id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-4100 | Identity Element of Natural Number Multiplication is One | Let $1$ be the element one of $\N$.
Then $1$ is the identity element of multiplication:
:$\forall n \in \N: n \times 1 = n = 1 \times n$ | Firstly, by definition of multiplication:
{{begin-eqn}}
{{eqn | l = n \times 1
| r = \paren {n \times 0} + n
}}
{{eqn | r = n
}}
{{end-eqn}}
Next, recall that multiplication is recursively defined as:
:$\forall m, n \in \N: \begin{cases}
m \times 0 & = 0 \\
m \times \paren {n + 1} & = m \times n + m
\end{cases}$
... | Let $1$ be the element [[Definition:One|one]] of $\N$.
Then $1$ is the [[Definition:Identity Element|identity element]] of [[Definition:Natural Number Multiplication|multiplication]]:
:$\forall n \in \N: n \times 1 = n = 1 \times n$ | Firstly, by definition of [[Definition:Natural Number Multiplication|multiplication]]:
{{begin-eqn}}
{{eqn | l = n \times 1
| r = \paren {n \times 0} + n
}}
{{eqn | r = n
}}
{{end-eqn}}
Next, recall that [[Definition:Natural Number Multiplication|multiplication]] is [[Definition:Recursively Defined Mapping|recu... | Identity Element of Natural Number Multiplication is One | https://proofwiki.org/wiki/Identity_Element_of_Natural_Number_Multiplication_is_One | https://proofwiki.org/wiki/Identity_Element_of_Natural_Number_Multiplication_is_One | [
"Examples of Identity Elements",
"Natural Number Multiplication"
] | [
"Definition:One",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Multiplication/Natural Numbers"
] | [
"Definition:Multiplication/Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Recursively Defined Mapping",
"Principle of Recursive Definition",
"Definition:Mapping",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Multiplication/Natural Numbers"
] |
proofwiki-4101 | Secant Secant Theorem | Let $C$ be a point external to a circle $ABED$.
Let $CA$ and $CB$ be straight lines which cut the circle at $D$ and $E$ respectively.
Then:
:$CA \cdot CD = CB \cdot CE$ | :320px
Draw $CF$ tangent to the circle.
From the Tangent Secant Theorem we have that:
:$CF^2 = CA \cdot CD$
:$CF^2 = CB \cdot CE$
from which the result is obvious and immediate.
{{qed}} | Let $C$ be a [[Definition:Point|point]] external to a [[Definition:Circle|circle]] $ABED$.
Let $CA$ and $CB$ be [[Definition:Straight Line|straight lines]] which cut the [[Definition:Circle|circle]] at $D$ and $E$ respectively.
Then:
:$CA \cdot CD = CB \cdot CE$ | :[[File:SecantSecantTheorem.png|320px]]
[[Construction of Tangent from Point to Circle|Draw $CF$ tangent to the circle]].
From the [[Tangent Secant Theorem]] we have that:
:$CF^2 = CA \cdot CD$
:$CF^2 = CB \cdot CE$
from which the result is obvious and immediate.
{{qed}} | Secant Secant Theorem | https://proofwiki.org/wiki/Secant_Secant_Theorem | https://proofwiki.org/wiki/Secant_Secant_Theorem | [
"Secant Secant Theorem",
"Circles",
"Named Theorems"
] | [
"Definition:Point",
"Definition:Circle",
"Definition:Line/Straight Line",
"Definition:Circle"
] | [
"File:SecantSecantTheorem.png",
"Construction of Tangent from Point to Circle",
"Tangent Secant Theorem"
] |
proofwiki-4102 | Converse of Tangent Secant Theorem | Let $D$ be a point outside a circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Let $DB$ intersect the circle at $B$ such that $DB^2 = AD \cdot DC$.
Then $DB$ is tangent to the circle $ABC$.
{{:Euclid:Proposition/III/37}} | :320px
Let $DE$ be drawn tangent to the circle $ABC$.
Let $F$ be the center of $ABC$ and join $FB, FD, FE$.
From Radius at Right Angle to Tangent, $\angle FED$ is a right angle.
We have that $DE$ is tangent to the circle $ABC$ and $DA$ cuts it.
So from the Tangent Secant Theorem $AD \cdot DC = DE^2$.
But we also have {... | Let $D$ be a [[Definition:Point|point]] outside a [[Definition:Circle|circle]] $ABC$.
Let $DA$ be a [[Definition:Straight Line|straight line]] which cuts the [[Definition:Circle|circle]] $ABC$ at $A$ and $C$.
Let $DB$ intersect the circle at $B$ such that $DB^2 = AD \cdot DC$.
Then $DB$ is [[Definition:Tangent to C... | :[[File:Euclid-III-37.png|320px]]
Let $DE$ be drawn [[Definition:Tangent to Circle|tangent]] to the [[Definition:Circle|circle]] $ABC$.
Let $F$ be the [[Definition:Center of Circle|center]] of $ABC$ and join $FB, FD, FE$.
From [[Radius at Right Angle to Tangent]], $\angle FED$ is a [[Definition:Right Angle|right ang... | Converse of Tangent Secant Theorem | https://proofwiki.org/wiki/Converse_of_Tangent_Secant_Theorem | https://proofwiki.org/wiki/Converse_of_Tangent_Secant_Theorem | [
"Tangent Secant Theorem"
] | [
"Definition:Point",
"Definition:Circle",
"Definition:Line/Straight Line",
"Definition:Circle",
"Definition:Tangent Line/Circle",
"Definition:Circle"
] | [
"File:Euclid-III-37.png",
"Definition:Tangent Line/Circle",
"Definition:Circle",
"Definition:Circle/Center",
"Radius at Right Angle to Tangent",
"Definition:Right Angle",
"Definition:Tangent Line/Circle",
"Definition:Circle",
"Tangent Secant Theorem",
"Triangle Side-Side-Side Congruence",
"Defin... |
proofwiki-4103 | Fitting Chord Into Circle | Into a given circle, it is possible to fit a chord equal to a given line segment which is not greater than the diameter of the circle.
{{:Euclid:Proposition/IV/1}} | As $C$ is the center of $EFG$, $CG = CE$ and so $CG = D$.
But $CG$ is a chord of $ABC$.
{{Qed}}
{{Euclid Note|1|IV}} | Into a given [[Definition:Circle|circle]], it is possible to [[Definition:Euclid's Definitions - Book IV/7 - Fitted Into Circle|fit a chord]] equal to a given [[Definition:Line Segment|line segment]] which is not greater than the [[Definition:Diameter of Circle|diameter]] of the [[Definition:Circle|circle]].
{{:Eucli... | As $C$ is the [[Definition:Center of Circle|center]] of $EFG$, $CG = CE$ and so $CG = D$.
But $CG$ is a [[Definition:Chord of Circle|chord]] of $ABC$.
{{Qed}}
{{Euclid Note|1|IV}} | Fitting Chord Into Circle | https://proofwiki.org/wiki/Fitting_Chord_Into_Circle | https://proofwiki.org/wiki/Fitting_Chord_Into_Circle | [
"Circles"
] | [
"Definition:Circle",
"Definition:Euclid's Definitions - Book IV/7 - Fitted Into Circle",
"Definition:Line/Segment",
"Definition:Circle/Diameter",
"Definition:Circle"
] | [
"Definition:Circle/Center",
"Definition:Circle/Chord"
] |
proofwiki-4104 | Construction of Isosceles Triangle whose Base Angle is Twice Apex | It is possible to construct an isosceles triangle such that each of the angles at the base is twice that at the apex.
{{:Euclid:Proposition/IV/10}} | Join $CD$.
Circumscribe circle $ACD$ about $\triangle ACD$.
As $AC = BD$ we have that $AB \cdot BC = BD^2$.
We have that $B$ is outside the circle $ACD$.
From the converse of the Tangent Secant Theorem it follows that $BD$ is tangent to circle $ACD$.
Then from the Tangent-Chord Theorem:
:$\angle BDC = \angle DAC$
Add $... | It is possible to construct an [[Definition:Isosceles Triangle|isosceles triangle]] such that each of the [[Definition:Internal Angle|angles]] at the [[Definition:Base of Isosceles Triangle|base]] is twice that at the [[Definition:Apex of Isosceles Triangle|apex]].
{{:Euclid:Proposition/IV/10}} | Join $CD$.
[[Circumscribing Circle about Triangle|Circumscribe circle $ACD$ about $\triangle ACD$]].
As $AC = BD$ we have that $AB \cdot BC = BD^2$.
We have that $B$ is outside the [[Definition:Circle|circle]] $ACD$.
From the [[Converse of Tangent Secant Theorem|converse of the Tangent Secant Theorem]] it follows t... | Construction of Isosceles Triangle whose Base Angle is Twice Apex | https://proofwiki.org/wiki/Construction_of_Isosceles_Triangle_whose_Base_Angle_is_Twice_Apex | https://proofwiki.org/wiki/Construction_of_Isosceles_Triangle_whose_Base_Angle_is_Twice_Apex | [
"Circles",
"Isosceles Triangles"
] | [
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Polygon/Internal Angle",
"Definition:Triangle (Geometry)/Isosceles/Base",
"Definition:Triangle (Geometry)/Isosceles/Apex"
] | [
"Circumscribing Circle about Triangle",
"Definition:Circle",
"Converse of Tangent Secant Theorem",
"Definition:Tangent Line/Circle",
"Definition:Circle",
"Tangent-Chord Theorem",
"Sum of Angles of Triangle equals Two Right Angles",
"Isosceles Triangle has Two Equal Angles",
"Triangle with Two Equal ... |
proofwiki-4105 | Inscribing Regular Pentagon in Circle | In a given circle, it is possible to inscribe a regular pentagon.
{{:Euclid:Proposition/IV/11}} | We have that $\angle CDA = \angle DCA = 2 \angle CAD$.
As $\angle CDA$ and $ \angle DCA$ have been bisected, $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$.
From Equal Angles in Equal Circles it follows that the arcs $AB, BC, CD, DE, EA$ are all equal.
Hence from Equal Arcs of Circles Subtended by Equ... | In a given [[Definition:Circle|circle]], it is possible to [[Definition:Polygon Inscribed in Circle|inscribe]] a [[Definition:Regular Pentagon|regular pentagon]].
{{:Euclid:Proposition/IV/11}} | We have that $\angle CDA = \angle DCA = 2 \angle CAD$.
As $\angle CDA$ and $ \angle DCA$ have been [[Bisection of Angle|bisected]], $\angle DAC = \angle ACE = \angle ECD = \angle CDB = \angle BDA$.
From [[Equal Angles in Equal Circles]] it follows that the [[Definition:Arc of Circle|arcs]] $AB, BC, CD, DE, EA$ are al... | Inscribing Regular Pentagon in Circle | https://proofwiki.org/wiki/Inscribing_Regular_Pentagon_in_Circle | https://proofwiki.org/wiki/Inscribing_Regular_Pentagon_in_Circle | [
"Circles",
"Pentagons"
] | [
"Definition:Circle",
"Definition:Inscribe/Polygon in Circle",
"Definition:Pentagon/Regular"
] | [
"Bisection of Angle",
"Equal Angles in Equal Circles",
"Definition:Circle/Arc",
"Equal Arcs of Circles Subtended by Equal Straight Lines",
"Definition:Line/Straight Line",
"Definition:Pentagon",
"Definition:Polygon/Equilateral",
"Definition:Circle/Arc",
"Definition:Circle/Arc",
"Definition:Circle/... |
proofwiki-4106 | Circumscribing Regular Pentagon about Circle | About a given circle, it is possible to circumscribe a regular pentagon.
{{:Euclid:Proposition/IV/12}} | Join $FB, FK, FC, FL, FD$.
From Radius at Right Angle to Tangent, $HK, KL$ etc. are perpendicular to the radii they touch.
So $\angle KCF$ and $\angle LCF$ are right angles.
For the same reason, $\angle KBF$ and $\angle LDF$ are right angles.
By Pythagoras's Theorem, $FK^2 = FC^2 + CK^2$
For the same reason, $FK^2 = FB... | About a given [[Definition:Circle|circle]], it is possible to [[Definition:Polygon Circumscribed around Circle|circumscribe]] a [[Definition:Regular Pentagon|regular pentagon]].
{{:Euclid:Proposition/IV/12}} | Join $FB, FK, FC, FL, FD$.
From [[Radius at Right Angle to Tangent]], $HK, KL$ etc. are [[Definition:Perpendicular|perpendicular]] to the [[Definition:Radius of Circle|radii]] they touch.
So $\angle KCF$ and $\angle LCF$ are [[Definition:Right Angle|right angles]].
For the same reason, $\angle KBF$ and $\angle LDF$ ... | Circumscribing Regular Pentagon about Circle | https://proofwiki.org/wiki/Circumscribing_Regular_Pentagon_about_Circle | https://proofwiki.org/wiki/Circumscribing_Regular_Pentagon_about_Circle | [
"Circles",
"Pentagons"
] | [
"Definition:Circle",
"Definition:Circumscribe/Polygon around Circle",
"Definition:Pentagon/Regular"
] | [
"Radius at Right Angle to Tangent",
"Definition:Right Angle/Perpendicular",
"Definition:Circle/Radius",
"Definition:Right Angle",
"Definition:Right Angle",
"Pythagoras's Theorem",
"Triangle Side-Side-Side Congruence",
"Definition:Circle/Arc",
"Definition:Circle/Arc",
"Angles on Equal Arcs are Equa... |
proofwiki-4107 | Inscribing Circle in Regular Pentagon | In any given regular pentagon it is possible to inscribe a circle.
{{:Euclid:Proposition/IV/13}} | Join the straight lines $FB, FA, FE$.
We have that $BC = CD$, $CF$ is common and $\angle BCF = \angle DCF$.
So by Triangle Side-Angle-Side Congruence:
:$\triangle BCF = \triangle DCF$
and so:
:$BF = DF$
Thus $\angle CBF = \angle CDF$.
Since $\angle CDE = 2 \angle CDF$ and $\angle CDE = \angle CBF$, then $\angle CDF = \... | In any given [[Definition:Regular Pentagon|regular pentagon]] it is possible to [[Definition:Circle Inscribed in Polygon|inscribe]] a [[Definition:Circle|circle]].
{{:Euclid:Proposition/IV/13}} | Join the [[Definition:Straight Line|straight lines]] $FB, FA, FE$.
We have that $BC = CD$, $CF$ is common and $\angle BCF = \angle DCF$.
So by [[Triangle Side-Angle-Side Congruence]]:
:$\triangle BCF = \triangle DCF$
and so:
:$BF = DF$
Thus $\angle CBF = \angle CDF$.
Since $\angle CDE = 2 \angle CDF$ and $\angle CD... | Inscribing Circle in Regular Pentagon | https://proofwiki.org/wiki/Inscribing_Circle_in_Regular_Pentagon | https://proofwiki.org/wiki/Inscribing_Circle_in_Regular_Pentagon | [
"Circles",
"Pentagons"
] | [
"Definition:Pentagon/Regular",
"Definition:Inscribe/Circle in Polygon",
"Definition:Circle"
] | [
"Definition:Line/Straight Line",
"Triangle Side-Angle-Side Congruence",
"Bisection of Angle",
"Definition:Line/Straight Line",
"Bisection of Angle",
"Definition:Line/Straight Line",
"Definition:Right Angle/Perpendicular",
"Definition:Right Angle",
"Definition:Subtend",
"Triangle Side-Angle-Angle C... |
proofwiki-4108 | Circumscribing Circle about Regular Pentagon | About any given regular pentagon it is possible to circumscribe a circle.
{{:Euclid:Proposition/IV/14}} | From $F$ join the straight lines $FB, FA, FE$.
In a similar manner to Inscribing Circle in Regular Pentagon it can be shown that $\angle CBA, \angle BAE, \angle AED$ have been bisected by the straight lines $FB, FA, FE$ respectively.
Since $\angle BCD = CDE$ and $2 \angle FCD = \angle BCD$ and $2 \angle CDF = \angle CD... | About any given [[Definition:Regular Pentagon|regular pentagon]] it is possible to [[Definition:Circle Circumscribed around Polygon|circumscribe]] a [[Definition:Circle|circle]].
{{:Euclid:Proposition/IV/14}} | From $F$ join the [[Definition:Straight Line|straight lines]] $FB, FA, FE$.
In a similar manner to [[Inscribing Circle in Regular Pentagon]] it can be shown that $\angle CBA, \angle BAE, \angle AED$ have been [[Definition:Bisect|bisected]] by the [[Definition:Straight Line|straight lines]] $FB, FA, FE$ respectively.
... | Circumscribing Circle about Regular Pentagon | https://proofwiki.org/wiki/Circumscribing_Circle_about_Regular_Pentagon | https://proofwiki.org/wiki/Circumscribing_Circle_about_Regular_Pentagon | [
"Circles",
"Pentagons"
] | [
"Definition:Pentagon/Regular",
"Definition:Circumscribe/Circle around Polygon",
"Definition:Circle"
] | [
"Definition:Line/Straight Line",
"Inscribing Circle in Regular Pentagon",
"Definition:Bisection",
"Definition:Line/Straight Line",
"Triangle with Two Equal Angles is Isosceles",
"Definition:Circle",
"Definition:Circumscribe/Circle around Polygon",
"Definition:Pentagon/Regular"
] |
proofwiki-4109 | Inscribing Regular Hexagon in Circle | In a given circle, it is possible to inscribe a regular hexagon.
{{:Euclid:Proposition/IV/15}} | Since $G$ is the center of circle $ABCDEF$, it follows that $GE = GD$.
Since $D$ is the center of circle $EGCH$, it follows that $DE = GD$.
So $GE = GD = DE$ and so $\triangle EGD$ is equilateral and so equiangular.
By Sum of Angles of Triangle Equals Two Right Angles, $\angle EGD$ is one third of two right angles.
Sim... | In a given [[Definition:Circle|circle]], it is possible to [[Definition:Polygon Inscribed in Circle|inscribe]] a [[Definition:Regular Hexagon|regular hexagon]].
{{:Euclid:Proposition/IV/15}} | Since $G$ is the [[Definition:Center of Circle|center]] of [[Definition:Circle|circle]] $ABCDEF$, it follows that $GE = GD$.
Since $D$ is the [[Definition:Center of Circle|center]] of [[Definition:Circle|circle]] $EGCH$, it follows that $DE = GD$.
So $GE = GD = DE$ and so $\triangle EGD$ is [[Definition:Equilateral T... | Inscribing Regular Hexagon in Circle | https://proofwiki.org/wiki/Inscribing_Regular_Hexagon_in_Circle | https://proofwiki.org/wiki/Inscribing_Regular_Hexagon_in_Circle | [
"Circles",
"Hexagons"
] | [
"Definition:Circle",
"Definition:Inscribe/Polygon in Circle",
"Definition:Hexagon/Regular"
] | [
"Definition:Circle/Center",
"Definition:Circle",
"Definition:Circle/Center",
"Definition:Circle",
"Definition:Triangle (Geometry)/Equilateral",
"Definition:Polygon/Equiangular",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Right Angle",
"Definition:Line/Straight Line",
"Definit... |
proofwiki-4110 | Inscribing Regular 15-gon in Circle | In a given circle, it is possible to inscribe a regular 15-gon.
{{:Euclid:Proposition/IV/16}} | Consider the circumference of circle $ABCD$ as divided into $15$ equal arcs.
Of these, there will be $5$ in the shorter arc $AC$, and $3$ in the shorter arc $AB$.
So there are $2$ in the shorter arc $BC$.
