation of the rate of change of a function over a finite interval. As the interval between the two points becomes smaller and smaller (approaching zero), the secant line starts to resemble the tangent line at a specific point on the curve. This concept of narrowing the interval to capture instantaneous change leads to the fundamental idea of the derivative in calculus.
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Slope of the Secant Line Formula
We use the slope of a line formula to figure out how steep a secant line is because a secant line is like a regular line. There are different formulas for finding the steepness of a secant line, depending on what we know. Imagine a curve with the equation (y = f(x)), and think about a secant line we draw on this curve.
If ((x_1, y_1)) and ((x_2, y_3)) represent two points along the curve (y = f(x)), across which the secant line extends, then:
Slope of the secant line formula (= \frac{y_2 - y_1}{x_2 - x_1})
When the secant line traverses both the points (a, f(a)) and (b, f(b)), the following holds true:
Slope of the secant line formula (= \frac{f(b) - f(a)}{b - a})
This is commonly known as the "average rate of change of (f(x))" over the interval from (x = a) to (x = b).
When the secant line traverses two points, namely (P) and (Q), with (P) represented as ((x, f(x))) and (Q) as ((x + h, f(x + h))), the following scenario unfolds:
Slope of the secant line formula (= \frac{f(x + h) - f(x)}{x + h - x} = \frac{f(x + h) - f(x)}{h})
Here, (\frac{f(x + h) - f(x)}{h}) is also referred to as the "difference quotient".
Note: The slope of the secant line formula provides the slope of the tangent line formula (equivalent to the function's derivative at that specific point) when (x_{2}\to x_{1}), (or) (b\to a), (or) (h\to 0).
Let's explore a few selling price examples to help us better understand the Slope of the Secant Line Formula.
Slope of the Secant Line Formula Solved Examples
Example 1. Given the function (f(x) = 2x^{2} - 3x + 1), find the average rate of change of (f(x)) between (x = 1) and (x = 3).
Solution.
Using the formula: Slope of the secant line formula (= \frac{f(x + h) - f(x)}{h})
Substitute the values:
(h = 3 - 1 = 2)
Calculate (f(1) = 2(1)^{2} - 3(1) + 1 = 0)
Calculate (f(3) = 2(3)^{2} - 3(3) + 1 = 10)
\frac{f(x + h) - f(x)}{h} = \frac{(10 - 0)}{2} = 5
Thus, the average rate of change of (f(x)) is (5).
Example 2. Calculate the slope of a secant line for the function (f(x) = x^{2}) that joins the two points ((3, f(3))) and ((5, f(5))).
Solution.
Given: (f(x) = x^{2})
Calculate the value of (f(3)) and (f(5)).
(f(3) = 3^{2} = 9)
(f(5) = 5^{2} = 25)
Using the Slope of the secant line formula, we have:
(m = \frac{(f(b) - f(a))}{(b - a)})
(\Rightarrow m = \frac{(f(5) - f(3))}{(5 - 3)})
(\Rightarrow m = \frac{(25 - 9)}{2})
(\Rightarrow m = \frac{16}{2})
(\Rightarrow m = 8)
Therefore, the slope of a secant line is (8).
Example 3. The slope of a secant line that joins the two points ((x, 7)) and ((9, 2)) is (5). Find the value of (x).
Solution.
Given: ((x_1, y_1) = (x, 7)), ((x_2, y_2) = (9, 2)) and (m = 5)
Using the Slope of the secant line formula, we have:
(m = \frac{(y_2 - y_1)}{(x_2 - x_1)})
(\Rightarrow 5 = \frac{(2 - 7)}{(9 - x)})
(\Rightarrow 5 = \frac{-5}{(9 - x)})
(\Rightarrow 45 - 5x = -5)
(\Rightarrow 5x = 50)
(\Rightarrow x = 10)
Thus, the slope of a secant line is (10).
Example 4. Calculate the slope of a secant line for the function (f(x) = 4 - 3x^{3}) that joins the two points ((1, f(1))) and ((2, f(2))).
Solution.
Given: (f(x) = 4 - 3x^{3})
Calculate the value of (f(1)) and (f(2)).
(f(3) = 4 - 3(1)^{3} = 4 - 3 = 1)
(f(5) = 4 - 3(2)^{3} = 4 - 24 = -20)
Using the Slope of the secant line formula, we have:
(m = \frac{(f(b) - f(a))}{(b - a)})
(\Rightarrow m = \frac{(f(2) - f(1))}{(2 - 1)})
(\Rightarrow m = -20 - 1)
(\Rightarrow m = -21)
Hence, the slope of a secant line is (-21).
Example 5. The slope of a secant line that joins the two points ((1, 5)) and ((8, y)) is (9). Find the value of (y).
Solution.
Given: ((x_1, y_1) = (1, 5)), ((x_2, y_2) = (8, y)) and (m = 9)
Using the Slope of the secant line formula, we have:
(m = \frac{(y_2 - y_1)}{(x_2 - x_1)})
(\Rightarrow 9 = \frac{(y - 5)}{(8 - 1)})
(\Rightarrow 9 = \frac{(y - 5)}{7})
(\Rightarrow y - 5 = 63)
(\Rightarrow y = 68)
Hence, the slope of a secant line is (68).
We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
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Slope of the Secant Line Formula FAQs
What is the slope of the secant line formula?
The slope of the secant line formula calculates the average rate of change between two points on a curve.
What are the limitations of the slope of the secant line formula?
The slope of the secant line formula provides an average rate of change but doesn't precisely capture instantaneous changes, which is where the concept of the derivative comes in.
How to find the slope of the secant line through the points (x_1, y_1) and (x_2, ?
A3. If ((x_1, y_1)) and ((x_2, y_3)) represent two points along the curve (y = f(x)), across which the secant line extends, then slope of the secant line formula (= \frac{y_2 - y_1}{x_2 - x_1}).
How to find the slope of the secant line joining?
To find the slope of the secant line joining two points on a curve, divide the change in the y-coordinates by the change in the x-coordinates between those points.
Is the slope of the secant line formula a fundamental concept in calculus?
Yes, the slope of the secant line formula serves as a foundational concept bridging averages to instantaneous change, paving the way for more advanced calculus concepts.
What's the broader significance of understanding the slope of the secant line formula?
Understanding the slope of the secant line formula enriches our grasp of how things change over time, enabling us to better analyze and model real-world scenarios.
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190159 | https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry(Neils)/8%3A_Acids_and_Bases/8.09_Buffer_Capacity_and_Buffer_Range | Skip to main content
8.9 Buffer Capacity and Buffer Range
Last updated
: Jul 10, 2019
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8.8: Buffers: Solutions That Resist pH Change
8.10: Lewis Acids and Bases
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: 163841
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Buffer Capacity
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on K), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure 1, when NaOH is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
Figure 1: Effect of Buffer Concentration on the Capacity of a Buffer
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M NaOH to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of NaOH solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the NaOH solution results in only a relatively small change in pH.
Selecting proper components for desired pH
Buffers function best when the pKa of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the Ka for hydroflouric acid is 6.6 x 10-4 so its pKa= -log(6.6 x 10-4) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.
For the weak base ammonia (NH3), the value of Kb is 1.8x10-5, implying that the Ka for the dissociation of its conjugate acid, NH4+, is Kw/Kb=10-14/1.8x10-5 = 5.6x10-10. Thus, the pKa for NH4+ = 9.25, so buffers using NH4+/NH3 will work best around a pH of 9.25. (It's always the pKa of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pKb of the conjugate base, obviously.)
When the desired pH of a buffer solution is near the pKa of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.
Example 1: HF Buffer
In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the Henderson-Hasselbalch approximation to calculate the necessary ratio of F- and HF.
pH=pKa+log[Base]Acid
3.0=3.18+log[Base]Acid
log[Base][Acid]=−0.18(3)
[Base][Acid]=10−0.18(4)
[Base][Acid]=0.66(5)
This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)?
We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66. From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol. HF is a weak acid with a Ka = 6.6 x 10-4 and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F- we need to add.
The dissociation reaction is:
HF(aq)+H2O(l)⇌F−(aq)+H3O+(aq)(6)
We could use ICE tables to calculate the concentration of F- from HF dissociation, but, since Ka is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F- in the solution from HF dissociation will be negligible. Thus, the [HF] is about 1 M and the [F-] is close to 0. This will be especially true once we have added more F-, the addition of which will even further suppress the dissociation of HF.
We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66. Thus, [F-] should be about 0.66 M. For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F-. Since we are adding NaF as our source of F-, and since NaF completely dissociates in water, we need 0.066 moles of NaF. Thus, 0.066 moles x 41.99 g/mol = 2.767 g.
Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of "Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "moles of base over moles of acid." The pH of the solution does not, it turns out, depend on the volume! (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H+ or OH-. Such dilute solutions are rarely used as buffers, however.)
References
Brown, et al. Chemistry:The Central Science. 11th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2008.
Chang, Raymond. General Chemistry:The Essential Concepts. 3rd ed. New York: Mcgraw Hill, 2003
Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
Outside Links
Urbansky, Edward T.; Schock, Michael R. "Understanding, Deriving, and Computing Buffer Capacity." J. Chem. Educ. 2000 1640..
Contributors
Jose Pietri (UCD), Donald Land (UCD)
8.8: Buffers: Solutions That Resist pH Change
8.10: Lewis Acids and Bases |
190160 | https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOor5KrpjVF8aL8MYjOHAzIu6QlQ7Y789iJL9FeCz27cf2FcrmxoV | Art of Problem Solving
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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Algebra
Inequalities
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190161 | https://bmcnephrol.biomedcentral.com/articles/10.1186/s12882-020-01913-7 | Advertisement
The correlation analysis between the Oxford classification of Chinese IgA nephropathy children and renal outcome - a retrospective cohort study
BMC Nephrology
volume 21, Article number: 247 (2020)
Cite this article
2788 Accesses
22 Citations
Metrics details
Abstract
Background
The 2016 Oxford Classification’s MEST-C scoring system predicts outcomes in adults with IgA nephropathy (IgAN), but it lacks tremendous cohort validation in children with IgAN in China. We sought to verify whether the Oxford classification could be used to predict the renal outcome of children with IgAN.
Methods
In this retrospective cohort study, 1243 Chinese IgAN children who underwent renal biopsy in Jinling Hospital were enregistered from 2000 to 2017. The combined endpoint was defined as either a ≥ 50% reduction in estimated glomerular filtration rate (eGFR) or end-stage renal disease (ESRD). We probed into the relevance betwixt the Oxford classification and renal prognosis.
Results
There were 29% of children with mesangial proliferation(M1), 35% with endocapillary proliferation (E1), 37% with segmental sclerosis/adhesion lesion (S1), 23% with moderate tubular atrophy/interstitial fibrosis (T1 25–50% of cortical area involved), 4.3% with severe tubular atrophy/interstitial fibrosis (T2 > 50% of cortical area involved), 44% with crescent in< 25% of glomeruli(C1), and 4.6% with crescent in> 25% of glomeruli (C2). All children were followed for a medial of 7.2 (4.6–11.7) years, 171 children (14%) arrived at the combined endpoint. The multivariate COX regression model revealed that the presence of lesions S (HR2.7,95%CI 1.8 ~ 4.2, P<0.001) and T (HR6.6,95%CI 3.9 ~ 11.3, P<0.001) may be the reason for poorer prognosis in the whole cohort. In contrast, C lesion showed a significant association with the outcome only in children received no immunosuppressive treatment.
Conclusions
This study revealed that S and T lesions were useful as the long-term renal prognostic factors among Chinese IgAN children.
Peer Review reports
Background
IgA nephropathy (IgAN), being particularly frequent in Asia, is the primary reason for end-stage renal disease (ESRD) in all ages . Since IgAN does not exhibit a specific serologic profile, a percutaneous kidney biopsy remains a definitive tool to establish the diagnosis of IgAN . Additionally, the prognostic value of histological data has become increasingly recognized in the past decade . As a new pathological classification standard to judge the renal prognosis of IgAN, the Oxford classification has been put forward in recent years. The purpose of the Oxford classification was to consider the pathological features associated with clinical outcomes independently of clinical data and to improve the current ability to predict outcomes in IgAN patients. Although the classification standard has been formulated by rigorous methodology, its clinical application in children needs to be further verified. What’s more, IgAN has regional and ethnic differences, which determines that the Oxford classification needs to be refined in different populations and races. In this study, the clinical and pathological data of IgAN followed up for a long time in our department were retrospectively analyzed to assess the predictability of Oxford classification among Chinese children.
Methods
Inclusion criteria and clinical data set
Children with IgAN who underwent renal biopsy in Jinling Hospital were enrolled from January 1, 2000, to December 31, 2017, in this retrospective cohort study. The inclusion criteria were follow-up ≥12 months, age at the time of biopsy ≤18 years, a minimum of eight glomeruli and an initial eGFR≥15 ml/min /1.73m2. The exclusion criteria were secondary IgAN caused by Henoch-Schonlein purpura, liver cirrhosis, systemic lupus erythematosus, hepatitis B virus infection, tumour, ankylosing spondylitis and psoriasis. The following clinical data including age, gender, duration from onset to renal biopsy, estimate glomerular filtration rate (eGFR), mean arterial pressure (MAP), proteinuria and therapeutic regimen [Renin angiotensin aldosterone system blockade (RASB), glucocorticoid (GC) and other immunosuppressant agents] were collected during biopsy and follow-up. Immunosuppressive therapy after kidney biopsy was defined as treatment with any immunosuppressive agent, regardless of duration or dose.
Definitions
eGFR was calculated using the Schwartz formula , and the CKD-EPI equation was used in adolescents aged > 16 years at the time of biopsy. ESRD was defined as eGFR < 15 mL/min/1.73m2, initiation of dialysis or transplantation. Survival time was defined from the time of biopsy until the date of the last follow-up, the incidence of the event of interest, or March 2019 (end of the study). The combined endpoint was defined as either ≥50% reduction eGFR or ESRD or death.
Pathology review
Renal pathology was scored according to the Oxford classification of IgAN , assessed by the following parameters: mesangial hypercellularity (M), scored as absent (M0) or (M1) if≥50% of glomeruli had more than three cells per mesangial area; endocapillary hypercellularity (E), scored as E0 if absent or E1 if present; segmental glomerulosclerosis or adhesion (S), scored (S0) if absent or (S1) if present; tubular atrophy/interstitial fibrosis(T), scored as T0 (0–25% of cortical area), T1 (26–50% of cortical area), or T2(>50% of cortical area); cellular/ fibrocellular crescents scored as C0 (no crescents), C1 (crescents in at least one but < 25% of glomeruli), or C2 (crescents in more than 25% of glomeruli). Immunofluorescence studies were performed (IgG, IgA, IgM, C3) and showed at least 1+ (on scale from 0 to 3+) mesangial deposition of IgA, with IgA being the dominant immunoglobulin deposited in the glomeruli. The intensity of deposits as determined via immunofluorescence microscopy was graded semi-quantitatively on a scale from 0 to 3+, where 0 = no, 1+ = slight, 2+ = moderate, and 3+ = intense. At least two pathologists blinded to patient outcomes at the time of review confirmed the pathological results.
Statistical analyses
Continuous variables were presented as mean ± standard deviation (normally distributed variables) or median with interquartile range (IQR, non-normally distributed variables) for data distribution. Categorical variables are presented as percentages. The Kruskal-Wallis test and chi-squared test were used to compare continuous and categorical variables, respectively. The Kaplan–Meier curve was used to illustrate univariate differences between groups of pathological variables, and the log-rank test were used to test the two curves differences. The Cox proportional hazards regression model was conducted on univariable and multivariable analyses of Oxford classification in the IgAN children. Univariable Cox regression analysis was performed for each pathological variable. Multivariable Cox regression analysis (Backward: LR approach) adjusting for other clinically essential factors including initial eGFR, initial mean arterial pressure, and initial proteinuria was performed to appraise the function of MEST-C scoring system on the renal outcome. Hazard ratio (HR) with 95% confidence interval (CI) for each variable was estimated. All probabilities were two-tailed, and P-value< 0.05 was considered statistically significant.
Data analysis was executed using SPSS for windows version 26 (IBM Corporation, Armonk, NY).
Results
Clinical features
We enrolled 1243 Chinese children diagnosed with primary IgAN from 2000 to 2017 in Jinling Hospital. Table 1 outlines the clinical, treatment and follow-up characteristics of the study children. The average age of children at diagnosis was 14 ± 4 years, with male-dominated (68%). The average value of MBP was 89 ± 16 mmHg, initial proteinuria was 0.6 (interquartile ranges, 0.3–1.4) g/day per 1.73m2, and the initial eGFR was 102 ± 20 ml/min per 1.73m2. All children were followed for a medial of 7.2 (4.6–11.7) years, 171 children (14%) arrived at combined endpoint (ESRD, n = 82;≥50% eGFR decline, n = 89). During the follow-up period, 70% of children were treated with RASB, 45% were treated with GC, and 19% received GC combined other immunosuppressive drugs.
Pathological findings
The average of glomeruli was 20.4 ± 4.7 per biopsy. Based on Oxford classification,29% of the children showed M1, 35% showed E1, 37% showed S1, 23% showed T1, 4.3% showed T2, 44% showed C1 and 4.6% showed C2 (Table 2). The distribution of the percentage of crescents observed in every child was shown in Fig. 1. Of 48.6% children with any cellular/ fibrocellular crescents, 28% had crescents in<10% of glomeruli, whereas 9.4% had a fraction of glomeruli with crescents one-tenth or more, 6.6% had a fraction of glomeruli with crescents one-sixth or more, and only 4.6% had a fraction of glomeruli with crescents one-fourth or more. The percentage of immunoglobulins deposited only in the mesangial region was 68%, while 32% of immunoglobulins were deposited in both the mesangial and capillary loop regions. 25% of children showed positive glomerular staining for IgG, 44% showed positive glomerular staining for IgM, 84% showed positive glomerular staining for C3, and 1.1% showed positive glomerular staining for C4. The immunofluorescence intensity of IgA was between 1+ and 3+, including 5.6% of 1+, 13% of 2+ and 81% of 3 + .
Distribution of the percentage of glomeruli with crescents in biopsies with any crescents. Crescents were present in 599(48%) of 1243 total biopsies
Effects of different kidney biopsy time on the variables in Oxford classification
The median time (12 months) of onset to renal biopsy was selected as the cut-off point to analyze the effect of biopsy time on variables in the Oxford classification From Table 3. It showed that when the time of onset to renal biopsy was less than 12 months, the patient’s lesions were milder, dominated by S0(χ2 = 354.5, P<0.001), T0 (χ2 = 323.3, P<0.001), and C0(χ2 = 437.6, P<0.001). On the contrary, when the time of onset to renal biopsy was longer than 12 months, the lesions were corrected mainly by S1, T1–2 and C1–2. Concerning E and M lesions, there was no significant difference in time from onset to renal biopsy within available data.
Associations between clinical and histologic variables
Linear regression analysis of Oxford classification with the most robust indicators for estimating renal decline (eGFR, MAP and proteinuria) was performed to explore the correlation between Oxford classification and clinical signs. As shown in Table 4, Children with S1, T1–2 and C1–2 lesions were associated with a reduced initial eGFR at the time of biopsy. All histological lesions (M1, E1, S1, T1–2, and C1–2) were associated individually with higher initial proteinuria at the time of biopsy. All histological lesions were associated with more upper initial MAP at the time of biopsy.
Renal survival IgAN children according to Oxford classification
As presented in Fig. 2, the Kaplan-Meier analyses revealed that lesion S (Log-Rank, χ2 = 14.796, P<0.001; Fig. 2c) and T (χ2 = 48.976, P<0.001; Fig. 2d) were each correlation with renal survival. However, lesions M (χ2 = 1.459, P = 0.177, Fig. 2a), E (χ2 = 2.399, P = 0.331, Fig. 2b) and C (χ2 = 6.218, P = 0.054, Fig. 2e) were not associated with renal outcome.
Renal survival according to pathological variables. a Effect of the presence of mesangial hypercellularity on survival from a combined event in all patients. b Effect of the presence of endocapillary hypercellularity on survival from a combined event in all patients. c Effect of the presence of segmental sclerosis on survival from a combined event in all patients. d Effect of the presence of interstitial fibrosis/tubular atrophy on survival from a combined event in all patients. e Effect of the presence of cellular/fibrocellular crescents on survival from a combined event in all patients
Associations between pathologic features and renal outcome
The associations between pathological features and renal outcomes were analyzed in a COX regression model (Table 5). With univariate COX regression model, renal outcome from a combined event were both associated significantly with lesions S (HR3.5, 95%CI 2.3 ~ 5.3, P<0.001), T (HR 2.6, 95%CI 2.1 ~ 3.3, P<0.001) and C (HR 2.1, 95%CI 1.5 ~ 2.8, P<0.001). However, the lesion of M (HR 1.8, 95%CI 1.3 ~ 2.3, P = 0.115), and E (HR 1.4, 95%CI 0.9 ~ 2.1, P = 0.326) were not associated with renal outcome. When adjusted for clinically essential data in the multivariate COX regression model, only S (HR2.7, 95%CI1.8 ~ 4.2, P<0.001) and T (HR6.6, 95%CI 3.9 ~ 11.3, P<0.001) lesions remained as independent predictors of renal outcome.
Predictive value of lesion M, E and C between immunosuppressive and without immunosuppressive groups
We further assessed the predictive value of lesions M, E and C in children who received no immunosuppressive treatment to assess their natural predictive value. Children with crescent who didn’t receive immunosuppressive therapy experienced worse survival from the combined event (Fig. 3e), but this difference disappeared after received immunosuppression (Fig. 3f). The predictive value of lesion M and E were not changed by adding immunosuppressive treatment (Fig. 3a-d).
Predictive value of mesangial hypercellularity, endocapillary hypercellularity and cellular/fibrocellular crescents between immunosuppressive and without immunosuppressive groups. a Effect of the presence of mesangial hypercellularity on survival from a combined event in patients without immunosuppression. b Effect of the presence of mesangial hypercellularity on survival from a combined event in patients with immunosuppression. c Effect of the presence of endocapillary hypercellularity on survival from a combined event in patients without immunosuppression. d Effect of the presence of endocapillary hypercellularity on survival from a combined event in patients with immunosuppression. e Effect of the presence of cellular/fibrocellular crescents on survival from a combined event in patients without immunosuppression. f Effect of the presence of cellular/fibrocellular crescents on survival from a combined event in patients with immunosuppression
Discussion
This study investigated the clinical and histopathologic predictors of poor prognosis in pediatric patients with IgAN. The median duration from onset to renal biopsy was 12 months in our cohort. An early diagnosis seemed to be the primary reason for a low frequency of chronic and severe lesions such as lesion S, T and C. In our cohort, we confirmed that S and T were independent risk factors associated with renal outcomes. The lesion C enhanced the ability to predict progression only in those who did not receive immunosuppression. Lesion M and E were not significant variables, which may weaken their predictive values because of the low percentage in the cohort. The independent predictive value of pathology MEST-C score was reduced by immunosuppressive therapy.
Our results suggested that patients with severe pathological lesions (e.g. S、T、C) were associated with lower eGFR, higher blood pressure and higher proteinuria, which were consistent with other findings [7,8,9]. Glomerular hypertension may mediate progressive renal damage by leading to glomerular hyperfiltration and glomerular enlargement . For the control target of blood pressure in IgAN, the KDIGO guidelines pointed out that when proteinuria > 0.3 g/day, the recommended target blood pressure (BP) was < 130/80 mmHg, and when proteinuria > 1 g/d, the recommended target BP was < 125/75 mmHg. Bellur et al. showed that S was strongly associated with proteinuria and lower eGFR levels, which was consistent with our conclusion. Previous studies have shown that T was an independent risk factor for poor renal prognosis and associated with BP. Some scholars [11, 12] had found that the level of eGFR was lower in patients with IgAN with extensive crescent formation, and there was a negative correlation between eGFR and the proportion of crescents. Thus, it can be concluded that the most critical risk factors for the progression of IgAN (proteinuria, eGFR, MAP) are significantly correlated with the pathological damage found by renal biopsy, which reflects the value of the combination of clinical and pathological risk factors in judging the prognosis of IgAN children.
The lesion M was not a significant risk for renal outcome in our cohort. We speculated that it might be difficult to address its value because of its low prevalence in our study. But the value of lesion M as an independent risk for progression is debated. On the one hand, the VALIGA cohort and a Chinese adult cohort confirmed M lesion as a significant factor for progression, but on the other Shima et al. reported that M had lost predictive value in patients receiving immunosuppressive therapy. By and large, the presence of M lesion appears to have a negligible correlation with renal outcomes.
The lesion E, observed in 35% of the children, did not independently predict clinical outcome in the whole cohort. This was highly consistent with the findings in the Oxford classification cohort . However, two studies in which patients did not receive immunosuppressive therapy [17, 18] reported that E1 was independently associated with more rapid loss of renal function and worse renal survival, which indirectly suggested that proliferative lesions were treatment-responsive. We conjectured that the widespread use of immunosuppressants might have influenced the absence of correlation between E lesion and outcome.
Our data showed relevance between lesion S and renal prognosis, which further confirmed that S was a particular index to judge the prognosis. Many data from the children’s cohort have proved the independent predictive value of S lesion. Children’s group from France confirmed that lesion S was the only histological variable predicting a decline in renal function and was not associated with clinical data at the time of renal biopsy and whether they received immunosuppressive therapy . Studies have revealed S lesion develops from the organization of previous segmental necrotizing or endocapillary inflammatory lesions or in response to podocyte injury and detachment. Therefore, it has also been suggested that in children with active glomerular lesions, special attention needs to be paid to the relationship between lesion S and M and E.
The lesion T was confirmed to be associated with renal failure, which was accord with almost all previous adult validation studies [4, 8, 10]. This may not be surprising because T lesion represents a more chronic and late stage of IgAN renal damage. However, most children validation studies, such as Japan cohort , Sweden cohort and VALIGA cohort , failed to confirm that T lesion could maintain independent predictive value in children. Only the cohort from China by Le et al. and our cohort confirmed that T lesion has independent predictive value in children population. This difference may be due to only a small proportion of children arrived at a composite terminal during the follow-up in these child studies [13, 15, 21].
In our research, particular interest was given to children with lesion C as they had a predictive value (HR 2.1, 95% CI 1.5–2.8) in the univariate analysis, although it did not retain its significance in the multivariate analysis. C lesion was seen in 49% of the children, however, with 28% having crescents in<10% of glomeruli. At the same time, a higher percentage of children with C were treated with immunosuppression than children without this lesion. Overall, crescents predicted a higher risk of a combined event, although this remained significant only in children not receiving immunosuppression. Thus, crescents in a minority of glomeruli may represent a lesion reversible by immunosuppressive therapy. Our findings suggest that children whose biopsies show these active lesions should be considered for immunosuppressive treatment, which was consistent with a multicenter study .
The validation differences among the above different child cohorts are mainly related to the regional and ethnic differences in IgAN, the selection criteria, follow-up time and treatment measures of each study, which emphasizes the need to generate a large database for IgAN children to address the problem of insufficient statistical power due to the small number of progressive cases, especially the relatively short follow-up period.
Our cohort validated the significance of the Oxford classification in a large number of Chinese IgAN children. A comprehensive analysis of the renal pathological features and clinical conditions represented in the cohort suggests that Oxford classification must be considered in conjunction with clinical features (including proteinuria levels and eGFR values) and treatment given after renal biopsy. This also suggests that treatment operations after biopsy may regulate some pathological risk factors. To explore the risk factors and their impact on disease progression by studying the clinical and pathological features of IgAN, the level of diagnosis and treatment of IgAN will ultimately be improved.
The limitations of this study must be recognized. First, retrospective design makes the control of measured variables difficult. Second, our results may not be extrapolated to other ethnic groups due to geographical variability in IgAN outcomes. Final, due to the limitations of retrospective studies, not all children were treated with RASB, which may weaken the rigour of the study. However, some features of this study may increase the strength of these findings, including the broad set of data collected over many years and long-term follow-up by the same team with a well-established clinical protocol, as well as the careful re-evaluation of all renal biopsies by two expert pathologist blinded to clinical data and outcome.
Conclusions
In summary, this study showed that T and S lesions were independently linked to poor renal outcome in Chinese IgAN children. In contrast, C lesion showed significant association with prognosis only in children received no immunosuppressive treatment. M and E lesions appeared to be unrelated to renal prognosis.
Availability of data and materials
The datasets used and analyzed in this study are available from the first author and corresponding author on reasonable request.
Abbreviations
Crescent
Confidence intervals
Estimated glomerular filtration rate
Endocapillary proliferation
End-stage renal disease
Glucocorticoid
IgA Nephropathy
Hazard ratio
Interquartile range
Mesangial proliferation
Mean arterial pressure
The Kidney Disease Improving Global Outcomes
Renin angiotensin aldosterone system blockade
Segmental sclerosis/adhesion lesion
Tubular atrophy/interstitial fibrosis
European Validation Study Of The Oxford Classification Of IgA Nephropathy
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Acknowledgements
Not applicable.
Funding
The authors acknowledge support from the Clinical Advanced Techniques, Primary Research & Development Plan of Jiangsu Province (BE2017719) and the Pediatric Medical Innovation Team of Jiangsu Province (CXTDA2017022). The cost of publishing paper was funded in this study. The funders had no role in the design of the study, data collection, analysis and data interpretation, or writing of the manuscript.
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Department of Pediatrics, Jinling Hospital, The First School of Clinical Medicine, Southern Medical University, Nanjing, China
Heyan Wu, Zhengkun Xia, Chunlin Gao, Pei Zhang, Xiao Yang, Meiqiu Wang & Yingchao Peng
Department of Pediatrics, Nanfang Hospital, Southern Medical University, Guangzhou, China
Heyan Wu
Department of Pediatrics, Jinling Hospital, Nanjing Medical University, Nanjing, China
Ren Wang
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HW, ZX and CG performed the data collection and analysis and participated in manuscript writing. PZ, XY, RW, MW and YP performed the database setup and statistical analysis. HW, ZX and CG participated in the study design and coordination and helped to draft the manuscript. All of the authors have read and approved the final manuscript.
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Zhengkun Xia or Chunlin Gao.
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The study complied with the Declaration of Helsinki and was approved by the Medical Ethics Committee of Jinling Hospital (Approval number: 2019NZGKJ-266). The committee waived the requirement for informed consent in consideration of the retrospective nature of the study and anonymous data analyses.
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Wu, H., Xia, Z., Gao, C. et al. The correlation analysis between the Oxford classification of Chinese IgA nephropathy children and renal outcome - a retrospective cohort study.
BMC Nephrol 21, 247 (2020).
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190162 | https://brainly.com/question/32891996 | [FREE] A high school baseball player has a 0.267 batting average. In one game, he gets 8 at bats. What is the - brainly.com
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A high school baseball player has a 0.267 batting average. In one game, he gets 8 at bats. What is the probability that he will get at least 6 hits in the game?
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The probability that the high school baseball player will get at least 6 hits in the game is approximately 0.377, or 37.7%.
To calculate the probability that the high school baseball player will get at least 6 hits in the game, we can use the binomial distribution. The binomial distribution is appropriate here because each at-bat is independent, and there are only two outcomes (hit or no hit) for each at-bat.
The probability of getting a hit in a single at-bat is given by the batting average, which is 0.267. Therefore, the probability of not getting a hit in a single at-bat is 1 - 0.267 = 0.733.
We want to calculate the probability of getting at least 6 hits in 8 at-bats. This includes the probabilities of getting exactly 6, 7, and 8 hits. We can calculate these probabilities separately and sum them up.
To calculate the probability that the high school baseball player will get at least 6 hits in the game, let's use the binomial distribution.
P(X >= 6) = P(X = 6) + P(X = 7) + P(X = 8)
Where:
P(X = k) is the probability of getting exactly k hits
P(X >= 6) is the probability of getting at least 6 hits
X is the random variable representing the number of hits in the game
k is the number of hits
Let's calculate each probability:
P(X=6)=C(8,6)∗(0.26 7 6)∗(0.73 3 2)=28∗(0.26 7 6)∗(0.73 3 2)=0.242
P(X=7)=C(8,7)∗(0.26 7 7)∗(0.73 3 1)=8∗(0.26 7 7)∗(0.73 3 1)=0.123
P(X=8)=C(8,8)∗(0.26 7 8)∗(0.73 3 0)=1∗(0.26 7 8)∗(0.73 3 0)=0.012
Now, let's sum up these probabilities:
P(X >= 6) = P(X = 6) + P(X = 7) + P(X = 8)
= 0.242 + 0.123 + 0.012
= 0.377
Therefore, the probability that the high school baseball player will get at least 6 hits in the game is approximately 0.377, or 37.7%.
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The probability that the baseball player gets at least 6 hits in 8 at-bats is approximately 0.377, or 37.7%. This is calculated using the binomial distribution, considering the player's batting average of 0.267. The calculations involve finding the probabilities for getting exactly 6, 7, and 8 hits and summing them up.
Explanation
To find the probability that a high school baseball player with a batting average of 0.267 gets at least 6 hits in 8 at-bats, we will use the binomial distribution. This distribution is suitable here because each at-bat can result in either a hit or a miss, making it a binary outcome.
Identify probabilities:
The probability of getting a hit (success) in one at-bat is the batting average: 0.267.
Therefore, the probability of not getting a hit (failure) is: 1 - 0.267 = 0.733.
Set up the binomial probability formula: The probability of getting exactly k hits in n at-bats can be calculated with the formula:
P(X=k)=C(n,k)∗(p k)∗((1−p)n−k)
where:
C(n,k) is the binomial coefficient, calculated as k!(n−k)!n!
p is the probability of success (0.267)
n is the total number of trials (8 in this case)
k is the number of successes (hits).
Calculate probabilities for at least 6 hits: We need to calculate P(X ≥ 6), which is the sum of probabilities for getting exactly 6, 7, and 8 hits:
P(X≥6)=P(X=6)+P(X=7)+P(X=8)
To find P(X=6):
P(X=6)=C(8,6)∗(0.26 7 6)∗(0.73 3 2)
=28∗(0.26 7 6)∗(0.73 3 2)≈0.242
To find P(X=7):
P(X=7)=C(8,7)∗(0.26 7 7)∗(0.73 3 1)
=8∗(0.26 7 7)∗(0.73 3 1)≈0.123
To find P(X=8):
P(X=8)=C(8,8)∗(0.26 7 8)∗(0.73 3 0)
=1∗(0.26 7 8)≈0.012
Final calculation: Now we can sum these probabilities:
P(X≥6)=P(X=6)+P(X=7)+P(X=8)
≈0.242+0.123+0.012=0.377
Thus, the probability that the player will get at least 6 hits in the game is approximately 0.377, or 37.7%.
Examples & Evidence
For instance, if a player has 10 at-bats with the same batting average, you can use the same binomial approach to determine the probability for different outcomes, demonstrating how the formula works in varying scenarios.
The use of the binomial distribution is a standard statistical method to model situations with multiple independent trials, like batting performances in baseball. The calculations for probabilities reflect this method accurately.
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190163 | https://www.ck12.org/flexi/algebra-ii/conversion-between-degrees-and-radians/what-is-the-difference-between-positive-and-negative-angles/ | What is the difference between positive and negative angles?
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What is the difference between positive and negative angles?
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In trigonometry, the difference between positive and negative angles is based on the direction of rotation.
Positive Angle
A positive angle is generated by a counter-clockwise rotation from the initial side (usually the positive x-axis) to the terminal side.
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190164 | https://dcsl.gatech.edu/papers/iros11b.pdf | Solving Shortest Path Problems with Curvature Constraints Using Beamlets Oktay Arslan, Panagiotis Tsiotras and Xiaoming Huo Abstract— A new graph representation of a 2D environ-ment is proposed in order to solve path planning problems with curvature constraints. We construct a graph such that the vertices correspond to feasible beamlets and the edges between beamlets capture not only distance information but directionality as well. The A algorithm incorporated with new heuristic and cost function is implemented such that the curvature of the computed path can be constrained a priori.
The proposed Beamlet algorithm allows us to find paths with more strict feasibility guarantees, compared to other competing approaches, such as Basic Theta or A with post-smooth processing.
I. INTRODUCTION Planning a path for an autonomous vehicle in a 2D or 3D environment has been studied for many years , and many algorithms have been developed to solve this problem , , , , , , , , , , , to name just a few. From a system-theoretic view point, the path planning problem is to find a control input that will drive the state of the vehicle from a given initial state to a given goal state, while minimizing a cost function and satisfying the state and vehicle dynamic constraints. This is a complex problem since it requires searching a solution in an infinite dimensional space. One of the most common approach to overcome this difficulty is to decompose the original problem into two subproblems which address the geometric and dynamic parts separately. However, the feasibility of the computed path in the geometric level does not always imply that the computed path can be followed by the vehicle.
Since the geometric level planner does not have any prior information about the dynamic envelope of the vehicle, this approach may lead to either a suboptimal or a dynamically infeasible path. Generation of dynamically infeasible paths can be avoided by coupling the geometric and dynamic level planners via passing information about the allowable dynamic envelope of the vehicle during the geometric search.
One simple approach to capture the vehicle’s dynamic en-velope is to bound the curvature of the allowable paths, and pass this information to the geometric level planner .
Path planning algorithms that solve the problem at the geometric level can find obstacle-free paths (no kinematic Oktay Arslan is a graduate student in the D. Guggenheim School of Aerospace Engineering, Georgia Institute of Technology, Atlanta, GA 30332, USA, email: oktay@gatech.edu Professor Panagiotis Tsiotras is with the D. Guggenheim School of Aerospace Engineering, Georgia Institute of Technology, Atlanta, GA 30332, USA, email: tsiotras@gatech.edu Professor Xiaoming Huo is with the H. Milton Stewart School of Industrial and System Engineering, Georgia Institute of Technology, Atlanta, GA, 30332-0250, USA, email: huo@gatech.edu Fig. 1: Solution of the problem may be a self-intersecting curve due to constraints and dynamic constraints) very easily. These algorithms usu-ally search on a graph which is generated by discretizing the continuous environment into cells that are either blocked (black) or unblocked (white). A common decomposition is a grid-like partitioning to square cells. The corners of these cells and the links among them correspond to vertices and edges of the search graph, respectively. As a result of this abstraction, paths which are formed by grid edges are not smooth since the possible headings at each vertex are artificially constrained; furthermore, the graph formed by the grid is insufficient to express all possible paths which are embedded in the environment. For instance, consider a path planning problem for the vehicle shown in Figure 1, where the destination is shown with the cross and the vehicle can turn only to the left due to a malfunction of its actuators.
It is obvious that the trajectory of the vehicle should be a self-intersecting path, as shown in Figure 1. But this path cannot be computed by any search algorithm (Dijkstra, A or any variants) at this level of abstraction (i.e., one that searches on the graph formed by the topological grid of the environment , ). This occurs because these algo-rithms do not allow for self-intersecting paths. One solution to avoid such problems is to increase the graph dimension to also account for extra states (such as velocity) at each graph vertex, however this increases the dimensionality of the problem. Here we follow an alternative approach. We still perform the path-planning in the physical space (where the obstacles are more naturally described), but we also encode local directionality of the path in an efficient manner. It turns out that this is possible using beamlets . Beamlets are small line segments connecting points at the boundary of cells in a dyadic partitioning of the environment, organized in a hierarchical structure. They provide an efficient encoding of all smooth curves in the plane with an error that is no larger than the underlying resolution, in terms of Hausdorff distance [18, Lemma 1]. In other words, they offer sparse approximate representations of smooth curves in the plane; in a certain sense, they are optimally sparse. As shown in , the library of all beamlets is of order O(n2 log2 n), whereas all possible segments connecting the O(n2) vertices in the grid (i.e., the beams) are of order O(n4). The latter order of complexity is prohibitive for the development of an efficient algorithm. The reduced cardinality of beamlets, on the other hand, allows us to develop algorithms that have complexity only logarithmically larger than the given data. It follows that exhaustive searches through the collection of beamlets can be implemented much faster than exhaustive searches through the collection of beams.
In this paper we use the previous remarkable property of beamlets, to efficiently connect points between empty spaces in the plane1, and−most importantly−to include directionality information along the ensuing path. This will allow us to easily incorporate curvature constraints along the path. Such curvature constraints are meant to provide a “dynamic signature” for the resulting path so that can be followed by the actual vehicle.
II. RELATED WORK Several different algorithms have been proposed to find smooth trajectories in a continuous environment, notably among them are the A with Post Smoothed Paths and the Basic Theta . Both A with Post Smoothed Paths (A PS) and the Basic Theta algorithms perform some extra operations (line-of-sight check) in order to remove redundant vertices from the optimal path. The A PS algorithm first computes the shortest path using the classical A search algorithm, and then it performs a post-processing operation to smooth the computed path. During the post-processing operation, a vertex is removed from the computed path if its successor vertex is in the line-of-sight of its predecessor vertex. Its successor vertex is then assigned as the parent of its predecessor vertex. On the other hand, the Basic Theta algorithm performs a line-of-sight check during the update procedure of the each vertex. It assigns the parent of the current vertex as the parent of the neighbor vertex if the neighbor vertex is in the line-of-sight of the parent of the current vertex. Although both A PS and Basic Theta can find smooth paths, they perform the smoothing operation and the removal of the redundant vertices only around a neighborhood of the optimal shortest path found by A.
They cannot apply curvature constraints during the search and expansion of the vertices. The proposed approach, on the other hand, uses a new representation of the information of the environment in a way such that a priori curvature constraints can be imposed directly during each step of the search. Therefore, it can find smoother paths with a guaranteed curvature bound.
III. APPROACH The path planning problem we are interested in is to find a curvature bounded path from a start point (pstart) to a goal 1While two vertices in the nearest-neighbor graph corresponding to vertices at opposite sides of the environment can only be connected by a path with O(n) edges, no two vertices in the beamlet graph are ever more than 4log2(n)+1 edges apart.
point (pgoal) while minimizing a predefined cost function. It is assumed that the spatial information about the environment is given as an n×n square image, where n = 2smax and smax is a positive integer. There are several key elements which play an important role in the formation of the beamlet graph.
Their definition is given below.
Beamlets is a collection of line segments which are defined between certain points in the environment, selected from the boundaries of dyadic squares , . Unlike the points on the grid, beamlets may not necessarily have uniform properties and they occupy different scales and orientation on a given image as shown in Figure 2(a). A beamlet is feasible if all cells it passes through are unblocked. A dyadic square (d-square) is a set of points in the environment forming a square region. Each dyadic square is characterized by a 3-tuple (scale s and position (x,y)). Formally, it is defined as q(s;x,y) = {(i, j) : 2s(x−1) ≤i ≤2sx,2s(y−1) ≤ j ≤2sy}. Furthermore, q(s;x,y) can be rewritten as the union of four sub-d-squares of scale s −1; i.e., we have q(s;x,y) = q(s−1;2x−1,2y−1)∪q(s−1;2x−1,2y)∪q(s− 1;2x,2y −1) ∪q(s −1;2x,2y). This property of d-squares leads to a recursive partitioning scheme which is called dyadic partitioning.
(a) (b) Fig. 2: Sample of feasible beamlets on a 8×8 square image and the corresponding quadtree representation.
Due to the recursive nature of the d-squares, the set of all d-squares of a given dyadic partition can be stored as a quadtree . Figure 2(b) illustrates the corresponding quadtree of the dyadic partition given in Figure 2(a).
A. Multiresolution Representation of the Environment A quadtree decomposition is used to obtain a multiresolu-tion representation of the environment, as shown in Figure 3.
The algorithm basically counts the number of obstacles in each dyadic square in an efficient manner. If the counted number of obstacles is in the interval [1 + α(2s−1 −1),2s − 1 −α(2s−1 −1)], where s and α are the scale of the dyadic square and tuning parameter, respectively, it divides the current dyadic square further. This algorithm continues partitioning until there are no dyadic square which can be divided any further.
(a) α = 0.25 (b) α = 0.75 Fig. 3: Quadtree decomposition of the environment for two different values of obstacle distribution.
Typically, the nodes of the quadtree are labeled as white (free of obstacles), black (full of obstacles), or gray (mixture of both free cells and obstacles), and the leaves of the quadtree are allowed to be only either white or black. In our implementation, this condition is relaxed and the quadtree is allowed to have gray leaves. The tendency of the quadtree to have gray leaves is controlled by the parameter α ∈[0,1].
The greater the value of α, the more the tendency to have gray leaves in the quadtree. The algorithm creates a quadtree with no gray leaves if α = 0; the quadtree has only the root node if α = 1.
B. Construction of the Beamlet Graph The beamlet graph BG ≜(V ,E ) is defined such that each element in the set of vertices V corresponds to a feasible beamlet. Two beamlets are geometrically connected if they share a common point. The beamlet neighborhood of a given beamlet i includes all beamlets which are geometrically connected to the end points of beamlet i, as shown in Figure 4. The edge set E ⊂V × V consists of all pairs (i, j) where i, j ∈V , such that the beamlets i and j are geometrically connected. Each edge eij ∈E in BG includes two types of information: the length of the beamlet j and the angle between the two beamlets as shown in Figure 5.
Fig. 4: Beamlet connectivity.
Fig. 5: Edge between beamlets After a multiresolution representation of the environment is given, we find the set of all feasible beamlets that are geometrically attached to one of the end points of a given beamlet b. First, the algorithm finds the set of dyadic squares which include one of the end points of the beamlet b.
The black dyadic squares are ignored since they are full of obstacles and do not include any feasible beamlets. For each boundary point of a white dyadic square, a beamlet emanating from one of the related end points of the beamlet b is created. There is no need to check its feasibility, since there are no obstacles in a white dyadic square. For gray dyadic squares, in addition to the creation of the corresponding beamlets, a feasibility check is required since the dyadic square is mixture of free and blocked cells. This algorithm helps to build the beamlet graph incrementally as needed during the search.
C. Description of Search Algorithm Several functions are defined in order to apply the search algorithm on the beamlet graph. An edge cost function D : E →R+ × [−π,π] maps each edge in the beamlet graph to a pair of distance and an angle (ℓ,θ). A path-cost function G : V →R+ ×R+ ×[0,π] maps each vertex to a cost vector whose components are (∑ℓ, ∑|θ|, |θ|max). ∑ℓis the sum of the length of all beamlets corresponding to the path from the start vertex to the current vertex, ∑|θ| is the sum of the absolute value of the angles between the beamlets of the path, and |θ|max is the maximum of the absolute values of those angles. A path-cost function add(c,d) updates the path-cost vector c with the given edge cost d. It adds the edge cost d = (ℓ,θ) to the current value of the path-cost information c = (∑ℓ,∑|θ|,|θ|max) and updates |θ|max if |θ| is greater than its current value. This function is given in Algorithm 2.
Since the path-cost information is a vector, the function g(s) is defined in line 11 of Algorithm 2 in order to compare the “goodness” of different path-cost vectors. The function g maps each path-cost vector to a single non-negative cost value. The cost value is a linear combination of the distance ℓand angle θ , scaled appropriately.
The precedence relation ≺(c′,c) between two path-cost vectors c and c′ is defined as follows: c′ precedes c if and only if gscore of c′ is smaller than gscore of c. Finally, a heuristic h : V →R+ is defined in line 7 in Algorithm 2 to guide the search over the beamlet graph. The heuristic is admissible since it considers the cost information (ℓ,θ) of a beamlet formed by the second point of the current beamlet sp2 and the goal point pgoal. Simple geometry ensures that these cost value contributions will never overestimate the actual ones.
The A search algorithm with Fibonacci heap , as priority queue is used to search for the optimal path on the beamlet graph. For any given start and goal points (pstart,pgoal), the algorithm first calls Initialize(pstart, pgoal) in line 4 in Algorithm 1 to compute a multiresolution represen-tation of the environment and initialize internally used data structures. Then, the ComputeBeamletsEmanating(BG, p) function is called at lines 5 and 6 to create a set of start and goal vertices in the beamlet graph. This function first Algorithm 1: Beamlet 1 Beamlet(pstart,pgoal,|θ|max) 2 frontier ←/ 0 3 explored ←/ 0 4 BG ←Initialize(pstart, pgoal) 5 sstart ←ComputeBeamletsEmanating(BG, pstart) 6 sgoal ←ComputeBeamletsEmanating(BG, pgoal) 7 foreach s′ ∈sstart do 8 parent(s′) ←null 9 c′ ∑ℓ←∥s′ p2 −s′ p1∥ 10 c′ ∑|θ| ←c′ |θ|max ←0 11 G(s′ ) ←c′ 12 frontier.insert(s′,g(s′)+ h(s′)) 13 while frontier ̸= / 0 do 14 s ←frontier.pop() 15 if s ∈sgoal then 16 return “path found” 17 explored.insert(s) 18 foreach s′ ∈neighbor(s) do 19 if s′ ̸∈explored then 20 d ←D(edge(s,s′)) 21 if |dθ| ≤|θ|max then 22 if s′ ̸∈frontier then 23 g(s′) ←∞ 24 UpdateState(s,s′) 25 UpdateState(s,s′) 26 d ←D(edge(s,s′)) 27 c ←G(s ) 28 c′ ←G(s′ ) 29 cnew ←add(c,d ) 30 if cnew ≺c′ then 31 parent(s′) ←s 32 G(s′ ) ←cnew 33 if s′ ∈frontier then 34 frontier.remove(s′) 35 frontier.insert(s′,g(s′)+ h(s′)) finds the set of dyadic squares which include the point p by traversing the quadtree and it then computes a set of feasible beamlets by attaching the point p to the points located at the boundary of the dyadic squares. The path-cost vector and parent of the each vertex in the set of start vertices sstart are initialized in line 7 to 11. The first and second points of a given beamlet s′ are represented by s′ p1 and s′ p2, respectively in line 9. Those vertices are inserted into the frontier heap. The rest of the search procedure is the same as in the A algorithm, except that during the vertex expansion step, the algorithm checks whether the transition violates the given curvature constraint before updating a neighbor vertex at line 23. If the angle value θ of the edge cost d is greater than |θ|max, then that neighbor vertex is ignored. As a result, although two vertices are geometrically connected, the transition between them is not allowed due to the violation of the curvature constraint.
Algorithm 2: Heuristic, Cost and Other Functions 1 add(c,d) 2 c∑ℓ←c∑ℓ+ dℓ 3 c∑|θ| ←c∑|θ| + |dθ| 4 if c|θ|max < |dθ| then 5 c|θ|max ←|dθ| 6 return c 7 h(s) 8 s′ ←Beamlet(sp2, pgoal) 9 d ←D(edge(s,s′)) 10 return β ∗dℓ∗nℓ+ (1 −β)∗|dθ|∗nθ 11 g(s) 12 c ←G(s) 13 gscore ←β ∗c∑ℓ∗nℓ+ (1 −β)∗c∑|θ| ∗nθ 14 return gscore 15 ≺ ≺ ≺(c′,c) 16 g′ score ←β ∗c′ ∑ℓ∗nℓ+ (1 −β)∗c′ ∑|θ| ∗nθ 17 gscore ←β ∗c∑ℓ∗nℓ+ (1 −β)∗c∑|θ| ∗nθ 18 return g′ score < gscore IV. COMPARISON WITH EXISTING METHODS We compared the proposed Beamlet algorithm with the A, the A with Post Smoothed Paths (A PS) and the Basic Theta algorithms on several graphs formed by an eight-neighbor square grid. All search algorithms were im-plemented in C++ on a 2.5 GHz Core 2 with 3 GB of RAM laptop. All algorithms solved the same problems on several artificially created 2D cluttered environments which are characterized by two different grid sizes and 4 different per-centages of occurrence of obstacles. For a given percentage of obstacles, some cells in the environment were randomly blocked in order to create a cluttered environment. For each environment, 100 path planning problems were created by choosing the start and goal points randomly. The computed paths were compared with respect to the following criteria, averaged over 100 path planning problems: the length of the computed path ℓ, the absolute value of maximum heading angle |θ|max, the execution time of the search algorithm te, the number of expanded vertices during the search nexp, and the number of heading changes along the path nhc. We used quadtree decompositions where α = 0 for all cases, that is, there are no gray leaves in the final quadtree. Finally, the values of β, nℓ, and nθ were set to 1.0, √ 2n, and π, where n is the dimension of the square which represents environment, respectively. We tried to find the shortest possible path which satisfies the given heading angle constraint.
First, Beamlet was used to solve a path planning problem whose only possible solution is a self-intersecting curve owing to a constraint on the heading angle along the path.
This is shown in Figure 6, where only left turns (positive heading angles) are allowed along the path. For this problem, the start and goal points are the top-left and bottom-right green points, respectively, as shown in Figure 6. None of the A, A PS and Basic Theta algorithms can find the solution for this problem. This is due to the fact that these algorithms will start expanding all vertices at the top-left area of the map from the start point, but will never reach the goal point after all vertices are expanded. Since each vertex which corresponds to a point in the graph is never re-expanded once it has been inserted in the explored list, the path computed by A, A PS and Basic Theta algorithms will never pass through the same point twice.
Fig. 6: Beamlet searches over a larger space of curves and can find self-intersecting paths.
The comparison of these algorithms with respect to the path length and the maximum heading angle is summarized in Table I. Other metrics related to the computed paths are illustrated in Figures 7-9. Figure 7 shows the number of heading changes along the corresponding paths, Figure 8 shows the number of expanded vertices, and Figure 9 shows the execution times of the search algorithms.
As seen in Table I, Beamlet can find smoother paths than Basic Theta and the other algorithms for all cases tested, as expected. This is owing to the fact that Beamlet allows us to explicitly constraint the upper and lower bounds of the heading changes along the path. Therefore, although a vertex in the beamlet graph has too many geometrically connected neighbor vertices, only the ones that have an edge which does not violate the heading angle bounds are updated during the search.
Fig. 7: Heading Changes (grid size: 256x256) The number of heading changes on the paths computed by Beamlet is usually larger than those computed by other algorithms, as shown in Figure 7. The reason is that Beamlet computes multiple soft maneuvers instead of a single sharp maneuver due to the constraint on the maximum absolute value of the heading angle. Also, it is obvious that the number of heading changes on a path depends on how cluttered the environment is, and how the obstacles are distributed inside the environment. Therefore, in our numerical comparison, the constraint on the heading angle was relaxed on more cluttered environments in order to increase the chance of finding a path. We constrained |θ|max to 15◦, 20◦, 25◦and 30◦when executing the Beamlet algorithm for path planning problems in the environments of 5%, 10%, 20%, and 30% obstacles, respectively.
Fig. 8: Vertex Expansions (grid size: 256×256) The Beamlet algorithm usually expands more vertices than the Basic Theta algorithm, as shown in Figure 8. There are two main reasons for this overexpansion of vertices: First, the beamlet graph is larger than a graph formed by just the grid points, in terms of both the number of vertices and the number of edges. Second, although, the start and goal points may be spatially very close to each other in a path planning problem, and perhaps there exists a short path between the two, the solution path will be naturally longer if one imposes additional constraints on the allowable heading angle changes along the path. Computing such a longer path over a larger graph will inevitably require a larger number of vertex expansions during the search.
Fig. 9: Execution Time (grid size: 256x256) Despite the large number of vertex expansions in the Beamlet algorithm, the execution time of Beamlet is reasonable, and sometimes even better than the Basic Theta if the environment is less cluttered, as seen in Figure 9. This is because Beamlet and Basic Theta use different vertex update procedures during the search. Basic Theta requires line-of-sight checking for each neighbor vertex of the current Maps A∗ A∗PS Basic Theta∗ Beamlet∗ ℓ |θ|max ℓ |θ|max ℓ |θ|max ℓ |θ|max 128x128 %5 Obstacle 114.521 47.701 109.895 25.919 108.643 13.518 114.191 9.503 %10 Obstacle 113.443 52.202 108.906 36.491 107.618 25.354 112.886 13.492 %20 Obstacle 112.905 55.352 108.298 39.914 107.380 33.994 121.778 14.788 %30 Obstacle 107.509 90.253 102.601 73.511 101.373 70.507 105.544 26.771 256x256 %5 Obstacle 231.993 52.652 224.841 37.849 220.310 22.445 225.348 12.191 %10 Obstacle 231.341 56.702 224.023 44.627 219.430 33.305 236.584 13.182 %20 Obstacle 234.551 82.352 225.681 63.228 222.767 54.608 232.998 21.739 %30 Obstacle 241.613 92.253 231.788 71.666 228.910 61.383 230.576 27.240 TABLE I: Beamlet computes smoother paths than Theta as seen in |θ|max column.
vertex during vertex expansion step. This extra operation contributes a significant portion to the total execution time.
On the other hand, Beamlet only requires to check the feasibility of a beamlet if it is included by a gray dyadic square.
V. CONCLUSIONS We have proposed a new algorithm (called Beamlet) to compute shortest paths on a graph such that curvature constraints along the path can be easily incorporated. Such local curvature constraints serve as dynamic information surrogates to obtain feasible trajectories, and thus can bridge the gap between the geometric path-planning and feasible trajectory generation levels in the motion planning hierarchy.
Compared to other similar approaches, the most important advantage of Beamlet is that it guarantees an a priori upper bound of the heading angle changes along the resulting path. This tends to result in smoother paths. On the other hand, the result of the Beamlet algorithm is dependent on the multiresolution representation of the environment, and the algorithm may not find a solution given a bad dyadic partition. In this work, only a simple heuristic based on the Euclidean distance was used, and a one-directional A search algorithm was implemented. Significant improvements in performance can be achieved by incorporating different types of heuristics (e.g., landmark and search algorithms (e.g., bidirectional , , . These refinements of Beamlet can increase its numerical efficiency. Finally, the proposed approach can be easily extended to path planning problems in the 3D world. In this case, the multiresolution representation of the environment will consist of dyadic cubes (instead of squares) of different scales, and the computed feasible beamlets will be stored in an octree (instead of an quadtree) data structure.
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What is the periodic function?
by krishna30/10/2021
by krishna4,561 views
Overview(periodic function)
A function f(x) is said to be periodic function if f(x + T) = f(x) for all real x and some positive number T, here T is the period of f(x).
Suppose if we take sin x, then it repeats its value after the period of 2pi
such that, we write this as
We can say that sin x is a periodic function with the period of .
sin x, cos x, sec x, cosec x are the periodic functions with period , where tan x and cot x are the periodic functions with period
Note
Sin x is also known as sinusoidal periodic function.
The period of a sum of a number of periodic functions is the least common multiple of the periods.
A constant function is periodic for any positive T.
Suppose T is the period of f(x) then nT is also periodic of f for any integer n.
Fourier series
Trigonometric series-
This is a functional series of the form,
The constants a0, an and bn are the coefficients.
Fourier series of a periodic function f(x) with period is the trigonometric series with Fourier coefficients a0, an and bn
Any periodic function can be expanded in the form of Fourier series.
How to determine
We know that, the Fourier series,
To find
Intergrate equation (1) on both sides, from0 to 2π
That gives
To find
Multiply each side of eq. (1) by cos nx and integrate from 0 to 2π
We get,
Similarly we can findby, multiplying eq. (1) by sin nx and integrating from 0 to 2π
Solved example
Example: Find the fourier series of the function f(x) = x where 0 < x < 2π
Sol.We know that, from Fourier series,
First we will find a_0
Now a_n
And
Now put the value in Fourier series, we get
Example: Find the Fourier series for f(x) =in the interval
Sol.
Suppose
Then
And
So that
And then
Now put these value in equations (1), we get-
Interested in learning about similar topics? Here are a few hand-picked blogs for you!
What is Fourier series?
Taylor’s series.
What is maclaurin series?
What is homogeneous differential equation?
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190166 | https://webbook.nist.gov/cgi/cbook.cgi?Source=1969CAM%2FTHR32&Units=CAL&Mask=2811 | Nitrogen
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Nitrogen
Formula: N 2
Molecular weight: 28.0134
IUPAC Standard InChI:InChI=1S/N2/c1-2 Copy
IUPAC Standard InChIKey:IJGRMHOSHXDMSA-UHFFFAOYSA-N Copy
CAS Registry Number: 7727-37-9
Chemical structure:
This structure is also available as a 2d Mol file or as a computed3d SD file
View 3d structure (requires JavaScript / HTML 5)
Other names: Nitrogen gas; N2; UN 1066; UN 1977; Dinitrogen; Molecular nitrogen; Diatomic nitrogen; Nitrogen-14
Permanent link for this species. Use this link for bookmarking this species for future reference.
Information on this page:
Gas phase thermochemistry data
Henry's Law data
References
Notes
Other data available:
Phase change data
Reaction thermochemistry data: reactions 1 to 50, reactions 51 to 100, reactions 101 to 150, reactions 151 to 156
Gas phase ion energetics data
Ion clustering data
Mass spectrum (electron ionization)
Constants of diatomic molecules
Fluid Properties
Data at other public NIST sites:
Electron-Impact Ionization Cross Sections (on physics web site)
Gas Phase Kinetics Database
Options:
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Data at NIST subscription sites:
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NIST subscription sites provide data under the NIST Standard Reference Data Program, but require an annual fee to access. The purpose of the fee is to recover costs associated with the development of data collections included in such sites. Your institution may already be a subscriber. Follow the links above to find out more about the data in these sites and their terms of usage.
Gas phase thermochemistry data
Go To:Top, Henry's Law data, References, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
| Quantity | Value | Units | Method | Reference | Comment |
--- --- --- |
| S°gas,1 bar | 45.7957 ± 0.001 | cal/molK | Review | Cox, Wagman, et al., 1984 | CODATA Review value |
| S°gas,1 bar | 45.796 | cal/molK | Review | Chase, 1998 | Data last reviewed in March, 1977 |
Gas Phase Heat Capacity (Shomate Equation)
C p° = A + Bt + Ct 2 + Dt 3 + E/t 2
H° − H°298.15= At + Bt 2/2 + Ct 3/3 + Dt 4/4 − E/t + F − H
S° = Aln(t) + Bt + Ct 2/2 + Dt 3/3 − E/(2t 2) + G
C p = heat capacity (cal/molK)
H° = standard enthalpy (kcal/mol)
S° = standard entropy (cal/molK)
t = temperature (K) / 1000.
View plot Requires a JavaScript / HTML 5 canvas capable browser.
View table.
| Temperature (K) | 100. to 500. | 500. to 2000. | 2000. to 6000. |
| A | 6.927919 | 4.662006 | 8.489178 |
| B | 0.443111 | 4.753120 | 0.269772 |
| C | -2.305798 | -2.055099 | -0.046870 |
| D | 3.975949 | 0.327386 | 0.003504 |
| E | 0.000028 | 0.126100 | -1.088375 |
| F | -2.072637 | -1.179542 | -4.534157 |
| G | 54.11491 | 50.76243 | 53.77175 |
| H | 0.0 | 0.0 | 0.0 |
| Reference | Chase, 1998 | Chase, 1998 | Chase, 1998 |
| Comment | Data last reviewed in March, 1977; New parameter fit January 2009 | Data last reviewed in March, 1977; New parameter fit January 2009 | Data last reviewed in March, 1977; New parameter fit January 2009 |
Henry's Law data
Go To:Top, Gas phase thermochemistry data, References, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Data compiled by:Rolf Sander
Henry's Law constant (water solution)
k H(T) = k°H exp(d(ln(k H))/d(1/T) ((1/T) - 1/(298.15 K)))
k°H = Henry's law constant for solubility in water at 298.15 K (mol/(kgbar))
d(ln(k H))/d(1/T) = Temperature dependence constant (K)
| k°H (mol/(kgbar)) | d(ln(k H))/d(1/T) (K) | Method | Reference |
--- --- |
| 0.00060 | 1300. | X | N/A |
| 0.00065 | 1300. | L | N/A |
References
Go To:Top, Gas phase thermochemistry data, Henry's Law data, Notes
Data compilation copyright by the U.S. Secretary of Commerce on behalf of the U.S.A. All rights reserved.
Cox, Wagman, et al., 1984
Cox, J.D.; Wagman, D.D.; Medvedev, V.A., CODATA Key Values for Thermodynamics, Hemisphere Publishing Corp., New York, 1984, 1. [all data]
Chase, 1998
Chase, M.W., Jr., NIST-JANAF Themochemical Tables, Fourth Edition, J. Phys. Chem. Ref. Data, Monograph 9, 1998, 1-1951. [all data]
Notes
Go To:Top, Gas phase thermochemistry data, Henry's Law data, References
Symbols used in this document:
S°gas,1 bar Entropy of gas at standard conditions (1 bar)
d(ln(k H))/d(1/T)Temperature dependence parameter for Henry's Law constant
k°H Henry's Law constant at 298.15K
Data from NIST Standard Reference Database 69: NIST Chemistry WebBook
The National Institute of Standards and Technology (NIST) uses its best efforts to deliver a high quality copy of the Database and to verify that the data contained therein have been selected on the basis of sound scientific judgment. However, NIST makes no warranties to that effect, and NIST shall not be liable for any damage that may result from errors or omissions in the Database.
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190167 | https://www.quora.com/How-do-you-use-the-rational-root-theorem-to-solve-a-polynomial-with-no-constant-term | Something went wrong. Wait a moment and try again.
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5
How do you use the rational root theorem to solve a polynomial with no constant term?
Brand Burton
Technical Editor (2015–present)
·
3y
If there's no constant term then you haven't taken out a common factor. That is, a polynomial of the form:
f(x)=anxn+an−1xn−1+...+a2x2+a1x
still has an x in each term. Factoring that out, it is more conveniently expressed as:
f(x)=x(an−1xn−1+an−2xn−2+...+a1x+a0).
Then you can apply the rational root theorem to find further roots of the polynomial.
Edit note: This is the first time I’ve used Quora’s mathematics markup, so I think I must have edited it a couple dozen times.
John Calligy
Author has 2.2K answers and 1.6M answer views
·
3y
Nitpicky but still useful answer: there's no such thing as a polynomial with no constant term. Some polynomials have a constant term of 0. Of course, we usually don't bother to write +0 at the end. In this case, 0 is a root of the polynomial, and we can reduce all the other exponents by 1 and solve as usual. Repeat if necessary.
Wes Browning
Studied Mathematics at Cornell University
·
Author has 8.8K answers and 6M answer views
·
3y
Divide by the variable. If there's still no constant term, repeat. Keep repeating until there's a nonzero constant term or te polynomial is constant. You should be able to know what to do then.
Related questions
Where do we use the Rational root theorem in real life?
How can I prove that the polynomial
y
=
x
3
−
x
2
−
15
has no rational roots?
Where can I solve polynomials?
Let p be a polynomial with rational coefficients and degree greater than 1. Does there always exist an irrational number s such that p(s) is rational? I could answer that for certain types of polynomials but failed to generalise it. What do I miss?
Why do we teach the rational root theorem (secondary education, technology in education, solving polynomials, rational numbers, roots, math, education)?
Assistant
Bot
·
1y
The Rational Root Theorem states that if a polynomial has a rational root (a fraction pq), then p is a factor of the constant term and q is a factor of the leading coefficient. When dealing with a polynomial that has no constant term, the constant term is zero, which means we can simplify our approach.
Here’s how to use the Rational Root Theorem for a polynomial with no constant term:
Steps:
Identify the Polynomial: Consider a polynomial of the form:
P(x)=anxn+an−1xn−1+…+a1x
Here, the constant term is zero.
Apply the Rational Root Theorem: Since there is no
The Rational Root Theorem states that if a polynomial has a rational root (a fraction pq), then p is a factor of the constant term and q is a factor of the leading coefficient. When dealing with a polynomial that has no constant term, the constant term is zero, which means we can simplify our approach.
Here’s how to use the Rational Root Theorem for a polynomial with no constant term:
Steps:
Identify the Polynomial: Consider a polynomial of the form:
P(x)=anxn+an−1xn−1+…+a1x
Here, the constant term is zero.
Apply the Rational Root Theorem: Since there is no constant term, the only rational roots we can consider are those where p (the factor of the constant term) is zero. However, we can still look for integer roots, specifically x=0 and potentially other integer factors of the leading coefficient.
Check Possible Roots: The potential rational roots will be:
0 (since p=0)
Any integer factors of the leading coefficient an (for q=1)
Substitute and Simplify: Substitute these potential roots into the polynomial to check if they yield a value of zero:
Start with x=0. If P(0)=0, then x=0 is a root.
For other integer factors of an, substitute each into the polynomial and check if P(x)=0.
Factor the Polynomial: If you find a root r, you can factor the polynomial as:
P(x)=(x−r)Q(x)
where Q(x) is the resulting polynomial after dividing P(x) by (x−r).
Repeat: Continue applying the Rational Root Theorem to Q(x) until you find all roots or until Q(x) can no longer be factored using rational roots.
Example:
Consider the polynomial P(x)=2x3−4x2+6x.
Identify the polynomial: It has no constant term.
Possible rational roots: Check 0 and factors of 2 (the leading coefficient):
Possible roots: 0,1,−1,2,−2.
Check x=0:
P(0)=2(0)3−4(0)2+6(0)=0(root found)
Factor out x:
P(x)=x(2x2−4x+6)
Check 2x2−4x+6: Use the quadratic formula to find other roots:
x=−(−4)±√(−4)2−4⋅2⋅62⋅2=4±√16−484=4±√−324=1±i√2
Final Roots: The roots of the polynomial are 0,1+i√2,1−i√2.
Conclusion
Using the Rational Root Theorem with polynomials that have no constant term focuses primarily on finding roots at x=0 and checking integer factors of the leading coefficient. This method is effective for simplifying the polynomial and finding rational or complex roots efficiently.
Saibal Mitra
Author has 1.1K answers and 661.6K answer views
·
3y
In that case
x
=
0
is a zero, you can then divide by
x
=
0
and find the other zeroes. After diving by
x
there will either be a constant term or not. In the latter case, you can divide by
x
again and proceed in the same way (you then know that the zero at
x
=
0
has a multiplicity of
2
or larger).
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
·
Author has 3.9K answers and 4M answer views
·
3y
Related
How do you find the roots of an equation that has no constant term?
If there is no constant term, then factor out the lowest power of the variable. Then, factor the remaining polynomial.
For example, consider the equation:
x4−4x3−9x2+36x=0
Notice that there is no constant term and the lowest power of the variable is one. Next, factor out x:
x(x3−4x2−9x+36)=0
We see that x=0 is a solution. Next, we work on the newly created polynomial in the parentheses. Notice that this polynomial has a constant term. We can factor this polynomial using the rational root theorem and synthetic division. A more advanced approach would be to notice that this polynomial has the fo
If there is no constant term, then factor out the lowest power of the variable. Then, factor the remaining polynomial.
For example, consider the equation:
x4−4x3−9x2+36x=0
Notice that there is no constant term and the lowest power of the variable is one. Next, factor out x:
x(x3−4x2−9x+36)=0
We see that x=0 is a solution. Next, we work on the newly created polynomial in the parentheses. Notice that this polynomial has a constant term. We can factor this polynomial using the rational root theorem and synthetic division. A more advanced approach would be to notice that this polynomial has the form:
x3−ax2−bx+ab
where a=4, b=9 and ab=36 which factors as:
(x−a)(x2−b)
(x−4)(x2−9)
such that we get:
x(x−4)(x2−9)=0
Finally, we can factor the quadratic polynomial as:
x2−9=(x+3)(x−3)
to get:
x(x−4)(x+3)(x−3)=0
where we see the roots of the original equation:
x=0, x=4, x=−3, x=3
Related questions
What is the rational root theorem for finding the roots of a 3 degree polynomial equation?
Is there a method to work out if the roots of a polynomial (of order n) are rational without solving for the roots?
What is the most efficient method to solve quintic real roots (polynomials, roots, math)?
How do you find the zeros of a polynomial when there are no rational roots (by hand)? I tried factoring, the rational roots theorem, etc., or correct me if I missed something. The problem is: y=2x^4-7x^3+11x-4.
Why is the Rational Root Theorem not always applicable for finding all possible roots in a polynomial equation, even if there is at least one rational root?
Rik Bos
Ph.D. Mathematics from Utrecht University (Graduated 1979)
·
Author has 1.3K answers and 1.3M answer views
·
4y
Related
How can I prove that the polynomial
y
=
x
3
−
x
2
−
15
has no rational roots?
You can apply the rational root theorem (RRT). But assuming you don’t know that, we can also use very basic facts about integers. So assume we have a rational root
t
. Then we can write
t
=
m
/
n
where
m
,
n
are integers,
n
0
and
. Since
[math]t[/math]
is a root we have
[math]t^3=t^2+15[/math]
. After clearing denominators we get
[math]m^3=m^2n+15n^3[/math]
. But then any prime divisor of
[math]n[/math]
divides the RHS, so divides
[math]m[/math]
as well, contrary to
[math]\gcd(m,n)=1[/math]
. Therefore
[math]n[/math]
has no prime divisors, meaning
[math]n=1[/math]
. But then
[math]m^3-m^2=15[/math]
(and in particular
[math]m[/math]
divides
[math]15[/math]
; I mention this because that would be something you arrive at when applying the
You can apply the rational root theorem (RRT). But assuming you don’t know that, we can also use very basic facts about integers. So assume we have a rational root [math]t[/math]. Then we can write [math]t=m/n[/math] where [math]m,n[/math] are integers, [math]n>0[/math] and [math]\gcd(m,n)=1[/math]. Since [math]t[/math] is a root we have [math]t^3=t^2+15[/math]. After clearing denominators we get [math]m^3=m^2n+15n^3[/math]. But then any prime divisor of [math]n[/math] divides the RHS, so divides [math]m[/math] as well, contrary to [math]\gcd(m,n)=1[/math]. Therefore [math]n[/math] has no prime divisors, meaning [math]n=1[/math]. But then [math]m^3-m^2=15[/math] (and in particular [math]m[/math] divides [math]15[/math]; I mention this because that would be something you arrive at when applying the RRT). But that’s impossible, since [math]m^3-m^2[/math] is even.
Therefore, there is no rational root in this case.
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David Joyce
Ph.D. in Mathematics, University of Pennsylvania (Graduated 1979)
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When can I not use rational root theorem?
The rational root theorem says that if a rational number [math]r/s[/math] in lowest terms is a root of a polynomial with integer coefficients, then [math]s[/math] divides the leading coefficient and [math]r[/math] divides the constant.
It doesn’t imply the converse. If [math]s[/math] divides the leading coefficient and [math]r[/math] divides the constant, then you can’t conclude that [math]r/s[/math] is a root of the polynomial.
You can use this theorem to look for rational roots and find them all.
Example. Take the polynomial [math]f(x)=2x^3-3x^2+10x-15.[/math] The leading coefficient is [math]2[/math] and the constant is [math]-15.[/math] the only rational numbers that could be solutions are [math]\pm1,\pm3,\pm5,\p[/math]
The rational root theorem says that if a rational number [math]r/s[/math] in lowest terms is a root of a polynomial with integer coefficients, then [math]s[/math] divides the leading coefficient and [math]r[/math] divides the constant.
It doesn’t imply the converse. If [math]s[/math] divides the leading coefficient and [math]r[/math] divides the constant, then you can’t conclude that [math]r/s[/math] is a root of the polynomial.
You can use this theorem to look for rational roots and find them all.
Example. Take the polynomial [math]f(x)=2x^3-3x^2+10x-15.[/math] The leading coefficient is [math]2[/math] and the constant is [math]-15.[/math] the only rational numbers that could be solutions are [math]\pm1,\pm3,\pm5,\pm15,\pm\frac12,\pm\frac32,\pm\frac52,[/math] and [math]\pm\frac{15}2.[/math] Of these eight rational numbers, only [math]\frac32[/math] is a root. Since [math]f(x)=(2x-3)(x^2+5),[/math] therefore all the roots are [math]\frac32,i\sqrt 5,[/math] and [math]-i\sqrt5.[/math] Only the first of these three roots is rational.
In answer to your question, you can use the rational root theorem if you have a polynomial with integer coefficients, and you can’t if you don’t have a polynomial with integer coefficients.
The rational root theorem actually gives you a quick way to show that algebraic irrational numbers are irrational. For example, to show that [math]\sqrt2[/math] is irrational, it’s enough to note that it’s a root of [math]f(x)=x^2-2,[/math] then check that none of the four possible rational roots [math]\pm1,\pm2[/math] are actually roots of [math]f(x).[/math]
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Roberto Mastropietro
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David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 172 answers and 134.5K answer views
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How does the "Rational Root Theorem” and “Factor Theorem” helps you in solving polynomial equation? ?
The rational theorem suggests you possible (but not guaranteed) rational solutions for the polynomial, coming in the form [math]\pm\frac{p}{q}[/math], where [math]p[/math] is a divisor of the last term and [math]q[/math] is a divisor of the leading term of the polynomial.
Now, suppose you have a polynomial [math]p(x)[/math] and you find out that [math]\bar{x}[/math] is a solution, i.e. [math]p(\bar{x})=0[/math]. In this case, you can factor the polynomial as
[math]p(x)=(x-\bar{x})q(x)[/math]
where [math]q(x)[/math] is a polynomial whose degree is one less than [math]p(x)[/math]. Now you can look for solutions of [math]q(x)[/math], and so on until you reach degree zero or you have a polynomial with no solution. For example,
The rational theorem suggests you possible (but not guaranteed) rational solutions for the polynomial, coming in the form [math]\pm\frac{p}{q}[/math], where [math]p[/math] is a divisor of the last term and [math]q[/math] is a divisor of the leading term of the polynomial.
Now, suppose you have a polynomial [math]p(x)[/math] and you find out that [math]\bar{x}[/math] is a solution, i.e. [math]p(\bar{x})=0[/math]. In this case, you can factor the polynomial as
[math]p(x)=(x-\bar{x})q(x)[/math]
where [math]q(x)[/math] is a polynomial whose degree is one less than [math]p(x)[/math]. Now you can look for solutions of [math]q(x)[/math], and so on until you reach degree zero or you have a polynomial with no solution. For example, if we start with the polynomial
[math]p(x)=2 x^4 + x^3 + x^2 + 2 x - 6[/math]
using the rational root theorem, we can find you that [math]x=-\frac{3}{2}[/math] is a solution. So, we divide [math]p(x)[/math] by [math]x+\frac{3}{2}[/math], and we get
[math]\dfrac{2 x^4 + x^3 + x^2 + 2 x - 6}{x+\frac{3}{2}}=2 x^3 - 2 x^2 + 4 x - 4[/math]
Again, using the rational root theorem on [math]2 x^3 - 2 x^2 + 4 x - 4[/math], we find out that [math]x=1[/math] is a solution. So, we go ahead and divide
[math]\dfrac{2 x^3 - 2 x^2 + 4 x - 4}{x-1}=2x^2+4[/math]
We can rewrite this polynomial as [math]2x^2+4=2(x^2+2)[/math], which clearly has no real solutions. So, we can factor our original polynomial as
[math]\displaystyle p(x)=2 x^4 + x^3 + x^2 + 2 x - 6=2\left(x+\frac{3}{2}\right)(x-1)(x^2+2)[/math]
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Raymond Beck
Former Infantry Sergeant
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Author has 42.4K answers and 10.2M answer views
·
Jan 18
Related
How do you find the roots of an equation that has no constant term?
x^2 +4x = 0 has no constant term, so one root = 0
x(x+4) = 0, set each factor = 0, solve for x
x = 0, or -4
higher degree equations with no constant term will also have one root =0
x^3 +x = 0
x(x^2+1) = 0
x = 0, i, -i
Terry Moore
M.Sc. in Mathematics, University of Southampton (Graduated 1968)
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Author has 16.4K answers and 29.1M answer views
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2y
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Let P(x) be a polynomial of even degree and all of its coefficients are odd numbers. How do I prove that it is impossible for the polynomials to have a rational root?
Let P(x) be a polynomial of even degree and all of its coefficients are odd numbers. How do I prove that it is impossible for the polynomials to have a rational root?
By “odd number” I assume you mean odd integer (therefore not complex).
Let the degree be [math]n[/math]. A rational root corresponds to a factor of the form [math]mx-c[/math].
There is no loss of generality in assuming that the coefficient of the highest degree term is [math]1[/math] (monic polynomial). Why? If we multiply [math]a_nx^n[/math] by [math]a_n^{n-1}[/math] all coefficients remain odd and we may let [math]y=a_nx[/math]. So we have a monic polynomial in [math]y[/math] and [math]m=1[/math] so a rational root will be an intege
Let P(x) be a polynomial of even degree and all of its coefficients are odd numbers. How do I prove that it is impossible for the polynomials to have a rational root?
By “odd number” I assume you mean odd integer (therefore not complex).
Let the degree be [math]n[/math]. A rational root corresponds to a factor of the form [math]mx-c[/math].
There is no loss of generality in assuming that the coefficient of the highest degree term is [math]1[/math] (monic polynomial). Why? If we multiply [math]a_nx^n[/math] by [math]a_n^{n-1}[/math] all coefficients remain odd and we may let [math]y=a_nx[/math]. So we have a monic polynomial in [math]y[/math] and [math]m=1[/math] so a rational root will be an integer.
So we may take [math]P(x)[/math] to be a monic polynomial with an integer root. This root must divide [math]a_0[/math] and is therefore odd.
Let [math]P(x)=(x-c)Q(x)[/math] where [math]Q(x)=b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\dots+b_0[/math]. Then [math]b_{n-1}=1[/math], [math]a_0=-cb_0[/math], [math]a_1=b_0-cb_1[/math], [math]a_2=b_1-cb_2[/math], etc. So [math]b_0[/math] and [math]c[/math] must be odd, [math]b_1[/math] must be even, so [math]b_2[/math] must be odd, and so on. In other words odd numbered coefficients are even. But [math]Q(x)[/math] has odd degree and [math]b_{n-1}[/math] is even. But we know that [math]b_{n-1}=1[/math]. So this a contradiction.
So assuming there was a rational root leads to a contradiction so there is no rational root.
Howard Ludwig
Ph.D. in Physics, Northwestern University (Graduated 1982)
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Author has 3K answers and 10.2M answer views
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How do I solve the rational roots of a polynomial of x³ 12x² 20x 5?
“How do I solve the rational roots of a polynomial of x³ 12x² 20x 5?”
Your question cannot be answered for two reasons:
Your expression appears to be the product of four monomials: (𝑥³)(12𝑥²)(20𝑥)(5) = 1200𝑥⁶, which has 0 as a 6-fold zero. I suspect that you meant to put some mix of plus signs and minus signs between those monomials. What mix of signs you put in changes what the zeros/roots will be.
Polynomials do not have roots. Polynomials have zeros. Polynomial equations have roots. An equation has an equality sign [=], or at least corresponding words, with values on both sides.
Did you me
“How do I solve the rational roots of a polynomial of x³ 12x² 20x 5?”
Your question cannot be answered for two reasons:
Your expression appears to be the product of four monomials: (𝑥³)(12𝑥²)(20𝑥)(5) = 1200𝑥⁶, which has 0 as a 6-fold zero. I suspect that you meant to put some mix of plus signs and minus signs between those monomials. What mix of signs you put in changes what the zeros/roots will be.
Polynomials do not have roots. Polynomials have zeros. Polynomial equations have roots. An equation has an equality sign [=], or at least corresponding words, with values on both sides.
Did you mean to ask for the “rational zeros” of the polynomial, or did you mean to specify “polynomial equation” and to put an equality sign with some value [perhaps 0, perhaps something else]?
If you intend to solve for the zeros of the polynomial 𝑥³ ± 12𝑥² ± 20𝑥 ± 5, where at each “±” you pick either the plus sign or the minus sign, we can answer this—you ask merely how to solve, not what are the solutions. That is the much better type of question, so that you learn how to do the problem rather than having somebody else do the work for you.
The polynomial has all coefficients being integers, so the rational root theorem applies.
The constant coefficient is 5, whose only positive divisors are 1 and 5.
The leading coefficient is 1, whose only positive divisor is 1.
Now we take, one at a time, each divisor of the constant coefficient, divide it by each divisor of the leading coefficient, and record the quotient and the negative of the quotient:
1/1 = 1, so record 1 and −1; 5/1 = 5, so record 5 and −5.
Substitute each resulting recorded value (in this case four of them), one at a time, for 𝑥 in the polynomial, compute the resulting polynomial value, and record those values of 𝑥 for which the polynomial yields a value of 0—these are the rational zeros of the polynomial.
There are some shortcuts you might be able to take:
If all terms of the polynomial are added or all terms are subtracted (including a minus sign on the first term), then only negative values of 𝑥 can possibly yield 0, so do not try the positive values for 𝑥 [Descartes rule of signs]. Similarly, if all terms with an odd exponent applied to 𝑥 (including the 𝑥 term with implied exponent 1) have the same sign (all positive or all negative) and all terms with an even exponent applied to 𝑥 (including the constant term which has implied 𝑥⁰), then only positive values of 𝑥 can possibly yield 0, so do not try the negative values for 𝑥.
If any one term strongly dominates all of the other terms combined for any value of 𝑥, then neither 𝑥 nor −𝑥 can yield a polynomial value of 0. [Extension of the triangle inequality theorem].
This situation actually applies to 𝑥³ ± 12𝑥² ± 20𝑥 ± 5, no matter which signs you pick.
— For 𝑥 = ±1, look at the absolute value of each term: 1, 12, 20, 5 and notice that the sum of all terms except 20 is 1 + 12 + 5 = 18 < 20. It does not matter which terms are added and which terms are subtracted nor whether you are using +1 or −1 for 𝑥, the sign of the value of the polynomial is the same as the sign of that dominant term (including the minus if the dominant term is subtracted). Thus, neither 1 nor −1 can be a zero of the given polynomial regardless of sign mix.
— For 𝑥 = ±5, look at the absolute value of each term: 125, 300, 100, 5 and notice that the sum of all terms except 300 is 125 + 100 + 5 = 230 < 300, so, again one term (this time 12𝑥², unlike for 𝑥 = ±1). Thus, neither 5 nor −5 can be a zero of the given polynomial regardless of sign mix.
— We have worked through all possible rational zeros indicated by the rational root theorem, and one worked. Therefore, regardless of sign mix of the given terms of the polynomial, there is no rational zero.
Other “tricks” apply, but are challenging to describe with Quora’s layout.
Charles Armentrout
Former Retd Ph.D. Physicist (1992–2001)
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Author has 271 answers and 335.5K answer views
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1y
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How do I solve the rational roots of a polynomial of x³ 12x² 20x 5?
Please resubmit this question using standard notation.
It is possible that you are using a notation that your own class room developed, but in the usual “standard” notation I must translate this as “x_cubed times 12 times x_squared times 20 times x times 5” Really? this would not be a polynomial but isolated term with the variable to the sixth power … 1200x^6. I suspect that this is really not what you were taught but your own simplification.
I would be very surprised to find that you were never taught the nature of (+), (–), (×), (/). The multiply and divide symbols have many different ways to
Please resubmit this question using standard notation.
It is possible that you are using a notation that your own class room developed, but in the usual “standard” notation I must translate this as “x_cubed times 12 times x_squared times 20 times x times 5” Really? this would not be a polynomial but isolated term with the variable to the sixth power … 1200x^6. I suspect that this is really not what you were taught but your own simplification.
I would be very surprised to find that you were never taught the nature of (+), (–), (×), (/). The multiply and divide symbols have many different ways to write them. and the order of precedence is required to make sense of something like 6×2/3+4/2^2.
You allow your responders to use their own imaginations to give you an answer.
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190168 | https://www.youtube.com/watch?v=QYYWfmgooGk | [IMO Combinatorics] 2009 IMO shortlist using Invariants and Monovariants
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In the previous video, we saw how invariants and monovariants are useful tools in solving certain combinatorics problems in mathematical olympiads. We apply these techniques to solve an International Mathematical Olympiad shortlist problem from 2009.
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190169 | https://artofproblemsolving.com/wiki/index.php/Geometric_sequence?srsltid=AfmBOoocTkQiO257Nw9Rh8nhfgNoHQLHHlNtYnZzZTnqVexuA7TiCoeM | Art of Problem Solving
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Geometric sequence
In algebra, a geometric sequence, sometimes called a geometric progression, is a sequence of numbers such that the ratio between any two consecutive terms is constant. This constant is called the common ratio of the sequence.
For example, is a geometric sequence with common ratio and is a geometric sequence with common ratio ; however, and are not geometric sequences, as the ratio between consecutive terms varies.
More formally, the sequence is a geometric progression if and only if . A similar definition holds for infinite geometric sequences. It appears most frequently in its three-term form: namely, that constants , , and are in geometric progression if and only if .
Contents
[hide]
1 Properties
2 Sum
2.1 Finite
2.2 Infinite
3 Problems
3.1 Introductory
3.2 Intermediate
4 See also
Properties
Because each term is a common multiple of the one before it, every term of a geometric sequence can be expressed as the sum of the first term and a multiple of the common ratio. Let be the first term, be the th term, and be the common ratio of any geometric sequence; then, .
A common lemma is that a sequence is in geometric progression if and only if is the geometric mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of geometric sequences.
Sum
A geometric series is the sum of all the terms of a geometric sequence. They come in two varieties, both of which have their own formulas: finitely or infinitely many terms.
Finite
A finite geometric series with first term , common ratio not equal to one, and total terms has a value equal to .
Proof: Let the geometric series have value . Then Factoring out , mulltiplying both sides by , and using the difference of powers factorization yields Dividing both sides by yields , as desired.
Infinite
An infinite geometric series converges if and only if ; if this condition is satisfied, the series converges to .
Proof: The proof that the series convergence if and only if is an easy application of the ratio test from calculus; thus, such a proof is beyond the scope of this article. If one assumes convergence, there is an elementary proof of the formula that uses telescoping. Using the terms defined above, Multiplying both sides by and adding , we find that Thus, , and so .
Problems
Here are some problems with solutions that utilize geometric sequences and series.
Introductory
2025 AMC 8 Problem 20
Intermediate
1965 AHSME Problem 36
2005 AIME II Problem 3
2007 AIME II Problem 12
See also
Arithmetic sequence
Harmonic sequence
Sequence
Series
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190170 | https://www.aatbio.com/resources/faq-frequently-asked-questions/do-plant-cells-have-lysosomes | Do plant cells have lysosomes? | AAT Bioquest
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Do plant cells have lysosomes?
Posted November 30, 2022
Cell Structures and OrganellesCellular ProcessesEnzymesLysosomesPhysiological Probes
Answer
Plant cells do not have lysosomes, for a few reasons. First, plant cells have cell walls durable enough to keep foreign substances that lysosomes would typically digest out. Second, plant cells have vacuoles containing hydrolytic enzymes that function similarly to lysosomes in an animal cell. These reasons together make it structurally unnecessary for plant cells to need lysosomes. There is some debate whether plant cells could have lysosomes, as many processes are possible in biology, but scientists agree only in very rare cases could plant cells have lysosomes.