Once this has been bisected by the construction which produces $E$, we see that each of these parts is a copy of th... | In a given [[Definition:Circle|circle]], it is possible to [[Definition:Polygon Inscribed in Circle|inscribe]] a [[Definition:Regular Polygon|regular 15-gon]].
{{:Euclid:Proposition/IV/16}} | Consider the [[Definition:Circumference of Circle|circumference]] of [[Definition:Circle|circle]] $ABCD$ as divided into $15$ equal [[Definition:Arc of Circle|arcs]].
Of these, there will be $5$ in the [[Definition:Arc of Circle|shorter arc]] $AC$, and $3$ in the [[Definition:Arc of Circle|shorter arc]] $AB$.
So ther... | Inscribing Regular 15-gon in Circle | https://proofwiki.org/wiki/Inscribing_Regular_15-gon_in_Circle | https://proofwiki.org/wiki/Inscribing_Regular_15-gon_in_Circle | [
"Inscribing Regular 15-gon in Circle",
"Circles",
"Regular Polygons"
] | [
"Definition:Circle",
"Definition:Inscribe/Polygon in Circle",
"Definition:Polygon/Regular"
] | [
"Definition:Circle/Circumference",
"Definition:Circle",
"Definition:Circle/Arc",
"Definition:Circle/Arc",
"Definition:Circle/Arc",
"Definition:Circle/Arc",
"Bisection of Straight Line",
"Definition:Circle/Arc",
"Definition:Circle/Circumference"
] |
proofwiki-4111 | Multiplication of Numbers is Left Distributive over Addition | {{:Euclid:Proposition/V/1}}
That is, if $m a$, $m b$, $m c$ etc. be any equimultiples of $a$, $b$, $c$ etc., then:
:$m a + m b + m c + \cdots = m \paren {a + b + c + \cdots }$ | Let any number of magnitudes whatever $AB, CD$ be respectively equimultiples of any magnitudes $E, F$ equal in multitude.
Then we are to show that whatever multiple $AB$ is of $E$, then that multiple will $AB + CD$ be of $E + F$.
:520px
Since $AB$ is the same multiple of $E$ that $CD$ is of $F$, as many magnitudes as t... | {{:Euclid:Proposition/V/1}}
That is, if $m a$, $m b$, $m c$ etc. be any [[Definition:Equimultiples|equimultiples]] of $a$, $b$, $c$ etc., then:
:$m a + m b + m c + \cdots = m \paren {a + b + c + \cdots }$ | Let any number of [[Definition:Strictly Positive Real Number|magnitudes]] whatever $AB, CD$ be respectively [[Definition:Equimultiples|equimultiples]] of any [[Definition:Strictly Positive Real Number|magnitudes]] $E, F$ equal in multitude.
Then we are to show that whatever multiple $AB$ is of $E$, then that multiple ... | Multiplication of Numbers is Left Distributive over Addition | https://proofwiki.org/wiki/Multiplication_of_Numbers_is_Left_Distributive_over_Addition | https://proofwiki.org/wiki/Multiplication_of_Numbers_is_Left_Distributive_over_Addition | [
"Multiplication",
"Addition",
"Examples of Distributive Operations",
"Distributive Laws of Arithmetic"
] | [
"Definition:Equimultiples"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Equimultiples",
"Definition:Strictly Positive/Real Number",
"File:Euclid-V-1.png",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/R... |
proofwiki-4112 | Multiplication of Numbers is Right Distributive over Addition | {{:Euclid:Proposition/V/2}}
That is:
:$m a + n a + p a + \cdots = \paren {m + n + p + \cdots} a$ | Let a first magnitude, $AB$, be the same multiple of a second, $C$, that a third, $DE$, is of a fourth, $F$.
Let a fifth, $BG$, be the same multiple of $C$ that a sixth, $EH$, is of $F$.
:380px
We need to show that $AG = AB + BG$ is the same multiple of $C$ that $DH = DE + EH$ is of $F$.
We have that $AB$ is the same m... | {{:Euclid:Proposition/V/2}}
That is:
:$m a + n a + p a + \cdots = \paren {m + n + p + \cdots} a$ | Let a first [[Definition:Strictly Positive Real Number|magnitude]], $AB$, be the same [[Definition:Multiple|multiple]] of a second, $C$, that a third, $DE$, is of a fourth, $F$.
Let a fifth, $BG$, be the same [[Definition:Multiple|multiple]] of $C$ that a sixth, $EH$, is of $F$.
:[[File:Euclid-V-2.png|380px]]
We need... | Multiplication of Numbers is Right Distributive over Addition | https://proofwiki.org/wiki/Multiplication_of_Numbers_is_Right_Distributive_over_Addition | https://proofwiki.org/wiki/Multiplication_of_Numbers_is_Right_Distributive_over_Addition | [
"Multiplication",
"Examples of Distributive Operations",
"Distributive Laws of Arithmetic"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Multiple",
"Definition:Multiple",
"File:Euclid-V-2.png",
"Definition:Multiple",
"Definition:Multiple",
"Definition:Strictly Positive/Real Number"
] |
proofwiki-4113 | Multiples of Terms in Equal Ratios | Let $a, b, c, d$ be quantities.
Let $a : b = c : d$ where $a : b$ denotes the ratio between $a$ and $b$.
Then for any numbers $m$ and $n$:
:$m a : n b = m c : n d$
{{:Euclid:Proposition/V/4}} | Let a first magnitude $A$ have to a second magnitude $B$ the same ratio as a third $C$ to a fourth $D$.
Let equimultiples $E, F$ be taken of $A, C$, and let different equimultiples $G, H$ be taken of $B, D$.
We need to show that $E : G = F : H$.
:600px
Let equimultiples $K, L$ be taken of $E, F$ and other arbitrary equ... | Let $a, b, c, d$ be [[Definition:Physical Quantity|quantities]].
Let $a : b = c : d$ where $a : b$ denotes the [[Definition:Ratio|ratio]] between $a$ and $b$.
Then for any [[Definition:Number|numbers]] $m$ and $n$:
:$m a : n b = m c : n d$
{{:Euclid:Proposition/V/4}} | Let a first [[Definition:Strictly Positive Real Number|magnitude]] $A$ have to a second [[Definition:Strictly Positive Real Number|magnitude]] $B$ the same [[Definition:Ratio|ratio]] as a third $C$ to a fourth $D$.
Let [[Definition:Equimultiples|equimultiples]] $E, F$ be taken of $A, C$, and let different [[Definition... | Multiples of Terms in Equal Ratios/Euclid's Proof | https://proofwiki.org/wiki/Multiples_of_Terms_in_Equal_Ratios | https://proofwiki.org/wiki/Multiples_of_Terms_in_Equal_Ratios/Euclid's_Proof | [
"Ratios",
"Multiples of Terms in Equal Ratios"
] | [
"Definition:Physical Quantity",
"Definition:Ratio",
"Definition:Number"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Definition:Equimultiples",
"Definition:Equimultiples",
"File:Euclid-V-4.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Multiple",
"Associative Law of Multiplicati... |
proofwiki-4114 | Pólya-Vinogradov Inequality | Let $p$ be a positive odd prime.
Then:
:$\forall m, n \in \N: \ds \size {\sum_{k \mathop = m}^{m + n} \paren {\frac k p} } < \sqrt p \, \ln p$
where $\paren {\dfrac k p}$ is the Legendre symbol. | Start with the following manipulations:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = m}^{m + n} \paren {\dfrac k p}
| r = \frac 1 p \sum_{k \mathop = 0}^{p - 1} \sum_{x \mathop = m}^{m + n} \sum_{a \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{2 \pi i a \paren {x - k} / p}
| c =
}}
{{eqn | r = \frac 1 p \sum_... | Let $p$ be a [[Definition:Positive Integer|positive]] [[Definition:Odd Prime|odd prime]].
Then:
:$\forall m, n \in \N: \ds \size {\sum_{k \mathop = m}^{m + n} \paren {\frac k p} } < \sqrt p \, \ln p$
where $\paren {\dfrac k p}$ is the [[Definition:Legendre Symbol|Legendre symbol]]. | Start with the following manipulations:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = m}^{m + n} \paren {\dfrac k p}
| r = \frac 1 p \sum_{k \mathop = 0}^{p - 1} \sum_{x \mathop = m}^{m + n} \sum_{a \mathop = 0}^{p - 1} \paren {\dfrac k p} e^{2 \pi i a \paren {x - k} / p}
| c =
}}
{{eqn | r = \frac 1 p \sum... | Pólya-Vinogradov Inequality | https://proofwiki.org/wiki/Pólya-Vinogradov_Inequality | https://proofwiki.org/wiki/Pólya-Vinogradov_Inequality | [
"Legendre Symbol"
] | [
"Definition:Positive/Integer",
"Definition:Odd Prime",
"Definition:Legendre Symbol"
] | [
"Definition:Expression",
"Definition:Gauss Sum",
"Definition:Integer",
"Definition:Odd Integer",
"Definition:Function",
"Definition:Decreasing",
"Definition:Limit",
"L'Hôpital's Rule",
"Definition:Real Interval",
"Definition:Increasing"
] |
proofwiki-4115 | Multiplication of Real Numbers is Left Distributive over Subtraction | {{:Euclid:Proposition/V/5}}
That is, for any numbers $a, b$ and for any integer $m$:
:$m a - m b = m \paren {a - b}$ | Let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the part $AE$ subtracted is of the part $CF$ subtracted.
We need to show that the remainder $EB$ is also the same multiple of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:320px
Whatever multiple $AE$ is of $CF$, let $EB$ be made tha... | {{:Euclid:Proposition/V/5}}
That is, for any [[Definition:Number|numbers]] $a, b$ and for any [[Definition:Integer|integer]] $m$:
:$m a - m b = m \paren {a - b}$ | Let the [[Definition:Strictly Positive Real Number|magnitude]] $AB$ be the same [[Definition:Multiple|multiple]] of the [[Definition:Strictly Positive Real Number|magnitude]] $CD$ that the part $AE$ subtracted is of the part $CF$ subtracted.
We need to show that the remainder $EB$ is also the same multiple of the rema... | Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 1 | https://proofwiki.org/wiki/Multiplication_of_Real_Numbers_is_Left_Distributive_over_Subtraction | https://proofwiki.org/wiki/Multiplication_of_Real_Numbers_is_Left_Distributive_over_Subtraction/Proof_1 | [
"Real Multiplication",
"Real Subtraction",
"Examples of Distributive Operations",
"Multiplication of Real Numbers Distributes over Subtraction",
"Multiplication of Real Numbers is Left Distributive over Subtraction"
] | [
"Definition:Number",
"Definition:Integer"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Multiple",
"Definition:Strictly Positive/Real Number",
"File:Euclid-V-5.png",
"Distributive Laws/Arithmetic",
"Definition:Strictly Positive/Real Number"
] |
proofwiki-4116 | Multiplication of Real Numbers is Left Distributive over Subtraction | {{:Euclid:Proposition/V/5}}
That is, for any numbers $a, b$ and for any integer $m$:
:$m a - m b = m \paren {a - b}$ | {{begin-eqn}}
{{eqn | l = m \times \paren {a - b}
| r = m \times \paren {a + \paren {- b} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = m \times a + m \times \paren {- b}
| c = {{Real-number-axiom|D}}
}}
{{eqn | r = m \times a + \paren {- m \times b}
| c = Multiplication by Negative Real Num... | {{:Euclid:Proposition/V/5}}
That is, for any [[Definition:Number|numbers]] $a, b$ and for any [[Definition:Integer|integer]] $m$:
:$m a - m b = m \paren {a - b}$ | {{begin-eqn}}
{{eqn | l = m \times \paren {a - b}
| r = m \times \paren {a + \paren {- b} }
| c = {{Defof|Real Subtraction}}
}}
{{eqn | r = m \times a + m \times \paren {- b}
| c = {{Real-number-axiom|D}}
}}
{{eqn | r = m \times a + \paren {- m \times b}
| c = [[Multiplication by Negative Real N... | Multiplication of Real Numbers is Left Distributive over Subtraction/Proof 2 | https://proofwiki.org/wiki/Multiplication_of_Real_Numbers_is_Left_Distributive_over_Subtraction | https://proofwiki.org/wiki/Multiplication_of_Real_Numbers_is_Left_Distributive_over_Subtraction/Proof_2 | [
"Real Multiplication",
"Real Subtraction",
"Examples of Distributive Operations",
"Multiplication of Real Numbers Distributes over Subtraction",
"Multiplication of Real Numbers is Left Distributive over Subtraction"
] | [
"Definition:Number",
"Definition:Integer"
] | [
"Multiplication by Negative Real Number"
] |
proofwiki-4117 | Ratios of Equal Magnitudes | {{:Euclid:Proposition/V/7}}
That is:
:$a = b \implies a : c = b : c$
:$a = b \implies c : a = c : b$ | Let $A, B$ be equal magnitudes and let $C$ be any other arbitrary magnitude.
We need to show that $A : C = B : C$ and $C : A = C : B$.
:400px
Let equimultiples $D, E$ of $A, B$ be taken, and another arbitrary multiple $F$ of $C$.
We have that $D$ is the same multiple of $A$ that $E$ is of $B$, while $A = B$.
Therefore ... | {{:Euclid:Proposition/V/7}}
That is:
:$a = b \implies a : c = b : c$
:$a = b \implies c : a = c : b$ | Let $A, B$ be equal [[Definition:Strictly Positive Real Number|magnitudes]] and let $C$ be any other arbitrary [[Definition:Strictly Positive Real Number|magnitude]].
We need to show that $A : C = B : C$ and $C : A = C : B$.
:[[File:Euclid-V-7.png|400px]]
Let [[Definition:Equimultiples|equimultiples]] $D, E$ of $A, ... | Ratios of Equal Magnitudes | https://proofwiki.org/wiki/Ratios_of_Equal_Magnitudes | https://proofwiki.org/wiki/Ratios_of_Equal_Magnitudes | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"File:Euclid-V-7.png",
"Definition:Equimultiples",
"Definition:Multiple",
"Definition:Strictly Positive/Real Number",
"Definition:Equimultiples",
"Definition:Multiple",
"Definition:Strictly Positive/Real Number",... |
proofwiki-4118 | Relative Sizes of Ratios on Unequal Magnitudes | {{:Euclid:Proposition/V/8}}
That is:
:$a > b \implies a : c > b : c$
:$a > b \implies c : a < c : b$ | Let $AB, C$ be unequal magnitudes, and let $AB$ be greater.
Let $D$ be another arbitrary magnitude.
We are to show that $AB$ has to $D$ a greater ratio than $C$ has to $D$, and $D$ has to $C$ a greater ratio than it has to $AB$.
We have that $AB > C$, so let $BE = C$.
Then from {{EuclidDefLink|V|4|Existence of Ratio}},... | {{:Euclid:Proposition/V/8}}
That is:
:$a > b \implies a : c > b : c$
:$a > b \implies c : a < c : b$ | Let $AB, C$ be unequal [[Definition:Strictly Positive Real Number|magnitudes]], and let $AB$ be greater.
Let $D$ be another arbitrary [[Definition:Strictly Positive Real Number|magnitude]].
We are to show that $AB$ has to $D$ a greater [[Definition:Ratio|ratio]] than $C$ has to $D$, and $D$ has to $C$ a greater ratio... | Relative Sizes of Ratios on Unequal Magnitudes | https://proofwiki.org/wiki/Relative_Sizes_of_Ratios_on_Unequal_Magnitudes | https://proofwiki.org/wiki/Relative_Sizes_of_Ratios_on_Unequal_Magnitudes | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Definition:Strictly Positive/Real Number",
"File:Euclid-V-8a.png",
"Definition:Multiple",
"Multiplication of Numbers is Left Distributive over Addition",
"Definition:Equimultiples",
"Definiti... |
proofwiki-4119 | Magnitudes with Same Ratios are Equal | {{:Euclid:Proposition/V/9}}
That is:
:$a : c = b : c \implies a = b$
:$c : a = c : b \implies a = b$ | :350px
Let each of $A$ and $B$ have the same ratio to $C$, i.e. $A : C = B : C$.
Suppose $A \ne B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : C \ne B : C$.
But $A : C = B : C$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
Again, let $C$ have the same ratio to each of $A$ and $B$... | {{:Euclid:Proposition/V/9}}
That is:
:$a : c = b : c \implies a = b$
:$c : a = c : b \implies a = b$ | :[[File:Euclid-V-9.png|350px]]
Let each of $A$ and $B$ have the same [[Definition:Ratio|ratio]] to $C$, i.e. $A : C = B : C$.
Suppose $A \ne B$.
Then from [[Relative Sizes of Ratios on Unequal Magnitudes]] $A : C \ne B : C$.
But $A : C = B : C$ so therefore it is not the case that $A \ne B$.
Therefore $A = B$.
A... | Magnitudes with Same Ratios are Equal | https://proofwiki.org/wiki/Magnitudes_with_Same_Ratios_are_Equal | https://proofwiki.org/wiki/Magnitudes_with_Same_Ratios_are_Equal | [
"Ratios"
] | [] | [
"File:Euclid-V-9.png",
"Definition:Ratio",
"Relative Sizes of Ratios on Unequal Magnitudes",
"Definition:Ratio",
"Relative Sizes of Ratios on Unequal Magnitudes"
] |
proofwiki-4120 | Relative Sizes of Magnitudes on Unequal Ratios | {{:Euclid:Proposition/V/10}}
That is:
:$a : c > b : c \implies a > b$
:$c : b > c : a \implies b < a$ | Let $A$ have to $C$ a greater ratio than $B$ has to $C$.
:350px
Suppose $A = B$.
Then from Ratios of Equal Magnitudes $A : C = B : C$.
But by hypothesis $A : C > B : C$, so $A \ne B$.
Suppose $A < B$.
Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $A : C < B : C$.
But by hypothesis $A : C... | {{:Euclid:Proposition/V/10}}
That is:
:$a : c > b : c \implies a > b$
:$c : b > c : a \implies b < a$ | Let $A$ have to $C$ a greater [[Definition:Ratio|ratio]] than $B$ has to $C$.
:[[File:Euclid-V-10.png|350px]]
Suppose $A = B$.
Then from [[Ratios of Equal Magnitudes]] $A : C = B : C$.
But [[Definition:By Hypothesis|by hypothesis]] $A : C > B : C$, so $A \ne B$.
Suppose $A < B$.
Then from [[Relative Sizes of Rati... | Relative Sizes of Magnitudes on Unequal Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Magnitudes_on_Unequal_Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Magnitudes_on_Unequal_Ratios | [
"Ratios"
] | [] | [
"Definition:Ratio",
"File:Euclid-V-10.png",
"Ratios of Equal Magnitudes",
"Definition:By Hypothesis",
"Relative Sizes of Ratios on Unequal Magnitudes",
"Definition:By Hypothesis",
"Definition:Ratio",
"Ratios of Equal Magnitudes",
"Definition:By Hypothesis",
"Relative Sizes of Ratios on Unequal Mag... |
proofwiki-4121 | Equality of Ratios is Transitive | {{:Euclid:Proposition/V/11}}
That is:
:$A : B = C : D, C : D = E : F \implies A : B = E : F$ | Let $A : B = C : D$ and $C : D = E : F$.
:560px
Of $A, C, E$ let equimultiples $G, H, K$ be taken.
Of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.
We have that:
:$A : B = C : D$
:$G, H$ are equimultiples of $A, C$
:$L, M$ are equimultiples of $B, D$
So:
:$G > L \implies H > M$
:$G = L \implies H = M$... | {{:Euclid:Proposition/V/11}}
That is:
:$A : B = C : D, C : D = E : F \implies A : B = E : F$ | Let $A : B = C : D$ and $C : D = E : F$.