Additional resources
Cell Navigator® Lysosome Staining Kit Deep Red FluorescenceComparing Plant and Animal CellsLysoBrite™ NIR
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190171 | https://www.atlas.org/solution/12676d38-9a35-4162-85da-cf29da002194/convert-the-following-angle-in-degrees-to-radians-5-degrees | BlogSign inSign up free
Question
Convert the following angle in degrees to radians. 5 degrees
Answer
100% (1 rated)
π 36\frac{\pi}{36}36 π
Steps
1. To convert degrees to radians, we use the conversion formula: radians=degrees×π 180\text{radians} = \text{degrees} \times \frac{\pi}{180}radians=degrees×180 π Here, the given angle is 5 degrees.
2. Substituting 5 degrees into the formula: radians=5×π 180\text{radians} = 5 \times \frac{\pi}{180}radians=5×180 π This simplifies to: radians=5 π 180\text{radians} = \frac{5\pi}{180}radians=180 5 π
3. Now, we can simplify 5 π 180\frac{5\pi}{180}180 5 π: 5 π 180=π 36\frac{5\pi}{180} = \frac{\pi}{36}180 5 π=36 π Thus, 5 degrees is equal to π 36\frac{\pi}{36}36 π radians.
4. Finally, we provide the answer which is the value of the angle in radians as the solution to the original question.
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Solved: Convert the following angle in degrees to radians. 5 degrees
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190172 | https://brainly.com/question/42462919 | [FREE] Why is \Delta H_f^\circ 62.4 kJ/mol for I_2(g), but 0.0 kJ/mol for I_2(s)? A. Enthalpy changes occur only - brainly.com
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Why is Δ H f∘ 62.4 kJ/mol for I 2(g), but 0.0 kJ/mol for I 2(s)?
A. Enthalpy changes occur only when gases are formed.
B. I 2(s) is in its standard state.
C. Δ H f∘ is 0.0 kJ/mol for all solids.
D. The heat of formation for gases is different from that of solids.
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The standard enthalpy of formation (∆Hf°) is zero for I2 (s) because iodine is in its standard state and no energy is required to form it. However, for I2 (g), the ∆Hf° is 62.4 kJ/mol because it needs energy to be converted from solid to gas. Hence, the heat of formation for gases is different from that of solids.
Explanation
The standard enthalpy of formation, or ∆Hf°, is defined as the heat change that occurs when one mole of a compound is formed from its elements in their most stable states at 1 bar and 298.15 K. For elements in their standard state (which is solid for iodine, I2), the heat of formation is defined as 0 kJ/mol, simply because it is their most stable state and no heat is involved or required to create them; they already exist as they are. This is the case for I2 (s). However, if the substance in question is not in its most stable state (like iodine gas, I2 (g)), energy changes are required to convert it from the standard state to the new state. The enthalpy of formation, in such a case, would not be zero because it represents the energy changes (heat) required to form the substance in a non-standard state from its elements in their most stable states. This is why I2 (g) has a ∆Hf° of 62.4 kJ/mol: That's the amount of heat involved in its formation from solid iodine. The key takeaways here are: formation enthalpy pertains to how a substance forms (whether it be solid, liquid, or gas) and how this process involves energy changes. Furthermore, the most stable state of an element will always have a formation enthalpy of zero because no energy is needed to form it.
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The standard enthalpy of formation (
\Delta H_f^
\circ) for I₂(s) is 0.0 kJ/mol because it is in its standard state, while for I₂(g), it is 62.4 kJ/mol due to the energy required to convert the solid to gas. This difference illustrates that the enthalpy of formation depends on the phase of the substance. Therefore, the correct answer is B.
Explanation
The standard enthalpy of formation (
\Delta H_f^
\circ) is the heat change that occurs when one mole of a compound is formed from its elements in their standard states at a specified temperature and pressure, usually 25°C and 1 atm.
For iodine (I₂), the standard state is as a solid (I₂(s)), where it is most stable. Therefore, the
\Delta H_f^
\circ for I₂(s) is defined as 0.0 kJ/mol because it does not require energy to form its most stable state; it already exists as a solid at standard conditions.
In contrast, I₂ gas (I₂(g)) is not in its standard state at room temperature and pressure. For I₂(g), the
\Delta H_f^
\circ is 62.4 kJ/mol. This value indicates the energy required to convert iodine from its solid state to its gaseous state. The formation of a gas from a solid necessitates an input of energy to overcome intermolecular forces and convert to a more energetic state (gas), hence a positive
\Delta H_f^
\circ value.
Thus, the correct answer to why
\Delta H_f^
\circ is different for I₂(g) and I₂(s) is that
I₂(s) is in its standard state while I₂(g) requires energy to be formed from solid iodine, demonstrating that the standard enthalpy of formation is 0.0 kJ/mol for I₂(s) and non-zero for I₂(g). Therefore, this implies that:
B. I₂(s) is in its standard state.
Examples & Evidence
Consider other elements: the standard enthalpy of formation for solid carbon (graphite) is 0.0 kJ/mol because it is its most stable form. However, gaseous carbon (C(g)) would have a non-zero
\Delta H_f^
\circ since energy is needed to vaporize the solid into gas.
Chemistry textbooks and resources define that the standard enthalpy of formation for an element in its standard state is taken as 0.0 kJ/mol, confirming this concept.
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Community Answer At room temperature, iodine exists as a solid, and hydrogen exists as a gas. Consider the reaction: 2 HI(g) → H2(g) + 12 (s) AHO = -53.0 kJ What is the enthalpy of formation of HI(g)? a) -106 kJ/mol b) -53.0 kJ/mol c) -26.5 kJ/mol d) 26.5 kJ/mol e) 53.0 kJ/mol
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190173 | https://www.reddit.com/r/learnmath/comments/19d3e13/is_the_empty_set_an_element_of_every_set_or_is_it/ | Is the empty set an element of every set, or is it just a subset of every set? : r/learnmath
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Is the empty set an element of every set, or is it just a subset of every set?
Reading a book that says if A and B are disjoint, then neither is a subset of the other unless one is the empty set. I thought 'Isn't it impossible to be disjoint with the empty set since it's in every set?' but then I realised it's a subset of every set, but that doesn't necessarily mean it's an element. I'm not sure what this implies though, or I feel like I don't really understand the thought I just had or know if it's even correct
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190174 | https://www.mathlearningcenter.org/sites/default/files/pdfs/LTM_Multiplication.pdf | Learning to Think Mathematically About Multiplication A Resource for Teachers, A Tool for Young Children Jeffrey Frykholm, Ph.D.
Learning to Think Mathematically About Multiplication A Resource for Teachers, A Tool for Young Children by Jeffrey Frykholm, Ph.D.
Published by The Math Learning Center © 2018 The Math Learning Center. All rights reserved. The Math Learning Center, PO Box 12929, Salem, Oregon 97309. Tel 1 (800) 575-8130 www.mathlearningcenter.org Originally published in 2013 by Cloudbreak Publishing, Inc., Boulder, Colorado (ISBN 978-0-692-27478-1) Revised 2018 by The Math Learning Center.
The Math Learning Center grants permission to reproduce and share print copies or electronic copies of the materials in this publication for educational purposes. For usage questions, please contact The Math Learning Center.
The Math Learning Center grants permission to writers to quote passages and illustrations, with attribution, for academic publications or research purposes. Suggested attribution: “Learning to Think Mathematically About Multiplication,” Jeffrey Frykholm, 2013.
The Math Learning Center is a nonprofit organization serving the education community. Our mission is to inspire and enable individuals to discover and develop their mathematical confidence and ability. We offer innovative and standards-based professional development, curriculum, materials, and resources to support learning and teaching. ISBN: 978-1-60262-564-8 Learning to Think Mathematically About Multiplication A Resource for Teachers, A Tool for Young Children Authored by Jeffrey Frykholm, Ph.D.
This book is designed to help students develop a rich understanding of multiplication and division through a variety of problem contexts, models, and methods that elicit multiplicative thinking. Elementary level math textbooks have historically presented only one construct for multiplication: repeated addition. In truth, daily life presents us with various contexts that are multiplicative in nature that do not present themselves as repeated addition. This book engages those different contexts and suggests appropriate strategies and models, such as the area model and the ratio table, that resonate with children’s intuitions as they engage multiplication concepts. These models are offered as alternative strategies to the traditional multi-digit multiplication algorithm. While it is efficient, is not inherently intuitive to young learners. Students equipped with a wealth of multiplication and division strategies can call up those that best suit the problem contexts they may be facing. The book also explores the times table, useful both for strengthening students’ recall of important mathematical facts and helping them see the number patterns that become helpful in solving more complex problems. Emphasis is not on memorizing procedures inherent in various computational algorithms but on developing students’ understanding about mathematical models and recognizing when they fit the problem at hand. Learning to Think Mathematically about Multiplication 2 About the Author Dr. Jeffrey Frykholm has had a long career in mathematics education as a teacher in the public school context, as well as a professor of mathematics education at three universities across the United States. Dr. Frykholm has spent of his career teaching young children, working with beginning teachers in preservice teacher preparation courses, providing professional development support for practicing teachers, and working to improve mathematics education policy and practices across the globe (in the U.S., Africa, South America, Central America, and the Caribbean). Dr. Frykholm has authored over 30 articles in various math and science education journals for both practicing teachers, and educational researchers. He has been a part of research teams that have won in excess of six million dollars in grant funding to support research in mathematics education. He also has extensive experience in curriculum development, serving on the NCTM Navigations series writing team, and having authored two highly regarded curriculum programs: An integrated math and science, K-4 program entitled Earth Systems Connections (funded by NASA in 2005), and an innovative middle grades program entitled, Inside Math (Cambium Learning, 2009). This book, is part of his series of textbooks for teachers. Other books in this series include: Dr. Frykholm was a recipient of the highly prestigious National Academy of Education Spencer Foundation Fellowship, as well as a Fulbright Fellowship in Santiago, Chile to teach and research in mathematics education. Learning to Think Mathematically about Multiplication 3 Table of Contents LEARNING TO THINK MATHEMATICALLY: AN INTRODUCTION 4 The Learning to Think Mathematically Series 4 How to Use this Book 4 Book Chapters and Content 5 CHAPTER 1: THE NATURE OF MULTIPICATION AND DIVISION 7 Activity Sheet 1 10 Activity Sheet 2 12 CHAPTER 2: THE TIMES TABLE AND BASIC FACTS 13 Activity Sheet 3 15 Activity Sheet 4 18 Activity Sheet 5 19 Table for Activity Sheet 5 20 CHAPTER 3: THE AREA MODEL OF MULTIPLICATION 26 Activity Sheet 6 30 Activity Sheet 7 33 CHAPTER 4: THE RATIO TABLE AS A MODEL FOR MULTIPLICATION 38 Activity Sheet 8 44 Activity Sheet 9 45 CHAPTER : THE TRADITIONAL MULTIPLICATION METHOD 0 Activity Sheet 1 3 APPENDIX A: THE TIMES TABLE 5 Learning to Think Mathematically about Multiplication 4 Learning to think Mathematically: An Introduction The Learning to Think Mathematically Series One driving goal for K-8 mathematics education is to help children develop a rich understanding of numbers – their meanings, their relationships to one another, and how we operate with them. In recent years, there has been growing interest in mathematical models as a means to help children develop such number sense. These models (e.g., the number line, the rekenrek , the ratio table, the area model of multiplication, etc.) are instrumental in helping children develop structures – or ways of seeing – mathematical concepts. This textbook series has been designed to introduce some of these models to teachers – perhaps for the first time, perhaps as a refresher – and to help teachers develop the expertise to implement these models effectively with children. While the approaches shared in these books are unique, they are also easily connected to more traditional strategies for teaching mathematics and for developing number sense. Toward that end, we hope they will be helpful resources for your teaching. In short, these books are designed with the hope that they will support teachers’ content knowledge and pedagogical expertise toward the goal of providing a meaningful and powerful mathematics education for all children. How to Use this Book This is not a typical textbook. While it does contain a number of activities for students, the intent of the book is to provide teachers with a wide variety of ideas and examples that might be used to further their ability and interest in approaching the topics of multiplication and division from a conceptual point of view. The book contains ideas about how to teach multiplication through the use of mathematical models like the area model and the ratio table. Each chapter has a blend of teaching ideas, mathematical ideas, examples, and specific problems for children to engage as they learn about the nature of multiplication, as well as these models for multiplication. We hope that teachers will apply their own expertise and craft knowledge to these explanations and activities to make them relevant, appropriate (and better!) in the context of their own classrooms. In many cases, a lesson may be extended to a higher grade level, or perhaps modified for use with students who may need additional support. Ideas toward those pedagogical adaptations are provided throughout. Learning to Think Mathematically about Multiplication 5 Book Chapters and Content This book is divided into chapters. The first chapter, The Nature of Multiplication and Division, explores various contexts that are multiplicative in nature. While the idea of “repeated addition” is certainly a significant part of multiplicative reasoning, there are other equally important ways of thinking about multiplication. Contexts that promote these different ways of thinking about multiplication are presented in Chapter One. The second chapter, The Times Table, encourages students to discover and appreciate the many patterns that exist in the times table. When students are given the opportunity to investigate the times table deeply, they will discover interesting patterns and number relationships that ultimately help them develop intuitive strategies and conceptual understanding to help master the multiplication facts. For example… every odd number is surrounded by even numbers… the product of two odd numbers is always odd… the product of two even numbers is always even… the product of an even and an odd number is always even… diagonals in the times table increase and decrease in regular increments… there is a line of reflection from the top left to bottom right corner of the times table… etc. There are many number relationships in the times table, and if given the chance, students will make many discoveries about multiplication and division on their own. These findings are important for the development of their confidence and mastery of the basic facts, a topic that is also addressed in the second chapter. The third chapter of the book, The Area Model of Multiplication, explores the area model as a viable method not only to conceptualize multiplicative contexts, but also to find solutions to multiplication problems. Area representations for multiplication are common in geometry, but are rarely used to help students learn how to multiply. Hence, we lose a valuable opportunity to make mathematical connections between, in this case, geometric reasoning and arithmetic. Resting heavily on the important mathematical skill of decomposing numbers, the Area Model recognizes the connection between multiplication as an operation, and area models as representations of multiplication. Moreover, the area model allows student with strong spatial reasoning skills to visualize the product of two numbers as an area. The intent of this chapter is to present students with problems that will help them develop facility with this representational model for multiplication computation, as well as to use that model to better understand what multiplication really is. The fourth chapter of the book, The Ratio Table as a Model for Multiplication, illustrates how children may complete two and three-digit multiplication problems with the ratio table. (There is an entire book in the Thinking Mathematically series devoted to the ratio table: Learning to Think Mathematically with the Ratio Table.) Building on the mental math strategies developed more fully in the Ratio Table book noted above, students develop powerful techniques with the ratio table that they may choose as an alternative to the traditional multiplication algorithm. The benefit of the ratio table as a computational tool is its transparency, as well as its fundamental link to the very nature of multiplication. Multiplication is often described to young learners as repeated addition. Yet, this simple message is often clouded when students learn the traditional multiplication algorithm. A young learner would be hard pressed to recognize the link between the traditional algorithm and “repeated addition” as they split numbers, “put down the zero”, “carry”, and follow other steps in the standard algorithm that, in truth, hide the very simple multiplicative principle of repeated addition. In contrast, the ratio table builds fundamentally on the idea of “groups of…” and “repeated addition.” With time and practice, students develop Learning to Think Mathematically about Multiplication 6 remarkably efficient and effective problem solving strategies to multiply two and three-digit numbers with both accuracy, and with conceptual understanding. The final chapter of the book, The Traditional Multiplication Algorithm, is an important chapter. We must recognize the value of the traditional multiplication model – it has been taught almost exclusively in American schools for over a century. It s ubiquitous in elementary text books, and certainly is one of the most well-recognized and commonly used methods in all of arithmetic. And yet… research has indicated that the traditional algorithm is difficult for children to understand from a conceptual point of view. With practice, children memorize the steps of the traditional model, Without conceptual understanding, however, they are often unable to determine if they have used the algorithm correctly, or whether or not they have obtained a reasonable answer for the problem context. Hence, while we certainly should continue to teach the traditional model, it may not be the multiplication model of choice for many students if they are given the chance to learn other methods of multiplication in the same depth as we typically teach the traditional method. This chapter elaborates the traditional algorithm, drawing comparisons to other methods of multiplication when relevant. Learning to Think Mathematically about Multiplication 7 Chapter 1: The Nature of Multiplication and Division The primary goal for this chapter is to provide students with a conceptual introduction to multiplication (and by extension, division). In order for students to appropriate any method for multi-digit multiplication or division – and to understand what they are doing through the method – they must develop some basic understanding of the nature of multiplication. This would include, for example, what multiplication is, how various multiplication contexts can be represented with different models, and how these representations and subsequent models can lead to elegant solution strategies and, ultimately, answers to multiplication problems. Four primary kinds of multiplication problems are highlighted in this introduction. These include: • Multiplication as “repeated addition” (e.g., 3 groups of 4) • Comparison problems (e.g., “I have 4 times as many as you have.”) • Area representations • Combinations Brief conceptual explanations of these problem types are included below, with the intent that they are presented to students as well. Subsequently, an activity designed to be distributed to students is presented which supports understanding of these initial multiplication contexts. Multiplication as Repeated Addition The first, and most common, representation of multiplication is often thought of as “repeated addition.” Indeed, the first multiplication algorithms were created to help people solve addition problems of this nature. That is, people sought more efficient methods for adding the same number to itself over… and over… and over. The most common type of multiplication problem in traditional elementary textbooks is of this variety. For example: Manuel has 4 packs of gum. Each pack of gum has 6 individual sticks. How many individual sticks of gum does Manuel have? This problem is most often conceptualized, and therefore solved, by repeated addition. The solution may be found by adding the following: 6 + 6 + 6 + 6 = 24 Again, most of the multiplication problems we have typically presented to young learners in the traditional elementary classroom come in this form. While there is nothing wrong with doing so – indeed, many problem in real life are of this nature – we would be remiss if we did not present other multiplicative contexts such as the following.
Multiplication as a Comparison Comparison problems are solved similarly to repeated addition problems: we employ simple addition techniques to resolve the problem. What makes them fundamentally different, Learning to Think Mathematically about Multiplication 8 x 2 x 3 x 4 10 2 10 4 12 14 100 40 20 8 however, is that we conceptualize a comparison problem in a different manner. In other words, the comparison problem elicits a different kind of thinking than the typical “repeated addition” problem. Therefore, we must offer students ample opportunities to engage comparison problems, to think about them conceptually, and to consequently solve them with an appropriate tool. The “comparison” problem often includes language such as, “… times as many as…” For example, Jenny’s found six shells at the beach. Sarah found 4 times as many. How shells did Sarah find? This problem can also be solved by adding, but we imagine the problem in a different way than the traditional “repeated addition” problem. This envisioning includes the notion of a comparison. In this case… we compare Jenny’s six shells with Sarah’s collection – 4 times as many. Multiplication as Area One of the first “formulas” that young children learn in the math classroom is that the area of a rectangle may be found by multiplying the length of the rectangle by its width. We can use this common understanding to provide children with a viable method for multiplication. The product of two numbers can be shown as a visual representation, in the form of a rectangle. For example, the answer to 14 x 12 may be found through the illustration shown below – a rectangle with sides of lengths 14 and 12. This method is extremely effective so long as students understand two primary mathematical ideas: 1) numbers (just like physical areas) can be decomposed into the sum of smaller numbers (or the sum of smaller areas); and 2) the distributive property can be used to break down one large multiplication problem into several smaller ones. With minimal practice, students grasp these ideas, as well as the area model for multiplication, with confidence. Example: Solve 14 x 12 with the Area Model Find the sum of the individual areas: 100 + 40 + 20 + 8 = 168 Jenny Sarah x 1 14 x 12 Learning to Think Mathematically about Multiplication 9 Multiplication as a Combination Combination problems are probably the least often represented multiplication problems in the elementary curriculum. Yet, they do offer children a unique way to conceptualize multiplication that will be visited later in their mathematics careers as they explore combinations and permutations. For younger children, however, this idea of different “combinations” can be a helpful tool for representing multiplication, particularly when multiplying more than two numbers together. The “combination” form of multiplication often requires students to imagine the total number of distinct combinations that can be made with two or more unique items. For example: Nikki has 2 shorts, and 3 shirts. How many outfits can she wear? Solution: How many distinct outfits? 3 with red shorts, 3 with blue shorts. Red-Blue Blue-Blue Red-Red Blue-Red Red- Green Blue-Green The problems on Student Activity Sheet #1 will encourage students to begin thinking about multiplication in these four different forms. As you begin to engage students in divergent thinking about multiplication, be sure they are comfortable with these four problem types. Help your students make connections between verbal descriptions of the problems with their visual representations. Cues in the language of the problem should help students correctly match the description with the model. These pictorial models themselves are suggestive of various ways in which these problems can be solved.
Learning to Think Mathematically about Multiplication 10 C) Student Activity Sheet 1 NAME: ____ 1.
Match the problem statements below with the diagram that best fits the solution.
Statement 1: I rode my bike 10 miles an hour for 6 hours. How many miles? Statement 2: I have 3 fish tanks, each holding 4 fish. How many fish? Statement 3: The table is 4 feet long, and three feet wide. How many tiles do we need to cover the table? Statement 4: The soccer team has 2 jerseys, 2 shorts, and 2 pairs of socks. How many uniform combinations? 10 10 10 10 10 10 A) B) D) Learning to Think Mathematically about Multiplication 11 Multiplication and Division: Related Operations As students begin to develop conceptual understanding of multiplication – what it is, and how we use the operation to solve problems – it is important to help children see the close connection between multiplication and division. This relationship is waiting to be explored, and we must take advantage of readily available opportunities to encourage students to see the connections both conceptually, as well as operationally. The activities that appear on Student Activity Sheet #2 will help young learners recognize the inverse relationship shared by multiplication and division. The key point to illustrate at this juncture in their learning about these operations is that in both division and multiplication, we are always seeking to discover the unique relationship that exists between three numbers. For example, consider the various ways we might explore the following fact: 3 x 4 = 12. In one case, we may wish to find the answer (the unknown result) of the product: 3 x 4 = ??? In a second case, we may know the result, but not the starting value of the relationship: ??? x 4 = 12. In this form, we are likely to view the problem as one of division, and thereby seek to solve it with a division process as well. Of course, it might just as easily be solved by multiplication, a strategy many students will take if appropriately prepared to engage problem contexts intuitively. The idea of seeking the relationship between three numbers is significant: students should be fluent in expressing these number relationships in both multiplication and division contexts. The activity sheet asks students to develop this relational view of multiplication and division by asking them to change a multiplication problem to one of division, and vice versa. As students engage these problems, emphasize that the three numbers in each problem are in relationship with one another, and that the relationship can be expressed either by division, or by multiplication. Learning to Think Mathematically about Multiplication 12 Student Activity Sheet 2 NAME: ____ Introduction: It can be helpful to think about multiplication and division at the same time. In other words, any multiplication problem can also be thought of as a division problem. It just depends on how you look at it. For example… I rode my bike 10 miles every hour, for 3 hours. How many miles did I ride? Could be changed to a division problem in this way… I rode my bike for 3 hours. I traveled a total of 30 miles. How fast (miles per hour) was I riding? ----- 1.
Change the two following multiplication problems into division problems by reworking the statement. (There is more than one correct answer!) a) I have 3 fish tanks, each holding 4 fish. How many fish?
Rewrite as a division problem: __________ b) The table is 4 feet long, and 3 feet wide. How many square tiles (each tile is !"! ) do we need to cover the table?
Rewrite as a division problem: _________ 2. Remember the four different kinds of multiplication problems from an earlier activity? Come up with your own multiplication problem for each kind below.
a) Repeated addition b) Comparison c) Area d) Combination Learning to Think Mathematically about Multiplication 13 Chapter 2: The Times Table and Basic Facts Before moving into the various models for multi-digit multiplication that comprise the bulk of this book, we would be remiss if we were not first thoughtful about the necessary building blocks for multi-digit multiplication. Specifically, in order to multiply larger numbers accurately and with meaning, students must first have command of the single digit multiplication facts. There are various perspectives on what this “command of the facts” means. As adults, we can probably remember our first explorations with multiplication facts, which likely included an emphasis on memorization, flash cards, and instant recall. For many years, the informal standard has been that students should be able to recall any single-digit multiplication fact accurately within three seconds. One might ask the question, however, “What makes three seconds the magic length of time to recall a number fact?” One might also ask if it is indeed necessary to have every multiplication fact memorized. If a student is able to reason her way to the solution of 6 x 8 quickly and accurately (as opposed to recalling the fact from rote memory), might we agree that she is proficient with that fact? Imagine, for example, this sort of thinking: Q: What is 6 x 8? A: Ok.. 6 groups of 8. Well, I know that 5 groups of 8 is 40. So, one more group of 8 must be 48. So, 6 groups of 8, or 6 x 8, equals 48. Q: What is 5 x 6? A: Ok.. 5 groups of 6 would be half as much as 10 groups of 6. I know that 10 times 6 is 60. Half of 60 is 30. So… 5 groups of 6 is equal to 30. The confident mathematical thinker – one with good number sense, with an understanding of what it means to multiply, with the wherewithal to link one fact to another – could complete either of these particular chains of reasoning in 3-4 seconds. One might argue that this child, in fact, has a richer “command” of 6 x 8 than a child who is able to recite the answer from memory, but does not fundamentally know what this number fact means, or how it could be used to determine a different fact. It is not our intent in this book to take up the argument of whether number facts should be memorized or derived. The point is that before children can multiply large numbers, they must be confident multiplying single digits. The purpose of this chapter is to introduce a decidedly conceptual approach to the teaching and learning of the times table. We invite you to use as many of these ideas and strategies as you find useful in the context of your classroom. Several student activity sheets are included throughout this chapter that we have found to be helpful for young learners who are still wrestling to master the multiplication facts. Patterns in the Times Table Learning the times table is one of the most memorable aspects of the elementary math experience. Of course… sometimes the memory is not always positive! Based perhaps on the expectation that children leaving the elementary classroom should have a firm command on their Learning to Think Mathematically about Multiplication 14 multiplication facts, we often jump right into the teaching of the times table without pausing to appreciate the richness of this collection of math facts as they appear in order, in the times table. In truth, there are many patterns that exist in the times table that can help students not only develop a more nuanced sense of multiplication, but also master the multiplication facts much more easily than they might otherwise. Appendix A contains a completed multiplication fact table that you may use for various activities that follow in this chapter. We begin this exploration of the times table – and the many facts it contains – with a search for meaningful patterns that exist in the table itself. If students are given opportunity to investigate the times table and to discover the many interesting patterns that exist within it, there is a much greater chance that they will be able to develop intuitive strategies that will help them master the multiplication facts. Inquisitive learners will enjoy the pursuit of unique patterns in the table. Given time and perhaps some basic directions, they will discover, for example, that every odd number is surrounded by even numbers… that the product of two odd numbers is always odd… the product of two even numbers is always even… the product of an even and an odd number is always even… diagonals in the times table increase and decrease in regular increments… there is a line of reflection from the top left to bottom right corner of the times table… etc. There are many number relationships like these in the times table. If given the chance, students will make many discoveries about multiplication and division on their own. These findings are important for the development of their confidence and mastery of the basic facts. On Student Activity Sheet #3, students are given a completed times table, and asked to look for patterns. It will be helpful to give the students an introduction to the table as some students may be seeing it for the first time. Others will have varying levels of experience with the table and the facts contained therein. Be sure to offer an appropriate level of introduction which should include a basic explanation of what the table is, how each cell is determined, etc. To prepare students for the subsequent activity, you might highlight a pattern or two with the class, and then as a whole group solicit suggestions for other obvious patterns the children might see. The activity sheet is somewhat self-guided, and is designed both to be helpful in a large group setting, or as an individual exploration as might best fit the needs of your classroom. Allow students to linger on this problem for at least 15 minutes. They will make many interesting discoveries! Students will notice that some rows or columns increment by certain amounts, or end in a particular number, or have a special relationship with adjacent cells. Some students will notice patterns with odd and even numbers. Some will notice interesting patterns on the diagonals. As students reveal these “discoveries,” push them to explain why that particular phenomenon occurs. Learning to Think Mathematically about Multiplication 15 Student Activity Sheet 3 NAME: ____ Directions: Does this table of multiplication facts look familiar? Maybe you have used a table like this one to help learn your times facts. That is a lot to remember! Or… is it? There are many interesting patterns in the table that will help you learn and recall the number facts. Take a few minutes to explore the table. Circle any patterns you see, and then describe them below. Two examples have been given for you. 1) The second row increases by 2 all the way across: 2, 4, 6, 8, … 2) Start with the 6 on the first row (just below the yellow row). Go down one, and over one to the left. Keep doing that. The numbers are like a mirror: 6, 10, 12, 12, 10, 6.
3) X 1 2 3 4 5 6 7 8 9 10 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 12 14 16 18 20 3 3 6 9 12 15 18 21 24 27 30 4 4 8 12 16 20 24 28 32 36 40 5 5 10 15 20 25 30 35 40 45 50 6 6 12 18 24 30 36 42 48 54 60 7 7 14 21 28 35 42 49 56 63 70 8 8 16 24 32 40 48 56 64 72 80 9 9 18 27 36 45 54 63 72 81 90 10 10 20 30 40 50 60 70 80 90 100 Learning to Think Mathematically about Multiplication 16 Teaching Tip: Examining the thinking of students You may wish to have students write about their understandings of various patterns in the times table. The following two examples are reflective of qualitatively different thinking strategies and sophistication that are common among students. By asking students to describe their thinking in writing, you are encouraging powerful metacognitive skills that will serve them well throughout their mathematics education. Samples of Student Thinking: Patterns in the Times Table Sixth Grade Thinking Fourth Grade Thinking Learning to Think Mathematically about Multiplication 17 More about Multiplicative Patterns In the following exercise (Activity Sheet 4), students are asked to shade in all multiples of either 4, or 8. While they may choose between the two options, doing both will of course benefit their understanding. As students complete this task, they may recognize that the multiples of 4 and the 8 overlap – that is, every multiple of 8 (e.g., 8, 16, 24, 32, etc.) is also a multiple of 4, although the reverse is not true). If students do not recognize this on their own, have them partner with a peer who chose to shade the opposite set of multiples (or have the students shade multiples of both 4 and 8). In fact, it is well worth the time to duplicate the times table provided in the activity, and have them find patterns in the multiples of other numbers as well. Be sure to help them recognize patterns that occur horizontally (jumping by an interval), as well as the patterns that occur up and down the vertical columns as well. The time invested in uncovering patterns with this exercise will pay dividends later for all students, particularly those students that may continue to struggle with the multiplication facts. In fact, it is a good exercise for students to shade the multiples of every number between 2-10. The more they can recognize and predict patterns of multiples, the more they will be able to draw on their intuitions as they learn and use the multiplication facts. Activity Sheet 5 presents some challenges for students who may be ready to think more deeply about the times table, and multiplication in general. Depending on the grade and level of your students, these problems may not be accessible to every child. They examine a few intriguing mathematical concepts that are embedded in the times table. While these patterns may not be particularly practical or helpful toward the mastery of individual number facts, they nevertheless illustrate interesting nuances in the table that are likely lost on most individuals that casually peruse the times table, or focus exclusively on memorization. Encourage students (as appropriate) to grapple with the questions that exist on Activity Sheet 5. Particularly talented mathematical thinkers will find these interesting contexts to be naturally intriguing. Learning to Think Mathematically about Multiplication 18 Student Activity Sheet 4: Patterns NAME: ____ Let’s look at some patterns. You get an interesting design when you shade in all multiples of 6 (all the numbers you list if you started counting by sixes, like… 6, 12, 18, 24… and so on). X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 What patterns exist if you shade by 4’s? What about shading by 8’s? Choose one of these multiples (4 or 8), and shade in the table below. Your Choice: Shade in the Multiples of _ X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 Learning to Think Mathematically about Multiplication 19 Student Activity Sheet 5: Challenge NAME: ____ Use the times table on the second page to answer the following questions. 1) Did you know…that every odd number is surrounded by even numbers? Check it out for yourself, and then write an explanation for why that is the case.
2) Did you know…that if you start in any box in the top row and go down and to the left diagonally, it will look like a mirror? That is, you will see a group of the same numbers, repeated in order, “reflected” over the diagonal that starts in the upper left corner, and goes to the lower right corner of the table. Please explain why that pattern exists.
3) Did you know…that if you choose any four boxes that make up the corners of a rectangle, and then multiply the opposite corners together, you will get the same answer? (See the shaded box in the table below.) Find another example that works, and then explain why the process will always work.
Example: Explore the shaded box. Why does 10 x 24 = 20 x 12 ? 4) Did you know …that this table can be used to multiply even bigger numbers?
• What is 2 x 23? (Check out the striped box outlined in black: 46!!) • How about 4 x 12? (check out the striped box outlined in Black 48!) • Here are two tough ones. Can you explain the answer to 10 x 78? Or, 8 x 56? (These cells are highlighted in the table.) Learning to Think Mathematically about Multiplication 20 Table for Student Activity Sheet 5 X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 Learning to Think Mathematically about Multiplication 21 Teacher Tips, Activity Sheet 5 These problems points specifically to several unique characteristics of the times table. Explanations for each specific problem appear below. 1) Every odd number is surrounded by even numbers… Why? This phenomenon can be traced to something that students may have noticed in their informal examinations of the times table. Specifically, when multiplying, three options exist: 1) an odd times an odd; 2) an even times and even; or 3) an odd times and even. The only case in which the product of two numbers is odd is when the two numbers being multiplied are themselves odd.
This goes back to the first conceptualization of multiplication – repeated addition. The only way to get an odd number is to add an even with an odd. Hence, when we multiply an odd number an odd number of times – say 3 x 5 for example – what we are really doing is repeatedly adding 5, in this case three times. Any number plus itself is even (e.g., 5 + 5 = 10). Hence, if we continue, adding an odd (5) to an even (10) results in an odd answer (10 + 5 = 15). So… back to the original question: Why is every odd number surrounded by even numbers? Because the only way to get an odd is to multiply an odd by an odd. No matter which adjacent cell one selects will be the product of an odd times an even, or an even times an even. To illustrate this phenomenon, select one or two odd entries in the table, and look at the combinations of numbers that lead to the products of the adjacent cells.
2) The diagonal line from upper left to lower right is the “squares” line. This is a great opportunity to introduce some important mathematical concepts. First, the “squares” diagonal is a line of reflection, which points to the notion of symmetry. The portion of the table to the left of the “squares” line is identical to the portion of the table to the right. The reason for this is that multiplication is commutative. This is a significant concept for students to embrace. 4 x 3, (4 groups of 3) is the same quantity as 3 x 4 (three groups of four). To begin this discussion, ask students to locate a particular cell – say the cell containing the answer to 8 x 4. Then, ask them to find the cell containing the answer to 4 x 8. Do this with several other number pairs as well. They soon will see the relationship – that these pairs of numbers are on the same diagonal, perpendicular to the squares line.
3) This interesting relationship intrigues many students. It can quite easily be explained when students look at the smaller products that comprise the larger problem. Take the given example: the rectangle outlined by the corners 10, 20, 24, and 12. The problem suggests that for any rectangle, the products of the opposite corners will be equal. So, in this case, 10 x 24 = 20 x 12. Help the students explore where each of these particular “corners” came from: 10 is found by multiplying 2 x 5; 24 is found by multiplying 6 x 4; 20 is found by multiplying 5 x 4; 12 is found by multiplying 2 x 6. When you combine these products according to the definition of the rule (the products of the opposite corners are equal to each other), you get: 2 x 5 x 6 x 4 = 5 x 4 x 2 x 6 – the same product of numbers!
4) The number relationships explored in this problem are quite interesting. The times table can be used as a calculator for larger computations assuming that students are comfortable with base-10 principles. The first example highlighted asks students to Learning to Think Mathematically about Multiplication 22 consider the product of 2 x 23. This problem is solved by using the distributive property. That is, 2 x 23 is the same as (2 x 20) + (2 x 3). In the table, we first look at the cell containing the product of 2 x 2, which is really 2 x 20, or 40 (4 groups of 10). The second cell we consider the product of 2 x 3 = 6 (ones). So, our answer is 40 +3 = 43. The same logic can be used anywhere in the table. For example, consider the example of 10 x 78 (highlighted in green). When broken down with the distributive property, this can be expressed as (10 x 70) + (10 x 8). When expanded, we can find each of these products in the table. 10 x 7(0) = 70 (groups of 10). Likewise, 10 x 8 = 80 (ones). Hence, the product is 70 tens plus 80 ones: 780. Learning to Think Mathematically about Multiplication 23 Duplicate facts, to the right of the Doubles Line Teaching The Multiplication Facts The times table can be very helpful in learning the multiplication facts. For many years, teachers and text books were inclined to begin the process of memorizing the times facts by starting with the 1’s, and gradually progressing toward 10, or beyond. Knowing much more now about how children think, and how they use intuition to solve problems, it is far more effective to consider the big picture – the whole table – when teaching children the multiplication facts. Children probably know more than they think they do. For starters, one doesn’t have to memorize every box – only half of the table! The right half of the table (shaded) is identical to the left half of the table. The diagonal line down the middle (called the “doubles” line) acts as a mirror, separating two identical sets of multiplication facts. X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 There are some other helpful facts that students probably know already as well. As an exercise with students, shade in the boxes that correspond to the following number facts: The ones facts… (one times any number is the number itself) The twos facts… (doubles… 2x2, 2x3, 2x4, 2x5, etc.) The fives facts… (five, ten, fifteen, twenty…) The tens facts… (ten, twenty, thirty, forty…) In the table below, building on what the students already know, the following facts have been shaded: The 1’s… The 2’s… The 5’s… The 10’s… The “Doubles” Line Learning to Think Mathematically about Multiplication 24 The Times Table… what we already know! X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 Now… if a student knows these facts, she can probably figure out the related “next door neighbor” facts. That is, if you know that 5 groups of 5 is 25, then… 6 groups of 5 (6x5) would be five more than 25 … or, 30. As you are encouraging students to think in this way, return to the times table and shade in the following boxes: • The twos (+1)… for example, if 2 groups of 6 = 12, then 3 groups of 6 would be… 2 x 6 = 12… plus one more 6 18! • The fives (+1 and -1)… for example, if 5 groups of 8 = 40, then 4 groups of 8 would be 5 x 8 = 40 … minus one group of 8 32! • The 10’s (+1 and -1) … for example, if 10 x 6 = 60, then 11 x 6 would be 10 x 6 = 60… plus one more 6 66! Shade in these “next door neighbor” facts with the class. When students begin to view the number facts in this way – building on what they already know, and using their mathematical intuitions to construct one fact from another known relationship – the times table quickly becomes a manageable task. A Question for Students: Are there other number facts that you can quickly solve based on what you already know? Shade them in! Learning to Think Mathematically about Multiplication 25 Teacher notes: Summarizing the teaching of the multiplication facts After the informal explorations with the times table completed previously, students who may not be comfortable with all of their multiplication facts can turn to the table for that purpose – to guide their use of intuition and informal explorations to help them recall the multiplication facts. Before progressing, the question must be asked: What does it mean to have mastery of the times table? Of course, instant recall of the multiplication facts is desirable. With time and proper exposure, most students develop such facility with the times table. But, how soon should we expect students to have instant recall of the times facts? There is notable research to suggest that over-emphasizing recall of the facts found in the times tables at an early age can actually inhibit students’ understanding of the mathematics of multiplication. So, then, what is mastery? A good rule of thumb is that if students are able to discern a correct answer to a single digit multiplication problem within 3-5 seconds, that is probably quick enough for the moment (assuming that greater efficiency naturally comes with time). Truly, there are few situations in life in which it is crucial that young learners be able to recall multiplication facts in less than 3 seconds. Taking the pressure off students in this regard will lead them to greater confidence in their intuitive ability to discern these facts, as well as their overall efficacy as learners and doers of mathematics. The 60-second multiplication facts tests of the past have long been known to leave indelible and negative marks on students – and for what reason? If students can solve single digit multiplication problems in a few seconds using any number of intuitive strategies, they will be ahead in the long run when compared to students who are forced to simply memorize 100 number facts. So… the moral of the story here is to relax. Let your students use informal strategies to solve multiplication problems. With repeated practice and exposure, they will become more efficient –without the scars that many adults carry around today because of the way in which they were intimidated by the number facts. The strategies outlined previously in this section of the book will help students toward this end. Learning to Think Mathematically about Multiplication 26 Chapter 3: The Area Model of Multiplication Overview The area model of multiplication is often thought of as the most conceptually comprehensible multiplication strategy for young learners. In addition to the fact that this strategy builds on the learner’s spatial reasoning (a preferred learning style for more than half of all young children), this strategy also clearly isolates the partial products of the multiplication problem, an essential component of learning any multi-digit multiplication strategy. Three examples are provided below to illustrate the learning path that is elaborated in this chapter. Example 1: Single digit multiplication with the area model If teachers introduce the area model with small numbers (e.g., 3 x 4), students can literally count tiles or cells, further solidifying their conceptual understanding of area (i.e., “covering” a given space) as representative construct for multiplication and division. For example: Question: What is the answer to 3 times 4? Response: It might be helpful for learners to think of this problem in terms of rows and columns. That is, we might recommend children to restate the problem in these words: “How many tiles are there in 3 rows, where each row has 4 tiles?” Example 2: Single x Double-digit multiplication with the area model With practice, students can apply an area model strategy with larger numbers. The ability to decompose numbers will be crucial to the successful implementation of this strategy. Decomposition is the ability for the learner “see inside” a number, and to think of it in terms of its component parts. That is, the number 23 might be thought of as one group of 20, plus 3 more. The number 18 might be thought of as one group of 10, and 8 ones. This ability to decompose is essential for successful use of the area model with multi-digit numbers. For example: Question: What is the answer to 23 times 3? Response: I need 3 rows of 23. To make it easier, I can think of 23 as one group of 20, plus 3 more. Therefore, I need three rows of 20, plus 3 rows of 3. + 20 3 Example 3: Multi-Digit multiplication As students become comfortable with the use of partial products and various representations that make use of them, they will be able to tackle larger, multi-digit problems with confidence 1 2 3 4 5 6 7 8 9 10 11 12 3 60 tiles 9 tiles = 63 Learning to Think Mathematically about Multiplication 27 such as the example provided below. A significant portion of this chapter is devoted to this pursuit. It is crucial to note that, whether using area models, ratio tables, or the traditional multiplication algorithm, the degree to which students can see the partial products inherent in the problem will likely determine their success with the strategies. Question: What is 24 x 32? Response: I can think of 24 as 20 (10 + 10), plus 4 more. Likewise, I can think of 32 as 30 (10 +10 + 10), plus 2 more. Arranging these component parts in a rectangle, I get the following: 10 10 10 2 100 100 100 20 100 100 100 20 40 40 40 8 Area = (100 + 100 + 100 + 100 + 100 + 100) + (40 + 40 + 40) + (20 + 20) + (8) = (600) + (120) + (40) + (8) = 768 This chapter outlines a progression of activities that will build students’ confidence with the area model as illustrated in the previous examples. If students are given multiple opportunities to practice this method, they will quickly gravitate toward nuanced (and rapid) application of the area model, even with larger numbers. A student who is comfortable with this method will, for example, solve the previous problem with a diagram like the following, indicating a sophisticated understanding of decomposition, and indeed, multiplication itself. 24 32 10 10 4 60 40 120 8 20 4 30 2 Learning to Think Mathematically about Multiplication 28 The Area Model: Beginning Steps At the conclusion of this chapter on the Area Model, students should be able to: • Recognize the connection between multiplication as an operation, and area models as representations of multiplication; • Use the area model to visualize products of two numbers; • Use the area model to help understand both decomposition of numbers, as well as the distributive property of multiplication.
Fundamental to understanding the area model as a computational tool is that, 1) numbers can be decomposed into the sum of smaller parts; and 2) the distributive property can be used to break down one large multiplication problem into several smaller ones. Begin to teach the area model with these two ideas in mind. Decomposition and Partial Products To begin this chapter, students must engage the idea that multi-digit numbers can be represented as the sum of a set of smaller parts. This ideas is known as “number decomposition.” Students must be encouraged to think of 15 as the sum of 10 + 5. Or, 36 might be represented as the sum of 10 + 10 + 10 + 5 + 1. Once this has been established, the next question to consider is whether or not this process of decomposing numbers can actually assist us with other operations and actions on numbers. In the case of multiplication, it turns out that it can. Begin by using base-10 blocks to represent a series of numbers. Engage students in a discussion of the following (and additional) representations. They should be encouraged to represent numbers, decomposed into component parts, with their own base-10 blocks. Represent 5 with base-10 blocks Represent 15 with base-10 blocks Represent 25 with base-10 blocks Represent 33 with base-10 blocks Represent 18 with base-10 blocks Learning to Think Mathematically about Multiplication 29 What makes decomposition of numbers important is the way that it can be combined with the distributive property in order to flexibly and powerfully compute with large numbers. Fundamental to the area model is the understanding that 3 x 15, for example, can be thought of in the following way: 3 x 15 3 groups of 15 15 + 15 + 15 = 45 Building on the previous representations with the base-10 blocks (or other manipulatives), it would be relatively straight forward to rewrite this problem (3 piles of 15 counters) by rearranging the 15 counters strategically. For example, we could separate each pile of 15 into two piles of 10 and 5. Illustrate this problem with base-10 blocks. Therefore… 3 x (10 + 5) 3 groups of 10 and 3 groups of 5 10 + 10 + 10 + 5 + 5 + 5 = 45 This concept is fundamentally important to the area model of multiplication. Instead of using piles of counters, however, we can represent the quantity of each number to be multiplied as the dimensions of a rectangle. From this point, students are led to the idea that the whole area of the rectangle is simply the sum of the areas of smaller regions within the rectangle. For example: 10 1 2 3 4 5 10 1 2 3 4 5 10 + 1 2 3 4 5 3 x 15 (10 + 10 + 10) + (5 + 5 + 5) 30 + 15 = 45 The problems on Activity Sheet 6 will help student develop understanding of these important prerequisite concepts for the area model of multiplication. 15 15 15 10 5 10 5 10 5 Learning to Think Mathematically about Multiplication 30 Student Activity Sheet 6: Decomposition NAME: ____ Use Base-10 Blocks to help you answer the following questions. Show each number with as few blocks as possible. Draw in your answer with a pencil. 23 (Example) 12 42 14 18 52 34 122 Learning to Think Mathematically about Multiplication 31 Beginning with the Area Model The intent of the problems found in Activity Sheet 7 is help students not only develop facility with a representational model for multiplication computation, but also to use that model to better understand what multiplication really is. To set the stage for the problems in Activity Sheet 7, begin by motivating the idea that multiplication can be thought of as finding the “area” of a rectangular space. Using tiles on the floor, measurements of a rectangular rug, tiles on the ceiling, an arrangement of blocks, etc., model the connection between area and multiplication. Teaching Example 1: What is the area of a rug that is 2 feet wide, and 5 feet long? Model the problem with tiles or blocks, and have students count the number of squares required to cover the rug. 1 2 3 4 5 6 7 8 9 10 Teaching Example 2: What is the area of a rug that is 4 feet wide, and 3 feet long? 1 2 3 4 5 6 7 8 9 10 11 12 Teaching Example 3: What is the area of a rug that is 4 feet wide, and 12 feet long? Do we really want to count each square? What if the dimensions were 24 feet wide, by 48 feet long? Is there a better strategy to find the answer than counting each square? Solicit and record students’ suggestions about how to more efficiently find the number of tiles without counting each one. 4 12 Learning to Think Mathematically about Multiplication 32 A Teaching Example Pose the following question to students: How can we use the area of a rectangle to find the answer to 23 x 18? Proceed to explain the following steps to students. We can use some basic number principles to make a helpful diagram to solve this program by computing its area. We know, for example, that 23 can be expressed as 23 = 10 + 10 + 3. We also know that 18 can be expressed as: 10 + 8. By making a rectangle with these dimensions in mind, we can begin to make progress toward finding a solution – the area of a rug that is 23 feet long, and 18 feet wide. With this new drawing, we can now compute our original problem (23 x 18) using the smaller rectangles. That is, we can compute the area in each of the smaller squares quite easily, and then add the totals together. For example, this first square has dimensions of 10 x 10, which means it has an area of 100. The same can be done for each of the other squares. So… when we add all of the squares together, we get: 23 x 18 100 + 100 + 100 + 100 + 30 + 24 = 454 100 100 30 10 8 100 100 24 23 x18 10 10 3 23 18 8 10 10 10 3 Learning to Think Mathematically about Multiplication 33 10 10 10 1 4 14 21 Student Activity Sheet 7: Area Model NAME: ____ Use the Area Model to complete the following multiplication problems. 1) 14 x 21 2) 12 x 15 3) 11 x 22 4) 17 x 42 5) 13 x 12 ANSWER: __ ANSWER: __ ANSWER: __ ANSWER: __ Learning to Think Mathematically about Multiplication 34 ANSWER: ___ ANSWER: ___ ANSWER: ___ Student Activity Sheet 7: Area Model NAME: ____ Continued 6) 18 x 34 7) 26 x 24 8) 12 x 300 100 100 100 10 2 9) 26 x 24 10) 15 x 211 (Challenge!) Learning to Think Mathematically about Multiplication 35 12 x 23 23 x 12 Teacher notes for Activity Sheet 7 The progression of problems on Activity Sheet 7 is intended to guide students through gradually more sophisticated problems that require subtle nuances in the use of the area model diagrams. The starting diagrams provided in some of the problems are intended to be helpful scaffolds for students to lean on as they engage this method of multiplication on their own. Be aware that some students may see the possibility of setting up different rectangular area models that might also be used to solve the problem successfully. For example, an area model for 12 x 23 might be set up in one of two representations: Teachers might challenge students to use both representations to solve the problem which will not only help students develop confidence with this method, but also more fully come to understand the nature (and definition) of multiplication. As students work the problems, continue to help them recognize the value in deliberately choosing to decompose numbers so as to later take advantage of multiplication “facts” that may come more easily than others. Multiplying by 10, for example, is one of the first conventions of multiplication that students cling to. Multiplying by 100 is also a powerful strategy that is helpful when computing with larger numbers. On the final challenge problem, for instance, students that are grasping this method may, on their own, be able to diagram the problem by using 1’s, 10’s, and 100’s in the following way: 15 x 211 100 100 10 1 10 5 Solution: 15 x 211 = 1000 + 1000 + 500 + 500 + 100 + 50 + 10 + 5 = 3165 Please note that this diagram is not drawn to scale. Such is the nature of the use of some mathematical models. At first, some students may wish to represent each problem to scale. It may be helpful, in fact, to give students standard graph paper that they can use to model simple problems. (Base-10 blocks may be used with similar effect.) The purpose of using models to represent and solve problems, however, is to provide students with a powerful method that can effectively enable them to represent and illuminate mathematical relationships and meaning, and 1000 1000 100 10 500 500 50 5 Learning to Think Mathematically about Multiplication 36 ultimately find correct answers to problems with a high degree of confidence. In fact, in this specific case of area models for multiplication, the representations of students will tell you a great deal about the depth of their understanding. While at first many students will draw their area models with a great deal of care and precision, with time, those children who begin to grasp the method will spend less time on the diagram itself. This is a great sign that they are understanding the decomposition of numbers, conservation of area, and indeed, the nature of multiplication itself. The model should facilitate the development of more sophisticated understanding of the problem itself, if not multiplication more generally. For example, consider how three students might solve the following problem: 22 x 25 = ? Example 1: 22 x 25 10 10 5 10 10 2 Example 2: 22 x 25 20 5 10 10 2 Example 3: 22 x 25 20 5 20 2 Finally, please note the connection between the area models illustrated above, and the traditional multi-digit algorithm that is so commonly taught in the upper elementary grade levels. Each step 100 100 50 100 100 50 20 20 10 200 50 200 50 40 10 400 100 40 10 Learning to Think Mathematically about Multiplication 37 400 + 100 + 40 + 10 = 550 in the traditional method is shown below. When color coded and compared to the area model, the links between the model and the algorithm become transparent. Traditional Method: 22 x 25 22 x 25 10 (5 x 2) 100 (5 x 20) 40 (20 x 2) 400 (20 x 20) 550 Area Model: 22 x 25 20 5 20 2 Up for consideration is the degree to which each model allows not only for the correct solution to be found, but also, and perhaps more importantly, a chance to fundamentally understand what it means to multiply. One might argue that the mathematical meaning of multiplication is somewhat veiled in the traditional method, whereas the area model quite clearly illustrates the ways in which the partial products may be combined to determine the final answer of the problem. In summary, central to the area model are a few key mathematical ideas: 1) the notion of number decomposition (e.g., 23 can be thought of as a group of 20, plus 3 more), 2) the distributive property (i.e., 3 x 23 = 3 x 20 + 3 x 3), as well as 3) the more abstract notion of the conservation of area – a rectangle can be broken into smaller parts that, when added together, represent the same area as the original rectangle. With repeated use of the area model – tying these constructs together – students develop a rich understanding of multiplication, both a useful and efficient method to solve multiplication problems, but also more fundamentally, a nuanced understanding of what it means to multiply. 400 100 40 10 Learning to Think Mathematically about Multiplication 38 Chapter 4: The Ratio Table as a Model for Multiplication In the ongoing debate in the mathematics education community about the relative merits of teaching for algorithmic proficiency versus teaching for understanding, the ratio table has emerged as a mathematical model that can be viewed as a conduit between these two sides of the debate. Given its rise to popularity in contemporary math education curriculum, a stand-alone book in the Thinking Mathematically series has been devoted entirely to an explanation of the ratio table as a model for multiplication and division. The information shared in this chapter is only a summary of the full range of applications of the ratio table that can be found in the full book (Learning to Think Mathematically with the Ratio Table). Please understand that this chapter is a proxy to the power and applicability of the ratio table as a mathematical model that has great potential to influence the thinking and mathematical understanding of children. What is a ratio table? And, why should we use it? The ratio table is an excellent computational tool that, when understood well by students, can be used quickly, efficiently, and accurately to multiply and divide, calculate percentages, explore ratios and proportions, etc. It is particularly effective in problem contexts that present themselves in a way that elicits proportional reasoning. For example, consider the following two problems: Example 1: What is the square footage of a garage that is 15 feet wide and 12 feet long? Example 2: Markers come in packs of 12. The teacher ordered 15 packs. How many markers are there in 15 packs? These problems are fundamentally the same: 12 x 15 = 180. Yet, if students are encouraged to let the context inform their progress toward a solution, they will approach each problem in very different ways. Students that have only experienced the traditional algorithm will likely solve both problem contexts in the following way, which is likely familiar to most all of us: Solution 1: Traditional Model (12 x 15) 15 x 12 30 150 180 Students who are comfortable with the area model as articulated in the previous chapter, however, would likely solve Example ! (i.e., square foot of a garage) with an area diagram such as the following: Learning to Think Mathematically about Multiplication 39 Solution 2: Area Model (12 x 15) 10 5 10 .
0 8 1 = 0 1 + 0 2 + 0 5 + 0 0 1 = a e r A 2 Example 2 (i.e., packs of markers), in contrast, is stated in a way that elicits proportional reasoning, and therefore is a great candidate for the use of a ratio table as a solution strategy. The ratio table below is a representation of the following thinking as a solution to the problem of 15 packs of 12 markers. Step 1: “One pack contains 12 markers.” Step 2: “Okay… if one pack contains 12 markers, then two packs contain 24.” Step 2: “Likewise… if one pack contains 12 markers, then ten packs would contain 120 markers.” Step 3: “Ok… if ten packs contain 120 markers, then 5 packs must have 60 markers.” Step 4: “So… if ten packs contain 120, and if 5 packs contain 60, then 15 packs must contain 180 markers.” Expressed in a table, this thinking might look like the following: Solution 3: Ratio Table (12 x 15) Packs 1 2 10 5 15 Markers 12 24 120 60 180 As suggested in this abbreviated chapter on the model, ratio tables like the one above are extremely powerful tools. As you and your students explore this chapter, opportunities will be presented for students to embrace ratio tables not only as a tool for calculations, but also as a way of thinking about mathematical relationships. Bear in mind that a full explanation of the nuanced uses of the model articulated below may be found in the ratio table book that is part of this Thinking Mathematically series. Common Ratio Table Strategies Strategy 1: Multiplication by 10 The number relationship inherent in multiplication by ten is comforting to young learners, and it becomes one of the most important and well-used strategies with ratio tables. In ratio table form, the “tens” strategy looks like the following: Example 1: Example 2: 100 50 20 10 Learning to Think Mathematically about Multiplication 40 Doubling Strategy Students should be encouraged to adopt the following language to accompany the mental strategy: Example 1: “If one group has 15, then 10 groups would have 150.” Example 2: “If 3 groups have 12, then 30 groups would have 120.” Strategy 2: Multiplication by Any Number Related to the 10’s multiplication strategy above, students may also be led to recognize that multiplication by any factor is a viable ratio table strategy. For example: Quite readily, students realize that multiplication can be a helpful tool in quickly scaling any particular ratio. Moreover, recognize that students may choose various multipliers that they feel comfortable using in this process. Consider the following two examples as illustrations of these important considerations. Example 2: There are 5 pencils in a box. How many pencils are in 18 boxes? To find the solution for this problem, the student first multiplied by 6, followed by multiplication by 3. Consider an alternative strategy for the same problem: Example 3: There are 5 pencils in a box. How many pencils are in 18 boxes? The strategy employed by this student was to first multiply by 9, and then multiply by 2. Strategy 3: Doubling A particular subset of the multiplication strategy outlined above is the special case of “doubling” a column in a ratio table. The strategy of doubling is quite common among children and, once again, is a wonderful mental math strategy that will serve them well throughout their many mathematical explorations in contexts both in and outside of school. Consider the following example: Doubling Example 1: I get paid $8 per hour. How much did I earn after working 20 hours? Boxes 1 9 18 Pencils 5 45 90 Hours 1 2 4 5 10 20 Wages 8 16 32 40 80 160 Boxes 2 6 Nails 50 150 x 3 Student thinking: “I know 2 x 3 = 6. So… I could also multiply 50 by 3: 50 x 3 = 150. That is … if 2 boxes contained 50 nails, then 6 boxes would contain 150 nails.” Learning to Think Mathematically about Multiplication 41 Halving Strategy A “doubling” strategy is used repeatedly toward the solution for this problem. Each of the arrows above the table indicates that a doubling calculation between adjacent columns has taken place. On the top row, then, the student has completed the following doubles facts: 1 x 2 = 2; 2 x 2 = 4; 5 x 2 = 10; 10 x 2 = 20. Across the bottom row, we see the following numbers were likewise doubled: 8 to 16; 16 to 32; 40 to 80; and 80 to 160. Doubling is a powerful strategy with wide reaching application and, again, easily motivated and developed through the use of ratio tables. Strategy 4: Halving The inverse operation to doubling the entries in one column to another is to “halve” a column. Again, this is a powerful mental math strategy that children use intuitively in many contexts. Given that halving makes a quantity smaller, it should be noted therefore that this halving strategy is often used to reduce quantities in a ratio table. Hence, as discussed later in the book, halving strategies are instrumental in using the ratio table for division. Consider the following example: Halving Example 1: 120 new baseballs must be split evenly between 8 teams. How many new baseballs does each team get? A “halving” strategy is used repeatedly toward the solution for this problem. Each of the arrows above the table indicates that a halving calculation between adjacent columns has been performed. On the top row, then, the student has completed the following halving calculations: Half of 8 is 4; half of 4 is 2; half of 2 is 1. Likewise, across the bottom row we see the following calculations: half of 120 is 60; half of 60 is 30; half of 30 is 15. Therefore, each team receives 15 baseballs. Be aware that halving strategies are not exclusive to division problems. Sometimes a student will use a halving strategy as part of a string of calculations toward a desired end. Consider the following example in which the student is trying to determine how many eggs are in 6 cartons: Cartons 1 10 5 6 Eggs 12 120 60 72 In the context of the solution strategy for this problem, the student has “halved” 10 to get 5, and subsequently taken half of 120 to arrive at 60. The student is trying to arrive at 6 groups of 12 (for a total of 72). The thinking might go as follows: “I know that one group is 12. I am trying to find the total for 6 groups. Well, 10 groups would be 120. Half of that is 5 groups, or, 60. So, if 5 groups is 60, then one more group, six in all, would be 60 + 12 = 72 eggs.” In this example we see how a halving strategy might very well be appropriately applied in a problem that is, by nature, multiplicative. Teams 8 4 2 1 Baseballs 120 60 30 15 Learning to Think Mathematically about Multiplication 42 + = 1 + 2 + 4 = 7 = Strategy 5: Combining Columns (Adding and Subtracting) The final strategy consists of combining columns – either by addition, or by subtraction. The basic idea that we want children to understand is that we are, in a sense, “combining buckets.” For example, imagine the following ratio table that indicates the number of cherries that may be found in various combinations of baskets (15 cherries per basket). Baskets 1 2 4 7 Cherries 15 30 60 105 We begin with our initial ratio: 1 basket holds 15 cherries. Using our multiplication by two strategy, our next column indicates that 2 baskets would therefore hold 30 cherries. The third column simply doubles the second column: if two baskets hold 30 cherries, 4 baskets would hold 60. Now, at this point, we are ready to combine columns. Observe the 4th column. We see 7 in the top row, and 105 in the bottom row. How did we arrive at those figures? We did so by combining the previous columns. 1 + 2 + 4 = 7 baskets. Correspondingly, 15 + 30 + 60 = 105 cherries. As noted previously, one of the positive features of the ratio table is the extent to which it fosters students’ mental math ability. There are several key arithmetic strategies that are essential to successful work with the ratio table. These strategies are highlighted below, followed by a series of activities that encourage the development of these mental math skills. As you teach these methods, it is essential that students become comfortable with each of the following strategies. The ratio table can become an instrumental tool in helping children develop confidence with mental math strategies. This will occur, however, only if students are given ample opportunity to experiment with the ratio table, to “play” with different strategies, and to create their own pathways to solutions. Though guidance is needed at first, eventually the students should take ownership of their chosen strategies. Teachers should not force students to use a prescribed set of steps with the ratio table, even if those steps are more efficient than the path being taken by the child. With time, students gravitate toward efficiency; they derive great satisfaction in determining their own solution strategy, and will naturally seek to complete the table with as few steps as possible. Please bear in mind the following: Pushing students toward efficiency too early will stunt the development of their native mathematical intuitions and flexibility with mental math strategies. STRATEGY SUMMARY In the previous pages, we have explored a number of unique strategies: Baskets 1 2 4 7 Cherries 15 30 60 105 Learning to Think Mathematically about Multiplication 43 • Multiplying by 10 • Multiplying by any number • Doubling • Halving • Adding or Subtracting While students will naturally gravitate to some strategies over others, it is important that they understand how the various strategies work, and how they may be used together to solve a given problem. Teachers should make every attempt to compare and share the various solution strategies used by students to solve the same problem. Encourage a diversity of strategy use. Be sure to regularly model multiple approaches to a given problem. This will pay dividends down the road. As students begin to trust their own intuitions and native strategies, they develop an expanded vision of mathematical thinking, of their own efficacy as young mathematicians, of the ratio table itself as a mathematical model for multiplication and division, and powerful mental math strategies. Learning to Think Mathematically about Multiplication 44 Student Activity Sheet 8: NAME: ____ Ratio Table Strategies Here are some common strategies for solving problems with a ratio table. Multiply Multiplying by 10 Doubling Halving Adding Subtracting Find the missing numbers in the shaded boxes in the ratio tables below. Then write which of the above strategies you used. 1. 2. Strategy:____ Strategy:___ 3. 4. Strategy:__ Strategy:__ 5. 6. Strategy:__ Strategy:__ 7. 8. Strategy:__ Strategy:__ 1 10 15 150 2 6 50 150 4 8 15 30 1 2 3 25 50 75 20 10 30 15 1 10 9 12 120 108 1 10 11 12 120 1 10 5 12 120 1 10 20 12 120 2 12 5 4 40 18 2 10 12 8 40 4 8 10 12 8 4 Learning to Think Mathematically about Multiplication 45 x 10 x 10 Student Activity Sheet 9: NAME: _____ Ratio Table Practice Problems Directions: Solve each problem with a ratio table. Example: There are 5 pieces of gum in a pack. How many pieces of gum are in 10 packs? 1) There are 6 chairs per row. How many chairs are there in 10 rows?
2) There are 3 eggs per nest. How many eggs are there in 10 nests?
3) There are 4 legs per table. How many legs are there in 10 tables?
4) Two trays contain 12 ice cubes. How many ice cubes are there in 20 trays?
5) There are 4 balloons on each string. How many balloons are there on 20 strings?
6) There are 12 eggs in every carton. How many eggs are there in 8 cartons?
Pack 1 10 Pieces 5 50 Rows 1 10 Chairs 6 Nests 1 Eggs 3 Tables Legs Trays 2 Ice Cubes 12 Strings 1 Balloons 4 Cartons 1 Eggs 12 Learning to Think Mathematically about Multiplication 46 7) There are three gallons in each container. How many gallons are in 12 containers?
8) 5 students need 20 pencils. How many pencils do 40 students need?
9) 60 eggs fit into 4 baskets. How many eggs fit into one basket?
10) You can buy 12 peaches for $4. How much does it cost to buy 6 peaches?
11) It takes 1 minute to travel 2 miles on the high speed train. How many miles can you travel in 12 minutes?
12) It takes 2 minutes to run 1 lap around the track. How long would it take to run 5 laps?
13). A tube contains 3 tennis balls. How many tennis balls are there in 32 tubes? Tube Tennis Balls 14) Comic books cost $8. How much will it cost to buy 12 comic books?
Comic Books Cost Containers Gallons Students Pencils Baskets 4 Eggs 60 Peaches Cost Minute 1 2 10 12 Miles 2 4 Laps 1 2 3 5 Minutes 2 Learning to Think Mathematically about Multiplication 47 15) At the market, 2 apples sell for $3. How much would it cost to buy 12 apples?
16) At the zoo, the seals eat 25 pounds of fish a day. How much would they eat in one week (7 days)?
17) At the same zoo, 25 visitors are allowed in the gates each hour. How many visitors are allowed into the zoo in one day? The zoo is open for 8 hours each day.
18) It costs $12 per student to get into the movie theater. How much would it cost for a group of 21 students to go to the movies?
19) Every 4 hours, the dentist sees 8 patients. How many patients does the dentist see after 11 hours?
Hours Patients Seen 20) Use a ratio table to solve this problem: What is 6 x 15? (Think… six groups of 15.) Groups of 15 Total Learning to Think Mathematically about Multiplication 48 x 2 x 2 x 2 x4 x2 Teacher notes for Activity Sheet 9 The problems on this activity sheet reflect various strategies highlighted in Activity Sheet #8. Problems 1-4, for example, may be solved most efficiently with the “multiply by 10” strategy. Problems 5-8 can be solved by using various multiplication strategies – multiplying by 2, by 4, or by various combinations of multiplication factors. For example, on problem 8, students are asked to determine how many pencils will be needed to supply 40 students, assuming that for every 5 students, 20 pencils are required. Solution strategies will vary. While one student may choose to use a doubling strategy (Solution 1), another student may use a variety of multiplication facts (Solution 2). Both strategies are correct, and teachers should take advantage of the opportunity to compare the thinking of various students as captured in their ratio tables. These are powerful moments in the classroom in which students, when challenged to compare their own thinking to that of their peers, can make leaps in their understanding of the nature of multiplication. Problem 8: 5 students need 20 pencils. How many pencils do 40 students need? Solution 1: Doubling Doubling strategy use: 5 x 2 = 10; 10 x 2 = 20; 20 x 2 = 40 20 x 2 = 40; 40 x 2 = 80; 80 x 2 = 160 Solution 2: Combination of strategies Strategy use: Multiply by 4; Double 5 x 4 = 20… and therefore… 20 x 2 = 40 20 x 4 = 80… and therefore… 80 x 2 = 160 Problems 9 and 10 are unique in the sense that they are division problems. One of the great benefits of the ratio table is that if students understand proportional reasoning as developed through the ratio table model, then they are not inclined (nor need) to make a distinction between multiplication and division problems; they may be solved with the same ratio table strategies. The remaining problems on the page require students to apply various strategies to resolve the problems. Bear in mind that students will need to practice with ratio tables well beyond the problems that exist on this activity sheet. Indeed, an entire book in this series is devoted to the use of ratio tables as a tool for multiplicative reasoning. Teachers may need to adapt the problems on this activity sheet depending on the level of understanding exhibited by students. In every case, however, the power of the ratio table is that it allows students to apply unique strategies toward the solution of the problem. Teachers should make use of every opportunity Students 5 10 20 40 Pencils 20 40 80 160 Students 5 20 40 Pencils 20 80 160 Learning to Think Mathematically about Multiplication 49 that arises to compare strategies among and between students. Observing the thinking patterns of peers who may have solved a problem with a different strategy is a powerful teaching and learning tool. Learning to Think Mathematically about Multiplication 0 Chapter : The Traditional Multiplication Method Overview The intent of a model-based approach to mathematics education is to provide a scaffold for students to approach and solve problems. Effective models typically have a structure that is inherently reflective of the essential features of the mathematics embedded in the problem. In the first chapter of the book, several types of multiplication problems were presented (e.g., multiplication as repeated addition, as combinations, as an area, etc.). These models were explored in subsequent chapters. The intent of a model-based approach is, once again, to represent the mathematics of the problem in a form that reflects the inherent structure of the mathematics at hand, as well as to use that structure as an avenue toward a solution. As we conclude this book by exploring the traditional method of multiplication, it is important to note that one of the criticisms of the traditional algorithm that has been taught almost exclusively in public schools for decades is that, while efficient, it is not inherently conceptual. There is little in the method itself that would be intuitively obvious to the student. Simply put, steps must be memorized and applied. The algorithm was promoted heavily in school textbooks in an era of American education that modeled itself on the industrial revolution in which efficiency and decontextualization were preferred over conceptual understanding. Hence, while the algorithm is efficient and, when memorized, quite dependable, it provides little scaffolding for students to understand the very discrete steps inherent in the method. As the most common method for multiplication in the American educational context, however, it is important for teachers and students alike to grapple with the algorithm, hopefully understanding not just the steps themselves, but the mathematics behind each stage of the algorithm. Further, when fully understood, it is possible for students to make connections between the algorithm itself, and other strategies for multi-digit multiplication. The intent of this chapter is to uncover the inner-workings of the traditional method. For example, why do we put a zero down in the right hand column when we compute the second row of computation? Why do we “carry” as we multiply? These and other question are briefly explored in this section. Unpacking the Traditional Method of Multiplication It is important for students to understand that there are several viable ways to multiply. No one method is better than another, as long as the method of choice is both reliable for the student, and is understood well enough such that the student has an idea when the result of a given computation is reasonable (and equally importantly, unreasonable). In the steps below, the traditional algorithm is dissected for students. The explanation below might be replicated with students, proceeding from simple problems to those that are more complex. Consider the following problem, solved with the traditional algorithm. What are the steps used in solving this problem, and how might we teach them to students? Solve: 86 x 24 with the Traditional Model Learning to Think Mathematically about Multiplication 1 Let’s take it one step at a time, and begin with a slightly easier problem: 86 x 4. The key to understanding the traditional method is to recognize the use of partial products. That is, we can think of 86 x 4 in the following way: “We need 86 groups of 4. Well… we could start with 80 groups of 4 to begin with, and then we could add in 6 more groups of 4. That gives us a total of 86 groups of 4.” In mathematical symbols, we get the following: 86 x 4 = (80 x 4) + (6 x 4). The key to understanding the traditional model is to recognize this use of number decomposition (e.g., 86 = 80 + 6), as well as the use of the distributive property. Below, these concepts are reflected in the sequence of computations that comprise the traditional method. Recall that the traditional method was designed to be as efficient as possible, combining steps when available. One might argue that understanding is lost at the expense of efficiency. If we were to help students see the entire set of calculations for this problem, we might represent the solution to the problem in the following way: In step #1, we multiply 4 times 86. Only, we do it in two steps. First, we multiply the “ones” of the problem: 4 x 6 (6 groups of 4). 4 x 6 = 24, or… 2 tens, and 4 ones. So, we put the “4” in the ones column. We have not yet recorded the 2 tens below the line, but we keep track of them until we can tally them together with the other tens we get as we continue the problem. In step #2, we continue toward the goal of finding the answer to 86 groups of 4. So far, we have recorded the total for 6 groups of 4. Now we need to determine the answer to the remaining 80 groups of 4. For this step, we will be counting tens and hundreds. 4 x 8(0) = 32(0). That means 3 hundreds, and 2 tens. Recall that we still had the 2 extra tens from the first step, which are collected in the tens column. In all, we have three hundreds, 4 tens, and 4 ones. So, that means we have 4 tens (we record them), and 3 hundreds (we record them). Hundreds Tens Ones Learning to Think Mathematically about Multiplication 2 6 groups of 4 = 24 80 groups of 4 = 320 Together, 86 groups of 4 = 344 Notice how the traditional algorithm reduces this process by a step as it requires one to “carry the two” (i.e., collect 4 ones below the line, and record the 2 tens above the problem) on the first step rather than to represent the 24 as one number below the line. We might also consider the way students are taught to “put a zero down under the 4” prior to multiplying 4 x 32. Hidden in this step is that the calculation being made is not really 4 x 8, but rather 4 x 80. The following activity sheet asks students to provide explanations for various steps in the problem, starting with a single digit multiplier, and then moving toward a double-digit multipliers. It may be necessary to walk through several practice problems prior to assigning the tasks below. Take time to illuminate the meaning behind these steps with students as they develop both facility with, and comprehension of, the traditional multi-digit multiplication algorithm. Learning to Think Mathematically about Multiplication 3 Student Activity Sheet 1 : Traditional Method NAME: ____ Directions: Provide explanations for the steps highlighted below. 1. Can you explain each of the steps involved in the strategy shown below?
Where did these 5 (ones) come from? Where did these 2 (tens) come from? Where did these 2 (hundreds) come from? 2. Can you explain each of these steps in this problem?
Where did the 0 (ones) come from? Where did the 8 (tens) come from? Where did the 3 (hundreds) come from? How did we arrive at a final answer of 456? Learning to Think Mathematically about Multiplication 4 Student Activity Sheet 1 Continued NAME: ______ 3. Solve the following problems with the traditional method.
a) 13 x 22 b) 34 x 25 c) 45 x 32 Learning to Think Mathematically about Multiplication 5 Appendix A The Multiplication Table X 1 2 3 4 5 6 7 8 9 10 11 12 1 1 2 3 4 5 6 7 8 9 10 11 12 2 2 4 6 8 10 12 14 16 18 20 22 24 3 3 6 9 12 15 18 21 24 27 30 33 36 4 4 8 12 16 20 24 28 32 36 40 44 48 5 5 10 15 20 25 30 35 40 45 50 55 60 6 6 12 18 24 30 36 42 48 54 60 66 72 7 7 14 21 28 35 42 49 56 63 70 77 84 8 8 16 24 32 40 48 56 64 72 80 88 96 9 9 18 27 36 45 54 63 72 81 90 99 108 10 10 20 30 40 50 60 70 80 90 100 110 120 11 11 22 33 44 55 66 77 88 99 110 121 132 12 12 24 36 48 60 72 84 96 108 120 132 144 Jeffrey Frykholm, Ph.D.
An award winning author, Dr. Jeffrey Frykholm is a former classroom teacher who now focuses on helping teachers develop pedagogical expertise and content knowledge to enhance mathematics teaching and learning. In his Learning to Think Mathematically series of textbooks for teachers, he shares his unique approach to mathematics teaching and learning by highlighting ways in which teachers can use mathematical models (e.g., the rekenrek, the ratio table, the number line, the area model) as fundamental tools their classroom instruction. These books are designed to support teachers' content knowledge and pedagogical expertise toward the goal of providing a menaingful and powerful mathematics education for all children.
• • • • • Understand the nature of multiplication Recognize multiplicative contexts Develop fluency with various multiplication models Make the connection between multiplication models See the connection between multiplication and division Example Problem 12 x14 with the Area Model 10 10 2 20 100 40 4 8 12 14 This book is designed to help students develop a rich understanding of multiplication and division through a variety of problem contexts, models, and methods that elicit multiplicative thinking. Elementary level math textbooks have historically presented only one construct for multiplication: repeated addition. In truth, daily life presents us with various contexts that are multiplicative in nature that do not present themselves as repeated addition. This book engages those different contexts and suggests appropriate strategies and models that resonate with children’s intuitions as they engage multiplication concepts. The book also addresses common approaches to multiplication, including a close look at the multiplication facts, as well as the traditional multiplication alogorithms. Students are also led to see connections between multiplication and division. |
190175 | https://www.scirp.org/journal/paperinformation?paperid=133242 | Standard Model Fermion Masses and Charges from Holographic Analysis
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Journal of Modern Physics>Vol.15 No.6, May 2024
Standard Model Fermion Masses and Charges from Holographic Analysis ()
T. R. Mongan
Sausalito, CA, USA.
DOI:10.4236/jmp.2024.156035PDFHTMLXML 129 Downloads 500 ViewsCitations
Abstract
The Standard Model of particle physics involves twelve fundamental fermions, treated as point particles, in four charge states. However, the Standard Model does not explain why only three fermions are in each charge state or account for neutrino mass. This holographic analysis treats charged Standard Model fermions as spheres with mass 0.187 g/cm 2 times their surface area, using the proportionality constant in the holographic relation between mass of the observable universe and event horizon radius. The analysis requires three Standard Model fermions per charge state and relates up quark and down quark masses to electron mass. Holographic analysis specifies electron mass, to six significant figures, in terms of fundamental constants α,ℏ,G,Λ α,ℏ,G,Λ and Ω Λ Ω Λ. Treating neutrinos as spheres and equating electron neutrino energy density with cosmic vacuum energy density predicts neutrino masses consistent with experiment.
Keywords
Electron Mass, Up Quark Mass, Down Quark Mass, Neutrino Masses
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Mongan, T. (2024) Standard Model Fermion Masses and Charges from Holographic Analysis. Journal of Modern Physics, 15, 796-803. doi: 10.4236/jmp.2024.156035.
1. Introduction
This holographic analysis
1) Explains why three Standard Model fermions are in each charge state e,2 3 e e,2 3 e and −1 3 e−1 3 e ;
2) Relates electron mass to up and down quarks masses;
3) Specifies electron mass in terms of fine structure constant α, Planck’s constant ℏ ℏ , gravitational constant G, cosmological constant Λ, and vacuum energy fraction Ω Λ Ω Λ .
2. Background for Holographic Analysis
Holographic analysis is based on quantum mechanics, general relativity, thermodynamics, and Shannon information theory. Bousso’s review of holographic analysis indicates only about 5 × 10 122 of bits of information on the event horizon will ever be available to describe physics in our universe where the cosmological constant is Λ=1.088×10−56 cm 2 Λ=1.088×10−56 cm 2 .
The radius of the event horizon is R H=3/Λ−−−√=1.66×10 28 cm R H=3/Λ=1.66×10 28 cm . With Hubble constant H 0=67.4 km/(sec⋅Mpc)H 0=67.4 km/(sec⋅Mpc) , critical energy density ρ c r i t=3 H 2 0 8 π G ρ c r i t=3 H 0 2 8 π G , gravitational constant G=6.67430×10−8 cm 3/(g⋅sec 2)G=6.67430×10−8 cm 3/(g⋅sec 2) , and vacuum energy fraction Ω Λ=0.685 Ω Λ=0.685 , the mass of the observable universe within the event horizon is M H=4 3 π(1−Ω Λ)ρ c r i t R 3 H=(0.187 g/cm 2)R 2 H M H=4 3 π(1−Ω Λ)ρ c r i t R H 3=(0.187 g/cm 2)R H 2 . So M H M H is the total mass of the bits of information necessary to describe all physics within the event horizon, indicating the bits of information describing a particle with definite mass m within the universe are available on a spherical surface around the particle with radius r=m M H−−−√R H r=m M H R H .
3. Input Data
Discussion of charged Standard Model fermion masses must begin with PDG 2023 experimental data on those masses. PDG 2023 lists electron mass m e=0.511 MeV/c 2=9.11×10−28 g m e=0.511 MeV/c 2=9.11×10−28 g , up quark mass m u p=2.16+.49−.26 MeV/c 2 m u p=2.16−.26+.49 MeV/c 2 , and down quark mass m d o w n=4.67+.49−.26 MeV/c 2 m d o w n=4.67−.26+.49 MeV/c 2 where c is the speed of light. Up and down quark holographic radii r u p=2 r e r u p=2 r e , r d o w n=3 r e r d o w n=3 r e and quark masses m u p=4 m e m u p=4 m e , m d o w n=9 m e m d o w n=9 m e used in this analysis are consistent with PDG data. Table 1 then lists masses and holographic radii of the nine charged Standard Model fermions as used in this analysis.
Fermion Charge Mass (MeV/c 2)Holographic radius r (fm)
electron−e 0.510999 0.698031
muon−e 105.7 10.04
tau−e 1777 41.16
up quark 2 e/3 2.04340 1.39606
charm quark 2 e/3 1270 34.80
top quark 2 e/3 173,000 406
down quark−e/3 4.59899 2.09409
strange quark−e/3 93.4 9.44
bottom quark−e/3 4180 6.31
Table 1. Charged standard model fermion masses and holographic radii.
4. Charged Standard Model Fermion Structure and Masses
Standard Model fermions described as spheres with holographic radius r, mass m=(0.187 g/cm 2)r 2 m=(0.187 g/cm 2)r 2 , and matter density ρ=m/(4 3 π r 3)ρ=m/(4 3 π r 3) have a surface mass component and an axial mass component along their rotation axis. Holographic analysis is based in part on general relativity, and general relativity is not valid at distances less than the Planck length
l P=ℏ G c 3−−−√=1.61625×10−33 cm=1.61625×10−20 fm l P=ℏ G c 3=1.61625×10−33 cm=1.61625×10−20 fm , so fermion surface mass components are treated as spherical shells with mass m S m S , thickness l P l P , and matter density ρ S l P ρ S l P per unit area, while axial mass components with mass m A m A are treated as cylinders with diameter l P l P and matter density ρ A l 2 P ρ A l P 2 per unit length. A cubic equation for fermion holographic radius in each charge state is
4 3 π ρ r 3=m S+m A=ρ S l P 4 π r 2+ρ A π l 2 P( 2 r )4 3 π ρ r 3=m S+m A=ρ S l P 4 π r 2+ρ A π l P 2( 2 r )
or
ρ r 3−3 ρ S l P r 2−3 2 ρ A l 2 P r=a r 3+b r 2+c r=0 ρ r 3−3 ρ S l P r 2−3 2 ρ A l P 2 r=a r 3+b r 2+c r=0
with a=ρ a=ρ , b=−3 ρ S l P b=−3 ρ S l P and c=−3 2 ρ A l 2 P c=−3 2 ρ A l P 2 . Discriminant b 2 c 2−4 a c 3 b 2 c 2−4 a c 3 of the cubic is positive and three real roots of the equation correspond to holographic radii of three fermions per charge state. Parameters ρ S ρ S and ρ A ρ A (determined by holographic radii r 1, r 2, and r 3) using Nickalls’ solutions to cubic equations involving
r N=r 1+r 2+r 3 3 r N=r 1+r 2+r 3 3 and δ 2=(r 1−r N)2 4+(r 2−r 3)2 12 δ 2=(r 1−r N)2 4+(r 2−r 3)2 12
are ρ S l P=r N ρ ρ S l P=r N ρ , from r N=−b/3 r N=−b/3 , and ρ A l 2 P=2 ρ(r 2 N−δ 2)ρ A l P 2=2 ρ(r N 2−δ 2) , from δ 2=b 2−3 a c 9 a 2 δ 2=b 2−3 a c 9 a 2 .
Tangential velocity v T v T of points on fermion surfaces are found from I ω=ℏ/2 I ω=ℏ/2 , where fermion moment of inertia I=I S+I A I=I S+I A , with shell moment of inertia I S=2 3 m S r 2 I S=2 3 m S r 2 , axial moment of inertia I A=m A 2(l P 2)2 I A=m A 2(l P 2)2 , and I S≫I A I S≫I A . The electron is the only Standard Model fermion with v T=ℏ 4 m r N>c v T=ℏ 4 m r N>c , but points on the electron surface are not particles and cannot send signals, so no particle or signal travels with speed >c.
5. Electron, Up Quark, and Down Quark Masses
All persistent structures in the universe are composed of electrons, protons, and neutrons. Protons and neutrons are composed of up and down quarks. So the lowest mass Standard Model fermions in each charge state (electrons, up quarks, and down quarks) are constituents of all persistent structures in the universe. Holographic analysis then provides succinct explanations relating lowest mass Standard Model fermions in each charge state.
Electrons, the only Standard Model fermions that can persist in isolation in the universe, have the smallest mass and holographic radius of the nine charged Standard Model fermions. Up quarks, with twice the electron holographic radius, have four times the electron mass. Down quarks, with three times the electron holographic radius, have nine times the electron holographic mass.
Protons are composed of two up quarks and one down quark, and neutrons are composed of two down quarks and one up quark. Isolated neutrons decay to protons, so up quarks must have lower mass than down quarks, consistent with experimental data.
6. Electron Mass from α,ℏ,G,Λ α,ℏ,G,Λ and Ω Λ Ω Λ
Using electron holographic radius r e=m e M H−−−√R H r e=m e M H R H , holographic analysis specifies electron mass to six significant figures in terms of fundamental constants α,ℏ,G,Λ α,ℏ,G,Λ and Ω Λ Ω Λ .
Our universe is so large it is almost flat, and Friedmann’s equation H 2 0=8 π G 3 ρ c r i t+Λ c 2 3 H 0 2=8 π G 3 ρ c r i t+Λ c 2 3 identifies Ω Λ=Λ c 2 3 H 2 0 Ω Λ=Λ c 2 3 H 0 2 . Since M H=4 3 π(1−Ω Λ)ρ c r i t R 3 H M H=4 3 π(1−Ω Λ)ρ c r i t R H 3 , M H=(1−Ω Λ)c 2 2 G Ω Λ 3 Λ−−√M H=(1−Ω Λ)c 2 2 G Ω Λ 3 Λ is constant in time.
Electrostatic potential energy of electron charge e and positron charge −e separated by 2 r e 2 r e is V=−e 2 2 r e=−α ℏ c 2 r e V=−e 2 2 r e=−α ℏ c 2 r e , with Planck’s constant ℏ=1.05457×10−27 g⋅cm 2/sec ℏ=1.05457×10−27 g⋅cm 2/sec . Adjacent spheres with holographic radii r e r e , a precursor for electron-positron pair production, have total energy E=2 m e c 2−α ℏ c 2 r e=0 E=2 m e c 2−α ℏ c 2 r e=0 when r e=α ℏ c 4 m e c 2 r e=α ℏ c 4 m e c 2 .
Two equations for r e r e give α ℏ c 4 m e c 2=m e M H−−−√R H α ℏ c 4 m e c 2=m e M H R H and electron mass m e=[(α ℏ 2 32)(1−Ω Λ G Ω Λ Λ 3−−√)]1/3 m e=[(α ℏ 2 32)(1−Ω Λ G Ω Λ Λ 3)]1/3 .
If Λ=1.08800×10−56 cm 2 Λ=1.08800×10−56 cm 2 and Ω Λ=0.6853855 Ω Λ=0.6853855 (within PDG 2023 error bars) electron mass is specified to six significant figures, since gravitational constant G is only known to six significant figures.
7. Neutrino Masses
Each lepton has a corresponding neutrino, but neutrinos oscillate between mass states when propagating through space and are not persistent structures in the universe. So neutrinos are not consistently related to holographic radii. Characteristic lengths of neutrinos are Compton wavelengths λ=ℏ m c λ=ℏ m c . Electron neutrinos with radius 1 4 ℏ m c 1 4 ℏ m c and the lowest energy density in the universe (cosmic vacuum energy density ρ v=5.83×10−30 g/cm 3 ρ v=5.83×10−30 g/cm 3 ) have mass m 1=[π 6 ρ v(ℏ c)3]1 4=2.02×10−36 g=0.0013 eV m 1=[π 6 ρ v(ℏ c)3]1 4=2.02×10−36 g=0.0013 eV . Neutrino oscillation data predict m 2=m 2 1+7.37×10−5 eV 2−−−−−−−−−−−−−−−−−√=0.00866 eV m 2=m 1 2+7.37×10−5 eV 2=0.00866 eV and m 3=0.5(m 1+m 2)2+2.5×10−3 eV 2−−−−−−−−−−−−−−−−−−−−−−−−−√=0.0505 eV m 3=0.5(m 1+m 2)2+2.5×10−3 eV 2=0.0505 eV . Neutrino mass sum 0.0603 eV is then 50% of Vagnozzi’s experimental upper limit of 0.12 eV on neutrino mass sum.
8. Results
This holographic analysis based on quantum mechanics, general relativity, thermodynamics, and Shannon information theory:
1) Requires three Standard Model fermions in each charge state e,2 3 e e,2 3 e , and −1 3 e−1 3 e ;
2) Relates masses of the electron, up quark, and down quark constituents of all permanent structures in the universe to electron mass;
3) Specifies electron mass in terms of fundamental constants α,ℏ,G,Λ α,ℏ,G,Λ and Ω Λ Ω Λ ;
4) Accounts for observed matter dominance in the universe, if the universe is closed.
9. Conclusion
This analysis, specifying electron, up quark, down quark, and three neutrino masses, reduces by six the number of free parameters in the Standard Model of particle physics. These results can provide insight to develop physical theories to supplement the Standard Model.
Appendix: Matter Dominance in a Closed Universe
Charged bits on Standard Model fermion surfaces must be at the rotation axis, to avoid radiation from accelerated charge. Bits of information on the horizon can indicate presence of a charged Standard Model fermion somewhere along the axis between diametrically opposed bits of information on opposite hemispheres of the horizon, so charge ±e/6 must be associated with each bit of information. One, two, or three bit pairs on opposite surfaces of spherical Standard Model fermions at their rotation axis specify charge −e/3 or 2 e/3 quarks, or charge e leptons.
A closed universe beginning by a quantum fluctuation from nothing must be charge neutral, with equal numbers of e/6 and −e/6 bits. Regardless of details of how bits of information specify protons or anti-protons, configurations specifying protons differ in 6 bits from configurations specifying anti-protons. In any physical system, energy is transferred to change bits from one state to another, and e/6 bits with lower energy than −e/6 bits result in more matter than anti-matter in a closed universe, as discussed below.
Temperature at time of baryon formation was T B=2 m p c 2 k=2.18×10 3 K T B=2 m p c 2 k=2.18×10 3 K , with
Boltzmann constant k=1.38×10−16(g⋅cm 2/sec 2)/K k=1.38×10−16(g⋅cm 2/sec 2)/K and proton mass m p=1.67×10−24 g m p=1.67×10−24 g . Radius of the universe at baryogenesis was
R B=(2.725 2.18×10 3)R 0≈10 15 cm R B=(2.725 2.18×10 3)R 0≈10 15 cm , where 2.725 K is today’s microwave background
temperature and radius of the universe today is R 0≈10 28 cm R 0≈10 28 cm . Baryogenesis time t B t B in seconds after the end of inflation is determined by Friedmann’s equation
(d R d t)2−(8 π G 3)ε(R)(R c)2=−κ c 2(d R d t)2−(8 π G 3)ε(R)(R c)2=−κ c 2 . After inflation, in a closed universe so large
it is almost flat, curvature parameter κ≈0 κ≈0 . Energy density
ε(R)=ε r(R 0 R)4+ε m(R 0 R)3+ε v ε(R)=ε r(R 0 R)4+ε m(R 0 R)3+ε v , where ε r,ε m ε r,ε m and ε v ε v are today’s radiation,
matter, and vacuum energy densities. Matter energy density
ε m≈9×10−9(g⋅cm 2⋅sec−2/cm 3)ε m≈9×10−9(g⋅cm 2⋅sec−2/cm 3) , and vacuum energy density was negligible in the early universe, so radiation dominated when R≪10−5 R 0 R≪10−5 R 0 before radiation/
matter equality. Integrating (d R d t)2−(8 π G 3 c 2)ε r R 4 0 R 2=(d R d t)2−A 2 R 2=0(d R d t)2−(8 π G 3 c 2)ε r R 0 4 R 2=(d R d t)2−A 2 R 2=0 where A=8 π G ε r R 4 0 3 c 2−−−−−−√A=8 π G ε r R 0 4 3 c 2 , from the end of inflation at t=0 t=0 to t, determines A t=1 2(R(t)2−R 2 i)A t=1 2(R(t)2−R i 2) , where R i R i is radius of the universe at the end of inflation and R B≫R i R B≫R i . So t B=R 2 B−R 2 i 2 A≈R 2 B 2 A≈10−7 t B=R B 2−R i 2 2 A≈R B 2 2 A≈10−7 seconds. Distance from any point in the universe at baryogenesis to the horizon for that point is d B=c∫t B 0 d t′R(t′)=c R B A[R 2 i+2 A t B−−−−−−−−−√−R i]≈c R 2 B A≈10 4 cm d B=c∫0 t B d t′R(t′)=c R B A[R i 2+2 A t B−R i]≈c R B 2 A≈10 4 cm .
Surface gravity on the horizon at baryogenesis is
g H B=G 4 π 3 ε(R B)c 2 d B≈4 π G 3 c ε r R 4 0 A R 2 B g H B=G 4 π 3 ε(R B)c 2 d B≈4 π G 3 c ε r R 0 4 A R B 2 , and associated horizon temperature is T H B=ℏ 2 π c k g H B≈6×10−7 K T H B=ℏ 2 π c k g H B≈6×10−7 K . Occupation probabilities of bits on the horizon at
baryogenesis are proportional to their Boltzmann factors. If energy of e/6 bits is E−Δ E−Δ and energy of −e/6 bits is E+Δ E+Δ , proton-antiproton ratio at baryogenesis
is (e−E−Δ k T H B/e−E+Δ k T H B)6=e 12 Δ k T H B≈1+12 Δ k T H B(e−E−Δ k T H B/e−E+Δ k T H B)6=e 12 Δ k T H B≈1+12 Δ k T H B and proton excess is 12 Δ k T H B 12 Δ k T H B . Energy released when a −e/6 bit on the horizon changes to an e/6 bit raises another bit from e/6 to −e/6, ensuring charge conservation. Energy to change the state of bits on the horizon must be transferred by massless quanta with wavelength related to the scale of the horizon, and the only macroscopic length characteristic of the horizon at baryogenesis is the circumference 2 π R B 2 π R B . If energy 2Δ to change the state of bits on the horizon (and corresponding bits within the universe) is energy of massless quanta with wavelength characteristic of a closed Friedmann universe with radius R B R B at baryogenesis, 2 Δ=ℏ c R B 2 Δ=ℏ c R B . Substituting from above, proton excess at baryogenesis is 12 Δ k T H B=(24 π c 2 R 0)(2.725 T B)3 8 π G ε r−−−−√≈1.8×10−9 12 Δ k T H B=(24 π c 2 R 0)(2.725 T B)3 8 π G ε r≈1.8×10−9 .
WMAP found (baryon density)/(microwave background photon density) = 6.1 × 10−10. At baryogenesis, the number of protons, anti-protons, and photons were approximately equal. When almost all protons and anti-protons annihilated to two photons, the baryon to photon ratio became 1 3(1.8×10−9)=6×10−10 1 3(1.8×10−9)=6×10−10 , in agreement with WMAP results.
Conflicts of Interest
The author declares no conflicts of interest regarding the publication of this paper.
References
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Capozzi, F., Lisi, E., Marrone, A., Montanino, D. and Palazzo, A. (2016) Nuclear Physics B, 908, 218-234.
Vagnozzi, S. (2019) Cosmological Searches for the Neutrino Mass Scale and Mass Ordering. Ph.D. Thesis, Stockholm University, Stockholm.
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190177 | http://jingweizhu.weebly.com/uploads/1/3/5/4/13548262/forced_convection_correlations.pdf | Part B: Heat Transfer Principals in Electronics Cooling 75 MPE 635: Electronics Cooling 9. Forced Convection Correlations Our primary objective is to determine heat transfer coefficients (local and average) for different flow geometries and this heat transfer coefficient (h) may be obtained by experimental or theoretical methods. Theoretical methods involve solution of the boundary layer equations to get the Nusselt number such as explained before in the previous lecture. On the other hand the experimental methods involve performing heat transfer measurement under controlled laboratory conditions and correlating data in dimensionless parameters. Many correlations for finding the convective heat transfer coefficient are based on experimental data which needs uncertainty analysis, although the experiments are performed under carefully controlled conditions. The causes of the uncertainty are many. Actual situations rarely conform completely to the experimental situations for which the correlations are applicable. Hence, one should not expect the actual value of the heat transfer coefficient to be within better than 10%of the predicted value. 9.1 Flow over Flat Plate With a fluid flowing parallel to a flat plate we have several cases arise: • Flows with or without pressure gradient • Laminar or turbulent boundary layer • Negligible or significant viscous dissipation (effect of frictional heating) 9.1.1 Flow with Zero Pressure Gradient and Negligible Viscous Dissipation When the free-stream pressure is uniform, the free-stream velocity is also uniform. Whether the boundary layer is laminar or turbulent depends on the Reynolds number ReX (ρu∞x/µ) and the shape of the solid at entrance. With a sharp edge at the leading edge the boundary layer is initially laminar but at some distance downstream there is a transition region and downstream of the transition region the boundary layer becomes turbulent. For engineering applications the transition region is usually neglected and it is assumed that the boundary layer becomes turbulent if the Reynolds number, ReX, is greater than the critical Reynolds number, Recr. A typical value of 5 x105 for the critical Reynolds number is generally accepted. The viscous dissipation and high-speed effects can be neglected if Pr1/2 Ec<1. The Eckert number Ec is defined as Ec = u2 ∞/Cp (TS-T∞) with a rectangular plate of length L in the direction of the fluid flow. 9.1.1.1 Laminar Boundary Layer (Rex ≤ 5x105) With heating or cooling starting from the leading edge the following correlations are recommended. Note: in all equations evaluate fluid properties at the film temperature (Tf) defined as the arithmetic mean of the surface and free-stream temperatures unless otherwise stated. Tf = (TS + T∞)/2 Part B: Heat Transfer Principals in Electronics Cooling 76 MPE 635: Electronics Cooling Flow with uniform surface temperature (TS = constant) - Local heat transfer coefficient The Nusselt number based on the local convective heat transfer coefficient is expressed as 2 / 1 Pr Re X X f Nu = The expression of ƒPr depend on the fluid Prandtl number For liquid metals with very low Prandtl number liquid metals (Pr ≤ 0.05) 2 / 1 Pr Pr 564 .
0 = f For 0.6 < Pr < 50 3 / 1 Pr Pr 332 .
0 = f For very large Prandtl number 3 / 1 Pr Pr 339 .
0 = f For all Prandtl numbers; Correlations valid developed by Churchill (1976) and Rose (1979) are 4 / 1 3 / 2 3 / 1 2 / 1 Pr 0468 .
0 1 Pr Re 3387 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = x x Nu - Average heat transfer coefficient The average heat transfer coefficient can be evaluated by performing integration along the flat plate length, if Prandtl number is assumed constant along the plate, the integration yields to the following result: x x Nu Nu 2 = Flow with uniform heat flux (q// = constant) - Local heat transfer coefficient Churchill and Ozoe (1973) recommend the following single correlation for all Prandtl numbers 4 / 1 3 / 2 2 / 1 2 / 1 0207 .
0 Pr 1 Pr Re 886 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = X X Nu (9.1) (9.3) (9.2) (9.4) Part B: Heat Transfer Principals in Electronics Cooling 77 MPE 635: Electronics Cooling 9.1.1.2 Turbulent Boundary Layer (Rex > 5x105) With heating or cooling starting from the leading edge the following correlations are recommended. - Local heat transfer coefficient Recr< Rex ≤107 3 / 1 5 / 4 Pr Re 0296 .
0 x x Nu = Rex >107 3 / 1 584 .
2 Pr ) Re (ln Re 596 .
1 − = x x x Nu Equation 9.6 is obtained by applying Colburn’s j factor in conjunction with the friction factor suggested by Schlichting (1979). With turbulent boundary layers the correlations for the local convective heat transfer coefficient can be used for both uniform surface temperature and uniform heat flux. - Average heat transfer coefficient If the boundary layer is initially laminar followed by a turbulent boundary layer some times called mixed boundary layer the Following correlations for 0.6 < Pr < 60 are suggested: Recr < Rex ≤ 107 3 / 1 5 / 4 Pr ) 871 Re 037 .
0 ( − = x x Nu Rex >107 3 / 1 584 .
2 Pr ] 871 ) Re (ln Re 963 .
1 − = − x x x Nu 9.1.1.3 Unheated Starting Length Uniform Surface Temperature & Pr > 0.6 If the plate is not heated or (cooled) from the leading edge where the boundary layer develops from the leading edge until being heated at x = xo as shown in Figure 9.1, the correlations have to be modified. . xo δ δT x Leading edge Figure 9.1 Heating starts at x = xo - Local heat transfer coefficient (9.5) (9.6) (9.7) (9.8) Part B: Heat Transfer Principals in Electronics Cooling 78 MPE 635: Electronics Cooling Laminar flow (Rex < Recr) 3 / 1 4 / 3 3 / 1 2 / 1 1 Pr Re 332 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = x x Nu o x x Turbulent flow (Rex > Recr) 9 / 1 10 / 9 5 / 3 5 / 4 1 Pr Re 0296 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = x x Nu o x x - Average heat transfer coefficient over the Length (L – xo) Laminar flow (ReL < Recr) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = − 4 / 3 3 / 2 4 / 3 3 / 1 2 / 1 1 ) / ( 1 2 1 Pr Re 664 .
0 L x L x h L x x L k h o o L x o L o x L o Evaluate hx=L from Equation 9.9. Turbulent flow (ReL > Recr) L x o o o o L x L h L x L x x L k L x h o = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = 1 1 25 .
1 1 Pr Re 037 .
0 10 / 9 9 / 8 10 / 9 5 / 3 5 / 4 Evaluate hx=L from Equation 9.10. Example 9.1: Experimental results obtained for heat transfer over a flat plate with zero pressure gradients yields 3 / 1 9 .
0 Pr Re 04 .
0 x x Nu = Where this correlation is based on x (the distance measured from the leading edge). Calculate the ratio of the average heat transfer coefficient to the local heat transfer coefficient. Schematic: (9.9) (9.10) (9.11) (9.12) Part B: Heat Transfer Principals in Electronics Cooling 79 MPE 635: Electronics Cooling x Solution: It is known that k x h Nu x x = Local heat transfer coefficient 3 / 1 9 .
0 Pr Re 04 .
0 x x x k h = The average heat transfer coefficient is: 9 .
0 Pr Re 9 .
0 04 .
0 9 .
0 Pr 04 .
0 Pr 04 .
0 1 Pr 04 .
0 1 3 / 1 9 .
0 9 .
0 9 .
0 3 / 1 0 1 .
0 9 .
0 3 / 1 0 9 .
0 3 / 1 0 x x x x x x x h x k x v u x k dx x v u x k dx x v x u x k dx h x h = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = = ∞ − ∞ ∞ ∫ ∫ ∫ The ratio of the average heat transfer coefficient to the local heat transfer coefficient is: 9 .
0 1 = x x h h Example 9.2: Air at a temperature of 30 oC and 6 kPa pressure flow with a velocity of 10 m/s over a flat plate 0.5 m long. Estimate the heat transferred per unit width of the plate needed to maintain its surface at a temperature of 45 oC. Schematic: x T∞=30 oC P∞=6 kpa u∞ =10 m/s L=0.5 m TS=45 oC Part B: Heat Transfer Principals in Electronics Cooling 80 MPE 635: Electronics Cooling Solution: Properties of the air evaluated at the film temperature Tf = 75/2 = 37.5 oC From air properties table at 37.5 oC: ν = 16.95 x 10-6 m2/s. k = 0.027 W/m. oC Pr= 0.7055 Cp= 1007.4 J/kg. oC Note: These properties are evaluated at atmospheric pressure, thus we must perform correction for the kinematic viscosity ν) act. = ν)atm.x (1.0135/0.06) = 2.863 x10-4 m2/s. Viscous effect check, Ec =u2 ∞/Cp(TS-T∞) = (10)2/(1007.4x15) = 6.6177x10-3 It produce Pr1/2 Ec<1 so that we neglect the viscous effect Reynolds number check ReL = u∞ L / ν = 10 x 0.5 / (2.863 x10-4) = 17464.2 The flow is Laminar because ReL ≤ 5x105 Using the Equation 9.3 3 / 1 2 / 1 Pr Re 664 .
0 L L Nu = = 0.664 (17464.2)1/2(0.7055)1/3 = 78.12 Now the average heat transfer coefficient is k L h Nu L = h =78.12x 0.027 /0.5 = 4.22 W/m2. oC Then the total heat transfer per unit width equals q = h L (TS -T∞) = 4.22 x 0.5 (45 – 30) = 31.64 W/m Example 9.3: Water at 25 oC is in parallel flow over an isothermal, 1-m long flat plate with velocity of 2 m/s. Calculate the value of average heat transfer coefficient. Schematic: x T∞=25 oC u∞ =2 m/s L=1 m Solution: Assumptions Part B: Heat Transfer Principals in Electronics Cooling 81 MPE 635: Electronics Cooling - neglect the viscous effect - The properties of water are evaluated at free stream temperature From water properties table at 25 oC: ν =8.57x10-7 m/s. k = 0.613W/m. oC Pr= 5.83 Cp= 4180 J/kg. oC Reynolds number check, ReL = u∞ L / ν = 2 x 1 / (8.57x10-7) = 2.33x106 The flow is mixed because ReL ≥ 5x105 By using the Equation 9.7 k L h x x Nu L x x = = − = − = 78 .
6704 ) 83 .
5 10 33 .
2 ( 037 .
0 [ Pr ) 871 Re 037 .
0 ( 3 / 1 5 / 4 6 3 / 1 5 / 4 The average heat transfer coefficient is 2 4110 W/m K L h = ⋅ 9.3 Flow over Cylinders, Spheres, and other Geometries The flow over cylinders and spheres is of equal importance to the flow over flat plate, they are more complex due to boundary layer effect along the surface where the free stream velocity u∞ brought to the rest at the forward stagnation point (u = 0 and maximum pressure) the pressure decrease with increasing x is a favorable pressure gradient (dp/dx<0) bring the pressure to minimum, then the pressure begin to increase with increasing x by adverse pressure gradient (dp/dx>0) on the rear of the cylinder. In general, the flow over cylinders and spheres may have a laminar boundary layer followed by a turbulent boundary layer. The laminar flow is very weak to the adverse pressure gradient on the rear of cylinder so that the separation occurs at θ = 80o and caused wide wakes as show in Figure9.2.a. The turbulent flow is more resistant so that the separation occurs at θ = 120o and caused narrow wakes as show in Figure9.2.b. θ u∞ ,T∞ ,P∞ separation wide wakes stagnation point (a) θ u∞ ,T∞ ,P∞ separation stagnation point (b) narrow wakes Figure 9.2 Flow over cylinder Because of the complexity of the flow patterns, only correlations for the average heat transfer coefficients have been developed. Part B: Heat Transfer Principals in Electronics Cooling 82 MPE 635: Electronics Cooling 9.3.1 Cylinders The empirical relation represented by Hilpert given below in Equation 9.13 is widely used, where the constants c, m is given in Table 9.1, all properties are evaluated at film temperature Tf. 3 / 1 Pr Re m D D c Nu = Table 9.1 Constants of Equation 9.13 at different Reynolds numbers ReD c m 0.4-4 0.989 0.33 4-40 0.911 0.385 40-4000 0.683 0.466 4000-40,000 0.193 0.618 40,000-400,000 0.027 0.805 Other correlation has been suggested for circular cylinder. This correlation is represented by Zhukauskas given below in Equation 9.14 , where the constants c ,m are given in Table 9.2 , all properties are evaluated at Free stream temperature T∞, except PrS which is evaluated at TS which is used in limited Prandtl number 0.7 < Pr < 500 4 / 1 Pr Pr Pr Re ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = S n m D D c Nu If Pr ≤ 10, n= 0.37 If Pr > 10, n= 0.36 Table 9.2 Constants of Equation 9.14 at different Reynolds numbers ReD c m 1-40 0.75 0.4 40-1000 0.51 0.5 1000-2 x105 0.26 0.6 2 x105-106 0.076 0.7 For entire ranges of ReD as well as the wide ranges of Prandtl numbers, the following correlations proposed by Churchill and Bernstein (1977): ReD Pr > 0.2. Evaluate properties at film temperature Tf: Re 400, 000 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 2 / 1 000 , 282 Re 1 Pr 4 .
0 1 Pr Re 62 .
0 3 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = D D D Nu 10, 000 Re 400, 000 (9.13) (9.14) (9.15) Part B: Heat Transfer Principals in Electronics Cooling 83 MPE 635: Electronics Cooling ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = 2 / 1 4 / 1 3 / 2 3 / 1 2 / 1 000 , 282 Re 1 Pr 4 .
0 1 Pr Re 62 .
0 3 .
0 D D D Nu Re 10, 000 4 / 1 3 / 2 3 / 1 2 / 1 Pr 4 .
0 1 Pr Re 62 .
0 3 .
0 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + = D D Nu For flow of liquid metals, use the following correlation suggested by Ishiguro et al. (1979): 1 Re d Pr 100 413 .
0 Pr) (Re 125 .
1 D d Nu = 9.3.2 Spheres Boundary layer effects with flow over circular cylinders are much like the flow over spheres. The following two correlations are explicitly used for flows over spheres, 1. Whitaker (1972): All properties at T except s at Ts 3 .5 Re d 76,000 0.71 Pr 380 1 s 3.2 ( ) 4 / 1 5 / 2 3 / 2 2 / 1 Pr Re 06 .
0 Re 4 .
0 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = S D D D Nu µ µ 2. Achenbach (1978): All properties at film temperature 100 Re d 2 x105 Pr = 0.71 ( ) 2 / 1 5 / 8 4 Re 10 3 Re 25 .
0 2 D D D x Nu − + + = 4 x105 Re d 5 x106 Pr = 0.71 3 17 2 9 3 Re 10 1 .
3 Re 10 25 .
0 Re 10 5 430 D D D D x x x Nu − − − − + + = For Liquid Metals convective flow, experimental results for liquid sodium, Witte (1968) proposed, 3.6 x104 Re d 1.5 x105 2 / 1 Pr) (Re 386 .
0 2 D D Nu + = 9.3.3 Other Geometries For geometries other than cylinders and spheres, use Equation 9.24 with the characteristic dimensions and values of the constants given in the Table 9.3 for different geometries, all properties are evaluated at film temperature Tf. 3 / 1 Pr Re m D D c Nu = Equation 9.24 is based on experimental data done for gases. Its use can be extended to fluids with moderate Prandtl numbers by multiplying Equation 9.24 by 1.11. (9.16) (9.17) (9.18) (9.20) (9.21) (9.22) (9.23) (9.24) Part B: Heat Transfer Principals in Electronics Cooling 84 MPE 635: Electronics Cooling Table 9.3 Constants for Equation 9.24 for non circular cylinders external flow Geometry ReD C m D u∞ 5×103-105 0.102 0.675 D u∞ 5×103-105 0.246 0.588 D u∞ 5×103-105 0.153 0.638 D u∞ 5×103-1.95×104 1.95×104-105 0.16 0.0385 0.638 0.782 D u∞ 4 ×103-1.5×104 0.228 0.731 Example 9.4: Atmospheric air at 25 oC flowing at velocity of 15 m/s. over the following surfaces, each at 75 oC. Calculate the rate of heat transfer per unit length for each arrangement. a) A circular cylinder of 20 mm diameter , b) A square cylinder of 20 mm length on a side c) A vertical plate of 20 mm height Schematic: d u∞ (a) u∞ (b) (c) u∞ Solution: From the film temperature Tf = (25+75)/2 = 50 oC Air properties are: ν =1.8x10-5 m/s. k = 0.028W/m. oC Pr= 0.70378 Calculation of Reynolds number for all cases have the same characteristic length=20mm ReD = u∞ L / ν = 15 x 0.02 / 1.8x10-5 = 16666.667 By using Equation 9.13 for all cases 3 / 1 Pr Re m D D c Nu = Part B: Heat Transfer Principals in Electronics Cooling 85 MPE 635: Electronics Cooling Case (a) a circular cylinder From Table 9.1 C = 0.193 and m= 0.618 79 .
69 ) 70378 .
0 ( ) 667 .
16666 ( 193 .
0 3 / 1 618 .
0 = = D Nu The average heat transfer coefficient is 2 o 69.79 0.028 97.7 W/m . C 0.02 D D Nu k x h D = = = The total heat transfer per unit length is q/ = 97.7 (π x 0.02)50 = 306.9 W/m Case (b) a square cylinder From Table 9.3 C = 0.102 and m= 0.675 19 .
64 ) 70378 .
0 ( ) 667 .
16666 ( 102 .
0 3 / 1 675 .
0 = = D Nu The average heat transfer coefficient is 2 o 64.19 0.028 89.87 W/m . C 0.02 D D Nu k x h D = = = The total heat transfer per unit length is q/ = 89.87 (4 x 0.02)50 = 359.48 W/m Case (c) a vertical plate From Table 9.3 C = 0.228 and m= 0.731 316 .
247 ) 70378 .
0 ( ) 667 .
16666 ( 228 .
0 3 / 1 731 .
0 = = D Nu The average heat transfer coefficient is 2 o 247.316 0.028 346.24 W/m . C 0.02 D D Nu k x h D = = = The total heat transfer per unit length is q/ = 346.24 x (2 x 0.02) x 50 = 692.48 W/m 9.4 Heat Transfer across Tube Banks (Heat Exchangers) Heat transfer through a bank (or bundle) of tubes has several applications in industry as heat exchanger which can be used in many applications. When tube banks are used in heat exchangers, two arrangements of the tubes are considered aligned and staggered as shown in Figure 9.3. If the spacing between the tubes is very much greater than the diameter of the tubes, correlations for single tubes can be used. Correlations for flow over tube banks when the spacing between tubes in a row and a column are not much greater than the diameter of the tubes have been developed for use in heat-exchanger applications will be discussed as follows. Part B: Heat Transfer Principals in Electronics Cooling 86 MPE 635: Electronics Cooling T∞ , u∞ ST SL row 1 row 2 row 3 row 4 (a) SL ST SD T∞ , u∞ row 1 row 2 row 3 row 4 (b) NL NT Figure 9.3 Arrangements of the tubes (a) In-line arrangement, (b) staggered arrangement For the average convective heat transfer coefficient with tubes at uniform surface temperature, experimental results carried by Zukauskas (1987) recommended the following correlation: 25 .
0 ) Pr (Pr/ Pr Re ) / ( s n m D P D b a c Nu = a = ST /D; b = SL /D. ST = Transverse pitch SL = Lateral pitch D = tube diameter All properties are evaluated at the arithmetic mean of the inlet and exit temperatures of the fluid (T∞i + T∞o)/2, except PrS which is evaluated at the surface temperature TS. The values of the constants c, p, m, and n are given in Table 9.4 for in-line arrangement and in Table 9.5 for staggered arrangement. The maximum average velocity between tubes is used to calculate ReD. The maximum velocity can be calculated as follows: For in-line arrangement: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ∞ D S S u u T T max For Staggered arrangement: If ∞ − = ∴ + > u D S S u D S S T T T D max 2 If not ( ) ∞ − = ∴ u D S S u D T 5 .
0 max Where 2 / 1 2 2 2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = T L D S S S Table 9.4 In-Line arrangement values of constants in Equation 9.25 (p = 0 in all cases) ReD C m n 1-100 0.9 0.4 0.36 100-1000 0.52 0.5 0.36 1000-2x105 0.27 0.63 0.36 2x105-2x106 0.033 0.8 0.4 (9.25) Part B: Heat Transfer Principals in Electronics Cooling 87 MPE 635: Electronics Cooling Table 9.5 Staggered arrangement values of constants in Equation 9.25 ReD c P m n 1-500 1.04 0 0.4 0.36 500-1000 0.71 0 0.5 0.36 1000-2x105 0.35 0.2 0.6 0.36 2x105-2x106 0.031 0.2 0.8 0.36 The temperature of the fluid varies in the direction of flow, and, therefore, the value of the convective heat transfer coefficient (which depends on the temperature-dependent properties of the fluid) also varies in the direction of flow. It is a common practice to compute the total heat transfer rate with the assumption of uniform convective heat transfer coefficient. With such an assumption of uniform convective heat transfer coefficient, uniform surface temperature and constant specific heat, the heat transfer rate to the fluid is related by the heat balance. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = − = ∞ ∞ ∞ ∞ • 2 ) ( i o S i o P T T T A h T T C m q T∞i = inlet free stream temperature T∞o= outlet free stream temperature • m = out side tubes gaseous flow rate = ρ∞i(NTSTL)u∞ A = Total heat transfer area = NTNLπDL NT = number of tubes in transverse direction NL = number of tubes in lateral direction L = length of tube per pass. Pressure drop across tube banks: Pressure drop is a significant factor, as it determines the power required to move the fluid across bank. The pressure drop for flow gases over a bank may be calculated with Equation 9.27. 14 .
0 2 max / 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∆ µ µ ρ S T N G f p ∆p = pressure drop in pascals. ƒ/ = friction factor is given by Jacob Gmax= mass velocity at minimum flow rate, kg/m2.s ρ = density evaluated at free stream conditions, kg/m3 NT = number of transverse rows. µS = fluid viscosity evaluated at surface temperature. µ = average free stream viscosity. For in-line: (9.27) (9.26) Part B: Heat Transfer Principals in Electronics Cooling 88 MPE 635: Electronics Cooling 15 .
0 max / 13 .
1 43 .
0 / Re / 08 .
0 044 .
0 − + ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = L S D T L D D S D S f For staggered: 16 .
0 max 08 .
1 / Re 118 .
0 25 .
0 − ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = D D S f T Example 9.5: A heat exchanger with aligned tubes is used to heat 40 kg/sec of atmospheric air from 10 to 50 °C with the tube surfaces maintained at 100 °C. Details of the heat exchanger are Diameter of tubes 25 mm Number of columns (NT) 20 Length of each tube 3 m SL = ST 75 mm -Determine the number of rows (NL) required Solution: Properties of atmospheric air at average air temperature = (T∞i + T∞o)/2 = 30 C. ρ = 1.151 kg/ m3 CP= 1007 J/ kg.oC k = 0.0265 W/ m.oC µ = 186x10-7 N.s/ m2 Pr = 0.7066 At surface temperature Ts=100 oC Prs = 0.6954 At inlet free stream temperature T∞i =10 oC ρ∞i = 1.24 kg/ m3 To find u∞ • m = ρ∞i(NTSTL)u∞ 40 = 1.24(20x0.075x3) u∞ u∞ = 7.168 m/s For in-line arrangement: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ∞ D S S u u T T max = 7.168 ×0.075 (0.075 - 0.025) (9.28) (9.29) Part B: Heat Transfer Principals in Electronics Cooling 89 MPE 635: Electronics Cooling = 10.752 m/ s Reynolds number based on maximum velocity ReD = ρumaxD/µ = 1.151×10.752×0.025/(186×10-7 ) = 16633.8 From Table 9.4 p = 0 C = 0.27 m = 0.63 n = 0.36 From Equation 9.25 D Nu = 0.27 (16633.8)0.63 (0.7066)0.36(0.7066/0.6954)0.25 = 109.15 =h D/ k The average heat transfer coefficient is h = 109.15x0.0265/0.025 = 115.699 W/m2.k From Equation 9.26.The Total heat transfer is ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = − = ∞ ∞ ∞ ∞ • 2 ) ( i o S i o P T T T A h T T C m q 40x1007 (50 -10) =115.699A [100-30] Total heat transfer area A = 198.94 m2 Number of rows (NL) A = NTNLπDL 198.94 = 20NL (π x 0.025 x 3) NL = 43 rows 9.5 Heat Transfer with Jet Impingement The heat transfer with jet impingement on a heated (or cooled) surface results in high heat transfer rates, and is used in a wide range of applications such as cooling of electronic equipments. Usually, the jets are circular with round nozzle of diameter d, or rectangular with width w. They may be used individually or in an array. The jets may impinge normally to the heated surface as shown in Figure 9.4 or at an angle. The liquid coolant under pressure inside a chamber is allowed to pass through a jet and directly into the heated surface, there are two modes of operation possible with liquid jet impingement, namely single phase cooling, and two phases cooling. And the jet may be free or submerged. If there is no parallel solid surface close to the heated surface, the jet is said to be free as shown in Figure 9.4, while Jets may be submerged if the cavity is completely filled with the fluid (from the nozzle into a heated surface). In this lecture only single, free jets (round or rectangular) impinging normally to the heated surface are considered. Part B: Heat Transfer Principals in Electronics Cooling 90 MPE 635: Electronics Cooling d Figure 9.4 Jet impinge normally to the heated surface Cooling analysis: Free single phase jet impingement cooling is affected with many variables such as: - Jet diameter (d) - Fluid velocity (v) - Jet to heated surface distance (H ) - Size of heated surface area (L x L) - Coolant properties The average heat transfer coefficient correlation is given by Jigi and Dagn ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 1 008 .
0 Pr Re 84 .
3 33 .
0 5 .
0 d L Nu d L The properties are evaluated at mean film temperature (TS + T∞)/2 This correlation is experimented for FC-77 and water and also valid for 3< H/d <15 - 3< H/d <15 - d = 0.508 to 1.016 mm - v < 15 m/s - small surface dimensions L<12.7 mm (microelectronic devices) Example 9.6: A single phase free jet impingement nozzle is placed in the center of an electronic heated surface 12x12 mm2 and the heated surface is placed at 4 mm from the jet. The working medium is FC-77 passing through 1mm tube diameter at a rate of 0.015 kg/s. To cool the plate, if the supply coolant is at 25 oC and the heat load is 20 W Determine the average heat transfer coefficient and the surface temperature of the heated surface. Solution: To get the properties of the FC-77 we need to assume the surface temperature: let the surface temperature equal 45 oC as a first approximation. Tf = (45 + 25)/2 = 35 oC From FC-77 property tables at 35 oC: ρ = 1746 kg/m3 µ = 1.198x10-3 N.s/m2 k = 0.0623 W/m.oC Pr= 20.3 We must check on H/d ratio: H/d = 4/1 Calculation of Reynolds number (9.30) H Part B: Heat Transfer Principals in Electronics Cooling 91 MPE 635: Electronics Cooling Re = ρvd / µ = 4m' / πdµ = 4×0.015 / π ×1×10-3×1.198×10-3 = 15942 From Equation 9.30 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = 1 008 .
0 Pr Re 84 .
3 33 .
0 5 .
0 d L Nu d L = 3.84 (15942)0.5 (20.3)0.33(0.008×12/1 +1) = 1435.12 Knowing that k L h Nu l = And so the average heat transfer coefficient is: h = (1435.12×0.0623)/ (12×10-3) = 7450.656 W/m2.k But since q = h A ∆T Then the temperature difference is: ∆T = 20/ (7450.656 ×12×12×10-6) = 18.64 oC Therefore the surface temperature is: TS = 43.64 oC For more accuracy we may make another trial at new film temperature and reach more accurate results. 9.5.1 Case Study The use of impinging air jets is proposed as a means of effectively cooling high-power logic chips in a computer. However, before the technique can be implemented, the convection coefficient associated with jet impingement on a chip surface must be known. Design an experiment that could be used to determine convection coefficients associated with air jet impingement on a chip measuring approximately 10 mm by 10 mm on a side. 9.6 Internal Flows (Inside Tubes or Ducts) The heat transfer to (or from) a fluid flowing inside a tube or duct used in a modern instrument and equipment such as laser coolant lines ,compact heat exchanger , and electronics cooling(heat pipe method). Only heat transfer to or from a single-phase fluid is considered. The fluid flow may be laminar or turbulent, the flow is laminar if the Reynolds number ( umDH/ ) is less than 2300 (Re ≤ 2300), based on the tube hydraulic diameter (DH = 4A / P) where A, P is the cross sectional area and wetted perimeter respectively and um is average velocity over the tube cross section. Also the hydraulic diameter should be used in calculating Nusselt number. And If the Reynolds number is greater than 2300 the flow is turbulent (Re >2300). 9.6.1 Fully Developed Velocity Profiles When a fluid enters a tube from a large reservoir, the velocity profile at the entrance is almost uniform because the flow at the entrance is almost inviscid while the effect of the viscosity increase with the length of tube (in x-direction) which have an effect on the shape of the velocity profile as shown in Figure 9.5. Part B: Heat Transfer Principals in Electronics Cooling 92 MPE 635: Electronics Cooling At some location downstream of the pipe inlet, the boundary layer thickness ( reaches its maximum possible value, which is equal to the radius of the tube at this point and the velocity profile does not change. The distance from the entrance to this point is called fully developed region or entrance region, and it is expressed as the fully developing length (LD), this length depends on the flow which may be laminar or turbulent. Through the developing length (LD) the heat transfer coefficient is not constant and after the developing length the heat transfer coefficient is nearly constant as shown in Figure 9.5. um δ LD(Fully developing length) Fully developed flow R h ( W/m2.K ) X Figure 9.5 Flow inside the pipe The fully developing length is given by: LD/D = 0.05Re Pr -For laminar flow And LD/D = 10 -For turbulent flow 9.6.2 Heat Transfer Correlations Through the entrance region and after this region (fully developed flow) both regions have different heat transfer correlations, given in Table 9.6. Part B: Heat Transfer Principals in Electronics Cooling 93 MPE 635: Electronics Cooling Table 9.6 Summery for forced convection heat transfer correlations - internal flow Correlations Conditions Sieder and Tate (1936) 14 .
0 3 / 1 / Pr Re 86 .
1 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = S D D D L Nu µ µ (9.31) - Laminar , entrance region - uniform surface temperature - L/D < 42 .
0 8 Pr Re ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ S D µ µ - 0.48 < Pr <16,700 - 0.0044 < µ /µS < 9.75 NuD =3.66 (9.32) - Laminar , fully developed - Uniform surface temperature - Pr ≥ 0.6 NuD = 4.36 (9.33) - Laminar , fully developed - Uniform Heat Flux - Pr ≥ 0.6 Dittus–Boelter (1930) n d D Nu Pr Re 023 .
0 5 / 4 = (9.34) - Turbulent flow, fully developed - 0.7 ≤ Pr≤ 160 , L/D ≥10 - n = 0.4 for heating (Ts > Tm) and - n = 0.3 for cooling (Ts < Tm). Sleicher and Rouse (1976) 93 .
0 85 .
0 , , Pr Re 0156 .
0 8 .
4 S f D m D Nu + = 93 .
0 85 .
0 , , Pr Re 0167 .
0 3 .
6 S f D m D Nu + = (9.35) (9.36) - For liquid metals , Pr «1 - Uniform surface temperature - For liquid metals , Pr «1 - Uniform heat flux Properties of the fluid for each equation: For Equation 9.31 at the arithmetic mean of the inlet and exit temperatures (Tmi+Tmo)/2 For Equation 9.32, Equation 9.33 Equation 9.34 and at the mean temperature Tm For Equation 9.35 and Equation 9.36 the subscripts m, f, and s indicate that the variables are to be evaluated at the mean temperature Tm, film temperature Tf (arithmetic mean of the mean temperature and surface temperatures (TS +Tm)/2, and surface temperature, respectively. 9.6.3 Variation of Fluid Temperature(Tm) in a Tube By taking a differential control volume as shown in Figure 9.6 considering that the fluid enters the tube at Tmi and exits at Tmo with constant flow rate m' , convection heat transfer occurring at the inner surface (h), heat addition by convection q, neglecting the conduction in the axial direction, and assuming no work done by the fluid. Then applying heat balance on the control volume we can obtain an equation relating the surface temperature at any point. Part B: Heat Transfer Principals in Electronics Cooling 94 MPE 635: Electronics Cooling Tm Tm i Tm o X L dx dTm dq Figure9.6 differential control volume inside tube Case 1: - At uniform surface temperature TS = constant dq = m' Cp dTm (9.37) = h (Pwdx) (TS – Tm) (9.38) Where Pw = out side tube perimeter By equating the both Equations 9.37, 9.38 ) ( , m S p w m T T C m P h dx dT − = dx T d dx dT T T T m m S ∆ − = ∆ = − ) ( , T C m P h dx T d p w ∆ = ∆ − dx C m P h T T d p w , = ∆ ∆ − By integration of Equation 9.40 as shown bellow: ∫ ∫ ∫ ∫ ∆ ∆ ∆ ∆ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ∆ ∆ = ∆ ∆ X i X i T T x p w T T x p w dx h x C m P x T T d dx C m P h T T d 0 , 0 , 1 (9.39) (9.40) Part B: Heat Transfer Principals in Electronics Cooling 95 MPE 635: Electronics Cooling ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ∆ ∆ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∆ ∆ − x C m h P i x p w i x p w e T T x C m h P T T , , ln Part B: Heat Transfer Principals in Electronics Cooling 96 MPE 635: Electronics Cooling At x = L ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = ∆ ∆ L C m h p i o p w e T T , So that the temperature distribution along the tube is as shown in Figure 9.7 X L TS = constant Tm i Tm o ∆T= TS - Tm x ∆Ti ∆To temperature Figure 9.7 Temperature distributions along the tube at uniform surface temperature The total heat transfer is: ) ( , mi mo p t T T C m q − = But also Tmo- Tmi = ∆Ti – ∆To So that the Equation 9.42 can be written as follows ) ( , o i p t T T C m q ∆ − ∆ = Also for total heat transfer by average heat transfer coefficient average m S w t T T LP h q ) ( − = From Equations 9.43, 9.44 produce: w p o i average average w o i p LP h C m T T T T LP h T T C m , , ) ( ) ( ) ( ∆ − ∆ = ∆ ∆ = ∆ − ∆ Where ∆Taverage = (TS -Tm)average Substitute the Equation 9.41 in Equation 9.45 produces: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∆ ∆ ∆ − ∆ = ∆ o i o i average T T T T T ln ∆Taverage called logarithmic mean temperature difference (LMTD) (9.42) (9.43) (9.44) (9.45) (9.41) (9.46) Part B: Heat Transfer Principals in Electronics Cooling 97 MPE 635: Electronics Cooling Case 2: - At uniform heat flux q// = constant From Equation 9.37 dq = m' Cp dTm And dq = q// Pwdx Combining both Equations 9.37, 9.47 produces: p w m C m P q dx dT ' // = By integration: ∫ ∫ = x p w T T m dx C m P q dT mx mi 0 ' // // mx mi ' T - T = x w p q P mC And q// = h (TS – Tm) TS – Tm = q// / h ≈ constant along the tube length From the Equations 9.49, 9.50 the temperature distribution as shown in Figure 9.8. L TS Tm Tm o x Temperature Tm i Entrance Region Fully developed flow Figure 9.8 Temperature distributions along the tube at uniform heat flux For along the tube length Tmo = Tmi + p w C m P q ' // L (9.47) (9.49) (9.48) (9.50) |
190178 | https://www.sciencedirect.com/science/article/abs/pii/S0261561421005008 | Skip to article
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Clinical Nutrition
Volume 41, Issue 1, January 2022, Pages 33-39
Original article Clinical diagnosis, outcomes and treatment of thiamine deficiency in a tertiary hospital
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Summary
Background
Acute thiamine deficiency can occur in patients with or without history of alcohol abuse and can lead to life-threatening complications. Clinical diagnosis is challenging, often resulting in delayed recognition and treatment. Patients may present with heterogenous symptoms, more diverse than the historical neurological description. Cerebral MRI can contribute to the diagnosis in patients with neurological signs but it is not always feasible in emergency settings. Prompt parenteral supplementation is required to obtain the improvement of symptoms and avoid chronic complications.
Aims
To describe the clinical presentation of reported cases of thiamine deficiency, assess prescription and results of cerebral imaging, review treatments that had been prescribed in accordance or not with available guidelines, and study the short-term outcome of these patients.
Methods
This is a monocentric retrospective analysis of all reported cases of thiamine deficiency in a French tertiary hospital between January 1st 2008 and December 31st 2018.
Results
Fifty-six cases were identified during the study period. Forty-five (80%) patients had a history of alcohol abuse. Most patients were diagnosed based on neurological symptoms but non-specific and digestive symptoms were frequent. Thirty-four percent of patients fulfilled clinical criteria for malnutrition. A brain MRI was performed in 54% of patients and was abnormal in 63% of these cases. Eighty-five percent of patients were treated by parenteral thiamine administration and the supplementation was continued orally in 55% of them. The majority of patients initially received 1000 mg daily of IV thiamine but the dose and duration of thiamine supplementation were variable. At the time of discharge, partial or complete improvement of symptoms was noted in 59% of patients.
Conclusion
This study highlights the clinical and radiological heterogeneity of thiamine deficiency. These observations should encourage starting thiamine supplementation early in patients with risk factors or suggestive symptoms even in non-alcoholic patients, and underline the importance of early nutritional support.
Introduction
Thiamine is an important cofactor for several enzymes that are essential for energy metabolism including the Krebs cycle. It also has a critical protective role in neuronal structure and function, both in the central nervous system and in peripheral nerves . In case of insufficient intake, decrease of intestinal absorption or increased renal excretion, clinical manifestations of thiamine deficiency can develop in only a few weeks especially in situations of high carbohydrate intake, chronic alcohol consumption, or when the metabolic demand is high . Thiamine deficiency can rapidly lead to irreversible neurological damage, heart disease and life-threatening complications [3,4]. Prompt diagnosis and supplementation are thus required to limit short- and long-term consequences. However, diagnosis of thiamine deficiency is challenging and several autopsy studies suggest that it often remains undiagnosed during life . The most frequent form of acute or subacute thiamine deficiency is Wernicke's encephalopathy (WE), historically defined by a triad of gait ataxia, nystagmus or ophthalmoplegia and global confusion. However in a 1986 study, only 16% of patients with confirmed WE presented with the classical clinical triad and 19% had none of them . To improve the identification of patients with WE, Caine and al. Modified the classical triad to propose an operational criteria with better sensitivity and specificity : this criteria require only two of the four following signs: 1- dietary deficiency or malnutrition; 2- oculomotor abnormalities; 3- cerebellar dysfunction; 4- either an altered mental state or mild memory impairment. As measurement of blood thiamine does not necessarily reflect the vitamin store and is not performed urgently, it is recommended to rely on clinical criteria to start treatment and then assess clinical evolution .
There is currently no formal evidence on the optimal administration of thiamine in patients with thiamine deficiency. Because of the frequent decreased intestinal absorption and limited biodisponibility of orally administered thiamine, it is widely recommended to use the IV route, at least during the initial phase . However, there is no evidence available regarding the optimal dosage and duration of treatment . A Cochrane review concluded that a dose of 200 mg per day was more effective than lower doses on the neurologic and cognitive symptoms [11,12]. Many authors recommend higher doses of 501000 mg per injection per day, and based on the 96 min half-life of free thiamine in the blood multiple daily injections are now recommended, but their superiority has not been demonstrated to date . The benefit of a switch to an oral thiamine supplementation and the total duration of treatment has never been proven but it is considered good practice to continue the latter for at least several weeks, as the tolerance is generally excellent [18,19]. Given the difficulty of the diagnosis and the uncertainties in treatment modalities, the aim of this study was to review all cases of thiamine deficiency reported in a tertiary hospital in France over a 10 years period. The objectives were to describe the clinical presentation of cases as well as imaging results, review treatments that had been prescribed in accordance or not with available guidelines, and study the short-term outcome of these patients.
This is a retrospective analysis of all cases of thiamine deficiency reported in health medical records of the Georges Pompidou European Hospital in Paris (France) between January 1st 2008 and December 31st 2018. Cases were identified by searching electronic medical charts over this period for the corresponding ICD-10 diagnosis encoded by physicians. All medical charts were then manually reviewed by an investigator to confirm diagnosis and collect data from medical history, physical examination, lab and imaging results, treatment and clinical evolution during the hospital stay. Nutritional status was determined by body mass index (kg/m2) and weight loss percent criteria. The diagnosis of malnutrition was established retrospectively based on the Global Leadership Initiative on Malnutrition (GLIM) that requires at least one etiologic criterion among reduced food intake or assimilation, inflammation or disease burden, and at least one phenotypic criterion among non-volitional weight loss, low BMI and reduced muscle mass . Phenotypic criteria were assessed using BMI and percent weight loss. Moderate malnutrition was defined by weight loss of 510 percents within the past 6 months or 10 to 20 percents over more than 6 months and/or BMI <20 kg/m2 (if age <70 years) or BMI <22 kg/m2 (if age >70). Severe malnutrition was defined by weight loss greater than 10 percents within the past 6 months or greater than 20 percents over more than 6 months and/or BMI <18.5 kg/m2 (if age <70 years) or BMI <20 kg/m2 (if age >70) . The study protocol was approved by the local ethic committee (CERAPHP.5).
The collected characteristics are reported using median, 25th (Q1) and 75th (Q3) percentiles for quantitative variables. For categorical variables counts and percentages are reported. Differences between groups were tested with a Fischer's exact test for categorial variables and with the MannWhitney test for quantitative variables. The association between studied variables and improvement of symptoms was tested by univariate logistic regression; odds ratios and their confidence intervals were calculated to quantify their prognostic impact. Quantitative variables were dichotomized to plot the corresponding odds ratios along with binary variables' odds ratios. The cut-off was the median value for age, number of infusions and duration of supplementation. For BMI, oral and IV doses the cut-offs were respectively ¥20 kg/m2 and ¥1000 mg. The absence of significant association with symptoms improvement was also verified when these variables were analyzed as continuous. All statistical tests were performed in R statistical software at a 95% confidence level .
Section snippets
Results
The electronic search identified 86 patients with a diagnosis of thiamine deficiency encoded during the study period. In 30 patients, individual review of the medical charts found no justification for this encoding and did not confirm suspicion of thiamine deficiency. Fifty-six cases of thiamine deficiency were identified, from 8 different departments including the intensive care unit. The total number of reported cases varied from 2 to 8 per year and was stable over the study period (
Discussion
We present a retrospective analysis of all reported cases of thiamine deficiency in a French tertiary hospital. All reported patients had neurological or psychiatric symptoms. As only diagnosed and electronically encoded cases were retrieved, this study does not allow to estimate the prevalence of thiamine deficiency in this population. Given that the prevalence of WE was estimated around 1% on autopsy studies [5,22], the annual number of diagnosed cases was low in comparison with the 253,659
Conclusion
This retrospective review of thiamine deficiency cases highlights the clinical and radiological heterogeneity of this under-diagnosed but life-threatening condition. The difficulty of diagnosis should lead to starting thiamine supplementation in patients with risk factors or suggestive symptoms, especially in patients with an history of alcohol abuse, as an early supplementation may allow a regression of symptoms. Evidence-based guidelines widely spread across medical specialties are strongly
Funding statement
This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.
Author contribution
François Mifsud: Investigation, Formal analysis, Writing - Original Draft. Diane Messager: Investigation, Data Curation. Anne-Sophie Jannot: Investigation. Benoît Védie: Resources, Writing - Review & Editing. Nadia Aissaoui Balanant: Resources, Writing - Review & Editing. Tigran Poghosyan: Resources, Writing - Review & Editing. Edouard Flamarion: Resources, Writing - Review & Editing. Claire Carette: Resources, Writing - Review & Editing. Léa Lucas-Martini: Supervision, Writing - Review &
Conflict of interest
The authors declare that they have no competing interests.
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Treatment variability and its relationships to outcomes among patients with Wernicke's encephalopathy: A multicenter retrospective study
2023, Drug and Alcohol Dependence
Citation Excerpt :
These findings reinforce the key roles of early diagnosis and treatment, which clearly appears to improve the prognosis of WE. It has been previously highlighted the need for maximal clinical suspicion in high-risk patients and the potential benefit of early treatment, which far outweighs possible risks (Mateos-DÃaz et al., 2022; Mifsud et al., 2022). We also found that more patients with AUDs and those presenting with oculomotor abnormalities received early WE diagnoses, whereas an atypical presentation with altered mental status and reduced consciousness were linked to late diagnosis and greater risks of incomplete recovery and mortality.
Despite guidelines and recommendations, Wernicke's encephalopathy (WE) treatment lacks evidence, leading to clinical practice variability.
Given the overall lack of information on thiamine use for WE treatment, we analyzed data from a large, well-characterized multicenter sample of patients with WE, examining thiamine dosages; factors associated with the use of different doses, frequencies, and routes; and the influence of differences in thiamine treatment on the outcome.
This retrospective study was conducted with data from 443 patients from 21 centers obtained from a nationwide registry of the Spanish Society of Internal Medicine (from 2000 to 2012). Discharge codes and Caine criteria were applied for WE diagnosis, and treatment-related (thiamine dosage, frequency, and route of administration) demographic, clinical, and outcome variables were analyzed.
We found marked variability in WE treatment and a low rate of high-dose intravenous thiamine administration. Seventy-eight patients out of 373 (20.9%) received > 300 mg/day of thiamine as initial dose. Patients fulfilling the Caine criteria or presenting with the classic WE triad more frequently received parenteral treatment. Delayed diagnosis (after 24 h hospitalization), the fulfillment of more than two Caine criteria at diagnosis, mental status alterations, and folic acid deficiency were associated significantly with the lack of complete recovery. Malnutrition, reduced consciousness, folic acid deficiency, and the lack of timely thiamine treatment were risk factors for mortality.
Our results clearly show extreme variability in thiamine dosages and routes used in the management of WE. Measures should be implemented to ensure adherence to current guidelines and to correct potential nutritional deficits in patients with alcohol use disorders or other risk factors for WE.
### Clinical diagnosis, outcomes and treatment of thiamine deficiency in a tertiary hospital
2022, Clinical Nutrition
### Case Report: Favorable outcome in a patient with simultaneous dengue meningoencephalitis and Wernickes thiamine deficiency
2025, Frontiers in Medicine
### Recognizing Thiamine Deficiency: Keeping Patients Safe and Clinicians Out of Court
2024, Practical Gastroenterology
### The importance of early recognition of extraintestinal manifestations of digestive tract dysfunction following gastrointestinal surgery
2024, Journal of Surgical Case Reports
### Association Between Vitamin Deficiencies and Ophthalmological Conditions
2023, Clinical Ophthalmology
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190179 | https://math.berkeley.edu/~mhaiman/math172-spring10/ordinary.pdf | Math 172—Spring 2010—Haiman Notes on ordinary generating functions 1. How do we count with generating functions?
Many enumeration problems which are not so easy to handle by elementary means can be solved using generating functions. In these notes I’ll describe the general set-up in which ordinary gener-ating functions apply. Exponential generating functions are useful in a somewhat different set-up, which we will discuss later.
Our basic problem will be to count objects of a given size, or weight, n. The specific meaning of the weight will vary with the context, but here are some typical possibilities: n might be the length of a word, or the size of a subset or multisubset of a given set, or we might be counting compositions of n, or partitions of n. We can also have situations in which there is more than one relevant weight. For example in counting partitions λ = (λ1, . . . , λk), we may want to keep track of both the size of the partition n = w(λ) = λ1 + λ2 + · · · + λk, and the number of parts l(λ) = k, to arrive at the number p(n, k) of partitions of n with k parts.
What often happens is that if we don’t fix the weight in advance, but rather try to count objects of all weights simultaneously, it allows us extra freedom to make independent choices and thereby simplify the problem. But we have to do this in a way that keeps track of the weights, so we can solve our original problem. We will also have to deal with the fact that the set of all objects of all possible weights is typically infinite. Both of these difficulties can be handled by generalizing our usual idea of the number of elements of a finite set to a weight enumerator for elements of a possibly infinite set with weights.
Definition Let S be a set equipped with a function w: S →N, called the weight function. Assume that for each n, the subset {s ∈S : w(s) = n} is finite, and let sn = |{s ∈S : w(s) = n}| be its number of elements (the whole set S is allowed to be infinite). The weight enumerator for S is the generating function S(x) = ∞ X n=0 snxn = X s∈S xw(s).
Note that if S is finite, then S(1) is just the number of elements of S. In general, S(x) should be thought of as a kind of weighted count of the elements of S, where an element of weight n counts as xn instead of 1.
The sum principle for weight enumerators is just like that for elementary counting, and follows immediately from the definition: Sum Principle. If S is the union S = A ∪B of two disjoint subsets A ∩B = ∅, then S(x) = A(x) + B(x), where S(x), A(x), B(x) denote the weight enumerators for S, A and B respectively (taking the weight functions on A and B to be the restrictions of the given one on S).
There is also an analog for weight enumerators of the product principle from elementary counting.
For this, however, we have to be careful about how weights on the elements of the product are related to those on the components: the weights are required to be additive.
Product Principle. If S is a Cartesian product S = A1 × · · · × Ak of weighted sets, and if the weights satisfy w(s) = w(a1) + · · · + w(ak) for s = (a1, . . . , ak), then S(x) = A1(x)A2(x) · · · Ak(x) where S(x), Ai(x) denote the weight enumerators for S and the sets Ai.
1 To see why the product principle works, observe that for two sets S = A × B, we have S(x) = X (a,b)∈S xw((a,b)) = X (a,b)∈S xw(a)+w(b) = X (a,b)∈S xw(a)xw(b) = X a∈A xw(a) X b∈B xw(b) = A(x)B(x).
The general case of a product of k sets follows from the two-set case by induction on k.
If our set S is equipped with more than one weight function, we will have a corresponding weight enumerator which is a generating function in more than one variable, for example, if we have two weight functions w and z, we would have a weight enumerator S(x, y) = X s∈S xw(s)yz(s) = ∞ X n=0 ∞ X k=0 sn,kxnyk, where sn,k is the number of elements s ∈S such that w(s) = n and z(s) = k.
2. Words, compositions.
In many instances, typified by the examples in this section, we can assemble an object from smaller pieces by first deciding on the number of pieces to use, say j, and then choosing the j pieces independently, in a definite order. This leads to situations in which we use the product principle for each j, and then sum over all j, leading to a geometric series involving a previously known generating function.
As a first example, let us determine the number gn of words w of length n in the symbols {0, 1} which do not contain consecutive 1’s. Before considering the generating function, we might observe that gn satisfies the same recurrence as the Fibonacci numbers. To see this, note that the word w must either start with a 0 or a 1. If it starts with a 0, the rest of the word can be any word of length n −1 without consecutive 1’s, giving gn−1 choices. Provided that n ≥2, if our word starts with 1, then it must in fact start with 10. This can be followed by anything, giving gn−2 choices.
Hence gn = gn−1 + gn−2, (n ≥2).
Since we also clearly have g0 = 1 and g1 = 2, we can deduce that gn is equal to the Fibonacci number Fn+1. We could in turn obtain the generating function by using the recurrence, as explained in your textbook. Instead we’ll take a different approach.
As a little thought will convince you, any word without consecutive 1’s can be built up in a unique way by concatenating a sequence of subwords of the forms 0 or 10, followed by an optional 1 at the end. Assigning each word a weight equal to its length, their weight enumerator is G(x) = P∞ n=0 gnxn. Let us try to evaluate this directly.
Suppose our word is built from j subwords 0 or 10. We can choose each of these independently and multiply together the weight enumerators for each of the j choices. The weights are additive, since the total length of the word we build up will be the sum of the lengths of the subwords we build it from.
Now for each choice of a subword 0 or 10, we have a corresponding weight enumerator (x+x2), so we get (x + x2)j as the generating function enumerating all concatentations of j of these subwords.
Summing this over all j, we get ∞ X j=0 (x + x2)j = 1 1 −(x + x2) as the generating function for all words (including the empty word) built by concatenating any sequence of any number of subwords 0 or 10. Finally, we can choose to add a 1 at the end or not, either contributing nothing or one extra to the weight, so we should multiply the above generating function by 1 + x, obtaining G(x) = 1 + x 1 −x −x2 .
Note, by the way, that xG(x) + 1 = 1/(1 −x −x2). The coefficient of xn in xG(x) + 1 is gn−1 = Fn for n ≥1, and is 1 = F0 for n = 0, so the Fibonacci number generating function is given by F(x) = ∞ X n=0 Fnxn = 1 1 −x −x2 , in agreement with the formula you would obtain by using the recurrence (see Bona, Chapter 8, Ex.
4).
What we have done in this example is an instance of a more general method, which we may formalize as a separate enumeration principle for generating functions.
Sequence principle.
Let A be a set with weighted elements and generating function A(x).
Assume there are no elements of weight zero, so A(0) = 0.
Then the generating function for ordered sequences α1, α2, . . . , αj of of elements of A (of any length, including the empty sequence), with weight defined to be the sum of the weights of the elements αi, is given by 1 1 −A(x).
Some caution is due, when using the sequence principle, to be sure the set A has no elements of weight zero. This is essential for two reasons. First of all, the expression 1/(1 −A(x)) makes no sense as a formal series if A has non-zero constant term. Secondly, it makes no sense to enumerate all sequences if there is an element α of weight zero in A, since we could pad any sequence with an unlimited number of copies of α without changing its weight, giving infinitely many sequences of the same weight. (These two difficulties are related to each other.) Here are some more examples.
Example. Let us find the generating function for all words from the alphabet [k] = {1, . . . , k}. In this case A is just [k], and each of its k elements has weight 1, so A(x) = kx. Hence the generating function for all words is 1 1 −kx = X n knxn, which shows that there are kn words with n letters from [k]. Of course we knew this already. This example was just a warm-up.
Example. A composition of n is an expression n = λ1 + λ2 + · · · + λk of n as a sum of positive integers. In a composition of n, as a opposed to a partition of n, the order of the parts matters. Thus 2+3+2 and 3+2+2, for instance, count as two different compositions of 7. Let m(n, k) denote the number of compositions of n with k parts. We will find the generating function M(n, k) = P n,k m(n, k)xnyk, and use it to solve for m(n, k). In this case, we will use the sequence principle for a two-variable generating function, which works just like the one-variable case. Note that we are not fixing n or k in advance, so a composition is just an arbitrary sequence of positive integers. The set A here is therefore the set of all positive integers. We have assigned each composition two weights. One is its length, which is the exponent of y, while the other is its size, which is the exponent of x in our generating function. Each integer i in the composition contributes one to the length and i to the size, so is counted by the monomial xiy in the generating function for A. Summing over all i ≥1 this gives A(x, y) = yx/(1 −x), and hence M(x, y) = 1 1 −yx/(1 −x) = 1 −x 1 −x −xy.
To solve for m(n, k) it is convenient to rewrite this as M(x, y) = 1 + xy 1 1 −x(1 + y) = 1 + xy ∞ X j=0 xj(1 + y)j = 1 + xy X j xj X i j i yi.
Extracting the coefficient of xnyk, we see that this monomial occurs only in the term with j = n−1, i = k −1, so m(n, k) = n −1 k −1 .
Stricly speaking, this derivation is not correct for n = 0, although the answer is correct, since −1 k = 0 for k > 0, and −1 0 = 1.
Compositons of n with k parts are really the same thing as n-element multisets from [k] that use every element at least once. From our table of distribution problems, their number is the same as the number of (n −k)-element multisets from [k], or k+(n−k)−1 n−k = n−1 n−k = n−1 k−1 , in agreement with the answer above.
Exercise: modify the above example to allow compositions of n in which some of the parts can be zero. Note that as long as we use a two-variable generating function, counting the compositions by length as well as size, this does not put an element of weight zero in A. The integer zero has weight 1 for the length, and is counted by the monomial y, so the modified A(x, y) will still have no constant term.
3. Catalan numbers We define the Catalan number Cn to be the number of words formed from n left and n right parentheses, with all parentheses balanced. We may tabulate the first few Catalan numbers by listing all the possibilities.
n Cn 0 1 ∅ 1 1 () 2 2 ()(), (()) 3 5 ()()(), (())(), ()(()), (()()), ((())) Let C(x) = ∞ X n=0 Cnxn be the generating function for Catalan numbers. We will use the sequence principle to evaluate it. Observe that each balanced parenthesis word, or Catalan word, can be uniquely expressed as a sequence of outermost groups ( w1 )( w2 ) · · · ( wk ), where the words wi inside each outer pair of parentheses are themselves Catalan words. Therefore, let us apply the sequence principle taking A to be the set of words of the form ( w ) with w a Catalan word. Note that the weight of ( w ) is one more than that of w, since the weight is the number of left parentheses, or equivalently, the number of right parentheses. In particular, there are no elements of weight zero in A, even though w can be the empty word. We get each element of A by choosing a Catalan word, and surrounding it with an extra pair of parentheses. We can think of this extra pair as a forced “choice,” with generating function x because it adds one to the weight. Thus by a trivial application of the product principle, we have the generating function A(x) = xC(x).
However, by the sequence principle, we also have C(x) = 1 1 −A(x).
Substiuting xC(x) for A(x) and clearing the denominator yields the quadratic equation xC(x)2 −C(x) + 1 = 0 for the Catalan number generating function. Hence C(x) = 1 −√1 −4x 2x .
An interesting point is how we know whether to take +√1 −4x or −√1 −4x here. To see that the −sign must be right, note that 2xC(x) must equal the numerator in the above fraction, so this numerator must have zero constant term as a formal series. Since 1 + √1 −4x does not vanish at x = 0, it is the wrong numerator. Now we can use the extended binomial theorem to solve for the Catalan numbers Cn themselves. We have 2xC(x) = 1 −(1 −4x)1/2 = 1 − ∞ X n=0 1/2 n (−4)nxn.
To extract Cn, compare coefficients of xn+1 on both sides, obtaining the rather messy formula Cn = −(−4)n+1 2 1/2 n + 1 = (−1)n22n−1 1/2 n + 1 .
Writing out the binomial coefficient 1/2 n+1 = (1/2)(−1/2)(−3/2) · · · (−(2n −1)/2)/(n + 1)!, multi-plying both numerator and denominator by n!, and simplifying, we eventually get Cn = 1 n + 1 2n n .
There is another interpretation of Catalan numbers in terms of lattice paths, as follows.
The condition for a parenthesis word to be balanced is that every initial segment of it have at least as many left parentheses as right parentheses. If we translate the word into a lattice path that takes one step north for each left parenthesis and one step east for each right parenthesis, then we get all paths from (0, 0) to (n, n) that never go below the diagonal line x = y. Hence the Catalan number Cn is equal to the number of these special lattice paths (known as Dyck paths). There are many other interesting combinatorial objects whose number turns out to be Cn.
4. Ordered rooted trees A tree is a connected graph without cycles. If we fix the set of vertices of our tree to be, say, [n] = {1, . . . , n}, then we call the tree a labelled tree, the elements of the vertex set being the labels.
It is a famous theorem of Cayley that there are nn−2 different labelled trees on n vertices. A rooted tree is a tree in which one of the vertices is distinguished and called the root. It is conventional to draw a rooted tree with the root at the bottom and the path from the root to each vertex going up, like this.
Since there are n choices for the root, there are nn−1 different rooted labelled trees on n vertices.
The vertex directly below a vertex v in a rooted tree is the parent of v (the root has no parent).
The vertices directly above v, if any, are its children.
Besides labelled trees, we can also count various sorts of unlabelled trees. Technically, an unla-belled tree is an equivalence classes of labelled trees, two trees being equivalent if they differ only by a permutation of the labels. In this section we will only consider ordered, rooted, unlabelled trees. By ordered, we mean that in addition to the tree itself, there is given a linear ordering of the children of each vertex. In practice, the ordering is indicated implicitly by drawing the tree so that the children of each vertex are ordered left-to-right. This means, for instance, that the drawing above represents a different ordered tree than the tree .
However, these two drawings would represent the same unordered, unlabelled tree.
We can use generating functions to count ordered rooted trees. Let tn denote the number of these trees with n vertices, including the root, and let T(x) = P n tnxn be their generating function.
To build any tree, we can first place the root, and then choose the (ordered) sequence of branches based on each child of the root. Each branch is itself an ordered rooted tree, and there can be any number of them. The generating function for the collection of branches is then 1/(1 −T(x)) by the sequence principle. We must multiply this by a factor of x to account for the additional vertex at the root, giving T(x) = x 1 −T(x).
As with the Catalan number generating function, this leads to a quadratic equation for T(x): T(x)2 −T(x) + x = 0, whose solution is T(x) = 1 −√1 −4x 2 .
Note that T(x) = xC(x).
Hence, comparing coefficients of xn, we see that tn = Cn−1 for n > 0, while t0 = 0, since by definition a rooted tree is not empty.
Incidentally, for ordered trees, it is easy to switch between the labelled and unlabelled versions of the problem. The ordering of the tree effectively “names” every vertex, so that on a labelled ordered rooted tree, every permutation of the labels gives a different tree. Hence the number of ordered, rooted labelled trees on n vertices is simply n!tn = n!Cn−1. |
190180 | https://www-users.cse.umn.edu/~adams/FM5011/proutynotes.pdf | Notes on Measure Theory Definitions and Facts from Topic 1500 • For any set M, 2M := {subsets of M} is called the power set of M.
The power set is the ”set of all sets”.
• Let A ⊆2M.
A function µ : A →[0, ∞] is finitely additive if, ∀integer n ≥1, ∀pw −dj A1, ..., An ∈A, F Aj ∈A ⇒µ(F Aj) = P µ(Aj).
Finite additivity states that if we have pairwise-disjoint sets in some space, the measure of the union of those disjoint sets is equal to the sum of the measures of the individual sets.
• Let A ⊆2M. A function µ : A →[0, ∞] is σ-additive if µ is finitely additive & ∀pw −dj A1, ..., An ∈A, F Aj ∈A ⇒µ(F Aj) = P µ(Aj).
σ-additive differs from finitely additive in that you can add infinitely many things.
Specifically, you can add countably many things.
σ-additive improves finitely-additive by making it ”infinite”.
• Let A ⊆2R. A function µ : A →[0, ∞] is translation invariant if ∀A ∈A, ∀c ∈R, we have: A + c ∈A and µ(A + c) = µ(A).
Translation invariance states that if we measure some interval, we should be able to move it and the measure should not change.
• I := intervals ⊆2R. λ0 : I →[0, ∞] defined by λ0(I) = [sup I] − [inf I] is called length.
For intervals based in the reals, the size of the interval is defined as the supremum of the interval minus the infimum.
• Let A ⊆B ⊆2M. Let µ : A →[0, ∞], ν : B →[0, ∞]. We say that ν extends µ if: ∀A ∈A, µ(A) = ν(A).
If you have two sigma algebras, one larger than the other, and a mea-sure coincides in the sets from the smaller one, then it is said that the measure on the bigger sigma algebra extends the smaller measure.
1 • Let A ⊆B ⊆2M. A function µ : A →[0, ∞] is σ-finite (on M ) if ∃A1, A2, ... ∈A s.t. M = S Aj and ∀j, µ(Aj) < ∞.
A function is σ-finite if all of the constituent sets have measure less than infinity.
Fact: Length is σ-additive, σ-finite, and translation invariant.
Fact: I is a near algebra on R, length σ-additive and σ-finite.
• A is an algebra (on M ) if M ∈A and ∀A, B ∈A, we have: A\B ∈A.
Algebras exist to allow the formation of unions, intersections, and com-plements. If we have some measure space, we want to be able to par-tition it into pieces and measure those pieces; algebras allow that to happen.
• A is a σ-algebra (on M ) if A is an algebra in M and S A ⊆A.
A σ-algebra extends the definition of an algebra by allowing countably many unions and intersections. Like other σ definitions, the σ-algebra allows the notion of ”infinite” instead of ”finite” operations.
Fact: A is an algebra (on M) iffA ̸= null and A is closed under: finite union, finite intersection, and complement in M.
Fact: A is a σ-algebra (on M) iffA ̸= null and A is closed under: countable union, countable intersection, and complement in M.
• The σ-algebra (on M) generated by S ⊆2M is the intersection of all of the σ-algebras on M that contain S. It is denoted < S >σ.
Generation is a process that creates a σ-algebra with the smallest num-ber of elements that still captures all relevant information about the σ-algebra.
• < µ0 >σ:=µ is called the measure generated by µ0.
The generated measure extends our definition of length to allow for measuring σ-algebras.
2 • A subset of R is Borel if it’s an element of < I >σ. The unique exten-sion of length to {Borel sets in R} is called Lebesgue measure on R.
A subset is Borel if it is an element of the sigma algebra generated by intersecting all intervals. Lebesgue measure can be thought of as the analogue to length on Borel sets. Virtually everything is Borel.
Definitions and Facts from Topic 1600 • P, Q ⊆M are essentially equal (written P .
= Q) if ∃null sets Z, Z′ ⊆ M s.t. (P\Z) S Z′ = Q.
Essentially equal sets can be visualized by imagining an interval [0,2].
Imagine sets P:=[0,0.5)(0.5,2] and Q:=[0,1)(1,2]. Although set P does not cover the point 0.5 and set Q does not cover the point 1, since both points have measure zero we say that the two are essentially equal.
• A subset C ⊆M is conull in M (or µ-conull) if M\C is null.
If the complement of a subset is null, then the subset is said to be conull.
• A subset of P ⊆M is measurable (or µ-measurable) if ∃A ∈A s.t.
P .
= A.
If we wish to measure a set P, but do not have a measure to do so, but another essentially equal set A is measurable, then we can measure P also.
• The completion of (w.r.t. µ) is ¯ A := {µ −measurable sets}.
• The completion of µ is the function ¯ µ : ¯ A →[0, ∞]; defined by: ¯ µ(P) = µ(A), ∀A ∈A s.t. A .
= P.
Measurable sets are one step beyond Borel sets because one may need to add things of measure zero. We use the completion of the measure to measure such things, since regular measure cannot be applied. We need the definition of ”essentially equal” to make the leap.
Fact: A, B ∈A, A .
= B ⇒µ(A) = µ(B) Fact: ¯ µ : ¯ A →[0, ∞] is σ-additive.
3 Fact: {Borel sets in R} is countably generated.
< {(a, b)|a, b ∈Q, a < b} >σ • A subset of R is measurable if it’s an element of the completion of {Borel sets in R} w.r.t. Lebesgue measure.
It’s hard to make non-measurable sets, or even non-Borel sets. This follows from the earlier definition of a Borel set, which was essentially defined as the sigma algebra generated by intersecting all intervals.
• A σ-algebra A on M is countably generated if ∃a countable set C ⊆A s.t. A =< C >σ.
If A is a countably generated σ-algebra on M, then the elements of A are called Borel sets. If, furthermore, µ : A →[0, ∞] is σ-finite, then the elements of the completion of A w.r.t. µ are called measurable sets.
• A Borel space is a set with a countably generated σ-algebra on it.
In an abstract space, if a countably-generated σ-algebra can be defined, we call it a Borel space.
• A measure on a Borel space (M, B) is a σ-additive function µ : B → [0, ∞].
• A measure space is a Borel space with a σ-finite measure on it.
• A measure µ on a Borel space (M, B) is a probability measure if µ(M) = 1.
• A measure µ on M is finite if µ(M) < ∞.
• monotonicity: ∀A, B ∈A, A ⊆B ⇒µ(A) ≤µ(B) Fact: µ is finite iff, ∀A ∈A, µ(A) < ∞ Fact: measures are monotone.
4 • For any countable set M, counting measure on M is the measure µ : 2M →[0, ∞] defined by µ(S) = #S.
A measure on the (countably generated) Borel space (M, 2M).
• ∀set M, 2M is the discrete σ-algebra on M and {null, M} is the in-discrete σ-algebra on M.
If M is a countable set, then the power set of M is the discrete σ-algebra and the most coarse σ-algebra is the indiscrete.
Definitions and Facts from Topic 1700 • The σ-algebra inherited (or restricted) from A to W is A|W := {A T W|A ∈A}.
TBD • A Borel space is a set with a countably generated σ-algebra on it.
Sometimes called a measure space. A Borel space (M,A) is discrete is A = 2M.
I always had the impression that a Borel space and a measure space were different things...
• ∀z ∈C, ∀δ > 0, D(z, δ) := {w ∈C||w −z| < δ} • The standard σ-algebra on C is BC :=< {D(z, δ)|z ∈C, δ > 0} >σ.
TBD • ∀A ⊆C, the standard σ-algebra on A is BA := BC|A.
TBD • ∀finite F, standard σ-algebra on F is BF := 2F.
TBD • ∀A ⊆C, ∀finite F, the standard σ-algebra on A S F is BA S F :=< BA S BF >σ.
TBD 5 • Let A and B be σ-algebras on M and N, respectively.
Let C := {A × B|A ∈A, B ∈B}.
A × B :=< C >σ is called the product of A and B.
TBD • µ × ν :=< ω >σ is called product of µ and ν.
TBD • Let B := {Borel sets in R}. A subset of R2 is Borel iffit’s an element of B ×B. A subset of R3 is Borel iffit’s an element of B ×B ×B. etc...
TBD • Let λ be Lebesgue measure on R. λ × λ is Lebesgue measure on R2.
λ × λ × λ is Lebesgue measure on R3. etc...
TBD • A subset of R2 is measurable if it’s an element of ¯ B ¯ × ¯ B.
this principle applies for R n • Let M be a set. Let (N, B) be a Borel space. Let f : M →N be a function.
f ∗(B) := {f −1(B)|B ∈B} is the pull back σ-algebra on M (from B via f ).
TBD • Let (M, A) be a Borel space.
Let (N, B) be a Borel space.
Let f : M →N be a function.
f is (A, B)-Borel (or just Borel) if f ∗B ⊆A.
The expression f ∗B ⊆A can be alternatively expressed as ∀B ∈B, f −1(B) ∈ A • Let (M, A, µ) be a measure space. Let (N, B) be a Borel space. Let f : M →N be a function.
f is (µ, B)-measurable if f ∗B ⊆¯ A (or µ-measurable or just mea-surable).
6 The expression f ∗B ⊆A can be alternatively expressed as ∀B ∈B, f −1(B) ∈ A.
• The measure f B ∗: B →[0, ∞] defined by (f B ∗(µ))(B) = ¯ µ(f −1(B)) is the push forward measure on N (on B from µ via f ). (We usually just write f∗(µ).) TBD Fact: Functoriality of push-forward: (g ◦f)∗(µ) = (g∗(f∗(µ))) = (g∗◦f∗)(µ).
• Let M be a set. Let x ∈M.
The delta mass (or point mass) at x (in M ) is the measure δx : 2M →[0, ∞] (or δM x ) defined by δx(A) = {1, if x ∈A; 0, if x / ∈A}.
TBD • Let S ∈¯ B, i.e., let S be measurable. We say µ is concentrated offS if ¯ µ(S) = 0. We say µ is concentrated on S if ¯ µ(M\S) = 0.
TBD • C ⊆Rd closed, ν a measure on C. Let S ⊆C be closed. We say ν is supported on S if ν(C\S) = 0.
TBD • C ⊆Rd closed, ν a measure on C. The support of ν the intersection of all of the closed sets that support ν.
TBD Definitions and Facts from Topic 2300 • We say f is simple if both f : M →R is measurable and f(M is finite, in which case, R M f dµ := P y∈f(M) y[µ(f −1(y))] TBD • For any set S, for any R ⊆S, the function 1S R : S →{0, 1} defined by 1S R(s) = {1 if s ∈R; 0 if s ∈S\R is the indicator function of R (in 7 S).
The indicator function is a simple switching technique, whereby the function equals 1 if s is contained in R, and 0 if s is not contained in R.
• Definitions and Facts from Topic 2330 • We say f is integrable or L1 if both R M f+dµ < ∞and R M f−dµ < ∞.
We can say a function is L1 if the integral of both the positive and negative parts of the function are finite.
• We say f is integrable on A or L1 on A if χf is integrable.
In this case, χ represents an indicator function that returns 1 if A is contained in M and 0 if it is not.
• Let (M, B) be a Borel space. Let µ and ν be two measures on (M, B).
We say ν is absolutely continuous w.r.t. µ if ∀Z ∈B, µ(Z) = 0 ⇒ ν(Z) = 0.
One measure is absolutely continuous with respect to another measure if, for some Z contained in the Borel set, Z’s measure is the same when measured using both measures. An earlier definition of absolutely con-tinuous used the notation ν << µ to express ”ν has at least as many null sets as µ”.
• We say µ and ν are equivalent if both ν << µ and µ << ν, i.e. if {Z ∈B|µ(Z) = 0} = {Z ∈B|ν(Z) = 0}.
If both measures of Z yield the same results, we say the measures are equivalent.
Definitions and Facts from Topic 2360 • Measures ρ and σ on (M, B) are mutually singular if ∃Z ∈B s.t. ρ(Z) = 0 and σ(M\Z) = 0.
TBD 8 Fact: hµ << µ, ∀measurable h : M →[0, ∞) • Let (M, B, µ) be a probability space.
A change of measure (for (M, B, µ)) is a probability measure µ on (M, B) s.t. µ ≈ν.
Suppose a measure exists on some Borel set. A change of measure may take place if there is a new measure on the same set which is equivalent (by the above definition of equivalent).
Fact: Sophisticated change of variables formula: Let (M, A, µ) be a measure space.
Let (N, B) be a Borel space.
Let f : M →N be measurable.
Let g : N →¯ R be Borel.
R M(g ◦f)dµ = R N gd(f∗µ) Fact: How to integrate against h: R M(fh)dµ = R M fd(hµ) R M[f(x)][h(x)]dµ(x) = R M f(x)d(hµ)(x) Fact: h := dν dµ Fact: φ∗(φ′ · λ) = λ Definitions and Facts from Topic 2400 Fact: Monotone Convergence Theorem Let (M, A, µ) be a measure space. Let f1, f2, f3, ... : M →[0, ∞] be a non-decreasing sequence of measurable functions. Assume limn→∞fn ≥C on M.
Then limn→∞ R M fndµ = R M limn→∞fndµ.
Fact: Fatou’s Lemma Let (M, A, µ) be a measure space. Let g1, g2, g3, ... : M →[0, ∞] be measur-able. Then R M lim infn→∞gndµ ≤lim infn→∞ R M gndµ.
• A sequence f1, f2, f3, ... : M →[−∞, ∞] is L1-minorized if ∃L1 func-tion g : M →[0, ∞] s.t. ∀integers n ≥1, −g ≤fn.
9 TBD • A sequence f1, f2, f3, ... : M →[−∞, ∞] is L1-majorized if ∃L1 func-tion g : M →[0, ∞] s.t. ∀integers n ≥1, fn ≤g.
TBD • A sequence f1, f2, f3, ... : M →[−∞, ∞] is L1-enveloped if ∃L1 func-tion g : M →[0, ∞] s.t. ∀integers n ≥1, −g ≤fn ≤g.
TBD Fact: Bounded Convergence Theorem Let (M, A, µ) be a measure space.
Assume µ(M) < ∞. Let K > 0. Let f1, f2, f3, ... : M →[−K, K] be mea-surable and ptwise convergent. Then R M limn→∞fndµ = limn→∞ R M fndµ.
Fact: Dominated Convergence Theorem Let (M, A, µ) be a measure space.
Let f1, f2, f3, ... : M →[−∞, ∞] be measurable, L1-enveloped and ptwise convergent. Then R M limn→∞fndµ = limn→∞ R M fndµ.
Definitions and Facts from Topic 2450 Fact: Let I := [0, 1]. ∀measurable f : M →I, ∃simple s : M →I, ∀ε > 0, s.t. f −ε ≤s ≤f on M.
Fact: ∀measurable f : M →[0, ∞], ∃simple s1, s2, .... : M →[0, ∞] s.t. s1, s2, ... ≤ f and s.t. limn→∞sn = f • A Borel space (M, A) is standard if ∀x, y ∈M, x ̸= y ⇒∃A ∈ A s.t. x ∈A and s.t. y ̸∈A Imagine a space M and a σ-algebra A. One point, x, may lie within A. If the Borel space is standard, point y does not lie within A. The σ-algebra ”separates points”.
• A measure space (M, A, µ) is standard if (M, A) is standard.
Definitions and Facts from Topic 2500 10 • For any set S, let idS : S →S be the identity function on S, defined by: idS(s) = s.
Definition of an identity function.
• Two Borel space (M, A) and (N, B) are isomorphic (or Borel iso-morphic) if ∃f : M →N Borel ∃g : N →M Borel s.t. g ◦f = idM and f ◦g = idN.
In this case, we say that f : M →N and g : N →M are Borel isomorphisms.
• Two measure spaces (M, A, µ) and (N, B, ν) are isomorphic (or mea-sure isomorphic) if ∃f : M →N Borel ∃g : N →M Borel s.t. g◦f = idM and f ◦g = idN and f∗(µ) = ν and g∗(ν) = µ.
In this case, we say that f : M →N and g : N →M are measure isomorphisms.
• A measure space (M, A, µ) is standard if (M, A) is standard.
See above definition of a Borel space being standard.
Definitions and Facts from Topic 2700 • Let (Ω, B, µ) be a probability space. A random variable (or RV) on (Ω, B, µ) is a measurable map Ω→R.
This is the generalized version of a PCRV. Whereas a PCRV is a func-tion from the unit interval to the reals, we now have a map from a measure space to the reals.
• Let X : Ω→R be a RV on (Ω, B, µ).
The distribution fof X is δX := X∗(µ) Here X is defined as a random variable (by the definition above). Since X is a random variable, it must have a distribution. That distribution is given by the notation defined here.
• A measure on R is proper if every bounded interval has finite measure.
An example of this is Lebesgue measure on R.
11 • For any measure µ on R, the cumulative distribution function (or CDF) of µ is the function CDFµ : R →[0, ∞] defined by CDFµ(s) = µ((−∞, s]).
The cumulative distribution function describes the probability that a variable with a given distribution will be found at a value less than or equal to x. It is cumulative in the sense that as the value x increases, the total value returned by the function increases.
• f : R →R is cadlag if ∀a ∈R, limx→a+(f(x)) = f(a) and limx→a−(f(x)) exists.
Cadlag is an acronym for a French phrase that translates to ”contin-uous from right and limit from left,” which adequately describes the process taking place.
• f : R →R is CDF type if f is nondecreasing, bounded, cadlag and f(−∞) = 0.
TBD • f : R →R is a regular CDF if f is increasing, continuous, and f(−∞) = 0, f(∞) = 1.
TBD • µ is a regular distribution if µ is a probability measure, ∀x, µ({x}) = 0 and ∀(a < b), µ((a, b)) > 0.
TBD • Let µ be a measure on R. Then a measurable function h : R →[0, ∞] is a probability density function (PDF) for µ if µ = hλ.
TBD • A RV is standard normal if φ′ is a PDF of its distribution.
TBD • CB := {continuous, bounded functions R →R} TBD 12 • CE := {continuous, exponentially −bounded functions R →R}.
TBD • Let µ1, µ2, ... and ν be probability measures on R. Then µ1, µ2, ... →ν means: ∀f ∈CB, R Rn fdµn → R Rn fdν.
The same expression holds ∀f ∈CE (continuous, exponentially-bounded).
TBD • For any probability measure µ on R, the Fourier transform of µ is the function Fµ : R →C defined by Fµ := R ∞ −∞e−itxdµ(x).
TBD Fact: Fµn →Fν, ∀t ⇒µn →ν.
Fact: F(φ′λ) = e−t2/2 Corollary: A RV X is standard normal iffFδX = e−t2/2, i.e. the Fourier transform of the distribution of X is e−t2/2.
Fact: ∀probability measures µ and ν on R, F(µ ∗ν) = [Fµ][Fν].
• The standard normal distribution on R is φ′λ.
TBD • A RV is standard normal if φ′ is a PDF on its distribution. i.e.: A RV X is standard normal iffδX = φ′λ, i.e., the distribution of X is the standard normal distribution.
TBD • Define A : R×R →R by A(x, y) = x+y. For any two measures µ and ν on R, the convolution of µ and ν is the measure µ ∗ν on R defined by µ ∗ν := A∗(µ × ν).
TBD 13 • A measure space (M, B, µ) is nonatomic if ∀x ∈M, µ({x}) = 0.
A measure space is nonatomic if all the points in M have measure zero.
Definitions and Facts from Topic 2800 • A σ-finite signed measure on (M, B) is a function ω : B →[−∞, ∞] s.t., for some pair µ, ν of σ-finite measures on (M, B) we have (µ(M), ν(M)) ̸= (∞, ∞) and ω = µ −ν, i.e., ∀B ∈B, ω(B) = (µ(B)) −(ν(B)).
TBD • v : [a, b] →R has bounded variation if ∃nondecreasing f, g : [a, b] → R s.t. v = f −g.
TBD • Let v : [a, b] →R have bounded variation, and let f, g : [a, b] →R be nondecreasing functions such that v = f −g. We define dv := d f −dg.
TBD Fact: Let v : [a, b] →R be continuous, differentiable on (a,b). Then v has bounded variation. Then [f(ν(b))] −[f(ν(a))] = R b a f ′(ν(t))dν(t).
Definitions and Facts from Topic 2900 • An event (in Ωor in (Ω, A, µ)) is a measurable subset of Ω.
We define an event to be something measurable in a set.
• The probability of an event E is Pr[E] := ¯ µ(E). We write E a.s. if Pr[E]=1.
The probability of an event is defined as the completed measure of the event.
• Let (M, B) be a standard Borel space. An (M, B)-RV or M -RV (on (Ω, A, µ)) or on Ωis a measurable map Ω→M.
Just like a PCRV is a function from [0, 1] →R, an RV works in a similar way, mapping values from Ωto a σ-algebra.
14 • Let X : Ω→M be an M -RV. For all Borel S ⊆M, the event X ∈S is {ω ∈Ω|X(ω) ∈S} = X−1(S).
TBD • Referencing the definition above, X is deterministic if ∃p ∈M s.t.
X=p a.s.
TBD • A RV (on (Ω, A, µ) or on Ω) is a measurable map Ω→R.
Like the definition of an M-RV from above, the definition of a random variable is a direct analogue to a PCRV mapping from the unit interval to the reals.
• Let X : Ω→R be a RV on Ω.
The event a ≥X is the event X ∈(−∞, a]; the event a < X ≤b is the event X ∈(a, b];, etc.
If we are trying to measure the probability of an event, we are essentially testing whether the event falls within a certain interval. For example: Pr[1 ≤X] = ¯ µ(X−1([1, ∞)).
• Let X : Ω→C be a C-RV on Ω. The expectation or mean of X is Eµ[X] := E[X] := R ΩXdµ.
The definition of expectation for a complex random variable is sim-ilar to the definition of expectation for other random variables. We integrate X over the space.
• X is integrable or L1 if E[|X|] is finite; in this case we define: X◦:= X −(E[X]), so that X◦has mean zero.
If X is integrable, we define a new X◦such that we normalize the new X to have mean zero.
• X is square integrable or L2 if E[|X|2] < ∞.
If the expectation of the square of X is finite, then X is square inte-grable.
15 Fact: X is L2 ⇒X is L1 and X◦is L2 Fact: X is deterministic iffVar[X]=0 Fact: X is square integrable if V ar[X] < ∞ Fact: X and Y are uncorrelated if Cov[X,Y]=0 • Let X : Ω→R be an L2 RV on Ω. The variance of X is V ar[X] := E[(X◦)2].
TBD • Let X : Ω→R be square integrable. X is standard if both E[X]=0 and Var[X]=1.
• The standard deviation of X is SD[X] := p V ar[X].
• Let X, Y : Ω→R be square integrable. The covariance of X,Y is Cov[X,Y]:=((Var[X+Y]-Var[X]-Var[Y])/2).
• Let X, Y : Ω→R be square integrable and non-deterministic. The correlation of X,Y is Corr[X,Y]:=Cov[X,Y]/(SD[X]SD[Y]).
• For all Borel space (N, C), for all Borel f : M →N, f(X) := f ◦X.
Definition of composition. The traditional notation for a function f(X) is defined as f composed with X.
• The distribution of X is δµ[X] = δ[X] = δµ X = δX := X∗(µ).
Various alternative notations for a distribution.
Fact: For all Borel space (N, C), for all Borel f : M →N, δf(X) = f∗(δX) Fact: For all Borel S ⊆M, Pr[X ∈S] = δX(S) Fact: Say M = R. Then E[f(X)] = R ∞ −∞fdδX 16 • The joint variable of X and Y is the (M × N) −RV (X, Y ) : Ω→ M × N defined by (X, Y )(ω) = (X(ω), Y (ω)).