:[[File:Euclid-V-11.png|560px]]
Of $A, C, E$ let [[Definition:Equimultiples|equimultiples]] $G, H, K$ be taken.
Of $B, D, F$ let other arbitrary [[Definition:Equimultiples|equimultiples]] $L, M, N$ be taken.
We have that:
:$A : B = C : D$
:$G, H$ are [[Definition:Equimultipl... | Equality of Ratios is Transitive | https://proofwiki.org/wiki/Equality_of_Ratios_is_Transitive | https://proofwiki.org/wiki/Equality_of_Ratios_is_Transitive | [
"Ratios"
] | [] | [
"File:Euclid-V-11.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples"
] |
proofwiki-4122 | Sum of Components of Equal Ratios | {{:Euclid:Proposition/V/12}}
That is:
:$a_1 : b_1 = a_2 : b_2 = a_3 : b_3 = \cdots \implies \left({a_1 + a_2 + a_3 + \cdots}\right) : \left({b_1 + b_2 + b_3 + \cdots}\right)$ | Let any number of magnitudes $A, B, C, D, E, F$ be proportional, so that:
:$A : B = C : D = E : F$
etc.
:450px
Of $A, C, E$ let equimultiples $G, H, K$ be taken, and of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.
We have that $A : B = C : D = E : F$.
Therefore:
:$G > L \implies H > M, K > N$
:$G = L... | {{:Euclid:Proposition/V/12}}
That is:
:$a_1 : b_1 = a_2 : b_2 = a_3 : b_3 = \cdots \implies \left({a_1 + a_2 + a_3 + \cdots}\right) : \left({b_1 + b_2 + b_3 + \cdots}\right)$ | Let any number of [[Definition:Strictly Positive Real Number|magnitudes]] $A, B, C, D, E, F$ be [[Definition:Proportion|proportional]], so that:
:$A : B = C : D = E : F$
etc.
:[[File:Euclid-V-12.png|450px]]
Of $A, C, E$ let [[Definition:Equimultiples|equimultiples]] $G, H, K$ be taken, and of $B, D, F$ let other arbi... | Sum of Components of Equal Ratios | https://proofwiki.org/wiki/Sum_of_Components_of_Equal_Ratios | https://proofwiki.org/wiki/Sum_of_Components_of_Equal_Ratios | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Proportion",
"File:Euclid-V-12.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Multiplication of Numbers is Left Distributive over Addition",
"Definition:Equimultiples",
"Definition:Equimultiples"
] |
proofwiki-4123 | Relative Sizes of Proportional Magnitudes | {{:Euclid:Proposition/V/13}}
That is:
:$a : b = c : d, c : d > e : f \implies a : b > e : f$ | Let a first magnitude $A$ have to a second $B$ the same ratio as a third $C$ to a fourth $D$.
Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.
:450px
We have that $C : D > E : F$.
From {{EuclidDefLink|V|7|Greater Ratio}}, there will be some equimultiples of $C, E$ and other ... | {{:Euclid:Proposition/V/13}}
That is:
:$a : b = c : d, c : d > e : f \implies a : b > e : f$ | Let a first [[Definition:Strictly Positive Real Number|magnitude]] $A$ have to a second $B$ the same [[Definition:Ratio|ratio]] as a third $C$ to a fourth $D$.
Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.
:[[File:Euclid-V-13.png|450px]]
We have that $C : D > E : F$.
... | Relative Sizes of Proportional Magnitudes | https://proofwiki.org/wiki/Relative_Sizes_of_Proportional_Magnitudes | https://proofwiki.org/wiki/Relative_Sizes_of_Proportional_Magnitudes | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"File:Euclid-V-13.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Multiple",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equim... |
proofwiki-4124 | Relative Sizes of Components of Ratios | {{:Euclid:Proposition/V/14}}
That is, if $a : b = c : d$ then:
:$a > c \implies b > d$
:$a = c \implies b = d$
:$a < c \implies b < d$ | Let a first magnitude $A$ have the same ratio to a second $B$ as a third $C$ has to a fourth $D$.
:300px
Let $A > C$.
Then from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.
But $A : B = C : D$.
So from Relative Sizes of Proportional Magnitudes $C : D > C : B$.
But from Relative Sizes of Magnitudes on... | {{:Euclid:Proposition/V/14}}
That is, if $a : b = c : d$ then:
:$a > c \implies b > d$
:$a = c \implies b = d$
:$a < c \implies b < d$ | Let a first [[Definition:Strictly Positive Real Number|magnitude]] $A$ have the same [[Definition:Ratio|ratio]] to a second $B$ as a third $C$ has to a fourth $D$.
:[[File:Euclid-V-14.png|300px]]
Let $A > C$.
Then from [[Relative Sizes of Ratios on Unequal Magnitudes]] $A : B > C : B$.
But $A : B = C : D$.
So from... | Relative Sizes of Components of Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Components_of_Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Components_of_Ratios | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"File:Euclid-V-14.png",
"Relative Sizes of Ratios on Unequal Magnitudes",
"Relative Sizes of Proportional Magnitudes",
"Relative Sizes of Magnitudes on Unequal Ratios"
] |
proofwiki-4125 | Ratio Equals its Multiples | {{:Euclid:Proposition/V/15}}
That is:
:$a : b \implies ma = mb$ | Let $AB$ be the same multiple of $C$ that $DE$ is of $F$.
:300px
So as many magnitudes as there are in $AB$ equal to $C$, so many are there also in $DE$ equal to $F$.
Let $AB$ be divided into the magnitudes $AG, GH, HB$ equal to $C$.
Let $DE$ be divided into the magnitudes $DK, KL, LE$ equal to $F$.
Then the number of ... | {{:Euclid:Proposition/V/15}}
That is:
:$a : b \implies ma = mb$ | Let $AB$ be the same [[Definition:Multiple|multiple]] of $C$ that $DE$ is of $F$.
:[[File:Euclid-V-15.png|300px]]
So as many [[Definition:Strictly Positive Real Number|magnitudes]] as there are in $AB$ equal to $C$, so many are there also in $DE$ equal to $F$.
Let $AB$ be divided into the [[Definition:Strictly Posit... | Ratio Equals its Multiples | https://proofwiki.org/wiki/Ratio_Equals_its_Multiples | https://proofwiki.org/wiki/Ratio_Equals_its_Multiples | [
"Ratios"
] | [] | [
"Definition:Multiple",
"File:Euclid-V-15.png",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Ratios of Equal Magnitudes",
"Su... |
proofwiki-4126 | Proportional Magnitudes are Proportional Alternately | {{:Euclid:Proposition/V/16}}
That is:
:$a : b = c : d \implies a : c = b : d$ | Let $A, B, C, D$ be four proportional magnitudes, so that as $A$ is to $B$, then so is $C$ to $D$.
We need to show that as $A$ is to $C$, then $B$ is to $D$.
:420px
Let equimultiples $E, F$ be taken of $A, B$.
Let other arbitrary equimultiples $G, H$ be taken of $C, D$.
We have that $E$ is the same multiple of $A$ that... | {{:Euclid:Proposition/V/16}}
That is:
:$a : b = c : d \implies a : c = b : d$ | Let $A, B, C, D$ be four [[Definition:Proportion|proportional]] [[Definition:Strictly Positive Real Number|magnitudes]], so that as $A$ is to $B$, then so is $C$ to $D$.
We need to show that as $A$ is to $C$, then $B$ is to $D$.
:[[File:Euclid-V-16.png|420px]]
Let [[Definition:Equimultiples|equimultiples]] $E, F$ be... | Proportional Magnitudes are Proportional Alternately | https://proofwiki.org/wiki/Proportional_Magnitudes_are_Proportional_Alternately | https://proofwiki.org/wiki/Proportional_Magnitudes_are_Proportional_Alternately | [
"Ratios"
] | [] | [
"Definition:Proportion",
"Definition:Strictly Positive/Real Number",
"File:Euclid-V-16.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Multiple",
"Ratio Equals its Multiples",
"Equality of Ratios is Transitive",
"Definition:Equimultiples",
"Ratio Equals its Multiples",
... |
proofwiki-4127 | Magnitudes Proportional Compounded are Proportional Separated | {{:Euclid:Proposition/V/17}}
That is:
:$a : b = c : d \implies \left({a - b}\right) : b = \left({c - d}\right) : d$ | Let $AB, BE, CD, DF$ be magnitudes which are proportional ''componendo'', so that $AB : BE = CD : DF$.
We need to show that they are proportional ''separando'', that is, that $AE : EB = CF : DF$.
:500px
Let equimultiples $GH, HK, LM, MN$ be taken of $AE, EB, CF, FD$.
Let $KO, NP$ be other arbitrary equimultiples of $EB... | {{:Euclid:Proposition/V/17}}
That is:
:$a : b = c : d \implies \left({a - b}\right) : b = \left({c - d}\right) : d$ | Let $AB, BE, CD, DF$ be [[Definition:Strictly Positive Real Number|magnitudes]] which are [[Definition:Proportion|proportional]] ''componendo'', so that $AB : BE = CD : DF$.
We need to show that they are [[Definition:Proportion|proportional]] ''separando'', that is, that $AE : EB = CF : DF$.
:[[File:Euclid-V-17.png|5... | Magnitudes Proportional Compounded are Proportional Separated | https://proofwiki.org/wiki/Magnitudes_Proportional_Compounded_are_Proportional_Separated | https://proofwiki.org/wiki/Magnitudes_Proportional_Compounded_are_Proportional_Separated | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Proportion",
"Definition:Proportion",
"File:Euclid-V-17.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Multiple",
"Multiplication of Numbers is Left Distributive over Addition",
"Multiplication of Numbers is Left Dist... |
proofwiki-4128 | Magnitudes Proportional Separated are Proportional Compounded | {{:Euclid:Proposition/V/18}}
That is:
:$a : b = c : d \implies \paren {a + b} : b = \paren {c + d} : d$ | Let $AE, EB, CF, FD$ be magnitudes which are proportional ''separando'', that is:
:$AE : EB = CF : FD$
We need to show that they are also proportional ''componendo'', that is:
:$AB : BE = CD : FD$
:300px
Suppose $CD : DF \ne AB : BE$.
Then as $AB : BE$ so will $CD$ be to some magnitude less than $DF$ or greater.
Suppos... | {{:Euclid:Proposition/V/18}}
That is:
:$a : b = c : d \implies \paren {a + b} : b = \paren {c + d} : d$ | Let $AE, EB, CF, FD$ be [[Definition:Strictly Positive Real Number|magnitudes]] which are [[Definition:Proportion|proportional]] ''separando'', that is:
:$AE : EB = CF : FD$
We need to show that they are also [[Definition:Proportion|proportional]] ''componendo'', that is:
:$AB : BE = CD : FD$
:[[File:Euclid-V-18.png|... | Magnitudes Proportional Separated are Proportional Compounded | https://proofwiki.org/wiki/Magnitudes_Proportional_Separated_are_Proportional_Compounded | https://proofwiki.org/wiki/Magnitudes_Proportional_Separated_are_Proportional_Compounded | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Proportion",
"Definition:Proportion",
"File:Euclid-V-18.png",
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Definition:Strictly Positive/Real Number",
"Magnitudes Proportional Compounded are Proportional Separated",
"Definiti... |
proofwiki-4129 | Proportional Magnitudes have Proportional Remainders | {{:Euclid:Proposition/V/19}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$. | As the whole $AB$ is to the whole $CD$, so let the part $AE$ subtracted be to the part $CF$ subtracted.
That is:
:$AB : CD = AE : CF$
We need to show that $EB : FD = AB : CD$.
:200px
We have that :$AB : CD = AE : CF$.
So from Proportional Magnitudes are Proportional Alternately we have that $BA : AE = DC : CF$.
From Ma... | {{:Euclid:Proposition/V/19}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the [[Definition:Ratio|ratio]] of $a$ to $b$. | As the whole $AB$ is to the whole $CD$, so let the part $AE$ subtracted be to the part $CF$ subtracted.
That is:
:$AB : CD = AE : CF$
We need to show that $EB : FD = AB : CD$.
:[[File:Euclid-V-19.png|200px]]
We have that :$AB : CD = AE : CF$.
So from [[Proportional Magnitudes are Proportional Alternately]] we have... | Proportional Magnitudes have Proportional Remainders | https://proofwiki.org/wiki/Proportional_Magnitudes_have_Proportional_Remainders | https://proofwiki.org/wiki/Proportional_Magnitudes_have_Proportional_Remainders | [
"Ratios"
] | [
"Definition:Ratio"
] | [
"File:Euclid-V-19.png",
"Proportional Magnitudes are Proportional Alternately",
"Magnitudes Proportional Compounded are Proportional Separated",
"Proportional Magnitudes are Proportional Alternately",
"Definition:By Hypothesis",
"Equality of Ratios is Transitive"
] |
proofwiki-4130 | Relative Sizes of Successive Ratios | {{:Euclid:Proposition/V/20}}
That is, let:
:$a : b = d : e$
:$b : c = e : f$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$ | Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two are in the same ratio:
:$A : B = D : E$
:$B : C = E : F$
and let $A > C$ ''ex aequali''.
We need to show that $D > F$.
Similarly, we need to show that $A = C \implies D = F$ and $A < C \implies D < F$.
:300... | {{:Euclid:Proposition/V/20}}
That is, let:
:$a : b = d : e$
:$b : c = e : f$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$ | Let there be three [[Definition:Strictly Positive Real Number|magnitudes]] $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two are in the same [[Definition:Ratio|ratio]]:
:$A : B = D : E$
:$B : C = E : F$
and let $A > C$ ''[[Definition:Ratio Ex Aequali|ex aequali]]''.
We need to show th... | Relative Sizes of Successive Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Successive_Ratios | https://proofwiki.org/wiki/Relative_Sizes_of_Successive_Ratios | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Definition:Ratio Ex Aequali",
"File:Euclid-V-20.png",
"Relative Sizes of Ratios on Unequal Magnitudes",
"Relative Sizes of Proportional Magnitudes",
"Relative Sizes of Magnitudes on Unequal Ratios"
] |
proofwiki-4131 | Relative Sizes of Elements in Perturbed Proportion | {{:Euclid:Proposition/V/21}}
That is, let:
:$a : b = e : f$
:$b : c = d : e$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$ | Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio.
Let the proportion of them be perturbed, that is:
:$A : B = E : F$
:$B : C = D : E$
Let $A > C$.
Then we need to show that $D > F$.
:350px
We have that $A > C$.
So from Relat... | {{:Euclid:Proposition/V/21}}
That is, let:
:$a : b = e : f$
:$b : c = d : e$
Then:
:$a > c \implies d > f$
:$a = c \implies d = f$
:$a < c \implies d < f$ | Let there be three [[Definition:Strictly Positive Real Number|magnitudes]] $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same [[Definition:Ratio|ratio]].
Let the [[Definition:Proportion|proportion]] of them be [[Definition:Perturbed Proportion|perturbed]], that... | Relative Sizes of Elements in Perturbed Proportion | https://proofwiki.org/wiki/Relative_Sizes_of_Elements_in_Perturbed_Proportion | https://proofwiki.org/wiki/Relative_Sizes_of_Elements_in_Perturbed_Proportion | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Definition:Proportion",
"Definition:Proportion/Perturbed",
"File:Euclid-V-21.png",
"Relative Sizes of Ratios on Unequal Magnitudes",
"Relative Sizes of Proportional Magnitudes",
"Relative Sizes of Magnitudes on Unequal Ratios"
] |
proofwiki-4132 | Equality of Ratios Ex Aequali | {{:Euclid:Proposition/V/22}}
That is, if:
:$a : b = d : e$
:$b : c = e : f$
then:
:$a : c = d : f$ | Let there be any number of magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio, so that:
:$A : B = D : E$
:$B : C = E : F$
Then we need to show that:
:$A : C = D : F$
:550px
Let equimultiples $G, H$ be taken of $A, D$.
Let other arbitrary equimul... | {{:Euclid:Proposition/V/22}}
That is, if:
:$a : b = d : e$
:$b : c = e : f$
then:
:$a : c = d : f$ | Let there be any number of [[Definition:Strictly Positive Real Number|magnitudes]] $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same [[Definition:Ratio|ratio]], so that:
:$A : B = D : E$
:$B : C = E : F$
Then we need to show that:
:$A : C = D : F$
:[[File:Euc... | Equality of Ratios Ex Aequali | https://proofwiki.org/wiki/Equality_of_Ratios_Ex_Aequali | https://proofwiki.org/wiki/Equality_of_Ratios_Ex_Aequali | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"File:Euclid-V-22.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Multiples of Terms in Equal Ratios",
"Definition:Strictly Positive/Real Number",
"Definition:Ratio",
"Relative Sizes of Succe... |
proofwiki-4133 | Equality of Ratios in Perturbed Proportion | {{:Euclid:Proposition/V/23}}
That is, if:
:$a : b = e : f$
:$b : c = d : e$
then:
:$a : c = d : f$ | Let there be three magnitudes $A, B, C$, and others equal to them in multitude, which taken two and two together are in the same proportion, namely $D, E, F$.
Let the proportion of them be perturbed, that is:
:$A : B = E : F$
:$B : C = D : E$
then we need to show that:
:$A : C = D : F$
:500px
Let equimultiples $A, B, D... | {{:Euclid:Proposition/V/23}}
That is, if:
:$a : b = e : f$
:$b : c = d : e$
then:
:$a : c = d : f$ | Let there be three [[Definition:Strictly Positive Real Number|magnitudes]] $A, B, C$, and others equal to them in multitude, which taken two and two together are in the same [[Definition:Proportion|proportion]], namely $D, E, F$.
Let the [[Definition:Proportion|proportion]] of them be [[Definition:Perturbed Proportion... | Equality of Ratios in Perturbed Proportion | https://proofwiki.org/wiki/Equality_of_Ratios_in_Perturbed_Proportion | https://proofwiki.org/wiki/Equality_of_Ratios_in_Perturbed_Proportion | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Proportion",
"Definition:Proportion",
"Definition:Proportion/Perturbed",
"File:Euclid-V-23.png",
"Definition:Equimultiples",
"Definition:Equimultiples",
"Ratio Equals its Multiples",
"Equality of Ratios is Transitive",
"Proportional Magnitude... |
proofwiki-4134 | Sum of Antecedent and Consequent of Proportion | {{:Euclid:Proposition/V/25}}
That is, if $a : b = c : d$ and $a$ is the greatest and $d$ is the least, then:
:$a + d > b + c$ | Let the four magnitudes $AB, CD, E, F$ be proportional, so that $AB : CD = E : F$.
Let $AB$ be the greatest and $F$ the least.
We need to show that $AB + F > CD + E$.
:260px
Let $AG = E, CH = F$.
We have that $AB : CD = E : F$, $AG = E, F = CH$.
So $AB : CD = AG : CH$.
So from Proportional Magnitudes have Proportional ... | {{:Euclid:Proposition/V/25}}
That is, if $a : b = c : d$ and $a$ is the greatest and $d$ is the least, then:
:$a + d > b + c$ | Let the four [[Definition:Strictly Positive Real Number|magnitudes]] $AB, CD, E, F$ be [[Definition:Proportion|proportional]], so that $AB : CD = E : F$.
Let $AB$ be the greatest and $F$ the least.
We need to show that $AB + F > CD + E$.
:[[File:Euclid-V-25.png|260px]]
Let $AG = E, CH = F$.