TBD • δX,Y := δ(X,Y ) is the joint distribution of X and Y ; it’s a measure on M × N.
TBD • Let (M, B), (N, C) be standard Borel spaces. Define p : M × N → M, q : M × N →N by p(x, y) = x and q(x, y) = y. ∀measure ω on M × N, the marginals of ω are p∗(ω) and q∗(ω); they are measures on M and N, respectively.
When dealing with joint variables and joint distributions, the individ-ual variables and distributions that make up the joint are known as the marginals of the joint.
Fact: For all measures µ on M and ν on N, p∗(µ×ν) = µ and q∗(µ×ν) = ν. ”The marginals of a product measure are the factor measures.” Fact: For all standard probability space (Ω, B, µ), ∀M −RV X : Ω→ M, ∀N −RV Y : Ω→N, p∗(δX,Y ) = δX and q∗(δX,Y ) = δY ”The marginals of a joint distribution are the individual distributions.” Fact: E[f(X, Y )] = R R2 f(x, y)dδX,Y (x, y) • ∀E ∈A, ∀F ∈A,E and F are info-equivalent if E∆F is null, where A is an event.
The definition essentially states that if we know whether or not ω ∈E, then we also know whether or not ω ∈F.
• For all sets E,F, the symmetric difference of E and F is E∆F := (E\F) S(F\E).
TBD • Let X : Ω→M be an M -RV. The σ-algebra of X is SB X := SX := X∗(B) := {X−1(B)|B ∈B}.
17 TBD • For any σ-subalgebra S of ¯ A, we say that X is S-measurable if ¯ SX ⊆S.
TBD • Two events E, F ∈¯ A are independent if ¯ µ(E T F) = [¯ µ(E)][¯ µ(F)] This definition of independence is analogous to the definition by ex-pectation: E[XY]=E[X]E[Y]. This definition adapts the traditional def-inition to include events.
• Two subsets E, F ⊆¯ A are independent if, ∀E ∈E, F ∈F, E and F are independent.
If all the events present in a subset are independent from all the events in another subset, we say the two subsets are independent.
• An M -RV X and a subset E ⊆¯ A are independent if SX and ⊆E are independent.
A random variable on some Borel space M and a subset are independent if the σ-algebra and the subset are independent. Refer to the definition a few rows above to review the σ-algebra definition.
• An M -RV X and an N -RV Y are independent if SX and SY are independent.
Two random variables on respective Borel spaces are independent if their σ-algebras are independent.
Fact: X and Y are independent iffδX,Y = δX × δY .
Fact: X, Y independent, (P, D), (Q, E) standard Borel spaces ⇒∀Borel f : M →P, ∀Borel g : N →Q, f(X) and g(Y) are independent.
Fact: Say M = N = R and X, Y, L2. Then if X,Y are independent then they are uncorrelated. Cov[X,Y]=(E[XY])-(E[X])(E[Y]).
Fact: F(µ ∗ν) = (Fµ)(Fν) 18 • Define: A : R × R →R by A(x, y) = x + y.
For all measures µ, ν on R, µ ∗ν := A∗(µ × ν) is the convolution of µ and ν.
Convolution is the process by which we take the distributions of two random variables and multiply them instead of summing them.
• The joint variable of X1, ..., Xn is the M -RV (X1, ..., Xn) : Ω→M defined by (X1, ..., Xn)(ω) = (X1(ω), ..., Xn(ω)). For all standard Borel spaces (P, D), for all Borel f : M →P, f(X1, ..., Xn) := f((X1, ..., Xn)) = f ◦(X1, ..., Xn); it’s a P-RV on Ω.
TBD • δX1,...,Xn := δX1,...,Xn is the joint distribution of X1, ..., Xn; it’s a mea-sure on M.
TBD • X1, ..., Xn are (jointly) independent if δX1,...,Xn = δX1 × ... × δXn.
TBD • For all measure τ on M, the marginals of τ are (p1)∗(τ), ..., (pn)∗(τ); they are measures on M1, ..., Mn, respectively.
TBD Fact: For all measures µ1 on M1, ..., µn on Mn, ∀k, (pk)∗(µ1 × ... × µn) = µk.
Fact: For all standard probability space (Ω, A, µ), ∀M1−RV X1, ..., ∀Mn− RV Xn, all on Ω, ∀k, (pk)∗(δX1,...,Xn) = δXk Fact: For all Borel S1 ⊆M1, ..., Sn ⊆Mn, Pr[(X1 ∈S1)&...&(Xn ∈ Sn)] = (Pr[X1 ∈S1])...(Pr[Xn ∈Sn]) Fact: For all standard Borel spaces (P1, D1), ..., (Pn, Dn), ∀f1 : M1 → P1, ..., ∀fn : Mn →Pn, all Borel, f1(X1), ..., fn(Xn) are jointly independent.
19 Fact: δX1+...+Xn = δX1 ∗... ∗δXn Corollary: FδX1+...+Xn = (FδX1)...(FδXn) • Let X be a random variable. Let F : R →[0, 1] be the CDF of (the distribution of) X. The grade of X is gr[X ]:=F(X ).
TBD Fact: If X has no values of positive probability, i.e., if, ∀c ∈R, Pr[X = c] = 0, then δ[gr[X]] is Lebesgue measure on [0,1].
• The joint distribution of X1, ..., Xn is δ[X1, ..., Xn] := (X1, ...Xn)∗(µ), a probability measure on Rn.
TBD • The copula of X1, ..., Xn is cop[X1, ..., Xn] := δ[gr[X1], ..., gr[Xn]].
TBD • X1, X2, ... M -RVs,... (all on Ω) X1, X2, ... are iid means both X1, X2, ...
are (jointly) independent and for all integers j, k, ≥1, δXj = δXk TBD Definitions and Facts from Topic 3000 • Let (Ω, A, µ) be a standard probability space. Let E be an event, i.e., a measurable subset of Ω, so E ∈¯ A. The probability of E is defined by Pr[E] := ¯ µ(E).
TBD • Let E and F be events. The conditional probability of E given F is defined by Pr[E|F] := ¯ µ(E T F) ¯ µ(F) .
TBD 20 • Let E be an event and let X be an L1 RV. The conditional expec-tation of X given E is E[X|E] := 1 ¯ µ(E) R E Xdµ.
TBD • The conditional expectation of V given P is the RV E[V |P] : Ω→ R defined by (E[V |P])(ω) = E[V |Pω]. Here, P is a finite partition of Ω.
TBD Fact: Let U := E[V |P]. Then U is < Pσ >-measurable, and, ∀P ∈< P >σ of positive measure, E[U|P] = E[V |P].
• TBD Definitions and Facts from Topic 3100 • Let P and Q be partitions of Ω. We say that P is finer than Q if: ∀P ∈P, ∃Q ∈Q s.t. P ⊆Q.
If the question ”Which set in P contained ω?” gives enough info to answer ”Which set in Q contained ω?”, then we say that P is finer than Q.
Fact: P finer than Q ⇒∀Q ∈Q, ∃P1, ..., Pk ∈P s.t. Q = P1 ⊔... ⊔Pk.
Fact: P finer than Q implies any Q-measurable RV is P-measurable.
• The Tower Law: Let V be a L1 RV. Let P and Q be finite, posi-tive measure partitions of Ω. Assume that P is finer than Q. Then E[E[V |P]|Q] = E[V |Q].
Forcing P-measurability is weaker than forcing Q-measurability, so do-ing both is redundant.
• Let S and T be σ-subalgebras on Ω. We say that S than T if T ⊆S.
TBD 21 • The Power Tower Law: Let V be an L1 RV. Let S be a σ-algebra on Ω. Then E[E[V |S]] = E[V ].
TBD Definitions and Facts from Topic 3200 • For all functions f, g : R →R, the convolution of f and g is the function f ∗g defined by (f ∗g)(x) = R ∞ −∞[f(t)][g(x −t)]dt.
This definition was used on the final exam from last year in one of the first computation problems • Γ(s) := [ R ∞ −∞zxe−exdx]z:→es This is the definition of the gamma function, which is used to calcu-late the PDF for the chi-squared distribution. There is a simpler way to define the gamma function in a practical way, as given in the two definitions below.
• For integer values of n, the result of the gamma function is given by (n −1)!
• For non-integer values of n, the result of the gamma function is given by (2n)!
4nn!
√π. Note that Γ(1 2) = √π.
• (∗nψ)(x) = x(n/2)−1e−x/2 2n/2Γ(n/2) This function yields the PDF for a chi-squared distribution with n degrees of freedom. Note that it relies on the above definition of the gamma function.
Fact: Let (M, B), (N, C), (P, D) be Borel spaces. Let F : M × N →P be Borel. Let x ∈M and let ν be a measure on N. Then F∗(δx × ν) = (F(x, •))∗(ν).
Fact: Let (M, B), (N, C) be Borel spaces. Let F : M →N be a Borel isomorphism. Let µ be a measure on M. Let g : M →[0, ∞] be measurable.
Then F∗(gµ) = [g ◦F −1][F∗(µ)].
Fact: If f is a PDF for µ and if g is a PDF for ν then f ∗g is a PDF for µ ∗ν.
22 Definitions and Facts from Topic 3300 • TBD Definitions and Facts from Topic 3400 • TBD Definitions and Facts from Topic 3500 • The exponential distribution describes the time between events that occur continuously and independently at a constant average rate.
• The CDF for the exponential distribution is as follows: CDFδX(x) := {1 −e−αx, if x ≥0; 0, if x ≤0 • The PDF for the exponential distribution is as follows: PDFδX(x) := {αe−αx, if x > 0; 0, if x < 0 Fact: X, Y independent ⇒δX+Y = δX ∗δY • TODO: Gamma, Poisson, and empirical distributions Definitions and Facts from Topic 3600 • Fix a probability space (Ω, B, µ).
For all Borel spaces (T, A), a T-process is a function X : T →{RV s on Ω} s.t. (ω, t) 7→(X(t))(ω) : Ω× T →R is measurable.
A process is a series of random variables together in a sequence that describe the evolution of a path of some kind.
• A process is a [0, ∞)-process.
By default, a process is defined on the positive real numbers.
• A spacetime-process is an (R × [0, ∞))-process.
A process may be defined in terms of both space and time. For instance, a Brownian motion has parameters that describe both the position of a particle and a related time index.
23 • A process X• (Xt or X(t)) is continuous if ∀ω ∈Ω, t 7→Xt(ω) : [0, ∞) →R is continuous.
Some processes are continuous, like Brownian motion, and others are not, like a Levy process.
• For any set T ⊆R, a T-filtration is a function F• : T →{σ − subalgebras} s.t. t, u ∈T, t ≤u ⇒Ft ⊆Fu.
A filtration can be thought of a σ-algebra that becomes increasingly finer as time passes. The σ-algebra allows for the measurement of the process. If one thinks of data appearing on a screen and that data is changing, a filtration is a collection of those data.
• A RV X : Ω→R is S-measurable if for all Borel B ⊆R, X−1(B) ∈S.
TBD • X• is F•-adapted means: ∀t ∈T, Xt is Ft-measurable.
TBD • The filtration of X• is the filtration FX • defined by FX t :=< S s≤t SXs >σ.
As mentioned in the definition of filtration above, the filtration is an increasingly finer σ-algebra.
• The (t1, ..., td)-marginal of X• is the joint distribution δ[Xt1, ..., Xtd].
TBD • X• = Y• in finite dimensional (f.d.) marginals, written X• δ = Y• means: for all integers d ≥1, ∀t1, ...td ∈[0, ∞), δ[Xt1, ..., Xtd] = δ[Yt1, ..., Ytd].
TBD Fact: Any process is adapted to its filtration.
Various Useful Facts You Should Know • E[cA] = c(E[A]) 24 • V ar[cA] = c2V ar[A] • SD[cA] = |c|SD[A] • E[c + A] = c + E[A] • V ar[c + A] = V ar[A] • SD[c + A] = SD[A] • E[Pn A] = n × E[A] • V ar[Pn A] = n × V ar[A] • SD[Pn A] = √n × SD[A] Finite is to algebra as countable is to sigma algebra sigma-additive is a more ”robust” form of additivity than finitely-additive??
sigma is another way to say infinite 25 |
190181 | https://en.wikipedia.org/wiki/Mallory%E2%80%93Weiss_syndrome | Contents
Mallory–Weiss syndrome
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Mallory–Weiss syndrome
Other names | Gastro-esophageal laceration syndrome, Mallory-Weiss tear
Mallory–Weiss tear affecting the esophageal side of the gastroesophageal junction
Specialty | Gastroenterology
Mallory–Weiss syndrome is a condition where high intra-abdominal pressures causes laceration and bleeding of the mucosa called Mallory-Weiss tears. Additionally, Mallory–Weiss syndrome is one of the most common causes of acute upper gastrointestinal bleeding, counting of around 1-15% of all cases in adults and less than 5% in children. It has been found that tears are up to 2 to 4 times more prevalent in men than women. The tears can cause upper gastrointestinal bleeding and predominantly occur where the esophagus meets the stomach (gastroesophageal junction). However, the tears can happen anywhere from the middle of the esophagus to the cardia of the stomach. Mallory–Weiss syndrome is often caused by constant vomiting and retching from alcoholism or bulimia. Gastroesophageal reflux disease (GERD) is another risk factor that is often linked with Mallory–Weiss syndrome. However, not every individual with Mallory–Weiss syndrome will have these risk factors. Individuals with Mallory–Weiss syndrome will have hematemesis (vomiting up blood), however the symptoms can vary.
History
Before 1929, there were cases reported with similar symptoms of bleeding in the esophagus, the first being Johann Friedrich Hermann Albers reporting ulcer in the lower esophagus in 1833 via autopsy; however those were caused by ulcers and not lacerations. Another instance of Mallory–Weiss syndrome was from 1879 when Dr. Heinrich Quincke discovered 3 cases of bleeding from the formation of ulcers in the gastroesophageal tube; 2 of the cases were fatal due to vomiting of blood. This was followed by 2 cases reported by Dieulafoy to witness death from the phenomenon via vomiting of blood and 100 more cases in later literature before the findings in 1929.
Mallory–Weiss syndrome was named after G. Kenneth Mallory and Soma Weiss who accurately characterized the condition as a lower esophageal laceration in 1929 in 15 patients afflicted with alcoholism who presented with signs and symptoms of vomiting and retching. It was hypothesized that repeated vomiting would lead to the formation of tears if the body was not able to coordinate the cardiac opening of the stomach with the contraction of the abdominal muscles to induce the vomiting. Years later, Weiss and Mallory performed autopsies on 4 patients that died due to the complications of the syndrome caused by the hemorrhage. With the autopsies, it was noted that patients had lesions that were present on the esophagus down to the junction of the esophagus that meets the stomach. These particular lesions had signs of continual exposure to gastric juices from the stomach caused by the vomiting due to the pressure imbalances from the stomach; as a result the acutely formed lesions developed into chronic ulcerative lesions that ran deep into the layers of the esophagus up until the muscle fibers. In addition, there were signs of small arterioles that ruptured and small veins that were near the lesions which explains the hemorrhage that was present in these patients.
Dr. John Decker also examined patients afflicted with Mallory–Weiss syndrome via autopsy to note that many of patients did not have a history of alcoholism unlike the initial study conducted by the physicians the syndrome is named after; though Decker did comment that patients could be examined via gastroscope alongside Dr. Palmer who specifically mentioned the use of an endoscopy for diagnosis of Mallory–Weiss syndrome, so clinicians would not have to wait for a patient to die before performing an autopsy. However, a common finding between the patients with Decker's analysis is the exacerbation of the lesions caused by vomiting with atrophic gastritis being an underlying factor that to those formations; though atrophic gastritis is a condition that is common with the elderly population which most of the 11 patients undergoing an autopsy were above 60 years of age. Moving forward to 1955, advances in surgery allow for a patient afflicted with Mallory–Weiss syndrome to be identified, then treated with the surgical procedure with Dr. E. Gale Whiting & Dr. Gilbert Baronne, when the only way in the past was to perform an autopsy when a patient is deceased. The following year, Hardy per the recommendations of Palmer and Decker was able to complete the first diagnosis of the syndrome via endoscopy, leading to an increased incidence of Mallory–Weiss syndrome as shown with over 200 cases being mentioned in the literature as of 1973, and eventually the standard to make use of endoscopy to diagnosis the condition to witness lacerations along the esophageal lining and the signs of hemorrhage.
Signs and symptoms
Mallory–Weiss syndrome often presents as an episode of vomiting up blood (hematemesis) after violent retching or vomiting, but may also be noticed as old blood in the stool (melena), and a history of retching may be absent. Oftentimes, hematemesis is accompanied by chest, back, or epigastric pain. Additional symptoms can occur depending on how severe the condition is. Some individuals have experienced dizziness, loss of consciousness, and upper abdomen pain.
The condition is rarely fatal since in 90% of cases the tears heal on their own and the bleeding will stop spontaneously within 48 to 72 hours. However, endoscopic or surgical treatment may be necessary for severe bleeds. In cases of more severe bleeding, the typical symptoms of Mallory-Weiss Syndrome are those typical found in shock, which can be life-threatening. If a patient does happen to go into shock it may be reversed if discovered early. Although there are multiple types of shock, hemorrhagic hypovolemic shock is most commonly associated with gastrointestinal bleeding. Furthermore, gastrointestinal losses, such as those incurred from prolonged vomiting or diarrhea are associated with non-hemorrhagic hypovolemic shock. Both hemorrhagic and non-hemorrhagic hypovolemic shock can occur when there are decreases in intravascular volume, such as when the body is hemorrhaging (bleeding) or significant fluid loss. This decrease in intravascular volume causes subsequent reflex mechanism produced by the body to activate SANs (sympathetic nervous system) in the later stages of hypovolemic shock. SANs is activated in response to the drop in mean arterial pressure that is brought on by the loss of fluid.
Causes
The causes of Mallory–Weiss syndrome is often associated with alcoholism, eating disorders such as bulimia nervosa, and gastroesophageal reflux disease (GERD). Specifically, up to 75% of patients have been observed with a heavy alcohol use associated with emesis.
It is also thought that Mallory–Weiss syndrome can be caused by actions that cause sudden increases in intra-abdominal pressure, such as repeated severe vomiting or coughing. There is some conflicting evidence that the presence of a hiatal hernia could be a predisposing condition to developing Mallory–Weiss syndrome. There is conflicting data suggesting the association between hiatal hernias and Mallory–Weiss syndrome. In 1989, a study conducted in Japan set out to determine if there was a link to Mallory–Weiss syndrome and hiatal hernias, this study found that hiatial hernias were found in 75% of patients with Mallory–Weiss syndrome. On the contrary, a case-control study in 2017 found there was no association between hiatal hernias and Mallory–Weiss syndrome. Forceful vomiting causes tearing of the mucosa at the junction. Additionally, the use of NSAIDs (non-steroidal anti-inflammatory drugs) such as ibuprofen, are known to increase the risk of upper gastrointestinal bleeding. NSAIDs can increase the risk of upper gastrointestinal bleeding because they can cause further damage to the intestinal submucosa by inhibiting prostaglandin synthesis. NSAID abuse is also a rare association. In rare instances some chronic disorders like Ménière's disease that cause long term nausea and vomiting could be a factor. Other potential risks for GI bleeds are usage of anticoagulants and older age.
Additionally, bleeding from Mallory-Weiss tears is often associated with individuals who have a history of portal hypertension and esophageal varices. Portal hypertension is where there is increased pressure within the venous portal system. Additionally, studies that were performed in patients with cirrhosis (scaring/fibrosis of the liver) who also had portal hypertension have shown that an increase in portal pressure can cause an increase in intra-abdominal pressure. These increases in intra-abdominal pressure are associated with Mallory-Weiss Syndrome. More severe upper gastrointestinal bleeds are associated with concurrent portal hypertension and esophageal varices. The formation of esophageal varices (dilated veins) is linked to the presence of portal hypertension. Additionally, esophageal varices can rupture which can be fatal.
The tear involves the mucosa and submucosa but not the muscular layer (contrast to Boerhaave syndrome which involves all the layers). Most patients are between the ages of 30 and 50 years, although it has been reported in infants aged as young as 3 weeks, as well as in older people. Hyperemesis gravidarum, which is severe morning sickness associated with vomiting and retching in pregnancy, is also a known cause of Mallory–Weiss tear.
There have been a few complications from invasive procedures such as trans-esophageal echocardiography and upper gastrointestinal endoscopy that cause Mallory-Weiss tears called iatrogenic Mallory–Weiss syndrome. However, it is infrequent since it only occurs in 0.07% to 0.49% of individuals who have received the upper gastrointestinal endoscopy procedure.
Furthermore, there were some cases reported of individuals developing Mallory-Weiss tears after cardiopulmonary resuscitation (CPR). The individuals did not have a history of alcoholism, hiatal hernia, or gastrointestinal diseases, but woke up vomiting blood. A GI endoscopy was performed and tears were found on the esophagogastric junction and lesser curvature of the stomach. An increase in intragastric pressure during CPR caused the Mallory-Weiss tears.
Diagnosis
Definitive diagnosis of Mallory-Weiss tears is by upper GI endoscopy of the esophagus and stomach. Typically, the tear is located near the top of the stomach's lesser curvature and below the gastroesophageal junction. In the majority of patients, tears usually range from approximately 2 to 4 cm in length. The findings may include indications of non-bleeding, active bleeding, or the presence of clot over the tear. Furthermore, an upper GI endoscopy can reveal underlying conditions that lead to the signs of bleeding secondary to the tears, including varices and ulcers along the upper GI tract. To determine if the patient has active bleedings or signs of chronic alcoholism that can precede Mallory–Weiss syndrome, the patient's lab values would be obtained to get a complete blood count (CBC) including hematocrit & hemoglobin levels alongside platelet count. Additionally, diagnosis of Mallory-Weiss Syndrome includes elimination of other causes of an upper gastrointestinal bleed and/or bleeding in general. For example, a patient should undergo more labs to determine kidney function via measuring blood urea nitrogen and creatinine as a patient with chronic kidney disease can be mistaken to have active bleeding due to anemia induced by chronic kidney disease or if both the esophageal lacerations and chronic kidney disease are contributing to the low hematocrit & hemoglobin levels.
Proper history taking by the medical doctor to distinguish other conditions that cause haematemesis but definitive diagnosis is by conducting esophagogastroduodenoscopy, which is a procedure that allows the oropharynx, esophagus, stomach, and proximal duodenum (beginning of the small intestine) to be visualized.
Treatment
The course of treatment and management of Mallory–Weiss syndrome depends on the amount of bleeding or hematemesis. Although blood transfusion is ultimately needed for many patients with Mallory–Weiss syndrome, 90% of Mallory-Weiss tears can heal on their own spontaneously. If the bleeding is mild and localized, the condition can be managed with conservative treatment methods such as intravenous antacids, antiemetics, fasting, and bedrest. Antiemetics are medications used to help with nausea and vomiting. However, if constant bleeding is observed upon endoscopy, endoscopic hemostasis techniques are necessary as the first-line treatment. Four examples of endoscopic hemostasis techniques are hemoclipping, heat probe thermocoagulation, injection therapy, and band ligation.
Hemoclipping is an effective method for treating Mallory-Weiss tears because it uses small metal clips, which cause minimal tissue damage and stop the bleeding by clipping the affected blood vessels. Although hemoclip placement is a convenient procedure for nonfibrotic tissue, such as Mallory–Weiss syndrome, placing a hemoclip can be challenging at the typical location of Mallory-Weiss tears at the gastroesophageal junction.
Heat probe thermocoagulation is one of the endoscopic therapies used to stop bleeding by simultaneously applying heat and pressure directly on the area of the active bleed to start the coagulation. Thermocoagulation with bipolar or multipolar electrocautery can be employed to cauterize tissue. It is most appropriate for small and localized lesions that require minimal cauterization. However, it should be avoided in patients with esophageal varices as the absence of a serosal layer in the esophagus increases susceptibility to perforation and could exacerbate bleeding, posing significant risks. It should also be avoided in individuals with portal hypertension because more bleeding can occur. It is also noted that repeated coagulation therapy can lead to risk of transmural injury. Treatment is usually supportive as persistent bleeding after endoscopic treatment or esophagogastroscopy is uncommon.
Injection of epinephrine or cauterization to stop the bleeding through vasoconstriction may be undertaken during the index endoscopy procedure in the case of active and recurrent bleeding. Because it is easy to implement and widely available, such injection methods to stop bleeding are commonly used. However, this method requires close monitoring due to the possibility of causing ventricular tachycardia when administered submucosally. Thus, epinephrine injections should not be used in patients who have existing cardiovascular conditions.
Band ligation stops the bleeding by applying a direct pressure from a transparent ligation cap. The role of transparent cap is to stabilize the bleeding site and reduce the effects of peristalsis. Band ligation technique is relatively simple compared to other hemostatic techniques. Band ligation is recommended for individuals with esophageal varices or portal hypertension.
Some other options to stop bleeding include ethanol injections, ε-aminocaproic acid, or Argon plasma coagulation (APC). When endoscopy is ineffective, angiography or embolization of the arteries supplying the region may be required to stop the bleeding. If all other methods fail, high gastrostomy can be used to ligate the bleeding vessel. A Sengstaken-Blakemore tube will not be able to stop bleeding as here the bleeding is arterial and the pressure in the balloon is not sufficient to overcome the arterial pressure. After patient receives appropriate intervention, bleeding must be observed for at least 48 hours as a follow-up.
If a patient is thought to have shock, intravenous (IV) fluid resuscitation should begin immediately. In the case of hypovolemic shock, patients are typically placed in the Trendelenburg position where the feet are above the head. Additionally, if there is found to be an active bleed treatment with PRBCs (packed red blood cells) is typical.
In pharmacological treatment, proton pump inhibitors (such as omeprazole, pantoprazole) and H2 receptor antagonist (such as famotidine) are utilized to manage and lower gastric acidity. Decreasing the acidity through use of proton pump inhibitors and H2 receptor antagonists allows there to be time for healing. Proton pump inhibitors are preferred over H2 receptor antagonists because they are more potent and can keep gastric pH under control for a longer period of time. Furthermore, proton pump inhibitors have a decreased recurrent bleeding rate and do not lose their efficacy as a side effect when taken regularly over time (tachyphylaxis) compared to H2 receptor antagonists. It is recommended that individuals are given proton pump inhibitors within 72 hours of an endoscopy to prevent further GI bleeds. Additionally, antiemetics such as promethazine are given to control nausea and vomiting as part of the treatment regimen.
See also
References
External links
Classification | DICD-10:K22.6ICD-9-CM:530.7MeSH:D008309DiseasesDB:7803
External resources | MedlinePlus:000269eMedicine:ped/1359Patient UK:Mallory–Weiss syndrome
vteDiseases of thehuman digestive system
Upper GI tract | EsophagusEsophagitisCandidalEosinophilicHerpetiformRuptureBoerhaave syndromeMallory–Weiss syndromeZenker's diverticulumBarrett's esophagusEsophageal motility disorderNutcracker esophagusAchalasiaEsophagogastric junction outflow obstructionDiffuse esophageal spasmGastroesophageal reflux disease(GERD)Laryngopharyngeal reflux(LPR)Esophageal strictureInlet patchMegaesophagusEsophageal intramural pseudodiverticulosisAcute esophageal necrosisStomachGastritisAtrophicMénétrier's diseaseGastroenteritisPeptic (gastric) ulcerCushing ulcerDieulafoy's lesionDyspepsiaFunctional dyspepsiaPyloric stenosisAchlorhydriaGastroparesisGastroptosisPortal hypertensive gastropathyGastric antral vascular ectasiaGastric dumping syndromeGastric volvulusBuried bumper syndromeGastrinomaZollinger–Ellison syndrome | Esophagus | EsophagitisCandidalEosinophilicHerpetiformRuptureBoerhaave syndromeMallory–Weiss syndromeZenker's diverticulumBarrett's esophagusEsophageal motility disorderNutcracker esophagusAchalasiaEsophagogastric junction outflow obstructionDiffuse esophageal spasmGastroesophageal reflux disease(GERD)Laryngopharyngeal reflux(LPR)Esophageal strictureInlet patchMegaesophagusEsophageal intramural pseudodiverticulosisAcute esophageal necrosis | Stomach | GastritisAtrophicMénétrier's diseaseGastroenteritisPeptic (gastric) ulcerCushing ulcerDieulafoy's lesionDyspepsiaFunctional dyspepsiaPyloric stenosisAchlorhydriaGastroparesisGastroptosisPortal hypertensive gastropathyGastric antral vascular ectasiaGastric dumping syndromeGastric volvulusBuried bumper syndromeGastrinomaZollinger–Ellison syndrome
Esophagus | EsophagitisCandidalEosinophilicHerpetiformRuptureBoerhaave syndromeMallory–Weiss syndromeZenker's diverticulumBarrett's esophagusEsophageal motility disorderNutcracker esophagusAchalasiaEsophagogastric junction outflow obstructionDiffuse esophageal spasmGastroesophageal reflux disease(GERD)Laryngopharyngeal reflux(LPR)Esophageal strictureInlet patchMegaesophagusEsophageal intramural pseudodiverticulosisAcute esophageal necrosis
Stomach | GastritisAtrophicMénétrier's diseaseGastroenteritisPeptic (gastric) ulcerCushing ulcerDieulafoy's lesionDyspepsiaFunctional dyspepsiaPyloric stenosisAchlorhydriaGastroparesisGastroptosisPortal hypertensive gastropathyGastric antral vascular ectasiaGastric dumping syndromeGastric volvulusBuried bumper syndromeGastrinomaZollinger–Ellison syndrome
Lower GI tractEnteropathy | Small intestine(Duodenum/Jejunum/Ileum)EnteritisDuodenitisJejunitisIleitisPeptic (duodenal) ulcerCurling's ulcerMalabsorption:CoeliacTropical sprueBlind loop syndromeSmall intestinal bacterial overgrowthWhipple'sShort bowel syndromeSteatorrheaMilroy diseaseBile acid malabsorptionLarge intestine(Appendix/Colon)AppendicitisColitisPseudomembranousUlcerativeIschemicMicroscopicCollagenousLymphocyticDysenteryFunctional colonic diseaseIBSIntestinal pseudoobstruction/Ogilvie syndromeMegacolon/Toxic megacolonDiverticulitis/Diverticulosis/SCADLarge and/or smallEnterocolitisNecrotizingGastroenterocolitisIBDCrohn's diseaseVascular:Abdominal anginaMesenteric ischemiaAngiodysplasiaBowel obstruction:IleusIntussusceptionVolvulusFecal impactionConstipationFunctionalDiarrheaInfectiousIntestinal adhesionsRectumProctitisRadiation proctitisProctalgia fugaxRectal prolapse(Internal rectal prolapse)AnismusSolitary rectal ulcer syndromeRectal strictureAnal canalAnal fissure/Anal fistulaAnal abscessHemorrhoidAnal dysplasiaPruritus aniAnal stricture | Small intestine(Duodenum/Jejunum/Ileum) | EnteritisDuodenitisJejunitisIleitisPeptic (duodenal) ulcerCurling's ulcerMalabsorption:CoeliacTropical sprueBlind loop syndromeSmall intestinal bacterial overgrowthWhipple'sShort bowel syndromeSteatorrheaMilroy diseaseBile acid malabsorption | Large intestine(Appendix/Colon) | AppendicitisColitisPseudomembranousUlcerativeIschemicMicroscopicCollagenousLymphocyticDysenteryFunctional colonic diseaseIBSIntestinal pseudoobstruction/Ogilvie syndromeMegacolon/Toxic megacolonDiverticulitis/Diverticulosis/SCAD | Large and/or small | EnterocolitisNecrotizingGastroenterocolitisIBDCrohn's diseaseVascular:Abdominal anginaMesenteric ischemiaAngiodysplasiaBowel obstruction:IleusIntussusceptionVolvulusFecal impactionConstipationFunctionalDiarrheaInfectiousIntestinal adhesions | Rectum | ProctitisRadiation proctitisProctalgia fugaxRectal prolapse(Internal rectal prolapse)AnismusSolitary rectal ulcer syndromeRectal stricture | Anal canal | Anal fissure/Anal fistulaAnal abscessHemorrhoidAnal dysplasiaPruritus aniAnal stricture
Small intestine(Duodenum/Jejunum/Ileum) | EnteritisDuodenitisJejunitisIleitisPeptic (duodenal) ulcerCurling's ulcerMalabsorption:CoeliacTropical sprueBlind loop syndromeSmall intestinal bacterial overgrowthWhipple'sShort bowel syndromeSteatorrheaMilroy diseaseBile acid malabsorption
Large intestine(Appendix/Colon) | AppendicitisColitisPseudomembranousUlcerativeIschemicMicroscopicCollagenousLymphocyticDysenteryFunctional colonic diseaseIBSIntestinal pseudoobstruction/Ogilvie syndromeMegacolon/Toxic megacolonDiverticulitis/Diverticulosis/SCAD
Large and/or small | EnterocolitisNecrotizingGastroenterocolitisIBDCrohn's diseaseVascular:Abdominal anginaMesenteric ischemiaAngiodysplasiaBowel obstruction:IleusIntussusceptionVolvulusFecal impactionConstipationFunctionalDiarrheaInfectiousIntestinal adhesions
Rectum | ProctitisRadiation proctitisProctalgia fugaxRectal prolapse(Internal rectal prolapse)AnismusSolitary rectal ulcer syndromeRectal stricture
Anal canal | Anal fissure/Anal fistulaAnal abscessHemorrhoidAnal dysplasiaPruritus aniAnal stricture
GI bleeding | Blood in stoolUpperHematemesisMelenaLowerHematochezia
Accessory | LiverHepatitisViral hepatitisAutoimmune hepatitisAlcoholic hepatitisCirrhosisPBCFatty liverMASLDVascularBudd–Chiari syndromeHepatic veno-occlusive diseasePortal hypertensionNutmeg liverAlcoholic liver diseaseLiver failureHepatic encephalopathyAcute liver failureLiver abscessPyogenicAmoebicHepatorenal syndromePeliosis hepatisMetabolic disordersWilson's diseaseHemochromatosisGallbladderCholecystitisGallstone / CholelithiasisCholesterolosisAdenomyomatosisPostcholecystectomy syndromePorcelain gallbladderBile duct/Otherbiliary treeCholangitisPrimary sclerosing cholangitisSecondary sclerosing cholangitisAscendingCholestasis/Mirizzi's syndromeBiliary fistulaHaemobiliaCommon bile ductCholedocholithiasisBiliary dyskinesiaSphincter of Oddi dysfunctionPancreaticPancreatitisAcuteChronicHereditaryPancreatic abscessPancreatic pseudocystExocrine pancreatic insufficiencyPancreatic fistula | Liver | HepatitisViral hepatitisAutoimmune hepatitisAlcoholic hepatitisCirrhosisPBCFatty liverMASLDVascularBudd–Chiari syndromeHepatic veno-occlusive diseasePortal hypertensionNutmeg liverAlcoholic liver diseaseLiver failureHepatic encephalopathyAcute liver failureLiver abscessPyogenicAmoebicHepatorenal syndromePeliosis hepatisMetabolic disordersWilson's diseaseHemochromatosis | Gallbladder | CholecystitisGallstone / CholelithiasisCholesterolosisAdenomyomatosisPostcholecystectomy syndromePorcelain gallbladder | Bile duct/Otherbiliary tree | CholangitisPrimary sclerosing cholangitisSecondary sclerosing cholangitisAscendingCholestasis/Mirizzi's syndromeBiliary fistulaHaemobiliaCommon bile ductCholedocholithiasisBiliary dyskinesiaSphincter of Oddi dysfunction | Pancreatic | PancreatitisAcuteChronicHereditaryPancreatic abscessPancreatic pseudocystExocrine pancreatic insufficiencyPancreatic fistula
Liver | HepatitisViral hepatitisAutoimmune hepatitisAlcoholic hepatitisCirrhosisPBCFatty liverMASLDVascularBudd–Chiari syndromeHepatic veno-occlusive diseasePortal hypertensionNutmeg liverAlcoholic liver diseaseLiver failureHepatic encephalopathyAcute liver failureLiver abscessPyogenicAmoebicHepatorenal syndromePeliosis hepatisMetabolic disordersWilson's diseaseHemochromatosis
Gallbladder | CholecystitisGallstone / CholelithiasisCholesterolosisAdenomyomatosisPostcholecystectomy syndromePorcelain gallbladder
Bile duct/Otherbiliary tree | CholangitisPrimary sclerosing cholangitisSecondary sclerosing cholangitisAscendingCholestasis/Mirizzi's syndromeBiliary fistulaHaemobiliaCommon bile ductCholedocholithiasisBiliary dyskinesiaSphincter of Oddi dysfunction
Pancreatic | PancreatitisAcuteChronicHereditaryPancreatic abscessPancreatic pseudocystExocrine pancreatic insufficiencyPancreatic fistula
Other | HerniaDiaphragmaticCongenitalHiatusInguinalIndirectDirectUmbilicalParaumbilicalFemoralObturatorSpigelianLumbarPetit'sGrynfeltt–LesshaftUndefined locationIncisionalInternal herniaRichter'sPeritonealPeritonitisSpontaneous bacterial peritonitisHemoperitoneumPneumoperitoneum | Hernia | DiaphragmaticCongenitalHiatusInguinalIndirectDirectUmbilicalParaumbilicalFemoralObturatorSpigelianLumbarPetit'sGrynfeltt–LesshaftUndefined locationIncisionalInternal herniaRichter's | Peritoneal | PeritonitisSpontaneous bacterial peritonitisHemoperitoneumPneumoperitoneum
Hernia | DiaphragmaticCongenitalHiatusInguinalIndirectDirectUmbilicalParaumbilicalFemoralObturatorSpigelianLumbarPetit'sGrynfeltt–LesshaftUndefined locationIncisionalInternal herniaRichter's
Peritoneal | PeritonitisSpontaneous bacterial peritonitisHemoperitoneumPneumoperitoneum
Esophagus | EsophagitisCandidalEosinophilicHerpetiformRuptureBoerhaave syndromeMallory–Weiss syndromeZenker's diverticulumBarrett's esophagusEsophageal motility disorderNutcracker esophagusAchalasiaEsophagogastric junction outflow obstructionDiffuse esophageal spasmGastroesophageal reflux disease(GERD)Laryngopharyngeal reflux(LPR)Esophageal strictureInlet patchMegaesophagusEsophageal intramural pseudodiverticulosisAcute esophageal necrosis
Stomach | GastritisAtrophicMénétrier's diseaseGastroenteritisPeptic (gastric) ulcerCushing ulcerDieulafoy's lesionDyspepsiaFunctional dyspepsiaPyloric stenosisAchlorhydriaGastroparesisGastroptosisPortal hypertensive gastropathyGastric antral vascular ectasiaGastric dumping syndromeGastric volvulusBuried bumper syndromeGastrinomaZollinger–Ellison syndrome
Small intestine(Duodenum/Jejunum/Ileum) | EnteritisDuodenitisJejunitisIleitisPeptic (duodenal) ulcerCurling's ulcerMalabsorption:CoeliacTropical sprueBlind loop syndromeSmall intestinal bacterial overgrowthWhipple'sShort bowel syndromeSteatorrheaMilroy diseaseBile acid malabsorption
Large intestine(Appendix/Colon) | AppendicitisColitisPseudomembranousUlcerativeIschemicMicroscopicCollagenousLymphocyticDysenteryFunctional colonic diseaseIBSIntestinal pseudoobstruction/Ogilvie syndromeMegacolon/Toxic megacolonDiverticulitis/Diverticulosis/SCAD
Large and/or small | EnterocolitisNecrotizingGastroenterocolitisIBDCrohn's diseaseVascular:Abdominal anginaMesenteric ischemiaAngiodysplasiaBowel obstruction:IleusIntussusceptionVolvulusFecal impactionConstipationFunctionalDiarrheaInfectiousIntestinal adhesions
Rectum | ProctitisRadiation proctitisProctalgia fugaxRectal prolapse(Internal rectal prolapse)AnismusSolitary rectal ulcer syndromeRectal stricture
Anal canal | Anal fissure/Anal fistulaAnal abscessHemorrhoidAnal dysplasiaPruritus aniAnal stricture
Liver | HepatitisViral hepatitisAutoimmune hepatitisAlcoholic hepatitisCirrhosisPBCFatty liverMASLDVascularBudd–Chiari syndromeHepatic veno-occlusive diseasePortal hypertensionNutmeg liverAlcoholic liver diseaseLiver failureHepatic encephalopathyAcute liver failureLiver abscessPyogenicAmoebicHepatorenal syndromePeliosis hepatisMetabolic disordersWilson's diseaseHemochromatosis
Gallbladder | CholecystitisGallstone / CholelithiasisCholesterolosisAdenomyomatosisPostcholecystectomy syndromePorcelain gallbladder
Bile duct/Otherbiliary tree | CholangitisPrimary sclerosing cholangitisSecondary sclerosing cholangitisAscendingCholestasis/Mirizzi's syndromeBiliary fistulaHaemobiliaCommon bile ductCholedocholithiasisBiliary dyskinesiaSphincter of Oddi dysfunction
Pancreatic | PancreatitisAcuteChronicHereditaryPancreatic abscessPancreatic pseudocystExocrine pancreatic insufficiencyPancreatic fistula
Hernia | DiaphragmaticCongenitalHiatusInguinalIndirectDirectUmbilicalParaumbilicalFemoralObturatorSpigelianLumbarPetit'sGrynfeltt–LesshaftUndefined locationIncisionalInternal herniaRichter's
Peritoneal | PeritonitisSpontaneous bacterial peritonitisHemoperitoneumPneumoperitoneum |
190182 | https://asombro.org/wp-content/uploads/EvaporationInvestigationEducatorGuide.pdf | © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) COMMON CORE STATE STANDARDS English Language Arts Standards » Science & Technical Subjects » Grade 6-8 CCSS.ELA-LITERACY.RST.6-8.3. Follow precisely a multistep procedure when carrying out experiments, taking measurements, or performing technical tasks.
CCSS.ELA-LITERACY.RST.6-8.4. Determine the meaning of symbols, key terms, and other domain-specific words and phrases as they are used in a specific scientific or technical context relevant to grades 6-8 texts and topics.
English Language Arts Standards » Science & Technical Subjects » Grade 9-10 CCSS.ELA-LITERACY.RST.9-10.3. Follow precisely a complex multistep procedure when carrying out experiments, taking measurements, or performing technical tasks, attending to special cases or exceptions defined in the text.
CCSS.ELA-LITERACY.RST.9-10.4. Determine the meaning of symbols, key terms, and other domain-specific words and phrases as they are used in a specific scientific or technical context relevant to grades 9-10 texts and topics.
English Language Arts Standards » Science & Technical Subjects » Grade 11-12 CCSS.ELA-LITERACY.RST.11-12.3. Follow precisely a complex multistep procedure when carrying out experiments, taking measurements, or performing technical tasks; analyze the specific results based on explanations in the text.
Grade 6 » Statistics & Probability CCSS.MATH.CONTENT.6.SP.B.5. Summarize numerical data sets in relation to their context, such as by: CCSS.MATH.CONTENT.6.SP.B.5.C. Giving quantitative measures of center (median and/or mean) and variability (interquartile range and/or mean absolute deviation), as well as describing any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered.
NEXT GENERATION SCIENCE STANDARDS Science & Engineering Practices Disciplinary Core Ideas Crosscutting Concepts Planning and Carrying Out Investigations (MS, HS) Analyzing and Interpreting Data (MS, HS) Constructing Explanations and Designing Solutions (MS, HS) ESS2.A Earth Materials and Systems (MS, HS) ESS2.C The Roles of Water in Earth's Surface Processes (MS) Systems and System Models (MS, HS) © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) EVAPORATION INVESTIGATION GRADE LEVEL 6 – 12 PHENOMENON What factors have the greatest effect on evaporation?
DESCRIPTION Students conduct an experiment to investigate the effects of various factors on the rate of evaporation.
OBJECTIVES Students will: • Make a prediction using prior knowledge and experience • Model evaporation of water from soil under various environmental conditions • Synthesize results of an experiment • Apply understanding of experimental results to make further predictions about Earth’s systems TIME 1 HOUR TOTAL OVER 4 DAYS DAY 1:20 MINUTES DAY 2:10 MINUTES DAY 3:10 MINUTES DAY 4:20 MINUTES © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) D-02 CLIMATE CHANGE AND THE WATER CYCLE EVAPORATION INVESTIGATION BACKGROUND With increasing temperatures and changing weather patterns, climate change will affect evaporation rates in many areas. Evaporation is a major part of the earth’s water cycle. It is the process of water molecules gaining enough energy to escape the surface of a water layer. In the water cycle, water from lakes, ponds, rivers, streams, and oceans is heated by the sun and converted into water vapor. This vapor rises into the air and may result in the development of clouds.
Many factors can affect evaporation. Heat makes water evaporate more quickly because water molecules move faster when they are warm. Since the molecules are moving faster, more of them can leave the surface of the water at one time. Humidity, the amount of water vapor in the air, also affects evaporation rates. For evaporation to occur, the humidity of the atmosphere must be less than the evaporation surface. At 100% relative humidity, there is no evaporation. Additionally, wind increases the rate of evaporation by blowing away moist air from the water surface, thus bringing in less humid air with room for more water molecules.
• Evaporation Investigation handout [1 per student] • Small, sandwich-sized, clear plastic storage containers, approximately 6.5” x 6.5” [10 per class] • Permanent markers • Masking tape • Metric rulers • Soil, well mixed (from the schoolyard or other location or bagged potting soil) • Trowels [1-4] • 100-mL graduated cylinders [5-10] • Metric scales (Figure 1) [2-10] • Heat lamps, can be metal clamp lamps or desktop lamps (Figure 2) [2 per class] • Heating pads (Figure 2) [2 per class] • Small desktop fans (Figure 2) [2 per class] • Dried Spanish moss [1-2 bags, approximately 2-L] • Automatic timers [2-6 per class, depending on location of outlets] • Power strips [2-6 per class, depending on location of outlets] • Optional: extension cords (if needed) • Optional: document camera and projector (if using for the “Class Mean Data Table”) • Evaporation Investigation instructional video, optional introduction to the experiment for the educator MATERIALS Figure 1. Example metric scale and container Figure 2. Five experimental containers with treatment materials (left to right): heat lamp, heating pad, desktop fan, Spanish moss, and control © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) D-03 CLIMATE CHANGE AND THE WATER CYCLE EVAPORATION INVESTIGATION PREPARATION 1. If possible, watch the Evaporation Investigation instructional video for an introduction to the experiment.
2. Plan to divide the class into 10 groups. This allows for two replicates of each treatment (control, solar radiation, infrared ground radiation, wind, and ground cover). To save space, you can split each class into only five groups if you have more than one class conducting this experiment. Each class will share their data with other classes, thereby increasing replication.
3. Plan locations for five experimental treatments. Treatment locations can be simple surfaces, such as counters or tables, that are away from direct sunlight. Three locations should be near an outlet.
4. Set up supplies for each treatment. Ideally, the two replicates of each treatment will not be located right next to each other to avoid confounding the results with site-specific factors (e.g., a heating or cooling vent, etc.). However, space constraints may prevent placing the replicate containers far apart.
a. Control (only soil and water): needs a surface away from direct sunlight.
b. Solar radiation (heat lamp): needs a surface away from direct sunlight, near an outlet, and that can accommodate the heat lamp(s). Plug an automatic timer into a nearby outlet, and program the timer to be on during the day and off during the times when classes are not in session. Plug a power strip and/or extension cord (if needed) into the timer. Set up the heat lamp(s), and plug them into the power strip that is connected to the timer.
c. Infrared ground radiation (heating pad): needs a surface away from direct sunlight, near an outlet, and that can accommodate the heating pad(s). Plug an automatic timer into a nearby outlet, and program the timer to be on during the day and off during the times when classes are not in session. Plug a power strip and/or extension cord (if needed) into the timer. Set up one or two heating pads; two small containers may fit on one heating pad, depending on the size of the containers and heating pad. Plug the heating pad(s) into the power strip that is connected to the timer.
d. Wind (fan): needs a surface away from direct sunlight, near an outlet, and that can accommodate the fan(s). Plug an automatic timer into a nearby outlet, and program the timer to be on during the day and off during the times when classes are not in session. Plug a power strip and/or extension cord (if needed) into the timer. Set up the fan(s), and plug into the power strip that is connected to the timer.
e. Ground cover (moss): needs a surface away from direct sunlight. Place the bag of Spanish moss nearby. Note: automatic timers can be shared among treatments, depending on the location of outlets and surface space, by using extension cords and/or power strips.
5. Place scales on level surfaces around the room that will be convenient for student access.
6. Place the container of soil in a central place that will be convenient for student access, and leave 1 – 4 trowels near the container.
7. Attach a small strip of masking or lab tape to the side of each container to be used as a label.
8. On the board or a large piece of paper, draw the “Class Mean Data Table” from page 3 of the handout, or prepare to show it with a document camera.
PROCEDURES DAY 1 1. Divide the class into 10 groups.
2. Pass out an Evaporation Investigation handout to each student.
3. Explain that students will conduct an experiment to investigate evaporation because it is an important process in the water cycle. As the climate changes and temperature increases, evaporation will occur at a greater rate. In this experiment, we will examine some of the factors that affect evaporation and hypothesize about how the results relate to climate change.
4. Inform students that they will be determining which variable results in the most water loss through evaporation. Their experiment will examine the effects of the following variables on water loss from a container of soil: control (nothing is added), solar radiation (under a heat lamp), infrared ground radiation (on top of a heating pad), wind (in front of a fan), and ground cover (with moss on top).
5. Instruct students to complete the prediction at the top of page 1 of the handout.
6. Give each group a plastic container, a permanent marker, and a ruler, and have them measure 2 cm from the bottom of the container and place a mark with the marker.
7. Assign each group to one of the treatments: control, solar radiation, infrared ground radiation, wind, or ground cover. Assign two groups to each treatment. The first group assigned to the control treatment will be group 1 and the second will be group 2. If you are conducting the experiment with more than one class, assign sequential numbers (3, 4, etc.) to groups in subsequent classes.
8. Instruct students to write their treatment name and group number on page 3 of the handout and on the masking tape on their container.
9. Have each group fill the container with dry soil to the 2 cm mark. Tell them not to pat down the soil. They will then take the mass of the container with the dry soil and record the mass on page 3 of the handout.
a. Soil for the entire class should © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) D-04 CLIMATE CHANGE AND THE WATER CYCLE EVAPORATION INVESTIGATION © Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) be very well mixed in a large container. Groups should not collect their own soil because it may differ considerably in texture.
10. Ask each group to measure 100 mL of water in a graduated cylinder and carefully sprinkle the water on top of the soil in the container. Tell students to make sure that water is sprinkled evenly across the top and not poured into one area of the container.
11. Instruct each group to take the mass of the container immediately after sprinkling the water. Students will record the mass on page 3 of the handout.
a. The mass of 100 mL of water is 100 g. A quick check of student handouts should show that the mass of the container after sprinkling water is 100 g more than the mass of the container with dry soil.
12. Explain to each group how to set up their container.
a. Control: do not add any other variables.
b. Solar radiation (heat lamp): place the container under the heat lamp, approximately 3-6 inches away. Rotate the container 180° daily. The lamp simulates radiation from the sun.
c. Infrared ground radiation (heating pad): place the container on top of the heating pad. Rotate the container daily. The heating pad simulates ground surface radiation.
d. Wind (fan): set the container in front of the fan, and position the fan so that it blows level with the container. Put the fan on the lowest setting. Make sure that no other containers are in the line of the fan. Rotate the container daily. The fan simulates the wind.
e. Ground cover (moss): place moss on top of the soil in patches. Do not cover the soil completely. Make sure all containers have the same percentage of the surface covered. The moss simulates plants covering the soil surface.
DAYS 2 - 4 1. On a daily basis for the following three days, instruct students to take the mass of the container and record it in “Your Group’s Data Table” on page 3 of the handout. If possible, take the mass at the same time each day.
2. Direct students to calculate the amount of water lost by subtracting the current day’s mass from the previous day’s mass.
DAY 4 1. At the end of the experiment, have students calculate the total mass of water lost in the table on page 3 of the handout by adding the water lost on days two, three, and four.
2. Collect the total mass of water loss for each group in the “Class Mean Data Table” on the board (or displayed on the document camera). Either direct students to write their data on the displayed table themselves, or ask each group for their total verbally, and write it in the table.
3. Instruct students to write the class data in the “Class Mean Data Table” on page 3 of the handout and calculate the mean mass of water loss for each treatment.
4. Have students make a bar graph of the mean mass of water loss for each treatment in the graph on page 4 of the handout.
5. Students draw conclusions from the “Class Mean Data Table” and bar graph and use them to answer the conclusion questions on page 5 of the handout. It may be helpful to lead a discussion of experiment conclusions using the following questions and direct students to use some of the answers to write their own conclusions for conclusion question 2.
a. Which treatments had the most evaporation?
b. Which treatments had the least evaporation?
c. Do the results support your prediction? (This is conclusion question 1.) d. What may be the reasons that some treatments had more evaporation than others?
e. What might be the result of two or more of these variables acting at the same time in an ecosystem?
f. Why is it important to replicate the experiment (i.e., to have more than one container for each treatment)?
6. After students answer conclusion question 3, discuss how increased evaporation rates affect the water cycle under climate change conditions. Earth is experiencing global warming, and the average global surface temperature is increasing. In this experiment, global warming is equivalent to turning up the heat on the heating pad. As temperatures increase around the world, more areas are experiencing increased evaporation rates, and there is more water in the atmosphere. More water in the atmosphere results in changes to precipitation patterns; some areas receive more precipitation and some receive less. It also results in increased frequency of extreme weather events, such as severe storms. In addition, water vapor is a greenhouse gas and further enhances the greenhouse effect.
EXTENSIONS 1. For older or advanced students, conduct this activity as a student-directed experiment instead of providing students with the procedures. Present the following question to students, “Which factor has the greatest impact on the rate of evaporation?” Explain the available materials, and ask students to develop a hypothesis and design an experiment that will test it. Have students design the needed data tables and determine how to analyze their results if possible.
2. Challenge students to think of at least one more variable that may affect the rate of evaporation and design and conduct an experiment to test its effects.
© Southwest Regional Climate Hub · Developed by the Asombro Institute for Science Education (www.asombro.org) D-05 CLIMATE CHANGE AND THE WATER CYCLE EVAPORATION INVESTIGATION ADDITIONAL RESOURCES Website with information about the effects of climate change on the processes of the water cycle: Environmental Protection Agency (EPA), Climate Change and Water. Water Impacts of Climate Change. Updated 13 Mar. 2014. Web. Accessed 26 May 2015. |
190183 | https://www.pbsciences.org/fulltext/8-1486104927.pdf?1707300580 | 47 Journal of Mood Disorders V olume: 7, Number: 1, 2017 - www.jmood.org
Case Report DOI: 10.5455/jmood.20170204031431
Onychophagia (Nail Biting): A Body Focused Repetitive Behavior due to Psychiatric Co-morbidity Javed Ather Siddiqui 1, Shazia Farheen Qureshi 1, Waseem M Marei 2, Talal Abdullah Mahfouz 31Psychiatrist, 2Consultant Psychiatrist, 3Consultant Psychiatrist and Head, Mental Health Hospital, Department of Psychiatry, TAIF
Corresponding Author:
Javed Ather Siddiqui, Mental Health Hospital, Department of Psychiatry, TAIF P.O.BOX.2056, TAIF -21944 (KSA)
E-ma il add ress:
javedsiddiqui2000@gmail.com
Date of received:
February 3, 2017
Da te of ac cep tan ce:
February 14, 2017
Declaration of interest:
J.A.S., S.F.Q., W.M.M., T.A.M.: The authors reported no conflicts of interest related to this article.
ABS TRACT:
Onychophagia (nail biting): a body focused repetitive behavior due to psychiatric co-morbidity Onychophagia is an habit of biting one’s nails and finger tips. It is also called nail biting (NB). It is a stress relieving oral habit adopted by many children and adults. People usually do it when they are nervous, stressed, hungry or bored. All the above situations are having a common phenomenon between them which is anxiety. Here, we present a case of onychophagia (NB), who had psychiatric comorbidity. Onychophagia cannot be managed without considering some related factors such as comorbidities, precedent and consequences of the behavior. The best way to treat a nail biter is to educate them, encourage good habits and should provide emotional support and encouragement. Behavior modification therapy has proved to be a successful means of treatment along with drug management.
Keywords: onychophagia, nail biting, multidisciplinary approach, psychiatric comorbidity
Journal of Mood Disorders (JMOOD) 2017;7(1):47-9
INTRODUCTION
Onychophagia (NB) is a behavior with a spectrum. It is characterized by putting the nail into the mouth in such a manner that contact occurs between a finger nail and one or more teeth. This could also it results in physical damage and is considered as a self-mutilative behavior (1,2). It is a stress removing habit common among children and adolescents. It is a habit of biting one’s nails and described as a habit to release tension, self-mutilation behavior, or an impulse control disorder. In the he Diagnostic and Statistical Manual of Mental Disorders Fifth Edition (DSM–5), NB is classified as an obsessive-compulsive and related disorder (3). NB has been less published area in the literature of both psychiatry and dermatology (4,5) also medicine, psychology, and dentistry but have been unable to resolve the problem (6).
There is lot of controversies about the causes of NB. Research has produced theories about NB that includes it being a sign of self-hostility that result in self-mutilation. While some studies relate NB to behavioral problem (7), some theory portrays NB as a symptom of anxiety or a nervous habit (8,9). This condition is believed to be induced by nervousness, stress, or boredom and can be tied to emotional and psychological issues. Anxiety in children with NB is not a trait; it is a state (10). The trait which is accompanied with NB is oral aggression (10). Recent studies did not support the anxiety theory to NB (11), but supported that two of the largest contributing factors to NB were boredom and frustration.
NB usually does not start until age of 3 or 4 years. The prevalence of NB increases from childhood to adolescence, and then decrease in adulthood (6). Studies have found that between 28 percent and 33 percent of children between 48 Journal of Mood Disorders V olume: 7, Number: 1, 2017 - www.jmood.org Onychophagia (nail biting): a body focused repetitive behavior due to psychiatric co-morbidity
ages 7 and 10 years, 44 percent of adolescents, and 19 to 29 percent of young adults engage in NB (12). It is not clear what percentage of the children with NB behavior stops it, and will not suffer from it later. NB is age-related, and its prevalence decrease with the increase of age (13). It is observed significantly higher in boys than girls as nail biters (14). However, NB, especially its benign forms, can also present without any accompanying psychiatric disorders but most of the comorbid psychiatric disorders in these are attention deficit hyperactivity disorder (74.6%), oppositional defiant disorder (36%), separation anxiety disorder (20.6%), enuresis (15.6 %), tic disorder (12.7%), and obsessive compulsive disorder (11.1%). The rates of major depressive disorder, mental retardation, and pervasive developmental disorder were 6.7, 9.5, and 3.2 percent respectively (15). The only study that investigated the parents of children with NB reported that about 56.8 percent of the mothers and 45.9 percent of the fathers were suffering from at least one psychiatric disorder. The most common psychiatric disorder found in these parents was major depressive disorder (15). More than half of the children with nail biting (65.1%) had at least one stereotypic behavior. The most common co-morbid stereotypic body focused repetitive behaviors were lip biting (33.3%) and head banging (12.7%) (15). NB in adults is under-recognized because patients often fail to seek psychiatric help due to feeling of shame and embarrassment, and, consequently the disorder has received little attention in the psychiatric literature.
In most cases, NB seems to be only a cosmetic issue, and no treatment is required. But, in severe cases need pharmacological treatment but it is sparse and limited to anti-depressant such as fluoxetine (16) and clomipramine (17) and non-pharmacological approach, best way to treat a nail biter is to educate them, stimulate good habits, develop conscious awareness, and thus guarantee effective results. During treatment, the child should be given emotional support and encouragement. A multidisciplinary approach should focus on efforts to build up the child’s self confidence and self-esteem (6), sometimes also needs cognitive behavioral techniques for management of children’s behaviors, such as using habit reversal techniques. Any treatment should be accompanied by educating the afflicted children as well as their parents, siblings, and teachers.
Case Description A 20-year old, well-built and socially active male student was referred by his family physicians for psychiatric evaluation. He has no past history of psychiatric illness but had family history of depression in his father. He is a 12 th grade high school student. His family has high expectation from him to get admission into professional field.
On psychiatric examination; he had depressed symptoms and anxiety features for one year, when he was in 11 th grade. He was always ruminating how he would meet his family’s expectations. Therefore, he was unable to concentrate when studying, became anxious and sad, and started nail biting. Overall examination revealed habit of NB whenever he was in stress. Although initially, this habit was associated with his reaction to a stressful condition, gradually it had become more habitual; wherein he would revert back to his habit, whenever he would be in process of thinking and feeling satisfaction after nail biting, sometime he injured his fingers. The diagnosis of onychophagia (NB) was made on basis of structured clinical interview according to DSM-5 criteria and also comorbid diagnosis depression was made. There was no personality traits observed. His routine laboratory finding, including complete blood cell count, serum electrolyte, and renal and liver function tests, were all within normal range.
He was started receiving antidepressant fluoxetine 20 mg per day, and we referred him to our clinical psychologist for behavioral therapy to start cognitive behavioral therapy (CBT). CBT is a common type of psychotherapy. This is done to change behavior by becoming aware of negative emotions and related habits so that they can be dealt with more effective ways. It may involve one of the following methods or combination of them; competing response: with this technique, the person is provided with an alternative to nail biting such as chewing gum to satisfy orally-motivated urges, habit reversal therapy (HRT): this four-step process teaches an individual how to breath and feel grounded, achieve relaxation, and to complete muscle-response exercises, this self-control intervention tried to build self-confidence and self-esteem. Stimulus control: a behavioral treatment that helps to identify, get rid of, or transform the environmental circumstances, or emotions that trigger nail biting. The goal of this therapy is to control triggers through conscious behavior modification, and to channel the unhealthy urges into behaviors that are non-destructive also self-monitoring: since this is often an unconscious act, taking notes can create more awareness of 49 Journal of Mood Disorders V olume: 7, Number: 1, 2017 - www.jmood.org Siddiqui JA, Qureshi SF, Marei WM, Mahfouz TA
the behavior. many sessions of CBT were done and advised for his follow-up after two weeks but patient attended after one month with continuing his treatment and attended all session of psychotherapy. His mood was lifted resulting in a decrease in the habit of NB.
DISCUSSION
NB is a habit that cannot be managed without considering some related factors such as comorbidities, precedent and consequences of the behavior, and management is much more complicated without non-pharmacological approach.
Here, this case report has psychiatric comorbidity like depression. Patient has lot of expectation from parents, due to which he is always preoccupied in thought and in stressful situation; he used to bite nail and feeling relaxed after nail biting. In such cases, need to treat the patient, to educate them, stimulate good habits and develop conscious awareness. It is also very important patient should be given emotional support and encouragement. Multidisciplinary approach should focus on efforts to build up the child’s self-confidence and self-esteem. Punishment is not effective and its effect is not more than placebo.
CONCLUSION
Nail biting is not an isolated symptom. It can be one symptom from a cluster of symptoms, all of which as well as the motivation behind NB should be evaluated, assessed, and managed. Behavioral modification techniques, positive reinforcement, and regular follow-ups are important for the treatment of nail biting or onychophagia with multidisciplinary approach, when necessary.
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American Psychiatric Association. Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition. Washington, DC: American Psychiatric Association; 2013. 4. Bohne A, Keuthen N, Wilhelm S. Pathologic hairpulling, skin picking, and nail biting. Ann Clin Psychiatry. 2005;17:227-32.
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Foster LG. Nervous habits and stereotyped behaviors in preschool children. J Am Acad Child Adolesc Psychiatry. 1998;37:711-7.
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Leonard HL, Lenane MC, Swedo SE, Rettew DC, Rapoport JL. A double-blind comparison of clomipramine and desipramine treatment of severe onychophagia (nail biting). Arch Gen Psychiatry. 1991;48:821-7. [CrossRef] |
190184 | https://cms.cern/book/export/html/1618 | Detector
About CMS
The 27-km Large Hadron Collider (LHC) is the largest and most powerful particle accelerator ever built. It accelerates protons to nearly the velocity of light -- in clockwise and anti-clockwise directions -- and then collides them at four locations around its ring. At these points, the energy of the particle collisions gets transformed into mass, spraying particles in all directions.
The Compact Muon Solenoid (or CMS) detector sits at one of these four collision points. It is a general-purpose detector; that is, it is designed to observe any new physics phenomena that the LHC might reveal.
CMS acts as a giant, high-speed camera, taking 3D “photographs” of particle collisions from all directions up to 40 million times each second. Although most of the particles produced in the collisions are “unstable”, they transform rapidly into stable particles that can be detected by CMS. By identifying (nearly) all the stable particles produced in each collision, measuring their momenta and energies, and then piecing together the information of all these particles like putting together the pieces of a puzzle, the detector can recreate an “image” of the collision for further analysis.
How CMS Works
The 14,000-tonne detector gets its name from the fact that:
The CMS detector is shaped like a cylindrical onion, with several concentric layers of components. These components help prepare “photographs” of each collision event by determining the properties of the particles produced in that particular collision. This is done by:
1. Bending Particles
A powerful magnet is needed to bend charged particles as they fly outwards from the collision point. Bending the trajectories of the particles serves two purposes:
The solenoid magnet, which gives CMS its last name, is formed by a cylindrical coil of superconducting fibres. When electricity (18,500 amps!) is circulated within these coils, they encounter no resistance -- the magic of superconductivity! -- and can generate a magnetic field of around 4 tesla, which is about 100,000 times the strength of the Earth’s magnetic field. This high magnetic field must be confined to the volume of the detector and is done by the steel “yoke” that forms the bulk of the detector’s mass. The magnet coils and its return yoke weigh in at 12,500 tonnes, by far CMS’s heaviest component!
This solenoid is the largest magnet of its type ever constructed and allows the Tracker and Calorimeters (see below) to be placed inside the coil, resulting in a detector that is, overall, “compact”, compared to detectors of similar weight.
[Read more about our Solenoid]
2. Identifying Tracks
Merely bending particles is not enough -- CMS must identify the paths taken by these bent charged particles with a very high precision. This is done by a silicon Tracker made of around 75 million individual electronic sensors arranged in concentric layers. When a charged particle flies through the Tracker layer, it interacts electromagnetically with the silicon and produces a hit -- these individual hits can then be joined together to identify the track of the traversing particle.
[Read more about the Tracker and its components]
3. Measuring Energy
Information about the energies of the various particles produced in each collision is crucial to understanding what occurred at the collision point. This information is collected from two kinds of “calorimeters” in CMS. The Electromagnetic Calorimeter (ECAL) is the inner layer of the two and measures the energy of electrons and photons by stopping them completely. Hadrons, which are composite particles made up of quarks and gluons, fly through the ECAL and are stopped by the outer layer called the Hadron Calorimeter (HCAL).
[Read more about the ECAL] / [Read more about the HCAL]
4. Detecting Muons
The final particle that CMS observe directly is the muon. Muons belong to the same family of particles as the electron, although they are around 200 times heavier. They are not stopped by the calorimeters, so special sub-detectors have to be built to detect them as they traverse CMS. These sub-detectors are interleaved with the return yoke of the solenoid. The large magnet of CMS also allows us to measure each muon’s momentum both inside the superconducting coil (by the tracking devices) and outside of it (by the muon chambers).
[Read about the various muon sub-detectors of CMS]
An unusual feature of the CMS detector is that instead of being built in-situ like the other giant detectors of the LHC experiments, it was constructed in 15 sections at ground level before being lowered into an underground cavern near Cessy in France and then reassembled. Engineers found that building sections above ground, rather than constructing them in the cavern with all its access and safety issues, saved valuable time. Another important conclusion was that sub-detectors should be made more easily accessible to allow for easier and faster maintenance.
Thus CMS was designed in fifteen separate sections or “slices” that were built on the surface and lowered down ready-made into the cavern. Being able to work in parallel on excavating the cavern and building the detector saved valuable time. This slicing, along with the careful design of cabling and piping, also ensures that the sections can be fully opened and closed with minimum disruption, and each piece remains accessible within the cavern.
These considerations, along with the unique conditions of the LHC, affected the design of each layer of the detector.
3D Print your own CMS detector!
Find the files to 3D print your own version of the detector here in full detail!
If you need something a little smaller and simpler, try these mini detectors that can be printed in one or four colours:
Build your own CMS detector out of lego!
Originally designed by a team at the University of Maryland, you can find an article about the model here, full instructions here, and a detailed description here.
See the CMS detector in 3D
Download the mobile application Alaro and scan the photo below. Find instructions on how to do it here. You can also print this photo from here and scan it in a similar way.
CMS Detector Design
Detectors consist of layers of material that exploit the different properties of particles to catch and measure the energy and momentum of each one. CMS needed:
With these priorities in mind, the first essential item was a very strong magnet. The higher a charged particle’s momentum, the less its path is curved in the magnetic field, so when we know its path we can measure its momentum. A strong magnet was therefore needed to allow us to accurately measure even the very high momentum particles, such as muons. A large magnet also allowed for a number of layers of muon detectors within the magnetic field, so momentum could be measured both inside the coil (by the tracking devices) and outside of the coil (by the muon chambers).
The magnet is the “Solenoid” in Compact Muon Solenoid (CMS). The solenoid is a coil of superconducting wire that creates a magnetic field when electricity flows through it; in CMS the solenoid has an overall length of 13m and a diameter of 7m, and a magnetic field about 100,000 times stronger than that of the Earth. It is the largest magnet of its type ever constructed and allows the tracker and calorimeter detectors to be placed inside the coil, resulting in a detector that is, overall, “compact”, compared to detectors of similar weight.
The design of the whole detector was also inspired by lessons learnt from previous CERN experiments at LEP (the Large Electron Positron Collider). Engineers found that building sections above ground, rather than constructing them in the cavern with all its access and safety issues, saved valuable time. Another important conclusion was that sub-detectors should be made more easily accessible to allow for easier and faster maintenance.
Thus CMS was designed in fifteen separate sections or “slices” that were built on the surface and lowered down ready-made into the cavern. Being able to work in parallel on excavating the cavern and building the detector saved valuable time. This slicing, along with the careful design of cabling and piping, also ensures that the sections can be fully opened and closed with minimum disruption, and each piece remains accessible within the cavern.
These considerations, along with the unique conditions of the LHC, affected the design of each layer of the detector. To read about the design of each sub-detector, click on the links in the left-hand menu.
Civil Engineering
The CMS cavern, at Point 5 of the LHC near Cessy, was excavated from scratch in an old LEP (a previous CERN accelerator, the Large Electron Positron Collider) access point. The work, which took six and a half years, finishing in February 2005, consisted of creating two caverns 100m underground, including the 53-metre long, 27-metre wide and 24-metre high experiment cavern, as well as two new shafts.
CMS is like a cylindrical onion built around the beam pipe, but cut up into 15 slices. The experiment employed a unique method of construction: assembling and testing each of the slices of detector on the surface before lowering them underground, rather than building the components within the cavern itself. This meant having to transport and lower huge but delicate pieces of detector, weighing as much as 2000 tonnes, down a 100-metre shaft but also gave a number of advantages. Not only did it enable civil engineering work on the cavern to go on in parallel with detector construction, but also the ‘slicing’ means that each piece remains accessible for ease of maintenance within the cavern.
Because of this method, CMS was in a unique position to be able to excavate where geology was difficult, without any delays hampering the progress of the detector itself. And at Cessy the ground is indeed difficult: 75 metres of “moraine”, a glacial mixture of sand and gravel containing two water tables, followed by “molasse” rock, a form of soft sandstone. To excavate through this loose earth and soft rock and to build a cavern without it collapsing was an enormous challenge. To find out how they did it, click here
Facts about CMS civil engineering…
CMS assembly
The decision to employ a method that involves hoisting thousands of tonnes of expensive machinery, the result of endless hours of work, down a 100 metre gaping hole was never going to be taken lightly. But CMS’s unique method of slicing up the detector and lowering each piece into the cavern ready-made, has been nothing short of a success.
“It’s a huge amount of responsibility; you have all the physicists waiting for an element to arrive, and if you make a serious mistake, it could be the end of CMS,” says Hubert Gerwig, whose task it was to ensure each sub-detector’s safe arrival.
But lowering CMS by means of heavy lifting was a decision taken at the very beginning, some 16 years ago, inspired by experiences with LEP. “The concept of building large objects on the surface and transferring them completed to the underground area was the clear way to go,” says Alain Hervé, CMS’s original technical coordinator. Though not in LEP’s initial plans, success in lowering two pieces into position in L3, weighing 300 and 350 tonnes, suggested the method could work.
Being able to work in parallel on the civil engineering and the detectors was clearly a huge advantage, and the ‘slicing’ also meant that, through complicated loops of cable chains, water, gas and cooling leads, each piece remained accessible within the cavern, a contrast with experiments built in a ‘Russian dolls’ style, from the inside, out.
But the implications of using this method needed careful consideration too: “As soon as I saw the sketch I had a clear vision of how such an experiment had to be organized,” says Alain. “We were already two years inside LEP running, and lessons from the construction and the first shut-downs were clear to me. The way the experiment had to be sectioned and installed was the first priority, not developing the design,” says Alain.
And so the experimental area, pits and surface hall all had to be designed around the lowering method. Foundations for the gantry and a massive strong plug to hold the pieces over the shaft before the cranes took the strain were incorporated. Anchor points for lifting were integrated in the design of the yoke as well as HCAL (Hadron Calorimeter) and HF (Forward Calorimeter) platforms and cradles.
The first lowering took place in November 2006, and the last on 22 January 2008. “Everything has been calculated and calculated again, but that’s still not the real thing,” says Hubert. “In the end we were successful and nothing was damaged. So there is a sense of relief; it’s a bit like an exam, you feel better once it is over and you can celebrate!”
Heavy lowering
Trials and Tribulations of Heavy Lowering
Ten companies internationally were capable of taking on such a project as CMS, but VSL Heavy Lifting, a Swiss company, were the eventual winners of the 2 million CHF contract.
Though VSL were used to lifting such heavy loads as the 7500-tonne roof of the Airbus Assembly Hall, and the walkway between the Kuala Lumpur Twin Towers, CMS presented unique challenges. For a start, the equipment needed to lower, not lift, and a long way. Adding to this was the fact that the scientific cargo was far more delicate than its industrial counterparts.
The gantry support system was rented and made from existing towers, but the 3.4m high horizontal beam, that took the strain of the load, had to be custom made. Even such an enormous beam still bent 5cm under the weight of the heaviest element, the 2000 tonne central barrel ring known as YB0.
During each lowering, the element is supported at four points, but by many more cables as each bunch is made up of 55 individual 15.7mm diameter strands of cold-drawn steel, each with a capacity to hold 28.4 tonnes. Within each bundle, half of the strands are twisted one way and half the other to make overall turning force zero and preventing any turning of the load during the descent.
The lowering in fact happens in 200 small steps as the cables are fed through the hydraulic system half a metre at a time. This works on a grip and release basis, similar to your own hands lowering a bucket down a well, and at all times the load is safely supported by one or other of the clamping mechanisms. Once the cargo reaches the ground, it comes to a soft landing on hovercraft-like airpads that take the load off the cables and later on allow easy horizontal movement within the experiment hall.
Each lowering was a big event, requiring a week’s preparation to manoeuvre the plug, element and attachments in place, and each was a success. Though there were also, of course, a few challenges that arose along the way: YB0, for instance, the largest segment to enter the cavern, at one point cleared the sides of the shaft by just 10cm either side, and the team needed surveillance video cameras to carefully monitor its movements and ensure that the element passed the bottleneck safely.
The slight slope in the CMS cavern presented another issue. Whilst unnoticeable underfoot, the 1.23% gradient meant that when one side of any element touched down, the other still had 3cm to go. “Initially we tried to compensate for the slope, so that there was an even landing,” explains Hubert Gerwig, the engineer in charge of heavy lifting, “but soon we realised it wasn’t necessary because under the load, the metal strands actually elongate by 10cm, making them somewhat elastic.” The elasticity meant they could simply carry on lowering the piece until both ends were on the ground, also giving rise to the slight “bouncing” effect visible as each piece finally came to rest on the experiment hall floor.
That some of the elements couldn’t be supported on just four attachment points presented another issue and the endcap disks , for example, were instead carried down on a special crate. But being supported nine metres below their centre of gravity made them naturally unstable. The team solved this dilemma by using stabilisers at the top of the endcap such that if it tilted, it would simply lean against the lowering cables.
In the end each slice of CMS found its way to the experiment hall safely, thanks to the hard work and dedication of all involved, and the group and their equipment can now be disbanded. The VSL team will now take their gantries and hydraulic equipment to Durban in South Africa to build a stadium for the 2010 World Cup. Some parts, such as the cables that have carried down every piece of CMS, are now too fatigued to be used again. Finishing the lowering is a sad event for some, but the VSL mechanics like to travel the world and are eager to move on. “I think some of the workers get the opposite of home sickness,” says Hubert, “they cannot wait to get away again!” And we wish them good luck.
Massive underground excavation
Between 1998 and 2005 a total of almost 250,000 m3 of soil and rock was excavated from the CMS site at “Point 5” of the LHC, in Cessy. The contractors - Seli, an Italian company and Dragados, Spanish - had previously worked on the Madrid metro but here faced some unique surprises.
It started with an unexpected find: “We did trial pits around the site, because the archaeology of an area is always something we have to consider, and what we found was great – a Roman villa, complete with pots, tiles and coins,” explains John Osborne, project manager of CMS civil engineering. “And even better for us was that it wasn’t anywhere near any planned shafts or buildings.”
They were able to leave archaeologists to spend a year thoroughly investigating the site whilst the project got underway. But it wasn’t an easy task. “Funnily enough Point 5 was one of he worst places possible to build the cavern in terms of geology, though best from a physics point of view. And CMS was pretty much the only experiment that could cope with being built here,” remarks John.
The first challenge came in the form of underground water tables approximately ten to twenty metres below the surface. To dig down through them the team had to first freeze the ground around the shafts to act as a barrier to the water. By drilling holes around each shaft’s circumference and pumping down brine, cooled to -25oC, the water froze into a 3-metre wall of ice. But the water coming from Cessy was moving even faster than predicted, and combined with the channelling effect of water between the two shafts, pressure built up until it penetrated the walls.
Injecting the holes with an even colder substance: liquid nitrogen, at less than at -195oC, finally solved the problem. With its help engineers formed a wall of ice around the shafts that was solid-enough for teams to keep on digging.
However, the problems were not over yet. Once the shafts were dug out, work had to begin on the caverns, but because the ground materials were so soft, with no intervention any cavern excavated would collapse. “At Point 5 there is only 15 metres of solid rock. For the first 75 metres of digging down it’s just a type of glacial deposit called moraine - basically a mixture of sand and gravel,” explains John. “And the rock we did have was a kind of soft sandstone called Molasse. A large cavern built in this would just collapse.”
The solution was to build a large supporting structure underground that could hold up the caverns and withstand the mass of the soil above it. Engineers envisaged a concrete pillar as the divider between the two caverns that could do just this.
“Knowing we would have to build this structure anyway, we asked radiation teams how thick we’d need it to be to protect people from radiation in the cavern next-door to the experiment,” explains John. “They gave the figure at 7 metres. The width needed to ensure adequate support for the caverns was 7.2 metres, so this worked out very well, and in fact the second cavern can now be safely used when the machine is on.”
After these delays, engineers experimented with using explosives to clear away rock at a faster rate, but in the end it had little benefit for a big disruption. Instead the best trick seemed to be to excavate small sections at a time and immediately install “shotcrete”, a sprayed concrete that sets as soon as it hits the walls, and drilling steel anchors that reached 12 metres into surrounding rock: “If we didn’t do this, the whole thing would collapse whilst we were building it. We constantly monitored any movement with a host of instruments and adjusted the support as necessary.”
For the workers on the site, another exciting yet daunting nature of working on the excavation was descending into the CMS cavern in the lift. During the construction period, access underground was via a lift cage being lowered down the shaft on a system of ropes. “This was good as it meant you could always be lowered to exactly the right level, as opposed to fixed lifts,” explains John. “But being on a rope also meant that it had a tendency to sway as you went down. And 100 metres is a long way!”
As the environment 100 metres below the surface is also full of water, the final stages involved waterproofing, installing drainage systems and painting the cavern, as water could otherwise turn soft rock into mud. “Once everything was in place we could seal off the cavern with waterproofing and put in a permanent concrete wall up to 4 metres thick, reinforced with steel bars.”
Throughout the whole construction, consideration for the environment was at the forefront of people’s minds. In order to keep noise levels to a minimum, sound barriers were built all around the site. And instead of removing the many tonnes of earth and rock excavated from the site, causing noise and road disruption, it was deposited right by the buildings, covered with fresh soil and planted with vegetation, creating a now flourishing artificial hill.
Whilst this was going on, thanks to the “pre-packaged” design of CMS, split up into 15 slices, where each is assembled and as near to complete as possible on the surface, work on building the detector could carry on this whole time. And once the cavern was complete, in 2005, pieces of CMS could be lowered underground and installed.
But though the civil engineering stage of building CMS is long over, the underground nature of the experiment must always be taken into account. Though the detector weighs almost as much as the soil and rock it replaced, the caverns within the spongy earth are like bubbles in water, and they could potentially rise by as much as one millimetre (mm) per year. Geologists predict that CMS, along with the immediately surrounding sections of the LHC machine, will rise 15 mm over 10 years. This may seem small, but when you think that the detector’s tracker and muon systems, for instance, must align to within 0.15 mm, one hundredth of the distance, this small amount becomes significant. In fact the ability to adjust by this amount is built into the detector and movement will be monitored throughout the life of the experiment, so civil engineering continues to play a part in the running of CMS.
Computing Grid
Even when whittled down by the trigger system, CMS still produces a huge amount of data that must be analysed, more than five petabytes per year when running at peak performance. To meet this challenge, the LHC employs a novel computing system, a distributed computing and data storage infrastructure called the Worldwide LHC Computing Grid (WLCG). In ‘The Grid’, tens of thousands of standard PCs collaborate worldwide to have much more processing capacity than could be achieved by a single supercomputer, giving access to data to thousands of scientists all over the world.
The “Tier 0” centre at CERN first reconstructs the full collision events and analysts start to look for patterns; but the data has a long way to go yet. Once CERN has made a primary backup of the data it is then sent to large “Tier 1” computer centres in seven locations around the world: in France, Germany, Italy, Spain, Taiwan, the UK and the US. Here events are reconstructed again, using information from the experiment to improve calculations using refined calibration constants.
Tier 1 starts to interpret and make sense of the particle events and collate the results to see patterns emerging. Meanwhile each sends the most complex events to a number of “Tier 2” facilities, which total around 40, for further specific analysis tasks. In this way information braches out from each tier across the world so that, on a local level, physicists and students whether in Rio de Janeiro or Oxford, can study CMS data from their own computer, updated on a regular basis by the LHC Computing Grid.
For a more detailed account of CMS Computing see:
CMS: The Computing Project Technical Design Report
Detector Overview
The CMS experiment is 21 m long, 15 m wide and 15 m high, and sits in a cavern that could contain all the residents of Geneva; albeit not comfortably.
The detector is like a giant filter, where each layer is designed to stop, track or measure a different type of particle emerging from proton-proton and heavy ion collisions. Finding the energy and momentum of a particle gives clues to its identity, and particular patterns of particles or “signatures” are indications of new and exciting physics
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The detector is built around a huge solenoid magnet. This takes the form of a cylindrical coil of superconducting cable, cooled to -268.5oC, that generates a magnetic field of 4 Tesla, about 100,000 times that of the Earth.
Particles emerging from collisions first meet a tracker, made entirely of silicon, that charts their positions as they move through the detector, allowing us to measure their momentum. Outside the tracker are calorimeters that measure the energy of particles. In measuring the momentum, the tracker should interfere with the particles as little as possible, whereas the calorimeters are specifically designed to stop the particles in their tracks.
The Electromagnetic Calorimeter (ECAL) - made of lead tungstate, a very dense material that produces light when hit – measures the energy of photons and electrons whereas the Hadron Calorimeter (HCAL) is designed principally to detect any particle made up of quarks (the basic building blocks of protons and neutrons). The size of the magnet allows the tracker and calorimeters to be placed inside its coil, resulting in an overall compact detector.
As the name indicates, CMS is also designed to measure muons. The outer part of the detector, the iron magnet “return yoke”, confines the magnetic field and stops all remaining particles except for muons and neutrinos. The muon tracks are measured by four layers of muon detectors that are interleaved with the iron yoke. The neutrinos escape from CMS undetected, although their presence can be indirectly inferred from the “missing transverse energy” in the event.
Within the LHC, bunches of particles collide up to 40 million times per second, so a “trigger” system that saves only potentially interesting events is essential. This reduces the number recorded from one billion to around 100 per second.
Bending Particles
The CMS magnet is the central device around which the experiment is built, with a 4 Tesla magnetic field that is 100,000 times stronger than the Earth’s.
Its job is to bend the paths of particles emerging from high-energy collisions in the LHC. The more momentum a particle has the less its path is curved by the magnetic field, so tracing its path gives a measure of momentum. CMS began with the aim of having the strongest magnet possible because a higher strength field bends paths more and, combined with high-precision position measurements in the tracker and muon detectors, this allows accurate measurement of the momentum of even high-energy particles.
The CMS magnet is a “solenoid” - a magnet made of coils of wire that produce a uniform magnetic field when electricity flows through them. The CMS magnet is “superconducting”, allowing electricity to flow without resistance and creating a powerful magnetic field. In fact at ordinary temperatures the strongest possible magnet has only half the strength of the CMS solenoid.
The tracker and calorimeter detectors (ECAL and HCAL) fit snugly inside the magnet coil whilst the muon detectors are interleaved with a 12-sided iron structure that surrounds the magnet coils and contains and guides the field. Made up of three layers this “return yoke” reaches out 14 metres in diameter and also acts as a filter, allowing through only muons and weakly interacting particles such as neutrinos. The enormous magnet also provides most of the experiment’s structural support, and must be very strong itself to withstand the forces of its own magnetic field.
The CMS magnet…
How big can you build a magnet?
More coils give a stronger field, a stronger field gives more precise results, and with more precise results we can spot more physics. But whilst getting the best magnetic field possible was the most important consideration in designing the detector, its size was also limited. For the sake of efficiency the magnet was to be built offsite and transported to CMS by road, which meant it physically could not be more than 7 metres in diameter, else it would not fit through the streets on its way to Cessy.
Visit the link for magnet updates:
Detecting Muons
As the name “Compact Muon Solenoid” suggests, detecting muons is one of the most important tasks of CMS. Muons are charged particles that are just like electrons and positrons, but 200 times heavier. We expect them to be produced in the decay of a number of potential new particles; for instance, one of the clearest "signatures" of the Higgs Boson is its decay into four muons.
Because muons can penetrate several metres of material losing little energy, unlike most particles, they are not stopped by any of CMS calorimeters. Therefore, chambers to detect muons are placed in the outer part of the experiment where they are the only particles likely to produce a clear signal.
A particle is measured by fitting a curve to the hits registered in the four muon stations, which are located outside of the magnet coil, interleaved with iron "return yoke" plates. The particle path is measured by tracking its position through the multiple active layers of each station; for improved precision, this information is combined with the CMS silicon tracker measurements. Measuring the trajectory provides a measurement of particle momentum. Indeed, the strong magnetic field generated by the CMS solenoid bends the particle's trajectory, with a bending radius that depends on its momentum: the more straight the track, the higher the momentum.
In total there are 1400 muon chambers: 250 drift tubes (DTs) and 540 cathode strip chambers (CSCs) track the particle positions and provide a trigger, while 610 resistive plate chambers (RPCs) and 72 gas electron multiplier chambers (GEMs) form a redundant trigger system, which quickly decides to keep or discard the acquired muon data. Because of the many layers of detector and different specialities of each type, the system is naturally robust and able to filter out background noise.
A muon, in the plane perpendicular to the LHC beams, leaves a curved trajectory in four layers of muon detectors ("stations")
DTs and square-shaped RPCs are arranged in concentric cylinders around the beam line (‘the barrel region’), whilst CSCs, trapezoidal RPCs, and GEMs form the ‘endcap’ disks that “close” the ends of the barrel.
The muon system…
Making Tough Chambers
Most of the muon chambers were built in different laboratories around the world before being shipped all the way to CERN, so the chambers needed to be robust as they are meant to act as precision detectors. Their mechanical robustness and performance was extensively tested in several phases of prototyping, design, test beams, assembly and commissioning, both at the respective labs and at CERN.
As a demonstration of their robustness, when physicists overseeing the beam test decided to reposition a chamber, one side of its support structure suddenly fell by more than 30 cm. Such an unprecedented test of robustness had the physicists worry about the fate of that unlucky chamber. Instead, they found it was still "taking data" in spite of its hard drop.
Visit the link for muon system updates:
Cathode Strip Chambers
Cathode strip chambers (CSC) are used in the endcap disks where the magnetic field is uneven and particle rates are higher than in the barrel.
CSCs consist of arrays of positively-charged anode wires crossed with negatively-charged copper cathode strips within a gas volume. When muons pass through, they knock electrons off the gas atoms, which flock to the anode wires creating an avalanche of electrons. Positive ions move away from the wire and towards the copper cathode, also inducing a charge pulse in the strips, at right angles to the wire direction.
Because the strips and the wires are perpendicular, we get two position coordinates for each passing particle.
In addition to providing precise spatial and timing information, the closely spaced wires make the CSCs fast detectors suitable for triggering. Each CSC module contains six layers making it able to accurately identify muons and match their tracks to those in the tracker.
Visit the link for CSC updates:
GEMs (Gas Electron Multiplier)
Gas electron multipliers (GEMs) are a new addition to the CMS muon system. They complement existing detectors in the forward regions close to the beampipe, where large radiation doses and high event rates will increase during Phase-2 of the LHC. The harsh environment imposes strict requirements on detector characteristics, requiring technologies with high rate capabilities–hence the choice of GEMs. In the endcaps, the GEM system will improve the measurement of the polar muon bending angle, allowing the trigger of the muon system to cope with the high rates. Additionally, GEMs will further extend the muon system coverage in the very forward regions.
Like the other CMS muon subsystems, GEMs are gaseous detectors. Their key feature is the GEM foil, which consists of a 50 micrometer-thick insulating polymer (polyimide) surrounded on the top and bottom with copper conductors. Throughout the foil, microscopic holes are etched in a regular hexagonal pattern. The CMS GEM chambers consist of two PCBs, containing the gas volume, and a stack of three GEM foils in between. The chambers are filled with an Ar/CO2 gas mixture, which is ionized by incident ionizing particles. A potential difference applied across the foils generates sharp electric fields in the holes. The electrons created during the ionisation process drift towards the foils and are multiplied in the holes. The resulting electron avalanche induces a readout signal on the finely spaced strips.
The CMS GEMs are the first GEM chambers with such a large size (an area of about 0.5 m^2) and the largest GEM system ever installed. A first batch of 144 chambers was installed during Long Shutdown 2 on the first disk of the two endcaps. These chambers will contribute to data-taking from Run-3 of the LHC. Additionally, two more disks of GEM chambers will be installed in each endcap during 2024-2026, before Phase-2 of the LHC.
Visit the link for GEM updates:
Muon Drift Tubes
The drift tube (DT) system measures and identifies muon tracks in the barrel part of the detector. Each 4cm-wide tube contains a stretched wire within a gas volume. When a muon or any charged particle passes through the volume, it knocks electrons off the atoms of the gas. These electrons “drift” towards the anode following the electric field, ending up at the positively-charged wire where they are amplified and produce a measurable charge pulse. Each unit or “superlayer” contains four layers of staggered drift cells.
By registering the time taken by the electrons to reach the wire, and since the drift velocity along the cell is a rather constant value, we can identify the exact crossing point of the muon along the cell. Each superlayer provides 4 points in 2D for the muons position.
The sizes of DT chambers range from 2m x 2.5m to 4m x 2.5m, approximately. Each chamber consists of 8 or 12 aluminium layers, arranged in two or three superlayers, each up with up to 90 tubes: the middle superlayer measures the coordinate along the direction parallel to the beam and the two outside superlayers measure the perpendicular coordinate, thus providing a combined 3D measurement of the muon track.
Visit the link for Drift Tubes (DTs) updates:
Resistive Plate Chambers
Resistive Plate Chambers (RPCs) are fast gaseous detectors that provide a muon trigger system parallel to those of the DTs and CSCs.
RPCs consist of two parallel plates, a positively-charged anode and a negatively-charged cathode, both made of a very high resistivity plastic material and separated by a thin gas volume.
When a muon passes through the chamber, electrons are knocked out of the atoms of the gas. These electrons in turn hit other atoms causing an avalanche of electrons. The electrodes are transparent to the signal (the electrons), which are instead picked up by external metallic strips after a small but precise time delay. The pattern of hit strips gives a quick measure of the muon's momentum, which is then used by the trigger to make immediate decisions about whether the data are worth keeping. RPCs combine a good spatial resolution with a time resolution of just one nanosecond (one billionth of a second).
Visit the link for RPC updates:
Energy of Electrons and Photons (ECAL)
In order to build up a picture of events occurring in the LHC, CMS must measure the energies of emerging particles. Of particular interest are electrons and photons, because of their use in discovering the Higgs boson and other new physics.
The energies of electrons and photons are measured using the CMS electromagnetic calorimeter (ECAL). Measuring their energies with the necessary precision in the very strict conditions of the LHC - a high magnetic field, high levels of radiation, and only 25 nanoseconds between collisions - requires dedicated detector materials.
Lead tungstate crystal is made primarily of metal and is heavier than stainless steel, but with a touch of oxygen in this crystalline form, it is highly transparent and “scintillates” when electrons and photons pass through it. This means it produces light in proportion to the impinging particle’s energy. These high-density crystals produce light in fast, short, well-defined photon bursts that allow for a precise, fast, and fairly compact detector.
Photodetectors, which have been especially designed to work within the high magnetic field, are glued onto the back of each of the crystals to detect the scintillation light and convert it to an electrical signal that is amplified and sent for analysis.
The ECAL, made up of a “barrel” section and two “endcaps”, forms a layer between the tracker and the HCAL. The cylindrical barrel consists of 61,200 crystals formed into 36 “supermodules”, each weighing around three tonnes and containing 1700 crystals. The flat endcaps seal off the barrel at each end and are made up of almost 15,000 more crystals.
For extra spatial precision, the ECAL also contains a preshower detector that sits in front of the endcaps. These allow CMS to distinguish between single high-energy photons (often signs of exciting physics) and the less interesting close pairs of low-energy photons.
The CMS ECAL…
CMS ECAL Technical Design Report
Production Story ....
A Russian factory in a former military complex took on the job of producing most of the crystals, while the remainder were produced in China. It took about ten years to grow all 78,000 crystals to stringent specifications, taking around two days to grow each one. Inside the detector, the crystals face high radiation, so the lead tungstate material had to be carefully chosen and developed.
Crystal Calorimeter
CMS uses lead tungstate (PbWO4) for the almost 80,000 crystals: a material with high density that produces scintillation light in fast, small, well-defined photon showers. This means the calorimeter system can be very precise and very compact, fitting within the CMS superconducting magnet. Lead tungstate is also relatively easy to produce from readily available raw materials, and the experience and capability to manufacture them already existed in China and Russia.
Of course, the material also has drawbacks: firstly that the yield of light depends strongly on temperature, a problem given how much heat is released by the readout electronics. To solve the cooling problem a custom-built system maintains the temperature of the 100 tonnes of crystal to within 0.1oC.
Additionally, because the yield of light output is low,the scintillation light [see explanation below] must be captured by photodetectors, converted to an electrical signal, and then amplified. The stronger and now digitised electrical signals are then transmitted through optic fibres. Because the photodetectors also need to be radiation hard and operate within a strong magnetic field, avalanche photodiodes or APDs [see explanation below] were selected for the crystal barrel, and vacuum phototriodes (VPTs) for the endcaps.
Finally, though already fairly radiation resistant after intense research to purify the material, researchers found that lead tungstate would still suffer limited radiation damage inside CMS. However the R&D programme led to a much better understanding of how this damage occurs and found that its main effect is to colour the crystal, affecting how light travels through it.
Knowing this, physicists can ensure that they allow for it in their analysis using a “light monitoring system” during operation that sends pulses of light through each crystal to measure its “optical transmission”. The crystals can reverse radiation damage, or “anneal”, when the accelerator stops (when they are at room temperature) as the warmer temperature naturally shakes atoms back into their ordered structure.
Getting the material right was only one of the challenges for the ECAL team; each crystal had to be cut, machined, polished, tested,and attached to a photodetector. Groups of crystals were then assembled side-by-side in glass-fibre or carbon-fibre “pockets” to form larger structures known as “supercrystals”, “modules”, and “supermodules”.
Crystal Facts
What is scintillation?
When an atom is excited, that is, given energy, an electron goes into a higher orbit and then when it falls back, it releases this energy as a photon. In CMS, when a high-energy electron or photon collides with the heavy nuclei of the ECAL crystals, it generates a shower of electrons, positrons, and photons, and atoms in the material take energy from the passing particles to excite their electrons. The atoms quickly ‘relax’ and the electrons each emit the extra energy as a photon of blue light. A photo device [see below] then picks up these ‘scintillation’ photons and the amount of light generated is proportional to the energy that was deposited in this crystal. This tells us the energy of the incoming electron or photon.
What are photodetectors?
Avalanche photodiodes (APDs) are photodetectors made of semiconducting silicon with a strong electric field applied to them. When a scintillation photon strikes the silicon and knocks an electron out of an atom, the electron is accelerated in the electric field, quickly striking other atoms and knocking electrons off those too. As these are also accelerated, this method will produce an avalanche of electrons with their numbers increasing exponentially. Through this method, APDs are able to produce a very high current in a short time, which is necessary as the lead tungstate crystals give a relatively low light yield for each incident particle. The signal is then amplified, digitised and immediately transported away by fibre optic cables, so that the analysis can be done away from the radiation area.
Crystals in the endcaps use vacuum phototriodes (VPTs) because here the radiation is too high to use silicon photodiodes. As opposed to an APD, a VPT contains three electrodes within a vacuum. Initially, electrons are released when the light strikes atoms in the first electrode in the normal way. Then the voltage difference between electrodes accelerates the electrons into the second electrode (the anode), producing several more electrons, that are all then accelerated to the third (the dynode), releasing a second batch of electrons. Thus this also creates a large current from the initial tiny amount of scintillation light, which is turned into a digital signal and sent along the optic fibres to the upper level readout.
All the photodetectors were glued to the crystals here at CERN, but the finished products are the sum of contributions made by several hundred people and many institutes.
ECAL Preshower
One way the Higgs boson might decay is into high-energy photons and detecting them is one of the ECAL’s primary functions. However, short-lived particles called neutral pions, also produced in collisions, can inadvertently mimic high-energy photons when they decay into two closely-spaced lower energy photons that the ECAL picks up together.
In the endcap regions, where the angle between the two emerging photons from the decay of a neutral pion is likely to be small enough to cause this problem, a preshower detector sits in front of the ECAL to prevent such false signals. The preshower has a much finer granularity than the ECAL with detector strips 2 mm wide, compared to the 3 cm-wide ECAL crystals, and can measure each of the pion-produced particles as a separate photon.
The preshower is made of two planes of lead followed by silicon sensors, similar to those used in the CMS tracker. When a photon passes through the lead layer it causes an electromagnetic shower, containing electron-positron pairs, which the silicon sensors then detect and measure. From this we get a measure of the photon’s energy, while having two detector layers gives us two measurements, allowing us to pinpoint the particle’s position.
When seemingly high-energy photons are then found in the ECAL, we can extrapolate their paths back to the centre of the collision and look for their “hits” in the preshower along the way, adding the energy deposited there to the total energy from the ECAL, and deducing if they really were individual high-energy photons or photon pairs.
Each endcap preshower uses 8 square metres of silicon (a material especially chosen for its accuracy, compactness, ability to deal with radiation, and easiness to handle). The silicon sensors, each measuring about 6.3cm x 6.3cm x 0.3mm and divided into 32 strips, are arranged in a grid in the endcaps to form an approximately circular shape covering most of the area of the crystal endcap. For optimal performance during the lifetime of the experiment, as in the tracker, the silicon detectors must be kept at a temperature of between -10oC and -15oC. However, the nearby ECAL is very sensitive and must be kept within precisely 0.1oC of its (higher) optimal temperature. The preshower must therefore be cold on the inside but warm on the outside, achieved using both heating and cooling systems.
The complete preshower system forms a disc, about 2.5m in circumference with a 50cm diameter hole in the middle (through which the beam pipe passes). This disc is only 20cm thick but manages to squeeze inside two layers of lead, two layers of sensors (and their electronics) as well as the cooling and heating systems – another example of the “compact” nature of CMS.
Energy of Hadrons (HCAL)
The Hadron Calorimeter (HCAL) measures the energy of “hadrons”, particles made of quarks and gluons (for example protons, neutrons, pions and kaons). Additionally it provides indirect measurement of the presence of non-interacting, uncharged particles such as neutrinos.
Measuring these particles is important as they can tell us if new particles such as the Higgs boson or supersymmetric particles (much heavier versions of the standard particles we know) have been formed.
As these particles decay they may produce new particles that do not leave records of their presence in any part of the CMS detector. To spot these the HCAL must be “hermetic”, that is make sure it captures, to the extent possible, every particle emerging from the collisions. This way if we see particles shoot out one side of the detector, but not the other, with an imbalance in the momentum and energy (measured in the sideways “transverse” direction relative to the beam line), we can deduce that we’re producing “invisible” particles.
To ensure that we’re seeing something new, rather than just letting familiar particles escape undetected, layers of the HCAL were built in a staggered fashion so that there are no gaps in direct lines that a familiar particle might escape through.
The HCAL is a sampling calorimeter [see explanation below] meaning it finds a particle’s position, energy and arrival time using alternating layers of “absorber” and fluorescent “scintillator” materials that produce a rapid light pulse when the particle passes through. Special optic fibres collect up this light and feed it into readout boxes where photodetectors amplify the signal. When the amount of light in a given region is summed up over many layers of tiles in depth, called a “tower”, this total amount of light is a measure of a particle’s energy.
As the HCAL is massive and thick, fitting it into “compact” CMS was a challenge, as the cascades of particles produced when a hadron hits the dense absorber material (known as showers) are large, and the minimum amount of material needed to contain and measure them is about one metre.
To accomplish this feat, the HCAL is organised into barrel (HB and HO), endcap (HE) and forward (HF) sections. There are 36 barrel “wedges”, each weighing 26 tonnes. These form the last layer of detector inside the magnet coil whilst a few additional layers, the outer barrel (HO), sit outside the coil, ensuring no energy leaks out the back of the HB undetected. Similarly, 36 endcap wedges measure particle energies as they emerge through the ends of the solenoid magnet.
Lastly, the two hadronic forward calorimeters (HF) are positioned at either end of CMS, to pick up the myriad particles coming out of the collision region at shallow angles relative to the beam line. These receive the bulk of the particle energy contained in the collision so must be very resistant to radiation and use different materials to the other parts of the HCAL.
The CMS HCAL…
For a detailed account of the HCAL detector see:
CMS Technical Design Report for the Phase 1 Upgrade of the Hadron Calorimeter
CMS HCAL Technical Design Report (1997)
HCAL Sampling Calorimeter
The CMS barrel and endcap sampling calorimeters are made of repeating layers of dense absorber and tiles of plastic scintillator. When a hadronic particle hits a plate of absorber, in this case brass or steel, an interaction can occur producing numerous secondary particles. As these secondary particles flow through successive layers of absorber they too can interact and a cascade or “shower” of particles results. As this shower develops, the particles pass through the alternating layers of active scintillation material causing them to emit blue-violet light. Within each tile tiny optical “wavelength-shifting fibres”, with a diameter of less than 1mm, absorb this light. These shift the blue-violet light into the green region of the spectrum, and clear optic cables then carry the green light away to readout boxes located at strategic locations within the HCAL volume.
A megatile is a layer of tiles whose sizes depend on their spatial location and orientation relative to the collision, chosen so that each receives roughly the same number of particles. Optic fibres fit into grooves cut into the individual tiles. Because the light picked up gives a measure of energy, the gaps between tiles must be filled with a reflective paint to ensure that light produced in each tile cannot escape into others and vice versa.
HCAL module showing sampling layers
The optical signals arrive at the readout boxes from megatile layers. There, signals from successive tiles, one behind the other, are then added optically to form “towers”. This optical summation covers the path of the particle through the HCAL and is a measure of its energy and/or can be an indicator of particle type.
These summed optical signals are converted into fast electronic signals by photosensors called Silicon Photomultipliers (SiPMs) . Special electronics then integrates and encodes these signals and sends them to the data acquisition system for purposes of event triggering and event reconstruction.
SiPMs are photodetectors configured especially for CMS that can operate in any magnetic field and give an amplified response, in proportion to the original signal, for a large range of particle energies. The SiPMs are housed in special readout boxes within the calorimeter volume. Light signals from the calorimeter megatiles are delivered to the SiPMs by special fibre-optic waveguides.
Using Russian navy shells
Constructing the endcap HCAL (HE) was the responsibility of the Russian and Dubna (RDMS) groups. The endcaps consist of 17 layers of dense absorber plates that take energy from hadrons and muons emerging from collisions and, together with plastic scintillator plates, give a measure of the particles’ energies.
Because the plates are so large and heavy the teams knew that within the experiment they would be under high levels of stress; yet, to maintain their position relative to the beam pipe for as long as 15 years, they couldn’t bend. The material eventually chosen for the plates was brass, in layers 50 mm thick.
But they needed brass of such high quality that it was practically impossible to find and was an expensive material. When Russian engineers discussed the question, one remembered a study carried out on the properties of brass made for military artillery use that sounded promising.
In Russian military storage there were thousands of shells made of brass that would fit such stringent requirements – all 50 years old, made by the Navy and designed to stand internal high stress and sea storage aboard a 1940s Navy Vessel. The shells could be melted for use in the HCAL.
The first hurdle involved obtaining permission of the Commander of the Navy, never easy, but who in fact agreed fairly quickly. Fifteen navy arsenals began recycling their shells, many being off-loaded from battleships in Murmansk to begin their new scientific life as absorber plates within CMS.
First the shells were sent to a plant in the North of Russia to be disarmed of their explosive contents. The shell casings were then stored at JSC Krasnyj Vyborzhets in St. Petersburg before being melted down and rolled into plates, a difficult process given that the brass was already semi-processed.
Over a million World War II brass shell casings were melted down to make the detector components. But when even this fell short of the massive 600 tonnes of brass needed to make the endcaps, the US agreed to provide a further $1 million in copper (brass being an alloy of copper and zinc) to complete the endcaps. As an example of the international cooperation and mutual trust fostered through this work, even this deal was agreed to on an amicable "handshake" basis.
In Minsk, Belarus, the brass was machined into absorber plates and finally pre assembled into parts of the endcaps, weighing 300 tonnes each, before being sent to CERN. Dollezhal Research and Development Institute of Power Engineering (NIKIET) coordinated the efforts of the Northern Navy and the companies involved, and ran 1400 tests to ensure the material was of high quality and suitable for constructing the endcap HCALs.
The shells and additional copper became a total of 1296 brass pieces, which were delivered to CERN in 2002 and 2003 and have since been installed and tested within CMS.
This project was part of an international agreement to convert Russian military industry into peaceful technology. The recycling of stocks of ship artillery into materials for fundamental research was also symbolic; instead of being used to destroy, the weapons could actively benefit humankind through their contribution to enhanced knowledge and technology.
The crystal story
A crystal’s journey from creation to its place within the CMS electromagnetic calorimeter is an arduous one: its trials include travelling thousands of miles, being sawed and polished, as well as being heated to 1,165oC and cooled to 18oC.
The journey starts in either a disused military complex in Bogoroditsk, Russia, or the Shanghai Institute of Ceramics (SIC) in China. These two institutions together took on the job of producing the 78,000 crystals to stringent specifications. Crystals are grown around a tiny “seed” - an existing piece of the crystal with the required properties - from a 1,165oC melt of tungsten oxide, lead oxide and “doping” materials, small amounts of other materials that refine the crystals’ properties.
In the Russian method, the seed is pulled from the scalding mixture at a very slow, precise speed that allows the atoms of the melt to arrange into a crystal structure around it. “In 2 or 3 days you have your crystal, which may seem slow but it’s quick compared to geology,” explains Michel Lebeau, an engineer in the crystals team.
Dusting with diamond
But the raw form of the crystal still needs a makeover before it can become one of the sparkling gems we see in the experiment. First it must be cut, ground and polished with diamond. Diamond is the hardest naturally occurring material and can scratch or cut through other crystals, allowing them to be shaped.
A disc covered with fine diamond grains first cuts the crystals’ prismatic accurate shape from its original round form. The second stage, called lapping, involves grinding the rough sides of the crystal as if with diamond sandpaper - a sprayed diamond suspension coats a rotating table, similar to a potter’s wheel, and as the crystals pass over it their surfaces are sanded to a fine matte finish, the flatness of which is controlled to less than a fraction of the width of a human hair.
Finally, to replace the crystals’ cloudy surfaces with a transparent finish, they must now pass over an even finer suspension of diamond, as finer abrasive gives an even smoother surface. The grains used here are in fact so tiny that the polished surface pattern is smaller than the wavelength of light.
To ensure a perfect match with the neighbouring crystals, and to give accurate measures of particle energies, each crystal must comply with very accurate dimensions so that they react in exactly the same way to the light. To make sure of this, the machines that transform the crystals into their sleek final form must also be fine-tuned and precise. After extensive research and development, machines that could give these perfect finishes were designed, fabricated locally and finally shipped, installed and ran at production sites in Russia and China.
Into the detector
For the crystals arriving at the regional centres in CERN and Rome, the welcome was warm but testing was thorough, as each had to be vetted before becoming part of the detector.
The arriving crystals were first unpacked and identified with a barcode before a visual inspection checks for any visible defects. Finally they were registered into a database so they could be followed throughout production and their lifetime in CMS.
Then the dimensions, light transmission and scintillation properties of the crystals were measured, 30 at a time, in an automatic machine (ACCOS) developed in both regional centres. “Usually with a crystal, the beauty of it is that it is coloured. But in our case if crystals are coloured would mean defects and reduced performance,” explains Etiennette Auffray, leader of the crystal assembly at CERN. But in fact only a very small number (less than 1%) of the 1200 crystals processed every month proved unsuitable for use
The vast majority of crystals made it through and were then carefully glued with a photodetector, which will collect and amplify their scintillation light and convert the light into an electrical signal. Each photodetector was then housed in one of 22 slightly different varieties of capsules and glued to one of 34 categories of crystal, in total making 146 different possible combinations: “So there’s a lot of scope to go wrong! We have to check it very carefully after each stage,” explains Auffray.
Then, like building blocks, the crystals were grouped in increasing numbers – for the barrel section they are first packed into lightweight glass-fibre boxes in groups of ten, making a ‘sub-module’. A ‘module’ was then constructed out of 40 or 50 sub-modules.
These modules were then joined by those assembled at the INFN/ENEA Laboratory near Rome, which also carried out the crystal tests and the glueing of photodetectors, in tandem with CERN. These were packed in groups of four, making one of the 36 “supermodules”, each of which weighs 2.5 tonnes.
For the endcaps, 25 crystals are grouped in five-by-five blocks named “supercrystals” and inserted into one of four “Dees”, structures named for their resemblance to the letter D, that fit around the beam allowing easy installation.
Finally the monitoring, cooling system and final electronics were added to the supermodules before they entered the experiment cavern. The cooling system is a record-breaking achievement that keeps the 75,848 crystals within 0.1°C of their optimum temperature to ensure they give a stable and equal response. The flow of water needed to achieve this, recycled through the system, equals 1/10th of the output of Geneva’s Jet d’Eau (Geneva’s fountain).
Inside CMS the supermodules are supported on a strong but lightweight system, designed to add only a minimum amount of material in front of the detector. The structure and crystals are so densely and precisely packed that the remaining space is only 1% of the total. The 2.5-tonne weight of each supermodule is cantilevered from one end so that two-thirds of it is taken at the back and the remaining third on a slender arm of aluminium and glass fibre composite. The maximum expected sag, including in the event of (very improbable) earthquakes, is about half a millimetre. This structure accounts itself for just 10% of the total weight - like a car that carries four passengers weighing just 30kg!
The CRISTAL database
Each CMS crystal is unique even though they are all made from the same material. Its physical characteristics therefore needed to be measured and recorded, not only for the construction phase but also for data collecting itself. Every crystal was labelled with a barcode to identify it and track it through the production and operation stages.
The long time-frame of producing the crystals meant that the means of production as well as the associated electronics evolved over time as improvements were made. Themethods of characterising the crystals themselves and the data formats used for storage also evolved with time. The database into which the information was to be stored would itself need to adapt to these evolving workflows. At the same time, the database must be able to fetch all of the data ever recorded –– i.e. it must be capable of handling legacy datasets –– and must be in a situation to do so in the future when the implementation might potentially be very different. It would also have to handle frequent changes, and so be flexible in design.
No such system existed in the mid-'90s, and so a consortium of scientists from CERN, CNRS (the French National Centre for Scientific Research) and UWE (the University of the West of England) set about developing their own database, giving birth to CRISTAL.
CRISTAL stands for Cooperative Repositories and Information System for Tracking Assembly Lifecycles, and was developed at CERN to manage the data of all CMS ECAL components at each stage of construction.
The CRISTAL database and its underlying technology is extremely novel, and has applications far beyond the world of tracking detector components. CRISTAL is description-driven; that is, the database structures can be modified at any time without affecting any of the collected information. The structures themselves are stored as data too, meaning they can be copied between different databases with the data they define. As a result, any large-scale project with rapidly evolving workflows can benefit from using CRISTAL. It was thus one of the first knowledge-transfer projects undertaken by CERN, and was spun-off into Agilium, a private company that handles business-process management. A second spin-off company called Technoledge is providing database solutions for e-governance, accounting firms, and more.
CRISTAL is another excellent example of technology developed in the course of particle physics research that has since found wide-spread use in industry and elsewhere.
Tracking
The measurement of the momentum of particles is crucial in helping us to build up a picture of what happens at the heart of the collision. One method to calculate the momentum of a particle is to track its path through a magnetic field; the more curved the path, the less momentum the particle had. The CMS tracker records the paths taken by charged particles by measuring their positions at a number of points.
The tracker can reconstruct the paths of high-energy muons, electrons, and charged hadrons (particles made up of quarks) as well as see tracks coming from the decay of very short-lived particles such as beauty or “b quarks” that are for example being used to study the differences between matter and antimatter.
The tracker needs to record particle paths accurately yet be lightweight so as to disturb the particles as little as possible. It does this by taking position measurements so accurate that tracks can be reliably reconstructed using just a few measurement points. Each measurement is accurate to 10 µm, a fraction of the width of a human hair. It is also the innermost part of the CMS detector and so receives the highest volume of particles: the construction materials were therefore carefully chosen to resist radiation.
The final design consists of a tracker made entirely of silicon detection elements: the pixels, at the very core of the detector and dealing with the highest intensity of particles, and the silicon microstrip detectors that surround it. As particles travel through the tracker the pixels and microstrips produce tiny electric signals that are amplified and detected. The signals are stored in chips’ memory for several microseconds and then processed before being sent to a laser to be converted into infrared pulses. These are then transmitted over a 100m fibre optic cable for analysis in a radiation-free environment. The tracker uses 40,000 such fibre optic links providing a low power, lightweight way of transporting the signal. Much of the technology behind the tracker electronics came from innovation in collaboration with industry.
The tracker employs sensors covering an area the size of a tennis court, with 135 million separate electronic readout channels: in the pixel detector there are some 6600 connections per square centimetre. Silicon sensors are highly suited to receive many particles in a small space due to their fast response and good spatial resolution.
The CMS strip tracker was installed in CMS in late 2007 and has been in continuous operation since 2008. The original CMS pixel detector was operated during the years 2009 – 2016 and was replaced by the CMS Phase-1 pixel detector in an extended year-end stop 2016/2017.
Visit the link for Pixel Tracker updates:
More details about the CMS Tracker can be found in the following selection of documents:
The CMS experiment at the CERN LHC (paper written in ~2008 describing the original CMS experiment as build for LHC Run 1)
The CMS Phase-1 Pixel Detector Upgrade (paper from 2020 describing the Phase-1 pixel detector)
The very original description of the CMS tracking detectors can be found in the Technical Design Report (TDR) from 1998 and 2000, respectively
CMS: The Tracker Project Technical Design Report
The CMS tracker : addendum to the Technical Design Report
While many design aspects have evolved from the original concept, they still provide interesting insights about the challenges for the detectors and how they were thought to be overcome. A few more words on this are in the following paragraph.
Some design history ...
In the very first design sketch of CMS, the tracker section was left blank because it was thought that with the intensity of particles experienced in the LHC it would be impossible to make a tracker that could withstand it. In particular, making readout electronics that could work in this radiation would be a challenge. Hints that the US military, needing such electronics for space travel and nuclear warheads, might have solved some of the potential problems encouraged the team to explore this field.
However, the technology actually came from a most unexpected source; as the military market shrank at the end of the cold war and commercial electronics began to excel, the team took a gamble on a very fine feature manufacturing process, produced for a commercial market, which was not formally in the realm of “radiation hard electronics”. With some modifications, a few simple design tricks and selection of the right technology, it could be as radiation hard as needed. And it was a remarkable success. Eighteen months later the teams had rebuilt the old chips. They had low power consumption, low noise, and high overall performance and on top of this could be easily produced on a large scale, were relatively cheap and thoroughly tested. It was a big breakthrough. The same technology was then also used for the ECAL as well as CMS and ATLAS pixels.
Silicon Pixels
The pixel detector contains 124 million pixels, allowing it to track the paths of particles emerging from the collision with extreme precision. It is also the closest detector to the beam pipe, with cylindrical layers roughly at 3cm, 7cm, 11cm, and 16cm and disks at either end, and so is vital in reconstructing the tracks of very short-lived particles.
Each of the four layers is composed of individual silicon modules, split into little silicon sensors, like tiny kitchen tiles: the pixels. Each of these silicon pixels is 100µm by 150µm, about two hairs' width. When a charged particle passes through a pixel, it disturbs the electrons in the silicon pixel, which results in an electric pulse, or as we call it, a 'hit'. These hits are used to draw the track or trajectory of the charged particle.
By applying a voltage to the sensor, these pulses collect into a small electric signal, which is then amplified by a readout chip for a total of 16 chips per module.
CMS silicon pixel detector
The detector is made of 2D tiles and has four layers, meaning that we can create a three-dimensional picture of the particle track through the layers.
However, being so close to the collision means that the number of particles passing through is huge: the rate of particles received at 3cm from the beamline is around 600 million particles per square centimetre per second! The pixel detector is able to disentangle and reconstruct all the tracks particles leave behind, and withstand such a pummeling over the duration of the experiment.
Because there are 124 million pixels, the power for each must be kept to a minimum. Even with only around 50 microwatts per pixel, the total power output is 7.5kW- around the same as the energy produced by two electric ovens. So as not to overheat the detector, the pixels are mounted on cooling tubes and kept at -20°C degrees. Operating at this temperature also helps to reduce the effects of the damage to the silicon modules inflicted by the continuous stream of particles.
Visit the link for Pixel Tracker updates
Silicon Strips
After the pixels and on their way out of the tracker, particles pass through ten layers of silicon strip detectors, reaching out to a radius of 130 centimetres.
The silicon strip detector consists of four inner barrel (TIB) layers assembled in shells with two inner endcaps (TID), each composed of three small discs. The outer barrel (TOB) – surrounding both the TIB and the TID – consists of six concentric layers. Finally two endcaps (TEC) close off the tracker on either end. Each has silicon modules optimised differently for its place within the detector.
This part of the tracker contains 15,200 highly sensitive modules with a total of about 10 million detector strips read by 72,000 microelectronic chips. Each module consists of three elements: one or two silicon sensors, its mechanical support structure and readout electronics.
The silicon detectors work in much the same way as the pixels: as a charged particle crosses the material it knocks electrons from atoms giving a very small pulse of current lasting a few nanoseconds. This small amount of charge is then amplified by Analogue Pipeline Voltage (APV25) chips, giving us “hits” when a particle passes, allowing us to reconstruct its path. Four or six such chips are housed within a “hybrid”, which also contains electronics to monitor key sensor information, such as temperature, and provide timing information in order to match “hits” with collisions.
Due to the nature of their job, the tracker and its electronics are pummelled by radiation but they are designed to withstand it. To minimise disorder in the silicon this part of the detector is kept at -20oC, to “freeze” any damage and prevent it from perpetuating.
CMS Tracker layers shown in the plane perpendicular to the beam.
Triggering and Data Acquisition
When CMS is performing at its peak, about one billion proton-proton interactions will take place every second inside the detector. There is no way that data from all these events could be read out, and even if they could, most would be less likely to reveal new phenomena; they might be low-energy glancing collisions for instance, rather than energetic, head-on interactions.
We therefore need a “trigger” that can select the potentially interesting events, such as those which will produce the Higgs particle, and reduce the rate to just a few hundred “events” per second, which can be read out and stored on computer disk for subsequent analysis.
However, with groups of protons colliding 40 million times per second there are only ever 25 nanoseconds (25 billionths of a second) before the next lot arrive. New waves of particles are being generated before those from the last event have even left the detector! The solution is to store the data in pipelines that can retain and process information from many interactions at the same time. To not confuse particles from two different events, the detectors must have very good time resolution and the signals from the millions of electronic channels must be synchronised so that they can all be identified as being from the same event.
Triggering
Level 1 of the trigger is an extremely fast and wholly automatic process that looks for simple signs of interesting physics, e.g. particles with a large amount of energy or in unusual combinations. This is like a reader simply scanning the headlines of a newspaper to see if anything catches their eye. This way we select the best 100,000 events or “issues” each second from the billion available. For the next test, the higher level trigger, we assimilate and synchronise information from different parts of the detector to recreate the entire event - like collating the different pages to form the full newspaper - and send it to a farm of more than 1000 standard computers.
Here the PCs are like speed readers, who with more detailed information review the information for longer, less than a tenth of a second. They run very complex physics tests to look for specific signatures, for instance matching tracks to hits in the muon chambers, or spotting photons through their high energy but lack of charge. Overall they select 100 events per second and the remaining 99,900 are thrown out. We are left with only the collision events that might teach us something new about physics. Despite the trigger system, CMS still records and analyses several petabytes of data, that’s millions of gigabytes, around the same amount of information as in 10,000 Encyclopaedia Britannica every second, although not all these data are stored.
For a more detailed account of CMS Triggering see:
CMS The TriDAS Project Technical Design Report
The History of CMS Data Acquisition and Triggering
Early History and Key Decisions
The CMS Trigger Group was originally led by Fritz Szoncsó. Wesley Smith was appointed Trigger Project Manager in 1994. One of the earliest decisions made by the group was not to follow existing three-level trigger-system designs.
Just a few years prior to this, in 1985, the ZEUS experiment was the first to decide to use a three-level trigger system which consisted of: pure hardware (L1), mostly custom hardware (L2) and the computer farm (L3). The Tevatron experiments also adopted this architecture.
At CMS, TriDAS (Trigger and Data Acquisition System) Project Manager Sergio Cittolin, TriDAS Institution Board Chair Paris Sphicas and Smith decided not to have a second level trigger. They would take the output of L1 straight to the computer farm for software processing. The main reason for doing this was that the L2 hardware was too restrictive. It was not fully programmable, and was only used at the time because there was no telecom switch that could convey the full L1 output of 100 kHz of 1 MB events to the farm.
However, Cittolin, studying technology trends and extrapolating world-wide computing network infrastructure, was convinced that a switch with the required bandwidth would be available and affordable by the year 2000. So, when the technical proposal was written in ’94-’95, a plan to go from L1 to the computer farm was laid out.
When the technical proposal was presented to the LHC Experiments Committee (LHCC) in 1994, a different approach was adopted to deal with the bandwidth problem, in order to pass the reviews: The proposal said that 10% of the data would go through the switch for processing and would be used as a basis to knock down the rate by a factor of 10 — sufficient for the switches of 1994 to handle! This 10% — including main calorimeter information, summary of tracking information and so on — would then be used to decide whether to keep the event or not. The group hoped, though, that they would not have to face the problem at all.
Prototyping and testing followed until the Technical Design Report (TDR) was published in 1999/2000 and defended in front of the LHCC. By 2002, network switches with the required bandwidth were available and the problem was solved.
Consequences of Two-Level Trigger
One consequence of the decision was that L1 has to be more efficient than in previous systems in order to perform certain tasks traditionally performed by L2.
Prior to the CMS Trigger, triggers were designed to count objects: the number of electrons/muons over a certain threshold and the like, providing a histogram. In CMS, characteristics of objects, including their energy and co-ordinates would be retained, which required sorting of the objects so that only the prime candidates would be selected.
However, sorting consumes time. Which is why the latency — trigger processing time — is longer than it would have been for the L1 trigger in a three-level trigger system. Custom chips were built for the calorimeter trigger to perform the sorting fast enough.
The upside to this, Smith says, is that instead of asking how many jets or electrons we have, we can ask, “Is there an electron above an ET threshold back-to-back with a jet above that threshold?” and other such questions. This had never been done before and almost allows CMS to perform offline operations at the L1 stage itself.
Wesley Smith adds: “I think it’s been proven that we really built it almost optimally. It was a long extrapolation but I think in retrospect it’s a fairly successful design. The higher level triggers worked out quite well, are well designed, and showed the flexibility of the system. In the end, if you look back at the history, we did make the right decisions. Although at the time, as always, it was not clear.”
How CMS detects particles
Each particle that emerges from an LHC collision is like a piece of a puzzle, with some of these pieces breaking up further as they travel away from the collision. Each leaves a trace in the detector and CMS’s job is to gather up information about every one - perhaps 20, 100 or even 1000 puzzle tracks - so that physicists can put the jigsaw back together and see the full picture of what happened at the heart of the collision.
This event shows the production of a supersymmetric particle. The decay of the particle is such that every single layer of CMS is needed to detect the full range of emerging particles: electrons, muons, neutrinos and jets, produced by quarks.
To do this, CMS consists of layers of detector material that exploit the different properties of particles to catch and measure the energy or momentum of each one. New particles discovered in CMS will be typically unstable and rapidly transform into a cascade of lighter, more stable and better-understood particles.
NOTE: A Powerpoint version of the above slice animation can be found here.
A particle emerging from the collision and travelling outwards will first encounter the tracking system, made of silicon pixels and silicon strip detectors. These accurately measure the positions of passing charged particles allowing physicists to reconstruct their tracks. Charged particles follow spiralling paths in the CMS magnetic field and the curvature of their paths reveal their momenta.
The energies of the particles will be measured in the next layer of the detector, the so-called calorimeters. Electrons, photons and jets (sprays of particles produced by quarks) will all be stopped by the calorimeters, allowing their energy to be measured.
The first calorimeter layer is designed to measure the energies of electrons and photons with great precision. Since these particles interact electromagnetically, it is called an electromagnetic calorimeter (ECAL).
Particles that interact by the strong force, hadrons, deposit most of their energy in the next layer, the hadronic calorimeter (HCAL). The only known particles to penetrate beyond the HCAL are muons and weakly interacting particles such as neutrinos. Muons are charged particles, which are then tracked further in dedicated muon chamber detectors. Their momenta are also measured from the bending of paths in the CMS magnetic field. Neutrinos, however, are neutral and since they hardly interact at all they will escape detection. Their presence can nevertheless be inferred. By adding up the momenta of all the detected particles, and assigning the missing momentum to the neutrinos, CMS physicists will be able to tell where these particles were.
Particles travelling through CMS thus leave behind characteristic patterns, or ‘signatures’, in the different layers, allowing them to be identified (see figure). The data is passed from CERN to centres around the world where analysts then reconstruct the “event” and the presence of any new particles can be inferred.
Medical Imaging
From Detecting Particles to Detecting Tumours
The CMS’s electromagnetic calorimeter uses scintillating crystal technology that is similar to the technology used in Positron Emission Tomography (PET) scans.
PET is a medical imaging device that uses antimatter to observe metabolic processes in the body, such as cancerous tumours. In this process, a radioactive “tracer” is injected into the body where it is absorbed by the targeted tissue. An isotope in the tracer will decay and produce antimatter particles called positrons, which will annihilate with an electron (its antimatter counterpart) in the body and release a pair of photons that fly in opposite directions. The ring-shaped PET detector surrounding the body detects the photons and uses their positions to reconstruct a 3D image of the target area. This image reveals information about the tissue by mapping the tracers concentrated there.
Credit: Imperial College London
The Crystal Clear Collaboration (CCC) was founded at CERN in 1990 to search for a suitable scintillator for use in the Large Hadron Collider’s electromagnetic calorimeter (ECAL). Among the various studied crystals, lead tungstate was found to be the most suitable and was then used by CMS and ALICE.
The CMS’s ECAL measures the energies of photons and electrons that are produced by high-energy collisions in the LHC. When electrons or high-energy photons pass through the ECAL, the scintillating crystals produce bursts of optical photons in proportion to the particles’ energy. These collisions in the LHC are like high-energy versions of the electron-positron collisions that occur in the body during a PET scan.
Since 1996, the CCC has been working in the development of high-performance PET systems in addition to their generic research to better understand scintillator properties. One of these systems, ClearPET, is a small animal PET scanner that allows researchers to study the effects of radiopharmaceuticals on the brains of small animals.
In continuation of their PET research, the CCC launched the ClearPEM (Positron Emission Mammography) project in 2002 to develop a dedicated breast PET scanner. ClearPEM can detect cancer lesions as small as 1.3 millimetres, hopefully leading to earlier detection and subsequent lower breast cancer mortality. Its photodetector (the device that converts light from the scintillating crystals into electrical signals) is the same avalanche photodiode as the one developed for CMS’s ECAL! The ClearPEM system has gone through successful clinical trials in hospitals in Coimbra, Portugal, Marseille, France, and Monza, Italy.
More recently, some members of the CCC within an FP7 European project, EndoTOFPET-US (Endoscopic Time-Of-Flight PET & Ultrasound), developed a “time-of-flight” PET for the improved detection of pancreas and prostate cancer. The time-of-flight PET takes into account an additional measurement to more accurately triangulate the photons’ point of annihilation: the difference in photon arrival times to each side of the detector, or “time-of-flight”. This means it can achieve better-quality images in less time or reduce the dose of the radiopharmaceutical injected in the patient.
The development of faster detection being carried out by CCC would not only be valuable for medical imaging but also holds promise for future components of the CMS detector.
Find out more:
Contact person: Etiennette Auffray, CERN
What is CMS?
About CMS
The Large Hadron Collider (LHC) at CERN smashes protons together at close to the speed of light with seven times the energy of the most powerful accelerators built up to now. Some of the collision energy is turned into mass, creating new particles which are observed in the Compact Muon Solenoid (CMS) particle detector. CMS data is analyzed by scientists around the world to build up a picture of what happened at the heart of the collision. This will help us answer questions such as: "what is the Universe really made of and what forces act within it?" and "what gives everything substance?" Such research increases our basic understanding and may also spark new technologies that change the world we live in.
The LHC smashes groups of protons together at close to the speed of light: 40 million times per second and with seven times the energy of the most powerful accelerators built up to now. Many of these will just be glancing blows but some will be head on collisions and very energetic. When this happens some of the energy of the collision is turned into mass and previously unobserved, short-lived particles – which could give clues about how Nature behaves at a fundamental level - fly out and into the detector.
CMS is a particle detector that is designed to see a wide range of particles and phenomena produced in high-energy collisions in the LHC. Like a cylindrical onion, different layers of detectors measure the different particles, and use this key data to build up a picture of events at the heart of the collision.
Scientists then use this data to search for new phenomena that will help to answer questions such as: What is the Universe really made of and what forces act within it? And what gives everything substance? CMS will also measure the properties of previously discovered particles with unprecedented precision, and be on the lookout for completely new, unpredicted phenomena.
Why is CMS so big?
It may seem strange that to record the Universe’s tiniest constituents we need the world’s most powerful machines and detectors. But the detector needs to be big because the particles flying out of the collisions have such high energies that it takes big distances to absorb them. And the higher the energy, the greater the amount of material needed.
A bigger detector also means the possibility of obtaining more accurate measurements. To measure the momentum of a particle we track its path through a magnetic field, as the greater the momentum the less it bends within it. The bigger the detector, the more measurements can be taken in “tracking” the particle, meaning more accuracy in the momentum calculation.
As this measurement gets harder with more energetic particles, to maintain the required accuracy we need a strong magnetic field to bend the paths as much as possible. This brings with it other size requirements, as the field needs to be guided and contained by an iron “return yoke”. The amount of iron grows in size in proportion with the magnetic field, so for the 4 Tesla CMS magnetic field, 100,000 times stronger than that of the Earth, we use 12,000 tonnes of iron.
As a result of all of this, CMS is 15 metres in diameter and weighs around the same as 30 jumbo jets or 2,500 African elephants. And though it is the size of a cathedral, it contains detectors as precise as Swiss watches.
CMS detector in a cavern 100 m underground at CERN's Large Hadron Collider. |
190185 | https://www.math.mcgill.ca/pprzytyc/mnpc/math348notes2018.pdf | TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE TAUGHT BY PIOTR PRZYTYCKI.
NOTES BY DYLAN CANT.
1. Introduction The Euclidean plane is a collection of points. Lines are subsets of the Euclidean plane.
We will typically refer to points by capital letters A, B, C etc and lines by lower case letters l, k, m etc.
k C A B In the figure above, the point C lies on the line k, and we have drawn the line segment AB between the points A and B. If a point X lies on the segment AB, then we say X is between A and B.
X A B We take the point of view that lines, circles, and any other geometric figures are sets of points. Consequently, we will often use the notation X ∈k (read: X is in/on k) when X is a point and k is a geometric figure.
Another important concept is the notion of a transformation of the plane. A transfor-mation of the plane should be thought of as a “movement” of points. The first chapter is devoted to exploring reflections, and the second chapter explores rotations, both of which are transformations of the plane.
Definition 1.1. Parallel lines Two lines l and l′ are parallel if they do not intersect l l′ Before we begin proving theorems, let us note a two intuitive facts: 1 2 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Fact 1.2. For any two distinct points A and B there exists one and only one line k = AB which contains these two points.
Fact 1.3. For any line k there is at least one point P not contained in that line.
Given a pair of points A, B we denote the distance between them as |AB|. The next fact about Euclidean geometry is less obvious then the previous facts, so take some time to convince yourself that it is reasonable.
Fact 1.4 (Triangle Inequality). If A, B, C are three points in the plane, then their distances satisfy |AC| ≤|AB| + |BC| , and equality holds if and only if B lies on the line segment AC.
A |AB| B |BC| C |AC| Definition 1.5 (angles). If A, B, C are three distinct points in the plane, then we may construct the angle ∠ABC, which has a degree-measure in between 0◦and 180◦.
B C A ∠ABC = 0◦ a right angle 90◦ B C A ∠ABC = 180◦ For now we are only considering unoriented angles (when we consider oriented angles, we will need to allow angles with degree-measure in between 180◦and 360◦).
Given three points A, B, C, we say they form a triangle if they do not lie on a common line L. If three points A, B, C lie on a straight line then we say they form a degenerate triangle or are collinear.
Exercise 1.6. Let ABC be a triangle. Using the triangle inequality, prove that |AB| −|BC| < |AC| < |AB| + |BC| ; why is this inequality strict? Hint: non-degeneracy of the triangle ABC implies that the above inequalities are strict.
Given two lines k, l and points A, C on k and B, D on l such that line segments AC and BD intersect in a point E, we can form the four angles ∠AEB, ∠BEC, ∠CED, and ∠DEA.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 3 If all four of these angles are equal, then the lines l, k are perpendicular.
l k E D B C A Remark. When any two lines AC and BD intesect in a point E as shown above, four angles are formed.
Even if the lines are not perpendicular, we still have two equalities ∠AEB = ∠CED and ∠AED = ∠CEB. What distinguishes a perpendicular pair of lines is that all four angles are equal; one might say that a perpendicular pair of lines has more symmetry than an arbitrary pair of lines.
Fact 1.7. Given a line k, and any point A, there is a unique perpendicular l to k which contains A. Note: this fact is true whether or not A lies on k (see the figure below).
k A l dropping the perpendicular from A to k k A l raising the perpendicular to k from A 1.1. Reflections. Using Fact 1.7, we can define reflections through lines; reflections will be an indispensable tool when we start proving theorems.
Definition 1.8. Let k be a line. The reflection through k is a transformation of the Euclidean plane which sends a point A to its reflection A′.
This reflection satisfies two properties.
(i) If A lies on k, then A′ = A (in other words, the reflection fixes the line k).
(ii) If A does not lie on k, then A′ is the point such that the line AA′ is perpendicular to k in a point O such that |AO| = |OA′| (see figure below).
4 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
k A A′ B′ B C C′ O There is an important fact related to this definition.
Fact 1.9. The reflection through any line k preserves distances and angles. In other words, if A, B, C are any two points whose reflections through k are A′, B′ and C′, then |AB| = |A′B′|, and ∠ABC = ∠A′B′C′.
Exercise 1.10. Prove that the reflection of a line is a line (hint: use the equality case in the triangle inequality). Prove that if l is parallel to k, then its reflection l′ through k is also parallel to k. Prove that l is perpendicular to k if and only if l is its own reflection (through k).
Example. If we give our plane x and y coordinates (something we will almost never do), then the reflection of A = (x, y) through the y axis is easily seen to be A′ = (−x, y).
x y A′ = (−x, y) A = (x, y) Definition 1.11. The bisector (or perpendicular bisector) of a segment AB is the line k perpendicular to AB at its centre.
A B k k is the bisector of AB TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 5 Exercise 1.12. Let k be the bisector of AB. Prove that the reflection of A through k is B.
Theorem 1.13 (characterization of points on the bisector). A point X lies on the bisector of a segment AB if and only if |AX| = |XB|.
Proof. Throughout the proof, let k denote the bisector of AB, and let Z′ denote the reflec-tion of Z through the bisector of AB: A B k Z Z′ This theorem is an “if and only if” statement, so we must prove two implications. First we prove the = ⇒direction, namely, we will prove that if X lies on the bisector of AB then |AX| = |XB|. Since X lies on the bisector, X = X′. By the result of Exercise 1.12, B = A′, and since reflections preserve distances, |AX| = |A′X′| = |BX|, as desired.
Now we prove the ⇐ = direction of the proof. We assume that |AX| = |XB|. We prove by contradiction, and suppose that X does not lie on k. Then either AX or BX intersects k, and so without loss of generality, we can assume that AX intersects k at a point Y , as in the following figure.
A B k X Y Since Y lies on the bisector through AB, we know from the = ⇒direction of this theorem (which we already proved) that |AY | = |Y B|. Applying the triangle inequality twice yields |BX| < |BY | + |Y X| |AX| = |AY | + |Y X| , where we know the first inequality is strict since Y cannot lie on BX, since X and B lie on the same half-plane determined by k, while Y lies on k. We also know the second inequality is actually an equality because Y lies on AX. But since |AY | = |BY |, we conclude that |BX| < |AX| , which contradicts our assumption. Since we began our argument by assuming 6 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
that X did not lie on k, the contradiction |BX| < |AX| forces us to conclude that X must lie on k, and this completes the proof.
□ For the next proof we require some preliminary results.
Fact 1.14. If l is a line and X is a point not on l, then there is a unique line k containing X which is parallel to l.
Exercise 1.15. Prove the following: (a) If l, k are parallel lines, and m intersects k, then m also intersects l. Hint: use the preceeding fact.
(b) If k is a line containing distinct points A and B, then the perpendiculars to k raised from A and B are parallel lines. Hint: Let X lie on the perpendiculars to k raised from A and from B, then use uniqueness of perpendiculars to k dropped from X.
(c) If l, k are parallel, and m is perpendicular to l at A, then there is a point B on k such that m is perpendicular to k at B. Hint: by (a), m intersects k at some point B, and by (b) the perpendicular to m raised from B is parallel to l. Then use the fact stated before this exercise.
(d) If l, k are parallel lines, and m ̸= n are such that m is perpendicular to l and n is perpendicular to k, then m and n are parallel. Hint: use (c) to conclude that m is perpendicular to k, and so n and m are both perpendicular to k. Then use (b).
l k m n Theorem 1.16. In any triangle ABC, the bisectors of AB, AC and BC all intersect in a single point.
Proof. Since AB and BC are not parallel, their perpendicular bisectors are also not parallel (by Exercise 1.15.d), and so they intersect at some point O. By Theorem 1.13, |OA| = |OB| and |OB| = |OC|. But then |OC| = |OA|, so O lies on the bisector of AC (using Theorem 1.13 again). Thus O lies on all three bisectors, which is what we wanted to show.
O A B C □ TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 7 Remark. From the proof, |AO| = |BO| = |CO|, and so there is circle passing through ABC centred at O. This circle is unique, since any other circle with centre O′ passing through ABC would have |AO′| = |BO′| = |CO′|, and so O′ would also lie on all three bisectors of ABC. Since three lines intersect in (at most) a unique point, we conclude that O′ = O, so the circle is unique. This circle is called the circumscribed circle of the triangle ABC.
Theorem 1.17. Let X be a point outside the line AB. Then X lies on the bisector of AB if and only if ∠XAB = ∠XBA.
Remark. The assumption that X lies outside the line AB is crucial, for it is easy to show that if X is anywhere inside the segment AB, then ∠XAB = ∠XBA (even if it is not on the bisector).
Proof. Let k be the bisector of AB. We begin with the = ⇒direction of the theorem, so we assume that X ∈k. If we reflect in k, then X′ = X and A′ = B, and since reflections preserve angles, ∠XAB = ∠X′A′B′ = ∠XBA, as desired.
Now we prove the ⇐ = direction of the theorem. As in the proof of Theorem 1.13, we prove by contradiction, and so we suppose that ∠XAB = ∠XBA, but X does not lie on k.
Without loss of generality, suppose that X and B lie on the same half plane determined by k. Let Y be the intersection point of AX and k, as shown below.
A B Y X k Since the lines AX and AY are equal, the angles ∠XAB and ∠Y AB are equal. By the = ⇒ part of this theorem, we know that ∠Y AB = ∠Y BA, and thus ∠XAB = ∠Y BA. However, we assume that ∠XAB = ∠XBA, so we conclude that ∠XBA = ∠Y BA (if we look at the above figure, this equality is obviously not true, but we need to use logical arguments to finish the proof). Since ∠XBA = ∠Y BA, we conclude that the lines BX and BY are equal (since X and Y lie on the same half-plane determined by line AB), and so the line Y X is equal to the line BX. However, we defined Y to lie on the line AX, so the line AX is equal to the line Y X. But then we have equality of lines BX = AX, and so X lies on the line AB, which we assume is not true. This is a contradiction, and so we must have that X lies on k, and so we have completed the proof.
□ Corollary 1.18. In a triangle ABC, the following three conditions are equivalent: (i) |AC| = |CB|.
(ii) ∠CAB = ∠CBA.
8 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
(iii) C lies on the bisector of AB.
Proof. Since ABC is a triangle, we know that C does not lie on the line AB. The statement (i) ⇐ ⇒(iii) is Theorem 1.13, (ii) ⇐ ⇒(iii) is Theorem 1.17, and so these three statements are equivalent.
□ Definition. If ABC is a triangle which satisfies any of the equivalent properties in the above corollary, then ABC is called an isosceles triangle. (Note: by our definition, equilateral triangles are also technically isosceles triangles).
A B C Exercise 1.19. Let T = ABC be a triangle. Prove that T is isosceles if and only if there exists a line k such that T is its own reflection when we reflect through k. (we say that the reflection through k is a symmetry of the triangle T).
Prove that T is equilateral if and only if there exist two distinct lines k and l such that T is its own reflection through k and through l.
Remark: we could summarize this exercise by saying that non-isosceles triangles have no symmetry, isoceles triangles have some symmetry, and equilateral triangles have the most symmetry.
Theorem 1.20. Let ABC be a triangle with ∠CAB < ∠CBA. Then |BC| < |AC|. (this theorem says that, opposite a smaller angle, you have a smaller side).
Proof. Since ∠CAB < ∠CBA, we know that there exists a point X in the segment AC such that ∠XBA = ∠CAB. This is shown in the figure below.
A B C X Then we apply the triangle equality to conclude (∗) |AX| + |XC| = |AC| , and the triangle inequality to conclude that (∗∗) |BX| + |XC| > |BC| .
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 9 We can conclude a strict inequality above because non-degeneracy of ABC implies that BXC cannot lie on a common line. By Theorem 1.17 we know that |AX| = |BX|, and so we combine (∗) and (∗∗) to conclude that |BC| < |AC|, as desired.
□ 1.2. Some applications of reflections.
Problem 1.21. Given a line k and points A, B ∈k on the same side of k, find X ∈k such that |AX| + |XB| is minimized.
A B X k To solve this problem, we will consider the reflection through k. The idea for the solution is summarized nicely in the following figure. Let X be any point on k, and let A′ denote the reflection of A through k. Then |AX| = |A′X|, so the problem of minimizing the distance |AX| + |XB| is the same as the problem of minimizing the distance |A′X| + |XB|. Suppose that X1 ∈k does not lie on the segment AB. Then the triangle inequality tells us that |A′X1| + |X1B| > |A′B|. Now let X2 ∈k be the intersection of segment A′B with k (X2 exists because A and B lie on the same side of k, by assumption, therefore A′ and B lie on opposite sides of k). Since X2 lies on the segment AB, |A′X2| + |X2B| = |A′B|, and so we deduce that for all X1 ̸= X2 on k, we have |AX2| + |X2B| < |AX1| + |X1B| .
This inequality tells us that the point X2 is the unique minimizer of |AX| + |XB|, and this completes the solution.
A B X1 X2 k A′ 10 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Problem 1.22. Show that two distinct circles can intersect in at most two points.
A O1 O2 A′ Before we solve this problem, let us recall the definition of a circle. If O is a point and r a positive real number, then the circle centred at O with radius r is the collection of all points X such that |OX| = r.
Now suppose that we have two distinct circles o1 and o2 with centres O1 and O2, respec-tively. If O1 = O2, then the radii of o1 and o2 must be different (or else they would be the same circle!), in which case it is easy to show that o1 and o2 do not intersect.
Supposing now that O1 ̸= O2, suppose that A ∈o1 ∩o2 (read: A is in the intersection of o1 and o2). If A′ is any other point on o1 ∩o2, then |AO1| = |A′O1| and |AO2| = |A′O2|, by the definition of a circle. But by Theorem 1.13, this implies that the line O1O2 bisects the segment AA′. Thus the reflection through O1O2 sends A to A′. Therefore, A′ is uniquely determined by A. This completes the solution.
Exercise 1.23. Show that if two circles o1 and o2 intersect in a single point A, then A must lie on the line O1O2 (therefore, in the notation of the previous example, A′ = A).
Exercise 1.24 (Congruence of circles). If o1 and o2 are two circles with the same radius, construct a line k so that o1 and o2 are reflections of each other through k.
o1 o2 k Exercise 1.25. Let k be a line and o a circle. Prove that k and o intersect in at most two points. Hint: Follow Example 1.2.2; if A is one intersection point, any other intersection point is uniquely determined by A.
Exercise 1.26 (Preparation for Example 1.2.3). Prove that s = ABCD is a square if and only if A is the reflection of C through BD, B is the reflection of D through AC, and TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 11 |AC| = |BD|.
C B A D Example 1.27. Given circles o1 and o2 and a line k, find necessary conditions under which a square ABCD such that A ∈o1, C ∈o2 and B, D ∈k exists.
k o1 o2 C A B D Let us begin our solution by supposing that such a square ABCD exists. Thanks to Exercise 1.26, we know that the reflection through BD = k sends A to C. Therefore o′ 1 intersects o2, where o′ 1 denotes the reflection of o1 through k, (since C is on o′ 1 and o2). Thus a necessary condition for the existence of such a square is that o′ 1 and o2 must intersect.
Now we prove that this condition on o1, o2 and k is sufficient; that is, if o′ 1 and o2 intersect, then a square ABCD satisfying our problem exists. Let o′ 1 and o2 intersect in a point C, and let A be the reflection of C through the line k. Let O be the midpoint of AC; clearly O lies on k. Now let B ̸= D be the (only) two points on k such that |OB| = |OD| = |OA| = |OC|.
This construction is summarized in the figure below.
k o1 o′ 1 o2 C O A := C′ B D k o1 o′ 1 o2 O C A B D 12 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Then since BD is the bisector of AC and AC is the bisector of BD (prove this!), and |BD| = |AC| (by our construction), we can apply Exercise 1.26 to conclude that ABCD is a square. This completes the solution.
Definition 1.28. Here we introduce two concepts which will be used in the next example.
Let X be a point and k a line. The distance from X to k (written |X, k|) is the length of the perpendicular line segment from X to k.
k X Exercise 1.29. Prove that the distance |X, k| is the minimum distance between X and a point on k.
Exercise 1.30. If s = ABCD is a square of length 1 and l and k are lines such that AD = l and BC = k, then any point X on the segment AB satisfies |X, l| + |X, k| = 1.
k l A B C D Example 1.31. Suppose the vertices of a quadrilateral A, B, C, D lie on distinct sides of a square s of side length 1. Prove that the perimeter of ABCD is at least than 2 √ 2.
C A B D The solution is nicely summarized in the following figure. Let l1 be the line coinciding with the side of s which contains the point B. Reflecting s in l1 defines a new square s′ containing TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 13 the reflections A′B′C′D′ of our original quadrilateral ABCD.
C A B′ B D C′ A′ D′ l1 We repeat this process with the line l2 coinciding with the side of s′ containing the point C′; this defines the points A′′, B′′, C′′ and D′′. Repeating the process one last time, we reflect in the line l3 coinciding with the side of s′′ containing the point D′′. This construction is summarized in the figure below. Now note that |AB| + |BC| + |CD| + |DA| is equal to |AB| + |BC′| + |C′D′′| + |D′′A′′′|, and by the triangle inequality, this sum of distances is bounded below by |AA′′′|.
C A B D C′ A′ D′ A′′ B′′ D′′ C′′′ A′′′ B′′′ l1 l2 l3 l4 O Luckily, we can calculate |AA′′′|. Let O be the reflection of A′′′ through l2. Since l3 bisects A′′A′′′ and the reflection through l2 fixes l3 (since l3 ⊥l2) we deduce that l3 bisects A′O (since the l2 reflection sends A′′ to A′ and A′′′ to O). Therefore the l3 reflection sends A′ to O and so the distance from A′ to O is twice the distance from A′ to l3. Since the l1 reflection sends A to A′, the distance from A to A′ is twice the distance from A′ to l1. We obtain |AO| = |AA′| + |A′O| = 2 |l1A′| + 2 |A′l3| = 2(|l1A′| + |A′l3|) = 2, since l1 and l3 coincide with parallel sides of the square s′ (see Exercise 1.30). An easier argument shows that |OA′′′| = 2, and thus |AA′′′| is the hypotenuse of a right triangle with both non-hypotenuse side lengths equal to 2. We apply Pythagoras’ theorem (which we will prove later on!) to deduce |AA′′′|2 = 22 + 22, so |AA′′′| = 2 √ 2.
14 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Exercise 1.32. Let s be a unit square and let A be a point on one side of s (suppose A is not a corner). Prove that there is exactly one set of points B, C, D on the other sides of s such that the perimeter of ABCD satisfies the lower bound of 2 √ 2.
A Hint: referring to the figure in Example 1.2.4, draw a straight line from A to A′′′ and consider “folding” this straight line back into a quadrilateral.
Exercise 1.33 (continuation of previous exercise). Consider a point A and a unit square s as in the previous exercise. Suppose we “play billiards” in s, where we can hit a billiard ball initially located at A and let it bounce around the square. The rule of the game is that when the ball bounces offof an edge of the square the angle of incidence must the same as the angle of reflection.
A (This rule is a very good approximation to how a billiard ball actually moves, as most billiard players know.) Prove that if you hit the ball towards the adjacent wall in such a way that it comes back to A after hitting each of the other three walls exactly once, then the trajectory of the ball will be the unique quadrilateral from Exercise 1.32.
A TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 15 2. Rotations In this section we introduce rotations as a new tool we can use to prove theorems.
Definition 2.1. The rotation through an (oriented) angle α at a point O is the unique transformation of the plane satisfying two properties: (i) The point O is the only point fixed by the rotation (this is the centre of the rotation).
(ii) If A ̸= O, then A gets rotated to the unique point A′ such that ∠AOA′ = α and |OA′| = |OA|.
Theorem 2.2. Pick points A, B distinct from O such that ∠AOB = α/2; then the rotation is the reflection through k := OA followed by the reflection through l := OB.
Proof. The reason this theorem is true is best seen in the following figure.
X A B X′ X′′ k l α/2 Since O is clearly fixed under both reflections, it is clear that this composition of two reflec-tions is a transformation of the plane which satisfies (i). It remains to prove (ii), namely if X ̸= O then ∠XOX′′ = α, and |OX| = |OX′′|.
The angle ∠XOX′ is twice the angle ∠AOX′ since reflecting in k proves the equality ∠AOX′ = ∠XOA and clearly ∠XOX′ = ∠AOX′ + ∠XOA. Similarly, the angle ∠X′OX′′ is twice the angle ∠X′OB. Thus ∠XOX′′ = ∠XOX′ + ∠X′OX′′ = 2∠AOX′ + 2∠X′OB = 2∠AOB = α.
The second part is easy since reflections preserve distances so we know |OX| = |OX′| = |OX′′|. We have proved this composition of two reflections is a transformation which satisfies (i) and (ii) in Definition 2.1, so it must be the reflection through α at O.
□ Corollary 2.3. Rotations preserve distances and angles.
Proof. This follows from Fact 1.9 (which states that reflections preserve distances and angles), since any rotation is simply a composition of two reflections.
□ 16 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Fact 2.4. Let k, l be lines in the plane intersecting another line m. Then ℓand k are parallel if and only if α = β (in the figure, α is the red angle and β is the blue angle).
k m ℓ β α Exercise 2.5. We can actually prove this Fact, using the previous Fact 1.14.
(a) Let O be a point and ℓa line not containing O. Prove that the 180◦rotation around O sends ℓto a line ℓ′ parallel to ℓ. Hint: the 180◦rotation is a composition of reflection through any pair of perpendicular lines intersecting at O; choose a special pair.
(b) Suppose that ℓ∥k. In the context of Fact 2.4, Let m intersect ℓat a point A and k at a point B, and let O be the midpoint of the segment AB. Then the 180◦rotation through O sends A to B, and by part (a) sends ℓto a line ℓ′ parallel to ℓwhich contains B. Use Fact 1.14 to prove that ℓ′ = k.
(c) Use this rotation to conclude α = β. Hint: pick point C on ℓsuch that ∠OAC = α, and argue that ∠OBC′ = β, where C′ is the rotated image of C. Then use the fact that rotations preserve angles.
(d) For the converse, suppose the angles are equal. As in part (b), consider the rotation around O, and picking C ∈ℓsuch that ∠OAC = α = β, show that ∠OBC′ = β implies C′ must lie on k, so that the rotation of 180◦around O sends ℓto k. Conclude that ℓ∥k.
Corollary 2.6. The sum of interior angles in a triangle is 180◦.
D E C A B k ℓ Proof. Let k be the unique parallel to ℓ= AB passing through C. As in the figure, pick points D and E on k such that C is between D and E. Without loss of generality, we may TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 17 assume that the three angles ∠ACD, ∠BCA and ∠ECB do not intersect (by choosing D and B on opposite sides of AC). Then ∠ACD +∠BCA+∠ECB = ∠BCD = 180◦. Since k and ℓare parallel, we may apply Fact 2.4 to conclude ∠CAB = ∠ACD and ∠ABC = ∠ECB.
Then we conclude ∠CAB + ∠BCA + ∠ABC = 180◦, as desired.
□ Definition 2.7. A quadrilateral ABCD is a collection of four distinct segments called edges AB, BC, CD and DA; the four points A, B, C, D are called vertices. We require that no line contains three (or more) of the four points (i.e. ABCD is “non-degenerate”). If two edges intersect at a vertex (e.g. segments AB and AD intersect at A) we say that two edges are adjacent. We require that non-adjacent edges do not intersect.
In the figure below, the green figure is a genuine quadrilateral, and the red figures are not quadrilaterals.
C A B D C A B D C A B D Corollary 2.8. For any quadrilateral ABCD, the sum of the interior angles is 360◦.
Proof. As in the figure below, split the quadrilateral ABCD into two triangles ABC and ADC.
C A B D Then, using Corollary 2.6, we know that ∠DAC + ∠ACD + ∠CDA = 180◦and ∠CAB + ∠ABC + ∠BCA = 180◦. Since ∠DAB = ∠DAC + ∠CAB and ∠BCD = ∠BCA + ∠ACD, we conclude ∠DAB + ∠ABC + ∠BCD + ∠CDA = 360◦, as desired.
□ Theorem 2.9. Let k′ be the rotation of k around a point O by angle 0◦< α < 180◦. Then the angle between k and k′ is α. Here the angle between k and m is computed by starting 18 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
at any half-line of k and going clockwise until you hit one of the two half-lines of l.
O k k′ Proof. Let A′ be the rotation of A around O by the angle α.
If O ∈k, then the theorem follows immediately from the definition of a rotation. If not, since 0◦< α < 180◦, we claim that the two lines are not parallel and intersect at a point P.
To prove this claim, suppose not (so that k and k′ are parallel); then A, O and A′ all lie on a common line, so α must have been 180◦.
If O ̸∈k, then pick A ∈k such that OA is perpendicular to k, as shown below: P O k A A′ k′ Referring to the colours in the figure above, we note that since OA is perpendicular to k, the green angle between OA and k is 90◦. Since OA′ is the rotation of OA and k′ is the rotation of k, we conclude (since rotations preserve angles) that the second green angle is also 90◦. Since the angles inside a quadrilateral add up to 360◦, we know that the red angle plus the blue angle at O is 180◦. But we also know that the red angle plus the blue angle at the intersection of k and k′ is equal to 180◦(since they join to make a straight line) and so we conclude that the two blue angles are in fact equal. Since the blue angle at O is ∠AOA′, we deduce that it is the angle of rotation, namely α. This completes the proof.
□ TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 19 Problem 2.10. Let ABCD be a square and suppose E ∈DC and F ∈AD are such that |DE| = |AF|. Prove that AE and BF are perpendicular.
A B C D F E Solution. The trick is to rotate by 90◦through the centre of the square.
A B C D F E Then A 7→D 7→C 7→B 7→A, and since F is between AD, F ′ is between DC. Since E and F ′ are both between DC and located at a distance of |DE| from D, we conclude they are equal E = F ′. Hence the line FB is rotated to the line AE. We can thus apply Theorem 2.9 to conclude the angle between lines FB and AE is 90◦; in other words, they are perpendicular.
Definition 2.11 (Congruence). We say that two figures are congruent if one can be ob-tained from the other by a sequence of of reflections.
Exercise 2.12.
(1) Prove that all points are congruent.
(2) Prove that all lines are congruent.
(3) Prove that all circles of the same radius are congruent. Can two circles of differing radii be congruent?
(4) Consider a figure F defined by the union of two intersecting perpendicular lines, as shown below.
F F′ Prove that if F′ is another figure made out of two intersecting perpendicular lines, then F is congruent to F′.
20 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Exercise 2.13 (if you know what “equivalence relations” are). Prove that “congruence” is an equivalence relation on the set of all figures.
Definition 2.14 (Congruence of Triangles). For triangles ABC we modify the definition of congruence a little bit. We say ABC and A′B′C′ are congruent if there is a sequence of reflections taking A to A′, B to B′ and C to C′. In words, this modification of the definition of congruence takes the ordering of the vertices into account.
Theorem 2.15 (Triangle Congruence). ABC and A′B′C′ are congruent if and only if one of the following condition holds.
(i) (SSS) the three sides are equal; |AB| = |A′B′|, |AC| = |A′C′| and |BC| = |B′C′|.
(ii) (SAS) one of the pairs of corresponding angles are equal, and the two pairs of sides adjacent to this angle have equal length.
For example: ∠CAB = ∠C′A′B′ and |AC| = |A′C′| and |AB| = |A′B′|.
(iii) (ASA): two of the pairs of corresponding angles are equal, and the pair of segments which are adjacent to these angles is equal. For example: ∠CAB = ∠C′A′B′ and ∠CBA = ∠C′B′A′ and |AB| = |A′B′|. Note that equality in two angle pairs auto-matically implies the third angle pair is equal, so one does not care which ones they are.
In the figure below, we have shown an instance when the (SAS) criterion would apply: A B C A′ B′ C′ Remark. The following figure shows non-congruent triangles ABC and ABC′ with side-side-angle equality |AB| = |AB|, |AC| = |AC′| and ∠ABC = ∠ABC′.
A C′ B C While this remark demonstrates that “side-side-angle” equality is, in general, not sufficient to guarantee congruence of two triangles ABC and A′B′C′, there is one case where the “side-side-angle” equalities do guarantee congruence (see the Lemma below).
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 21 Definition 2.16. let ABC be an angle. The bisector of ∠ABC is a line k passing through B such that the reflection through k interchanges the two half-lines defining the angle ∠ABC.
A B C k A′ C′ Exercise 2.17. Prove that the bisector to an angle ABC always exists. Hint: pick e A on the ray BA so that ∠e ABC = ∠ABC, and such that e AB = |BC|. Let k be the bisector of the segment e AC. Prove that k bisects ∠ABC.
Proof (Proof of Theorem 2.15). Obviously, if ABC and A′B′C′ are congruent then (i), (ii) and (iii) all hold. The point of the theorem is that (i), (ii) and (iii) contain (a priori) less information than full-blown congruence.
In class we proved that the (SAS) criterion implies congruence. Here we will prove that the (SSS) criterion implies congruence. Suppose we have our two triangles ABC and A′B′C′, as shown below, and suppose that |AB| = |A′B′|, |BC| = |B′C′| and |AC| = |A′C′|.
A B C A′ B′ C′ If we reflect through the bisector of AA′, then A gets sent to A′, and B and C get sent to B′′ and C′′, respectively: B′′ C′′ A′ B′ C′ Assuming |AB| = |A′B′|, then |A′B′′| = |A′B′|, and so the bisector through B′B′′ contains A′. It follows that reflecting through the bisector of B′B′′ sends B′′ to B′, A′ to A′ (and C′′ to C′′′). If C′′′ = C′ then we got lucky, and the triangles are congruent. If C′′′ ̸= C′. then |A′C′′′| = |AC| = |A′C′| and |B′C′′′| = |BC| = |B′C′|, and so A′B′ is the bisector to the segment C′C′′′ (by Theorem 1.13). It follows that reflecting in A′B′ sends C′ to C′′′ (and 22 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
doesn’t move A′ or B′). This proves that ABC and A′B′C′ are congruent.
A′ B′ C′ C′′′ We remark that we needed, at most, three reflections to move ABC to A′B′C′.
□ Exercise 2.18. Prove that the (ASA) criterion for two triangles ABC and A′B′C′ implies congruence of ABC and A′B′C′. Hints: suppose |AB| = |A′B′| and, following the above proof, use two reflections to move the pair (A, B) to the pair (A′, B′).
Refering to the notation of the proof of Theorem 2.15, assume that C′ and C′′ are on opposite sides of A′B′.
Consider A′C′C′′ (see below). Prove that ∠A′C′C′′ = ∠A′C′′C′ (by reflecting one triangle through A′B′, and showing that the line B′C′′ must be reflected to B′C′ and A′C′′ must be reflected to A′C′, so that C′′ must be reflected to C′) and conclude (by Theorem 1.17) that A′ lies on the bisector through C′C′′. Similarly, conclude that B′ lies on the bisector through C′C′′, and thus A′B′ is the bisector through C′C′′. It follows that the reflection through A′B′ sends C′ to C′′ (as in the proof of Theorem 2.15, this is exactly what we want!).
A′ B′ C′ C′′ Problem 2.19. Let ABC be a right-angled triangle with |AB| = 3 |AC| and let D, E lie on side AB be chosen so that |AD| = |DE| = |EB|. Prove that ∠CDA+∠CEB+∠CBA = 90◦.
A D E B C Solution. It helps to build a 3 × 3 grid to house our triangle as in the figure below. With the labels from the figure below, we remark that the counterclockwise rotation of 90◦around C sends the rectangle ACE′E to KCF ′F. From this rotation we deduce two things: (1) the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 23 diagonal CE has the same length as the corresponding diagonal CF and (2) ∠ECF = 90◦.
Therefore the triangle ECF is a right angled isosceles triangle, and so we conclude ∠FEC = ∠CFE = 45◦= ∠CDA.
It is also clear from the figure that triangle EGF is congruent to triangle BAC, since are both right angled triangles with equal side lengths |AB| = |EG| and |FG| = |AC| (here we are using “side-angle-side” criterion to deduce congruence). By congruence, ∠GEF = ∠CBA. Combining everything we have concluded so far, we obtain ∠CBA + ∠CEA + ∠CDA = ∠GEF + ∠FEC + ∠CEA = ∠GEA = 90◦, and this completes the solution.
G A D E B C K E′ F ′ F Proposition 2.20. If |CA| = |C′A′|, |AB| = |A′B′| and ∠ABC = ∠A′B′C′ = 90◦, then ABC and A′B′C′ are congruent.
A C′ B C Proof. Since |AB| = |A′B′|, we can follow the proof of Theorem 2.15 and move A′B′C′ so that A′B′ coincides with AB.
Using another reflection if necessary, we may assume that C and C′ are on opposite sides of the line AB = A′B′.
Since ∠CBA = 90◦and ∠ABC′ = 90◦, we conclude that ∠CBC′ = 180◦, and thus B lies between CC′.
Since |AC| = |AC′|, we know that ACC′ forms an isosceles triangle. Then by Corollary 1.18, we 24 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
know that ∠ACC′ = ∠AC′C and consequently ∠ACB = ∠AC′B. Thus we may now use the “angle-side-angle” criterion for triangle congruence to deduce ABC is congruent with ABC′.
□ Definition 2.21. A parallelogram is a quadrilateral ABCD such that AB ∥CD and BC ∥AD and |AB| = |DC|, |AD| = |BC|.
A B C D Proposition 2.22. Let ABCD be a quadrilateral such that any of the following three conditions holds.
(a) |AB| = |DC| and |BC| = |AD|.
(b) |AB| = |DC| and AB ∥DC.
(c) AB ∥DC and BC ∥AD.
Then ABCD is a parallelogram.
Proof. Without loss of generality, assume that the diagonal segment AC is contained inside the quadrilateral.
A B C D First we note that if we can show that ACB is congruent to CAD, then we will be done.
Congruence easily gives us equality of all sides, and it also gives us parallel edges, as the following argument shows. If ∠CAB = ∠ACD, then the line AC cuts the two lines AB and CD in equal angles. Previously we have shown that this implies AB ∥CD. A similar argument shows that BC ∥AD. Therefore if ACB is congruent to CAD then ABCD is a parallelogram.
Case (a): If |AB| = |CD| and |BC| = |AD|, then the “side-side-side” criterion for con-gruence holds between triangles ACB and CAD.
Case (b): If |AB| = |CD| and AB ∥CD, then AC cuts AB and CD in equal angles so that ∠ACD = ∠CAB. Then we may use “side-angle-side” to conclude that triangle BAC is congruent to DCA.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 25 Case (c): If AB ∥CD and BC ∥AD, then AC cuts CD and AB in equal angles, and it cuts AD and BC in equal angles. Since triangles ACD and CAB share a side AC, we can use the “angle-side-angle” criterion to conclude ACD and CAB are congruent.
□ Exercise 2.23. Find a quadrilateral ABCD such that |AD| = |CB| and AB ∥CD yet ABCD is not a parallelogram.
3. Circular arcs Definition 3.1. Let A, B be points on a circle o with centre O. The arc AB is the arc of o going clockwise from A to B.
The central angle corresponding to the arc AB is the oriented angle AOB.
An inscribed angle is the oriented angle ASB with S ∈o outside the arc AB.
We say that the arc AB subtends the central angle AOB and the inscribed angle ASB (i.e. arcs subtend angles).
In the figure below we have shown an arc AB, the central angle ∠AOB it subtends and two incribed angles it subtends. We emphasize that, as shown, there is more than one inscribed angle corresponding to the arc AB.
A S S′ O B Theorem 3.2. Let A, B lie on a circle o. The measure of any inscribed angle subtended by the arc AB is half the measure of the central angle subtended by arc AB.
Corollary 3.3. Any two inscribed angles subtended by arcs of the same length have the same measure.
Remark. The length of an arc is proportional to the measure of the central angle it subtends, with proportionality constant equal to 2πR/360, where R is the radius of the circle. One can easily prove this whenever the measure θ of the central angle α is equal to m360◦/n, where m and n are natural numbers (by subdividing the circle into n congruent arcs and using a rotation to show that the arc subtending α has the same length as m of these n arcs). The case for general angle measure follows by an approximation by rational angle measures.
26 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
As a corollary to the proportionality between the subtended central angle’s measure and the arc’s length, we know that two arcs have equal length if and only if they subtend central angles with the same measure (assuming that the arcs lie on the same circle).
Corollary 3.4. Any inscribed angle subtended by arc AB, where the segment AB is a diameter of o, is 90◦.
A S O B Exercise 3.5. Prove Corollary 3.3 and Corollary 3.4 using Theorem 3.2.
Exercise 3.6. Let ASB be a right-angled triangle, with the right angle at S. If O is the midpoint of AB, prove that |AO| = |OB| = |OS|.
Proof of Theorem 3.2. We will consider three cases.
Case 1: assume that ∠AOB < 180◦, and that SA forms a diameter of the circle (see figure). This gives us ∠BOS + ∠AOB = 180◦, and since OSB is an isosceles triangle, we have ∠OSB = 1 2(180◦−∠BOS) = 1 2∠AOB, and since ∠ASB = ∠OSB, we have shown 2∠ASB = ∠AOB, as desired.
A S O B Case 2: The angle ∠ASB contains O. This case follows relatively easily from the figure.
From Case 1, we see that the cyan (cyan = light blue) angle is half the dark blue angle, and the orange angle is half the red angle. In other words, if we choose C so that SC form a diameter of the circle, then 2∠ASC = ∠AOC and 2∠CSB = ∠COB. Since ∠AOB = ∠AOC + ∠COB, and ∠ASB = ∠ASC + ∠CSB, adding our previous equalities gives us TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 27 2∠ASB = ∠AOB, as desired.
A S O B C Case 3: The angle ∠ASB does not contain O. This case is very similar to Case 2, although we replace the addition of angles used in Case 2 by subtraction of angles. The details are left to the reader.
A S O B C □ Corollary 3.7. Given a circle, any two inscribed angles with the same measure are sub-tended by arcs of the same length.
Proof. This follows immediately from Theorem 3.2, since it tells us that the central angle’s measure is twice the inscribed angle’s measure. We mentioned above that the arc’s length is proportional to the measure of the central angle it subtends.
□ Corollary 3.8. Given points A ̸= B, let σ be the half plane bounded by AB containing all points P such that ∠BPA < 180◦(here we mean the clockwise oriented angle).
A B P σ Then for any α > 180◦, the set of points P such that ∠BPA = α is the arc AB of the circle o with centre O on the bisector of AB satisfying ∠BOA = 2α. Remark: note that O may 28 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
lie on either side of AB, depending on the size of α.
A B P O o A B P O o Proof. By Theorem 3.2, it is clear that if P lies on the circle o, then ∠BPA = α, since ∠BPA is the inscribed angle of the arc AB subtending a central angle of measure 2α.
Conversely, we need to show that ∠BPA = α implies P lies on the circle o. In search of a contradiction, suppose that P lies offof the circle. We consider two cases.
Case 1: P lies inside the circle o. Then extend the line AP until it intersects o in a point X, as shown below.
A B X P O o Now the idea is to look at the angles in the triangle BPX. We know that ∠XPB+∠BPA = ∠XPA = 180◦(XPA lie on a straight line). But ∠BPA = α, and so ∠XPB = 180◦−α.
However, ∠BXP = α, since X lies on the circle o (we proved this in the first part of this proof). Since ∠BXP + ∠XPB + ∠PBX = 180◦, we conclude that ∠PBX = 0, which is a contradiction (since it implies the lines BX and BP are equal, and since BX intersects AP in a single point, we deduce that X = P, but we assumed P ̸∈o).
Case 2: P lies outside the circle o. Then pick X ∈o inside the angle ∠BPA, and let D be the intersection between AX and BP, as shown in the figure below.
A B X D P O o TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 29 Note that ∠BDX + ∠DXB < 180◦(since these are two angles in a triangle). However ∠DXB + ∠BXA = 180◦, and ∠BXA = α by the first part of this proof. Comparing the inequality with the equality yields ∠BDX < α. A similar argument shows that ∠BPA < ∠BDX < α. This contradicts our assumption that ∠BPA = α. This completes the proof.
□ Problem 3.9. Suppose that point P lies inside a parallelogram ABCD with ∠ABP = ∠PDA. Prove that ∠DAP = ∠PCD.
P A B C D Solution. The idea is to translate the triangle DCP along the parallel edges AD and BC so that edge DC coincides with the parallel edge AB, and let P be translated to a point E, as shown below.
P A B C D E It is clear that AEB and DPC are congruent triangles, and so, in particular |AE| = |DP|.
Then since ∠CDP = ∠BAE, we know that AE is parallel to DP. By the characterization of parallelograms in Proposition 2.22 we conclude that DPAE is a parallelogram. In particular we conclude that the opposite angles ∠PDA and ∠AEP are equal. By our assumption, ∠PDA = ∠ABP, and so ∠ABP = ∠AEP. By Corollary 3.8, we conclude that AEPD lie on a circle, as shown below.
A E P D B C Then we apply Corollary 3.8 again to conclude ∠EBA = ∠EPA. Since the line AP cuts the parallel lines AD and PE in equal angles we deduce ∠EPA = ∠DAP. Combining our 30 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
results so far, we have ∠EBA = ∠DAP. Since triangle DCP was congruent to ABE, we know ∠EBA = ∠PCD, so ∠PCD = ∠DAP, which was what we wanted to show.
Definition 3.10. A line k is tangent to a circle o if k intersects o in a single point.
O o k A Proposition 3.11. If A lies on the circle o centred at O, then a line k is tangent to o at A if and only if k is perpendicular to the radius OA.
O k A Proof. First suppose k contains A and is perpendicular to OA. Then the reflection through line OA fixes k and o (when we say the reflection “fixes” k we mean k is its own reflection).
It follows that the reflection sends intersection points of k and o to intersection points of k and o. If there is another point B ̸= A which lies on k and o, then B cannot lie on OA, so B gets reflected to some other point B ̸= B′ ̸= A. But then A, B, B′ all lie on the intersection of o and k, which is impossible since we proved a line intersects a circle in at most 2 points (Exercise 1.25). Therefore the perpendicular to OA passing through A is tangent to o. It remains to prove that this is the only tangent through A.
Suppose that k is not perpendicular to OA, then let B be the projection of O onto k, as shown below.
O k A B Consider the reflection in line OB; as in the first part of this proof, this reflection fixes k (since k is perpendicular OB) and fixes o (since any reflection through a line passing through the centre of a circle fixes that circle). Since A is the only intersection point of o and k, the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 31 reflection through OB must send A to itself, so OB and k intersect in two points A and B.
It follows that k = OB, which is a contradiction, since it is obvious that any line passing through the centre of a circle o intersects o in two points. This completes the proof.
□ Problem 3.12. Let k be a line tangent to a circle o at A. Let B, C be distinct points on o distinct from A. Let D ∈k so that the angle DAB contains C. Then ∠ABC = ∠DAC.
k A B C D Remark. If we let E be a point lying on the same side of AC as B, as shown below, then ∠DAC can be thought of ∠AEC in the limit as E →A. This motivational remark is not meant to be a proof.
k A B C D E Solution. Assume that ∠DAC < 90◦(see the remark below). Pick B′ such that AB′ is a diameter and so that the arc AC subtends the inscribed angle AB′C (so that ∠ABC = ∠AB′C).
k A B C D B′ Then, since the angle B′CA is subtended by a diameter, ∠B′CA = 90◦. Therefore ∠CAB′+ ∠AB′C = 90◦. Since AD is tangent to the circle, and AB′ is a diameter (so it contains the radius through A), we conclude by Proposition 3.11 that AD and AB′ are orthogonal, so that ∠DAB′ = 90◦.
The angle ∠DAB′ contains C; this requires the assumption that ∠DAC < 90◦, in addition to the assumption that angle DAB contains C. Then we may write ∠DAC + ∠CAB′ = ∠DAB′ = 90◦= ∠CAB′ + ∠AB′C, so ∠DAC = ∠AB′C, as desired.
32 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Remark. The proof we give below is the case when that ∠DAC ≤90◦. The problem still has a solution when ∠DAC ≥90◦, but it will require a modified argument, since we will not be able to conclude that ∠ABC = ∠AB′C (see the figure below).
B′ k A B C D D′ Exercise 3.13. Adapt the solution of Problem 3.12 to prove the case ∠DAC ≥90◦. Hint: prove the case ∠DAC = 90◦separately, and in the case ∠DAC > 90◦, consider a point D′ on line k as in the figure shown in the above Remark, and then apply the solution of Problem 3.12 with B and C interchanged.
Problem 3.14. Let ABC be a triangle with |AB| ̸= |BC|. Show that the bisector of angle ABC intersects the bisector of segment AC in a point X lying on the circumscribed cricle.
X A B C Solution. A good place to start when solving a problem which asks you to prove three figures intersect in a single point X is to guess what the point is, and then (if you guessed right) show that the three figures all contain it. In our problem, the three figures are the bisector of angle ABC, the bisector of AC and the circle. After staring at our figure for a while, we guess that X should be the point on the circle which splits the arc AC (not containing B) into two equal length arcs.
X A B C With this choice of X, we immediately realize that ∠XBC = ∠ABX since the angles XBC and ABX are subtended by arcs of the same length (recall Corollary 3.7).
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 33 Furthermore, it is intuitively obvious that |AX| = |CX|. This can be rigourously proved by noting that AOX and COX are congruent by “side-angle-side” criterion, since the central angles subtended by arcs AX and XC are equal since the arc lengths are equal (as shown below), and the sides OA, OX and OC all have equal length (since they are all radii of the circle). Thus the third side pair is also equal |AX| = |CX|.
X A B O C But then since |AX| = |XC| we know that X lies on the bisector of AC.
X A B C We have shown that X lies on both the bisector through angle ABC and through segment AC, and this completes the solution.
Definition 3.15. Let ABCD be a quadrilateral. We say that ABCD is convex if both diagonals AC and BD lie in the interior of ABCD.
A B C D Theorem 3.16. Let ABCD be a convex quadrilateral. Then the following are equivalent.
(i) ABCD lie on a common circle.
(ii) ∠BDA = ∠BCA.
(iii) ∠DAB + ∠BCD = 180◦.
34 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Proof. We begin by proving (i) and (ii) are equivalent. By convexity of ABCD, the points CD lie on the same side of AB, and so the equivalence of (i) and (ii) follows from a direct application of Corollary 3.8.
A B C D Next we will prove that (i) implies (iii). Assuming the points ABCD lie on a circle, as shown in the following figure, we have 2∠DAB = ∠DOB, since the angle DOB is the central angle corresponding to the inscribed angle DAB. Similarly 2∠BCD = ∠BOD. Therefore 2 (∠DAB + ∠DOB) = ∠DOB + ∠BOD = 360◦, which is what we wanted to show.
A B C D We show that (iii) implies (i). Let O be the centre of the circumscribed circle of ABD.
Since ∠DOB is twice ∠DAB and ∠BOD+∠DOB = 360◦, we conclude 2∠BCD = ∠BOD.
However, if we choose any other C′ on the arc DB of the circle through ABD we will also have 2∠BC′D = ∠BOD. By Corollary 3.8, this implies that ∠BCD is the correct value for C lies on the circle centred on O, and so C lies on the circle through ABD. This completes the proof.
□ TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 35 Corollary 3.17. Let ABCD be a quadrilateral made out of two right angles, as shown below. (So ∠ABC and ∠CDA are 90◦).
A B C D Then ABCD has a circumscribed circle, and furthermore, the circumscribed circle is centred on the midpoint of AC.
A B C D O Proof. Left to the reader.