We have that $AB : CD =... | Sum of Antecedent and Consequent of Proportion | https://proofwiki.org/wiki/Sum_of_Antecedent_and_Consequent_of_Proportion | https://proofwiki.org/wiki/Sum_of_Antecedent_and_Consequent_of_Proportion | [
"Ratios"
] | [] | [
"Definition:Strictly Positive/Real Number",
"Definition:Proportion",
"File:Euclid-V-25.png",
"Proportional Magnitudes have Proportional Remainders"
] |
proofwiki-4135 | Areas of Triangles and Parallelograms Proportional to Base | {{:Euclid:Proposition/VI/1}}
Let $ABC$ and $ACD$ be triangles.
Let $EC, CF$ be parallelograms under the same height.
Then:
:$AC : CD = \triangle ABC : \triangle ACD = \Box EC : \Box CF$
where:
:$AC : CD$ denotes the ratio of the length of $AC$ to that of $CD$
:$\triangle ABC : \triangle ACD$ denotes the ratio of the ar... | :300px
Let $BD$ be produced in both directions to the points $H, L$ and let any number of straight lines, for example, $BG, GH$ be made equal to the base $BC$, and any number of straight lines, for example, $DK, KL$ be made equal to the base $CD$.
Let $AG, AH, AK, AL$ be joined.
Since $CB = BG = GH$ it follows from Tri... | {{:Euclid:Proposition/VI/1}}
Let $ABC$ and $ACD$ be [[Definition:Triangle (Geometry)|triangles]].
Let $EC, CF$ be [[Definition:Parallelogram|parallelograms]] under the same [[Definition:Height of Triangle|height]].
Then:
:$AC : CD = \triangle ABC : \triangle ACD = \Box EC : \Box CF$
where:
:$AC : CD$ denotes the [[D... | :[[File:Euclid-VI-1.png|300px]]
Let $BD$ be [[Definition:Production|produced]] in both directions to the points $H, L$ and let any number of [[Definition:Straight Line|straight lines]], for example, $BG, GH$ be made equal to the base $BC$, and any number of [[Definition:Straight Line|straight lines]], for example, $DK... | Areas of Triangles and Parallelograms Proportional to Base | https://proofwiki.org/wiki/Areas_of_Triangles_and_Parallelograms_Proportional_to_Base | https://proofwiki.org/wiki/Areas_of_Triangles_and_Parallelograms_Proportional_to_Base | [
"Area of Parallelogram",
"Areas of Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Quadrilateral/Parallelogram",
"Definition:Triangle (Geometry)/Height",
"Definition:Ratio",
"Definition:Linear Measure",
"Definition:Ratio",
"Definition:Area",
"Definition:Ratio",
"Definition:Area",
"Definition:Quadrilateral/Parallelogram",
"Definitio... | [
"File:Euclid-VI-1.png",
"Definition:Production",
"Definition:Line/Straight Line",
"Definition:Line/Straight Line",
"Triangles with Equal Base and Same Height have Equal Area",
"Definition:Multiple",
"Triangles with Equal Base and Same Height have Equal Area",
"Definition:Equimultiples",
"Definition:... |
proofwiki-4136 | Angle Bisector Theorem | Let $\triangle ABC$ be a triangle.
Let $D$ lie on the base $BC$ of $\triangle ABC$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$AD$ is the angle bisector of $\angle BAC$}}
{{item|(2):|$BD : DC {{=}} AB : AC$}}
{{end-itemize}}
where $BD : DC$ denotes the ratio between the lengths $BD$ and $DC$. | === $(1)$ implies $(2)$ ===
:300px
Let $CE$ be drawn through $C$ parallel to $DA$.
Let $BA$ be produced so as to meet $CE$ at $E$.
From {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Alternate Angles}} we have that:
:$\angle ACE = \angle CAD$
But {{hypothesis}}:
:$\angle CAD = \angle BAD$
and so:... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
Let $D$ lie on the [[Definition:Base of Triangle|base]] $BC$ of $\triangle ABC$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$AD$ is the [[Definition:Angle Bisector|angle bisector]] of $\angle BAC$}}
{{item|(2):|$BD : DC {{=}} AB : AC$}}
{{end-itemize}... | === $(1)$ implies $(2)$ ===
:[[File:Euclid-VI-3.png|300px]]
Let $CE$ be drawn through $C$ [[Definition:Parallel Lines|parallel]] to $DA$.
Let $BA$ be [[Definition:Production|produced]] so as to meet $CE$ at $E$.
From {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Alternate Angles}} we have th... | Angle Bisector Theorem | https://proofwiki.org/wiki/Angle_Bisector_Theorem | https://proofwiki.org/wiki/Angle_Bisector_Theorem | [
"Angle Bisector Theorem",
"Triangles",
"Named Theorems",
"Angle Bisectors"
] | [
"Definition:Triangle (Geometry)",
"Definition:Triangle (Geometry)/Base",
"Definition:Angle Bisector",
"Definition:Ratio",
"Definition:Linear Measure"
] | [
"File:Euclid-VI-3.png",
"Definition:Parallel (Geometry)/Lines",
"Definition:Production"
] |
proofwiki-4137 | Equiangular Triangles are Similar | Let two triangles have the same corresponding angles.
Then their corresponding sides are proportional.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/4}} | Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that:
{{begin-eqn}}
{{eqn | l = \angle ABC
| r = \angle DCE
}}
{{eqn | l = \angle BAC
| r = \angle CDE
}}
{{eqn | l = \angle ACB
| r = \angle CED
}}
{{end-eqn}}
:300px
Let $BC$ be placed in a straight line with $CE$.
From Two Angles of T... | Let two [[Definition:Triangle (Geometry)|triangles]] have the same corresponding [[Definition:Angle|angles]].
Then their corresponding [[Definition:Side of Polygon|sides]] are [[Definition:Proportion|proportional]].
Thus, by definition, such triangles are [[Definition:Similar Figures|similar]].
{{:Euclid:Propositio... | Let $\triangle ABC, \triangle DCE$ be [[Definition:Equiangular Geometric Figures|equiangular]] [[Definition:Triangle (Geometry)|triangles]] such that:
{{begin-eqn}}
{{eqn | l = \angle ABC
| r = \angle DCE
}}
{{eqn | l = \angle BAC
| r = \angle CDE
}}
{{eqn | l = \angle ACB
| r = \angle CED
}}
{{end-e... | Equiangular Triangles are Similar | https://proofwiki.org/wiki/Equiangular_Triangles_are_Similar | https://proofwiki.org/wiki/Equiangular_Triangles_are_Similar | [
"Similar Triangles",
"Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Angle",
"Definition:Polygon/Side",
"Definition:Proportion",
"Definition:Similar Figures"
] | [
"Definition:Equiangular Geometric Figures",
"Definition:Triangle (Geometry)",
"File:Euclid-VI-4.png",
"Definition:Line/Straight Line",
"Two Angles of Triangle are Less than Two Right Angles",
"Definition:Right Angle",
"Definition:Right Angle",
"Axiom:Parallel Postulate",
"Definition:Production",
"... |
proofwiki-4138 | Triangles with Proportional Sides are Similar | Let two triangles have corresponding sides which are proportional.
Then their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/5}} | Let $\triangle ABC, \triangle DEF$ be triangles whose sides are proportional, so that:
:$ AB : BC = DE : EF$
:$ BC : CA = EF : FD$
:$ BA : AC = ED : DF$
We need to show that
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$
:$\angle BAC = \angle EDF$
:400px
On the straight line $EF$, and at the points $E, F$ on it,... | Let two [[Definition:Triangle (Geometry)|triangles]] have corresponding [[Definition:Side of Polygon|sides]] which are [[Definition:Proportion|proportional]].
Then their corresponding [[Definition:Angle|angles]] are equal.
Thus, by definition, such [[Definition:Triangle (Geometry)|triangles]] are [[Definition:Similar... | Let $\triangle ABC, \triangle DEF$ be [[Definition:Triangle (Geometry)|triangles]] whose [[Definition:Side of Polygon|sides]] are [[Definition:Proportion|proportional]], so that:
:$ AB : BC = DE : EF$
:$ BC : CA = EF : FD$
:$ BA : AC = ED : DF$
We need to show that
:$\angle ABC = \angle DEF$
:$\angle BCA = \angle EFD$... | Triangles with Proportional Sides are Similar | https://proofwiki.org/wiki/Triangles_with_Proportional_Sides_are_Similar | https://proofwiki.org/wiki/Triangles_with_Proportional_Sides_are_Similar | [
"Similar Triangles",
"Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Polygon/Side",
"Definition:Proportion",
"Definition:Angle",
"Definition:Triangle (Geometry)",
"Definition:Similar Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Polygon/Side",
"Definition:Proportion",
"File:Euclid-VI-5.png",
"Definition:Line/Straight Line",
"Construction of Equal Angle",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Angle",
"Definition:Equiangular Geometric Figures",
"Equian... |
proofwiki-4139 | Triangles with One Equal Angle and Two Sides Proportional are Similar | Let two triangles have two corresponding sides which are proportional.
Let the angles adjacent to both of these sides be equal.
Then all of their corresponding angles are equal.
Thus, by definition, such triangles are similar.
{{:Euclid:Proposition/VI/6}} | Let $\triangle ABC, \triangle DEF$ be triangles having $\angle BAC = \angle EDF$ and two sides proportional, so that $BA : AC = ED : DF$.
We need to show that $\triangle ABC$ is equiangular with $\triangle DEF$, such that $\angle ABC = \angle DEF$ and $\angle ACB = \angle DFE$.
:350px
Let $\angle FDG$ be constructed eq... | Let two [[Definition:Triangle (Geometry)|triangles]] have two corresponding [[Definition:Side of Polygon|sides]] which are [[Definition:Proportion|proportional]].
Let the [[Definition:Angle|angles]] [[Definition:Adjacent (in Triangle)|adjacent]] to both of these sides be equal.
Then all of their corresponding [[Defin... | Let $\triangle ABC, \triangle DEF$ be [[Definition:Triangle (Geometry)|triangles]] having $\angle BAC = \angle EDF$ and two sides [[Definition:Proportion|proportional]], so that $BA : AC = ED : DF$.
We need to show that $\triangle ABC$ is [[Definition:Equiangular Geometric Figures|equiangular]] with $\triangle DEF$, s... | Triangles with One Equal Angle and Two Sides Proportional are Similar | https://proofwiki.org/wiki/Triangles_with_One_Equal_Angle_and_Two_Sides_Proportional_are_Similar | https://proofwiki.org/wiki/Triangles_with_One_Equal_Angle_and_Two_Sides_Proportional_are_Similar | [
"Similar Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Polygon/Side",
"Definition:Proportion",
"Definition:Angle",
"Definition:Triangle (Geometry)/Adjacent",
"Definition:Angle",
"Definition:Similar Figures"
] | [
"Definition:Triangle (Geometry)",
"Definition:Proportion",
"Definition:Equiangular Geometric Figures",
"File:Euclid-VI-6.png",
"Construction of Equal Angle",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Equiangular Geometric Figures",
"Equiangular Triangles are Similar",
"Definit... |
proofwiki-4140 | Triangles with One Equal Angle and Two Other Sides Proportional are Similar | Let two triangles be such that one of the angles of one triangle equals one of the angles of the other.
Let two corresponding sides which are adjacent to one of the other angles, be proportional.
Let the third angle in both triangles be either both acute or both not acute.
Then all of the corresponding angles of these ... | Let $\triangle ABC, \triangle DEF$ be triangles such that:
:$\angle BAC = \angle EDF$
:two sides adjacent to $\angle ABC$ and $\angle DEF$ proportional, so that $AB : BC = DE : EF$
:$\angle ACB$ and $\angle DFE$ either both acute or both not acute.
We need to show that $\angle ABC = \angle DEF$ and $\angle BCA = \angle... | Let two [[Definition:Triangle (Geometry)|triangles]] be such that one of the [[Definition:Angle|angles]] of one triangle equals one of the angles of the other.
Let two corresponding [[Definition:Side of Polygon|sides]] which are [[Definition:Adjacent (in Triangle)|adjacent]] to one of the other angles, be [[Definition... | Let $\triangle ABC, \triangle DEF$ be [[Definition:Triangle (Geometry)|triangles]] such that:
:$\angle BAC = \angle EDF$
:two sides [[Definition:Adjacent (in Triangle)|adjacent]] to $\angle ABC$ and $\angle DEF$ [[Definition:Proportion|proportional]], so that $AB : BC = DE : EF$
:$\angle ACB$ and $\angle DFE$ either bo... | Triangles with One Equal Angle and Two Other Sides Proportional are Similar | https://proofwiki.org/wiki/Triangles_with_One_Equal_Angle_and_Two_Other_Sides_Proportional_are_Similar | https://proofwiki.org/wiki/Triangles_with_One_Equal_Angle_and_Two_Other_Sides_Proportional_are_Similar | [
"Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Angle",
"Definition:Polygon/Side",
"Definition:Triangle (Geometry)/Adjacent",
"Definition:Proportion",
"Definition:Acute Angle",
"Definition:Acute Angle",
"Definition:Angle",
"Definition:Similar Figures"
] | [
"Definition:Triangle (Geometry)",
"Definition:Triangle (Geometry)/Adjacent",
"Definition:Proportion",
"Definition:Acute Angle",
"Definition:Acute Angle",
"Definition:Acute Angle",
"File:Euclid-VI-7a.png",
"Construction of Equal Angle",
"Sum of Angles of Triangle equals Two Right Angles",
"Definiti... |
proofwiki-4141 | Zeckendorf's Theorem | Every positive integer has a unique Zeckendorf representation.
That is:
Let $n$ be a positive integer.
Then there exists a unique increasing sequence of integers $\sequence {c_i}$ such that:
{{begin-eqn}}
{{eqn | q = \forall i \in \N
| l = c_i
| o = \ge
| r = 2
| c =
}}
{{eqn | q = \forall i \i... | First note that every Fibonacci number $F_n$ is itself a '''Zeckendorf representation''' of itself, where the sequence $\sequence {c_i}$ contains $1$ term. | Every [[Definition:Positive Integer|positive integer]] has a [[Definition:Unique|unique]] [[Definition:Zeckendorf Representation|Zeckendorf representation]].
That is:
Let $n$ be a [[Definition:Positive Integer|positive integer]].
Then there exists a [[Definition:Unique|unique]] [[Definition:Increasing Sequence|incre... | First note that every [[Definition:Fibonacci Number|Fibonacci number]] $F_n$ is itself a '''[[Definition:Zeckendorf Representation|Zeckendorf representation]]''' of itself, where the [[Definition:Integer Sequence|sequence]] $\sequence {c_i}$ contains $1$ [[Definition:Term of Sequence|term]]. | Zeckendorf's Theorem | https://proofwiki.org/wiki/Zeckendorf's_Theorem | https://proofwiki.org/wiki/Zeckendorf's_Theorem | [
"Zeckendorf Representation"
] | [
"Definition:Positive/Integer",
"Definition:Unique",
"Definition:Zeckendorf Representation",
"Definition:Positive/Integer",
"Definition:Unique",
"Definition:Increasing/Sequence",
"Definition:Integer",
"Definition:Fibonacci Number",
"Definition:Unique"
] | [
"Definition:Fibonacci Number",
"Definition:Zeckendorf Representation",
"Definition:Integer Sequence",
"Definition:Term of Sequence",
"Definition:Integer",
"Definition:Zeckendorf Representation",
"Definition:Zeckendorf Representation",
"Definition:Zeckendorf Representation",
"Definition:Zeckendorf Re... |
proofwiki-4142 | Sum of Non-Consecutive Fibonacci Numbers | Let $S$ be a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$.
Let the largest element of $S$ be $F_j$.
Then:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
That is, the sum of all the elements of $S$ is strictly less than the next largest Fibonacci number.
That is, given some ... | The proof proceeds by induction on $j$ for $j \ge 2$.
For all $j \in \N_{>0}$, let $\map P j$ be the proposition:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Let the term '''allowable set''' be used to mean a non-empty set of distinct non-consecutive Fibonacci numbers not containing $F_0$ or $F_1$. | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] of [[Definition:Distinct|distinct]] non-consecutive [[Definition:Fibonacci Numbers|Fibonacci numbers]] not containing $F_0$ or $F_1$.
Let the largest element of $S$ be $F_j$.
Then:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
That is, the ... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $j$ for $j \ge 2$.
For all $j \in \N_{>0}$, let $\map P j$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{F_i \mathop \in S} F_i < F_{j + 1}$
Let the term '''allowable set''' be used to mean a [[Definition:Non-Empty Set|non-empty]... | Sum of Non-Consecutive Fibonacci Numbers | https://proofwiki.org/wiki/Sum_of_Non-Consecutive_Fibonacci_Numbers | https://proofwiki.org/wiki/Sum_of_Non-Consecutive_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Distinct",
"Definition:Fibonacci Number",
"Definition:Summation",
"Definition:Element",
"Definition:Fibonacci Number",
"Definition:Increasing/Sequence"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Distinct",
"Definition:Fibonacci Number",
"Definition:Fibonacci Number",
"Principle of Mathematical Induction"
] |
proofwiki-4143 | Construction of Third Proportional Straight Line | Given any two straight lines of given length $a$ and $b$, it is possible to construct a third straight line of length $c$ such that $a : b = b : c$.
{{:Euclid:Proposition/VI/11}} | :300px
We have that:
:$BC \parallel DE$
So from Parallel Transversal Theorem:
:$AB : BD = AC : CE$
But $BD = AC$ and so:
:$AB : AC = AC : CE$
as required.
{{qed}}
{{Euclid Note|11|VI|It is a special case of {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}.}} | Given any two [[Definition:Straight Line|straight lines]] of given length $a$ and $b$, it is possible to construct a third [[Definition:Straight Line|straight line]] of length $c$ such that $a : b = b : c$.
{{:Euclid:Proposition/VI/11}} | :[[File:Euclid-VI-11.png|300px]]
We have that:
:$BC \parallel DE$
So from [[Parallel Transversal Theorem]]:
:$AB : BD = AC : CE$
But $BD = AC$ and so:
:$AB : AC = AC : CE$
as required.
{{qed}}
{{Euclid Note|11|VI|It is a special case of {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportiona... | Construction of Third Proportional Straight Line | https://proofwiki.org/wiki/Construction_of_Third_Proportional_Straight_Line | https://proofwiki.org/wiki/Construction_of_Third_Proportional_Straight_Line | [
"Lines"
] | [
"Definition:Line/Straight Line",
"Definition:Line/Straight Line"
] | [
"File:Euclid-VI-11.png",
"Parallel Transversal Theorem"
] |
proofwiki-4144 | Construction of Fourth Proportional Straight Line | Given three straight lines of lengths $a, b, c$, it is possible to construct a fourth straight line of length $d$ such that $a : b = c : d$.
{{:Euclid:Proposition/VI/12}} | :450px
From Parallel Transversal Theorem:
:$DG : GE = DH : HF$
But:
:$DG = A, GE = B, DH = C$
Hence the result.
{{qed}}
{{Euclid Note|12|VI}} | Given three [[Definition:Straight Line|straight lines]] of [[Definition:Linear Measure|lengths]] $a, b, c$, it is possible to construct a fourth [[Definition:Straight Line|straight line]] of length $d$ such that $a : b = c : d$.
{{:Euclid:Proposition/VI/12}} | :[[File:Euclid-VI-12.png|450px]]
From [[Parallel Transversal Theorem]]:
:$DG : GE = DH : HF$
But:
:$DG = A, GE = B, DH = C$
Hence the result.