□ Problem 3.18. Let ABCD be a square, and pick E and F on sides AB and BC, respec-tively, so that |BE| = |BF|. Let P be the projection of B onto CE. Prove that ∠DPF is 90◦.
A B C D P F E Solution. We begin our solution by extending the segment BP until it intersects AD at a point Q.
A B C P Q E 36 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
We claim that triangles QAB and EBC are congruent. To prove this claim, we will use the “side-angle-angle” criterion for congruence.
Since ∠BCE = ∠BCP and BCP is a right angled triangle (by definition of P) we deduce ∠BCE + ∠QBC = 90◦. Since ABC is a right angle, we have ∠ABQ + ∠QBC = 90◦. From these two equalities, we deduce ∠BCE = ∠ABQ. Since ∠QAB = ∠EBC = 90◦, we now know that two angle pairs in triangles QAB and EBC are equal. Since the triangles also have an equal side length pair |AB| = ∠BC, we have satisfied the requirements for the “side-angle-angle” criterion, and we conclude QAB and EBC are congruent.
Thanks to this congruence, we conclude |QA| = |EB| = |BF|. Thus |DQ| = |CF|. Since DQ ∥CF, we conclude QFCD is a parallelogram (by Proposition 2.22) and since angle QDC is a right angle, we deduce QFCD forms a rectangle. Then, since the opposite angles at D and F add up to 180◦, we deduce by Corollary 3.17 that QFCD has a circumscribed circle. However, since ∠QPC = 90◦, and ∠QDC = 90◦, we deduce by Corollary 3.17 that QPCD also has a circumscribed circle. There is only one circle circumscribing the three points QCD, and so we conclude that the circumscribed circles of QFCD and QPCD are the same, so the five points QPFCD lie on a common circle. This is summarized in the figure below.
C D P F Q But then the quadrilateral DPFC is circumscribed by a circle.
C D P F Q so applying Corollary 3.17 once more, we deduce ∠DPF + ∠FCD = 180◦, and since ∠FCD = 90◦(since it is one of the corners of our original square) we deduce ∠DFP = 90◦, as desired. This completes the solution.
Exercise 3.19 (The Simson Line, hard exercise). Let ABC be a triangle and let P lie on the circumscribed circle of ABC. Let K, L, M be the projections of P onto the lines BC, AC and AB, respectively. Then the points K, L, M lie on a common line.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 37 4. Circles inscribed in angles and the “strongest theorem of geometry” Theorem 4.1. Let X be a point lying inside an acute angle AOB. Let P and Q be the projections of X onto the half-lines OA and OB, respectively. Note that P and Q exist because AOB is an acute angle.
P A Q B X O Then the following are equivalent.
(i) X lies on the bisector of the angle AOB.
(ii) |OP| = |OQ|.
(iii) |XP| = |XQ|.
Proof. To see that (i) implies (ii) and (iii), we use the “side-angle-angle” criterion for congruence to prove OXP and OXQ are congruent.
By the assumption (i) we know ∠POX = ∠QOX, and, since they are right triangles, we know ∠XPO = ∠XQO. They obviously share the side OX, and so we have enough information to apply the “side-angle-angle” criterion. Then (ii) and (iii) follow immediately from congruence of OXP and OXQ.
For an alternate proof, we simply reflect through the bisector of AOB. Since O and X lie on this bisector, O and X are fixed by the reflection. By definition1 of the bisector of an angle, the half lines OA and OB are interchanged by the reflection (since the projections must be interchanged). It follows that P and Q are interchanged by this reflection, and since reflections preserve distances we have |OP| = |OQ| and |XP| = |XQ|, as desired.
To prove (ii) implies (i) we use Proposition 2.20 which is the “side-side-angle” criterion for congruence when the equal angle pair is a pair of right angles. We can use this proposition to prove triangles OXP and OXQ are congruent, since we have equal side length pairs |OP| = |OQ| and |OX| = |OX|, and the equal angle pair ∠XPO = ∠XQO of right angles.
Then, by this congruence, we have ∠PXO = ∠OXQ, and so, indeed X lies on the bisector of the angle POQ (which is the same angle as the angle AOB).
A very similar argument (also using Proposition 2.20) proves the implication (iii) = ⇒(i), and we leave this to the reader.
□ Exercise 4.2. In the setting of Theorem 4.1, prove that as long as the projections P and Q of X onto the lines OA and OB lie on the half-lines OA and OB, then the theorem is true, 1some define the bisector of an angle as the line which interchanges the two half-lines defining the angle.
38 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
even if the angle AOB is not acute.
P A Q B X O Exercise 4.3. Let P lie outside a circle o centred at O. Prove that there are exactly two points A and B on o so that PA and PB are tangent to o. Hint: Consider the circle whose diameter is OP.
o A B O P Theorem 4.4 (strongest theorem of geometry). Let P lie outside a circle o centred at a point X, and pick A and B on o so that PA and PB are tangent to o. Then |PA| = |PB| o A B X P Proof. First we remark that the points A and B are well-defined thanks to Exercise 4.3.
Clearly |AX| = |BX| and since PA and PB are tangent to the circle o, we also have XA ⊥PA and XB ⊥PB (we proved this in Proposition 3.11). Thus we can use (iii) = ⇒(ii) in Theorem 4.1 to conclude |PA| = |PB|, as desired.
□ Remark. If we want to use Theorem 4.1, then we should require that ∠APB ≤90◦. How-ever, as mentioned in Exercise 4.2, as long as the projections remain on the angle, it does not matter if the angle is not acute. Since the points A B always lie on the angle APB in the statement of the “strongest theorem of geometry,” we can use Theorem 4.1 without worrying about acuteness of the angle APB.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 39 Theorem 4.5. In a triangle ABC the three bisectors of its angles intersect in a common point.
k l m A B C Proof. As in the figure above, let k, l, m be the bisectors of the angles at A, B, C, respec-tively. Let O be the point of intersection between k and l, and consider the projections P, Q, R of O onto the sides opposite A, B, C, respectively.
k l A B C O P Q R Since O lies on the bisector of CAB, we know ∠CAO and ∠OAB are both less than 90◦, and so the projections of O onto the lines AC and AB lie on the half-lines AC and AB, so we may apply Theorem 4.1 to the angle CAB and point O. Similarly, we may apply Theorem 4.1 to the angle ABC and point O (i.e. we do not need to worry about acuteness of angles).
Since O lies on the bisector of CAB we know from Theorem 4.1 that |QO| = |OR|, and (similarly) since O lies on the bisector of ABC we know |OR| = |OP|. But then |OQ| = |OP| so we conclude, again by Theorem 4.1, that O lies on the bisector of BCA. This completes the proof.
□ Corollary 4.6. In the setting of the above theorem, the circle o centred at O with radius |OR| = |OP| = |OQ| is the a circle contained in the triangle ABC which is tangent to the edges of ABC. In other words it is an inscribed circle of the triangle ABC. We have 40 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
therefore shown that every triangle admits an inscribed circle.
A B C O P Q R Exercise 4.7. We have shown above that every triangle admits an inscribed circle. However, we have not shown that the inscribed circle is unique. Prove that every triangle has only one inscribed circle. Hint: What must be the centre of any inscribed circle?
5. Altitudes of Triangles and Escribed Circles Definition 5.1. The altitude in a triangle ABC through the vertex C is the line perpen-dicular to AB through C. The intersection of the altitude with the line AB is called the foot of the altitude. In the figure below we have shown the altitude through C and its corresponding foot F in a few cases.
A B C F A B C F A B C F Theorem 5.2. The three altitudes in any triangle intersect in a common point.
A B C D E F Proof. We will only prove the theorem when all the feet lie inside the triangle, although the theorem is still true if we drop this assumption.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 41 Let D, E, F be the feet of the altitudes of A, B, C, respectively. Since BEA and BDA are both right angles, we conclude that E and D lie on the circle whose diameter is AB.
A B D E Then the quadrilateral ABDE is inscribed in a circle, so we can apply Theorem to conclude ∠EAB + ∠BDE = 180◦. Since BDC are collinear, we have ∠BDE + ∠EDC = ∠BDC = 180◦, and combining this with our previous equality, we deduce ∠EAB = ∠EDC.
A B C D E In a similar fashion, we deduce ACDF lie on a common circle.
A B C D F Since ∠CAF +∠FDC = 180◦, as we argued above this implies ∠CAB = ∠DFB. Combining this with what we have shown already yields ∠EDC = ∠BDF.
A B C D E F Since AD is perpendicular to BC we have ∠EDC+∠ADE = 90◦and ∠BDF+∠FDA = 90◦, so using our previous equality, we conclude ∠FDA = ∠ADE. This implies that AD is the 42 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
bisector of the angle FDE.
A B C D E F Then, relabeling A 7→B 7→C 7→A and D 7→E 7→F 7→D, the same argument we just gave will prove that each altitude of ABC is the bisector of the corresponding angle of the triangle DEF.
A B C D E F Since we proved in Theorem 4.5 that the bisectors of the angles of a triangle intersect in a single point, we conclude that the altitudes of ABC, which are the bisectors of the angles of DEF intersect in a single point, and this completes the proof.
□ Definition 5.3. Given a triangle ABC we define the bisector of the exterior angle at the vertex A as follows. The lines AC and AB intersect at A and make 4 angles, which split into two pairs of equal measure, as shown in the figure below.
A B C TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 43 As shown below, there is a unique bisector k of the first (interior) angle pair, and a bisector ℓof the second (exterior) angle pair.
A B C k ℓ Exercise 5.4. Let ABC be a triangle, and prove that the exterior bisector and interior bisector at the vertex C are perpendicular. Hint: show that the reflection in the interior bisector fixes the lines AB and lines AC, and hence fixes the exterior bisector.
Exercise 5.5. Prove that the exterior bisectors at A and B and the interior bisector at C intersect at a single point. (Obviously we have a similar result when we permute the vertices A 7→B 7→C 7→A).
Exercise 5.6. Let ABC be a triangle. Prove that the feet of A and B do not lie inside the triangle if and only if ∠BCA > 90◦. In thise case, let DEF be the triangle made out of the feet of ABC. Using Execise 5.5, let P be the intersection point of the interior angle bisector at F, and the exterior angle bisectors at D and E. Prove that the altitudes of ABC intersect at P.
A B C D E F P Definition 5.7. Let ABC be a triangle, and let O be the intersection point of exterior angle bisectors at B and C and the interior angle bisector at A (this intersection point is guaranteed by Exercise 5.5). A fairly straightforward application of Theorem 4.1 proves that O lies at an equal distance d from the sides AB, AC and BC. The circle centred at O with 44 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
radius d is called the escribed circle of ABC opposite A.
A C B O Problem 5.8. Let ABC be a triangle and let O be the centre of the escribed circle opposite A. Let P, Q, R be the projections of O onto BC, AC and AB. Then 2 |AQ| is equal to the perimeter of ABC.
A B C O P Q R Solution. Our solution uses the “strongest theorem of geometry” (Theorem 4.4). The first application of the theorem yields |AQ| = |AR|. Therefore, proving 2 |AQ| is equal to the perimeter of ABC is equivalent to proving |AR| + |AQ| = perimeter. Noting that |AQ| = |AC| + |CQ| and |AR| = |AB| + |BR| and that |CQ| = |CP| and |BR| = |BP| by the “strongest theorem of geometry,” we obtain |AQ| + |AR| = |AC| + |AB| + |CP| + |BP| = |AC| + |AB| + |BC| .
This completes the solution.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 45 Problem 5.9. Let ABC be a triangle and consider the inscribed circle and the escribed circle opposite A. Label the tangency points as in the following figure. Show that |CD| = |BP|.
A B C D E F P Q R Solution. A simple application of the “strongest theorem of geometry” shows that |AF| + |FB| + |CD| = |AB| + |CD| is half the perimeter of ABC.
A B C D E F However by the preceeding problem, we also have |AR| = |AB| + |BP| equal to half the perimeter. Since |AB| + |CD| = |AB| + |BP|, we deduce |CD| = |BP|, as desired.
Definition 5.10. A circle o is inscribed in a quadrilateral ABCD if o is tangent to each of the sides AB, BC, CD and DA.
A B C D 46 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
In the figure below we show a quadrilateral A′B′CD which does not have an inscribed circle.
A B A′ B′ C D E Theorem 5.11. A convex quadrilateral ABCD has an inscribed circle if and only if |AB| + |CD| = |AD| + |BC| .
Proof. We will only prove the = ⇒ direction, and the other direction is presented in Exercise 5.12. Suppose that ABCD has an inscribed circle o, and let K, L, M, N be the tangency points of edges AB, BC, CD, DA with o, respectively.
A B C D K L M N Then the “strongest theorem of geometry” (applied four times) yields |AK| = |AN| , |BK| = |BL| , |CL| = |CM| , |DM| = |DN| , and since |AB| = |AK| + |BK| |BC| = |BL| + |CL| |CD| = |CM| + |DM| |DA| = |DN| + |AN| we deduce |AB| + |CD| = |BC| + |AD|, as desired.
□ Exercise 5.12. The goal of this exercise is to establish the unproved direction in Theo-rem 5.11: If ABCD is a convex quadrilateral satisfying |AB| + |CD| = |AD| + |BC|, then ABCD has an inscribed circle.
(a) Prove that convexity is a necessary assumption.
(b) Suppose that |AD| is the smallest side. Prove that |BC| must be the largest side.
(c) If |AD| = |BC|, prove that all sides are equal, and so we have a rhombus. Show that, in this case, the diagonals AC and BD are also the bisectors of corresponding angles. Prove that there is a circle centred at the intersection point P of AC and BD TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 47 inscribed in ABCD. Hint: Consider the projections K, L, M, N of P onto the sides of ABCD.
(d) If |AD| = |DC|, then |AB| = |BC|, and so we have a deltoid (also known as a kite) A B C D Show that, in this case, DB is the the angle bisector of D and B, and the angle bisectors at A and C intersect on BD. As in part (c) prove that the intersection of all the angle bisectors is the centre of an inscribed circle. Note that (d) implies (c).
(e) Now we assume that |AD| is not equal to either adjacent side length |DC| or |AB|.
A B C D K L P Let K, L be points on BC and DC so that |DL| = |AD| and |BK| = |AB| (why is this possible?). Prove that |LC| = |KC|. Prove that, as shown in the figure above, the angle bisectors at B, C, D are the bisectors of the sides AK, KL and LA of the triangle AKL. Therefore the angle bisectors of B, C, D intersect in a single point P.
Prove that P is the centre of an inscribed circle of ABCD.
Problem 5.13. Consider a triangle EBF and a point D inside the triangle. Let C and A be the intesection points of FD and ED with BE and BF, respectively.
E F A B C D Prove that if |AB| + |DC| = |AD| + |CB|, then |EB| + |DF| = |DE| + |BF|.
48 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Solution. Using Theorem 5.11, we know that ABCD has an inscribed circle. Let K, L, M, N be the tangency points of the circle on the edges, as shown.
E F A B C D K L M N We compute |BE| = |BK| + |EK| = |BL| + |EM| = |BL| + |DE| + |DM| = |BL| + |DE| + |DN| , where we have used the “strongest theorem of geometry” in the second and fourth equalities.
We obtain |BE| + |DF| = |BL| + |DE| + |FN| = |BL| + |DE| + |FL| = |BF| + |DE| , where we have used the “strongest theorem of geometry” in the second equality to write |FN| = |FL|. This completes the solution.
6. Area and Thales’ Theorem Fact 6.1 (the existence of area function). There exists a non-negative area function for polygons, satisfying (i) The area of a rectangle ABCD is |AB| · |BC|.
A B C D (ii) If polygons P1 and P2 have disjoint interiors, then Area(P1 ∪P2) = Area(P1) + Area(P2).
(iii) If polygons P1 and P2 are congruent, then Area(P1) = Area(P2).
We will use the notation |P| := Area(P).
Remark. The usual “proof” of this fact simply defines the area of a triangle, and then defines the area of an arbitrary polygon P by “triangulating” the polygon (as shown below) and then defining the area of P as the sum of the areas of the triangles comprising the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 49 triangulation.
However, there is a major flaw with this “definition,” as it a priori depends on the partic-ular triangulation of P used. One must then prove that the area does not depend on the triangulation, and this is not easy to do.
There are more abstract proofs of the existence of the area function which use the ma-chinery of “measure theory” (a branch of real analysis).
Theorem 6.2. (a) The area of a parallelogram with side length a and height h is ah.
a h A B C D (b) The area of a triangle with base a and height h is ah/2.
a h A B C (c) The area of a trapezoid with parallel side lengths a, b and height h is (a + b)h/2.
a b h A B D C Proof. To prove (a), let K ∈AB be the foot of the perpendicular to AB through D, and suppose that K lies between AB. Drop a perpendicular to AB from C, intersecting the line 50 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
AB in a point L, as shown below.
a h A B C D K L Then ADK and BCL are congruent by side-angle-angle criterion, since |AD| = |BC| (by Definition 2.21, where we defined parallelograms) ∠KAD = ∠LBC (since AD ∥BC), and ∠AKD = ∠BLC = 90◦. It follows by property (iii) of the area function that |ADK| = |BCL|, so |ABCD| = |KBCD| + |AKD| = |KBCD| + |BCL| = |KDCL| = ah, where we have used part (ii) in the first and third equalities, and part (i) in the final equality.
We proceed in the same way of the projection of C lies in AB. Otherwise, after possibly interchanging C with D (and A with B) we can assume that the projection of D separates the projection of C from AB on the line AB. Let E be the point at distance |DC| from D distinct from C on the line CD. Note that AE ∥BD (show this!). Suppose, without loss of generality, that |BD| ≤|AC|.
a h A E B C D K L Triangles ADE and BCD are congruent by “side-side-angle” criterion with |ED| = |DC| , |AE| = |BC| and ∠DEA = ∠CDB (prove this) and so a similar argument to the one used in the first part of the proof shows |ABDE| = |ABCD|. Repeat this process until either D or C has its projection K inside the segment AB (convince yourself that this will eventually happen).
To prove part (b) of the theorem, we begin with a triangle ABC, and then we rotate 180◦through the midpoint of AC to obtain a parallelogram ABCB′, where B′ is the rotated TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 51 image of B, as shown below.
A B C B′ h By construction ABC and AB′C are congruent, so they have the same area. It follows from part (a) that |ABC| = ah/2.
For part (c), we let the diagonal AC cut the trapezoid into two triangles with the same height and sides lengths a and b, respecitvely. The result now follows from part (b).
□ Armed with the area function, we can give a straightforward proof of Thales’ Theorem.
Theorem 6.3. Thales’ Theorem If ℓ, k are two lines intersecting at a point A, points D, E and B, C lie on ℓand k, respectively, and DB ∥EC, ℓ k C A D B E then (i) |AB| |BC| = |AD| |DE| (ii) |AB| |AC| = |AD| |AE| (iii) |AB| |AC| = |BD| |CE| .
Remark. We note that B, C do need to lie on the same side of A as we have shown above.
The theorem is still true if they lie on opposite sides of A, but the figure is different in this case.
ℓ k D B E C A Proof of Thales’ Theorem. First we note that if B and C lie on the same side of A, then D and E must also lie on the same side of A. This is intuitively obvious, and it can be shown 52 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
by letting m be the line through A parallel to BD and CE.
ℓ k D B E C A m Clearly B, C are on the same half-plane determined by m (by assumption that they lie on the same side of A on k), and since BD and CE do not intersect m, we also conclude D, E are on the same side of m, and, in particular, this implies that A is not between D and E.
Now if A lies in between B and C and between D and E, then we can rotate both C and E by 180◦through A to obtain points C′ ∈k and E′ ∈ℓso that BC′ and DE′ do not intersect A, as shown. Furthermore, since 180◦rotations send parallel lines to parallel lines, we deduce that E′C′ ∥DB. Finally |AE| = |AE′|, |AC| = |AC′| and |EC| = |E′C′|.
ℓ k D B E′ C′ E C A Therefore, if we prove (ii) and (iii) in the case when A is not between B and C (and so also not between D and E) then we also prove (ii) and (iii) in the case when A is between B and C. Furthermore, if A is between B and C and between C and D then |CB| = |AB| + |AC| |DE| = |AD| + |AE| and so if we know (ii) holds, then we have |CB| |AB| = 1 + |AC| |AB| = 1 + |AE| |AD| = |DE| |AD|, which obviously implies (i).
This argument shows that it suffices to prove the theorem in the case when BC and ED do not intersect A.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 53 Consider triangles ABD and BCD.
ℓ k D B E C A It is clear the heights of these two triangles through vertex D are the same (since their bases lie on the same line BC) and so we deduce (∗) |ABD| |BCD| = |AB| |BC|.
A similar argument yields (∗∗) |ABD| |BDE| = |AD| |DE|.
Since triangles DBC and BDE share a height (namely, the distance between the parallel lines BD and CE) and a base BD, we deduce |BCD| = |BDE|, and so combining (∗) and (∗∗) we obtain |AB| |BC| = |AD| |DE|, which proves (i).
Part (ii) follows immediately from part (i), since we have |AC| |AB| = |AB| + |BC| |AB| = 1 + |BC| |AB| = 1 + |DE| |AD| = |AE| |AD|, where we have used part (i) in the third equality.
Part (iii) requires a bit more work. Pick K, L so that CKL is congruent to BAD, as shown.
D B E K L C A Since CE cuts KL and AE in equal angles (as shown), we deduce KL ∥AE. Now we apply Thales’ Theorem part (ii) to the angle ACE which intersects the parallel lines AE and LK.
We obtain |CK| |CA| = |CL| |CE|.
54 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Since CKL is congruent to BAD, the above equation implies |BA| |CA| = |BD| |CE| , which is part (iii) as desired. This completes the proof.
□ Theorem 6.4 (converse to Thales’ Theorem). Let k, ℓbe lines intersecting at a point A, let B, C and D, E lie on k and ℓ, respectively, and (1) suppose that BC and DE both do not contain A or (2) BC and DE both contain A. If |AB| |AC| = |AD| |AE| then BD and CE are parallel.
Remark. Note that our converse to Thales’ Theorem is not a complete converse, and it only treats the case when we have equality (i) from the statement of Thales’ Theorem.
Counterexamples to the converse to Thales’ when we have (ii) or (iii) are presented in the exercises below.
Exercise 6.5. Prove the converse to Thales’ Theorem. Hint: Using a 180◦rotation (as in the proof of Thales’ theorem) argue that it suffices to prove the case (1). In this case, argue that any point D′ ∈ℓsuch that (a) |AD′| / |AE| = |AB| / |AC| and (b) D′ lies on the same side of A as E must satisfy D′ = D, and argue that the parallel to BC through E intersects ℓin a point D′ satisfying (a) and (b).
Exercise 6.6. Suppose that A, B, C, D, E are as in Theorem 6.4, but instead of (i) from Thales’ theorem we have (iii), namely |EC| |BD| = |AC| |AB|, Show that BD need not be parallel to CE. Hint: A B = C D E Exercise 6.7. As in the previous exercise, suppose that A, B, C, D, E are as in Theorem 6.4, but suppose instead we have (ii). Prove that BD need not be parallel to CE. Hint: A C B D E TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 55 Remark. This counterexample shows that, in general (ii) does not imply BD ∥CE. This remark will show that if we assume that B is between A and C and D is between A and E (or C is between A and B and E is between A and D) then we actually do have (ii) = ⇒ parallelism. The details of the straightforward proof of this fact are left to the reader. Hint: use Theorem 6.4, and be careful when considering when we can write |AC| = |AB| + |BC| (versus |AC| = |AB| −|BC|).
Problem 6.8. Let ABCD be a trapezoid with AB ∥CD. Let P on segment AD and Q on segment BC be such that |AP| / |PD| = |BQ| / |QC|.
A B C D P Q R Prove that PQ ∥AB and that |PQ| = 1 1 + λ |AB| + λ 1 + λ |CD| , where λ is the common ratio |AP| / |PD| = |BQ| / |QC|.
Solution. We will use the converse to Thales’ Theorem (Theorem 6.4).
Pick R on the diagonal AC so that |AR| / |RC| = λ (see the figure above). Then, using the angle DAC with the converse to Thales’ Theorem, we have PR ∥DC. Similarly, using the angle BCA in the converse to Thales’ Theorem, we have RQ ∥AB. Since there is a unique parallel to AB ∥DC through R, we deduce the lines RQ and PR are equal and so the line PQ containing R is parallel to AB.
By Thales’ Theorem part (iii) (applied to both angles DAC and BCA), we know that |PR| = |AP| |AD| |DC| and |QR| = |CQ| |BC| |AB| .
Since |AD| = |AP| + |PD| and |BC| = |QC| + |QB|, we deduce |PQ| = |PR| + |QR| = λ 1 + λ |DC| + 1 1 + λ |AB| , as desired.
56 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Problem 6.9. Let ABCD be a convex quadrilateral with diagonals intersecting at a point P.
A B C D P Prove that |ABC| |ACD| = |BP| |DP|.
Solution. Since the problem concerns the area of triangles ABC and BCD we introduce the feet K and L of altitudes through B and D, respectively.
A B C D P L K Then |ABC| = |AC| |BK| /2 and |ACD| = |AC| |DL| /2. This implies (∗) |ABC| |ACD| = |BK| |DL| , Since lines DL and BK are parallel (since they are both orthogonal to the line AC) we can apply Thales’ Theorem to the angle formed at P. We obtain |BK| |DL| = |BP| |DP|, and combining this with (∗) gives us the desired result.
Problem 6.10 (Nine Point Circle). In a triangle ABC with orthocentre H (the orthocentre is the intersection point of the altitudes) there is a circle o containing (i) the feet K, L, M of altitudes through A, B, C, respectively, (ii) the midpoints D, E, F of the edges BC, AC, TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 57 AB, respectively, and (iii) the midpoints A′, B′, C′ of segments AH, BH, CH, respectively.
A B C A′ B′ C′ D E F K L M H Solution. This problem is fairly daunting without an idea where to begin, and so to start the solution we simply make a guess (the correct guess) that the nine-point circle has the segment EB′ as a diameter (we make this guess because it looks like it could be true).
Let o be the circle which has EB′ as a diameter. The first step of our solution is to prove that the midpoints D, F of the edges adjacent to B lie on o. By relabelling points it suffices to prove that D lies on o. Since |CE| / |CA| = 1/2 and |CD| / |CB| = 1/2, we can apply the Converse to Thales’ to the angle BCA, to conclude ED ∥AB. Similarly, since |BD| / |BC| = |BB′| / |BH|, we can apply the Converse to Thales’ to the angle HBC to conclude B′D ∥HC. Since the altitude HC is perpendicular to AB, and B′D ∥HC and ED ∥AB we conclude that ED is perpendicular to B′D, and so ∠B′DE = 90◦. By what we have shown in Corollary 3.8, we know that D lies on the circle o. As mentioned above, this argument also works to prove F lies on the circle o.
A B C B′ D E F H 58 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
The next step in our solution is to prove that A′ and C′ lie on o. To show this, we note that in the first step we proved that o is the circumscribed circle of EDF; in other words, we showed that EB′ is a diameter of the circumscribed circle of EDF. Since the circumscribed circle of EDF is independent of relabelling the vertices, we conclude (by symmetry) that DA′ and FC′ are also diameters o, and so, in particular A′ and C′ lie on o.
The final step in our solution is to prove that the feet of the altitudes lie on o. However this is easy to show using Corollary 3.8. Since ∠ELB′ = 90◦, we know that L lies on the circle o whose diameter is EB′. Again, by symmetry, K, M also lie on o.
A B C B′ E K L M This completes the solution.
Theorem 6.11 (Newton Theorem). Let ABCD be a convex quadrilateral with an inscribed circle centred at a point O, and let L and K be the midpoints of the diagonals AC and BD, respectively. Then L, O, and K lie on the same line.
A B C D K L O In the proof of Theorem 6.11, we will use a lemma (we will prove the lemma after we prove the theorem).
Lemma 6.12. Let ABCD be a convex quadrilateral with AB not parallel to CD, and choose some a > 0. Then the set of points P inside ABCD satisfying |PAB| + |PCD| = a lie on a common line. (In the figure below, we have shown points P, P ′ and P ′′ such that TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 59 |PAB| + |PCD| = |P ′AB| + |P ′CD| = |P ′′AB| + |P ′′CD|).
A B C D P P ′ P ′′ Proof of the Newton Theorem using Lemma 6.12. First suppose that AB ∥CD (this is the case when we cannot use the Lemma, and so we treat it with a different argument).
Let k be parallel to AB and CD so that k is equidistant from AB and CD (more formally, k is defined as all the points which are equidistant from AB and CD). Then the centre of the circle, call it O, lies on the line k, since O bisects the orthogonal segment connecting AB to CD through O.
k A B C D O We will show that K and L also lie on k. This follows from Thales’ theorem. Consider the angle at K formed between segments DB and the orthogonal segment to AB: k A B D K Since K bisects DB, we conclude (by Thales’) that K also bisects the orthogonal segment to AB through K, and so K lies on k. A similar argument shows L lies on k, and so we have 60 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
proved Newton’s theorem in the case when AB ∥CD.
k A B C D O L K In the case where CD is not parallel to AB, we can use Lemma 6.12. Let a = |ABCD| /2, and note that triangles ABK and ADK share a height and have a common base length |BK| = |KD|, so |ABK| = |AKD|.
A B C D K A similar argument proves that |CDK| = |CKB|. Therefore |ABK| + |CKD| = a.
A B C D K The same argument shows that |ALB| + |CLD| = a.
Let r be the radius of the inscribed circle. We claim that |AOB| = r |AB|. This is true because the perpendicular segment joining O to AB intersects AB at the tangency point of the circle (as we have shown in Proposition 3.11), and so the height of AOB is equal to the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 61 radius r. A similar argument gives |COD| = r |CD|, |AOD| = r |AD| and |BOC| = r |BC|.
A B C D O Since ABCD splits into the four disjoint triangles AOB, BOC, COD and DOA, we have |ABCD| = r (|AB| + |BC| + |CD| + |DA|) , but since |AB| + |CD| = |BC| + |DA| (Theorem 5.11), we conclude |ABCD| = 2r (|AB| + |CD|) = 2 (|AOB| + |COD|) , and so |AOB| + |COD| = a. By Lemma 6.12, we deduce that L, O, K lie on a common line.
□ Proof of Lemma 6.12. Since AB and CD are not parallel, we know that they intersect in a point E. Furthermore, suppose P is chosen so that |APB| + |CPD| = a. Then choose points L and K on the half-lines ED and EA so that |EL| = |DC| and |EK| = |AB|. We assume for simplicity that K and L lie outside the sides AB and DC, as shown below. The other cases are left as exercise for the reader.
A B C D P E K L Since triangles EPL and DPC share a base length |DC| = |LE| and share the height through P, they have the same area |EPL| = |DPC|. Similarly |EPK| = |APB|, and so |EKPL| = |EPL| + |EPK| = a.
Splitting the quadrilateral EKPL into two triangles EKL and KPL, we note that |EKPL| = a implies that |KPL| = a −|EKL|. Therefore, P lies on the set of all points (on the opposite side of LK to E) for which the triangle LPK has specified area a −|EKL|.
Conversely, it is easy to show that if we form the triangle LPK and it has area a −|EKL|, then |APB| + |CPD| = a.
62 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
In other words, the set of P such that |APB| + |CPD| = a is precisely the set of P such that |KPL| = a −|EKL|.
P E K L Let ℓbe parallel to LK passing through P, as shown. We claim that |KPL| = |KP ′L| if and only if P ′ lies on ℓ.
P E K L ℓ P ′ This is easy to see since, |KPL| = |KP ′L| if and only if the heights of triangles KPL and KP ′L through P and P ′ are equal, which happens if and only if PP ′ is parallel to base KL, so that PP ′ = ℓ. This completes the proof of the lemma.
□ Exercise 6.13. Modify the proof of Lemma 6.12 to prove the case when K and L do not lie outside the sides AB and DC.
Problem 6.14. Let ABCD be a parallelogram construct equilateral triangles BPC and DQC outside of ABCD. Prove that APQ is equilateral.
A B C D P Q TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 63 Solution. We begin by proving that ABP is congruent to ADQ; this follows from the “side-angle-side” criterion since (side) |AB| = |DC| = |DQ| , (angle) ∠ABP = ∠ABC + 60◦= ∠CDA + 60◦= ∠QDA, (side) |BP| = |BC| = |AD| .
In the (angle) computation, we have used the fact that opposite angles in a parallelogram are equal.
Therefore |AP| = |AQ| and so we deduce that PAQ is isosceles.
To prove PAQ is equilateral, it suffices to prove that ∠PAQ = 60◦. Our proof depends on ∠CDA: we assume that ∠CDA < 60◦, and we leave the case ∠CDA ≥60◦to the reader. Let α = ∠BPA = ∠DAQ and β = ∠PAB = ∠AQD.
A B C D P Q Then, since α + ∠PAQ + β = ∠DAB, α + β + ∠ABC + 60◦= 180◦(here we have simply added all the angles in triangle ABP), and ∠DAB+∠ABC = 180◦, we rearrange and obtain ∠PAQ = 60◦, as desired. This completes the solution.
Problem 6.15. Let ABCD be a square and ABE be the triangle in its interior satisfying ∠EAB = ∠ABE = 15◦. Prove that CDE is equilateral.
A B C D E 64 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Solution. Let F, G, H be the rotations of E by 90◦, 180◦, 270◦around the centre of the square inside ABCD, so that ∠GBC = ∠BCG = 15◦etc. Clearly EFGH is a square (by its 90◦rotational symmetry, it must have all sides and all interior angles equal). We claim that FCG and HDG are equilateral triangles.
A B C D E F G H Proving this claim is not hard. Since ∠BCD = 90◦and ∠BCF = 15◦and ∠GCD = 15◦, we deduce ∠FCG = 60◦. Furthermore, by “angle-side-angle” we know BFG and CGD are congruent (and both are isosceles), so |CG| = |CF|. Thus FCG is an isoceles triangle with vertex angle equal to 60◦, so FCG must be equilateral. Similarly GDH is equilateral.
Interpreting EFGH as a parallelogram with equilateral triangles FCG and GDH built outside of EFGH, we can use the previous problem to deduce EDC is equilateral. This completes the solution.
Exercise 6.16. Solve the previous problem without using Problem 6.14 by proving that triangles CGD, CFE and DHE are all similar, so that |CE| = |DE| = |CD|.
A B C D E F G H 7. Similarities and Similar Triangles Definition 7.1. Let k > 0. A similarity with scale k is a transformation of the plane A 7→A′ satisfying |A′B′| = |AB| for all pairs of points A, B in the plane.
We have seen that all reflections and rotations are similarities with scale k = 1. We give scale 1 similarities a special name: Definition 7.2. An isometry is a similarity with scale 1.
Example 7.3. In this example, we will construct the similarity known as a homothety with centre O. Fix a point O and a positive scale k. The homothety centred at O with scale k is the transformation of the plane which (a) fixes O and (b) maps each A ̸= O to TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 65 the unique point A′ on the half-line OA with |OA′| = k |OA|. In the figure below, we have shown the homothety centred at O with scale 2.
B B′ C C′ A A′ O We claim that the homothety centred at O with scale k is a similarity with scale k. To prove this, we use both Thales’ theorem and its converse. Let A, B be any two points so that A, O, B do not lie on the same line, and let A′, B′ be their images under the homothety.
Since |A′O| / |AO| = k and |B′O| / |BO| = k, we apply the converse to Thales’ to deduce A′B′ is parallel to AB.
B B′ A A′ O Then we can apply Thales’ to deduce |A′B′| / |AB| = |A′O| / |AO| = k, so that |A′B′| = k |AB|, as desired.
If A, O, B lie on a common line, then the proof that |A′B′| = k |AB| is an easy application of the equality case in the triangle equality; it is left as an exercise for the reader.
Exercise 7.4. Let O, A, B be three points on a line, and let A′, B′ be the image of A, B under the homothety with scale k, centred at O. Prove that |A′B′| = k |AB|. Hint: consider two cases: (i) when O is between A, B and (ii) when O is outside A, B. In case (i), prove that |A′B′| = |A′O|+|OB′| and |AB| = |AO|+|OB|, so that |A′B′| = k |AB|. In the second case, use a similar argument.
66 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Definition 7.5. Two triangles ABC and A′B′C′ are called similar if there is a similarity mapping ABC to A′B′C′.
B B′ C C′ A A′ Theorem 7.6. If ABC and A′B′C′ are similar then all corresponding angles are equal.
Proof. Suppose that ABC and A′B′C′ are similar with scale k. Let A′′, B′′ and C′′ be the images of A′, B′, C′ under the homothety centred at B′ with scale 1/k.
B B′ = B′′ C C′ A A′ C′′ A′′ Then A′′, B′′, C′′ is similar to ABC with scale 1, and so, by “side-side-side” criterion for congruence, A′′B′′C′′ and ABC are congruent. It follows that ∠ABC = ∠A′′B′′C′′. By definition of the homothety centred at B′, we know that the half lines B′′A′′ and B′A′ and B′′C′′ and B′C′ are equal, and so ∠A′′B′′C′′ = ∠A′B′C′. Thus ∠ABC = ∠A′B′C′. By relabeling the vertices of our triangles, we deduce all angle pairs are equal. This completes the proof.
□ Theorem 7.7. If ABC and A′B′C′ are two triangles satisfying any one of the following three conditions then ABC and A′B′C′ are similar.
(a) |AB| / |A′B′| = |BC| / |B′C′| = |AC| / |A′C′|.
(b) |AB| / |A′B′| = |BC| / |B′C′| and ∠ABC = ∠A′B′C′.
(c) All corresponding angle pairs between ABC and A′B′C′ are equal.
We refer to (a), (b) and (c) as the “side-side-side,” “side-angle-side” and “angle-angle-angle” criteria for similarity, respectively.
Proof. Let A′′B′′C′′ be the image of A′B′C′ under the homothety centred at B′ with scale |AB| / |A′B′|. If (a) holds, then A′′B′′C′′ is congruent to ABC by the “side-side-side” TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 67 condition for congruency. If (b) holds, then A′′B′′C′′ is congruent to ABC by the “side-angle-side” condition for congruency. If (c) holds, then A′′B′′C′′ is congruent to ABC by the “side-angle-angle” condition for congruency. In all three cases, there exists a sequence of reflections taking A′′B′′C′′ to ABC. Since reflections are isometries, the composition of the sequence of reflections taking A′′B′′C′′ to ABC with the initial homothety taking A′B′C′ to A′′B′′C′′ is a similarity taking A′B′C′ to ABC. This completes the proof.
□ Corollary 7.8 (Pythagoras). If ABC is a right-angled triangle (with right angle at C) then |AC|2 + |BC|2 = |AB|2 .
Proof. Let D be the projection of C onto the side AB. Let α = ∠CAB and β = ∠ABC.
Since the angles in any triangle add up to 180◦, we know α + β = 90. Adding the angles in triangle ADC, we deduce ∠DCA = β and, similarly, adding the angles in DBC we deduce ∠BCD = α.
B C A D Then, by the “angle-angle-angle” criterion for similarity, we know CDB, ADC and ACB are all similar. Thus |BC| |AB| = |BD| |BC| and |AC| |AB| = |DA| |AC| , whereby we obtain |BC|2 + |AC|2 |AB| = |BD| + |DA| = |AB| , which gives the desired result.
□ 68 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Corollary 7.9. Let ABC and AB′C′ be similar triangles, both labelled counterclockwise.
Then ABB′ and ACC′ are similar.
A B C B′ C′ A B C B′ C′ Proof. We will use the “side-angle-side” criterion for similarity presented in part (b) of Theorem 7.7. Because ABC and AB′C′ are similar, we have |AB′| / |AC′| = |AB| / |AC|, and rearranging we obtain |AB′| / |AB| = |AC′| / |AC|. Then to prove the angles ∠C′AC and ∠B′AB are equal, we compute ∠C′AC = ∠C′AB′ ± ∠B′AC = ∠B′AC ± ∠CAB = ∠B′AB, where we have used the fact that ∠C′AB′ = ∠CAB, which follows by similarity of ABC and AB′C′. The ± depends on whether or not the triangles are interlaced, i.e. we may have a figure of the form pictured below.
A B C B′ C′ A B C B′ C′ This completes the proof.
□ Problem 7.10 (Napoleon’s Theorem). Let ABC be a triangle, and build equilateral trian-gles ARB, BPC and CQA outside of ABC, and let M, K, L be their orthocentres. Prove TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 69 that KLM is equilateral.
A B C Q P R L M K Solution. Using the “angle-angle-angle” criterion for similarity (i.e. part (c) of Theorem 7.7) we deduce AMR is similar to ALC. Hence, by Corollary 7.9, we deduce AML is similar to ARC.
A C R L M A C R L M Therefore |ML| / |CR| = |AM| / |AR|.
Analogously, MBK is similar to RBC, and so |KM| / |CR| = |BM| / |BR|.
However, |AM| / |AR| = |BM| / |BR|, since AMR is con-gruent to BMR by the “side-side-angle” criterion for congruency, and so |ML| = |KM|.
But then, by symmetry, all sides are equal (i.e. |MK| = |KL| and |KL| = |ML|) and this completes the proof.
Exercise 7.11. In the solution of Problem 7.10 we used the fact that AMR is congruent to BMR. Prove this using elementary arguments by considering the reflection through the bisector of the segment AB.
70 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Ptolemy’s Theorem. Let ABCD be a convex quadrilateral. Then we have the following inequality |AB| |CD| + |AD| |BC| ≥|AC| |BD| , with equality holding if and only if ABCD lie on a circle.
A B D C Proof. Let E be the point inside the angle ABC such that ∠ABD = ∠EBC and such that |BE| / |BA| = |BC| / |BD|. Then, by the “side-angle-side” criterion for similarity, we know ABD is congruent to EBC.
A B E D C (We remark that it is possible for E to leave the interior of the quadrilateral ABCD). By similarity of ABD and EBC, we know (∗) |EC| / |AD| = |BC| / |BD| Using Corollary 7.9, we deduce ABE and DBC are similar, and hence (∗∗) |AE| / |DC| = |AB| / |DB| We note that we have found two equalities which involve all of the lengths present in Ptolemy’s inequality. Using the triangle inequality, we obtain (T) |AC| ≤|AE| + |EC| , and multiplying both sides of this equation by |DB|, we obtain |AC| |DB| ≤|AE| |DB| + |EC| |DB| = |AB| |DC| + |AD| |BC| , where we have used the equalities (∗) and (∗∗) in the second equality. This completes the proof of the inequality, and it only remains to characterize the case when equality holds.
Equality will hold in Ptolemy’s inequality if and only if equality holds in (T) which happens TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 71 if and only if E lies on the segment AC. We (cleverly) observe that E lies on AC if and only if ∠BCE = ∠BCA.
A B D C E If this is the case, then similarity of ABD and EBC implies that ∠BDA = ∠BCE, and so ∠BDA = ∠BCA. By Corollary 3.8 (a long time ago) we know that ABCD lie on a circle.
Conversely, if ABCD lie on a circle, then we have ∠BDA = ∠BCA. Since ABD and EBC are similar, We have ∠BDA = ∠BCE, and so ∠BCE = ∠BCA, so that E lies between A and C. Therefore, we have characterized the case when equality holds in Ptolemy’s inequality, and this completes the proof of theorem.
□ Example 7.12. Let ABC be an equilateral triangle with circumscribed circle o. Suppose P lies on the arc BA of o. Then |CP| = |AP| + |BP|.
A P B C Clearly APBC forms a convex quadrilateral whose vertices lie on a circle, and so we can apply the equality case in Ptolemy’s Theorem. We obtain |PC| |AB| = |AC| |PB| + |BC| |AP| .
However, |AB| = |AC| = |BC|, and so we can cancel these common terms in the above equality, and we obtain |PC| = |PB| + |AP|, as desired.
Problem 7.13. Let A1 · · · A7 be regular heptagon (a regular n-gon is a polygon whose vertices form an ordered collection of n points A1 · · · An on a circle so that the rotation of angle 360◦/n sends Aj to Aj+1 and An to A1. A heptagon is just a 7-gon). Let o be the 72 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
circumscribed circle of A1 · · · A7, and pick D on the arc joining vertices A1A7.
A1 A2 A3 A4 A5 A6 A7 D Show that (∗) |DA1| + |DA3| + |DA5| + |DA7| = |DA2| + |DA4| + |DA6| .
Solution. The idea for the solution is simple, we will use the equality case in Ptolemy’s Theorem seven times applied to various quadrilaterals inscribed in the circle, then we will add the seven resulting equalities together and, after some simplification, we will obtain (∗).
We remark that this problem can be seen as a generalization of Example 7.12.
We begin our solution by identifying three side lengths in the heptagon. Let a = |A1A2| and b = |A1A3|. Note that by the rotational symmetry of a regular heptagon, a = |AjAj+1| and b = |AjAj+2| (where we add integers modulo 7, so that 6 + 2 = 1 etc). Now we consider the quadrilaterals DAjAj+1Aj+2, where j = 1, · · · , 5 (j = 1, 3, 5 is shown on the left in the figure below, and j = 2, 4 is shown on the right).
D A1 A2 A3 A4 A5 A6 A7 D A1 A2 A3 A4 A5 A6 A7 Since each such quadrilateral is inscribed in a circle, we can apply the equality case in Ptolemy’s theorem, and in all cases we obtain (⋆) b |DAj+1| = a |DAj| + a |DAj+2| .
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 73 This gives us five equations. We still need to apply Ptolemy’s theorem twice more. We consider the two quadrilaterals DA1A6A7 and DA1A2A7, shown below.
D A1 A2 A3 A4 A5 A6 A7 D A1 A2 A3 A4 A5 A6 A7 Applying Ptolemy to these quadrilaterals yields (⋆⋆) a |DA6| = a |DA1| + b |DA7| and a |DA2| = b |DA1| + a |DA7|.
Now we add up the five equations in (⋆) and the 2 equations in (⋆⋆) in a clever way: we gather the “even” terms (terms containing |DAj| for j even) on the left, and move the “odd” terms (terms containing |DAj| for j odd) to the right: (⋆and ⋆⋆) b |DA2| = a |DA1| + a |DA3| a |DA2| + a |DA4| = b |DA3| b |DA4| = a |DA3| + a |DA5| a |DA2| + a |DA4| = b |DA5| b |DA6| = a |DA5| + a |DA7| a |DA6| = a |DA1| + b |DA7| a |DA2| = b |DA1| + a |DA7| Adding all of these up yields (2a + b)(|DA2| + |DA4| + |DA6|) = (2a + b)(|DA1| + |DA3| + |DA5| + |DA7|), and the common factor of (2a + b) cancels and we obtain (∗), as desired.
Theorem 7.14 (angle bisector theorem, cf. Definition 5.3 and Exercise 5.4). Let ABC be a triangle, and let P be the intersection of the interior bisector at C and AB, and suppose that the exterior bisector at C intersects AB in a point Q (note that this is only possible if 74 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
the exterior bisector at C is not parallel to AB).
A P Q B C Then P, Q satisfy |AP| |BP| = |AQ| |BQ| = |AC| |BC|.
Remark. If |AC| = |BC|, then P is the midpoint of AB (as expected) but then Q is actually not defined since the exterior bisector at C is parallel to AB. In this degenerate case we can think of Q as being a point “at infinity.” Proof. Draw a line through A parallel to BC intersecting CP in K and CQ in L.
A P Q B K C L The line CK cuts the parallel lines LK and CB in equal angles, so ∠AKC = ∠BCK = ∠KCA, where we have used the fact that CK bisects the angle ∠BCA. Similarly, the line CQ cuts the lines LK and CB in equal angles, so ∠CLA = ∠ACL. In particular, triangles ACK and ALC are isosceles, and so |AK| = |AC| = |AL|.
Now we use Thales’ Theorem at the angle at P intersecting the parallel lines LK and BC, and we obtain |AP| / |PB| = |AK| / |BC|. Using |AK| = |AC| = |AL|, this becomes |AP| / |PB| = |AC| / |BC| = |AL| / |BC|. Now we use Thales’ theorem at the angle at Q to conclude |AL| / |BC| = |AQ| / |BQ|, and combining everything, we obtain the desired result |AP| |BP| = |AC| |BC| = |AQ| |BQ|.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 75 □ Problem 7.15. Given points A ̸= B and a real number λ > 0, find the set of points such that |AX| = λ |BX|.
Remark. If λ = 1, we have characterized this set of points as the bisector of AB (Theorem 1.13).
Solution. The idea for the solution is to use the angle bisector theorem (Theorem 7.14).
Thanks to the above remark, we may assume λ ̸= 1. Then pick P inside the segment AB and Q outside so that |AP| / |BP| = |AQ| / |BQ| = λ. We claim that the set of points X satisfying |AX| = λ |BX| is precisely the circle with diameter PQ.
B A P Q X To prove this claim, we begin by supposing that X satisfies |AX| / |BX| = λ.
Since |AP| / |BP| = |AX| / |BX|, we know that XP must bisect the angle BXA (here we are using the angle bisector theorem and the fact that there is a unique point P between A and B satisfying |AP| / |BP| = |AX| / |BX|). Similarly XQ bisects the exterior angle of ABX at X. However, we have proved in Exercise 5.4 that the two angle bisectors at X are perpendicular, and so ∠PXQ = 90◦, and so X lies on the circle with diameter PQ (applying Corollary 3.8). Thus the set of points satisfying |AX| / |BX| = λ lies on the chosen circle.
It remains to prove that if X lies on the circle with diameter PQ, then X satisfies |AX| / |BX| = λ.
To show this, we construct two points M, N so that (i) AN ∥PX 76 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
and AM ∥XQ and (ii) M, N both lie on the line BX.
B A P Q M N X Now we apply Thales’ Theorem to the angle QBN intersecting parallel lines AN ∥PX and AM ∥QX and obtain the equalities (∗) |AP| |PB| = |NX| |XB| and |MX| |BX| = |AQ| |BQ|.
Since |AQ| / |BQ| = |AP| / |PB| = λ, we deduce |NX| = |MX|. Finally, since ∠PXQ = 90◦ (since X lies on the circle with diameter PQ) and AN ∥PX and AM ∥XQ, we deduce ∠NAM = 90◦, and consequently, A lies on the circle with diameter NM. Since X is the midpoint of segment NM, we conclude |AX| = |NX| = |MX| (this is proven in Exercise 3.6). Thus, returning to (∗), we obtain |AX| |BX| = |AQ| |BQ| = λ, as desired. This completes the proof.
8. The Power of a Point with Respect to a Circle Definition 8.1. The power of a point A with respect to a circle o is a number computed according to the following rule. Issue a line k from A intersecting o in two distinct points B and B′.
k o A B B′ Then the power of A with respect to o is given by the following formula.
power of A with respect to o = |AB| |AB′| if A is outside o 0 if A is on o −|AB| |AB′| if A is inside o TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 77 A priori, this definition depends on the line k. However, with a little bit of work, we see that it actually does not: Theorem 8.2. The definition given for the power of A with respect to o does not depend on the line k.
Proof. This result is called a “Theorem” because it is important, but not because it is hard to prove! The proof is an easy application of similar triangles. First suppose that A lies outside the circle o. Suppose k, ℓare two lines intersecting at A, and suppose B ̸= B′ and C ̸= C′ are the intersection points of k and ℓwith o respectively. As in the figure, we assume that B separates A from B′ and C separates A from C′.
k ℓ o A B B′ C C′ Since angles CC′B and CB′B are subtended by the same arc CB, we know that ∠AC′B = ∠CB′B, and since triangles AC′B and CB′A share a common angle at A, we deduce they are similar by the “angle-angle-angle” criterion. It follows that |AC| / |AB′| = |AB| / |AC′| and so |AC| |AC′| = |AB′| |AB|.
The case when A is inside the circle is similar, and it is left to the reader.
□ Proposition 8.3. Let A be a point lying outside of o. If D ∈o is such that AD is tangent to o, then the power of A with respect to o is |AD|2.
A D B′ B Proof. Let B, B′ be as in the definition of the power of a point. Applying Proposition 3.12, we deduce ∠BDA = ∠AB′D, and hence AB′D and ADB are similar by the “angle-angle-angle” criterion. In particular, |AD| / |AB′| = |AB| / |AD|, and hence |AD|2 = |AB| |AB′|, and this completes the proof.
□ Remark. Let A and D be as in the preceeding proposition, and let O be the centre of the circle o. Recall that AD is perpendicular to the radius OD, and hence we can apply Pythagoras’ theorem to write |AD|2 = |AO|2 −|DO|2. If we denote d := |AO| and r := |DO| 78 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
the radius of the circle, then we conclude the power of A with respect to O is given by d2−r2.
A D O d r In fact, this formula continues to hold if A is inside of o.
To show this, issue a line k passing through A and O, and consider the intersection points B and B′ between this line and the circle o. Without loss of generality, suppose that |AB| ≤|AB′|. Then we have the configuration shown in the following figure.
r −d r + d O A B B′ As described in the figure above, we have −|AB| |AB′| = −(r −d)(r + d) = d2 −r2, which is what we wanted to show. We remark that this formula is concise, but the definition given at the beginning of this section is much more useful in practice.
Problem 8.4. Let o1 and o2 be two circles intersecting in two points E and E′ lying on a line k. Let A, C lie on o1, and B, D lie on o2 so that AC intersects BD at a point P on k.
Prove that ABCD lie on a circle.
P A C B D k E E E′ Solution. In our solution we assume that P lies inside the intersection of the interiors of o1 and o2 (the other case is when P is outside of both o1 and o2). The idea for the solution is TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 79 simple: we let o3 be the circumscribed circle of ABC, and we will show that D lies on o3.
P A C B D o2 o1 o3 E E′ To prove D lies on o3, we will use an equality which will result from computing the power of P with respect to o1 and o2 in various ways.
By computing the (negative) power of P with respect to o1 in two different ways (using the lines AC and EE′) we obtain the equality |AP| |PC| = |EP| |E′P|. Similarly, by computing the power of P with respect to o2 in two different ways, we obtain the equality |PD| |BP| = |EP| |E′P|, and hence |AP| |PC| = |PD| |BP|. Now let o3 be the circumscribed circle of ABC. We can compute the (negative) power of P with respect to o3 in two ways, using the line AC and using the line BD. Let B and D′ be the intersection points of BD and o3, so that the power of P with respect to o3 is equal to |BP| |PD′|. However, this power is also equal to |AP| |PC| = |PD| |BP|. Hence, keeping track of the various equalities, we have |PD′| = |PD|. We claim that D and D′ are both on the opposite side of P to B.
This is true because P lies on the edge AC, so it must lie inside o3, and consequently P must separate B, D′, and (by our assumption that P lies inside o2) P separates B from D.
Therefore |PD′| = |PD| implies D = D′, and so D ∈o3, as desired. The case when P lies outside both o1 and o2 is similar and is left as an exercise.
80 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Exercise 8.5. Solve Problem 8.4 in the case when P lies outside of o1 and o2 (see figure below).
P A C B D E E′ Definition 8.6. If two circles o1 and o2 intersect in two distinct points E ̸= E′, then the line EE′ is called the radical axis of o1 and o2.
Theorem 8.7. Let o1 and o2 intersect in points E ̸= E′. Then P lies on the radical axis of o1 and o2 if and only if the power of P with respect to o1 is equal to the power of P with respect to o2.
P B′ B C′ C o1 o2 E E′ Proof. Let B ̸= B′ be on o1 and C ̸= C′ on o2 so that lines BB′ and CC′ both contain P.
First we show that if P lies on the radical axis EE′ then the powers of P with respect to o1 and o2 are equal. Case 1: if P lies outside o1 and o2. Proving equality of the powers in this case is easy to since |PE| |PE′| = |PB| |PB′| (by computing the power of P with respect to o1 in two different ways) and |PC| |PC′| = |PE| |PE′| (by computing the power of P with respect to o2 in two different ways). Thus |PC| |PC′| = |PB| |PB′|, as desired. Case 2: if P lies inside both o1 and o2. The same argument as in Case 1 shows −|PC| |PC′| = −|PB| |PB′|, as desired.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 81 We note that since P lies on the radical axis, it cannot be inside o1 and outside o2 (or vice-versa) and so we have exhausted all cases.
Now we assume that P has equal powers with respect to o1 and o2, and we want to show that P lies on the radical axis EE′. We will prove this by contradiction, so suppose that P does not lie on the radical axis. First we note that P cannot be inside o1 and outside o2 (or vice-versa), or else the powers with respect to o1 and o2 would have different signs.
Therefore there are only two cases: if P is outside both o1 and o2 or P is inside both o1 and o2 (the third case where P lies on the intersection of o1 and o2 is trivial, for then P is either E or E′). Assume that P lies outside both o1 and o2. Consider the line PE, and let PE intersect o1 in a point A1 and o2 in a point A2.
P o1 o2 A1 A2 E E′ Since P does not lie on EE′, we know that A1 ̸= A2 (since A1 = A2 implies A1 = A2 = E′ is the other intersection point between o1 and o2). But since the power of P with respect to o1 is the same as with respect to o2, we have |PE| |PA1| = |PE| |PA2|, and hence |PA1| = |PA2|.
Since A1 and A2 both lie on the ray PE, we deduce A1 = A2, which contradicts our assumption. A similar argument works when P lies inside both o1 and o2.
This completes the proof of the theorem.
□ Exercise 8.8. Prove that for any two circles o1, o2 with different centres, the set of points P for which its power with respect to o1 equals its power with respect to o2 forms a line.
Hint: let O1 and O2 be the centres of o1, o2 and issue a perpendicular to O1O2 through a point P on O1O2 which has equal powers with respect to o1 and o2.
Remark. By virtue of Exercise 8.8, we define the radical axis of two circles o1 and o2 with distinct centres to be the set of points P with equal power with respect to o1 and o2.
Problem 8.9. Let o1, o2 and o3 be circles so that o1 ∩o2 = {A, A′}, o2 ∩o3 = {B, B′} and o1 ∩o3 = {C, C′}.
82 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Then the three radical axes AA′, BB′ and CC′ either intersect in a point or are parallel.
A A′ B B′ C C′ o1 o2 o3 or A A′ B B′ C C′ Solution. In our solution we may obviously assume that the lines are not parallel, since if they are parallel, then we are already done!
Therefore one pair of the radical axes must intersect, and so, without loss of generality, we assume that AA′ and BB′ intersect in a point P. Then we apply Theorem 8.7 to deduce that the power of P with respect to o1 is equal to the power of P with respect to o2 (since P lies on AA′) and the power of P with respect to o2 is equal to the power of P with respect to o3 (since P lies on BB′). But then (applying Theorem 8.7 once again) we deduce that the power of P with respect to o1 equals its power with respect to o3, hence P lies on CC′.
This completes the proof.
Exercise 8.10. Show that the argument given above does not require the circles to intersect, and deduce that the three radical axes (as defined in the remark preceeding Exercise 8.8) which can be made out of three arbitrary circles o1, o2 and o3 intersect in a common point (or are parallel).
Problem 8.11. Let A, B, C, D, E be distinct points on a line k so that |AB| = |BC| = |CD| = |DE| = a.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 83 Prove that, given any circle o1 containing A, B and any circle o2 containing D, E so that o1 and o2 intersect in distinct points K, L, we have that K, L, C lie on a common line.
k A B C D E o2 o1 K L Solution. Without our knowledge of the power of a point, this problem seems like it would be difficult. However, it is easily solved using Theorem 8.7. Using the fact that |AC| = |CE| = 2a and |BC| = |CD| = a, we compute that the powers of C with respect to o1 and o2 are both equal to 2a2. Hence C lies on the radical axis of o1 and o2, and thus C lies on the line KL, as desired.
Problem 8.12 (Brianchon’s Theorem). Let ABCDEF be a convex hexagon with an in-scribed circle o; here convex means each diagonal is contained inside the hexagon. Then the lines AD, BE and CE intersect in a single point.
A B C D E F Solution. The idea for the solution is to find three circles o1, o2, o3 so that AD, BE and CF are the three radical axes constructed from o1, o2, o3. Then we can apply Exercise 8.10 to deduce AD, BE and CF intersect in a single point (a straightforward argument shows that they cannot be parallel). To do this, we introduce some convenient notation: let PAB be the tangency point between the line AB and the inscribed circle, and denote the other tangency points by PBC, PCD, PDE, PEF and PF A in a similar fashion. Let a denote the distance |APAB|, and recall that the “strongest theorem of geometry” states that a is also 84 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
equal to |APF A|. Define distances b, c, d, e, f similarly.
A B C D E F PAB PBC PCD PDE PEF PF A a a Let P be the point on the line AB at a distance of c + e from A on the opposite side of B, and let Q be the point on DE at a distance a + c from E on the opposite side of D. We claim that there is a circle o1 tangent to lines AB and DE at the points P and Q.
A B C D E F a e c + e a + c P Q o1 To prove this claim, we consider the intersection point S between lines AB and ED. Since S is equidistant to the tangency points PAB and PDE, and the distance from PAB is a + c + e and the distance from Q to PDE is also a + c + e, we deduce that P and Q are equidistant from S. A simple application of Theorem 4.1 proves that there is a circle tangent to the angle at DSB at points P and Q.
In a similar fashion, and by consulting the figure below, define points R, S, T and U.
Then circles o2 and o3 (as shown below) exist, and this details of the proof of this assertion TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 85 are left to the reader.
A B C D E F c + e a + c P Q o1 c + e a + e R S o2 a + e a + c T U o3 Now we claim that AD is the radical axis of o1 and o2. To prove this, note that |AP| = |AS| = c + e and AP and AS are tangent to o1 and o2 respectively. Therefore the power of A with respect to o1 is equal to its power with respect to o2. A similar computation proves that the power of D with respect to o1 equals its power with respect to o2, and so AD is the radical axis of o1 and o2. Similar arguments prove that BE is the radical axis of o1 and o3, and CF is the radial axis of o2 and o3. Therefore, by Exercise 8.10, the lines AD, BE and CF intersect in a common point or they are parallel. However, convexity of ABCDEF implies that the segments AD and BE intersect, and so we must have AD, BE and CF intersecting in a common point. This completes the solution.
Exercise 8.13 (challenging). Let ABCD be a quadrilateral with an inscribed circle o. Let E (respectively F, G, H) be the point of tangency between AB and o (respectively BC, CD, DA and o). Prove that AC, BD, EG and FH intersect in a common point. Hint: use the ideas in the proof of Brianchon’s Theorem by treating AEBCGD as a degenerate hexagon.
9. Ceva’s Theorem Theorem 9.1 (Ceva). Let ABC be a triangle and let D, E, and F be arbitrary points on the sides BC, CA, and AB, respectively. Then the lines AD, BE and CF intersect at a 86 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
single point if and only if (∗) |AF| |BD| |CE| = |BF| |CD| |AE| .
A B C D E F Before we give the proof of Ceva’s theorem, we demonstrate how useful it is with a corollary.
Corollary 9.2. If ABC is a triangle and D, E, F are the midpoints of the sides BC, AC and AB, respectively, then the medians AD, BE and CF intersect in a single point (recall that this point is called the centroid of the triangle ABC).
A B C D E F Proof. Using Ceva’s theorem, we see that it suffices to prove that |AF| |BD| |CE| = |BF| |CD| |AE|, but this is immediate, since we have equalities |AF| = |BF|, |BD| = |CD| and |AE| = |CE|!
□ Proof of Ceva’s Theorem. First we show that if CF, AD and BE intersect at a point then (∗) holds.
This is a fairly straightforward consequence of Thales’ theorem.
Let O denote the intersection of CF, AD and BE, let k be the line through C parallel to AB, and TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 87 consider the points P and R, obtained by extending BE to k and AD to k, as shown.
A B C D E F O P R k Then we apply Thales’ to the angle at D formed by the lines BC and AR to conclude (1) |BD| |CD| = |AB| |CR|, while applying Thales’ theorem to the analogous angle at E yields (2) |CE| |AE| = |CP| |AB|.
Applying Thales’ theorem to the two angles at O formed by the lines CF, AR and CF, BP we conclude (3) |AF| |CR| = |FO| |OC| = |BF| |CP|, hence |AF| |BF| = |CR| |CP|.
Taking the product of (1), (2) and (3) yields |AF| |BD| |CE| |BF| |CD| |AE| = 1, which is equivalent to (∗). This completes the first half of the proof.
Now we establish the converse, and so we assume that equality (∗) holds and wish to prove that AD, BE and CF intersect in a common point. Let O denote the point of intersection between CF and BE, and let D′ be the intersection of AO with the segment BC.
A B C D′ E F O D By applying the first part of Ceva’s theorem, we know that (4) |BD′| |D′C| = |BF| |AE| |AF| |CE| = |BD| |DC|, 88 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
where the second equality holds because we assume (∗) holds for the point D. But there is a unique point between B and C with the ratio given by (4), and so D = D′, which completes the proof.
□ 9.1. Applications of Ceva’s theorem. In this subsection we present an assortment of problems which we solve using Ceva’s theorem.
Problem 9.3. Let ABC be a triangle and let Let D, E be points on ABC, as shown. Prove that AD and BE intersect on the median from C if and only if ED is parallel to AB.
A B C D E Solution. First suppose that AD and BE intersect at a point O on the median through C. Let F be the midpoint of AB; then we apply Ceva’s theorem to the points D, E, F to conclude (1) |AF| |BD| |CE| = |FB| |CD| |AE| .
Since |AF| = |FB|, we conclude |CE| / |AE| = |CD| / |DB|, and hence Thales’ theorem implies ED is parallel to AB.
For the converse, we assume ED is parallel to AB, and use Thales’ theorem at the angle at C to conclude |CE| / |AE| = |CD| / |DB|, and since |AF| = |BF|, we conclude (1). Then Ceva’s theorem implies BE and AD intersect on CF, which is the median through C.
Problem 9.4. Let ABC be a triangle with inscribed circle o, and let D, E, F be the tangency points between o and the edges BC, CA and AB. Then AD, BE and CF intersect in a common point. Remark: this point of intersection is called the Gergonne point.
A B C D E F TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 89 Solution. By the “strongest theorem of geometry,” |AE| = |AF|, |BF| = |BD| and |CD| = |CE|, and multiplying these three equalities together yields |AE| |BF| |CD| = |AF| |BD| |CE|, hence Ceva’s theorem implies AD, BE and CF intersect in a common point.
Problem 9.5. Let ABC be a triangle, and let P, Q, R be the centres of the sides BC, CA and AB. Let S be a point inside the triangle PQR, and let K, L, M be the intersections of the lines PS, QS and RS with the sides QR, RP, PQ, respectively. Prove that AK, BL and CM intersect in a single point.
A B C P Q R K L M A B C P Q R S K′ L′ M ′ Solution. As in the above figure, let K′, L′, M ′ be the intersections of AK, BL and CM with the opposite side BC, AC and AB (respectively). Since Q, P are the midpoints of AC and BC, Thales’ theorem guarantees that QP is parallel to AB, and so we can apply Thales’ theorem to the angle at C formed by the lines CM ′ and CA to deduce 2 |QM| = |AM ′|.
M ′ A B C P Q M Similar argument yield 2 |AL′| = |RL|, 2 |RK| = |BK′|, 2 |PM| = |BM ′|, 2 |QK| = |CK′| and 2 |PL| = |CL′|. Therefore, by multiplying all these equalities we deduce |AM ′| |BK′| |CL′| = |BM ′| |CK′| |AL′| if and only if 8 |QM| |RK| |PL| = 8 |PM| |QK| |RL|, and the latter equality is true by Ceva’s theorem, since QL, RM and PK intersect at S.
Applying Ceva’s theorem once again allows us to conclude AK′, BL′ and CM ′ intersect in a common point. This completes the solution.
Problem 9.6. Let D, E, F be the tangency points of the escribed circles to the triangle A, B, C. Then the lines AD, BE and CF intersect in a common point.
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Solution. The idea for the solution is fairly simple: we will use the “strongest theorem of geometry” to produce the equalities needed to invoke Ceva’s theorem. By the “strongest theorem of geometry” we have (1) |AB| + |BD| = 1 2(perimeter of ABC) = |AB| + |AE| , so |AE| = |BD| (we proved (1) in the solution to Problem 5.8). Similarly |CD| = |AF| and |FB| = |CE|. Multiplying these three equalities yields |AF| |BD| |CE| = |CD| |AE| |FB|, as desired. This point of intersection is called the Nagel point.
A B C E D F Problem 9.7 (The intersection of symmedians). This application of Ceva’s theorem requires a new notion: If ABC is a triangle, then the symmedian through C is the reflection of C’s TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 91 median through the angle bisector at C.
median angle bisector symmedian A B C Prove that the three symmedians of a triangle intersect at a common point.
m b s m b s m b s A B C X Solution. Let X, Y, Z be points on sides BC, AC and AB, respectively, so that AX, BY and CZ are the symmedians through A, B, C, respectively. The key to the solution is the fact that (1) |BX| |CX| = |BA| |CA| 2 , |CY | |AY | = |CB| |AB| 2 , and |AZ| |BZ| = |AC| |BC| 2 .
To see why this implies the solution, we compute |BC| |CY | |AZ| |CX| |AY | |BZ| = 1, which, by Ceva’s theorem, implies that AX, BY and CZ intersect in a single point. To prove (1), it suffices to prove the first equality |BX| / |CX| = |BA|2 / |CA|2. To do this, we issue a line ℓthrough C parallel to AB, and let D denote the point of intersection of the 92 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
symmedian AX and ℓ, and let E denote the point of intersection of the median AM and ℓ.
ℓ A B C D E M X By Thales’ theorem, (i) |CE| = |AB|, and (ii) |BX| / |CX| = |AB| / |CD|.
Since the bisector through A bisects both CAB and DAE (by definition) we deduce that ∠CAD = ∠EAB. But since CE is parallel to AB, ∠EAB = ∠AEC. Thus triangles AEC and DAC share two angles, and so they are similar. Therefore |CE| / |AC| = |AC| / |DC|, and combining this with (i) and (ii) we deduce |BX| / |CX| = |BA|2 / |CA|2, as desired.
9.2. Menelaus’ Theorem. Suppose that we have a triangle ABC, points D, E on sides BC and AC, and a third point F lying on the line AB but not between A and B. In a similar spirit to Ceva’s theorem, Menelaus’ theorem gives a necessary and sufficient criterion for when D, E, F lie on a common line.
Theorem 9.8 (Menelaus). With ABC a triangle and D, E, F as described above, D, E, F are collinear if and only if the following equality holds |AF| |BD| |CE| = |BF| |CD| |AE| .
A B F C D E Exercise 9.9. Prove Menelaus’ theorem by considering the following figure A B F C D E K TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 93 Hint: Apply Thales’ theorem to the angles formed at E and D.
Problem 9.10 (The Simson line). Let ABC be a triangle and let S lie on the circumscribed circle of ABC. Let D, E, F be the projections of S onto the lines BC, AC and AB, respec-tively. Then the points D, E, F lie on a common line. Remark: this problem was already asked in Exercise 3.19, but here we will solve it as an application of Menelaus’ theorem.
A B F C D E S Solution. Without loss of generality, we can assume that S and C lie on opposite sides of AB. We assume that F lies on the segment AB (there is another case, where F does not lie on AB). Then, since ASBC admits a circumscribed circle, we know that ∠CAS+∠SBC = 180◦ (Theorem ). Thus, if ∠CAS = 90◦then ∠SBC = 90◦, so we have E = A and D = B, and so D, E, F lie on the line AB. Therefore, we may assume without loss that ∠CAS > 90◦, so that ∠SBC < 90◦. It follows that E lies outside AC while D lies inside BC (as in the figure above).
Therefore, we are in the setting of Menelaus’ theorem, since F, D lie on the edges of the triangle, while E lies outside of its corresponding segment. We will establish the desired equality (∗) |AE| |BF| |CD| |CE| |AF| |BD| = 1 by proving that various triangles are similar. We claim that SFA is similar to SDC.
A F C D B S 94 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
This follows by the “angle-angle” criterion for similarity, since ∠BAS = ∠BCS, since BAS and BCS both are subtended by the arc BS. Therefore, (i) |CD| / |AF| = |CS| / |AS|.
Similar arguments show that SEC is similar to SFB, so (ii) |BF| / |CE| = |BS| / |CS|, and SDB is similar to SEA, so (iii) |AE| / |BD| = |AS| / |BS|. Multiplying together (i), (ii) and (iii) we obtain (∗), and so we can apply Menelaus’ theorem to conclude D, E, F are collinear.
This completes the proof when F lies on AB. The proof of the other case is left to the reader (Exercise 9.11).
Exercise 9.11. Suppose that we are in the setting of the previous problem, with AB sep-arating C from S, but that F does lies outside of the segment AB. Prove that E, F, D still lie on a common line.
A B F C D E S If the reader wishes to prove solve this problem using the same arguments as the solution to Problem 9.10, she/he should prove an analogous theorem to Menelaus’ theorem which applies when the points D, E, F lie on the lines CB, AC, AB, respectively, but none of D, E, F lie inside their corresponding interval.
10. Isometries of the Euclidean plane Recall that an isometry is a similarity with scale 1, i.e. a transformation of the plane which preserves distances. Since isometries are similarities, they also preserves angles. The goal of this section is to try to classify all isometries.
We have already seen that reflections and rotations are isometries.
Another class of isometries are the “parallel translations,” which we will define shortly. The concept of a parallel translation is closely related to the concept of a “vector,” which is hopefully well known to the reader. For completeness, we now give the formal definition of a “vector.” Definition 10.1. We say that two pairs of points A, B and A′, B′ define the same vector, written − − − → A′B′ = − − → AB, if and only if the point reflection (i.e. 180◦rotation) through the midpoint of AB′ sends B to A′. The reader will show below that − − − → A′B′ = − − → AB defines an equivalence relation, and hence we can define a vector as an equivalence class of pairs of TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 95 points.
A B A′ B′ Exercise 10.2. Prove the following facts about vectors: (a) If − − → AB = − − − → A′B′, then |AB| = |A′B′|, and the point reflection through the midpoint of BA′ interchanges A and B′.
(b) If A ̸= B and A, B, A′ do not lie on a common line, then − − → AB = − − − → A′B′ implies ABB′A′ forms a parallelogram.
(c) − − → AB = − − − → A′B′ if and only if − − → AA′ = − − → BB′.
(d) If − − → AB = − − − → A′B′ and − − − → A′B′ = − − − → A′′B′′, then − − → AB = − − − → A′′B′′.
A B A′ B′ A′′ B′′ Hint for (d): first show this (i) when A, B, A′, B′, A′′, B′′ all lie on a common line, and (ii) when A′, B′, A′′, B′′ lie on a common line and A, B lie on a different line. Then using (ii), show that the general case can be reduced to the case when A, A′, A′′ lie on a common line.
Thus the relation − − → AB = − − − → A′B′ is an equivalence relation, and so we may define a vector as an equivalence class of this relation.
Exercise 10.3. The goal of this exercise is to prove that vectors form a vector space.
(a) For any vector v and any point A, prove there is unique B so that − − → AB = v.
(b) If v = − − → AB, and w = − − → BC, define v + w = − → AC, and show that this not depend on the choice of the pair A, B defining v.
(c) If v = − − → AB, λ ≥0 (resp. λ < 0) is a real number, and C is the unique point on AB so that |AC| = |λ| |AB| and C lies on the ray [A, B⟩(resp. C lies on the other ray AB−[A, B⟩), define λv = − → AC. Show this does not depend on the pair AB defining v.
(d) Prove that with these operations of addition and scalar multiplication, the set of vectors forms a vector space.
(e) Prove that this vector space is two-dimensional.
96 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Definition 10.4. If v is a vector, then the translation by v is a transformation of the plane which takes A to A′, where A′ is the unique point so that v = − − → AA′ (the proof of uniqueness was asked in Exercise 10.3).
v A A′ B B′ C C′ Proposition 10.5. The translation by a vector v is an isometry whose inverse is the trans-lation by the vector −v.
Proof. Let A, B be two points in the plane. Since − − → AA′ = − − → BB′, part (b) of Exercise 10.2 implies that − − → AB = − − − → A′B′, so part (a) of the same exercise guarantees that |AB| = |A′B′|.
Referring to the definition of scalar multiplication given in Exercise 10.3, a simple argu-ment shows − − → AA′ = v if and only if − − → A′A = −v, which proves that the inverse of the translation by v is the translation by −v.
□ Exercise 10.6. Let v and w be two vectors. If A′ is the translation of A by v, and A′′ is the translation of A′ by w, then A′′ is the translation of A by v + w. (This is easy, once we recall the definition of v + w given in Exercise 10.3).
Exercise 10.7. Let k, k′ be parallel lines, and let d be the distance between k and k′.
Prove that the composition of the reflection through k and the reflection through k′ is the translation by a vector perpendicular to k with length 2d.
k k′ v Groups of transformations. The main goal in this section is to classify all isometries of the Euclidean plane.
It is therefore useful to consider the set of all isometries.
The first observation one makes about the set of all isometries is that they form a group, which, informally, is a collection of transformations which can be composed and which have inverses.
Here are the formal definitions.
Definition 10.8. Recall that if f, g are two transformations of the plane, f ◦g is the com-posite transformation, which sends a point A to f(g(A)).
A transformation f is invertible if there is another transformation g (an inverse of f) such that f ◦g = g ◦f = id TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 97 where id is the identity transformation, which maps each point to itself. As a corollary to part (c) of Exercise 10.9 (see below), if f has an inverse, it has a unique inverse, which we denote by f −1.
If g satisfies f ◦g = id, then we say g is a right-inverse for f, while if g satisfies g◦f = id then we say g is left-inverse for f. Therefore, an inverse is both a right and left-inverse.
We say that f is onto (or surjective) if every point B is the image of some point A.
We say that f is one-to-one (or injective) if distinct points A ̸= B map to distinct points f(A) ̸= f(B).
Exercise 10.9. Let f be a transformation of the plane.
(a) Show that f has a right-inverse if and only if f is onto. Hint: to prove the existence of a right-inverse, for each point A define g(A) to be any point B so that f(g(A)) = B.
Prove that A 7→g(A) is the desired right-inverse. Why does this argument fail if f is not surjective?
(b) Show that f has a left-inverse if and only if f is one-to-one. Hint Pick a point Q. If B = f(A) for some A, let g(B) = A. If B ̸= f(A) for any A, let g(B) = Q. Prove that g is the desired left-inverse.
(c) Suppose that f has left-inverse h and right inverse g. Prove that g = h, and hence f has a unique inverse.
(d) Suppose that f is one-to-one and onto. Prove that f has a unique inverse.
Definition 10.10. A set G of transformations of the plane is a group if (i) id ∈G, (ii) f ∈G implies that f is invertible and f −1 ∈G, and (iii) f, g ∈G implies that f ◦g ∈G.
Example 10.11. Proposition 10.5 and Exercise 10.6 imply that the set of all translations forms a group. Indeed, (i) it is clear that the identity transformation equals the translation by the zero vector, (ii) if T is the translation by v, then T −1 is the translation by −v (Proposition 10.5) and (iii) if T, S are translations by v, w, respectively, then T ◦S is the translation by v + w.
Exercise 10.12. Prove that the group of all translations has the structure of a vector space.
Indeed, show that the group of all translations is isomorphic to the vector space of vectors.
Thanks to this result, we can ‘think’ of a vector as a translation, rather than an equivalence class of pairs of points.
Exercise 10.13. Fix a non-zero vector v.
(a) Consider the set Nv = {v, 2v, 3v, 4v, · · · }, and for each natural number n ∈N, let Tn be the translation by nv. Let G = {Tn : n ∈N}. Prove that G satisfies part (iii) of the definition of a group, but show that it does not satisfy (i) or (ii).
(b) Consider the set Zv = {· · · , −2v, −v, 0, v, 2v, 3v, · · · }, and for each integer n ∈Z, let Tn be the translation by nv. As in part (a), let G = {Tn : n ∈Z}. Prove that G is a group.
98 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
(c) Fix two non-zero vectors v, w, and consider the set Zv + Zw = {nv + mw : n, m ∈Z}.
Prove that the set of all translations by elements in Zv + Zw forms a group.
Example 10.14. Fix a point O, and consider the set G of all rotations around O.
A A′ O Then G is a group: to show this, we let Rα ∈G denote the rotation by angle α, and then note that (i) R0◦= id, (ii) Rα ◦R−α = R−α ◦Rα = R0◦= id, and (iii) Rα ◦Rβ = Rα+β.
Exercise 10.15. (a) Pick some natural number n > 0, and let α be the angle (360/n)◦. Let G be the set of all rotations by angle kα with k = 1, 2, 3, · · · (i.e. G = {Rα, R2α, R3α, · · · }).
Prove that G is a group. In fact, prove that G is a finite group.
(b) Pick an irrational number λ ∈(0, 1). Let α be the angle (λ 360)◦. Prove that G = {Rα, R2α, R3α, · · · } is not a group. However, prove that e G = {· · · , R−2α, R−α, id, Rα, R2α, R3α, · · · } is a group. Unlike the group in part (a), show that e G is not finite.
Exercise 10.16. Let G be the set of all reflections of the plane (there is one reflection for every line). Prove that G is not a group.
Proposition 10.17. Isometries form a group.
Before we give the proof, we ask the reader to solve the following exercise, whose conclusion will be used in the proof of the proposition.
Exercise 10.18. (a) Let A ̸= B be two points, and let d = |AB|. Suppose a, b are non-negative real numbers so that one of the following relations is satisfied: d = a + b or a = d + b or b = a + d.
Show there is a unique point C on the line AB so that |AC| = a and |CB| = b.
(b) Let f be an isometry, and suppose that C is a point lying on the line AB. Use the equality case in the triangle inequality to prove that f(C) lies on the line f(A)f(B).
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 99 (c) If Y is any point on f(A)f(B), use part (a) to show that there is a unique point C on AB so that |CA| = |Y f(A)| and |CB| = |Y f(B)|. Conclude that f(C) = Y .
(d) Combine parts (b) and (c) to conclude that any isometry f maps the line AB to the line f(A)f(B) in a one-to-one and onto fashion.
Proof of the Proposition. It is clear that the identity transformation is an isometry, so we have shown that isometries satisfy property (i) from the definition of a group.
Now we prove that isometries satisfy property (iii) for groups. Suppose f, g are both isometries, and we want to show that the distance between f ◦g(A) and f ◦g(B) equals |AB|. But this is easy: |f(g(A))f(g(B))| = |g(A)g(B)| = |AB| where the first equality follows since f is an isometry, and the second equality follows g is an isometry. Therefore f ◦g is an isometry.
The hardest part of the proposition is establishing that isometries satisfy property (ii) for groups. Our argument is broken into three steps: first we fix some isometry f, and show that it is one-to-one and onto, which guarantees the existence of an inverse f −1 (see Exercise 10.9). Then we show that this inverse f −1 is also an isometry.
If A ̸= B are distinct points, then |AB| > 0, and so |f(A)f(B)| > 0, so f(A) ̸= f(B).
Therefore f is one-to-one.
To prove that f is onto, we will use the result of Exercise 10.18. Let Y be an arbitrary point in the plane, and we want to find C so that f(C) = Y . Take a (non-degenerate) triangle A, B, D, and, without loss of generality, suppose that the line Y f(A) intersects the line f(B)f(D). By the triangle inequality, f(A)f(B)f(D) is also a non-degenerate triangle (e.g. no vertex lies on the line joining the other two).
Let the line Y f(A) intersect the line f(B)f(D) at a point C′. Since C′ lies on f(B)f(D), we know there is C on BD so that C′ = f(C). Then, since Y lies on the line f(A)f(C), we may apply Exercise 10.18 know Y = f(X) for some X on AC.
A B D C f f(A) f(B) f(D) f(C) Y This proves that f is onto. Therefore, by Exercise 10.9 we know that f −1 exists. Now it is easy to show that f −1 is an isometry, since for any A, B we have |AB| = f(f −1(A))f(f −1(B)) = f −1(A)f −1(B) , 100 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
as desired. Since we have shown that the set of isometries satisfies (i), (ii) and (iii), we have proved the proposition.
□ Theorem 10.19. If an isometry f fixes the vertices of a triangle ABC, then f is the identity.
Proof. First we will prove that f fixes every point D on the line AB. Since f fixes A, B, |f(D)A| = |DA|, |f(D)B| = |DB|, and so part (a) of Exercise 10.18 implies that f(D) = D.
Now suppose that Y is an arbitrary point. Without loss of generality, assume that the line Y C intersects the line AB in a point D.
A B C D Y By the first part of our proof, we know that f fixes C and D. Therefore, applying the same argument as in the first paragraph, we know that f fixes every point on the line CD, so f fixes Y . It follows that f = id.
□ Corollary 10.20. Let f, g be isometries, and assume there is a triangle ABC so that f(A) = g(A), f(B) = g(B) and f(C) = g(C).
Then f = g.
Proof. Apply Theorem 10.19 to the isometry g−1f.
□ 10.1. The classification of isometries of the plane. So far we have encountered three types of isometries: reflections, rotations and translations. There is one more kind of isometry we have not seen yet, the so-called glide-reflection.
Definition 10.21. Let k be a line and v a vector parallel to k.
The glide reflection through k along v is the composition of the reflection R through k and a translation by v.
It is easy to see that the image of A under the glide reflection does not depend on the order with which we translate and reflect (in other words, R commutes with the translation by v).
k v v A A′ Note that the reflection through k is the glide reflection with vector v = 0.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 101 Theorem 10.22 (classification of isometries). Any isometry f is a rotation or a translation or a glide reflection.
Proof. Let ABC be a triangle. Then the triangle f(A)f(B)f(C) is congruent to ABC by the “side-side-side” criterion. Hence, by definition of congruence, there is a sequence R1, R2, · · · , Rn of reflections taking ABC to f(A)f(B)f(C). By Corollary 10.20 we conclude that f = Rn ◦· · · ◦R1. In fact, the proof of Theorem 2.15 shows that there is a sequence of at most three reflections taking ABC to f(A)f(B)f(C), so we may assume that n = 1, 2 or 3.
If n = 1, so that f = R1, then we are obviously done.
If n = 2, then let R1 be the reflection through k and R2 be the reflection through ℓ. If k and ℓintersect at a point O, then R2 ◦R1 is a rotation through O (see Theorem 2.2). If k and ℓare parallel, then R2 ◦R1 is a translation (see Exercise 10.7). In either case, we have classified what f is.
The hardest part of the proof is analyzing the case n = 3. Let R1, R2 and R3 be the reflections through lines k, ℓand m, respectively.
We will prove the case when k and ℓ intersect in a point O (the case when they are parallel is left as an exercise, see Exercise 10.23). Suppose that the angle between k, ℓis α, so that R2 ◦R1 is the rotation around O of angle 2α.
ℓ k O R2 ◦R1 A A′ The key observation is that if we have any other pair of lines k′, ℓ′ so that (i) k′ and ℓ′ intersect at O and (ii) k′, ℓ′ are separated by an angle α, then the reflection R′ 1 through k′ and the reflection R′ 2 through ℓ′ satisfy R′ 2 ◦R′ 1 = R2 ◦R1 (since they are both the rotation around O by angle 2α). Therefore, without loss of generality, we may rotate the pair of lines k, ℓso that ℓis perpendicular to m. Let P denote the intersection point of ℓand m.
ℓ k O m P 102 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
As we did above with lines k and ℓ, we may now rotate the lines ℓand m to lines ℓ′ and m′ still intersecting at P so that R′ 3 ◦R′ 2 = R3 ◦R2 (where R′ 2, R′ 3 are the reflections through ℓ′ and m′). Therefore, replacing ℓand m by ℓ′ and m′ we may assume that m is perpendicular to k.
ℓ k O m P ℓ′ m′ Since m is perpendicular to both k and ℓ, k and ℓparallel, and so R3 ◦R2 ◦R1 is a glide reflection (R2 ◦R1 is a translation, and R3 is a reflection through a line parallel to the translation vector). This completes the proof of the theorem.
□ Exercise 10.23. Prove the above theorem in the case when k and ℓare parallel.
10.2. Bonus subsection on the linear part of an isometry. A key idea in the study of isometries is that isometries transform vectors in a natural way. If − − → AB is a vector and f is an isometry, we can consider the transformed vector − − − − − − → f(A)f(B). What is important about this transformation is that − − − − − − → f(A)f(B) only depends on the vector − − → AB - in other words, if − − → AB = − − − → A′B′, then − − − − − − → f(A)f(B) = − − − − − − − → f(A′)f(B′). In fact, the transformation − − → AB 7→− − − − − − → f(A)f(B) is a linear transformation of vectors. Exercise 10.24 presents the necessary results.
f Exercise 10.24. Let f be an isometry.
(a) If − − → AB = − − − → A′B′, show that − − − − − − → f(A)f(B) = − − − − − − − → f(A′)f(B′). Hint: − − → AB = − − − → A′B′ if and only if the midpoint of AB′ is the midpoint of A′B. The latter statement is preserved under isometries.
(b) Show that for each vector there is a corresponding vector f ♭(v) so that − − → AB = v implies − − − − − − → f(A)f(B) = f ♭(v). Hint: use part (a)!
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 103 Then v 7→f ♭(v) is a transformation of the set of vectors.
(c) Prove that for any positive real number λ > 0, f ♭(λv) = λf ♭(v). Hint: it may be convenient to prove the cases λ > 1 and λ < 1 separately.
(d) Suppose that v, w are two vectors. Prove that f ♭(v + w) = f ♭(v) + f ♭(w). Hint: Let A, B, C, D be chosen so that − − → AB = v and − − → BD = w, so − − → AD = v + w. Then − − − − − − − → f(A)f(D) = f ♭(v + w).
However, we also have − − − − − − → f(A)f(B) = f ♭(v) and − − − − − − − → f(B)f(D) = f ♭(w), so − − − − − − − → f(A)f(D) = f ♭(v) + f ♭(w).
(e) Use part (d) to show that f ♭(−v) = −f ♭(v), and then use part (c) to show that f ♭(λv) = λf ♭(v) for any real number λ.
Therefore, for each isometry f, f ♭is a linear operator on the set of all vectors (a “linear operator” is a transformation of a vector space which preserves addition and scalar multiplication).
(f) If g is another isometry, then (g ◦f)♭= g♭◦f ♭.
Definition 10.25. Following the notation introduced in Exercise 10.24, we define the linear part of an isometry f to be the linear operator f ♭. This operator acts on the space of vectors in such a way that if A, B are points in the plane, then f ♭(− − → AB) = − − − − − − → f(A)f(B).
Exercise 10.26. Give an example of two different isometries f ̸= g so that f ♭= g♭.