{{qed}}
{{Euclid Note|12|VI}} | Construction of Fourth Proportional Straight Line | https://proofwiki.org/wiki/Construction_of_Fourth_Proportional_Straight_Line | https://proofwiki.org/wiki/Construction_of_Fourth_Proportional_Straight_Line | [
"Lines"
] | [
"Definition:Line/Straight Line",
"Definition:Linear Measure",
"Definition:Line/Straight Line"
] | [
"File:Euclid-VI-12.png",
"Parallel Transversal Theorem"
] |
proofwiki-4145 | Construction of Mean Proportional | Given any two straight lines of length $a$ and $b$ it is possible to find a straight line of length $c$ such that $a : c = c : b$.
{{:Euclid:Proposition/VI/13}} | :250px
From Relative Sizes of Angles in Segments, $\angle ADC$ is a right angle.
So from the porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles, $DB$ is the mean proportional between $AB$ and $BC$.
{{qed}}
{{Euclid Note|13|VI|Also see {{EuclidPropLink|book=II|prop=2|title=Construction of Squar... | Given any two [[Definition:Straight Line|straight lines]] of [[Definition:Linear Measure|length]] $a$ and $b$ it is possible to find a [[Definition:Straight Line|straight line]] of [[Definition:Linear Measure|length]] $c$ such that $a : c = c : b$.
{{:Euclid:Proposition/VI/13}} | :[[File:Euclid-VI-13.png|250px]]
From [[Relative Sizes of Angles in Segments]], $\angle ADC$ is a [[Definition:Right Angle|right angle]].
So from [[Perpendicular in Right-Angled Triangle makes two Similar Triangles/Porism|the porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles]], $DB$ is the ... | Construction of Mean Proportional | https://proofwiki.org/wiki/Construction_of_Mean_Proportional | https://proofwiki.org/wiki/Construction_of_Mean_Proportional | [
"Lines"
] | [
"Definition:Line/Straight Line",
"Definition:Linear Measure",
"Definition:Line/Straight Line",
"Definition:Linear Measure"
] | [
"File:Euclid-VI-13.png",
"Relative Sizes of Angles in Segments",
"Definition:Right Angle",
"Perpendicular in Right-Angled Triangle makes two Similar Triangles/Porism",
"Definition:Geometric Mean/Mean Proportional",
"Definition:Geometric Mean/Mean Proportional"
] |
proofwiki-4146 | Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional | {{:Euclid:Proposition/VI/14}}
Note: in the above, ''equal'' is to be taken to mean ''of equal area''. | Let $\Box AB$ and $\Box BC$ be two equiangular parallelograms of equal area such that the angles at $B$ are equal.
Let $DB, BE$ be placed in a straight line.
By Two Angles making Two Right Angles make Straight Line it follows that $FB, BG$ also make a straight line.
We need to show that $DB : BE = GB : BF$, that is, th... | {{:Euclid:Proposition/VI/14}}
Note: in the above, ''equal'' is to be taken to mean ''of equal [[Definition:Area|area]]''. | Let $\Box AB$ and $\Box BC$ be two [[Definition:Equiangular Geometric Figures|equiangular]] [[Definition:Parallelogram|parallelograms]] of equal [[Definition:Area|area]] such that the [[Definition:Vertex of Polygon|angles]] at $B$ are equal.
Let $DB, BE$ be placed in a [[Definition:Straight Line|straight line]].
By [... | Sides of Equal and Equiangular Parallelograms are Reciprocally Proportional | https://proofwiki.org/wiki/Sides_of_Equal_and_Equiangular_Parallelograms_are_Reciprocally_Proportional | https://proofwiki.org/wiki/Sides_of_Equal_and_Equiangular_Parallelograms_are_Reciprocally_Proportional | [
"Parallelograms"
] | [
"Definition:Area"
] | [
"Definition:Equiangular Geometric Figures",
"Definition:Quadrilateral/Parallelogram",
"Definition:Area",
"Definition:Polygon/Vertex",
"Definition:Line/Straight Line",
"Two Angles making Two Right Angles make Straight Line",
"Definition:Line/Straight Line",
"Definition:Angle",
"Definition:Reciprocal ... |
proofwiki-4147 | Sides of Equiangular Triangles are Reciprocally Proportional | {{:Euclid:Proposition/VI/15}}
Note: in the above, ''equal'' is to be taken to mean ''of equal area''. | Let $\triangle ABC, \triangle ADE$ be triangles of equal area which have one angle equal to one angle, namely $\angle BAC = \angle DAE$.
We need to show that $CA : AD = EA : AB$, that is, the sides about the equal angles are reciprocally proportional.
:250px
Place them so $CA$ is in a straight line with $AD$.
From Two ... | {{:Euclid:Proposition/VI/15}}
Note: in the above, ''equal'' is to be taken to mean ''of equal [[Definition:Area|area]]''. | Let $\triangle ABC, \triangle ADE$ be [[Definition:Triangle (Geometry)|triangles]] of equal [[Definition:Area|area]] which have one [[Definition:Angle|angle]] equal to one angle, namely $\angle BAC = \angle DAE$.
We need to show that $CA : AD = EA : AB$, that is, the sides about the equal [[Definition:Angle|angles]] a... | Sides of Equiangular Triangles are Reciprocally Proportional | https://proofwiki.org/wiki/Sides_of_Equiangular_Triangles_are_Reciprocally_Proportional | https://proofwiki.org/wiki/Sides_of_Equiangular_Triangles_are_Reciprocally_Proportional | [
"Triangles"
] | [
"Definition:Area"
] | [
"Definition:Triangle (Geometry)",
"Definition:Area",
"Definition:Angle",
"Definition:Angle",
"Definition:Reciprocal Proportion",
"File:Euclid-VI-15.png",
"Definition:Line/Straight Line",
"Two Angles making Two Right Angles make Straight Line",
"Definition:Line/Straight Line",
"Ratios of Equal Magn... |
proofwiki-4148 | Rectangles Contained by Proportional Straight Lines | {{:Euclid:Proposition/VI/16}}
Note: in the above, ''equal'' is to be taken to mean ''of equal area''. | Let the four straight lines $AB, CD, E, F$ be proportional, that is, $AB : CD = E : F$.
What we need to show is that the rectangle contained by $AB$ and $F$ is equal in area to the rectangle contained by $CD$ and $E$.
:400px
Let $AG, CH$ be drawn perpendicular to $AB$ and $CD$.
Let $AG = F$, $CH = E$.
Complete the para... | {{:Euclid:Proposition/VI/16}}
Note: in the above, ''equal'' is to be taken to mean ''of equal [[Definition:Area|area]]''. | Let the four [[Definition:Straight Line|straight lines]] $AB, CD, E, F$ be [[Definition:Proportion|proportional]], that is, $AB : CD = E : F$.
What we need to show is that the [[Definition:Containment of Rectangle|rectangle contained]] by $AB$ and $F$ is equal in [[Definition:Area|area]] to the [[Definition:Containmen... | Rectangles Contained by Proportional Straight Lines | https://proofwiki.org/wiki/Rectangles_Contained_by_Proportional_Straight_Lines | https://proofwiki.org/wiki/Rectangles_Contained_by_Proportional_Straight_Lines | [
"Rectangles"
] | [
"Definition:Area"
] | [
"Definition:Line/Straight Line",
"Definition:Proportion",
"Definition:Quadrilateral/Rectangle/Containment",
"Definition:Area",
"Definition:Quadrilateral/Rectangle/Containment",
"File:Euclid-VI-16.png",
"Definition:Right Angle/Perpendicular",
"Definition:Quadrilateral/Parallelogram",
"Definition:Reci... |
proofwiki-4149 | Construction of Similar Polygon | On any given straight line it is possible to construct a polygon similar to any given polygon.
{{:Euclid:Proposition/VI/18}} | From Sum of Angles of Triangle Equals Two Right Angles $\angle CFD = \angle AGB$.
So $\triangle FCD$ is equiangular with $\triangle GAB$.
So from Equiangular Triangles are Similar, $\triangle FCD$ is similar to $\triangle GAB$.
So $FD : GB = FC : GA = CD : AB$.
Similarly from Sum of Angles of Triangle Equals Two Right ... | On any given [[Definition:Line Segment|straight line]] it is possible to construct a [[Definition:Polygon|polygon]] [[Definition:Similar Figures|similar]] to any given [[Definition:Polygon|polygon]].
{{:Euclid:Proposition/VI/18}} | From [[Sum of Angles of Triangle Equals Two Right Angles]] $\angle CFD = \angle AGB$.
So $\triangle FCD$ is [[Definition:Equiangular Geometric Figures|equiangular]] with $\triangle GAB$.
So from [[Equiangular Triangles are Similar]], $\triangle FCD$ is [[Definition:Similar Triangles|similar]] to $\triangle GAB$.
So ... | Construction of Similar Polygon | https://proofwiki.org/wiki/Construction_of_Similar_Polygon | https://proofwiki.org/wiki/Construction_of_Similar_Polygon | [
"Polygons"
] | [
"Definition:Line/Segment",
"Definition:Polygon",
"Definition:Similar Figures",
"Definition:Polygon"
] | [
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Equiangular Geometric Figures",
"Equiangular Triangles are Similar",
"Definition:Similar Triangles",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Equiangular Geometric Figures",
"Equiangular Triangles are Similar",
"D... |
proofwiki-4150 | Triangles with Two Equal Angles are Similar | Two triangles which have two corresponding angles which are equal are similar. | Let $\triangle ABC$ and $\triangle DEF$ be triangles such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$.
Then from Sum of Angles of Triangle Equals Two Right Angles $\angle ACB$ is equal to two right angles minus $\angle ABC + \angle BAC$.
Also from Sum of Angles of Triangle Equals Two Right Angles $\ang... | Two [[Definition:Triangle (Geometry)|triangles]] which have two corresponding [[Definition:Angle|angles]] which are equal are [[Definition:Similar Triangles|similar]]. | Let $\triangle ABC$ and $\triangle DEF$ be [[Definition:Triangle (Geometry)|triangles]] such that $\angle ABC = \angle DEF$ and $\angle BAC = \angle EDF$.
Then from [[Sum of Angles of Triangle Equals Two Right Angles]] $\angle ACB$ is equal to two [[Definition:Right Angle|right angles]] minus $\angle ABC + \angle BAC$... | Triangles with Two Equal Angles are Similar | https://proofwiki.org/wiki/Triangles_with_Two_Equal_Angles_are_Similar | https://proofwiki.org/wiki/Triangles_with_Two_Equal_Angles_are_Similar | [
"Similar Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Angle",
"Definition:Similar Triangles"
] | [
"Definition:Triangle (Geometry)",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Right Angle",
"Sum of Angles of Triangle equals Two Right Angles",
"Definition:Right Angle",
"Definition:Right Angle",
"Equiangular Triangles are Similar"
] |
proofwiki-4151 | Ratio of Areas of Similar Triangles | {{:Euclid:Proposition/VI/19}}
That is, the ratio of the areas of the similar triangles is the square of the ratio of the corresponding sides. | Let $\triangle ABC$ and $\triangle DEF$ be similar, such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ corresponds to $EF$.
:400px
Let $BG$ be constructed such that $EF : BG = BC : EF$, and join $AG$.
From Proportional Magnitudes are Proportional Alternately $AB : DE = BC : EF$.
So from Equality... | {{:Euclid:Proposition/VI/19}}
That is, the [[Definition:Ratio|ratio]] of the [[Definition:Area|areas]] of the [[Definition:Similar Triangles|similar triangles]] is the [[Definition:Square (Algebra)|square]] of the [[Definition:Ratio|ratio]] of the [[Definition:Corresponding Magnitudes|corresponding]] [[Definition:Side... | Let $\triangle ABC$ and $\triangle DEF$ be [[Definition:Similar Triangles|similar]], such that $\angle ABC = \angle DEF$ and $AB : BC = DE : EF$ such that $BC$ [[Definition:Corresponding Magnitudes|corresponds]] to $EF$.
:[[File:Euclid-VI-19.png|400px]]
Let $BG$ be constructed such that $EF : BG = BC : EF$, and join ... | Ratio of Areas of Similar Triangles | https://proofwiki.org/wiki/Ratio_of_Areas_of_Similar_Triangles | https://proofwiki.org/wiki/Ratio_of_Areas_of_Similar_Triangles | [
"Ratio of Areas of Similar Triangles",
"Areas of Triangles"
] | [
"Definition:Ratio",
"Definition:Area",
"Definition:Similar Triangles",
"Definition:Square/Function",
"Definition:Ratio",
"Definition:Corresponding Magnitudes",
"Definition:Polygon/Side"
] | [
"Definition:Similar Triangles",
"Definition:Corresponding Magnitudes",
"File:Euclid-VI-19.png",
"Proportional Magnitudes are Proportional Alternately",
"Equality of Ratios is Transitive",
"Definition:Angle",
"Definition:Reciprocal Proportion",
"Sides of Equiangular Triangles are Reciprocally Proportio... |
proofwiki-4152 | Similarity of Polygons is Equivalence Relation | Let $A, B, C$ be polygons.
If $A$ and $B$ are both similar to $C$, then $A$ is similar to $B$.
{{:Euclid:Proposition/VI/21}}
It is also worth noting that:
:$A$ is similar to $A$, and so similarity between polygons is reflexive.
:If $A$ is similar to $B$, then $B$ is similar to $A$, and so similarity between polygons is... | :500px
We have that $A$ is similar to $C$.
From {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the sides about the equal angles are proportional.
We also have that $B$ is similar to $C$.
Again, from {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is equiangular with it and the ... | Let $A, B, C$ be [[Definition:Polygon|polygons]].
If $A$ and $B$ are both [[Definition:Similar Figures|similar]] to $C$, then $A$ is [[Definition:Similar Figures|similar]] to $B$.
{{:Euclid:Proposition/VI/21}}
It is also worth noting that:
:$A$ is [[Definition:Similar Figures|similar]] to $A$, and so [[Definition:... | :[[File:Euclid-VI-21.png|500px]]
We have that $A$ is [[Definition:Similar Figures|similar]] to $C$.
From {{EuclidDefLink|VI|1|Similar Rectilineal Figures}}, it is [[Definition:Equiangular Geometric Figures|equiangular]] with it and the [[Definition:Side of Polygon|sides]] about the equal [[Definition:Angle|angles]] a... | Similarity of Polygons is Equivalence Relation | https://proofwiki.org/wiki/Similarity_of_Polygons_is_Equivalence_Relation | https://proofwiki.org/wiki/Similarity_of_Polygons_is_Equivalence_Relation | [
"Examples of Equivalence Relations",
"Polygons",
"Examples of Equivalence Relations"
] | [
"Definition:Polygon",
"Definition:Similar Figures",
"Definition:Similar Figures",
"Definition:Similar Figures",
"Definition:Similar Figures",
"Definition:Polygon",
"Definition:Reflexive Relation",
"Definition:Similar Figures",
"Definition:Similar Figures",
"Definition:Similar Figures",
"Definiti... | [
"File:Euclid-VI-21.png",
"Definition:Similar Figures",
"Definition:Equiangular Geometric Figures",
"Definition:Polygon/Side",
"Definition:Angle",
"Definition:Proportion",
"Definition:Similar Figures",
"Definition:Equiangular Geometric Figures",
"Definition:Polygon/Side",
"Definition:Angle",
"Def... |
proofwiki-4153 | Integral Multiple of an Algebraic Number | Let $K$ be a number field and $\alpha \in K$.
Then there exists a positive $n \in \Z$ such that $n \alpha \in \OO_K$.
In this context, $ \OO_K$ denotes the algebraic integers in $K$. | If $\alpha = 0$ then any integer works and the proof is finished.
Let $\alpha \ne 0$.
Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the minimal polynomial of $\alpha$ over $\Q$.
Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction for each $i$ such that $a_i \ne 0$.
Let $n$ be the least common ... | Let $K$ be a [[Definition:Number Field|number field]] and $\alpha \in K$.
Then there exists a [[Definition:Positive Integer|positive $n \in \Z$]] such that $n \alpha \in \OO_K$.
In this context, $ \OO_K$ denotes the [[Ring of Algebraic Integers|algebraic integers in $K$]]. | If $\alpha = 0$ then any [[Definition:Integer|integer]] works and the proof is finished.
Let $\alpha \ne 0$.
Let $\map f x = x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0$ be the [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $\Q$.
Suppose that $a_i = \dfrac {b_i} {c_i}$ is a reduced fraction fo... | Integral Multiple of an Algebraic Number | https://proofwiki.org/wiki/Integral_Multiple_of_an_Algebraic_Number | https://proofwiki.org/wiki/Integral_Multiple_of_an_Algebraic_Number | [
"Number Theory",
"Algebraic Number Theory"
] | [
"Definition:Number Field",
"Definition:Positive/Integer",
"Ring of Algebraic Integers"
] | [
"Definition:Integer",
"Definition:Minimal Polynomial",
"Definition:Lowest Common Multiple/Integers",
"Definition:Monic Polynomial",
"Definition:Algebraic Integer"
] |
proofwiki-4154 | Schönemann-Eisenstein Theorem | Let $\map f x = a_d x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0 \in \Z \sqbrk x$ be a polynomial over the ring of integers $\Z$.
Let $p$ be a prime such that:
:$(1): \quad p \divides a_i \iff i \ne d$
:$(2): \quad p^2 \nmid a_0$
where $p \divides a_i$ signifies that $p$ is a divisor of $a_i$.
Then $f$ is irreducible in $\... | By Gauss's Lemma on Irreducible Polynomials, it suffices to show that $f$ is irreducible in $\Z \sqbrk x$.
{{AimForCont}} that $f = g h$ where $g, h \in \Z \sqbrk x$ are both non-constant.
Let:
:$\map g x = b_e x^e + b_{e - 1} x^{e - 1} + \dotsb + b_0$
:$\map h x = c_f x^f + c_{f - 1} x^{f - 1} + \dotsb + c_0$
Then we ... | Let $\map f x = a_d x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0 \in \Z \sqbrk x$ be a [[Definition:Polynomial over Ring|polynomial]] over the [[Definition:Ring of Integers|ring of integers]] $\Z$.
Let $p$ be a [[Definition:Prime Number|prime]] such that:
:$(1): \quad p \divides a_i \iff i \ne d$
:$(2): \quad p^2 \nmid a... | By [[Gauss's Lemma on Irreducible Polynomials]], it suffices to show that $f$ is [[Definition:Irreducible Polynomial|irreducible]] in $\Z \sqbrk x$.
{{AimForCont}} that $f = g h$ where $g, h \in \Z \sqbrk x$ are both [[Definition:Constant Polynomial|non-constant]].
Let:
:$\map g x = b_e x^e + b_{e - 1} x^{e - 1} + \... | Schönemann-Eisenstein Theorem | https://proofwiki.org/wiki/Schönemann-Eisenstein_Theorem | https://proofwiki.org/wiki/Schönemann-Eisenstein_Theorem | [
"Schönemann-Eisenstein Theorem",
"Algebraic Number Theory"
] | [
"Definition:Polynomial over Ring",
"Definition:Ring of Integers",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Irreducible Polynomial"
] | [
"Gauss's Lemma on Irreducible Polynomials",
"Definition:Irreducible Polynomial",
"Definition:Constant Polynomial",
"Definition:Constant Polynomial",
"Degree of Product of Polynomials over Ring/Corollary 2",
"Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors",
"Definit... |
proofwiki-4155 | Isolated Point of Closure of Subset is Isolated Point of Subset | Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$ be a subset of $T$.
Let $\map \cl H$ denote the closure of $H$.
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
Then $x$ is also an isolated point of $H$. | Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
{{AimForCont}} that $x$ is not an isolated point of $H$.
Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
So by Limit Point of Subset is Limit P... | Let $\struct {T, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
Let $\map \cl H$ denote the [[Definition:Closure (Topology)|closure]] of $H$.
Let $x \in \map \cl H$ be an [[Definition:Isolated Point of Subset|isolated point]] of $\map \cl... | Let $x \in \map \cl H$ be an [[Definition:Isolated Point of Subset|isolated point]] of $\map \cl H$.
{{AimForCont}} that $x$ is not an [[Definition:Isolated Point of Subset|isolated point]] of $H$.
Then by the definition of a [[Definition:Limit Point of Set|limit point]], it follows that $x$ must be a [[Definition:L... | Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 1 | https://proofwiki.org/wiki/Isolated_Point_of_Closure_of_Subset_is_Isolated_Point_of_Subset | https://proofwiki.org/wiki/Isolated_Point_of_Closure_of_Subset_is_Isolated_Point_of_Subset/Proof_1 | [
"Isolated Points",
"Set Closures",
"Isolated Point of Closure of Subset is Isolated Point of Subset"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Isolated Point (Topology)/Subset"
] | [
"Definition:Isolated Point (Topology)/Subset",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set",
"Set is Subset of its Topological Closure",
"Limit Point of Subset is Limit Point of Set",
"Definition:Limit Point/Topology/Set",
... |
proofwiki-4156 | Isolated Point of Closure of Subset is Isolated Point of Subset | Let $\struct {T, \tau}$ be a topological space.
Let $H \subseteq T$ be a subset of $T$.
Let $\map \cl H$ denote the closure of $H$.
Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.
Then $x$ is also an isolated point of $H$. | Let $x \in \map \cl H$ be isolated in $\map \cl H$.
By definition of isolated point, $x$ is not a limit point of $\map \cl H$.
From Set is Subset of its Topological Closure:
:$H \subseteq \map \cl H$
We have that Limit Point of Subset is Limit Point of Set.
But $x$ is not a limit point of $\map \cl H$ .
So by the Rule ... | Let $\struct {T, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq T$ be a [[Definition:Subset|subset]] of $T$.
Let $\map \cl H$ denote the [[Definition:Closure (Topology)|closure]] of $H$.
Let $x \in \map \cl H$ be an [[Definition:Isolated Point of Subset|isolated point]] of $\map \cl... | Let $x \in \map \cl H$ be [[Definition:Isolated Point of Subset|isolated in $\map \cl H$]].
By definition of [[Definition:Isolated Point of Subset|isolated point]], $x$ is not a [[Definition:Limit Point of Set|limit point of $\map \cl H$]].
From [[Set is Subset of its Topological Closure]]:
:$H \subseteq \map \cl H$... | Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 2 | https://proofwiki.org/wiki/Isolated_Point_of_Closure_of_Subset_is_Isolated_Point_of_Subset | https://proofwiki.org/wiki/Isolated_Point_of_Closure_of_Subset_is_Isolated_Point_of_Subset/Proof_2 | [
"Isolated Points",
"Set Closures",
"Isolated Point of Closure of Subset is Isolated Point of Subset"
] | [
"Definition:Topological Space",
"Definition:Subset",
"Definition:Closure (Topology)",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Isolated Point (Topology)/Subset"
] | [
"Definition:Isolated Point (Topology)/Subset",
"Definition:Isolated Point (Topology)/Subset",
"Definition:Limit Point/Topology/Set",
"Set is Subset of its Topological Closure",
"Limit Point of Subset is Limit Point of Set",
"Definition:Limit Point/Topology/Set",
"Rule of Transposition",
"Definition:Li... |
proofwiki-4157 | Divisors obey Distributive Law | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$.
Let the (natural) number $D$ be the same aliquot part of another (natural) number $EF$ that $A$ is of $BC$.
We need to show that $A + D$ is the same aliquot part of $BC + EF$.
:250px
We have that whatever aliquot part $A$ is of $BC$, then $D$... | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | Let the [[Definition:Natural Number|(natural) number]] $A$ be an [[Definition:Aliquot Part|aliquot part]] of the [[Definition:Natural Number|(natural) number]] $BC$.
Let the [[Definition:Natural Number|(natural) number]] $D$ be the same [[Definition:Aliquot Part|aliquot part]] of another [[Definition:Natural Number|(n... | Divisors obey Distributive Law/Proof 1 | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law/Proof_1 | [
"Divisors obey Distributive Law",
"Divisors",
"Distributive Operations",
"Algebra"
] | [] | [
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"File:Euclid-VII-5.png"... |
proofwiki-4158 | Divisors obey Distributive Law | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | A direct application of the Distributive Property:
:$\dfrac 1 n b + \dfrac 1 n d = \dfrac 1 n \paren {b + d}$
{{qed}} | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | A direct application of the [[Distributive Property]]:
:$\dfrac 1 n b + \dfrac 1 n d = \dfrac 1 n \paren {b + d}$
{{qed}} | Divisors obey Distributive Law/Proof 2 | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law/Proof_2 | [
"Divisors obey Distributive Law",
"Divisors",
"Distributive Operations",
"Algebra"
] | [] | [
"Distributive Laws/Arithmetic"
] |
proofwiki-4159 | Divisors obey Distributive Law | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$.
Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$.
We need to show that $AB + DE$ is the same aliquant part of $C + F$.
:250px
We have that whatever aliquant part $AB$ is of $C$, $DE$... | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | Let the [[Definition:Natural Number|(natural) number]] $AB$ be an [[Definition:Aliquant Part|aliquant part]] of the [[Definition:Natural Number|(natural) number]] $C$.
Let the [[Definition:Natural Number|(natural) number]] $DE$ be the same [[Definition:Aliquant Part|aliquant part]] of another [[Definition:Natural Numb... | Multiples of Divisors obey Distributive Law/Proof 1 | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law/Proof_1 | [
"Divisors obey Distributive Law",
"Divisors",
"Distributive Operations",
"Algebra"
] | [] | [
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"File:Euclid-VII-6.p... |
proofwiki-4160 | Divisors obey Distributive Law | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | A direct application of the Distributive Property:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}} | {{:Euclid:Proposition/VII/5}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a + c = \dfrac 1 n \paren {b + d}$ | A direct application of the [[Distributive Property]]:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}} | Multiples of Divisors obey Distributive Law/Proof 2 | https://proofwiki.org/wiki/Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law/Proof_2 | [
"Divisors obey Distributive Law",
"Divisors",
"Distributive Operations",
"Algebra"
] | [] | [
"Distributive Laws/Arithmetic"
] |
proofwiki-4161 | Open Set Disjoint from Set is Disjoint from Closure | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $B$ be an open set of $T$ disjoint from $A$.
Then:
: $A^- \cap B = \O$
where $A^-$ denotes the closure of $A$. | Since $B \in \tau$, it follows by definition that $S \setminus B$ is closed.
Thus:
{{begin-eqn}}
{{eqn | l = A \cap B
| r = \O
| c = {{Defof|Disjoint Sets}}
}}
{{eqn | ll= \leadsto
| l = A
| o = \subseteq
| r = S \setminus B
| c = Empty Intersection iff Subset of Complement
}}
{{eqn ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$.
Let $B$ be an [[Definition:Open Set (Topology)|open set]] of $T$ [[Definition:Disjoint Sets|disjoint]] from $A$.
Then:
: $A^- \cap B = \O$
where $A^-$ denotes the [[Definition:Closure (Topology)|closure]] of ... | Since $B \in \tau$, it follows by definition that $S \setminus B$ is [[Definition:Closed Set (Topology)|closed]].
Thus:
{{begin-eqn}}
{{eqn | l = A \cap B
| r = \O
| c = {{Defof|Disjoint Sets}}
}}
{{eqn | ll= \leadsto
| l = A
| o = \subseteq
| r = S \setminus B
| c = [[Empty Interse... | Open Set Disjoint from Set is Disjoint from Closure | https://proofwiki.org/wiki/Open_Set_Disjoint_from_Set_is_Disjoint_from_Closure | https://proofwiki.org/wiki/Open_Set_Disjoint_from_Set_is_Disjoint_from_Closure | [
"Set Closures",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Disjoint Sets",
"Definition:Closure (Topology)"
] | [
"Definition:Closed Set/Topology",
"Empty Intersection iff Subset of Complement",
"Topological Closure of Subset is Subset of Topological Closure",
"Set is Closed iff Equals Topological Closure",
"Intersection with Complement is Empty iff Subset",
"Category:Set Closures",
"Category:Open Sets"
] |
proofwiki-4162 | Gauss's Lemma on Primitive Rational Polynomials | Let $\Q$ be the field of rational numbers.
Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.
Then their product $f g$ is also a primitive polynomial. | Let $f$ and $g$ be as follows:
{{begin-eqn}}
{{eqn | l = f
| r = \sum_{k \mathop \in \Z} a_k X^k
}}
{{eqn | l = g
| r = \sum_{k \mathop \in \Z} b_k X^k
}}
{{end-eqn}}
From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.
From the definition of polynomial product:
:$\... | Let $\Q$ be the [[Definition:Field of Rational Numbers|field of rational numbers]].
Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] over $\Q$ in one [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be [[Definition:Primitive Po... | Let $f$ and $g$ be as follows:
{{begin-eqn}}
{{eqn | l = f
| r = \sum_{k \mathop \in \Z} a_k X^k
}}
{{eqn | l = g
| r = \sum_{k \mathop \in \Z} b_k X^k
}}
{{end-eqn}}
From the definition of [[Definition:Primitive Polynomial (Ring Theory)|primitive polynomial]], the [[Definition:Polynomial Coefficient|coeff... | Gauss's Lemma on Primitive Rational Polynomials/Proof 1 | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Rational_Polynomials | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Rational_Polynomials/Proof_1 | [
"Polynomial Theory",
"Gauss's Lemma (Polynomial Theory)"
] | [
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Multiplication of Polynomials",
"Definition:Primitive Polynomial (Ring Theory)"
] | [
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Coefficient of Polynomial",
"Definition:Integer",
"Definition:Multiplication of Polynomials/Polynomial Forms",
"Definition:Coefficient of Polynomial",
"Definition:Integer",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Coeffic... |
proofwiki-4163 | Gauss's Lemma on Primitive Rational Polynomials | Let $\Q$ be the field of rational numbers.
Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.
Then their product $f g$ is also a primitive polynomial. | Recall Polynomial has Integer Coefficients iff Content is Integer.
By hypothesis $f$ and $g$ have content $1 \in \Z$.
Therefore:
:$f, g \in \Z \sqbrk X$
{{AimForCont}} that $f g$ is not primitive, say:
:$\cont {f g} = d \ne 1$
By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.
From Quotien... | Let $\Q$ be the [[Definition:Field of Rational Numbers|field of rational numbers]].
Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] over $\Q$ in one [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
Let $\map f X, \map g X \in \Q \sqbrk X$ be [[Definition:Primitive Po... | Recall [[Polynomial has Integer Coefficients iff Content is Integer]].
By hypothesis $f$ and $g$ have [[Definition:Content of Rational Polynomial|content]] $1 \in \Z$.
Therefore:
:$f, g \in \Z \sqbrk X$
{{AimForCont}} that $f g$ is not [[Definition:Primitive Polynomial (Ring Theory)|primitive]], say:
:$\cont {f g} ... | Gauss's Lemma on Primitive Rational Polynomials/Proof 2 | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Rational_Polynomials | https://proofwiki.org/wiki/Gauss's_Lemma_on_Primitive_Rational_Polynomials/Proof_2 | [
"Polynomial Theory",
"Gauss's Lemma (Polynomial Theory)"
] | [
"Definition:Field of Rational Numbers",
"Definition:Polynomial Ring",
"Definition:Polynomial Ring/Indeterminate",
"Definition:Primitive Polynomial (Ring Theory)",
"Definition:Multiplication of Polynomials",
"Definition:Primitive Polynomial (Ring Theory)"
] | [
"Polynomial has Integer Coefficients iff Content is Integer",
"Definition:Content of Polynomial/Rational",
"Definition:Primitive Polynomial (Ring Theory)",
"Fundamental Theorem of Arithmetic",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Quotient Epimorphism from Integers by Princ... |
proofwiki-4164 | Real Numbers under Addition Modulo 1 form Group | Let $S = \set {x \in \R: 0 \le x < 1}$.
Let $\circ: S \times S \to S$ be the operation defined as:
:$x \circ y = x + y - \floor {x + y}$
That is, $\circ$ is defined as addition modulo $1$.
Then $\struct {S, \circ}$ is a group. | First note that Modulo Addition is Well-Defined.
Taking the group axioms in turn: | Let $S = \set {x \in \R: 0 \le x < 1}$.
Let $\circ: S \times S \to S$ be the [[Definition:Binary Operation|operation]] defined as:
:$x \circ y = x + y - \floor {x + y}$
That is, $\circ$ is defined as [[Definition:Modulo Addition|addition modulo $1$]].
Then $\struct {S, \circ}$ is a [[Definition:Group|group]]. | First note that [[Modulo Addition is Well-Defined]].
Taking the [[Axiom:Group Axioms|group axioms]] in turn: | Real Numbers under Addition Modulo 1 form Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_Modulo_1_form_Group | https://proofwiki.org/wiki/Real_Numbers_under_Addition_Modulo_1_form_Group | [
"Examples of Groups"
] | [
"Definition:Operation/Binary Operation",
"Definition:Modulo Addition",
"Definition:Group"
] | [
"Modulo Addition is Well-Defined",
"Axiom:Group Axioms",
"Axiom:Group Axioms"
] |
proofwiki-4165 | Preimage of Cover is Cover | Let $\phi: S \to T$ be a mapping between the sets $S$ and $T$.
Let $\UU$ be a cover of $T$.
Then the set:
:$\set {\map {\phi^{-1} } U: U \in \UU}$
is a cover of $S$. | Let $x \in S$.
Then $\map \phi x \in T$.
Since $\UU$ is a cover of $T$:
:$\exists U \in \UU: \map \phi x \in U$
By definition of preimage, $x \in \map {\phi^{-1} } U$.
So:
:$\forall x \in S: \exists \map {\phi^{-1} } U \in S: x \in \map {\phi^{-1} } U$
That is, $\set {\map {\phi^{-1} } U: U \in \UU}$ is a cover of $S$.... | Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]] between the [[Definition:Set|sets]] $S$ and $T$.
Let $\UU$ be a [[Definition:Cover of Set|cover]] of $T$.
Then the set:
:$\set {\map {\phi^{-1} } U: U \in \UU}$
is a [[Definition:Cover of Set|cover]] of $S$. | Let $x \in S$.
Then $\map \phi x \in T$.
Since $\UU$ is a [[Definition:Cover of Set|cover]] of $T$:
:$\exists U \in \UU: \map \phi x \in U$
By definition of [[Definition:Preimage of Subset under Mapping|preimage]], $x \in \map {\phi^{-1} } U$.
So:
:$\forall x \in S: \exists \map {\phi^{-1} } U \in S: x \in \map {\p... | Preimage of Cover is Cover | https://proofwiki.org/wiki/Preimage_of_Cover_is_Cover | https://proofwiki.org/wiki/Preimage_of_Cover_is_Cover | [
"Preimages under Mappings",
"Covers"
] | [
"Definition:Mapping",
"Definition:Set",
"Definition:Cover of Set",
"Definition:Cover of Set"
] | [
"Definition:Cover of Set",
"Definition:Preimage/Mapping/Subset",
"Definition:Cover of Set",
"Category:Preimages under Mappings",
"Category:Covers"
] |
proofwiki-4166 | Compactness is Preserved under Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is compact, then $T_B$ is also compact. | Let $T_A$ be compact.
Take an open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\set {\phi^{-1} \sq... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
If $T_A$ is [[Definition:Compact Topological Space|compact]... | Let $T_A$ be [[Definition:Compact Topological Space|compact]].
Take an [[Definition:Open Cover|open cover]] $\UU$ of $T_B$.
From [[Preimage of Cover is Cover]], $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a [[Definition:Cover of Set|cover]] of $S_A$.
But $\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|... | Compactness is Preserved under Continuous Surjection | https://proofwiki.org/wiki/Compactness_is_Preserved_under_Continuous_Surjection | https://proofwiki.org/wiki/Compactness_is_Preserved_under_Continuous_Surjection | [
"Compactness Properties Preserved under Continuous Surjection",
"Compact Topological Spaces",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space"
] | [
"Definition:Compact Topological Space",
"Definition:Open Cover",
"Preimage of Cover is Cover",
"Definition:Cover of Set",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Compact Topological Sp... |
proofwiki-4167 | Sigma-Compactness is Preserved under Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is $\sigma$-compact, then $T_B$ is also $\sigma$-compact. | Let $T_A$ be $\sigma$-compact.
Then:
:$\ds S_A = \bigcup_{i \mathop = 1}^\infty S_i$
where $S_i \subseteq S_A$ are compact.
Since $\phi$ is surjective, we have from Image of Union under Relation:
:$\ds \phi \sqbrk {S_A} = S_B = \phi \sqbrk {\bigcup_{i \mathop = 1}^\infty S_i} = \bigcup_{i \mathop = 1}^\infty \phi \sqbr... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
If $T_A$ is [[Definition:Sigma-Compact Space|$\sigma$-compa... | Let $T_A$ be [[Definition:Sigma-Compact Space|$\sigma$-compact]].
Then:
:$\ds S_A = \bigcup_{i \mathop = 1}^\infty S_i$
where $S_i \subseteq S_A$ are [[Definition:Compact Topological Space|compact]].
Since $\phi$ is [[Definition:Surjection|surjective]], we have from [[Image of Union under Relation]]:
:$\ds \phi \sqbr... | Sigma-Compactness is Preserved under Continuous Surjection | https://proofwiki.org/wiki/Sigma-Compactness_is_Preserved_under_Continuous_Surjection | https://proofwiki.org/wiki/Sigma-Compactness_is_Preserved_under_Continuous_Surjection | [
"Compactness Properties Preserved under Continuous Surjection",
"Sigma-Compact Spaces",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Sigma-Compact Space",
"Definition:Sigma-Compact Space"
] | [
"Definition:Sigma-Compact Space",
"Definition:Compact Topological Space",
"Definition:Surjection",
"Image of Union under Relation",
"Compactness is Preserved under Continuous Surjection",
"Definition:Compact Topological Space",
"Definition:Set Union",
"Definition:Countable Set",
"Definition:Compact ... |
proofwiki-4168 | Countable Compactness is Preserved under Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is countably compact, then $T_B$ is also countably compact. | Let $T_A$ be countably compact.
Take a countable open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\VV := \set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
$\VV$ is a countable cover because there it is bijective with $\UU$.
{{explain|It needs to be demonstrated that "it" is bijective, and in this c... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
If $T_A$ is [[Definition:Countably Compact Space|countably ... | Let $T_A$ be [[Definition:Countably Compact Space|countably compact]].
Take a [[Definition:Countable Cover|countable]] [[Definition:Open Cover|open cover]] $\UU$ of $T_B$.
From [[Preimage of Cover is Cover]], $\VV := \set {\phi^{-1} \sqbrk U: U \in \UU}$ is a [[Definition:Cover of Set|cover]] of $S_A$.