Exercise 10.27. Show that f ♭(v) = v for all vectors v if and only if f is a translation. Hint: for the “only if” direction it suffices to show that, for all A, B, − − − − → Af(A) = − − − − → Bf(B).
Exercise 10.28. If ρ is any rotation by an angle α, then ρ♭(− − → AB) = − − → AB′, where B′ is the rotation of B around A by the angle α.
θ A B ρ(A) ρ(B) B′ θ Hint: Let O be the centre of rotation of ρ and let T be a translation taking A to O. Then show that T −1◦ρ◦T is a rotation around A by angle α. Finally, note that (T −1◦ρ◦T)♭= ρ♭.
10.3. Bonus subsection on orientation. The goal of this subsection is to make precise the statement: “the isometry f preserves (or reverses) orientation.” Intuitively, an isometry f “preserves orientation” if it takes a polygon A1 · · · An whose vertices are oriented “clockwise” to another polygon whose vertices are oriented “clockwise,” while f reverses orientation if the 104 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
vertices of f(A1) · · · f(An) are oriented “counter-clockwise.” In the figure below, f preserves orientation, while g reverses it.
A1 A2 A3 A4 A5 f A1 A2 A3 A4 A5 g A1 A2 A3 A4 A5 It is rather awkward to make the notion of “clockwise” precise, but it is possible to give a rigourous definition of orientation which agrees with our intuition.
First let us write down a few properties any reasonable definition of orientation should satisfy.
Requirements. Let f and g be isometries of the plane.
(I) If f preserves orientation and g preserves orientation, then f ◦g also preserves orien-tation.
(II) If f preserves orientation and g reverses orientation, then f ◦g and g ◦f reverse orientation.
(III) If f reverses orientation and g reverses orientation, then f ◦g preserves orientation.
(IV) All reflections reverse orientation.
We note that properties (I), (II), (III) can be packaged into the shorter statement: for every isometry f there is a corresponding number sgn(f) ∈{−1, +1}, so that sgn(f ◦g) = sgn(f)sgn(g), with the convention that sgn(f) = +1 means f preserves orientation and sgn(f) = −1 means f reverses orientation. Then property (IV) implies that sgn(R) = −1 whenever R is a reflection.
Therefore, the problem of giving a rigourous definition of orientation reduces to the prob-lem of constructing the function sgn.
Proposition 10.29. There is at most one function f 7→sgn(f) ∈{−1, +1}, defined on isometries, so that sgn(f ◦g) = sgn(f)sgn(g) and sgn(R) = −1 for all reflections R.
Proof. Let f be an isometry of the plane. Following the proof of the classification of all isometries, write f as a composition of reflections f = Rn ◦· · · ◦R1. Then sgn(f) = (−1)n, so the value of sgn(f) is already determined for us. This proves the proposition.
□ It is tempting to define sgn(f) = (−1)n whenever f = Rn ◦· · · ◦R1 for reflections R1, · · · , Rn, but this definition is, a priori, not well-defined, since it is certainly conceiv-able that we could also write f = R′ n+1 ◦· · ·◦R′ 1 for some different reflections R′ 1, · · · , R′ n+1.
In short, the value of sgn(f) seems to be over-determined.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 105 Theorem 10.30. There exists a function sgn : {isometries} →{±1} so that sgn(f ◦g) = sgn(f)sgn(g) and sgn(R) = −1 for all reflections R.
Proof. A key ingredient in the contruction is the linear part of an isometry. As we did above, we write f ♭for the linear part of an isometry f. Then f ♭is a linear operator on a (two-dimensional) vector space, and so its determinant det f ♭is well-defined. Furthermore, basic properties of f ♭and det imply that det(f ◦g)♭= det f ♭· det g♭. We define sgn(f) = det f ♭.
Then sgn(f ◦g) = sgn(f)sgn(g).
If R is a reflection through a line ℓ, pick v, w basis vectors so that v ∥ℓand w ⊥ℓ. By picking v = − − → AB with A, B ∈ℓwe clearly see that R♭(v) = v. Write w = − → AC with A ∈ℓ and C ̸∈ℓ. Then R♭(w) = − − → AC′ with C′ ̸= C. Since AC ⊥ℓ, A is the midpoint of C′C, so |CA| = |AC′|. Thus |AC′| + |AC| = 0, and so R♭(w) = −w.
ℓ v = R♭(v) w R♭(w) = −w The matrix of R♭with respect to the basis {v, w} is 1 0 0 −1 , so det R♭= −1. Since any isometry f is a composition of reflections, sgn(f) ∈{+1, −1}, and this completes the proof.
□ We will use orientation as a tool to distinguish isometries; the following exercise demon-strates how this can be done.
Exercise 10.31. Suppose that R1, · · · , Rn are rotations (whose centres of rotation may be different). Prove that R1 ◦· · · ◦Rn is either a translation or a rotation.
10.4. Applications of the classification of isometries.
Problem 10.32. Let f be a glide reflection parallel to the line k. Then for any point A, the centre of the segment Af(A) lies on k.
Solution. The case when the translation vector is zero is immediate, and so we suppose that the translation vector of the glide reflection is non-zero. Let A′ denote the reflection of 106 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
A through k, so that f(A) is the translation of A′ along a vector parallel to k.
k A A′ f(A) Then k is parallel to A′f(A), and so we may apply Thales’ theorem to the angle A′Af(A) to deduce that the intersection of Af(A) with k is the midpoint of Af(A).
Problem 10.33. Let ABCD and A′B′C′D′ be congruent quadrilaterals, but suppose the vertices of one quadrilateral are labelled clockwise, while the vertices of the other quadrilat-eral are labelled counter-clockwise. Then the centres of the segments AA′, BB′, CC′ and DD′ all lie on a common line.
A B C D A′ B′ C′ D′ Solution. This problem is a nice application of our classification of isometries. The fact that ABCD and A′B′C′D′ are congruent implies there is some isometry f which takes ABCD to A′B′C′D′. Since A′B′C′D′ has a different orientation to ABCD, we know that the isometry f must reverse orientation. Thus f cannot be a rotation or a translation. By our classification of isometries, f is therefore a glide reflection through some line k. Then Problem 10.32 implies that the midpoints of the segments AA′, BB′, CC′ and DD′ all lie on k, and this completes the solution.
Theorem 10.34. Let f = Tn◦Rn◦· · ·◦T1◦R1 be a composition of translations and rotations.
Suppose the rotation angle of Rj is αj. Then f is a translation if α1 + · · · + αn = 0 and otherwise f is a rotation by angle α1 + · · · + αn.
Proof. The idea of the proof is fairly simple: start with some line ℓ, analyze how ℓtransforms as we apply R1, T1, · · · , Rn, Tn.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 107 We have shown that R1(ℓ) intersects ℓat an angle α1 (see Theorem 2.9)); if α1 = 0◦ or = 180◦, then R1(ℓ) doesn’t actually intersect ℓ, but we will interpret “intersecting at an angle of 0◦or 180◦” to mean “parallel.” Then T1 ◦R1(ℓ) is parallel to R1(ℓ), and hence T1 ◦R1(ℓ) also intersects ℓat an angle of α1. Iterating this argument, we deduce that f(ℓ) = Tn ◦Rn ◦· · · ◦T1 ◦R1(ℓ) intersects ℓ at an angle of α = α1 + · · · + αn. If α is not 0◦or 180◦, then f(ℓ) actually intersects ℓin a unique point point P. It follows that f cannot be a translation, since translations send lines to parallel lines. Since f preserves orientation, f cannot be a glide reflection. By the classification of isometries, it follows that f is a rotation. Furthermore, since f(ℓ) intersects ℓat an angle α, we know (from Theorem 2.9 again) that the angle of rotation of f must be α or α+180◦(this indeterminacy is because there are always two angles formed between a pair of lines). To resolve this indeterminacy we will look at the linear part of f. Furthermore when α = 0◦or α = 180◦, then f(ℓ) is parallel to ℓand so we cannot detect whether f is rotation by 180◦or a translation simply by comparing f(ℓ) with ℓ. However, instead of keeping track of how the line ℓtransforms under R1, T1 · · · , Rn, Tn, if we keep track of how a vector v transforms, then we will be able to differentiate the angles α and α + 180◦and also classify the cases when α = 0◦and α = 180◦. In other words, we will look at the linear part of the composition Rn ◦Tn ◦· · · ◦R1 ◦T1. Since T ♭= identity for any translation T (Exercise 10.27) we have (Rn ◦Tn ◦· ◦R1 ◦T1)♭= R♭ n ◦T ♭ n ◦· · · ◦R♭ 1 ◦T ♭ 1 = R♭ n ◦· · · ◦R♭ 1.
Then, by Exercise 10.28, we know that R♭ n ◦· · · R♭ 1 simply takes a vector − − → AB and sends it to − − → AB′, where B′ is the rotation of B around A by an angle α1 + · · · + αn.
If α = α1 + · · · + αn ̸= 0◦, 180◦, then we already know that f is a rotation, so the fact that f ♭rotates by α implies f also rotates by α. This proves the case when α ̸= 0◦, 180◦.
If α1 + · · · + αn = 180◦, every vector v is mapped to its inverse −v. In this case, we conclude that Rn ◦Tn ◦·◦R1 ◦T1 is not a translation, and hence must be a rotation by angle 180◦. On the other hand, if α1 + · · · + αn = 0◦, then the above discussion implies that f ♭is the identity transformation, so f is a translation (by Exercise 10.27).
□ Problem 10.35. Let RA, RB and RC be rotations around the vertices of a triangle ABC by angles α, β and γ, respectively. Supposing that RC ◦RB ◦RA = id, find the interior angles of ABC.
A B C 108 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Solution. To begin, let us see how the point A transforms when we apply the rotations RA, RB, RC. Clearly RA(A) = A. Define A′ = RB(A); then RC ◦RB ◦RA = id implies RC(A′) = A.
A B C β γ A′ Then |A′C| = |AC| and |A′B| = |AB|, so that CB is the bisector of AA′. This fact uniquely determines A′, and also tells us that BC bisects the angles ∠ABA′ and ∠A′CA. Therefore ∠BCA = γ/2 and ∠ABC = β/2.
To find the interior angle at A, we use a similar argument. First we note that RC ◦RB ◦ RA = id implies RB ◦RA = R−1 C , and thus RB ◦RA ◦RC = id.
Then we may apply the same argument used in the first paragraph but starting with vertex C to conclude ∠CAB = α/2. This completes the solution.
A B C α/2 β/2 γ/2 Exercise 10.36. Let RA, RB be rotations by angles α, β around points A, B, respectively.
Assuming that α + β < 360◦, find a rotation RC by angle γ around a point C so that RB ◦RA = R−1 C . Hint: Use Problem 10.35.
Problem 10.37. Let E, C, D be three points lying on a common line with C between E and D. Suppose A and B are chosen on the same side of the line ED so that the triangles AEC and BDC are isosceles and right, and let M be the midpoint of AB. Prove that EMD TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 109 is isosceles and right.
A B M E C D Solution. Interestingly, we will solve this problem by invoking the result of Problem 10.35.
Let RD be the rotation around D sending B to C, and let RE be the rotation around E sending C to A. Then the composition RE ◦RD sends B to A. By Theorem 10.34, RE ◦RD must be a rotation by 180◦, and since RE ◦RD sends B to A, it must be the 180◦rotation centred at M, which we denote by RM. Then RM ◦RE ◦RD = id, and so, by Problem 10.35, ∠EMD = 90◦, ∠DEM = ∠MDE = 45◦, hence EMD is isosceles and right.
Problem 10.38. Let A, B, C, D be a square with side length a, and choose P on the side CD. Let PY QX be another square with diagonal QP ⊥DC and diagonal length a/2. Let K, L be constructed as in the figure below (so that LY B and AXK are corners of squares).
Then Q is the midpoint of the segment KL.
P A B K Q L X D Y C Solution. Let RY and RX be 90◦rotations around Y, X sending L to B and A to K, respectively, and let T be the translation by vector − − → BA. Then RX ◦T ◦RY (L) = K, and, further, by Theorem 10.34 we know RX◦T ◦RY is an 180◦rotation. We claim that RX◦T ◦RY 110 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
fixes the point Q. To see this, consider the following figure.
Q Q′ Q′′ X Y P T RY RX As above, let Q′ = RY (Q) and RX(Q′′) = Q.
As is clear in the above figure, QQ′ is orthogonal to QP, and hence QQ′ is parallel to T. Similarly QQ′′ is parallel to T, and |Q′Q′′| = |QQ′| + |QQ′′| = a, so T takes Q′ to Q′′. Thus RX ◦T ◦RY (Q) = Q, and so Q is the centre of the 180◦rotation taking L to K, which means Q is the midpoint of KL.
11. Homotheties with negative scale Definition 11.1 (Homothety with negative scale). In Example 7.3 we constructed homoth-eties with positive scale k > 0, and showed that they were similarities with scale k. We can extend this definition: the homothety with scale k < 0 centred at O is the transformation of the plane fixing O and sending each point A ̸= 0 to the point A′ lying on the line AO so that (a) O separates A from A′ and (b) |A′O| / |AO| = |k|.
A B O A′ B′ Using the fact that homotheties with positive scale |k| are similarities with scale |k|, it is straightforward to show that homotheties with scale k < 0 are also similarities with scale |k| (see Exercise 11.2 below).
Exercise 11.2. (a) Show that the homothety with scale −1 centred at O is the point reflection (i.e. 180◦rotation) around O.
(b) Using part (a), prove that the homothety of scale k < 0 centred at O is the composition of a homothety of scale |k| > 0 and an isometry. Conclude that homotheties of scale k < 0 are similarities with scale |k| > 0.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 111 Exercise 11.3. Prove that the inverse of a homothety of scale k (k may be positive or negative) centred at O is the homothety of scale 1/k centred at O.
Theorem 11.4. Homotheties map lines to parallel lines. In other words, if ℓis a line and ℓ′ is its image under a homothety, then ℓ′ is a line parallel to ℓ.
Proof. Let O and k denote the centre and scale of the homothety. We consider the case k > 0. If ℓcontains O, then the homothety fixes ℓ, so there is nothing to prove. Consider now a line ℓdisjoint from O. Pick a point A on the line ℓ, and let ℓ′ be the line parallel to ℓpassing through the image A′ of A under the homothety. If B lies on ℓ, we want to show that B′ lies on ℓ′.
A A′ B B′ ?
O ℓ′ ℓ To prove that B′ lies on ℓ′, we use the converse to Thales’ Theorem. Since |OA′| / |OA| = |OB′| / |OB|, and O does not separate A from A′ or B from B′, we deduce that AB and A′B′ are parallel, and so A′B′ must be the line ℓ′, hence B′ ∈ℓ′. We have shown that the homothety maps all points on ℓonto the line ℓ′.
We still need to prove that every point on ℓ′ is the image of a point from ℓ. To show this, fix C′ ∈ℓ′, and pick C on the ray OC′ so that |OC′| / |OC| = k (such a C′ exists by continuity). Applying the converse to Thales’ Theorem again, we deduce that ℓ′ = A′C′ is parallel to AC, and so C must lie on ℓ. By our construction, the image of C under the homothety is C′ (justifying our notation) and this completes the proof.
□ Exercise 11.5. Prove Theorem 11.4 in the case k < 0.
Problem 11.6. Given an acute triangle ABC, find an inscribed square A′B′F ′E′ such that the edge E′F ′ lies on the side AB, and so that A′ ∈AC and B′ ∈BC.
A A′ B C B′ E′ F ′ Solution. The idea for the solution is to begin by constructing a square which satisfies some of the requirements of the problem (but not all of the requirements) and then use a 112 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
homothety to transform the square into a new one which satisfies all the requirements of the problem.
We begin by defining E, F so that ABFE is a square and ABFE lies inside the angle BCA (this determines E and F uniquely), and we define E′ and F ′ to be the intersections of EC and FC with AB, respectively. Then we raise perpendiculars p and q to AB from the points E′ and F ′, respectively, and define A′ to be the intersection of p and AC and B′ the intersection of q with BC.
A A′ B C B′ E F E′ F ′ p q Consider the homothety centred at C of scale |CE′| / |CE|. We claim that this homothety sends A to A′, B to B′, E to E′ (obviously) and F to F ′. Since EF is parallel to E′F ′ (since E′F ′ = AB, and ABFE is a square), we can apply Thales’ Theorem to deduce that |CF ′| / |CF| = |CE′| / |CE|, and so we know that the homothety maps F to F ′.
Since homotheties preserve parallel lines, we deduce that the line EA is mapped to the line p (show this!), and so the image of A under the homothety must lie on p. However, since the homothety is centred at C, we also know that the image of A lies on the ray CA, and so the image of A must be the intersection point A′ of p and AC. A similar argument proves that the image of B is B′.
We claim that A′B′E′F ′ is a square. This is easy to see since ABEF is a square and homotheties are similarities (so they preserve angles and the ratios of sides). This completes the solution.
Problem 11.7 (Euler line). Let ABC be a triangle, and let O be the intersection point of the side bisectors (i.e. O is the centre of the circumscribed circle), let H be the orthocentre of ABC (recall that the orthocentre is the intersection of the altitudes), and let S be the centroid of ABC (recall that the centroid is the intersection of medians). Prove that S lies TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 113 on OH and divides it in a 1 : 2 ratio.
A B C S H O Solution. The idea is to consider the homothety with scale −1/2 with centre S, and show that H gets mapped to O, which will prove that S divides OH in a 1 : 2 ratio.
Let A′, B′, C′ be the midpoints of the edges BC, AC and AB, respectively. It can be shown that S divides A′A, B′B and C′C in 1 : 2 ratio (see Exercise 11.8 below). It follows that the homothety of scale −1/2 through S sends the triangle ABC to the triangle A′B′C′.
S A B C A′ B′ C′ Furthermore, by Theorem 11.4, the altitude ℓthrough C gets mapped to a parallel line ℓ′ passing through C′, and since ℓis perpendicular to AB, ℓ′ is also perpendicular to AB. Since C′ is the midpoint of AB and ℓ′ is perpendicular to AB, ℓ′ is the bisector of AB.
S A B C C′ ℓ ℓ′ 114 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
A similar argument shows that the altitude through A is mapped to bisector of BC. It follows that the intersection point H of the two altitudes is mapped to the intersection point O of the two bisectors. Thus HSO lie on a common line. Since the homothety has negative scale, we know S separates HO, and since it has scale −1/2, we know |SO| / |SH| = 1/2, so S splits OH in a 1 : 2 ratio, as desired.
Exercise 11.8. Let D, E, F be the centres of the sides BC, AC, AB of a triangle ABC.
Prove that the segments AD, BE, DF intersect at a common point that divides each of them in the ratio 2 : 1.
Hint: Consider the following figure, and prove that |AQ| / |DQ| = 2.
Q A B C F G E D Theorem 11.9. Let ABC and A′B′C′ be triangles with AB ∥A′B′, BC ∥B′C′ and AB ∥A′C′. Then there is a homothety or a translation mapping ABC to A′B′C′.
Before we prove this theorem, we remark that it has a nice corollary of independent interest.
Corollary 11.10. With ABC and A′B′C′ with parallel edges (as in Theorem 11.9), we deduce that AA′, BB′ and CC′ are parallel (corresponding to a translation in Theorem 11.9) or the intersect in a single point (corresponding to a homothety in Theorem 11.9).
A B C A′ B′ C′ O Exercise 11.11. Prove Corollary 11.10 using Theorem 11.9.
Proof of Theorem 11.9. Assume first that AA′ and BB′ intersect in a point O. Since AB and A′B′ are parallel, there are only two possibilities, either O lies outside of the segments TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 115 AA′ and BB′ or O separates A from A′ and B from B′ (for the proof of this assertion, refer to the proof of Thales’ theorem). Assume first that O lies outside of the segments AA′ and BB′. Consider the homothety with centre O and (positive) scale |A′B′| / |AB|.
A straightforward application of Thales’ theorem proves that the image of A is A′ and the image of B is B′. Since the line A′C′ is parallel to the line AC, and (by Theorem 11.4) the line AC is mapped to a parallel line under the homothety, we deduce that the image of the line AC is the line A′C′. Similarly, the image of the line BC is the line B′C′, and so the intersection point C of the lines AC and BC is mapped to the intersection point C′ of the lines A′C′ and B′C′. Therefore we have found a homothety taking ABC to A′B′C′.
In the case where O separates A from A′ and B from B′, we consider the homothety centred at O with scale −|A′B′| / |AB|. Using a similar argument to the one in the previous case, we deduce that this homothety takes C to C′. The details are left to the reader.
A B C A′ B′ C′ O Now we assume that AA′ and BB′ are parallel. Then ABB′A′ forms a parallelogram, and so the translation by vector − − → AA′ = − − → BB′ takes A to A′ and B to B′. Since translations map lines to parallel lines, we deduce that AC is mapped to A′C′ and BC is mapped to B′C′, and so C must be mapped to C′, as desired. This completes the proof of the theorem.
A B C A′ B′ C′ □ Problem 11.12. Let ABCDEF be a convex hexagon with area a, and so that each diagonal AD, BE and CF cuts ABCDEF into two quadrilaterals of equal area a/2. Prove that AD, 116 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
BE and CF intersect in a single point.
A B C D E F Solution. First we remark that |ABE| = |ADE|, since |ADEF| = |ABEF| = a/2, and |ADEF| = |AEF| + |ADE| and |ABEF| = |AEF| + |ABE|.
Then since ABE and ADE share the base AE, they must also share a height, so D and B are the same distance from AE, thus DB ∥AE.
A B C D E F Analogously, BF ∥CE and AD ∥DF. Therefore we may apply Corollary 11.10 to the triangles ACE and DFB (since they have parallel side pairs) so that AD, CF and EB intersect in a single point, as desired.
A B C D E F TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 117 Theorem 11.13. The homothety with centre S and scale k maps the circle o with centre O and radius r to the circle with centre O′ (the image of O) and radius |k| r.
O O′ S Proof. If X lies on o and X′ is the image of X under the homothety, then we know that |X′O′| = k |XO|, and so X′ lies on o′. This proves that every point on o is mapped onto o′.
If Y ′ lies on o′, then we let Y be the image of Y ′ under the homothety centred at S of scale 1/k, and note that Y must lie on o; this proves that every point on o′ is the image of a point from o, and this completes the proof.
□ Theorem 11.14. Let o1 and o2 be two circles with radii r1 ̸= r2. Then there are exactly two homotheties mapping o1 to o2 (one with scale r2/r1 and one with scale −r2/r1).
O1 O2 Sin Sout Proof. It is clear that the scale of any homothety sending o1 to o2 must be ±r2/r1, and so it remains to find the centre of the homotheties. Let O1 and O2 be the centres of o1 and o2. First we will find the centre of the homothety with scale −r2/r1. Pick S in the segment O1O2 so that |SO2| / |SO1| = r2/r1. Such an S always exists, by continuity (since the ratio |XO2| / |XO1| goes to 0 as X goes to O2 and to ∞as X goes to O1). Then the homothety of scale −r2/r1 around S sends O1 to O2, and, by Theorem 11.13, sends the circle of radius r1 to the circle of radius r2. Thus the chosen homothety sends o1 to o2. Note that we did not need to assume r1 ̸= r2 in this case.
Now we will find the centre of the homothety with scale r2/r1. Without loss of generality, assume that r1 < r2. Note that the ratio |XO2| / |XO1| goes to ∞as X approaches O1 and asymptotically decreases to 1 as X gets further from O1 (while remaining on the opposite side of O2). Therefore, by continuity, we can find an S on the line O1O2 so that O1 separates S from O2, and so that |SO2| / |SO1| = r2/r1 > 1. Then an analogous argument to the one 118 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
in the previous paragraph shows that the homothety of scale r2/r1 centered at O takes o1 to o2.
In Exercise 11.15 below, the reader will prove that the two homotheties we have con-structed are the only ones taking o1 to o2, and so we are justified in saying that there are exactly two homotheties taking o1 to o2.
□ Exercise 11.15. Prove that two homotheties we have constructed in the proof to Theorem 11.14 are unique using the following outline: Suppose there were a third homothety h sending o1 to o2, and let H be the centre of h.
(a) Prove that the scale of h is ±r2/r1.
(b) If the scale of h is −r2/r1, prove that H lies between O1 and O2, and furthermore that |O2H| / |O1H| = r2/r1. In this case, prove that H = Sin (using the notation in the figure accompanying Theorem 11.14).
(c) If the scale of h is r2/r1, and r2 > r1, prove that O1 lies between H and O2, and that |O1H| / |O2H| = r2/r1. In this case, prove that H = Sout.
(d) Repeat part (c) when r1 < r2.
(e) Conclude that h is one of the two homotheties we constructed in the proof to Theorem 11.14.
Exercise 11.16. Let o1 and o2 be two circles, and let ℓ1 be a line tangent to o1. Prove that there are exactly two lines, ℓ2,+ and ℓ2,−, which are parallel to ℓ1 and tangent to o2, and prove that they are the image of ℓ1 under the homotheties (or translations) from Theorem 11.14.
Remark. The proof of Theorem 11.14 uses a continuity argument to guarantee the existence of the two centres of homothety (these are denoted Sin and Sout in the accompanying figure).
It is natural to ask whether there is a more concrete description of Sin and Sout.
We will be able to describe Sin and Sout as the intersection of various tangent lines to o1 and o2. However, before we begin, we will need a preliminary exercise concerning tangent lines to circles.
Exercise 11.17. Given two circles o1, o2 with centres O1, O2, respectively, and two parallel lines ℓ1 and ℓ2 tangent to o1 at P1 and o2 at P2, respectively, we say that ℓ1 and ℓ2 are tangent to o1 and o2 from the same side if the vectors − − − → P1O1 and − − − → P2O2 differ by a positive scalar (since − − − → P1O1 and − − − → P2O2 are both perpendicular to ℓ1 ∥ℓ2, we know they differ by some scalar but it could be negative).
If − − − → P1O1 and − − − → P2O2 differ by a negative scalar then we say that ℓ1 and ℓ2 are tangent to o1 and o2 from opposite sides.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 119 ℓ2 O2 O1 ℓ1 P1 P2 Show that (a) Given a line ℓ1 tangent to a circle o1 and a second circle o2, there is a unique line ℓ2 tangent to o2 so that ℓ1 and ℓ2 are tangent to o1 and o2 from the same side.
(b) If there is a homothety with positive scale (resp. negative scale) taking the pair (o1, ℓ1) to the pair (o2, ℓ2), where ℓ1 is tangent to o1, then ℓ2 is tangent to o2 from the same (resp. opposite) side as ℓ1 is to o1. (Hint: consider how homotheties transform vectors.) With the results of this preliminary exercise at our disposal, we will describe Sin in various cases. First, if we assume that o1 and o2 are externally tangent at a point S, then Sin must be the tangency point S.
o1 o2 S ℓ To see why this is so, let ℓdenote the tangent line to o1 and o2 through S and consider the homothety with scale −r2/r1 centered at S. This homothety preserves ℓand S, and so, by part (b) of Exercise 11.17, we know that the image of o1 is tangent to ℓat S and lies on the opposite side as o1. Since o2 is the unique circle of radius r2 tangent to ℓat S lying on the opposite side of o1, the image of o1 must be o2.
120 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Now let us assume that o1 and o2 are tangent to opposite angles formed by a pair of lines k and ℓwhich intersect at a point S, as shown below.
o2 o1 k ℓ S Considering the homothety of negative scale mapping o1 to o2, part (b) of Exercise 11.17 guarantees that the image of k is a parallel line which is tangent to o2 from the opposite side as k is tangent to o1 - in this case, it is easy to see that this implies k is preserved by the homothety. Similarly ℓis preserved by the homothety, and so their intersection point must be the centre of the homothety.
Similar arguments allow us to find the centre of the positive scale homothety in the special cases when o1 and o2 are internally tangent, and when they are both tangent to the same angle (see Exercise 11.19).
Exercise 11.18. Let P and Q be two points on opposite sides of an angle (less than 180◦), both at the same distance from the centre S. Show that there is a unique circle tangent to angle with tangency points P and Q. (Hint: Consider the perpendicular raised from P).
S P Q Exercise 11.19. Let o1 and o2 be two circles with different radii.
(a) If o1 and o2 are internally tangent at a point S (this means that they are both tangent to a line ℓat a point S, and they both lie on the same side of the line), then S is the centre TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 121 of the positive scale homothety taking o1 to o2.
o1 o2 ℓ (b) If o1 and o2 are both tangent to an angle with vertex S (the vertex of an angle formed by two half-lines r, s is the common end point shared by the two half-lines), then S is the centre of the positive scale homothety taking o1 to o2.
o2 r s S o1 Problem 11.20. Let o1 and o2 be two circles internally tangent to a third circle o3 at points A and B, respectively. Let k be a common tangent line to o1 and o2 in points P and Q which does not separate o1 from o2. Prove that AP and BQ intersect on o3.
o3 o1 o2 P Q A B X k Solution. Consider the homothety h1 with positive scale taking o1 to o3. As proven in Exercise 11.19 h1, is centred at the point A. Let X be the image of P under h1, and let ℓ 122 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
be the image of k under h1. Since k is tangent to o1 at P, ℓis tangent to o3 at X. Since the homothety is centred at A, we know A, P, X lie on a common line.
o3 o1 o2 P Q A B X k ℓ Now consider the homothety h2 centred at B with positive scale taking o2 to o3, and let ℓ′ be the image k under h2. We claim that ℓ′ = ℓ. Using the notions and results introduced in Exercise 11.17, we know that ℓis tangent to o3 from the same side as k is to o1 and ℓ′ is tangent to o3 from the same side as k is to o2. By our assumption k is tangent to o1 from the same side as it is to o2. Therefore ℓis tangent to o3 from the same side as ℓ′ is to o3, and so ℓ= ℓ′. As in the previous paragraph, it follows that h2 sends Q to X, and so B, Q, X lie on a common line. Thus BQ and AP intersect at X ∈o3, and this completes the solution.
Theorem 11.21 (Classification of similarities). Every similarity is (i) a translation, (ii) a glide reflection, or a (iii) composition of a homothety with centre O and a rotation around O or a reflection through a line through O.
Exercise 11.22 (challenging). Prove Theorem 11.21. To make the exercise a bit easier, you may assume the fact that every similarity with scale 0 < λ < 1 has a fixed point.
12. Inversion in a circle In this section, we add to the plane one point “at infinity,” denoted ∞. By definition, every line contains ∞. The reason for this addition is that it allows us to define the following transformation of the plane: Definition 12.1. Let o be a circle with radius r and centre O. The inversion in o is the the transformation of the plane (with our added point!) which sends (a) O to ∞, (b) ∞to TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 123 O and (c) any point A ̸= O, ∞to the point A′ lying on the ray OA so that |OA| |OA′| = r2.
A A′ B′ B O Note that (i) If A ∈o, then A′ = A (i.e. the inversion fixes every point in the circle), (ii) the inside of o is swapped with the outside of o, and (iii) the inversion satisfies (A′)′ = A.
Theorem 12.2. If A′, B′ are the images of A, B, respectively, under the inversion through a circle o with centre O, then OBA is similar to OA′B′.
A A′ B B′ O Proof. We will use the “side-angle-side” criterion to show that OBA and OA′B′ are sim-ilar. Since the two triangles share the angle at O, it suffices to show that |OA′| / |OB| = |OB′| / |OA|, but this follows immediately from the fact that |OA| |OA′| = (radius of o)2 = |OB′| |OB| , and this completes the proof.
□ Corollary 12.3. In the setting of Theorem 12.2, we have equalities ∠ABO = ∠OA′B′ and ∠OAB = ∠A′B′O.
A A′ B B′ O Theorem 12.4. Inversions map circles and lines to circles and lines. (i.e. it is possible for a circle/line to be mapped to either a line or a circle, and these are the only possibilities.) 124 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Proof. Let o be the inversion circle. Consider first the case where a line k passes through the centre O. If A ∈k, then the entire ray OA lies in k, and hence A′ ∈k. Thus k is fixed by the inversion.
A A′ O k Now we consider the complementary case when k does not contain O.
Let A be the orthogonal projection of O onto k, and let A′ be the image of A under the inversion. We claim that the image of k under the inversion is the circle with diameter OA′. To see this, fix any point B ∈k, with B ̸= A, ∞, and let B′ be the image of B under the inversion through o. Since ∠OAB = 90◦, Corollary 12.3 implies ∠OB′A′ = 90◦, and so, by Corollary 3.8, B lies on the circle with diameter OA′.
A B B′ A′ O k Finally, it is clear that O and ∞are interchanged by the inversion, and so the entire line k is mapped into the circle with diameter OA′. We observe that the image of k is tangent to the line through O parallel to k.
Repeating the argument in reverse shows the the circle with diameter OA′ is mapped into the line k, and hence the line k and the circle with diameter OA′ are interchanged under the inversion.
The preceeding argument shows that a circle containing O is mapped to a line disjoint from O. The final case we consider is when we begin with a circle c disjoint from O. Let C be the centre of c, let A, B be the intersection points of the line OC with c, and let A′, B′ be their images under the inversion. We will prove that c′ is the circle with diameter A′B′.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 125 To see this, fix any point X ̸= A, B on c, and let X′ be its image.
A B X X′ B′ A′ O Now we apply Corollary 12.3 two times: ∠AXO = ∠OA′X′ and ∠BXO = ∠OB′X′ and using the fact that ∠BXA = ∠BXO −∠AXO and ∠OB′X′ = ∠A′X′B′ +∠OA′X′ (by adding the angles in triangle B′A′X′), we conclude ∠BXA = ∠OB′X′ −∠OA′X′ = ∠A′X′B′, and since ∠BXA = 90◦, we have ∠A′X′B′ = 90◦, and hence X′ lies on the circle with diameter A′B′. Therefore the entire circle c is mapped to the circle with diameter A′B′. By symmetry, we know that the circle with diameter A′B′ is mapped to c, and hence the two circles are interchanged by the inversion. This completes the proof of the theorem.
□ Exercise 12.5. Let o and c be circles with centres O and C, respectively, and suppose that O ̸∈c. Let C′ and c′ denote the images of C and c under the inversion through o (the proof of Theorem 12.4 establishes that c′ is indeed a circle). Show that the center of c′ is C′ if and only if c and o are concentric.
Problem 12.6. Let o1, o2, o3, o4 be circles so that o1, o2 are externally tangent at A, o2, o3 are externally tangent at B, o3, o4 are externally tangent at C and o4, o1 are externally tangent at D. Prove that A, B, C, D lie on a common circle.
A B C D o1 o2 o3 o4 126 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Solution. Without loss of generality we may assume that A, B, C, D are distinct, since we can always find a circle containing the vertices of a triangle. The idea for the solution is to apply an inversion in any circle centered at A. Since o1 and o2 are tangent at A, they get mapped to lines k1 = o′ 1 and k2 = o′ 2. Further, since o1 ∩o2 = A, k1 ∩k2 = A′ = ∞, so k1 and k2 are parallel. Since o3 is tangent to o2 at B, o2 cannot contain A, and similarly o4 also cannot contain A. Therefore, following the proof of Theorem 12.4, we conclude that o3 and o4 are mapped to circles o′ 3 and o′ 4.
k1 k2 o′ 3 o′ 4 C′ B′ D′ The idea now is to show that B′, C′, D′ lie on a common line, for then B′, C′, D′, ∞lie on a line ℓ, and when we reapply the inversion in the circle centred at A, we will deduce (by Theorem 12.4) that B, C, D, A lie on a common circle (or line, but then we will show that they cannot actually lie on a line).
To show that B′, C′, D′ lie on a common line, we consider the homothety with negative scale with centre C′ mapping o′ 4 to o′ 3. Since homotheties maps tangent lines to tangent lines (see Exercise 11.17), we conclude that the homothety maps k1 to k2 and D′ to B′. Therefore B′, C′, D′ lie on a common line, and, as described in the preceeding paragraph, this shows A, B, C, D lie on a common circle or line.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 127 However, if A, B, C, D lie on a common line ℓ, then the points A, B, C, D can be ordered on ℓ; without loss, suppose that A < B < C < D on ℓ.
ℓ o1 o2 o3 o4 A D B C Then since the diameter to the circle o2 is AB and the diameter of o1 is AD, and AB is entirely contained in AD, we conclude that o1 contains o2 in its interior, which means they are not externally tangent.
Therefore, A, B, C, D must lie on a common circle.
Problem 12.7. Let ABCD be a convex quadrilateral, and assume that the inscribed circles of ABC and ACD are tangent at a common point O ∈AC. Prove that the other tangency points of o1 and o2 lie on a common circle.
o2 o1 A B C D Z W X Y O Solution. Following the figure above, let X, Y and Z, W denote the other tangency points of o1 and o2 with ABC and ACD, respectively.
The idea for the solution is to apply the inversion through any circle centered at O, and then show that the images X′, Y ′, W ′, Z′ (of X, Y, W, Z) form a rectangle, and hence lie on a common circle. Then when we apply the inversion again, we will deduce that X, Y, Z, W lie on a circle or a line. It is impossible for a single line to intersect every edge on a non-degenerate quadrilateral, and hence X, Y, Z, W cannot lie on a common line, and we will be able to conclude that X, Y, W, Z lie on a common circle.
128 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Therefore, our task now is to show that X′Y ′W ′Z′ is a rectangle.
A′ B′ C′ D′ X′ Y ′ Z′ W ′ O The first observation we make is that, since o1, o2 and AC are tangent at O, o′ 1, o′ 2 and (AC)′ = AC become parallel lines (informally, thinking of lines as “circles” containing the point ∞, o′ 1, o′ 2 and (AC)′ are “circles” tangent at O′ = ∞, hence they are parallel lines).
Further, X′, Y ′ lie on o′ 1 and Z′, W ′ lie on o′ 2, and A′, C′ lie on AC.
A′ B′ C′ D′ X′ Y ′ Z′ W ′ O o′ 1 o′ 2 The second observation is that the perpendicular to o′ 1 raised from X′ bisects the segment A′O. To establish this claim, we observe that the line AB becomes a circle (AB)′ when we apply the inversion, and that (AB)′ is tangent to the line o′ 1 at X′ and contains A′ and O.
If S denotes the center of (AB)′, then the perpendicular bisector of the chord A′O intersects S (why?). Similarly the perpendicular to o′ 1 raised from X′ also intersects S (since (AB)′ is tangent to o′ 1 at X′).
Therefore the perpendicular to o′ 1 raised from X′ must be the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 129 perpendicular bisector of A′O (since they are both perpendicular to o′ 1 and intersect at S).
A′ B′ C′ D′ X′ Y ′ Z′ W ′ O o′ 1 o′ 2 S A similar argument shows that the perpendicular to o′ 2 raised from Z′ bisects A′O, and hence Z′X′ is perpendicular to o′ 1 and o′ 2. Similarly, Y ′W ′ is perpendicular to o′ 1 and o′ 2.
Therefore Z′X′ ∥Y ′W ′, and so X′Y ′W ′Z′ forms a rectangle, since it is a parallelogram with perpendicular edges.
Following the argument given in the second paragraph of this solution, we conclude that X, Y, W, Z lie on a circle.
12.1. More properties of the inversion. So far, all of our application of the inversion in a circle relied on the simple fact that it mapped circles/lines to circles/lines, and preserved “incidence,” i.e. if we could show that the images of some points lie on a circle/line, then we could deduce that the original points also lie on a circle/line. However, the inversion in a circle also preserves the “angles” between circles/lines! In order to make this precise, we need to define what the angle between two circles/lines is.
Definition 12.8. Let c1 and c2 be two circles intersecting at a point X. The angle between c1 and c2 at X is the angle between their respective tangent lines at X.
c1 c2 X Note that the ambiguity in the choice of angles between tangent lines forces us to consider α and 180◦−α as equivalent. If we wish to differentiate the angles α and 180◦−α, we could endow our circles with an “orientation,” but this is not necessary for what follows.
130 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Exercise 12.9. Give a similar definition for the angle between a circle c and a line ℓinter-secting at a point Y .
c ℓ Exercise 12.10. Suppose a circle/line a intersects another circle c in two points X, Y . Prove that the angle between a and c at X equals the angle between a and c at Y .
Exercise 12.11. Two lines ℓ1 and ℓ2 always intersect at ∞. Define the angle between ℓ1 and ℓ2 at ∞in such a way so that (i) the angle between parallel lines is 0 and (ii) Exercise 12.10 generalizes to the case when a line ℓ1 intersects a line ℓ2 in two points.
Theorem 12.12. The inversion in any circle preserves the angles between circles/lines, i.e.
if a, b are intersecting circles/lines, and a′, b′ denote their images under some inversion, then a′ and b′ intersect and the angle between a′ and b′ equals the angle between a and b.
Proof. In the following proof we fix an inversion, and we denote the image of any figure F by the primed symbol F ′.
Our first observation is that it suffices to check that the angle between ℓand k equals the angle between their images ℓ′ and k′, whenever ℓ, k are lines. To see this, we claim that if c is a circle and t is its tangent line at X, then c′ and t′ are tangent at X′; this is true because c′ and t′ being tangent at X′ is equivalent to them intersecting only at X′, which follows from c, t being tangent at X. Therefore, if c1, c2 intersect at X with respective tangent lines t1 and t2 at X, then the angle between t′ 1 and t′ 2 at X′ equals the angle between c′ 1 and c′ 2 at X′. Thus, without loss of generality, we will only prove that the angle between ℓ′ and k′ equals the angle between ℓand k whenever ℓand k are lines.
Let O denote the centre of the inversion circle, and suppose ℓ, k are two lines intersecting at a point P, with an angle of α between them. First suppose that neither ℓ, k contain O.
Then ℓ′ and k′ are circles intersecting at P ′ and O; we will calculate the angle between them at O. To do this, let ℓ, k denote the tangent lines to ℓ′, k′ at O. The discussion following the definition of the inversion established that ℓ∥ℓand k ∥k (Definition 12.1). Hence the angle TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 131 between k and ℓequals the angle between k and ℓ.
ℓ k P O inversion O ℓ′ k′ ℓ k P O The other cases to consider are when k or ℓcontains O. Both of these cases are easier than the one we proved, and so we leave them as exercises for the reader to complete. This completes the proof of the theorem.
□ Remark. One can define the angle between arbitrary differentiable curves intersecting at a point X, and one can show that the inversion is a differentiable map which preserves the angle between any two differentiable curves.
Problem 12.13. Consider four circles o1, o2, o3, o4, each three of which intersect in a com-mon point A ̸= B ̸= C ̸= D, as in the figure below. Prove that the angle between o1 and o2 equals the angle between o3 and o4.
o1 o2 o3 o4 A B C D Solution. The idea is to consider an inversion through any circle centered at A. Since the circles o1, o2 and o3 contain A, their images o′ 1, o′ 2 and o′ 3 are straight lines. Then, the points B′, C′, D′ form a triangle whose edges lie on the lines o′ 1, o′ 2 and o′ 3, and, in particular, the angle at vertex B′ equals the angle between o′ 1 and o′ 2. Further, since o4 is a circle which does not contain A, o′ 4 is still a circle, and since it contains B′, C′ and D′, o′ 4 is the circumscribed circle of B′C′D′.
132 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
We claim that ∠D′B′C′ equals the angle between the line C′D′ and the tangent line to o′ 4 at C′.
B′ C′ D′ A o′ 1 o′ 3 o′ 2 o′ 4 This is actually proved in a previous problem (Problem 3.12), and so we won’t repeat its proof. We conclude that the angle between o′ 3 and o′ 4 equals ∠D′B′C′ (which is the angle between o′ 1 and o′ 2).
The fact that inversions preserve angles implies that ∠D′B′C′ = angle between o1 and o2, and angle between o′ 3 and o′ 4 = angle between o3 and o4, and so we deduce that the angle between o3 and o4 equals the angle between o1 and o2, as desired.
Problem 12.14. Let o1, o2 and o3 be circles intersecting in a common point, and suppose they are pairwise non-tangent. Find a circle or line tangent to o1, o2 and o3.
o1 o2 o3 O A B C Solution. Let O be the single point of intersection. Since o1, o2, o3 are pairwise non-tangent, each pair intersects twice; label the other intersection points A ̸= B ̸= C ̸= O as in the figure above.
TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 133 As is customary with the problems in this section, the key step in the solution is to take the inversion in some circle - in this case, we will consider the inversion in any circle centered at O. If A′, B′, C′, o′ 1, o′ 2, o′ 3 denote the images of A, B, C, o1, o2, o3 under this inversion, then o′ 1, o′ 2, o′ 3 are lines and A′B′C′ is a triangle whose edges lie on o′ 1, o′ 2, o′ 3. Since the inscribed circle c′ of A′B′C′ is tangent to o′ 1, o′ 2, o′ 3, its image c′′ is a circle or line tangent to o1, o2, o3.
If c′′ is a circle, then we are done. If c′′ is a line, then O lies on c′, and hence O is contained in the interior of A′B′C′. Therefore O cannot lie on any escribed circle of A′B′C′, and so, if we take d′ an escribed circle of A′B′C′, then its image d′′ is a circle which is tangent to o1, o2, o3; either way we have completed the solution.
B′ C′ A′ o′ 1 o′ 2 o′ 3 c′ d′ Remark. (1) In the figure given in the preceeding problem statement, we chose the inscribed circles of A′B′C′. If we chose an escribed circle, the figure would look like this: o1 o2 o3 134 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
(2) Following the notation of the preceeding solution, if c′ is the inscribed circle of A′B′C′, it is possible for c′′ to be line, as the following figure shows: Problem 12.15 (Apollonius). Given any three circles o1, o2, o3 lying outside each other, find a common tangent circle or line. (Note: in the accompanying figure we have drawn multiple solutions to the problem.) o1 o3 o2 Solution. Let o1 be the smallest circle, and let r be its radius. Shrink o2 and o3 by radius r and turn o1 to a point - let o1, o2, o3 denote these shrunken circles. Note that o2 and o3 TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 135 may also be single points.
o1 o3 o2 We consider three cases: (i) If we can find a circle c internally tangent to o2 and o3 and containing o1,then expanding c by radius r produces a circle c tangent to o1, o2 and o3. (ii) If, instead we find a circle d so that d contains o1 but externally tangent to o2 and o3, then shrinking d by radius r produces a circle d tangent to o1, o2, o3 (why is the radius of d at least r?).
o1 o3 o2 c c d d Finally, (iii) if instead we find a line ℓcontaining o1 and tangent to both o2, o3 from the same side, then shifting ℓin a perpendicular direction by distance r produces a line tangent to o1, o2, o3.
136 TAUGHT BY PIOTR PRZYTYCKI. NOTES BY DYLAN CANT.
Therefore, we have reduced the problem to the case when one of the circles, o1, is actually just a single point O1. Now the key is to consider an inversion in some circle centered at O1 -in the figure below, an arc of the inversion circle is shown as a dashed curve. Let the images of o2 and o3 be circles o′ 2 and o′ 3. Then, finding a tangent circle/line to o2, o3 containing O1 is equivalent to finding a tangent line to o′ 2 and o′ 3. Let ℓbe a tangent line to o′ 2, o′ 3, tangent to them from the same side (as in the figure below). We consider three cases: (1) If ℓseparates O1 from o′ 2 and o′ 3, then the circle ℓ′ is a circle through O1 which contains both o2, o3 in its interior. (2) If ℓcontains O1, o′ 2 and o′ 3 all on one side, then ℓ′ is a circle through O1 which doesn’t contain o2 or o3. (3) If ℓcontains O1, then ℓ′ is a line tangent to o2 and o3 passing through O1. Since cases (1), (2) and (3) exhaust all possibilities, and in each case we have constructed a solution to the problem (corresponding to the cases (i), (ii), (iii) given above), we have completed the solution of the problem. Note that in the figure below, we have drawn two choices of ℓ, one which satisfies case (1) and one which satisfies case (2).
O1 o3 o2 o′ 2 o′ 3 ℓ1 ℓ2 Exercise 12.16. In the statement of Problem 12.15, we have drawn a figure presenting five different solutions of the problem. However, in our solution, we present only two solutions (corresponding to the two common tangent lines to o′ 2 and o′ 3 which are tangent from the TOPICS IN GEOMETRY: THE GEOMETRY OF THE EUCLIDEAN PLANE 137 same side). Which circles in the figure are not described by our solution? Explain how you would modify the solution so that it could describe all of the circles presented in the Problem statement’s figure. |
190186 | https://en.wikipedia.org/wiki/Union_(set_theory) | Jump to content
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Contents
(Top)
1 Union of two sets
1.1 Finite unions
2 Notation
2.1 Notation encoding
3 Arbitrary union
3.1 Formal derivation
4 Algebraic properties
5 History and etymology
6 See also
7 Notes
8 External links
Union (set theory)
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From Wikipedia, the free encyclopedia
Set of elements in any of some sets
In set theory, the union (denoted by ∪) of a collection of sets is the set of all elements in the collection. It is one of the fundamental operations through which sets can be combined and related to each other. A nullary union refers to a union of zero () sets and it is by definition equal to the empty set.
For explanation of the symbols used in this article, refer to the table of mathematical symbols.
Union of two sets
[edit]
The union of two sets A and B is the set of elements which are in A, in B, or in both A and B. In set-builder notation,
: .
For example, if A = {1, 3, 5, 7} and B = {1, 2, 4, 6, 7} then A ∪ B = {1, 2, 3, 4, 5, 6, 7}. A more elaborate example (involving two infinite sets) is:
: A = {x is an even integer greater than 1}
: B = {x is an odd integer greater than 1}
As another example, the number 9 is not contained in the union of the set of prime numbers {2, 3, 5, 7, 11, ...} and the set of even numbers {2, 4, 6, 8, 10, ...}, because 9 is neither prime nor even.
Sets cannot have duplicate elements, so the union of the sets {1, 2, 3} and {2, 3, 4} is {1, 2, 3, 4}.
Finite unions
[edit]
One can take the union of several sets simultaneously. For example, the union of three sets A, B, and C contains all elements of A, all elements of B, and all elements of C, and nothing else. Thus, x is an element of A ∪ B ∪ C if and only if x is in at least one of A, B, and C.
A finite union is the union of a finite number of sets; the phrase does not imply that the union set is a finite set.
Notation
[edit]
The notation for the general concept can vary considerably. For a finite union of sets one often writes or . Various common notations for arbitrary unions include , , and . The last of these notations refers to the union of the collection , where I is an index set and is a set for every . In the case that the index set I is the set of natural numbers, one uses the notation , which is analogous to that of the infinite sums in series.
When the symbol "∪" is placed before other symbols (instead of between them), it is usually rendered as a larger size.
Notation encoding
[edit]
In Unicode, union is represented by the character U+222A ∪ UNION. In TeX, is rendered from \cup and is rendered from \bigcup.
Arbitrary union
[edit]
The most general notion is the union of an arbitrary collection of sets, sometimes called an infinitary union. If M is a set or class whose elements are sets, then x is an element of the union of M if and only if there is at least one element A of M such that x is an element of A. In symbols:
This idea subsumes the preceding sections—for example, A ∪ B ∪ C is the union of the collection {A, B, C}. Also, if M is the empty collection, then the union of M is the empty set.
Formal derivation
[edit]
In Zermelo–Fraenkel set theory (ZFC) and other set theories, the ability to take the arbitrary union of any sets is granted by the axiom of union, which states that, given any set of sets , there exists a set , whose elements are exactly those of the elements of . Sometimes this axiom is less specific, where there exists a which contains the elements of the elements of , but may be larger. For example if then it may be that since contains 1 and 2. This can be fixed by using the axiom of specification to get the subset of whose elements are exactly those of the elements of . Then one can use the axiom of extensionality to show that this set is unique. For readability, define the binary predicate meaning " is the union of " or "" as:
Then, one can prove the statement "for all , there is a unique , such that is the union of ":
Then, one can use an extension by definition to add the union operator to the language of ZFC as:
or equivalently:
After the union operator has been defined, the binary union can be defined by showing there exists a unique set using the axiom of pairing, and defining . Then, finite unions can be defined inductively as:
Algebraic properties
[edit]
See also: List of set identities and relations and Algebra of sets
Binary union is an associative operation; that is, for any sets , Thus, the parentheses may be omitted without ambiguity: either of the above can be written as . Also, union is commutative, so the sets can be written in any order. The empty set is an identity element for the operation of union. That is, , for any set . Also, the union operation is idempotent: . All these properties follow from analogous facts about logical disjunction.
Intersection distributes over union and union distributes over intersection The power set of a set , together with the operations given by union, intersection, and complementation, is a Boolean algebra. In this Boolean algebra, union can be expressed in terms of intersection and complementation by the formula where the superscript denotes the complement in the universal set . Alternatively, intersection can be expressed in terms of union and complementation in a similar way: . These two expressions together are called De Morgan's laws.
History and etymology
[edit]
Further information: History of set theory
The english word union comes from the term in middle French meaning "coming together", which comes from the post-classical Latin unionem, "oneness". The original term for union in set theory was Vereinigung (in german), which was introduced in 1895 by Georg Cantor. The english use of union of two sets in mathematics began to be used by at least 1912, used by James Pierpont. The symbol used for union in mathematics was introduced by Giuseppe Peano in his Arithmetices principia in 1889, along with the notations for intersection , set membership , and subsets .
See also
[edit]
Mathematics portal
Algebra of sets – Identities and relationships involving sets
Alternation (formal language theory) − the union of sets of strings
Axiom of union – Concept in axiomatic set theory
Disjoint union – In mathematics, operation on sets
Inclusion–exclusion principle – Counting technique in combinatorics
Intersection (set theory) – Set of elements common to all of some sets
Iterated binary operation – Repeated application of an operation to a sequence
List of set identities and relations – Equalities for combinations of sets
Naive set theory – Informal set theories
Symmetric difference – Elements in exactly one of two sets
Notes
[edit]
^ Weisstein, Eric W. "Union". Wolfram Mathworld. Archived from the original on 2009-02-07. Retrieved 2009-07-14.
^ a b "Set Operations | Union | Intersection | Complement | Difference | Mutually Exclusive | Partitions | De Morgan's Law | Distributive Law | Cartesian Product". Probability Course. Retrieved 2020-09-05.
^ a b Vereshchagin, Nikolai Konstantinovich; Shen, Alexander (2002-01-01). Basic Set Theory. American Mathematical Soc. ISBN 9780821827314.
^ deHaan, Lex; Koppelaars, Toon (2007-10-25). Applied Mathematics for Database Professionals. Apress. ISBN 9781430203483.
^ Dasgupta, Abhijit (2013-12-11). Set Theory: With an Introduction to Real Point Sets. Springer Science & Business Media. ISBN 9781461488545.
^ "Finite Union of Finite Sets is Finite". ProofWiki. Archived from the original on 11 September 2014. Retrieved 29 April 2018.
^ a b Smith, Douglas; Eggen, Maurice; Andre, Richard St (2014-08-01). A Transition to Advanced Mathematics. Cengage Learning. ISBN 9781285463261.
^ "The Unicode Standard, Version 15.0 – Mathematical Operators – Range: 2200–22FF" (PDF). Unicode. p. 3.
^ Halmos, P. R. (2013-11-27). Naive Set Theory. Springer Science & Business Media. ISBN 9781475716450.
^ "MathCS.org - Real Analysis: Theorem 1.1.4: De Morgan's Laws". mathcs.org. Retrieved 2024-10-22.
^ Doerr, Al; Levasseur, Ken. ADS Laws of Set Theory.
^ "The algebra of sets - Wikipedia, the free encyclopedia". www.umsl.edu. Retrieved 2024-10-22.
^ "Etymology of "union" by etymonline". etymonline. Retrieved 2025-04-10.
^ Cantor, Georg (1895-11-01). "Beiträge zur Begründung der transfiniten Mengenlehre". Mathematische Annalen (in German). 46 (4): 481–512. doi:10.1007/BF02124929. ISSN 1432-1807.
^ Pierpont, James (1912). Lectures On The Theory Of Functions Of Real Variables Vol II. Osmania University, Digital Library Of India. Ginn And Company.
^ Oxford English Dictionary, “union (n.2), sense III.17,” March 2025,
^ "Earliest Uses of Symbols of Set Theory and Logic". Maths History. Retrieved 2025-04-10.
External links
[edit]
"Union of sets", Encyclopedia of Mathematics, EMS Press, 2001
Infinite Union and Intersection at ProvenMath De Morgan's laws formally proven from the axioms of set theory.
| v t e Set theory |
| Overview | Set (mathematics) |
| Axioms | Adjunction Choice + countable + dependent + global Constructibility (V=L) Determinacy + projective Extensionality Infinity Limitation of size Pairing Power set Regularity Union Martin's axiom Axiom schema + replacement + specification |
| Operations | Cartesian product Complement (i.e. set difference) De Morgan's laws Disjoint union Identities Intersection Power set Symmetric difference Union |
| Concepts Methods | Almost Cardinality Cardinal number (large) Class Constructible universe Continuum hypothesis Diagonal argument Element + ordered pair + tuple Family Forcing One-to-one correspondence Ordinal number Set-builder notation Transfinite induction Venn diagram |
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| Paradoxes Problems | Russell's paradox Suslin's problem Burali-Forti paradox |
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| General | Axiom + list Cardinality First-order logic Formal proof Formal semantics Foundations of mathematics Information theory Lemma Logical consequence Model Theorem Theory Type theory |
| Theorems (list) and paradoxes | Gödel's completeness and incompleteness theorems Tarski's undefinability Banach–Tarski paradox Cantor's theorem, paradox and diagonal argument Compactness Halting problem Lindström's Löwenheim–Skolem Russell's paradox |
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| Set theory | | | | Set + hereditary Class (Ur-)Element Ordinal number Extensionality Forcing Relation + equivalence + partition Set operations: + intersection + union + complement + Cartesian product + power set + identities | | Types of sets | Countable Uncountable Empty Inhabited Singleton Finite Infinite Transitive Ultrafilter Recursive Fuzzy Universal Universe + constructible + Grothendieck + Von Neumann | | Maps and cardinality | Function/Map + domain + codomain + image In/Sur/Bi-jection Schröder–Bernstein theorem Isomorphism Gödel numbering Enumeration Large cardinal + inaccessible Aleph number Operation + binary | | Set theories | Zermelo–Fraenkel + axiom of choice + continuum hypothesis General Kripke–Platek Morse–Kelley Naive New Foundations Tarski–Grothendieck Von Neumann–Bernays–Gödel Ackermann Constructive | |
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Deciding between the simple present and present continuous tenses will require you to look at the time period of the action. I use the present simple form when the action is regular and permanent, while I use the present continuous form for a temporary, ongoing action.
Keep reading as I show you the differences between the two English verb tenses in terms of uses and formulas. I also created an online exercise that will allow you to master the simple present and present continuous tenses.
What is the Present Simple Tense?
In English grammar, the present simple or simple present tense is a verb tense used when the action is happening at present or when the situation happens regularly. I’ve also seen this verb tense used for permanent situations.
Most regular verbs use the root form for the simple present tense unless the subject is in the third-person singular. In this case, we add -s or -es in the end. For example:
The present tense verb in the sentence is practices. It ends in -s because the subject is in the third-person singular.
The formula for creating a sentence in this verb tense is:
Present simple questions require you to change the order of the subject and auxiliary verb for the formula. Here is the correct question form:
The negative form is:
Here are more sentence examples.
What is the Present Continuous Tense?
The present continuous or progressive tense is one of the three progressive tenses that show ongoing action at the current period of time. I’d also use it when referring to a temporary action or an incomplete action in my writing.
The present progressive form is the auxiliary verb is, are, or am plus the present participle form or -ing form of the verb. Remember to use only stative verbs for the main verb. For example:
Affirmative sentences in the present continuous tense follow this formula:
Meanwhile, negative sentences have the following structure:
Here’s an example:
For present continuous questions, the correct formula is:
Here are more sentence examples:
What’s the Difference Between Present Simple and Present Continuous Tense?
Good question! There are specific rules to follow when using the simple tense and continuous tense. Let’s look at the differences between the functions and guidelines for present simple and present continuous tenses.
When to Use the Simple Present Tense
Use the simple present tense when presenting a narrative form for stories in the current period. For example:
You can also use it for permanent actions or regular habits. For example:
The most common time expressions to use with the simple present tense are always, generally, often, in the morning, on Fridays, every day, etc. Remember that the time expression is in a single form, which is day instead of days.
When to Use the Present Continuous Tense
Use the present continuous tense for temporary actions currently happening. For example:
You can make the verb form progressive when referring to annoying habits. Some time expressions you can use include constantly and always. For example:
Some verbs can be dynamic and stative. That means you can use some verbs in the stative form in the present continuous tense. But you can only use it if it takes on the dynamic meaning of the verb. For example:
Think is one of the mental process verbs aside from remember, wonder, and hope.
Summary of Present Simple vs. Present Continuous
This grammar guide thoroughly explained the differences between the simple present and present continuous tense. Remember that the simple present tense is for regular actions or permanent actions, while the present continuous tense is for a temporary situation.
Understanding the functions and formulas of these two English tenses will help you become a better speaker and writer. Keep practicing, and you’ll get there!
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190189 | https://people.math.harvard.edu/~barkley/qmmm21/VectorSpaces.pdf | QMMM Warm-up 1: Complex vector spaces Instructor: Grant T. Barkley 1 Linear algebra For now and forever, C denotes the field of complex numbers. In linear algebra over the complex numbers, a complex number λ ∈C is also called a scalar.
1.1 Vector spaces Definition 1.1.
A complex vector space (or just vector space) is a set V with two operations: An addition operation, which takes in two elements of V and outputs another element of V . If the inputs are v and w then the output of the addition operation is written as v + w and is called the sum of v and w.
A scalar multiplication operation, which takes in an element of C (the scalar) and an element of V (the vector) and outputs another element of V (the scaled vector). If λ is the scalar and v is the vector, then the scaled vector is written as λ · v, or even just λv.
We also pick a distinguished element of V , called the zero vector, which is denoted by 0.
These operations are required to satisfy certain properties. If u, v, w ∈V are any vectors and λ, λ1, λ2 ∈C are any scalars, then we require the following axioms: Properties of addition: u + (v + w) = (u + v) + w (Associativity) v + w = w + v (Commutativity) v + 0 = v (Identity) Properties of scalar multiplication: λ1 · (λ2 · v) = (λ1λ2) · v (Compatibility) 1 · v = v (Identity) 0 · v = 0 (Zero) Relating the two operations: λ(v + w) = λv + λw (Distributivity over vectors) (λ1 + λ2)v = λ1v + λ2v.
(Distributivity over scalars) 1 The axioms in Definition 1.1 are designed to encapsulate the core properties of the “standard vector spaces” Cn for n ≥0. Recall that Cn is the set of column vectors with n rows, with entries in C. For instance, a typical element of C3 might look like the vector v = 3.141 0 22 −7i .
Problem 1.
Check that Cn, with its usual addition and scalar multiplication operations, is a complex vector space.
(In other words, check that these operations satisfy the properties listed in Definition 1.1.) Problem 2.
Let V be a vector space. Fix any vector v ∈V . Prove that there is a unique vector v′ ∈V satisfying v + v′ = 0.
The unique vector v′ is called the negation or inverse of v, and is denoted −v.
The negation operation, which takes a vector v and outputs −v, is usually taken as an additional part of the structure of a vector space. If we did this, then we would add the following axiom to the list in Definition 1.1.
v + (−v) = 0.
(Inverses) Problem 3.
o Assume that V is a set with addition and scalar multiplication operations satisfying every axiom in Definition 1.1 except possibly (Zero). Prove that if V has a negation operation satisfying (Inverses), then V satisfies the (Zero) axiom.
Bonus: Can we prove the (Zero) axiom without using the negation operation?
Problem 4.
Let X be any set.
Define V to be the collection of all functions from X to C.
Define addition and scalar multiplication operations on V . Show that V is a vector space with these operations.
What if we look at the collection of all functions from X to W, where W is an arbitrary complex vector space?
1.2 Subspaces and spans 2 Definition 1.2.
Let W be a subset of a vector space V containing 0. Then W is a subspace of V if the sum of any two vectors in W is also in W and if any scalar multiple of a vector in W is also in W.
Problem 5.
Let X be a set and V be the vector space of functions from X to C defined in Problem 4.
Fix an element x ∈X and define W = {f ∈V | f(x) = 0}.
Check that W is a subspace of V .
Problem 6.
Let V be the vector space of functions from N (the positive integers) to C. Equivalently, V is the set of all sequences of complex numbers. Define W = {f ∈V | ∃N ∈N, ∀n > N, f(n) = 0}, the set of all sequences which are eventually 0. Check that W is a subspace of V .
Definition 1.3.
Let X be a subset of a vector space V . The (linear) span of X, denoted span X, is the set of all finite linear combinations of elements of X. In symbols, span X = {λ1v1 + ... + λrvr | r ∈N, λ1, ..., λr ∈C, v1, ..., vr ∈X}.
If X is the empty set ∅, then we define span ∅= {0}.
If W is a subspace of V , then we say that X spans W if span X = W.
Recall that a closure operature c on a set V is a map taking subsets of V to subsets of V , and satisfying the following three properties, for any subsets X, Y of V : (i) X ⊆c(X).
(ii) If X ⊆Y , then c(X) ⊆c(Y ).
(iii) c(c(X)) = c(X).
3 Problem 7.
Verify the following properties of the linear span.
(a) For any X ⊆V , the set span X is a subspace of V .
(b) The map taking a subset X of V to the set span X is a closure operator on V .
(c) If W is a subspace of V which contains a set X, then W also contains the subspace span X.
Problem 8.
♦ Let X be a nonempty subset of a vector space V . Assume v is a vector in the span of X, but v is not in the span of any proper subset of X. Show that X is finite, and if w is any element of X, then w is in the span of (X \ {w}) ∪{v}.
The span of a singleton {v} is often denoted Cv. If V1, V2 are two subspaces of the same vector space, then we write V1 + V2 to indicate the span of V1 ∪V2. Similarly, if {Vi}i∈I is any collection of subspaces of V (indexed by a set I), then P i∈I Vi denotes the span of S i∈I Vi.
Definition 1.4.
Let {Vi}i∈I be a collection of vector subspaces of a vector space V . We say that V is the (internal) direct sum of {Vi}i∈I, written V = M i∈I Vi if for all i ∈I, Vi ∩ X j∈I{i} Vj = {0}, and if P i∈I Vi = V .
We also say that V = L i∈I Vi is a direct sum decomposition of V if each Vi is a subspace of V and if V is the direct sum of {Vi}i∈I.
Direct sum notation follows the same conventions used for unions or intersections. For instance, if our index set I = {1, 2, . . . , r}, then we write V = V1 ⊕V2 ⊕... ⊕Vr = r M k=1 Vk, and if I = N, then we write V = V1 ⊕V2 ⊕· · · = ∞ M k=1 Vk.
4 Problem 9.
♦ Let V be a vector space and let V = V1 ⊕... ⊕Vr be a direct sum decomposition of V . Prove that for any vector v ∈V , there is a unique list of elements v1, ..., vr satisfying v = v1 + v2 + ... + vr, and such that vk ∈Vk for all k.
Generalize this statement to an arbitrary direct sum decomposition V = L i∈I Vi.
1.3 Linear maps Definition 1.5.
Let V and W be complex vector spaces and let L be a function from V to W. We say that L is a linear map if it satisfies the following, for all λ ∈C and v, v1, v2 ∈V : L(v1 + v2) = L(v1) + L(v2), and L(λv) = λL(v).
Problem 10.
Check the following properties of linear maps: (a) A linear map L : V →W takes 0 ∈V to 0 ∈W.
(b) The composition of two linear maps is a linear map.
(c) If a linear map L : V →W is a bijection, then its inverse function L −1 : W →V is also a linear map.
Describe the linear maps from C to C.
A linear map L : V →W is called invertible or an isomorphism if it is bijective. In this case we say that L is an isomorphism between V and W. If there is an isomorphism between V and W, then we say V and W are isomorphic.
Definition 1.6.
Let L : V →W be a linear map. The kernel of L is ker L .
= {v ∈V | L(v) = 0}.
The image of L is its image as a function. Explicitly, im L .
= {w ∈W | ∃v ∈V, L(v) = w}.
5 Problem 11.
Show the following properties of the kernel and image of a linear map L : V →W.
(a) ker L is a subspace of V and im L is a subspace of W.
(b) L is injective if and only if ker L = {0}.
(c) L is invertible if and only if ker L = {0} and im L = W.
(d) If L′ : W →U is a linear map to a third vector space U, then ker(L′ ◦L) ⊇ker L and im(L′ ◦L) ⊆im(L′).
Problem 12.
Let V = L i∈I Vi be a direct sum decomposition of a vector space. Fix an index j ∈I. As a result of Problem 9 , any vector v can be uniquely decomposed into the sum of an element vj ∈Vj and an element v′ ∈P i∈I{j} Vi.
Let Πj be the function from V to V which assigns to a vector v the vector vj. Show that Πj is a linear map and is idempotent (i.e. it satisfies Πj ◦Πj = Πj).
Any linear map Π : V →V satisfying Π ◦Π = Π is called a (linear) projection operator on V . The map Πj which is constructed in Problem 12 is called the (linear) projection onto Vj, relative to the decomposition L i∈I Vi. As the next problem shows, any projection operator is of this form.
Problem 13.
♦ Let Π be a projection operator on a vector space V . Show that there is a unique pair of subspaces V1, V2 ⊆V such that V = V1 ⊕V2 and so that Π is the projection onto V1, relative to V1 ⊕V2.
1.4 Quotient vector spaces If v is a vector in a vector space V , and K is a subset of V , then we set v + K .
= {v + z | z ∈K}.
Similarly, if X is a subset of V , then we set X + K .
= {v + z | v ∈X, z ∈K}.
Lastly, if λ is a scalar and X is a subset of V , then set λX .
= {λv | v ∈X}.
6 Problem 14.
Check the following properties of the operations defined above (set-sum and set-scaling).
(a) If X1, X2, X3 are any subsets of V , then (X1 + X2) + X3 = X1 + (X2 + X3).
(b) If X1, X2 are any subsets of V , and λ is any scalar, then λ(X1 + X2) = λX1 + λX2.
(c) If K is a subspace of V , then K + K = K and λK = K for all nonzero scalars λ.
Definition 1.7.
Let V be a vector space and K a subspace of V . The quotient (vector) space V/K is the vector space defined as follows. The underlying set of V/K is {v + K | v ∈V }.
So V/K is a collection of subsets of V . The addition and scalar multiplication operations on V/K are then given by the set-sum and set-scaling operations defined above (with the exception of scaling by 0, which is defined to always output the set K).
The function q : V →V/K v 7→v + K is called the quotient map.
Problem 15.
Check that V/K is a vector space with these operations and that q is a linear map with kernel equal to K.
Any linear map L from V/K to a vector space W lifts to a map L ◦q from V to W, with (by Problem 11 c) ker(L ◦q) ⊇K.
The converse also holds, as the following problem demon-strates.
Problem 16.
Let L : V →W be a linear map such that K ⊆ker L. Then there is a unique (induced) linear map L : V/K →W such that L ◦q = L.
7 Problem 17.
♦ Show, using Problem 16 , that any linear map L : V →W induces a linear map L : V/ ker(L) →im(L) which is an isomorphism.
Problem 18.
Let V = V1 ⊕V2 be a direct sum decomposition. Prove, using projection operators and the previous problem, that V/V2 is isomorphic to V1.
1.5 Linear dependence Definition 1.8.
Let v1, v2, ..., vr be a list of vectors in a vector space V . We say that these vectors are linearly independent if the only choice of scalars λ1, ..., λr ∈C such that λ1v1 + λ2v2 + ... + λrvr = 0 is λ1 = λ2 = ... = 0. The vectors are linearly dependent if there is some nontrivial linear combination of the vectors which is 0.
If B is a subset of V , then we say B is linearly independent if every finite list of distinct vectors from B is linearly independent.
B is a (linear) basis for V if it is linearly independent and spans V .
The vector space Cn comes with a standard ordered basis.
Problem 19.
Prove that a subset B of a vector space V is linearly dependent if and only if there exists a vector v in B, and vectors v1, ..., vr ∈B all distinct from v, such that v ∈span{v1, ..., vr}.
Problem 20.
Show the following properties of a basis B in a vector space V .
(a) If X ⊋B, then X is linearly dependent.
(b) V is the direct sum of {Cv | v ∈B}.
(c) Each vector v ∈V can be written as a unique sum of scalar multiples of elements of B.
8 Problem 21.
Let X be a finite set and V be the vector space of functions from X to C. Find a basis for V .
Problem 22.
Let W be the vector space of all sequences of complex numbers which are eventually 0. Find an (infinite) basis for W.
Can you write down a basis for the space of all sequences of complex numbers?
Problem 23.
♦ Let V be a vector space with basis {vi}i∈I, and let W be any vector space. Pick vectors wi ∈W for each i ∈I. Prove that there is a unique linear map L : V →W satisfying L(vi) = wi for all i ∈I.
Problem 23 is the most important in this section, and (even if not stated explicitly) is the corner-stone of most introductions to linear algebra. Consider its result when applied to V = Cn and W = Cm. In this case the problem shows that choosing a linear map L : Cn →Cm is equivalent to choosing a list of n vectors in Cm. If those vectors are w1, ..., wn, then we could organize them in a matrix ML, whose ith column is the column vector wi. So the matrix would look like: ML = | | ...
| w1 w2 ...
wn | | ...
| .
This is called the matrix representation of L. We can multiply matrices and vectors as usual.
Problem 24.
Prove that for any linear map L : Cn →Cm and any v ∈Cn, we have the identity ML · v = L(v).
Hint: Use the previous problem.
Problem 25.
Let L1 : Cn →Cm and L2 : Cm →Cℓbe linear maps. (So the matrix representation ML1 tells us where L1 sends the basis vectors of Cn and the matrix representation ML2 tells us where L2 sends the basis vectors of Cm.) Prove that for any v ∈Cn, we have the identity (ML2 · ML1) · v = (L2 ◦L1)(v).
9 In other words, ML2 · ML1 = ML2◦L1.
Remark 1.9. The lesson of this section is that linear maps can be completely described by their action on basis vectors. So we can write statements like the following.
Let W be the vector space of all sequences of complex numbers which are eventually zero, and let w1, w2, ... be the natural basis. Then define a linear map L1 : W →W by L1(wk) = wk+1 and a second linear map L2 : W →W by L2(wk) = e 2πik 3 wk.
Note that the kernel of L1 is Cw1, while L2 is invertible. This is lorem ipsum dolor sit amet....
Matrices, and all the rules we learned to manipulate them, are just a way of organizing this data (using coordinates).
1.6 Aside: Transfinite recursion Let’s prove that every vector space has a basis. Modern proofs usually use Zorn’s lemma, and you can look those up for another perspective1. I prefer the method of transfinite induction, which is an extension of the usual induction principle to handle sets which are “bigger” than N.
Definition 1.10.
Let ≺be a total ordering on a set S. Recall that a total ordering is a transitive relation ≺such that any x, y ∈S satisfy exactly one of the following: x ≺y, x = y, x ≻y.
We say ≺is a well-ordering if any nonempty subset of S has a minimum element.
So for instance, the usual ordering on N is a well-ordering, while the usual orders on Z and R are not. A more exotic example of a well-ordering is the lexicographic order on N × N: (1, 1) ≺(1, 2) ≺(1, 3) ≺... ≺(2, 1) ≺(2, 2) ≺(2, 3) ≺... ≺(3, 1) ≺(3, 2) ≺(3, 3) ≺... ≺....
There is a classification of well-orderings by ordinal numbers — the order on N corresponds to the ordinal ω and the order on N × N corresponds to ω2. We won’t need this, though.
Proposition 1.11 (Transfinite recursion). Let ≺be a well-ordering on a set S. We want to define a function f on S. Suppose for each t ∈S we uniquely specify a definition of f(t), possibly using the values f(s) for s ≺t in our definition. Then the principle of transfinite recursion says that f(t) is well-defined for all t ∈S.
10 For example, recursively defined sequences such as F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥2 satisfy the requirements of transfinite recursion (since N is well-ordered), and therefore give a well-defined sequence (the Fibonacci numbers). Meanwhile, you can’t write down a recursive function on the integers without flipping the ordering on the negative integers. The recursion needs a “base case” — that’s what well-ordering provides.
A special case of transfinite recursion, often used in conjunction with recursive definitions, is the principle of transfinite induction.
Proposition 1.12 (Transfinite induction). Let ≺be a well-ordering on a set S. Let ϕ(t) be a proposition about elements t of S. Assume, for each t ∈S, that we can prove ϕ(t) if we assume that ϕ(s) holds for all s < t. Then the principle of transfinite induction says that ϕ(t) holds for all t ∈S.
With these principles, we can easily prove that all vector spaces have a basis. Let V be any complex vector space. Assume for now that V has a well-ordering ≺. Using transfinite recursion, we will define sets Bw indexed by the elements w of V . The goal is to construct a basis by adding vectors from V one at a time, at each step adding the next vector which is linearly independent from the ones we have added so far. Our sets are defined as follows: If w is in the span of S v≺w Bv, then we set Bw .
= ∅.
Otherwise, set Bw .
= {w}.
Now define B to be [ w∈V Bw.
Then the span of B is all of V , since any element w ∈V is in the span of S v⪯w Bv, which is a subset of B. We also claim that B is linearly independent. Assume for a contradiction that the vectors in B are dependent. Then (by Problem 19 and Problem 8 ) there are vectors v1 ≺v2 ≺... ≺vr in B such that vr ∈span{v1, ..., vr−1}.
But then, by definition, Bvr does not contain vr, and so neither does B. But this is a contra-diction, since vr ∈B. Hence, B is a basis for V .
Now the only step missing from our proof that every vector space has a basis is justifying our assumption that every vector space has a well-ordering. For this we can use the following proposition (which, like Zorn’s lemma, is equivalent to the axiom of choice).
11 Proposition 1.13 (Well-ordering theorem). Any set has a well-ordering.
The following problem is harder than the ones so far.
Problem 26.
o Let B be a basis for a vector space V , and let X be a linearly independent set of vectors in V . Assume B and X are both well-ordered and, for simplicity, assume B ∩X = ∅.
Using transfinite recursion on X, construct a set Y ⊆B and a bijection φ : X →Y , such that (B \ Y ) ∪X is a basis of V .
Using this, prove that any two bases of V have the same size. (That is, if B and X are two bases of V , then there is a bijection from B to X.) Using transfinite induction on a set which is well-ordered using the axiom of choice, as we do here, usually gives an equivalent proof to one using Zorn’s lemma. But transfinite induction is more generally applicable, as we may see later.
12 |
190190 | https://study.com/skill/learn/how-to-inscribe-a-square-in-a-circle-explanation.html | How to Inscribe a Square in a Circle | Geometry | Study.com
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How to Inscribe a Square in a Circle
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Kathryn Boddie, Yannick Scarff
Instructors Kathryn Boddie
Kathryn has taught high school or university mathematics for over 10 years. She has a Ph.D. in Applied Mathematics from the University of Wisconsin-Milwaukee, an M.S. in Mathematics from Florida State University, and a B.S. in Mathematics from the University of Wisconsin-Madison.
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Yannick Scarff
Yannick Scarff has taught 7th/8th grade Math for over 10 years. She has a Bachelor of Arts in Interdisciplinary Studies with Concentration in 4th-8th grade Math-Science from the University of Texas at San Antonio. She holds a Texas State teacher certification in 4th-8th Math/Science, and EC-6th.
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Steps for Inscribing a Square in a Circle
Step 1: If the diameter or the radius of the circle is known, find the length of the diagonal of the square using the fact that the diameter of the circle equals the diagonal of the square, or equivalently, the radius of the circle is half the diagonal of the square. Then find the length of a side of the square using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle).
Step 2: If the side lengths of the square are known, find the diameter of the circle using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle). The diameter of the circle equals the diagonal of the square.
Step 3: Use the information from steps 1 and 2 to find the quantity asked for. This could be the area of the circle or the square, the circumference of the circle, the perimeter of the square, or other measurements.
Vocabulary and Equations for Inscribing a Square in a Circle
Inscribed: A square is said to be inscribed in a circle if all four corners of the square lie on the circle. The diagonal of an inscribed square is equal to the diameter of the circle. The diameter of the circle also divides the square into two equal right triangles.
Pythagorean Theorem: The Pythagorean theorem relates the lengths of the sides of a right triangle to the length of its hypotenuse. The theorem states that a 2+b 2=c 2 where a and b are the sides of the right triangle and c is the hypotenuse.
Area and Perimeter/Circumference: The area of a square is given by A=x 2, where x is the length of a side of the square and the area of a circle is given by A=π r 2, where r is the radius of the circle. The perimeter of a square is P=4 x, where x is the length of a side of the square and the circumference of a circle is given by C=2 π r, where r is the radius of the circle.
We will use these steps, definitions, and equations to find measurements involving a square inscribed in a circle in the following two examples.
Example Problem 1 - How to Inscribe a Square in a Circle
In the image below, the square is inscribed in the circle and the circle has a radius of 2 inches. Find the area and perimeter of the square.
Square inscribed in a Circle with Radius 2 in.
Step 1: If the diameter or the radius of the circle is known, find the length of the diagonal of the square using the fact that the diameter of the circle equals the diagonal of the square, or equivalently, the radius of the circle is half the diagonal of the square. Then find the length of a side of the square using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle).
Since we know that the radius is r=2 in. and the radius is half the diagonal of the square, we can find the diagonal of the square by multiplying the radius by 2. So the diagonal of the square is 2⋅2=4 in..
Now we can find the sides of the square using the Pythagorean theorem. The diagonal of the square forms a hypotenuse:
The Right Triangle Formed
Since we have a square, the two sides of the right triangle have to be equal, so we will call them x. Then, using the Pythagorean Theorem:
a 2+b 2=c 2 x 2+x 2=4 2 2 x 2=16 2 x 2 2=16 2 x 2=8 x=8
This can be simplified using the fact that 8=4⋅2.
x=8 x=4⋅2 x=4⋅2 x=2 2
So the square has sides of length x=2 2 in..
Step 2: If the side lengths of the square are known, find the diameter and radius of the circle using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle). The diameter of the circle equals the diagonal of the square.
We already know the radius of the circle is 2 inches, and the diameter of the circle is 4 inches.
Step 3: Use the information from steps 1 and 2 to find the quantity asked for. This could be the area of the circle or the square, the circumference of the circle, the perimeter of the square, or other measurements.
To find the area of the square, we need to use the formula A=x 2 where x is the length of a side of the square.
A=x 2 A=(2 2)2 A=2 2 2 2 A=4⋅2 A=8 in.2
To find the perimeter of the square, we need to use the formula P=4 x.
P=4 x P=4(2 2)P=8 2 in.
Therefore, the square has an area of 8 in.2 and a perimeter of 8 2 in.
Example Problem 2 - How to Inscribe a Square in a Circle
In the image below, the square is inscribed in the circle and has side lengths of 1 centimeter. Find the area and circumference of the circle.
Square with sides of Length 1 cm. inscribed in a Circle
Step 1: If the diameter or the radius of the circle is known, find the length of the diagonal of the square using the fact that the diameter of the circle equals the diagonal of the square, or equivalently, the radius of the circle is half the diagonal of the square. Then find the length of a side of the square using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle).
The diameter and radius of the circle are not known, so we need to move to step 2.
Step 2: If the side lengths of the square are known, find the diameter and radius of the circle using the Pythagorean theorem: a 2+b 2=c 2, where a and b are side lengths of the square and c is the diagonal of the square (the hypotenuse of a right triangle). The diameter of the circle equals the diagonal of the square.
The sides of the square measure 1 cm. and so we can find the diagonal of the square (which is also the diameter of the circle) using the Pythagorean theorem.
The Right Triangle Formed
a 2+b 2=c 2 1 2+1 2=c 2 2=c 2 2=c
So the diameter of the circle is d=2 cm., which makes the radius of the circle r=2 2 cm.
Step 3: Use the information from steps 1 and 2 to find the quantity asked for. This could be the area of the circle or the square, the circumference of the circle, the perimeter of the square, or other measurements.
To find the area of the circle, we need to use the formula A=π r 2 where r is the radius of the circle.
A=π r 2 A=π(2 2)2 A=π(2 4)A=1 2 π cm.2
To find the circumference of the circle, we need to use the formula C=2 π r.
C=2 π r C=2 π(2 2)C=π 2 cm.
Therefore, the circle has an area of 1 2 π cm.2 and a circumference of π 2 cm.
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Table of Contents
Steps for Inscribing a Square in a Circle
Vocabulary and Equations for Inscribing a Square in a Circle
Example Problem 1
Example Problem 2
Test your current knowledge
Practice
Inscribing a Square in a Circle
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190191 | https://math.stackexchange.com/questions/5001821/three-ways-of-computing-the-probability-of-getting-a-given-sequence-of-colors-wh | Three ways of computing the probability of getting a given sequence of colors when extracting balls from an urn - Mathematics Stack Exchange
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Three ways of computing the probability of getting a given sequence of colors when extracting balls from an urn
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I'm teaching myself some probability theory through Hans-Otto Georgii's beautiful Stochastics, and with this question I would like to clarify myself the meaning of some probabilistic models. My notation will diverge from that of the book, and anyway this post is self contained.
Suppose we're interested to model the following random experiment. Our system consists in an urn. The urn contains N N colored but otherwise indistinguishable balls. To take a measurement on the system means to draw n n times a ball from the urn (with always putting the extracted ball back with the others before extracting a new one), and record the sequence of colors we get. We want to compute the probability of obtaining a given sequence of colors.
A couple of ways to answer this question are the following. In any case, let's suppose that we have tagged the balls with numbers from 1 1 to N N, put Ω={1,…,N}Ω={1,…,N} and equip it with the uniform probability measure U Ω U Ω, and call Ω^Ω^ the set of all the colors. Define also an obvious random variable C:Ω→Ω^C:Ω→Ω^ that maps a ball (or better, it's unique identifier in Ω Ω) into its color.
The first way. We can equip Ω^Ω^ with the pushforward probability measure C∗U Ω C∗U Ω and consider the product space Ω^n Ω^n equipped with the product measure P 1=(C∗U Ω)⊗n P 1=(C∗U Ω)⊗n.
The second way. On the product space (Ω n,U⊗n Ω)(Ω n,U Ω⊗n) we define random variables C i C i with values in Ω^Ω^ by composing the maps Ω n−→p r i Ω−→C Ω^Ω n→p r i Ω→C Ω^, where p r i p r i is the obvious projection map. Each C i C i induces a probability measure on Ω^Ω^. So, we can multiply all of these and get another probability measure P 2=⨂n i=1(C i)∗U⊗n Ω P 2=⨂i=1 n(C i)∗U Ω⊗n on Ω^n Ω^n.
The third way. Almost same as before, but this time consider the pushforward measure P 2=(C 1,…,C n)∗U⊗n Ω P 2=(C 1,…,C n)∗U Ω⊗n, where (C 1,…,C n)(C 1,…,C n) is the combined random variable of the C i C i s.
I was wondering, since P 1=P 2=P 3 P 1=P 2=P 3 above, what's the semantic difference between P 1 P 1, P 2 P 2 and P 3 P 3? I mean, if were to build a dictionary like
tensor product of measures →→ iterated experiment
pushforward measure →→ ???
...
then what would the process of calculating the probability of a particular outcome (ω^1,…,ω^n)(ω^1,…,ω^n) using the function P i P i sound like in these words?
What's the difference --if there is one-- between P 1 P 1, P 2 P 2 and P 2 P 2? (For example, to me using P 1 P 1 sounds more natural, but the book employs P 3 P 3, which in turn would not be equal to P 2 P 2 if the events were not independent).
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edited Nov 22, 2024 at 9:28
N. F. Taussig
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asked Nov 22, 2024 at 2:05
GeometriaDifferenzialeGeometriaDifferenziale
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Not sure this is clear. Are you saying that the balls have different colors? How many different colors? Not sure why you are talking about "measure". That's a familiar mathematical term, but I think all you mean is observation. You just want to draw n n balls and record the sequence of colors, right?lulu –lulu 2024-11-22 02:18:07 +00:00 Commented Nov 22, 2024 at 2:18
I don't understand what you are asking. Suppose there are only two colors, black and white (one ball each, so two balls in total). Then you are just looking at binary strings of length n n. What is your question in that case?lulu –lulu 2024-11-22 02:19:12 +00:00 Commented Nov 22, 2024 at 2:19
I'm sorry, something got messed up when I was editing the post. I'll update it in a minute, but: 1) yes, the balls might have different colors; 2) I didn't specifies how many colors there are, but since I denoted with Ω^Ω^ the set of colors you can just denote with |Ω^||Ω^| this number; 3) the terminology "making a measure" comes from physics, i.e., I was just trying to explain that the set of "observations" of my system is the set of all the strings of n n colors; 4) yes, but (continues below)GeometriaDifferenziale –GeometriaDifferenziale 2024-11-22 02:27:10 +00:00 Commented Nov 22, 2024 at 2:27
1 But considering that many people consider probability to be a subset of measure theory, it seems unwise to insist on using the word "measure" in a non standard way. Just guaranteed to cause confusion.lulu –lulu 2024-11-22 02:35:02 +00:00 Commented Nov 22, 2024 at 2:35
1 Since trials are independent, you are just talking about a product measure here (using measure in the usual sense).lulu –lulu 2024-11-22 02:35:52 +00:00 Commented Nov 22, 2024 at 2:35
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190193 | https://www.whitman.edu/mathematics/calculus_online/section05.03.html | Home » Curve Sketching » The second derivative test
5.3 The second derivative test
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Introduction
1 Analytic Geometry
1. Lines
2. Distance Between Two Points; Circles
3. Functions
4. Shifts and Dilations
2 Instantaneous Rate of Change: The Derivative
1. The slope of a function
2. An example
3. Limits
4. The Derivative Function
5. Properties of Functions
3 Rules for Finding Derivatives
1. The Power Rule
2. Linearity of the Derivative
3. The Product Rule
4. The Quotient Rule
5. The Chain Rule
4 Transcendental Functions
1. Trigonometric Functions
2. The Derivative of $\sin x$
3. A hard limit
4. The Derivative of $\sin x$, continued
5. Derivatives of the Trigonometric Functions
6. Exponential and Logarithmic functions
7. Derivatives of the exponential and logarithmic functions
8. Implicit Differentiation
9. Inverse Trigonometric Functions
10. Limits revisited
11. Hyperbolic Functions
5 Curve Sketching
1. Maxima and Minima
2. The first derivative test
3. The second derivative test
4. Concavity and inflection points
5. Asymptotes and Other Things to Look For
6 Applications of the Derivative
1. Optimization
2. Related Rates
3. Newton's Method
4. Linear Approximations
5. The Mean Value Theorem
7 Integration
1. Two examples
2. The Fundamental Theorem of Calculus
3. Some Properties of Integrals
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18 Useful formulas
19 Introduction to Sage
1. Basics
2. Differentiation
3. Integration
The basis of the first derivative test is that if the derivative changes from positive to negative at a point at which the derivative is zero then there is a local maximum at the point, and similarly for a local minimum. If $f'$ changes from positive to negative it is decreasing; this means that the derivative of $f'$, $f''$, might be negative, and if in fact $f''$ is negative then $f'$ is definitely decreasing, so there is a local maximum at the point in question. Note well that $f'$ might change from positive to negative while $f''$ is zero, in which case $f''$ gives us no information about the critical value. Similarly, if $f'$ changes from negative to positive there is a local minimum at the point, and $f'$ is increasing. If $f''>0$ at the point, this tells us that $f'$ is increasing, and so there is a local minimum.