$\VV$ is a [[... | Countable Compactness is Preserved under Continuous Surjection | https://proofwiki.org/wiki/Countable_Compactness_is_Preserved_under_Continuous_Surjection | https://proofwiki.org/wiki/Countable_Compactness_is_Preserved_under_Continuous_Surjection | [
"Compactness Properties Preserved under Continuous Surjection",
"Countably Compact Spaces",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Countably Compact Space",
"Definition:Countably Compact Space"
] | [
"Definition:Countably Compact Space",
"Definition:Cover of Set/Countable",
"Definition:Open Cover",
"Preimage of Cover is Cover",
"Definition:Cover of Set",
"Definition:Cover of Set/Countable",
"Definition:Bijection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topolo... |
proofwiki-4169 | Sequential Compactness is Preserved under Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is sequentially compact, then $T_B$ is also sequentially compact. | Let $T_A$ be a sequentially compact space.
Take an infinite sequence $\sequence {x_n} \subseteq S_B$.
From the surjectivity of $\phi$, there exists another infinite sequence $\sequence {y_n} \subseteq S_A$ such that $\map \phi {y_n} = x_n$.
By the definition of sequential compactness, there exists a subsequence of $\se... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
If $T_A$ is [[Definition:Sequentially Compact Space|sequent... | Let $T_A$ be a [[Definition:Sequentially Compact Space|sequentially compact]] space.
Take an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n} \subseteq S_B$.
From the [[Definition:Surjection|surjectivity]] of $\phi$, there exists another [[Definition:Infinite Sequence|infinite sequence]] $\sequence... | Sequential Compactness is Preserved under Continuous Surjection | https://proofwiki.org/wiki/Sequential_Compactness_is_Preserved_under_Continuous_Surjection | https://proofwiki.org/wiki/Sequential_Compactness_is_Preserved_under_Continuous_Surjection | [
"Compactness Properties Preserved under Continuous Surjection",
"Sequentially Compact Spaces",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Sequentially Compact Space",
"Definition:Sequentially Compact Space"
] | [
"Definition:Sequentially Compact Space",
"Definition:Sequence/Infinite Sequence",
"Definition:Surjection",
"Definition:Sequence/Infinite Sequence",
"Definition:Sequentially Compact Space",
"Definition:Subsequence",
"Definition:Subsequence",
"Definition:Convergent Sequence/Topology",
"Definition:Conv... |
proofwiki-4170 | Lindelöf Property is Preserved under Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a continuous surjection.
If $T_A$ is a Lindelöf space, then $T_B$ is also a Lindelöf space. | Let $T_A$ be a Lindelöf space.
Take an open cover $\UU$ of $T_B$.
From Preimage of Cover is Cover, $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a cover of $S_A$.
But $\phi$ is continuous, and for all $U \in \UU$, $U$ is open in $T_B$.
It follows that $\forall U \in \UU: \phi^{-1} \sqbrk U$ is open in $T_A$.
So $\set {\phi... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
If $T_A$ is a [[Definition:Lindelöf Space|Lindelöf space]],... | Let $T_A$ be a [[Definition:Lindelöf Space|Lindelöf space]].
Take an [[Definition:Open Cover|open cover]] $\UU$ of $T_B$.
From [[Preimage of Cover is Cover]], $\set {\phi^{-1} \sqbrk U: U \in \UU}$ is a [[Definition:Cover of Set|cover]] of $S_A$.
But $\phi$ is [[Definition:Everywhere Continuous Mapping (Topology)|co... | Lindelöf Property is Preserved under Continuous Surjection | https://proofwiki.org/wiki/Lindelöf_Property_is_Preserved_under_Continuous_Surjection | https://proofwiki.org/wiki/Lindelöf_Property_is_Preserved_under_Continuous_Surjection | [
"Compactness Properties Preserved under Continuous Surjection",
"Lindelöf Spaces",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Surjection",
"Definition:Lindelöf Space",
"Definition:Lindelöf Space"
] | [
"Definition:Lindelöf Space",
"Definition:Open Cover",
"Preimage of Cover is Cover",
"Definition:Cover of Set",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Cover",
"Definition:Lindelöf Space",
"Definition:Su... |
proofwiki-4171 | Multiples of Divisors obey Distributive Law | {{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$ | Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$.
Let the (natural) number $DE$ be the same aliquant part of another (natural) number $F$ that $AB$ is of $C$.
We need to show that $AB + DE$ is the same aliquant part of $C + F$.
:250px
We have that whatever aliquant part $AB$ is of $C$, $DE$... | {{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$ | Let the [[Definition:Natural Number|(natural) number]] $AB$ be an [[Definition:Aliquant Part|aliquant part]] of the [[Definition:Natural Number|(natural) number]] $C$.
Let the [[Definition:Natural Number|(natural) number]] $DE$ be the same [[Definition:Aliquant Part|aliquant part]] of another [[Definition:Natural Numb... | Multiples of Divisors obey Distributive Law/Proof 1 | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law/Proof_1 | [
"Algebra",
"Divisors",
"Multiples of Divisors obey Distributive Law"
] | [] | [
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Natural Numbers",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"File:Euclid-VII-6.p... |
proofwiki-4172 | Multiples of Divisors obey Distributive Law | {{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$ | A direct application of the Distributive Property:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}} | {{:Euclid:Proposition/VII/6}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \paren {b + d}$ | A direct application of the [[Distributive Property]]:
:$\dfrac m n b + \dfrac m n d = \dfrac m n \paren {b + d}$
{{qed}} | Multiples of Divisors obey Distributive Law/Proof 2 | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law | https://proofwiki.org/wiki/Multiples_of_Divisors_obey_Distributive_Law/Proof_2 | [
"Algebra",
"Divisors",
"Multiples of Divisors obey Distributive Law"
] | [] | [
"Distributive Laws/Arithmetic"
] |
proofwiki-4173 | First-Countability is Preserved under Open Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is first-countable, then $T_B$ is also first-countable. | Let $\phi$ be surjective, continuous and open.
Let $T_A$ be first countable.
Let $b \in S_B$.
Since $\phi$ is surjective there is a point $a \in S_A$ such that:
:$\map \phi a = b$
From the first-countability of $T_A$, there is a local base $\BB$, say, of $a$ which is countable.
Let $\BB = \set {V_n: n \in \N}$.
We need... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Surjection|surjective]] [[Definition:Open Mapping|open mapping]] which is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
If $... | Let $\phi$ be [[Definition:Surjection|surjective]], [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and [[Definition:Open Mapping|open]].
Let $T_A$ be [[Definition:First-Countable Space|first countable]].
Let $b \in S_B$.
Since $\phi$ is [[Definition:Surjection|surjective]] there is a point $a \i... | First-Countability is Preserved under Open Continuous Surjection | https://proofwiki.org/wiki/First-Countability_is_Preserved_under_Open_Continuous_Surjection | https://proofwiki.org/wiki/First-Countability_is_Preserved_under_Open_Continuous_Surjection | [
"First-Countable Spaces",
"Open Mappings",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Surjection",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:First-Countable Space",
"Definition:First-Countable Space"
] | [
"Definition:Surjection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Mapping",
"Definition:First-Countable Space",
"Definition:Surjection",
"Definition:First-Countable Space",
"Definition:Local Basis",
"Definition:Countable Set",
"Definition:Local Basis",
"Definition:Op... |
proofwiki-4174 | Second-Countability is Preserved under Open Continuous Surjection | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is second-countable, then $T_B$ is also second-countable. | Let $\phi$ be surjective, continuous and open.
Let $T_A$ be second-countable.
By definition of second-countability $T_A$ has a countable basis, $\BB$, say.
Let $\BB = \set {V_n: n \in \N}$.
We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$.
Let $U$ be an open set of $T_B$.
$\phi$ is continuo... | Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be [[Definition:Topological Space|topological spaces]].
Let $\phi: T_A \to T_B$ be a [[Definition:Surjection|surjective]] [[Definition:Open Mapping|open mapping]] which is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
If $... | Let $\phi$ be [[Definition:Surjection|surjective]], [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and [[Definition:Open Mapping|open]].
Let $T_A$ be [[Definition:Second-Countable Space|second-countable]].
By definition of [[Definition:Second-Countable Space|second-countability]] $T_A$ has a [[Def... | Second-Countability is Preserved under Open Continuous Surjection | https://proofwiki.org/wiki/Second-Countability_is_Preserved_under_Open_Continuous_Surjection | https://proofwiki.org/wiki/Second-Countability_is_Preserved_under_Open_Continuous_Surjection | [
"Second-Countable Spaces",
"Open Mappings",
"Continuous Mappings",
"Surjections"
] | [
"Definition:Topological Space",
"Definition:Surjection",
"Definition:Open Mapping",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Second-Countable Space",
"Definition:Second-Countable Space"
] | [
"Definition:Surjection",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Open Mapping",
"Definition:Second-Countable Space",
"Definition:Second-Countable Space",
"Definition:Countable Basis",
"Definition:Basis (Topology)",
"Definition:Open Set/Topology",
"Definition:Continuous Map... |
proofwiki-4175 | Arens-Fort Space is Countable | Let $T = \struct {S, \tau}$ be the Arens-Fort space.
Then $S$ is countably infinite. | From the definition of the Arens-Fort space:
$S$ is the the Cartesian product $\Z_{\ge 0} \times \Z_{\ge 0}$ of the set of all positive integers:
:$S = \set {0, 1, 2, \ldots} \times \set {0, 1, 2, \ldots}$
We have by definition that $\Z_{\ge 0}$ is countable.
From Cartesian Product of Countable Sets is Countable it fol... | Let $T = \struct {S, \tau}$ be the [[Definition:Arens-Fort Space|Arens-Fort space]].
Then $S$ is [[Definition:Countable|countably infinite]]. | From the definition of the [[Definition:Arens-Fort Space|Arens-Fort space]]:
$S$ is the the [[Definition:Cartesian Product|Cartesian product]] $\Z_{\ge 0} \times \Z_{\ge 0}$ of the [[Definition:Set|set]] of all [[Definition:Positive Integer|positive integers]]:
:$S = \set {0, 1, 2, \ldots} \times \set {0, 1, 2, \ldots... | Arens-Fort Space is Countable | https://proofwiki.org/wiki/Arens-Fort_Space_is_Countable | https://proofwiki.org/wiki/Arens-Fort_Space_is_Countable | [
"Arens-Fort Space"
] | [
"Definition:Arens-Fort Space",
"Definition:Countable Set"
] | [
"Definition:Arens-Fort Space",
"Definition:Cartesian Product",
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Countable Set",
"Cartesian Product of Countable Sets is Countable",
"Definition:Countable Set",
"Category:Arens-Fort Space"
] |
proofwiki-4176 | Connected iff no Proper Clopen Sets | Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is connected
{{iff}}:
:there exists no non-empty proper subset of $S$ which is clopen in $T$. | === Sufficient Condition ===
Let $T$ be connected.
{{AimForCont}} there exists $H \subset S$ such that:
:$H$ is clopen in $T$
:$H$ is a non-empty proper subset of $S$, that is:
::$\O \ne H \ne S$
Then $H$ and $\relcomp S H$ are open sets whose union is $S$.
Thus $\set {H \mid \relcomp S H}$ form a partition of $S$.
Hen... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then:
:$T$ is [[Definition:Connected Topological Space|connected]]
{{iff}}:
:there exists no [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subset]] of $S$ which is [[Definition:Clopen Set|clopen]] in $T$. | === Sufficient Condition ===
Let $T$ be [[Definition:Connected Topological Space|connected]].
{{AimForCont}} there exists $H \subset S$ such that:
:$H$ is [[Definition:Clopen Set|clopen]] in $T$
:$H$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subset]] of $S$, that is:
::$\O \ne H \n... | Connected iff no Proper Clopen Sets | https://proofwiki.org/wiki/Connected_iff_no_Proper_Clopen_Sets | https://proofwiki.org/wiki/Connected_iff_no_Proper_Clopen_Sets | [
"Connected Topological Spaces"
] | [
"Definition:Topological Space",
"Definition:Connected Topological Space",
"Definition:Non-Empty Set",
"Definition:Proper Subset",
"Definition:Clopen Set"
] | [
"Definition:Connected Topological Space",
"Definition:Clopen Set",
"Definition:Non-Empty Set",
"Definition:Proper Subset",
"Definition:Open Set/Topology",
"Definition:Set Union",
"Definition:Separation (Topology)",
"Definition:Connected Topological Space",
"Definition:Contradiction",
"Proof by Con... |
proofwiki-4177 | Element of Principal Ideal Domain is Finite Product of Irreducible Elements | Let $R$ be a principal ideal domain.
Let $p \in R$ such that $p \ne 0$ and $p$ is not a unit.
Then there exist irreducible elements $p_1, \ldots, p_n$ such that $p = p_1 \cdots p_n$. | Let $S = \set {\ideal p: p \in R \setminus \set 0, p \text{ is not a unit, and } p \text{ is not a product of finitely many irreducible elements} }$.
Define a relation $\RR$ on $S$ by $\ideal a \mathrel \RR \ideal b$ {{iff}} $\ideal a \subsetneq \ideal b$.
For any $\ideal p \in S$, since $p$ is not irreducible, we can ... | Let $R$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Let $p \in R$ such that $p \ne 0$ and $p$ is not a [[Definition:Unit of Ring|unit]].
Then there exist [[Definition:Irreducible Element of Ring|irreducible elements]] $p_1, \ldots, p_n$ such that $p = p_1 \cdots p_n$. | Let $S = \set {\ideal p: p \in R \setminus \set 0, p \text{ is not a unit, and } p \text{ is not a product of finitely many irreducible elements} }$.
Define a [[Definition:Relation|relation]] $\RR$ on $S$ by $\ideal a \mathrel \RR \ideal b$ {{iff}} $\ideal a \subsetneq \ideal b$.
For any $\ideal p \in S$, since $p$ i... | Element of Principal Ideal Domain is Finite Product of Irreducible Elements/Proof 2 | https://proofwiki.org/wiki/Element_of_Principal_Ideal_Domain_is_Finite_Product_of_Irreducible_Elements | https://proofwiki.org/wiki/Element_of_Principal_Ideal_Domain_is_Finite_Product_of_Irreducible_Elements/Proof_2 | [
"Element of Principal Ideal Domain is Finite Product of Irreducible Elements",
"Principal Ideal Domains",
"Ideal Theory",
"Factorization"
] | [
"Definition:Principal Ideal Domain",
"Definition:Unit of Ring",
"Definition:Irreducible Element of Ring"
] | [
"Definition:Relation",
"Definition:Irreducible Element of Ring",
"Definition:Unit of Ring",
"Definition:Ideal of Ring",
"Definition:Left-Total Relation",
"Axiom:Axiom of Dependent Choice",
"Principal Ideal Domain fulfills Ascending Chain Condition"
] |
proofwiki-4178 | Principal Ideal Domain fulfills Ascending Chain Condition | Let $R$ be a principal ideal domain.
Then $R$ fulfils the ascending chain condition. | Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \dotsb$ be an ascending chain of ideals.
Build $\ds I = \bigcup_{i \mathop = 1}^\infty I_i$.
$I$ is an ideal.
Since $R$ is a principal ideal domain, $I = \ideal a$ for some $a \in R$.
Now, since $a \in I$, there is some $n$ such that $a \in I_n$.
Thus:
:$\ideal a \subsete... | Let $R$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
Then $R$ fulfils the [[Definition:Ascending Chain Condition|ascending chain condition]]. | Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \dotsb$ be an [[Definition:Ascending Chain Condition|ascending chain of ideals]].
Build $\ds I = \bigcup_{i \mathop = 1}^\infty I_i$.
$I$ is an [[Definition:Ideal of Ring|ideal]].
Since $R$ is a [[Definition:Principal Ideal Domain|principal ideal domain]], $I = \ideal ... | Principal Ideal Domain fulfills Ascending Chain Condition | https://proofwiki.org/wiki/Principal_Ideal_Domain_fulfills_Ascending_Chain_Condition | https://proofwiki.org/wiki/Principal_Ideal_Domain_fulfills_Ascending_Chain_Condition | [
"Ideal Theory",
"Principal Ideal Domains"
] | [
"Definition:Principal Ideal Domain",
"Definition:Ascending Chain Condition"
] | [
"Definition:Ascending Chain Condition",
"Definition:Ideal of Ring",
"Definition:Principal Ideal Domain",
"Category:Ideal Theory",
"Category:Principal Ideal Domains"
] |
proofwiki-4179 | Subspace of Product Space is Homeomorphic to Factor Space | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then fo... | From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty.
The conclusions follow immediately from Subspace of Product Space is Homeomorphic to Factor Space.
{{qed}} | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the [[Definition:Prod... | From [[Finite Cartesian Product of Non-Empty Sets is Non-Empty]] both $T_1 \times T_2$ and $T_2 \times T_1$ are both [[Definition:Non-Empty Set|non-empty]].
The conclusions follow immediately from [[Subspace of Product Space is Homeomorphic to Factor Space]].
{{qed}} | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 1 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton/Proof_1 | [
"Subspace of Product Space is Homeomorphic to Factor Space",
"Product Spaces",
"Topological Subspaces",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Non-Empty Set",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphi... | [
"Finite Cartesian Product of Non-Empty Sets is Non-Empty",
"Definition:Non-Empty Set",
"Subspace of Product Space is Homeomorphic to Factor Space"
] |
proofwiki-4180 | Subspace of Product Space is Homeomorphic to Factor Space | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then fo... | The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== Lemma ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Product with... | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the [[Definition:Prod... | The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is [[Definition:Homeomorphism|homeomorphic]] to the [[Definition:Topological Subspace|subspace]] $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== [[Subspace of ... | Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 2 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Product_with_Singleton/Proof_2 | [
"Subspace of Product Space is Homeomorphic to Factor Space",
"Product Spaces",
"Topological Subspaces",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Non-Empty Set",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphi... | [
"Definition:Homeomorphism",
"Definition:Topological Subspace",
"Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Lemma",
"Definition:Restriction/Mapping",
"Definition:Subspace",
"Definition:Projection",
"Projection from Product Topology is Continuous ",
"Definition:Cont... |
proofwiki-4181 | Subspace of Product Space is Homeomorphic to Factor Space | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then fo... | Let $z \in X$.
Let $i \in i$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
For all $j \in I$ let:
::<nowiki>$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$</nowiki>
=== Lemma 1 ===
{{:Subspace of Pro... | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the [[Definition:Prod... | Let $z \in X$.
Let $i \in i$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the [[Definition:Subspace Topology|subspace topology]] of $Y_i$ relative to $\tau$.
For all $j \in I$ let:
::<nowiki>$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$</nowik... | Subspace of Product Space is Homeomorphic to Factor Space/Proof 1 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Proof_1 | [
"Subspace of Product Space is Homeomorphic to Factor Space",
"Product Spaces",
"Topological Subspaces",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Non-Empty Set",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphi... | [
"Definition:Topological Subspace",
"Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1",
"Product Space of Subspaces is Subspace of Product Space ",
"Definition:Product Space (Topology)",
"Definition:Projection (Mapping Theory)/Family of Sets",
"Projection is Injection iff Factor is... |
proofwiki-4182 | Subspace of Product Space is Homeomorphic to Factor Space | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then fo... | Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Let $p_i = \pr_i {\restriction_{Y_i}}$.
Note that by definitions of a restriction and a projection then:
:$\forall y \in Y_i: \map {p_i} y = y_i... | Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the [[Definition:Prod... | Let $z \in X$.
Let $i \in I$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the [[Definition:Subspace Topology|subspace topology]] of $Y_i$ relative to $\tau$.
Let $p_i = \pr_i {\restriction_{Y_i}}$.