Example 5.3.1 Consider again $f(x)=\sin x + \cos x$, with $f'(x)=\cos x-\sin x$ and $ f''(x)=-\sin x -\cos x$. Since $\ds f''(\pi/4)=-\sqrt{2}/2-\sqrt2/2=-\sqrt2< 0$, we know there is a local maximum at $\pi/4$. Since $\ds f''(5\pi/4)=- -\sqrt{2}/2- -\sqrt2/2=\sqrt2>0$, there is a local minimum at $5\pi/4$. $\square$
When it works, the second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails, and sometimes the second derivative is quite difficult to evaluate; in such cases we must fall back on one of the previous tests.
Example 5.3.2 Let $\ds f(x)=x^4$. The derivatives are $\ds f'(x)=4x^3$ and $\ds f''(x)=12x^2$. Zero is the only critical value, but $f''(0)=0$, so the second derivative test tells us nothing. However, $f(x)$ is positive everywhere except at zero, so clearly $f(x)$ has a local minimum at zero. On the other hand, $\ds f(x)=-x^4$ also has zero as its only critical value, and the second derivative is again zero, but $\ds -x^4$ has a local maximum at zero. Finally, if $\ds f(x)=x^3$, $f'(x)=3x^2$ and $f''(x)=6x$. Again, zero is the only critical value and $f''(0)=0$, but $\ds x^3$ has neither a maximum nor a minimum at $0$.
$\square$
Exercises 5.3
Find all local maximum and minimum points by the second derivative test, when possible.
Ex 5.3.1 $\ds y=x^2-x$ (answer)
Ex 5.3.2 $\ds y=2+3x-x^3$ (answer)
Ex 5.3.3 $\ds y=x^3-9x^2+24x$ (answer)
Ex 5.3.4 $\ds y=x^4-2x^2+3$ (answer)
Ex 5.3.5 $\ds y=3x^4-4x^3$ (answer)
Ex 5.3.6 $\ds y=(x^2-1)/x$ (answer)
Ex 5.3.7 $\ds y=3x^2-(1/x^2)$ (answer)
Ex 5.3.8 $y=\cos(2x)-x$ (answer)
Ex 5.3.9 $\ds y = 4x+\sqrt{1-x}$ (answer)
Ex 5.3.10 $\ds y = (x+1)/\sqrt{5x^2 + 35}$ (answer)
Ex 5.3.11 $\ds y= x^5 - x$ (answer)
Ex 5.3.12 $\ds y = 6x +\sin 3x$ (answer)
Ex 5.3.13 $\ds y = x+ 1/x$ (answer)
Ex 5.3.14 $\ds y = x^2+ 1/x$ (answer)
Ex 5.3.15 $\ds y = (x+5)^{1/4}$ (answer)
Ex 5.3.16 $\ds y = \tan^2 x$ (answer)
Ex 5.3.17 $\ds y =\cos^2 x - \sin^2 x$ (answer)
Ex 5.3.18 $\ds y = \sin^3 x$ (answer) |
190194 | https://brainly.com/question/4756820 | [FREE] How many ways are there to put 8 beads of different colors on the vertices of a cube if rotations of the - brainly.com
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How many ways are there to put 8 beads of different colors on the vertices of a cube if rotations of the cube (but not reflections) are considered the same?
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Actually, this is the correct answer.
Consider one vertex of the cube. When the cube is rotated, there are 8 vertices which that vertex could end up at. At each of those vertices, there are 3 ways to rotate the cube onto itself with that vertex fixed. So, there are a total of 83=24 ways to rotate a cube. There are 8! ways to arrange the beads, not considering rotations. Since the arrangements come in groups of 24 equivalent arrangements, the actual number of ways to arrange the beads is 8!/24=1680.
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There are 1680 unique ways to place 8 beads of different colors on the vertices of a cube when accounting for rotations. This is done by dividing the total arrangements of 8! (40320) by the number of rotations (24). Thus, the formula used is 24 8!.
Explanation
To find how many ways there are to place 8 beads of different colors at the vertices of a cube, taking into account that rotations of the cube are considered the same, we can use concepts from group theory and combinatorics.
Understanding the Cube Rotation: The cube has 24 possible rotations. This is because it can be rotated around its axes and each rotation might be positioned in different ways. Specifically, for any fixed vertex, the cube can be rotated in 3 ways for each of the 8 vertices, leading us to:
8 vertices×3 rotations per vertex=24 total rotations
Arranging the Beads: Without considering the cube's rotational symmetry, the number of ways to arrange 8 distinct beads is given by 8 factorial, which is:
8!=40320
Counting Unique Arrangements: Since each configuration of beads repeats 24 times under the cube's rotations, we take the total arrangements and divide by the number of rotations:
Unique arrangements=24 8!=24 40320=1680
So, there are 1680 unique ways to place the 8 beads of different colors on the vertices of a cube, considering that rotations yield the same arrangement.
Examples & Evidence
For example, if you have 8 beads labeled A, B, C, D, E, F, G, and H, their arrangements can be counted despite the cube being rotated. Arranging A, B, C, D, E, F, G, and H with their positions affected by cube rotations, the arrangement ABCDEFGH is the same as any arrangement that results in those beads being in equivalent positions after rotation.
The solution draws from the basic principles of combinatorics and group theory related to symmetrical shapes like cubes, specifically the counting of arrangements by considering rotational symmetries of objects.
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What is being requested, if I'm not mistaken, is the number of permutations for placing each of the 8 beads on the vertices of the cubes;
In this case, we have 8 different beads and 8 possible locations for each of them;
So the number of permutations is:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
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190195 | https://artofproblemsolving.com/wiki/index.php/Triangle_Inequality?srsltid=AfmBOop2-Gp5-MbCPfeGUKFQB3knEyLzH9t9Y2YtK3XjX9AeEXXfrRCm | Page
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Triangle Inequality
The Triangle Inequality says that in a nondegenerate triangle :
That is, the sum of the lengths of any two sides is larger than the length of the third side.
In degenerate triangles, the strict inequality must be replaced by "greater than or equal to."
The Triangle Inequality can also be extended to other polygons. The lengths can only be the sides of a nondegenerate -gon if for . Expressing the inequality in this form leads to , where is the sum of the , or . Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter.
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Given , prove:
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190196 | https://www.math.pku.edu.cn/teachers/mwfy/Teaching/ProbabilityStatistics/slide1.pdf | .
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概率统计B 第一章随机事件与概率 原著：陈家鼎、刘婉如、汪仁官 制作：李东风，邓明华 2025 春季学期 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 1 / 102 .
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本章目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 2 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 3 / 102 .
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随机事件与概率 随机事件：在一定条件下，可能发生也可能不发生的事件。 例1.1 掷分币，结果“正面朝上” （记作A）是随机事件。 “正面朝 下” （记作B）也是随机事件。 例1.2 掷两枚分币。 A =“两个都是正面朝上” B =“两个都是正面朝下” C =“一个正面朝上，一个正面朝下” 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 4 / 102 .
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例1.3 10 件同类产品中有8 个正品，2 个次品。任意抽取3 个。 A =“3 个都是正品” B =“3 个中至少一个是次品” V =“3 个都是次品” U =“3 个中至少有一个是正品” A, B 是随机事件； V 是“不可能事件” ； U 是“必然事件” ； 不可能事件和必然事件也看作是随机事件。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 5 / 102 .
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概率 事件是否发生无法预知，但是其可能性大小可以定量描述。 比如，投掷一枚均匀硬币，正面朝上和正面朝下可能性大小相同。 投掷两枚均匀硬币，同时为正面和同时为背面可能性大小相同；一 个正面一个背面的可能性比都是正面的可能性大。 概率用来定量描述随机事件发生可能性大小。P(A)。 概率有“频率定义”、 “主观定义”、 “公理化定义” 。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 6 / 102 .
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频数 投掷一枚分币。条件组S。 条件组S 大量重复实现时，事件A 发生的次数，称为频数。约占总 试验次数的一半。 A 发生的频率= 频数 试验次数, 接近于1 2 长期经验积累所得的、所谓某事件发生的可能性大小，就是这个“频 率的稳定值” 。 见P.3 的多次投掷表格。次数越多，频率越接近0.5。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 7 / 102 .
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概率的频率定义 定义1.1 在不变的一组条件S 下，重复做n 次实验。记µ 是n 次试 验中事件A 发生的次数。当试验的次数n 很大时，如果频率µ/n 稳 定地在某一数值p 的附近摆动，而且一般说来随着试验次数的增多， 这种摆动的幅度越变越小，则称A 为随机事件，并称数值p 为随机 事件A 在条件组S 下发生的概率, 记作 P(A) = p 数值p 的大小是A 在S 下发生的可能性大小的数量刻画。例如0.5 是掷一枚分币出现“正面朝上”的可能性的数量刻画。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 8 / 102 .
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定义简述：频率具有稳定性的事件叫做随机事件，频率的稳定值叫 做该随机事件的概率。 随机事件简称事件。实际中遇到的事件一般都是随机事件。 频率µ/n 取值在[0, 1] 范围。所以概率 0 ≤P(A) ≤1.
对不可能事件V 和必然事件U, P(V) = 0, P(U) = 1.
概率的频率定义是近似值。许多测量值都是近似值，所以不必因为 只能求得近似值而怀疑真实概率的存在。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 9 / 102 .
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概率的主观定义 不能重复或不能大量重复的事件如何定义概率？ 定义1.2 一个事件的概率是人们根据已有的知识和经验对该事件发 生可能性所给出的个人信念，这种信念用[0, 1] 中的一个数来表示， 可能性大的对应较大的数。 称为概率的主观定义。 例：企业家对产品畅销可能性的预测；医生对某特定病人手术成功 的预测。 主观概率是当事人对事件作了详细考察并充分利用个人已有的经验 形成的“个人信念” ，而不是没有根据的乱说一通。也需要谨慎对待。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 10 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 11 / 102 .
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古典概型 某些概率问题可以根据问题本身所具有的“对称性” ，充分利用人类 长期积累的关于对称性的实际经验，分析事件的本质，就可以直接 计算其概率。 这是用数学模型求解概率的方法。 例如，投掷一枚分币，认为“正面朝上”和“正面朝下”概率相等（对 称性） ，各为0.5。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 12 / 102 .
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例2.1 例2.1 盒中5 个球， 3 白2 黑。 从中任取一个。 问： 取到白球的概率？ 直观看为3/5。 把5 个球编号，1—3 号为白球，4—5 号为黑球。 取到每个球的概率相同（对称性） 。事件互相排斥，概率各为1/5。 把3 个白球的概率加起来即可。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 13 / 102 .
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例2.2 例2.2 盒中5 个球，3 白2 黑。从中任取两个。问：两个都是白球 的概率？ 这时不能直观得出概率。 把5 个球编号，1—3 号为白球，4—5 号为黑球。 可能结果为： 1 + 2 1 + 3 1 + 4 1 + 5 2 + 3 2 + 4 2 + 5 3 + 4 3 + 5 4 + 5 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 14 / 102 .
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共10 个可能结果，且发生的机会相同，互斥，除此之外无其它可 能。 每个结果的概率为1/10，其中有1 + 2, 1 + 3, 2 + 3 共3 个结果为全 白球。所以全是白球的概率为3/10。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 15 / 102 .
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等概完备事件组 定义2.1 称一个事件组A1, A2, . . . , An 为一个等概完备事件组，如 果它具有下列三条性质： (1) A1, A2, . . . , An 发生的机会相同（等可能性） ； (2) 在任一次试验中，A1, A2, . . . , An 至少有一个发生（也就是所谓“除此 之外，不可能有别的结果” ） （完备性） ； (3) 在任一次试验中，A1, A2, . . . , An 至多有一个发生（也就是所谓“它们 是互相排斥的” ） （互不相容性） 。 等概完备事件组也称“等概基本事件组” ，其中任一事件 Ai(i = 1, 2, . . . , n) 称为基本事件。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 16 / 102 .
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例1.1，投掷一枚分币，等概基本事件组n = 2, 两个基本事件是“正 面朝上”和“正面朝下” 。 其它例子类似可求得等概基本事件组。 若A1, A2, . . . , An 是一个等概基本事件组，事件B 由其中的m 个基 本事件所构成，则 P(B) = m n .
(2.1) 古典概型就是用等概基本事件组和(2.1) 来计算事件的概率的模型。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 17 / 102 .
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例2.2（续） 例2.2（续）从三个白球和二个黑球中任取两个，共有C2 5 = 10 种 不同取法，出现机会相同。 每种取法为一个基本事件，构成等概完备事件组。 其中两个都是白球的事件为3 个，概率等于m/n = 3/10。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 18 / 102 .
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例2.3 例2.3 100 件产品，有5 件次品。任取50 件。求无次品的概率。 解共有C50 100 个结果构成等概基本事件组。 事件B: 任取50 件其中无次品，包括多少个基本事件？ 必须从95 个正品中取出50 件。取法有C50 95 种。 P(B) =C50 95/C50 100 = 95!/(50!45!) 100!/(50!50!) = 50!/45!
100!/95! = 50 · 49 · 48 · 47 · 46 100 · 99 · 98 · 97 · 96 =1 2 1 2 1 2 47 99 46 97 = 1081 38412 ≈2.8% 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 19 / 102 .
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例2.4 例2.4 同例2.3。问：恰好有2 件次品（记为事件A）的概率？ 在基本事件组中，符合条件的事件，必须是从5 个次品中任取2 个， 从95 个正品中任取48 个。 共有C2 5C48 95 种取法。 P(A) =C2 5C48 95 C50 100 = 5!
2!3!
95!
47!48!
100!
50!50!
= 0.32 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 20 / 102 .
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例2.5：无放回抽样 例2.5 设一批产品共N 个，其中次品共M 个。从中任取n 个。问： 恰好出现m 个次品的概率？ 0 ≤m ≤n, m ≤M, n −m ≤N −M。 这是例2.4 的一般化，所以 P(恰好出现m 个次品) = Cn−m N−MCm M Cn N (2.2) 记Ck n = 0，当k < 0 或k > n 时。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 21 / 102 .
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定理2.1 例2.5 的产品分为两类：次品和正品。考虑分为多类的情形。 定理2.1 设有N 个东西分成k 类，其中第i 类有Ni 个东西 （i = 1, 2, . . . , k） ，N1 + N2 + . . . Nk = N, 从这N 个东西中任取n 个， 而n = m1 + m2 + · · · + mk (0 ≤mi ≤Ni, i = 1, 2, . . . , k), 则事件A = “恰有m1 个属于第1 类, 恰有m2 个属于第2 类， · · · · · · ，恰有mk 个 属于第k 类”的概率为 P(A) = Cm1 N1Cm2 N2 . . . Cmk Nk Cn N (2.3) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 22 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 事件的包含与相等 事件的并与交 对立事件及事件的差 事件的运算规律 事件的互不相容性 概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 23 / 102 .
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事件的包含与相等 如果事件A 发生则事件B 一定发生，就称事件B 包含事件A，记作 A ⊂B 或B ⊃A.
如，投掷两枚硬币，A 表示“正好一个正面朝上” ，B 表示“至少一个 正面朝上” ，则A ⊂B。 如果A ⊂B 且B ⊂A，则称事件A 与事件B 相等，记作 A = B.
在概率的公理化定义中，事件等同于集合，事件的性质就是集合的 性质。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 24 / 102 .
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事件的并与交 定义3.1 事件“A 或B”称为事件A 与事件B 的并，记作A ∪B 或 A + B；某次试验中A ∪B 发生，即“A 或B”发生，意味着A, B 中只 要一个发生。 事件“A 且B”称为事件A 与B 的交，记作A ∩B 或AB 或A · B； A ∩B 发生，即“A 且B”发生，意味着A 和B 都发生。 例：投掷两枚硬币。A 表示“正好一个正面朝上” ，B 表示“正要两个 正面朝上, C 表示“至少一个正面朝上” ，则 A ∪B = C, AC = A, BC = B, AB = V(不可能事件) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 25 / 102 .
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对立事件 定义3.2 事件“非A”称为A 的对立事件，记作¯ A。 例如，投掷两枚硬币， “至少一个正面朝上”是“两个都是正面朝下”的 对立事件。 对立事件是相互的： (¯ A) = A 在一次试验中，A 和¯ A 互斥，且至少一个发生。即 A ∩¯ A =V A ∪¯ A =U (3.1) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 26 / 102 .
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事件的差 定义3.3 事件A 同事件B 的差表示A 发生而B 不发生的时间，记作 A\B。 A\B = A ∩¯ B (3.2) 事件及事件的运算用图形表示，见P.12 图1.1。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 27 / 102 .
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事件的运算规律 与集合运算规律相同。 并： (1)A ∪B = B ∪A (” 并” 的交换律) (2)A ∪(B ∪C) = (A ∪B) ∪C (” 并” 的结合律) (3)A ∪A = A (4)A ∪¯ A = U (5)A ∪U = U (6)A ∪V = A 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 28 / 102 .
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交： (7)A ∩B = B ∩A (” 交” 的交换律) (8)(AB)C = A(BC) (” 交” 的结合律) (9)A ∩A = A (10)A ∩¯ A = V (11)A ∩U = A (12)A ∩V = V 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 29 / 102 .
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分配律： (13)A(B ∪C) = (AB) ∪(AC) (分配律) (14)A ∪(B ∩C) = (A ∪B) ∩(A ∪C) (分配律) 交或并的对立事件： (15)A ∪B = ¯ A ∩¯ B (16)A ∩B = ¯ A ∪¯ B 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 30 / 102 .
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事件的互不相容性 定义3.4 如果事件A 与事件B 不能都发生，即 AB = V (不可能事件) 则称A 与B 是互不相容的事件。 例：两枚分币， “正好一个正面朝上”与“两个都是正面朝上”互不相 容。 A 与¯ A 互不相容。 多个事件互不相容是指两两互不相容。 等概完全事件组定义中“互相排斥”也是两两互不相容的意思。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 31 / 102 .
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概率的加法公式(1) 如果事件A, B 互不相容，则 P(A ∪B) = P(A) + P(B) (3.3) 其合理性和必要性可以用概率的频率定义解释。 推论： P(A) + P(¯ A) = P(A ∪¯ A) = P(U) = 1 从而得 P(A) = 1 −P(¯ A), P(¯ A) = 1 −P(A) (3.4) 这样，一个事件的概率难计算而其对立事件的概率容易计算时可用 (3.4) 计算。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 32 / 102 .
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概率的有限可加性：设n 个事件A1, A2, . . . , An 互不相容，则 P(A1 ∪A2 ∪· · · ∪An) = P(A1) + P(A2) + · · · + P(An) (3.5) 可以从(3.3) 归纳证明。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 33 / 102 .
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概率的加法公式(2) 对任意两个事件A, B, 有 P(A ∪B) = P(A) + P(B) −P(AB) (3.6) 证明易见A ∪B = A ∪(B¯ A), 且 A ∩(B¯ A) = AB¯ A = B(A¯ A) = BV = V 所以 P(A ∪B) = P(A ∪(B¯ A)) = P(A) + P(B¯ A) (3.7) 又因 B = B ∩U = B ∩(A ∪¯ A) = (BA) ∪(B¯ A) 且BA 与B¯ A 互不相容，所以 P(B) = P(BA) + P(B¯ A), P(B¯ A) = P(B) −P(AB) 代入(3.7) 即得(3.6)。证毕。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 34 / 102 .
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例3.1 例3.1 袋中有红、黄、白球各一个，每次任取一个，有放回地抽取 三次。 求：抽到的三个球中没有红球或没有黄球的概率。 记G =“三个球都不是红球，H =“三个球都不是黄球” 。要求 P(G ∪H)。 注意G 和H 不是互不相容。 P(G) = 8 27（共有27 种可能结果，其中没有红球的结果是2 × 2 × 2 个） ，P(H) = 8 27。 P(GH) = P(三个球都是白球) = 1 27(全是白球的结果只有一种)。 P(G ∪H) = P(G) + P(H) −P(GH) = 15 27 = 5 9.
原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 35 / 102 .
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无穷个事件的并和交 定义3.5 设A1, A2, . . . 是一系列事件，事件B 表示: 它的发生当且 仅当A1, A2, . . . 中至少一个发生，B 称为A1, A2, . . . 的并（或者和） ， 记作∪∞ k=1 Ak(或∑∞ k=1 Ak)，或A1 ∪A2 ∪. . . 。 ∩∞ k=1 Ak 表示这样的事件：当且仅当所有A1, A2, . . . 都同时发生时 此事件才发生。 例3.2 一射手向某目标连续射击，A1 ={第一次射击，命中}， Ak ={前k −1 次射击都未中，第k 次射击命中}(k = 2, 3, . . . )。 B ={终于命中}。则B = ∪∞ k=1 Ak。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 36 / 102 .
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概率的完全可加性 设A1, A2, . . . 是一系列事件，如果A1, A2, . . . 两两互不相容，则 P ( ∞ ∪ k=1 Ak ) = ∞ ∑ k=1 P(Ak) (3.8) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 37 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 集合 事件与集合的关系 概率的公理化定义介绍 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 38 / 102 .
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集合 事件是一种特殊集合，所以事件的运算就是集合的运算。 定义4.1 一个集合是指具有确切含义的若干个东西的全体。 集合A, B, C, . . . 。元素a, b, c, . . . 。 a ∈A。 a / ∈A。 空集∅。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 39 / 102 .
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集合的例子 例4.1 全体正整数的集合。 例4.2 不大于10 的正整数的集合。 例4.3 二维坐标平面上圆心在圆点的半径为1 的圆（称为单位圆） 内点的集合。 例4.5 红、黄、白三个球有放回抽取三次的结果集合。共33 = 27 个结果。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 40 / 102 .
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集合的关系 集合相等：两个集合的元素完全相同。记作A = B。 集合包含关系A ⊂B: A 的元素都是B 的元素。也记为B ⊃A。 A = B ⇐ ⇒A ⊂B 且B ⊂A。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 41 / 102 .
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集合的运算 并集A ∪B: 属于A 或者属于B 的元素的全体组成的集合。 交集A ∩B: 既属于A 也属于B 的元素的全体组成的集合。 只讨论某个非空集合Ω的自己的关系，Ω称为全集。 余集: Ac = {x : x ∈Ω但x / ∈A} (Ac)c = A.
集合运算可以用平面图形图示。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 42 / 102 .
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集合并的运算规则 1 A ∪B = B ∪A (交换律) 2 (A ∪B) ∪C = A ∪(B ∪C) (结合律) 3 A ∪A = A 4 A ∪Ac = Ω 5 A ∪Ω= Ω 6 A ∪∅= A 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 43 / 102 .
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集合交的运算规则 1 A ∩B = B ∩A 交换律 2 (A ∩B) ∩C = A ∩(B ∩C) 结合律 3 A ∩A = A 4 A ∩Ac = ∅ 5 A ∩Ω= A 6 A ∩∅= ∅ 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 44 / 102 .
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并与交的分配律 1 A ∩(B ∪C) = (A ∩B) ∪(A ∩C) 2 A ∪(B ∩C) = (A ∪B) ∩(A ∪C) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 45 / 102 .
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并、交、余的对偶律 1 (A ∪B)c = Ac ∩Bc 2 (A ∩B)c = Ac ∪Bc 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 46 / 102 .
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事件与集合的关系 事件是特殊的集合。事件的运算与集合的运算相同。 条件组S 下的所有可能不同结果的集合记作Ω，S 下的随机事件就 是Ω的子集。 Ω是必然事件，∅是不可能事件。 Ac = Ω\A 为A 的对立事件¯ A。 A, B 互不相容即A ∩B = ∅。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 47 / 102 .
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例4.6 例4.6 投掷两枚分币(条件S)，所有可能结果为 ω1 = “上，下” ω2 = “上，上” ω3 = “下，上” ω4 = “下，下” 全集Ω= {ω1, ω2, ω3, ω4}。 事件B 为“恰有一个正面朝上”, 则 B = {ω1, ω3} 事件C 为“至少有一个正面朝上” ，则 C = {ω1, ω2, ω3} 事件A 为“两个正面都朝上” ，则A = {ω2}。 C = B ∪A 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 48 / 102 .
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概率的公理化定义介绍 概率的频率解释直观，但数学严密性不足。 概率的主观定义则不易被接受，数学严密性不足。 用集合论、测度论可以严格定义概率，对需要作公理化假设。 为柯尔莫戈罗夫(Kolmogorov A. N., 1903-1987) 于1933 年建立。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 49 / 102 .
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概率的公理化定义介绍 设Ω为一个非空集合，叫做基本事件空间。 Ω的一些子集组成的集合F 叫做σ 代数，如果 (1)Ω∈F (2)若A ∈F, 则Ac = Ω−A ∈F (3)若An ∈F(n = 1, 2, . . . ), 则∪∞ n=1 An ∈F 事件是F 中的集合。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 50 / 102 .
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F 上有定义的函数P = P(·) 叫做概率测度(简称概率), 若 (1)P(A) ≥0(∀A ∈F) (4.17) (2)P(Ω) = 1 (4.18) (3)若An ∈F(n = 1, 2, . . . ), 且两两不相交, 则 P ( ∞ ∪ n=1 An ) = ∞ ∑ n=1 P(An) (完全可加性) (4.19) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 51 / 102 .
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σ 代数的性质 附有F, P 的Ω叫做概率空间。 σ 代数是可以合理定义概率的事件的全体，有些情况下不是所有Ω 的子集都可以合理定义概率。 若Ω为有限集或可数集则F 通常取为Ω的所有子集的集合。 F 关于基本集合运算封闭：有穷个或无穷个集合的并、交，两个集 合的差。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 52 / 102 .
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概率的性质 1 P(∅) = 0 2 若A ∈F 则P(Ac) = 1 −P(A) 3 若A1, . . . , An 都属于F 且两两不相交，则 P ( n ∪ i=1 Ai ) = n ∑ i=1 P(Ai) (有限可加性) (4.20) 4 若A ⊂B, A, B ∈F, 则P(A) ≤P(B) 且 P(B\A) = P(B) −P(A) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 53 / 102 .
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5 若An ⊂An+1, An ∈F(n = 1, 2, . . . )，则 P ( ∞ ∪ n=1 An ) = lim n→∞P(An) (单调上升事件的概率极限) 6 若An ⊃An+1, An ∈F(n = 1, 2, . . . )，则 P ( ∞ ∩ n=1 An ) = lim n→∞P(An) (单调下降事件的概率极限) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 54 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 条件概率 乘法公式 独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 55 / 102 .
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条件概率 条件概率在条件S 的基础上附加了条件，讨论则附加条件之后的概 率。 例5.1 16 个球，6 个玻璃球，10 个木球。玻璃球中有2 个红色，4 个蓝色；木球中有3 个红色，7 个蓝色。从16 个球中任取一个。 玻璃 木质 红 2 3 5 蓝 4 7 11 6 10 16 记A =“取到蓝球”, B = “取到玻璃球” 。P(A) = 11 16, P(B) = 6 16。 问：如果已知取到的是蓝球，则该球是玻璃球的概率？即事件A 已 经发生前提下事件B 发生的概率，记作P(B|A)。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 56 / 102 .
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可以用古典概型计算。蓝球共有11 个，其中4 个是玻璃球。 P(B|A) = 4 11 定义5.1 如果A, B 是条件组S 下的两个随机事件，P(A) ̸= 0，则称 在A 发生的前提下B 发生的概率为条件概率, 记作P(B|A)。 注意这不是严格数学定义。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 57 / 102 .
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例5.2 例5.2 5 个乒乓球，3 新2 旧。每次取一个，无放回取两次。 A=“第一次取到新球” ；B =“第二次取到新球” 。 求P(A), P(B), P(B|A)。 P(A) = 3 5。 P(B) 用古典概型，可以把5 个球编号，则两次抽取所有可能结果有 5 × 4 = 20 种，其中第二次抽取到新球的结果数为 3 × 2 + 2 × 3 = 12 种，P(B) = 12 20 = 3 5，即抽签是公平的。 若A 已经发生，则还剩2 新2 旧，于是第二次取到新球的概率为 2 4 = 1 2, 即P(B|A) = 1 2。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 58 / 102 .
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乘法公式 条件概率的等价定义为（多数教材这样定义条件概率） P(B|A) = P(AB) P(A) (5.1) (5.1) 改写为 P(AB) = P(A)P(B|A) (5.1’) 称为概率的乘法公式。 (5.1) 用来在已知P(A) 和P(AB) 时求条件概率； (5.1’) 用来在已知P(A) 和P(B|A) 时求P(AB)。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 59 / 102 .
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在古典概型下由定义5.1 可以证明(5.1)。 设条件组S 下一个等概完备事件组有n 个基本事件，A 由其中m 个 组成，B 由其中l 个组成，AB 由k 个组成。 则 P(B|A) =在A 发生的前提下B 中包含的基本事件数 在A 发生的前提下的基本事件总数 = k m = k/n m/n = P(AB) P(A) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 60 / 102 .
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独立性 例5.3 5 个乒乓球，3 新2 旧，每次取1 个，有放回取2 次。 A =“第一次取到新球” B =“第二次取到新球” 显然P(B|A) = P(B)，与A 是否发生无关。 这时 P(AB) = P(A)P(B|A) = P(A)P(B) 定义5.2 称两个随机事件A, B 是相互独立的, 如果 P(AB) = P(A)P(B) 定义中不要求P(A) > 0 或P(B) > 0。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 61 / 102 .
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独立性的直观解释 事件A 是否发生不影响事件B 的发生概率，事件B 是否发生也不影 响事件A 的发生概率。 在P(A) ̸= 0, P(B) ̸= 0 时，独立等价于 P(A|B) = P(A) 也等价于 P(B|A) = P(B) 即条件概率等于无条件概率为独立。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 62 / 102 .
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例5.4 例5.4 甲、乙同时向一敌机炮击。 甲击中概率0.6；乙击中概率0.5。 求被击中的概率。 解：记A =“甲击中” ，B ==“乙击中” ；C =“敌机被击中” 。 P(C) = P(A ∪B) = P(A) + P(B) −P(AB)。 可以认为A, B 独立。 P(AB) =P(A) × P(B) = 0.6 × 0.5 = 0.3 P(C) =0.6 + 0.5 −0.3 = 0.8 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 63 / 102 .
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另解： P(C) =1 −P(¯ C) = 1 −P(A ∪B) =1 −P(¯ A ∩¯ B) = 1 −P(¯ A)P(¯ B) =1 −(1 −0.6)(1 −0.5) = 0.8 把并的概率用交的概率来求解是常用的手法。 另解用到了：A, B 独立则¯ A, ¯ B 也独立。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 64 / 102 .
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对立事件与独立 定理5.1 若四对时间A, B, A, ¯ B, ¯ A, B, ¯ A, ¯ B 中有一对独立，则另外三 对也独立。 即这四对事件或者都独立，或者都不独立。 证明仅证明A, B 独立= ⇒A, ¯ B 独立。 P(A) =P(AU) = P(A ∩(B ∪¯ B)) =P((AB) ∪(A¯ B)) = P(AB) + P(A¯ B) 于是 P(A¯ B) =P(A) −P(AB) = P(A) −P(A)P(B) =P(A)[1 −P(B)] = P(A)P(¯ B) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 65 / 102 .
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多个事件相互独立 定义5.3 称A, B, C 是相互独立的，如果有 P(AB) = P(A)P(B) P(AC) = P(A)P(C) P(BC) = P(B)P(C) P(ABC) = P(A)P(B)P(C) (5.3) 定义5.4 称A1, A2, . . . , An 是相互独立的，如果对任意整数 k(2 ≤k ≤n) 以及从1, 2, . . . , n 中任意取出的k 个i1, i2, . . . , ik 都满足 P(Ai1Ai2 . . . Aik) = P(Ai1)P(Ai2) . . . P(Aik) (5.4) 其中一个要求是 P(A1A2 . . . An) = P(A1)P(A2) . . . P(An) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 66 / 102 .
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例5.5 例5.5 某型号高射炮单发击中飞机概率为0.6。若干门发射单发， 欲以99% 概率击中敌机。求高炮门数。 解：设需要n 门，Ai 为“第i 门高炮击中敌机” 。 A =“敌机被击中” 。A = A1 ∪A2 ∪· · · ∪An。 P(A) =P(A1 ∪A2 ∪· · · ∪An) = 1 −P(¯ A1 ∩¯ A2 ∩· · · ∩¯ An) =1 −P(¯ A1)P(¯ A2) . . . P(¯ An) (独立性) =1 −(1 −0.6)n ≥0.99 n ≥log 0.01 log 0.4 = 5.026 需要6 门高射炮。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 67 / 102 .
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例5.6 例5.6 三个事件两两独立不能保证三个事件独立的例子。 均匀正四面体，四面涂红色、黄色、蓝色、红黄蓝混杂。 投掷一次，考察底面出现的颜色。 A =“红色出现” ，B =“黄色出现” ，C =“蓝色出现” 。 基本事件：Ai =“第i 面在底面”, i = 1, 2, 3, 4, 构成等概基本事件组。 A = A1 ∪A4, B = A2 ∪A4, C = A3 ∪A4。 P(A) = P(B) = P(C) = 1 2。 AB = AC = BC = A4，P(AB) = P(AC) = P(BC) = 1 4, 按定义A, B 相 互独立，A, C 相互独立，B, C 相互独立。 但ABC = A4, P(ABC) = 1 4 ̸= P(A)P(B)P(C)。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 68 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 全概公式 逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 69 / 102 .
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例6.1 例6.1 5 个乒乓球，3 新2 旧。每次取一个，无放回取两次。求第二 次取到新球的概率。 A =“第一次取到新球” B =“第二次取到新球” B =BA ∪B¯ A (6.1) P(B) =P(BA) + P(B¯ A) =P(A)P(B|A) + P(¯ A)P(B|¯ A) =3 5 · 2 4 + 2 5 · 3 4 = 3 5 (6.1) 将复杂的事件（情况）分解为简单的事件（情况） 。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 70 / 102 .
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全概公式 定理6.1（全概公式）如果事件组A1, A2, . . . , An 满足： (1) A1, A2, . . . , An 互不相容，且P(Ai) > 0(i = 1, 2, . . . , n)。 (2) A1 ∪A2 ∪· · · ∪An = U(完备性)， 则对任一事件B 皆有 P(B) = n ∑ i=1 P(Ai)P(B|Ai).
(6.2) 证明B = BU = BA1 ∪BA2 ∪· · · ∪BAn, P(B) =P(BA1) + P(BA2) + · · · + P(BAn) =P(A1)P(B|A1) + P(A2)P(B|A2) + · · · + P(An)P(B|An).
原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 71 / 102 .
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满足条件（1）和（2）的事件组A1, A2, . . . , An 称为完备事件组。比 等概完备事件组少了等概条件。 更一般的全概公式中的完备事件组可以包含可数个事件。 运用全概公式关键在于求完备事件组。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 72 / 102 .
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例6.2 例6.2 甲、乙、丙三人射击敌机。击中概率: 甲：0.4 乙：0.5 丙：0.7 若只有一人击中，飞机坠毁概率0.2；若恰好二人击中，坠毁概率 0.6；三人全中，坠毁概率1。 求飞机坠毁概率。 解B =“飞机坠毁”, A0 =“三人都不中” ；A1 =“恰好一人击中” ；A2 = “恰好二人击中”; A3 =“三人都击中” 。 A0, A1, A2, A3 构成完备事件组。 已知P(B|A0) = 0, P(B|A1) = 0.2, P(B|A2) = 0.6, P(B|A3) = 1。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 73 / 102 .
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P(A0) =P(甲不中)P(乙不中)P(丙不中) =(1 −0.4)(1 −0.5)(1 −0.7) = 0.09 P(A1) =P(甲中)P(乙不中)P(丙不中) + P(甲不中)P(乙中)P(丙不中) + P(甲不中)P(乙不中)P(丙中) =0.4 × (1 −0.5) × (1 −0.7) + (1 −0.4) × 0.5 × (1 −0.7) + (1 −0.4) × (1 −0.5) × 0.7 =0.36 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 74 / 102 .
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P(A2) =P(甲不中)P(乙中)P(丙中) + P(甲中)P(乙不中)P(丙中) + P(甲中)P(乙中)P(丙不中) =(1 −0.4) × 0.5 × 0.7 + 0.4 × (1 −0.5) × 0.7 + 0.4 × 0.5 × (1 −0.7) =0.41 P(A3) =P(甲中)P(乙中)P(丙中) =0.4 × 0.5 × 0.7 = 0.14 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 75 / 102 .
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P(敌机坠毁) = P(B) = 3 ∑ i=0 P(Ai)P(B|Ai) =0.09 × 0 + 0.36 × 0.2 + 0.41 × 0.6 + 0.14 × 1 =0.458 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 76 / 102 .
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例6.3（赌徒输光问题） 例6.3 设甲有赌本M 元，乙有赌本N 元（M, N 是正整数） 。 每一局输赢为1 元，没有和局。 每局甲胜概率为p(0 < p < 1)。 问：甲输光的概率。 解记L = M + N, L ≥2。当L = 2 时M = N = 1, 甲输光概率为 1 −p (若第一局甲赢，则乙输光，赌局不能继续)。 只考虑L ≥3 的情形。 问题扩充为：若甲乙共有赌本L 元，甲有赌本i 元，乙有赌本L −i 元，则甲输光的概率pi 是多少？ （原问题求pM） 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 77 / 102 .
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记Ai =“甲有赌本i 元而最后输光” （i = 1, 2, . . . , L −1）, B =“甲赢了 第一局” 。 记q = 1 −p。 当2 ≤i ≤L −2 时，得递推公式 pi =P(B)P(Ai|B) + P(¯ B)P(Ai|¯ B) =ppi+1 + qpi−1 (6.3) p1 =pp2 + q pL−1 =p × 0 + qpL−2 记p0 = 1, pL = 0, 则(6.3) 对1 ≤i ≤L −1 成立。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 78 / 102 .
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由（6.3） pi =ppi + qpi = ppi+1 + qpi−1 p(pi+1 −pi) =q(pi −pi−1) (pi+1 −pi) =q p(pi −pi−1) = . . . . . .
= (q p )i (p1 −1) =ri(p1 −1), (i = 1, 2, . . . , L −1, 记r = q p) pi+1 −p1 = i ∑ k=1 (pk+1 −pk) = i ∑ k=1 rk(p1 −1) (6.4) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 79 / 102 .
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pi+1 −p1 = r −ri+1 1 −r (p1 −1) p ̸= 1 2 i(p1 −1) p = 1 2 i = L −1 时pi+1 = pL = 0, p1 = r −rL 1 −r (1 −p1) p ̸= 1 2 i(1 −p1) p = 1 2 = r −rL 1 −rL p ̸= 1 2 1 −1 L p = 1 2 (6.5) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 80 / 102 .
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由（6.4） ，p ̸= 1 2 时 pi =p1 + r −ri 1 −r (p1 −1) =ri −rL 1 −rL (2 ≤i ≤L −1) p = 1 2 时 pi =p1 + (i −1)(p1 −1) =1 −1 L + (i −1) ( −1 L ) =1 −i L (2 ≤i ≤L −1) 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 81 / 102 .
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甲输光的概率 pM = { rM−rM+N 1−rM+N p ̸= 1 2 N M+N p = 1 2 关键是根据第一局的输赢结果建立方程（6.3） 。叫做“首步（首局） 分析法” 。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 82 / 102 .
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例6.4（敏感问题的调查） 例6.4 在问卷调查时，某些敏感问题会遭到拒绝回答或谎报。 比如，要问卷调查运动员是否曾服用兴奋剂，直接问很难得到肯定 回答。 设计如下两个问题： 问题A：你的生日是否在7 月1 日之前（不含7 月1 日） ？ 问题B：你是否服用过兴奋剂？ 被调查者只需要回答其中一个问题，只需在只有“是”、 “否”的答卷上 选择其一。而回答哪一个是根据被调查者在其他人不能知道的情况 下随机抽取一个颜色决定的。 若抽出白球，则回答问题A；若抽出红球，则回答问题B。红球比 例π 已知。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 83 / 102 .
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样本量较大（如200 个受调查者）就可以统计, 估计服用兴奋剂比 例p。 设n 张答卷，k 张答“是” ，答“是”的比例φ = k/n。 全概公式： P(回答“是”) =P(抽到白球)P(生日在7 月1 日前|抽到白球) + P(抽到红球)P(服用过兴奋剂|抽到红球) =0.5(1 −π) + pπ ≈k n p ≈ k n −1−π 2 π 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 84 / 102 .
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例如，50 个球中有30 个红球，π = 0.6。 某国15 个项目n = 246 个运动员接受调查, 答“是”者k = 54。 p ≈0.0325 即约3.25% 服用兴奋剂。 思考：比例π 的选取有何影响？ 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 85 / 102 .
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例6.5（发报与接收） 例6.5（发报与接收）发报台分别以概率0.6 和0.4 发出信号“ · ”和 “−” 。 信号可能误码。正确接收与错误接收的概率如下表： 接收 · − 发出 · 0.8 0.2 − 0.1 0.9 求当收到信号“ · ” ，发报台真的发出“ · ”的概率。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 86 / 102 .
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记A =“发出信号‘·’” ，B =“收到信号‘·’” 。要求P(A|B)。 P(A|B) =P(AB) P(B) P(AB) =P(A)P(B|A) = 0.6 × 0.8 P(B) =P(A)P(B|A) + P(¯ A)P(B|¯ A) P(¯ A) =0.4 P(B|¯ A) =0.1 P(A|B) = P(A)P(B|A) P(A)P(B|A) + P(¯ A)P(B|¯ A) = 0.6 × 0.8 0.6 × 0.8 + 0.4 × 0.1 = 0.923 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 87 / 102 .
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逆概公式 逆概公式是全概公式的引申。 若有多个基本情况（事件）是完备事件组，则观测到一个结果后， 可以逆推原来到底是哪一个情况。 如：接收到“ · ”后，可以知道原来发出的是“ · ”的概率为0.923，即基 本可以判断原来是发出的“ · ” 。 定理6.2（逆概公式）设A1, A2, . . . , An 为一完备事件组，则对任一 事件B(P(B) ̸= 0) 有 P(Aj|B) = P(Aj)P(B|Aj) ∑n i=1 P(Ai)P(B|Ai) (6.6) 证明联合条件概率定义及全概公式。 逆概公式也称为贝叶斯（Bayes）公式。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 88 / 102 .
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例6.6（艾滋病检测） 例6.6（艾滋病检测）美国的艾滋病人比例保守估计1000 分之一。 是否应该对新婚夫妇进行艾滋病毒血液检测？ 血液检测方法各种结果及概率 检验结果 报告阳性 报告阴性 实际 AIDS 真阳性(0.95) 假阴性(0.05) 情况 非AIDS 假阳性(0.01) 真阴性(0.99) 如果报告阳性，则真正患病概率是多少？ A =“受试者带有艾滋病毒” ，T =“检测结果呈阳性” 。 求P(A|T)。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 89 / 102 .
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P(A) = 0.001, P(T|A) = 0.95, P(T|¯ A) = 0.01。 由逆概公式 P(A|T) = P(A)P(T|A) P(A)P(T|A) + P(¯ A)P(T|¯ A) = 0.001 × 0.95 0.001 × 0.95 + 0.999 × 0.01 = 0.087 即使检验报告阳性，真的患病的概率也只有8.7%，所以全面的检验 不太必要。 原因是P(A) 太小了。 P(A|T) = 0.95P(A) 0.95P(A) + 0.01(1 −P(A)) = 0.95 0.94 + 0.01 1 P(A) 是P(A) 的严格增函数。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 90 / 102 .
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本节目录 1 随机事件及其概率 2 古典概型 3 事件的运算及概率的加法公式 4 集合与事件、概率的公理化定义 5 条件概率、乘法公式、独立性 6 全概公式与逆概公式 7 独立试验序列概型 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 91 / 102 .
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独立试验序列概型：例7.1 例7.1 独立重复掷5 次分币。 求：恰有两次正面朝上的概率。 解古典概型。共有25 = 32 个等概基本事件。 其中恰有两次正面朝上的个数为C2 5 = 10。 p = 10 32。 p = C2 5 (1 2 )2 (1 2 )3 (7.1) （7.1）中，C2 5 是事件对应的等概基本事件个数，每个基本事件的概 率为 ( 1 2 )2 ( 1 2 )3。(7.1) 可以用加法公式说明。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 92 / 102 .
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若分币不均匀，每一次“正面朝上”概率为2 3, 则 P(恰有两次正面朝上) = C2 5 (2 3 )2 (1 3 )3 这已经不是古典概型，因为基本事件不等概。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 93 / 102 .
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例7.2 例7.2 某人打靶，命中率为0.7，独立重复射击5 次。 求恰好命中2 次的概率。 记p = 0.7, q = 1 −p = 0.3。 P(恰好命中2 次) = C2 5p2q3。 P(恰好命中3 次) = C3 5p3q2。 P(恰好命中4 次) = C4 5p4q1。 P(恰好命中5 次) = C5 5p5q0。 P(恰好命中1 次) = C1 5p1q4。 P(恰好命中0 次) = C0 5p0q5。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 94 / 102 .
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独立试验序列概型 定理（独立试验序列概型）设单次试验中，事件A 发生的概率为 p(0 < p < 1), 则在n 次重复试验中， P(A发生k 次) =Ck npkqn−k (q = 1 −p) (k = 0, 1, 2, . . . , n) 证明在n 次重复试验中，记B1, B2, . . . , Bm 为构成事件“A 发生k 次”的那些试验结果。 （1） “A 发生k 次”= B1 ∪B2 ∪· · · ∪Bm, 互不相容； （2）P(B1) = P(B2) = · · · = P(Bm) = pkqn−k; （3）m = Ck n（从n 次试验中选取k 个成功试验的方法数） 。 于是用加法公式（1）证明定理结论。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 95 / 102 .
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注意“重复”蕴含两重含义： （1）每次试验的条件相同，从而事件A 发生的概率（称为成功概率） 不变； （2）各次试验的结果独立。 当然，这只是理想化假设，实际情况只要比较近似满足就可以了。 反例: 已知80 个产品中有5 个次品，从中每次任取一个，无放回 地取20 次，求其中有2 个次品的概率。 这个例子： （1）每次抽取的试验条件不同，不能直接认为每次的成 功概率（这里是“取到次品”的事件概率）不变； （2）前后的抽取结果不是独立的。如果前5 次抽取到的都是次品， 则从第6 次起只能抽取到正品。 所以不适用独立试验序列概型。 产品批量特别大时， “无放回”抽取与“有放回”抽取结果相似，可以用 独立试验序列概型近似计算无放回抽样的概率。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 96 / 102 .
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例7.3 例7.3 设每次射击打中目标的概率等于0.001。如果射击5000 次， 求至少两次打中目标的概率。 p = 0.001, q = 0.999。 P(至少两次打中目标) = 5000 ∑ k=2 P(恰有k 次打中目标) =1 −P(都不中) −p(仅中一次) =1 −q5000 −5000 × pq4999 ≈1 −0.006721 −0.03364 ≈0.9596 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 97 / 102 .
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成功k 次概率近似公式一 当n 很大同时p 很小的时候，有近似公式 P(A 发生k 次) ≈(np)k k!
e−np (7.2) 称为泊松分布近似，见§2.2(P55)。 如， P(都不中) ≈e−5000×0.001 ≈0.006738 P(仅中一次) ≈5000 × 0.001 1!
e−5000×0.001 ≈0.03369 P(至少两次打中) ≈1 −0.006738 −0.03369 ≈0.9596 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 98 / 102 .
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成功k 次概率近似公式二 当n 很大但p 不是很小时，有第二近似公式 P(A 发生k 次) ≈ 1 √ np(1 −p) · 1 √ 2πe−1 2 x2 k xk = k −np √ np(1 −p) 参见§4.6(P153) 的中心极限定理。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 99 / 102 .
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例7.4 例7.4 设每次射击打中目标概率为1 6。如果射击6000 次，问：击中 次数在900 到1100 之间的概率？ 需要使用§4.6 的中心极限定理。记n = 6000, p = 1 6, P(击中次数在900 到1100 之间) =P(900 −0.5 < X < 1100 + 0.5) =P ( 900 −0.5 −np √ np(1 −p) < X −np √ np(1 −p) < 1100 + 0.5 −np √ np(1 −p) ) =Φ ( 1100 + 0.5 −np √ np(1 −p) ) −Φ ( 900 −0.5 −np √ np(1 −p) ) ≈0.99950 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 100 / 102 .
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例7.5（自由随机游动） 例7.5（自由随机游动）设一质点在数轴上运动，在时刻0 从原点 出发。 每隔单位时间位置向右或向左移动一个单位，向右移动概率 p(0 < p < 1), 向左移动概率q = 1 −p。 问：质点在时刻n 位于K 的概率？ （n 是正整数，K 是整数） 只考虑K 为正整数情况，负整数和零的情况类似。 为了质点在时刻n 位于K, 必须且只需在前n 步移动中向右移动的 次数比向左移动的次数多K。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 101 / 102 .
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设x 表示向右移动的次数，y 表示向左移动的次数，则 x + y = n, x −y = K x = n+K 2 。 因x, n, K 都是整数所以n 与K 有相同的奇偶性。当n 与K 奇偶性相 反时概率为0。 P(质点在时刻n 位于K) =P(质点在头n 次游动时有n + K 2 次向右，有n −K 2 次向左) =C n+K 2 n p n+K 2 q n−K 2 易见K ≤0 时也成立。 原著：陈家鼎、刘婉如、汪仁官制作：李东风，邓明华 概率统计B 第一章随机事件与概率 2025 春季学期 102 / 102 |
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Discrete Calculus on Graphs
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Text and Mathematica code
Introduction
This is an overview of the discrete differential calculus on graphs with an emphasis on the usage of Mathematica to perform related calculations. While the calculus is more generic and can also be applied to generic simplices, it’s more complex to program since it involves concepts which are not directly deduced from the algebraic properties of graphs (e.g. orientation). We, hence, focus mainly on graphs and the usage of matrix methods.
Much of what is presented here is described in many ways and within various application domains. It’s probably new however to see it presented within Mathematica. This offers tremendous possibilities due to the calculational power and in the ways the various knowledge domains of Mathematica can be combined.
Creating graphs
Mathematica can easily output a wide spectrum of different graph types and offers lots of ways to customize and manipulate graphs. We will use in what follows a collection of graphs to highlight the discrete calculus and the examples below describe the various commands and notations.
Note: each of the gothic symbols L, T, G… can be executed and really are just aliases to Mathematica functions.
Example: a small, random directed graph.
The function R[ ] generates a random Bernoulli graph with five vertices and probability 0.5.
You can change the default to produce something larger, e.g. R[12,0.3] produces a random graph with 12 vertices and probability of 0.3 for the edges;
Example: the cycle graph.
The gothic O will denote the cycle graph of ten vertices.
Example: the grid graph
The gothic G will denote the grid graph of five colums and three rows.
Example: the k-ary tree
The gothic T will denote the 3-ary tree of three levels.
Example: the linear graph of ten vertices.
The gothic L will denote the linear chain of ten vertices (9 edges).
Example: the complete graph
The gothic K will denote the complete graph over five vertices.
Matrices
The notation or A[g] will denote incidence matrix of the graph g. Note that our incidence matrix is the edge-vertex matrix and not the vertex-edge matrix of Mathematica (i.e. the transposed).
Example: visualizing the incidence matrix
g=R[];
array=A[g];
Row[{g,MatrixForm[array],ArrayPlot[array]}]
Note that the incidence matrix is a sparse array and that if you wish to display it in a standard list you need to convert it using Normal or using MatrixForm for a standard matrix.
It’s important to note that the incidence matrix is not rectangular in general and acts more as a conversion operator from vertices to edges.
The notation or W[g] will denote the adjacency matrix of the graph g. The Mathematica implementation takes the directed edges into account but for our purposes the adjacency matrix needs to be symmetric.
Example: the adjacency matrix of a tree
Row[{MatrixForm[W[T]], ArrayPlot[W[T], ImagePadding→12]}]
The vertex degrees casted in a diagonal matrix will be denoted by or D[g].
Example: the vertex degree matrix of a random graph
ArrayPlot[D[T],ColorRules->{1→Green,0→White,3→Orange, 4→Red}]
The face-edge matrix B representing the incidence of edges with a face depends on the orientation one applies by hand on a face, i.e. a graph or simplex does not have a natural orientation induced by the edges. We assign an orientation to the T, G,… graphs based on the right-hand rule with the thumb point outwards the plane in which the graph resides.
Example: B face-edge matrix
The graph has one face
and, hence, has a B matrix of one row
B = {{0,-1,1,-1,1,0}}
Note that the order of the B entries corresponds to the order of the edges defining the graph, which in this case is
{12,23, 24,35,45,56}
As a comparison, the is similar but has face edges all pointing in the opposite direction as the face orientation
leading to
B = {{0,-1,-1,-1,-1,0}}
Chains and cochains
A 0-chain is vector defined on the vertices of the graph and 1-chain is a vector defined on the edges of the graph. Note that this is not the same as a function on the graph elements, it’s simply an algebra over the graph elements. A function on the vertices is a 0-cochain and is also a vector but it exists in a different space.
We will use
• σ1, σ2… to denote the 0-chains or vertices
• τ1, τ2… to denote the 1-chains or edges
• τ to denote the whole edge space (the list of all edges)
• σ to denote the whole vertex space (the list of vertices)
For a given graph the makeChainBasis can be used to create the 0 and 1-chain basis elements. One can define higher order elments, like faces (2-chain) and volumes (3-chain) and we will use ocassionally the symbol φ1, φ2… to denote faces.
Example: chains
The 0-chains in the graph are generated (as a vector space) by the six basis elements {σ1, σ2, σ3, σ4, σ5, σ6} and the 1-chains by {τ1, τ2, τ3, τ4, τ5, τ6}.
The 0-chain
η = 2 σ1 + 4 σ3 + σ5
corresponds to
{2,0,4,0,1,0}
in the order of the vertices that the graph was defined.
The 1-chain
ζ = 3 τ3 + 7 τ6
corresponds to
{0,0,3,0,0,7}
in the order of the edges that the graph was defined. The linear combination of all vertices σ and the linear combination of all edges τ can be considered as the whole vertex space and edge space respectively.
Dual to the notion of chain are cochains which can be considered as discrete functions on the discrete elements. A 0-cochain is essentially a function on the vertices, a 1-cochain a function on the edges and so on.
We will use
• s1, s2… to denote the 0-cochains or vertex function
• t1, t2… to denote the 1-cochains or edge functions
To generate a symbolic (vertex and/or edge) function on a given graph the makeVertexFunction and makeEdgeFunction can be used.
Example: creating symbolic graph functions
A vertex function f has five entries
{f1,f2,f3,f4,f5}
and can be created by means of
Similarly, an edge function has four components
{g12,g23,g24,g45}
create using
Inner product, metric and volume
The inner product defines a pairing between chains and cochains. Conversely, one can look at cochains as duals of the chains. On the zero-th level this means that if η and ζ are chains
< ζ | η > = ∈ R or C
The inner product of the basis elements is defined through a metric
= .
On the edge level a similar inner product can be defined with a different metric. So, contrary to the continuous case, one can have a hierarchy of metrics and inner products (of different dimensions since the vertex count is not necessarily the same as the edge count).
We will denote by G the metric without specifying the level, the context making clear whether we deal with the metric on a vertex, edge or face level.
The metric can both be seen as a coupling factor in the inner product and as a weight on the vertices and edges. Especially on the edge level it’s interesting that the metric can be taken from the weight of the weighted graph. This means that the metric is usually taken as to be a diagonal matrix with orthogonality; only edge-edge and vertex-vertex coupling are considered.
Much like the continuous calculus you can then go back and forth between chains and cochains of the same level, i.e. for a given edge τ we define the cochain c
c = G τ
and conversely, with a cochain f we associate a chain π
π =
This implies that that cochains have an inner product defined as
< f | g > =
Example: metric and weighted graphs
Taking a random graph with random weights assigned to the edges we can form the edge-metric by means of the edgeMetric function, giving for instance the following
g = RandomGraph[{7,7},graphDesign,EdgeWeight→RandomInteger[100,7]]
edgeMetric[g]//MatrixForm
The vertex metric can be created similarly with the vertexMetric function.
Because the metric is diagonal and the duality of chains and cochains we can define the inner product of a cochain f with a chain π as follows
< f | π > =
which can also be seen as the inner product of two vectors since the chain and cochain spaces are finite.
Finally, the volume of a chain is defined as
Vol = G. 1
which corresponds in the continuous case to the Hodge dual of one.So, if an integraion of a 1-cochain over an unweighted graph corresponds to
while a weighted graphs yields
Note that if the metric is normalized to Tr(G) = 1 this amounts to the expectation value of f if the metric is seen as a probabilistic distribution.
Example: edge and vertex volume of a random graph
g = RandomGraph[{7,7},graphDesign,VertexWeight→RandomInteger[100,7]];
edgeVolume[g]
vertexVolume[g]
would give something like
{30,42,35,92,16,73,83}
{1,1,1,1,1,1,1}
Integral, boundary and differential operators
We define the integral of a cochain c over a chain η to be equal to the inner product of c and so that the integral of a cochain c over a chain η is
= < c | η >
The boundary operator is defined as
∂ = for a 1-chain
∂ = for a 1-chain
which are in fact the only way that the respective matrices can be used to transform one type of vector to another (of a different dimension in general).
Example: the boundary operator
Let g be the graph and recall that the B matrix for this graph is
{0,-1,1,0,0,0}
while the A matrix is
Because there is only one face it means that the boundary of this face is the same as B considered as a vector rather than a matrix. This indeed corresponds to the four edges with a sign indicating whether it’s traversed in the same or opposite direction.
Taking a 1-chain η = τ1+τ2+τ4 = {1, 1, 0, 1, 0, 0} the boundary is
∂η = {-1,0,0,0,1,0}
which corresponds to σ5 – σ1 and are indeed the endpoints of the line (sub)graph between vertices 1 and 5.
If we take the integral of a vertex function over the boundary of the full linear chain L we get the difference between the endpoints;
This is in essence the fundamental theorem of calculus in the graph context. A consequence of this is that any loop integral is zero. Of course, the general case is not always comparable to the coninuous calculus.
Example: integral over the grid graph
Since the cycle graph has no boundary, it’s obvious that the integral always gives zero. The discrete calculus is more general than just graphs and embraces simplices in general. If we would includes 2-chains or higher order elements we could show that in general the boundary of a boundary is zero. Mathematica does not have a simple function to output higher order boundary operators (like the IncidenceMatrix) and it would also require the concept of orientation.
The coboundary (derivative) d of a cochain is defined as the transpose of the boundary operator
∂ = for a 1-chain
∂ = for a 1-chain
which is again the only way one can, in fact, turn a vector from one space into another taking the underlying topology of the graph into account.
From a strict mathematical point of view the coboundary operator would be introduced in analogy to the continuous case:
< ∂σ | f > = < σ | df >
which is obviously true in this case since the respective matrices are transpose of each other.
Example: integral of a differential over L
Taking the linear L this leads to the usual difference
g=L;
makeChainBasis[g];
makeVertexFunction[“f”];
d f = {-f1+f2,-f2+f3,-f3+f4,-f4+f5,-f5+f6,-f6+f7,-f7+f8,-f8+f9,f10-f9}
and taking the integral of the linear chain gives
= -f1+f10
So, the integral over the linear chain of the differential of the function f is the difference between the endpoints, as should be.
Of course, this is by construction true for any graph and not just the linear chain. So, we have Stoke’s theorem for the discrete calculus
for arbitrary graphs. The extension to higher order chains and cochains is easily done theoretically but more difficult to implement in Mathematica, as said earlier, due to the need for orientations on a simplex or chain.
The differential of the 0-chains is not always a basis of the 1-chains, as is the case in the continuous calculus. For example, the graph has dσ1 + dσ2 + dσ3 = 0.This is also apparent from the fact that the incidence matrix as a rank one eigenspace. It can be easily shown that the egenspace has rank (m-n+1) with m edges, n vertices. So in this case we have effectively a cycle (when disgarding the direction of the edges).
The boundary and coboundary operators are nilpotent as should be
and
corresponding to
.
Div, Grad and Laplacian
The gradient of a scalar (0-cochain) u is defined as
∇ u ⇔ A u
which again is the only meaningful way in which one vector space element can be transformed to another. Also note that this corresponds to the differential d u which is identical in the continuous context.The divergence of a vector (1-cochain) v is similarly defined as
∇.v ⇔
Taking the divergence of a gradient leads to the Laplacian and is hence equal to
Δ ⇔
Example: diffusion on the triangle graph
The discrete diffusion equation on a graph with discrete time can be written as
u(t+1) = u(t) – α Δu(t)
where α acts as a diffusing coefficient from a vertex to its adjacent vertices. This can be taken as the edge weights or a global constant. If we take for example the following random initial state on the triangle graph
we get an equillibrium state after an initial oscillatory transition
but much depends on the initial state and the diffusion coefficient used.
It can be shown that the Laplacian satisfies
Δ = D – W
and hence that a stable state means that (
=
meaning that the value at a vertex is the average of the values of its neighbors.
Non-commutative differential algebra
Diifferential structures on discrete sets are never far off from non-commutative algebras and intriguing relationships with stochastics, KdV and other equations. Contrary to the continuous settings one can have many differential structures and topologies on a discrete ‘manifold’.
The vector space of 1-cochains can be turned into a bi-algebra by means of the multiplication ◦ of a 0-cochains f and g as
(f ◦ = =
( = =
which turns it into a non-commutative algebra with commutator
[dg, f] =
so that
∫ [f,df] =
which is often referred to as the energy of graph. In a more refined analysis one could proof that when one goes to a continuous limit the commutator vanishes and thus recover the de Rham calculus. Symbolically
Note that the Leibniz property and other differential properties are satisfied
d(fg) = df ◦ g + f ◦ dg = A f . g + f . A g
due to the fact that the incidence matrix transfers a vertex function to a difference across the edges.
References
•Discrete Calculus, J.Grady & J.Polimeni, Springer.
• Discrete Exterior Calculus,Hanil Hirani, PhD Thesis.
• Discrete Differential Manifolds and Dynamics on Networks, A.Dimakis, F.Muller-Hoissen & F.Vanderseypen, hep-th/9408114.
June 18, 2014Tags:Mathematica, Mathematics
309 539 Orbifold/wp-content/uploads/2021/04/OrbifoldLogo-300x78.png Orbifold 2014-06-18 07:05:00 2021-11-04 14:49:02 Discrete Calculus on Graphs
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Table Of Contents
List of Math Formulas for Binomial Expansion
Math Formula for Binomial Expansion
Importance of the Binomial Expansion Formula
Tips and Tricks to Memorize the Binomial Expansion Formula
Real-Life Applications of the Binomial Expansion Formula
Common Mistakes and How to Avoid Them While Using the Binomial Expansion Formula
Examples of Problems Using the Binomial Expansion Formula
FAQs on Binomial Expansion Formula
Glossary for Binomial Expansion Formula
Summarize this article:
129 Learners
Last updated on August 9, 2025
Math Formula for Binomial Expansion
In algebra, the binomial expansion formula is used to expand expressions that are raised to a power. It is particularly useful for expanding binomials expressed as (a+b)^n. In this topic, we will learn the formula for binomial expansion and how to apply it.
Math Formula for Binomial ExpansionforUSStudents
List of Math Formulas for Binomial Expansion
The binomial expansion allows us to express a binomial raised to a power in terms of a sum involving terms of the form C(n, k) a(n-k) bk. Let’s learn the formula for binomial expansion.
Math Formula for Binomial Expansion
The binomial expansion formula is a way to expand binomials raised to a power. It is expressed as: (a+b)n = Σ (C(n, k) a(n-k) bk) for k=0 to n, where C(n, k) is the binomial coefficient calculated as n!/(k!(n-k)!).
Importance of the Binomial Expansion Formula
The binomial expansion formula is crucial in algebra and calculus for simplifying expressions and solving problems involving higher powers.
It is used in probability theory, combinatorics, and calculus to simplify and solve problems involving polynomial expressions.
Tips and Tricks to Memorize the Binomial Expansion Formula
Students often find the binomial expansion formula challenging, but with some tips and tricks, it can be mastered.
Remember that the formula involves binomial coefficients and powers of the terms in the binomial.
Practice expanding simple binomials to get familiar, and use Pascal's triangle to determine coefficients easily.
Real-Life Applications of the Binomial Expansion Formula
The binomial expansion formula is used in various fields such as finance, physics, and computer science.
For example, in finance, it helps in calculating compound interest, and in physics, it is used in modeling phenomena where approximation of powers is needed.
Common Mistakes and How to Avoid Them While Using the Binomial Expansion Formula
Students make errors when using the binomial expansion formula. Here are some mistakes and the ways to avoid them, to master the formula.
Mistake 1
Misidentifying the binomial coefficients
Students sometimes incorrectly calculate binomial coefficients. To avoid this error, always use the formula C(n, k) = n!/(k!(n-k)!) and double-check your calculations, or use Pascal’s triangle for quick reference.
Mistake 2
Incorrectly applying powers to terms
When expanding the binomial, students often misapply powers to the terms. Remember that each term in the expansion has a specific power of a and b, following the pattern a(n-k) and bk.
Mistake 3
Ignoring negative signs in binomials
Students sometimes forget to correctly apply negative signs in binomials like (a-b)n. Always remember to apply the negative sign to the appropriate terms in the expansion.
Mistake 4
Confusing the order of terms
Students may confuse the order of terms in the binomial expansion. Ensure each term follows the pattern of coefficients and powers: C(n, k) a(n-k) bk.
Mistake 5
Forgetting to simplify terms
After expanding the binomial, students sometimes forget to simplify the terms. Always combine like terms and simplify the expression for the final result.
Hey!
Examples of Problems Using the Binomial Expansion Formula
Problem 1
Expand (x+2)^3 using the binomial expansion formula.
Okay, lets begin
The expansion is x3 + 6x2 + 12x + 8
Explanation
Using the binomial expansion formula, we have: (x+2)3 = Σ (C(3, k) x(3-k) 2k) for k=0 to 3 = C(3, 0)x3 + C(3, 1)x22 + C(3, 2)x22 + C(3, 3)23 = 1x3 + 3x22 + 3x4 + 18 = x3 + 6x2 + 12x + 8
Well explained 👍
Problem 2
Find the expansion of (a-b)^4 using the binomial expansion formula.
Okay, lets begin
The expansion is a^4 - 4a3b + 6a2b2 - 4ab3 + b4
Explanation
Using the binomial expansion formula, we have: (a-b)4 = Σ (C(4, k) a(4-k) (-b)k) for k=0 to 4 = C(4, 0)a4 + C(4, 1)a3(-b) + C(4, 2)a2(-b)2 + C(4, 3)a(-b)3 + C(4, 4)(-b)4 = 1a4 - 4a3b + 6a2b2 - 4ab3 + 1b4 = a4 - 4a3b + 6a2b2 - 4ab3 + b4
Well explained 👍
Problem 3
Use the binomial expansion to expand (3y+1)^2.
Okay, lets begin
The expansion is 9y2 + 6y + 1
Explanation
Using the binomial expansion formula, we have: (3y+1)2 = Σ (C(2, k) (3y)(2-k) 1k) for k=0 to 2 = C(2, 0)(3y)2 + C(2, 1)(3y)11 + C(2, 2)(1)2 = 1(9y2) + 2(3y) + 1 = 9y2 + 6y + 1
Well explained 👍
Problem 4
Expand (2x-3)^3 using the binomial expansion formula.
Okay, lets begin
The expansion is 8x3 - 36x2 + 54x - 27
Explanation
Using the binomial expansion formula, we have: (2x-3)3 = Σ (C(3, k) (2x)(3-k) (-3)k) for k=0 to 3 = C(3, 0)(2x)3 + C(3, 1)(2x)2(-3) + C(3, 2)(2x)(-3)2 + C(3, 3)(-3)3 = 18x3 - 34x23 + 32x9 - 127 = 8x3 - 36x2 + 54x - 27
Well explained 👍
Problem 5
Find the expansion of (x+y)^5 using the binomial expansion formula.
Okay, lets begin
The expansion is x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Explanation
Using the binomial expansion formula, we have: (x+y)5 = Σ (C(5, k) x(5-k) yk) for k=0 to 5 = C(5, 0)x5 + C(5, 1)x4y + C(5, 2)x3y2 + C(5, 3)x2y3 + C(5, 4)xy4 + C(5, 5)y5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Well explained 👍
FAQs on Binomial Expansion Formula
1.What is the binomial expansion formula?
The binomial expansion formula is used to expand expressions of the form (a+b)^n. It is expressed as: (a+b)^n = Σ (C(n, k) a(n-k) bk) for k=0 to n.
2.How do you calculate binomial coefficients?
Binomial coefficients are calculated using the formula C(n, k) = n!/(k!(n-k)!), where n is the power of the binomial, and k is the specific term in the expansion.
3.What is Pascal's Triangle?
Pascal's Triangle is a triangular array of numbers where each number is the sum of the two directly above it. It is used to quickly find binomial coefficients for the binomial expansion.
4.How can the binomial expansion formula be applied?
The binomial expansion formula is applied by identifying the binomial coefficients and corresponding powers of the terms in the binomial, then constructing the expanded expression.
5.What is the use of the binomial expansion in calculus?
In calculus, the binomial expansion is used for approximating functions and evaluating limits and series, particularly when dealing with expressions involving powers and roots.
Glossary for Binomial Expansion Formula
Binomial: An algebraic expression consisting of two terms, such as (a+b).
Binomial Coefficient: A numerical factor that multiplies the terms in the expansion; given by C(n, k) = n!/(k!(n-k)!).
Pascal's Triangle: A triangular array of numbers that provides binomial coefficients.
Polynomial: An expression consisting of variables and coefficients, involving terms with non-negative integer exponents.
Factorial: The product of all positive integers up to a certain number, denoted n!, used in combinations and permutations.
Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
Fun Fact
: He loves to play the quiz with kids through algebra to make kids love it.
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190199 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9549560/ | Dermlep Study Part 3: Post-RFT Events in Leprosy Patients Presenting to Dermatologists - PMC
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Indian Dermatol Online J
. 2022 May 5;13(3):340–345. doi: 10.4103/idoj.idoj_683_21
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Dermlep Study Part 3: Post-RFT Events in Leprosy Patients Presenting to Dermatologists
P Narasimha Rao
P Narasimha Rao
1 Department of Dermatology, Bhaskar Medical College, Hyderabad, Telangana, India
Find articles by P Narasimha Rao
1, Sujai Suneetha
Sujai Suneetha
1 Institute for Specialized Services in Leprosy (INSSIL), Nireekshana ACET, Hyderabad, Telangana, India
Find articles by Sujai Suneetha
1, Santoshdev P Rathod
Santoshdev P Rathod
2 Professor, Smt. NHL Municipal Medical College, Ahmedabad, Gujarat, India
Find articles by Santoshdev P Rathod
2,✉, Tarun Narang
Tarun Narang
3 Department of Dermatology, Venereology and Leprology, Postgraduate Institute of Medical Education and Research, Chandigarh, India
Find articles by Tarun Narang
3, Sunil Dogra
Sunil Dogra
3 Department of Dermatology, Venereology and Leprology, Postgraduate Institute of Medical Education and Research, Chandigarh, India
Find articles by Sunil Dogra
3, Archana Singal
Archana Singal
4 Director Professor, University College of Medical Sciences and GTB Hospital, Delhi, India
Find articles by Archana Singal
4, Sunilkumar Gupta
Sunilkumar Gupta
5 Department of Dermatology, AIIMS, Gorakhpur, Uttar Pradesh, India
Find articles by Sunilkumar Gupta
5, Rita Vora
Rita Vora
6 Department of Dermatology, Sree Krishna Hospital, Karamsad, Anand, Gujarat, India
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6
Author information
Article notes
Copyright and License information
1 Department of Dermatology, Bhaskar Medical College, Hyderabad, Telangana, India
1 Institute for Specialized Services in Leprosy (INSSIL), Nireekshana ACET, Hyderabad, Telangana, India
2 Professor, Smt. NHL Municipal Medical College, Ahmedabad, Gujarat, India
3 Department of Dermatology, Venereology and Leprology, Postgraduate Institute of Medical Education and Research, Chandigarh, India
4 Director Professor, University College of Medical Sciences and GTB Hospital, Delhi, India
5 Department of Dermatology, AIIMS, Gorakhpur, Uttar Pradesh, India
6 Department of Dermatology, Sree Krishna Hospital, Karamsad, Anand, Gujarat, India
✉
Address for correspondence: Dr. Santoshdev P. Rathod, Room No. 16, Skin OPD, Shardaben General Hospital, Saraspur, Ahmedabad, Gujarat, India. E-mail: Santosh_rathod85@yahoo.com
Received 2021 Nov 15; Revised 2021 Dec 1; Accepted 2021 Dec 11; Collection date 2022 May-Jun.
Copyright: © 2022 Indian Dermatology Online Journal
This is an open access journal, and articles are distributed under the terms of the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License, which allows others to remix, tweak, and build upon the work non-commercially, as long as appropriate credit is given and the new creations are licensed under the identical terms.
PMC Copyright notice
PMCID: PMC9549560 PMID: 36226023
Abstract
Introduction:
Presently the leprosy program has no defined surveillance protocols for patients who complete the fixed duration multidrug therapy and are released from treatment (RFT). Hence, the information about the post-RFT events in these patients is sparse and qualitative and quantitative data regarding their health care requirements is missing. During the DermLep survey carried out by the Indian Association of Dermatologists,Venereologists and Leprologists (IADVL), a number of patients presented to dermatologists during the post RFT period for a variety of symptoms. This paper analyses the events in these patients during the post RFT period.
Results:
Out of a total of 3701 leprosy patients who presented to 201 dermatologists across India during the DermLep survey, 708 (26.2%) were in the post RFT period (488 males; 220 females). Of these, 21% were PB and 79% MB patients as per their treatment records. Majority were in the age group of 31-59 years (55.5%); however, a significant proportion of them (20.7%) were elderly (>60 years). Majority of the patients (45.5%) presented within the first year of RFT with variable symptoms; 28% were between 1-5 years, 5.5% between 5-10 years; and 11.0% presented more than 10 years after RFT. Most common presenting complaint being persistent skin lesions as perceived by patients in 21.2%, followed by neuritis in 14.5%; trophic ulcers in 13.8%; deformities in 67 (11.8%); lepra reactions in 66 (11.6%); and recurrence of original symptoms in 6.7%.
Conclusion:
The DermLep Surve y highlights the importance of ‘post RFT’ patients as an important subset of leprosy patients who visit dermatologists for various health related issues. The most common complaints in this subset were active/persistent skin lesions, lepra reactions and neuritis. In these patients, who are a sub-group of ‘persons affected with leprosy’ the disease related issues can persist for many years post RFT. Hence, it is important to provide services in the programme to monitor and manage these complications for the prevention of impairments, disability and the related social issues.
Keywords:DermLep survey, grade 2 disability, lepra reactions, leprosy, post RFT period, surveillance
Introduction
In the global leprosy program, when a patient completes the required duration of standard multidrug therapy (MDT) regimens, the patient is “released from treatment” (RFT). Nonetheless, the responsibility of the health system towards the leprosy patient does not cease with the completion of MDT as the disabilities and deformities that develop during the active phase of the disease are not always reversible and the patient needs sustained care and surveillance post-RFT. Periodic surveillance of leprosy patients following RFT has three main objectives: the recognition and management of reactions occurring after MDT, differentiating reactions from relapse, and early identification of onset or progression of existing disability to institute appropriate measures to contain and reverse it.
In 1988, the WHO expert committee recommended that paucibacillary (PB) cases should be clinically examined once a year for a minimum of 2 years and that multibacillary (MB) cases be examined both clinically and bacteriologically once a year for a minimum of 5 years. However, 6 years later, in 1994, the WHO rolled back this recommendation citing the negligible risk of relapse after completion of WHO MDT regimens; thus, it was no longer necessary to continue annual surveillance. Instead, it was suggested that at the time of RFT, patients should be taught to recognize early signs of possible relapse and reactions and report for the treatment. This recommendation is being followed by the Indian National Leprosy Program till today.
It has been observed that a number of post-RFT leprosy patients, both PB and MB types, continue to experience varying signs and symptoms of the disease as well as complications, including disabilities for many years, requiring medical or surgical intervention. For these reasons, a considerable number of post-RFT patients with leprosy approach either government or private healthcare facilities. The objective of this survey is to analyze the profile and types of events in post RFT patients, observed and recorded by the dermatologists during the nationwide DermLep survey conducted by IADVL during the year 2017–18 with the premise to comprehend the need for follow up and plan methods to address the specific needs of post-RFT leprosy patients.
Materials and Methods
This study was part of the DermLep study, a nationwide survey carried out to estimate the number and profile of leprosy patients presenting to dermatologists in India and the leprosy services provided to them, the results and observations of which are already published.[3,4] A predesigned questionnaire was provided to dermatologists from all over the country between August 2017 and September 2018. It had 14 questions and was administered to the 3701 leprosy patients seen in their clinics/institutes during the study period. Nationwide, a total of 201 dermatologists took part in the DermLep survey. Ethical approval was obtained for the study, and informed written consent was obtained from all patients. Confidentiality of all patients and participating dermatologists was maintained, and the data was used only for the purpose of this study.
Of all the leprosy patients seen during the survey, details of the post-RFT patients were tabulated separately. The age-sex distribution, classification, presenting complaints, skin smear details, lepra reactions, and grade 2 disability status of all post RFT patients were recorded and analyzed.
Results
Out of the total of 3701 leprosy patients seen in the DermLep survey, 708 (26.2%) were post-RFT patients. Of these, 488 (68.9%) were males and 220 (31.1%) were females. The age of the patients ranged between 10 and 93 years. Most patients (n = 356; 55.5%) belonged to the age group of 31–59 years, followed by 147 (23.2%) in the age group of 16–30 years. Further, 133 (20.7%) were elderly (>60 years), and 4 were children (0.62%) below the age of 15 years.
The time period between RFT and the patient’s visit to the dermatologists is given in Table 1. Most of the patients [322 (45.5%)] presented within the first year after RFT, 185 (26.1%) between 1 and 2 years, 84 (11.9%) between 2 and 5 years, 39 (5.5%) between 5 and 10 years, and 78 (11.0%) more than 10 years after RFT. Based on their case records, the information regarding the type of MDT taken was available for 644/708 patients. It was found that 134 (21%) patients received PB-MDT and 510 (79%) received MB-MDT.
Table 1.
Duration of patients visit to dermatologist after RFT
| Duration of visit after RFT | No. of patients (%) |
:---: |
| <1 year | 322 (45.5%) |
| 1-2 years | 185 (26.1%) |
| 2-5 years | 84 (11.9%) |
| 5-10 years | 39 (5.5%) |
| >10 years | 78 (11.0%) |
| Total | 708 |
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Presenting complaints and reason for the visit to the dermatologist
The presenting symptom and reason for consulting the dermatologist were recorded for 566/708 patients [Figure 1]. Of them, the most frequent reason was active skin/persistent lesions in 120 (21.2%) and together with recurrence of symptoms in 38 (6.7%) was the most common complaint in 158 (27.9%) of the patients. Following closely were lepra reactions or neuritis in 148 (26.1%) patients and ulcers or deformity in 145 (25.6%) patients. This was followed by the “need for reassurance” about the complete regression of the disease or its infectiousness in 115 (20.3%).
Figure 1.
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Presenting complaints post-RFT
Lepra reaction and grade 2 deformity (G2D) in post-RFT patients
The survey included a question on the presence of grade 2 disability (G2D). Of the 684 responses recorded in the post-RFT group, 260 patients (37.9%) had G2D. Lepra reactions were observed in 148 (21.63%) of the patients. When the presence of lepra reaction and G2D was correlated, a high proportion of G2D 105 (70.94%) was found to be present in 225 post-RFT patients with lepra reactions. This was in contrast to only 155 (28.91%) patients with G2D recorded in the 440 patients without reaction. Furthermore, analysis of 260 post-RFT patients with G2D revealed that 96 (36.9%) of them had lepra reactions (T1R in 24 (9.2%) patients and T2R in 72 (27.7%) patients); the type of lepra reaction in a patient with neuritis could not be ascertained.
Registration of post RFT with NLEP
At the time of reporting to the dermatologist, 379 (58.5%) post-RFT patients were found to be registered with the NLEP, whereas 269 (41.5%) were not registered.
Discussion
The National Leprosy Eradication Program, India (NLEP) training manual mentions that the criteria for declaring a patient as cured/released from treatment (RFT) is the completion of 6 and 12-months doses of MDT for PB and MB patients, respectively. A leprosy patient is recorded as RFT in the program registers at the expected month of completion of treatment. It is an important task as the “number of patients RFT during a year” is one of the quality-of-service indicators of the program. Simultaneously, when patients are declared as RFT, their names are deleted from the leprosy registers to denote the “end of the treatment.”
While the patient is on MDT, the review of the “disability status” of patients initially and its follow-up is included in the operational guidelines for DPMR of 2012 by NLEP; however, the follow-up of RFT patients is not a part of the program. All the patients when being declared as RFT by health workers are to be explained that stopping chemotherapy does not mean stopping self-care by the patient. Only if the RFT patient needs treatment (e.g., for ulcers) or physiotherapy, the required services are arranged by the NLEP. With no provision for periodic follow-up of post-RFT patients, there is no reliable data on the issues faced by this important group of leprosy patients who have completed their MDT but are not entirely free of impediments caused by the disease.
The present study has attempted to record the magnitude and profile of problems faced by the post RFT patients attending dermatology clinics to highlight the need for medical attention and care even after the completion of MDT. Of the 708 post-RFT patients, 322 (45.5%) presented to dermatology clinics within the first year of RFT, whereas 507 (71.6%) presented within two years post RFT, which highlights the need for closer monitoring in immediate post-RFT years. The observation that 123 (17.4%) patients presented between 2 and 5 years after RFT and 78 (11.0%) presented more than ten years after RFT provides clear evidence that leprosy patients suffer from long-term disease complications warranting long-term care and support.
Age groups at risk of post-treatment events
The patients in this study included 4 children, but the predominant patient group was of young and middle-aged adults, who form the most productive age groups in the society. It was also observed that a significant proportion (20.7%; n = 133) of the patients were elderly patients above the age of 60 years. Leprosy in the geriatric group has not received much attention in the past. However, with increasing life expectancy in India, we are likely to encounter more leprosy patients in this age group as highlighted by a case report and a recent study. A study published last year from Brazil found that elderly patients have a higher likelihood of MB disease and are at greater risk of developing disabilities. Given the fact that the health system puts lesser emphasis on the issues of the geriatric age group, they carry a higher risk of delay in diagnosis, treatment failure, and becoming a stagnant pool of infection in the community.
Most common presenting complaint in post-RFT patients
The predominant reason for the post RFT patients to consult dermatologists was because the leprosy skin lesions continued to be active and the recurrence of original symptoms (n = 158, 27.9%). This shows that even after the completion of MDT in over 1/4 th of patients, the skin lesions may remain visible, and patients may have residual anesthesia/paresthesia, or these lesions could clinically be still active or in type 1 reaction requiring attention from a specialist healthcare provider to address their concerns. The persistence of visible skin lesions, either active or residual in post-RFT patients, can be a cause for apprehension among them that they are not completely free of the disease. Importantly, facial patches and lesions of leprosy on exposed parts of the body can lead to social stigma and discrimination, prompting patients to seek medical treatment until the complete disappearance of the lesions.
High-risk groups for post-treatment events
Patients of the entire spectrum presented with post-MDT issues, but MB leprosy patients formed 79.2% of them (510/644). MB leprosy patients can include BT and BB patients with more than 5 lesions as well as all BI positive and BL/LL patients. Some of these patients remain smear-positive at the time of RFT and are of concern both to the patients as well as to the health system. High-BI MB patients are identified as patients who are at the highest risk of relapse, T1R and T2R, neuritis, and grade 2 disabilities. With the disbanding of the vertical system and the integration of leprosy into the general health services, there is no clear mechanism to follow-up such MB patients or to perform periodic skin slit smear examinations to address their needs as well as adequately manage their complications. The fact that a large number of post-RFT patients sought the help of dermatologists points to their role and contribution to the healthcare system in our country. Although skin smears have been given up in the program, they continue to be a useful tool in leprosy diagnosis and in monitoring the progress of treatment. This study also brings out the fact that dermatologists rely on SSS for a full evaluation of their patients, with 57.23% of them using it routinely in the management of leprosy. Smears have great prognostic value in post-RFT patients in documenting the fall in BI, possible reactivation of the disease, or bacteriological relapse, as well as in alerting the physician to the possibility of drug resistance.
Leprosy patients with High BI are considered as cases of concern, and issues regarding their management, such as whether they should be treated for longer, for 24 months or until smear negativity, or with an alternate drug regimen is an important debate. Two studies of post-RFT patients carried out in India observed that a proportion of MB patients harbor viable M leprae even after completion of the full 12-month course of MDT.[16,17] This finding is of epidemiological concern as although they are considered “cured” by the health system, they can continue to transmit the disease in the family and community. This debate should be resolved for the long-term benefit of the patients as well as for mitigating leprosy in the country.
Lepra reactions, neuritis, and disabilities in post-RFT patients
Lepra reactions are closely linked to neuritis as both type 1 and type 2 reactions are known risk factors for and often precede neuritis. In the present study, 82 patients (14.5%) presented to the dermatologist because of neuritis and 66 (11.6%) with lepra reactions, together adding up to 26.1%, making it the second most common presenting complaint in the post-RFT period. A study from Brazil documented that T1R occurred in 37.1% of their post-RFT patients, T2R in 18.6%, and neuritis in 13.9% of their patients.
The WHO recognizes that lepra reactions and accompanying neuritis can occur both during treatment and in the post-MDT period. In the present study, 60% (n = 393) of the patients belonged to the borderline group (BT, BB, and BL), which is known to be immunologically unstable and prone to T1R and neuritis. Further, BL and LL patients comprised 319 (48.70%) patients, of whom 68 (21.31%) were high-BI patients, all of whom are at known higher risk for T2R and neuritis.
The strong association of lepra reactions and the development of G2D is highlighted from the data. A higher proportion of those who had lepra reactions developed G2D (45.7%) than those without lepra reactions (35.2%). This points to the importance of identifying reactions and managing them effectively to prevent the development of G2D. Interestingly, a reverse analysis of the 260 patients in the study with G2D revealed that 63.1% had no reaction but still developed G2D. This suggests that in the post-RFT period, patients can insidiously develop impairments. Other studies have also documented the occurrence of late T1R, recurrent and chronic ENL reactions, and neuritis/nerve function impairment (NFI) in the post-RFT period.[21,22]
It is known that in long-standing treated leprosy patients, the persistent bacterial antigen can incite nerve autoantibodies that can perpetuate neuropathy. A study from Brazil observed that the impairment status worsened in 40% of patients 10 years after RFT. Studies have also shown that the cumulative probability of worsening of the physical disability grade of persons treated and declared cured of leprosy increases with time.[25,26] It is noteworthy that multivariate analysis showed no significant difference between PB and MB cases with regard to the rate of disability progression. These observations emphasize the importance of post-RFT surveillance as a tool in the program to prevent development and worsening of disabilities. In addition, specialized tertiary centers need to be strengthened for the care of chronic ulcers, deformities, and to provide reconstructive surgery and occupational therapy as an integral part of the program as “care after cure” for post-RFT patients.
Persons affected by leprosy (PAL) are the best resource to identify their needs and problems and to set priorities to improve planning and management of leprosy services. There are recommendations globally for regional and national community-based organizations to involve PAL through meaningful engagements at all levels of decision-making processes regarding policies and programs that directly concern their lives. It is important to point out here that post-RFT individuals form a significant part of this PAL group. There are various initiatives to involve PAL as key stakeholders in addressing social issues of stigma and discrimination faced by them. However, a gap exists between their involvement in planning social remedial measures and the medical and physical challenges they face after RFT. Addressing only their social needs and ignoring their medical needs would be a disservice to this important group.
Conclusion
Although more than 16 million people affected by leprosy have received MDT and have been declared “cured,” it is estimated that 3–4 million people are living with visible impairments or deformities due to leprosy. The present study highlights the various complications or consequences that can occur in patients who have completed MDT. Some common issues were persisting skin lesions and anesthesia, lepra reactions, neuritis, deformity, and trophic ulcers, and all these need medical attention of high order.
Unfortunately, routine surveillance systems are yet to be put in place by most countries, including India, for post-treatment active monitoring of nerve damage and other disabling complications. The absence of a good “care after MDT” program for such patients will lead to “dehabilitation” and escalating stigma over time.
The WHO global leprosy document for 2021–30 also mentions “functioning post-treatment surveillance systems” as one of the key indicators of progress. It is important that this should be added to the national programs and implemented at all levels of care to ensure all leprosy patients get the required medical attention and care to overcome these post-RFT issues.
Declaration of patient consent
The authors certify that they have obtained all appropriate patient consent forms. In the form, the patient(s) has/have given his/her/their consent for his/her/their images and other clinical information to be reported in the journal. The patients understand that their names and initials will not be published and due efforts will be made to conceal their identity, but anonymity cannot be guaranteed.
Financial support and sponsorship
The Dermlep Study’ was carried out by IADVL Special Interest Group Leprosy as a fully funded research project of IADVL Academy.
Conflicts of interest
There are no conflicts of interest.
Acknowledgements
We sincerely thank all participating dermatologists who made this pan-India study possible.
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