Note that by definitions of a [[Definition:Restriction of Mapping|... | Subspace of Product Space is Homeomorphic to Factor Space/Proof 2 | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space | https://proofwiki.org/wiki/Subspace_of_Product_Space_is_Homeomorphic_to_Factor_Space/Proof_2 | [
"Subspace of Product Space is Homeomorphic to Factor Space",
"Product Spaces",
"Topological Subspaces",
"Homeomorphisms (Topological Spaces)"
] | [
"Definition:Indexing Set/Family",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Non-Empty Set",
"Definition:Topological Subspace",
"Definition:Homeomorphism/Topological Spaces",
"Definition:Topological Subspace",
"Definition:Homeomorphi... | [
"Definition:Topological Subspace",
"Definition:Restriction/Mapping",
"Definition:Projection",
"Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Injection",
"Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Surjection",
"Subspace of Product Space is Homeomorphic to Factor ... |
proofwiki-4183 | Closed Set of Countable Fort Space is G-Delta | Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Let $H \subseteq S$ be closed in $T$.
Then $H$ is a $G_\delta$ set. | By definition of Fort space, $H$ is finite or contains $p$.
Consider the set of sets defined by:
:$\DD = \set {S \setminus \set z: z \notin H}$
Because $H$ is finite, $S \setminus H$ is countably infinite.
From its method of construction, $\DD$ has the same cardinality as $S \setminus H$ and so is countably infinite.
W... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Fort Space|Fort space]] on a [[Definition:Countable|countably infinite set]] $S$.
Let $H \subseteq S$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
Then $H$ is a [[Definition:G-Delta Set|$G_\delta$ set]]. | By definition of [[Definition:Fort Space|Fort space]], $H$ is [[Definition:Finite Set|finite]] or contains $p$.
Consider the [[Definition:Set of Sets|set of sets]] defined by:
:$\DD = \set {S \setminus \set z: z \notin H}$
Because $H$ is [[Definition:Finite Set|finite]], $S \setminus H$ is [[Definition:Countably Infi... | Closed Set of Countable Fort Space is G-Delta | https://proofwiki.org/wiki/Closed_Set_of_Countable_Fort_Space_is_G-Delta | https://proofwiki.org/wiki/Closed_Set_of_Countable_Fort_Space_is_G-Delta | [
"Countable Fort Spaces",
"Examples of G-Delta Sets"
] | [
"Definition:Fort Space",
"Definition:Countable Set",
"Definition:Closed Set/Topology",
"Definition:G-Delta Set"
] | [
"Definition:Fort Space",
"Definition:Finite Set",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Countably Infinite/Set",
"Definition:Cardinality",
"Definition:Countably Infinite/Set",
"Definition:Open Set/Topology",
"Definition:G-Delta Set",
"Category:Countable Fort Spaces",
"Ca... |
proofwiki-4184 | Fundamental Theorem of Galois Theory | Let $L / K$ be a finite Galois extension.
Let $\Gal {L / K}$ denote the Galois group of the extension $L / K$.
Let $H$ denote a subgroup of $\Gal {L / K}$ and $F$ denote an intermediate field.
Let $L_H$ denote the fixed field of $H$.
The mappings:
:$H \mapsto L_H$, and
:$F \mapsto \Gal {L / F}$
are inclusion-reversing ... | First, we show that the maps are inclusion-reversing.
Let $K \subset F_1 \subset F_2 \subset L$.
Let $G_i = \Gal {L / F_i}$.
Let $\sigma \in G_2$.
Then $\sigma$ is an automorphism of $L$ which fixes $F_2$.
Since $F_1 \subset F_2$, it follows that $\sigma$ fixes $F_1$ and consequently $\sigma \in G_1$.
Let $H_1 \subset ... | Let $L / K$ be a [[Definition:Finite Field Extension|finite]] [[Definition:Galois Extension|Galois extension]].
Let $\Gal {L / K}$ denote the [[Definition:Galois Group of Field Extension|Galois group]] of the [[Definition:Field Extension |extension]] $L / K$.
Let $H$ denote a [[Definition:Subgroup|subgroup]] of $\Gal... | First, we show that the maps are [[Definition:Inclusion-Reversing|inclusion-reversing]].
Let $K \subset F_1 \subset F_2 \subset L$.
Let $G_i = \Gal {L / F_i}$.
Let $\sigma \in G_2$.
Then $\sigma$ is an [[Definition:Field Automorphism|automorphism]] of $L$ which [[Definition:Fixed Point|fixes]] $F_2$.
Since $F_1 \s... | Fundamental Theorem of Galois Theory | https://proofwiki.org/wiki/Fundamental_Theorem_of_Galois_Theory | https://proofwiki.org/wiki/Fundamental_Theorem_of_Galois_Theory | [
"Fundamental Theorems",
"Galois Theory"
] | [
"Definition:Field Extension/Degree/Finite",
"Definition:Galois Extension",
"Definition:Galois Group of Field Extension",
"Definition:Field Extension ",
"Definition:Subgroup",
"Definition:Intermediate Field",
"Definition:Fixed Field",
"Definition:Mapping",
"Definition:Inclusion-Reversing Mapping",
... | [
"Definition:Inclusion-Reversing Mapping",
"Definition:Field Automorphism",
"Definition:Fixed Point",
"Definition:Field (Abstract Algebra)",
"Definition:Intermediate Field",
"Definition:Subgroup",
"Definition:Field Extension/Degree",
"Definition:Minimal Polynomial",
"Degree Equation",
"Definition:C... |
proofwiki-4185 | Ring of Algebraic Integers | Let $K / \Q$ be a number field.
Let $\Z \sqbrk x$ denote the polynomial ring in one variable over $\Z$.
Let $\OO_K$ denote the set of all elements of $K / \Q$ which are a root of some monic polynomial $P \in \Z \sqbrk x$.
That is, let $\OO_K$ denote the algebraic integers over $K$.
Then $\OO_K$ is a ring, called the ''... | This is a special case of Integral Closure is Subring.
We have an extension of commutative rings with unity, $\Z \subseteq K$, and $\OO_K$ is the integral closure of $\Z$ in $K$.
The theorem says that $\OO_K$ is a subring of $K$.
{{qed}}
Category:Algebraic Number Theory
oo93c5tvo6i3bunkms20vt61z8taawl | Let $K / \Q$ be a [[Definition:Number Field|number field]].
Let $\Z \sqbrk x$ denote the [[Definition:Polynomial Ring|polynomial ring in one variable over $\Z$]].
Let $\OO_K$ denote the set of all elements of $K / \Q$ which are a [[Definition:Root of Polynomial|root]] of some [[Definition:Monic Polynomial|monic polyn... | This is a special case of [[Integral Closure is Subring]].
We have an [[Definition:Ring Extension|extension]] of [[Definition:Commutative and Unitary Ring|commutative rings with unity]], $\Z \subseteq K$, and $\OO_K$ is the [[Definition:Integral Closure|integral closure]] of $\Z$ in $K$.
The theorem says that $\OO_K$... | Ring of Algebraic Integers | https://proofwiki.org/wiki/Ring_of_Algebraic_Integers | https://proofwiki.org/wiki/Ring_of_Algebraic_Integers | [
"Algebraic Number Theory"
] | [
"Definition:Number Field",
"Definition:Polynomial Ring",
"Definition:Root of Polynomial",
"Definition:Monic Polynomial",
"Definition:Algebraic Integer",
"Definition:Ring (Abstract Algebra)"
] | [
"Integral Closure is Subring",
"Definition:Ring Extension",
"Definition:Commutative and Unitary Ring",
"Definition:Integral Closure",
"Category:Algebraic Number Theory"
] |
proofwiki-4186 | Subtraction of Divisors obeys Distributive Law | {{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$ | Let $AB$ be that aliquot part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted.
We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted.
:500px
Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$.
Then from {{Eucli... | {{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$ | Let $AB$ be that [[Definition:Aliquot Part|aliquot part]] of the [[Definition:Natural Number|(natural) number]] $CD$ which $AE$ subtracted is of $CF$ subtracted.
We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted.
:[[File:Euclid-VII-7.png|500px... | Subtraction of Divisors obeys Distributive Law/Proof 1 | https://proofwiki.org/wiki/Subtraction_of_Divisors_obeys_Distributive_Law | https://proofwiki.org/wiki/Subtraction_of_Divisors_obeys_Distributive_Law/Proof_1 | [
"Algebra",
"Subtraction of Divisors obeys Distributive Law"
] | [] | [
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Natural Numbers",
"File:Euclid-VII-7.png",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Divisor (Algebra)/Integer... |
proofwiki-4187 | Subtraction of Divisors obeys Distributive Law | {{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$ | A direct application of the Distributive Property:
{{begin-eqn}}
{{eqn | l = \frac 1 n b - \frac 1 n d
| r = \frac 1 n b + \frac 1 n \paren {-d}
| c =
}}
{{eqn | r = \frac 1 n \paren {b + \paren {-d} }
| c =
}}
{{eqn | r = \frac 1 n \paren {b - d}
| c =
}}
{{end-eqn}}
{{qed}} | {{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$ | A direct application of the [[Distributive Property]]:
{{begin-eqn}}
{{eqn | l = \frac 1 n b - \frac 1 n d
| r = \frac 1 n b + \frac 1 n \paren {-d}
| c =
}}
{{eqn | r = \frac 1 n \paren {b + \paren {-d} }
| c =
}}
{{eqn | r = \frac 1 n \paren {b - d}
| c =
}}
{{end-eqn}}
{{qed}} | Subtraction of Divisors obeys Distributive Law/Proof 2 | https://proofwiki.org/wiki/Subtraction_of_Divisors_obeys_Distributive_Law | https://proofwiki.org/wiki/Subtraction_of_Divisors_obeys_Distributive_Law/Proof_2 | [
"Algebra",
"Subtraction of Divisors obeys Distributive Law"
] | [] | [
"Distributive Laws/Arithmetic"
] |
proofwiki-4188 | Minimal Polynomial is Unique | Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique. | Let $f$ be a minimal polynomial of $\alpha$ over $K$ according to definition 1.
By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$.
Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.
By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\de... | Let $L / K$ be a [[Definition:Field Extension|field extension]].
Let $\alpha \in L$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
Then the [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $K$ is [[Definition:Unique|unique]]. | Let $f$ be a [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $K$ according to [[Definition:Minimal Polynomial/Definition 1|definition 1]].
By [[Minimal Polynomial is Irreducible]], we have that $f$ is [[Definition:Irreducible Polynomial|irreducible]] over $K$.
Let $g$ be another [[Definition:P... | Minimal Polynomial is Unique/Proof 1 | https://proofwiki.org/wiki/Minimal_Polynomial_is_Unique | https://proofwiki.org/wiki/Minimal_Polynomial_is_Unique/Proof_1 | [
"Minimal Polynomial is Unique",
"Minimal Polynomials"
] | [
"Definition:Field Extension",
"Definition:Algebraic Element of Field Extension",
"Definition:Minimal Polynomial",
"Definition:Unique"
] | [
"Definition:Minimal Polynomial",
"Definition:Minimal Polynomial/Definition 1",
"Minimal Polynomial is Irreducible",
"Definition:Irreducible Polynomial",
"Definition:Polynomial",
"Definition:Minimal Polynomial",
"Definition:Degree of Polynomial",
"Division Theorem for Polynomial Forms over Field",
"D... |
proofwiki-4189 | Minimal Polynomial is Unique | Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique. | Let $f, g \in K \sqbrk x$ be minimal polynomials for $\alpha$ according to definition 2.
That is, let $f$ and $g$ be irreducible monic polynomials in $K \sqbrk x$ with $\map f \alpha = \map g \alpha = 0$.
Suppose $f$ and $g$ are distinct.
Then $f$ and $g$ are coprime.
Thus there exist polynomials $a, b \in K \sqbrk x$ ... | Let $L / K$ be a [[Definition:Field Extension|field extension]].
Let $\alpha \in L$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
Then the [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $K$ is [[Definition:Unique|unique]]. | Let $f, g \in K \sqbrk x$ be [[Definition:Minimal Polynomial|minimal polynomials]] for $\alpha$ according to [[Definition:Minimal Polynomial/Definition 2|definition 2]].
That is, let $f$ and $g$ be [[Definition:Irreducible Polynomial|irreducible]] [[Definition:Monic Polynomial|monic polynomials]] in $K \sqbrk x$ with ... | Minimal Polynomial is Unique/Proof 2 | https://proofwiki.org/wiki/Minimal_Polynomial_is_Unique | https://proofwiki.org/wiki/Minimal_Polynomial_is_Unique/Proof_2 | [
"Minimal Polynomial is Unique",
"Minimal Polynomials"
] | [
"Definition:Field Extension",
"Definition:Algebraic Element of Field Extension",
"Definition:Minimal Polynomial",
"Definition:Unique"
] | [
"Definition:Minimal Polynomial",
"Definition:Minimal Polynomial/Definition 2",
"Definition:Irreducible Polynomial",
"Definition:Monic Polynomial",
"Definition:Coprime Polynomials",
"Definition:Polynomial",
"Definition:Polynomial Evaluation Homomorphism",
"Definition:Contradiction"
] |
proofwiki-4190 | Minimal Polynomial is Irreducible | Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial in $\alpha$ over $K$ is unique and irreducible. | Let $\map f x$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$.
From Minimal Polynomial is Unique we have that $\map f x$ is unique.
Since $K$ is a field, we may assume that the coefficient of $x^n$ is $1$.
{{AimForCont}} that $f$ is not irreducible.
Then there exist non-constant polynomials $g, h \in K \sqb... | Let $L / K$ be a [[Definition:Field Extension|field extension]].
Let $\alpha \in L$ be [[Definition:Algebraic over Field|algebraic]] over $K$.
Then the [[Definition:Minimal Polynomial|minimal polynomial]] in $\alpha$ over $K$ is [[Definition:Unique|unique]] and [[Definition:Irreducible Polynomial|irreducible]]. | Let $\map f x$ be a [[Definition:Minimal Polynomial|minimal polynomial]] of $\alpha$ over $K$ of [[Definition:Degree of Polynomial|degree]] $n$.
From [[Minimal Polynomial is Unique]] we have that $\map f x$ is [[Definition:Unique|unique]].
Since $K$ is a [[Definition:Field (Abstract Algebra)|field]], we may assume t... | Minimal Polynomial is Irreducible | https://proofwiki.org/wiki/Minimal_Polynomial_is_Irreducible | https://proofwiki.org/wiki/Minimal_Polynomial_is_Irreducible | [
"Minimal Polynomials"
] | [
"Definition:Field Extension",
"Definition:Algebraic Element of Field Extension",
"Definition:Minimal Polynomial",
"Definition:Unique",
"Definition:Irreducible Polynomial"
] | [
"Definition:Minimal Polynomial",
"Definition:Degree of Polynomial",
"Minimal Polynomial is Unique",
"Definition:Unique",
"Definition:Field (Abstract Algebra)",
"Definition:Coefficient of Polynomial",
"Definition:Irreducible Polynomial",
"Definition:Constant Polynomial",
"Definition:Minimal Polynomia... |
proofwiki-4191 | Equivalence of Definitions of Normal Extension | Let $L / K$ be an algebraic field extension.
{{TFAE|def = Normal Extension}} | === Definition $1$ implies Definition $2$ ===
Let $\alpha \in L$ be an arbitrary element.
Let $\sigma: L \mapsto \overline K$ be an arbitrary embedding of $L$ fixing $K$.
We wish to show that $\map \sigma \alpha \in L$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$, which exists because $L / K$ is algeb... | Let $L / K$ be an [[Definition:Algebraic Field Extension|algebraic field extension]].
{{TFAE|def = Normal Extension}} | === Definition $1$ implies Definition $2$ ===
Let $\alpha \in L$ be an arbitrary element.
Let $\sigma: L \mapsto \overline K$ be an arbitrary [[Definition:Embedding (Galois Theory)|embedding]] of $L$ fixing $K$.
We wish to show that $\map \sigma \alpha \in L$.
Let $m_\alpha$ be the [[Definition:Minimal Polynomial|m... | Equivalence of Definitions of Normal Extension | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Extension | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Normal_Extension | [
"Field Extensions"
] | [
"Definition:Algebraic Field Extension"
] | [
"Definition:Embedding (Galois Theory)",
"Definition:Minimal Polynomial",
"Definition:Algebraic Field Extension",
"Definition:Root of Polynomial",
"Definition:Minimal Polynomial",
"Definition:Embedding (Galois Theory)",
"Definition:Embedding (Galois Theory)"
] |
proofwiki-4192 | Abstract Model of Algebraic Extensions | Let $K$ be a field and $\alpha \in \overline{K}$ be an element of the algebraic closure of $K$ which is algebraic over $K$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Then:
:$K[\alpha] \cong K[x]/ \langle m_\alpha\rangle$ | Let $\phi: K[x] \to K [\alpha]$ be the homomorphism which fixes $K$ and maps $x$ to $\alpha$.
The kernel of $\phi$ consists of all polynomials which vanish at $\alpha$, which is precisely the ideal $\langle m_\alpha\rangle$.
The result follows by the First Fundamental Theorem on Ring Homomorphisms.
{{qed}}
Category:Fie... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] and $\alpha \in \overline{K}$ be an element of the [[Definition:Algebraic Closure|algebraic closure]] of $K$ which is [[Definition:Algebraic Element of Field Extension|algebraic]] over $K$.
Let $m_\alpha$ be the [[Definition:Minimal Polynomial|minimal polynomi... | Let $\phi: K[x] \to K [\alpha]$ be the [[Definition:Ring Homomorphism|homomorphism]] which fixes $K$ and maps $x$ to $\alpha$.
The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ consists of all polynomials which vanish at $\alpha$, which is precisely the [[Definition:Ideal of Ring|ideal]] $\langle m_\alph... | Abstract Model of Algebraic Extensions | https://proofwiki.org/wiki/Abstract_Model_of_Algebraic_Extensions | https://proofwiki.org/wiki/Abstract_Model_of_Algebraic_Extensions | [
"Field Extensions"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Algebraic Closure",
"Definition:Algebraic Element of Field Extension",
"Definition:Minimal Polynomial"
] | [
"Definition:Ring Homomorphism",
"Definition:Kernel of Ring Homomorphism",
"Definition:Ideal of Ring",
"First Isomorphism Theorem/Rings",
"Category:Field Extensions"
] |
proofwiki-4193 | Union of Local Bases is Basis | Let $T = \struct {X, \tau}$ be a topological space.
For each $x \in X$, let $\BB_x$ be a local basis at $x$ which consists entirely of open sets.
Then $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$. | Let $U \in \tau$ be any open set of $T$.
Consider any $x \in U$.
Then, by definition of local basis, there exists a $B_x \in \BB_x$ such that $B_x \subseteq U$.
Also by definition of local basis:
:$x \in B_x$
So by Set Union Preserves Subsets:
:$U = \ds \bigcup_{x \mathop \in U} \set x \subseteq \bigcup_{x \mathop \in ... | Let $T = \struct {X, \tau}$ be a [[Definition:Topological Space|topological space]].
For each $x \in X$, let $\BB_x$ be a [[Definition:Local Basis|local basis]] at $x$ which consists entirely of [[Definition:Open Set (Topology)|open sets]].
Then $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a [[Definition:Basis (Top... | Let $U \in \tau$ be any [[Definition:Open Set (Topology)|open set]] of $T$.
Consider any $x \in U$.
Then, by definition of [[Definition:Local Basis|local basis]], there exists a $B_x \in \BB_x$ such that $B_x \subseteq U$.
Also by definition of [[Definition:Local Basis|local basis]]:
:$x \in B_x$
So by [[Set Union ... | Union of Local Bases is Basis | https://proofwiki.org/wiki/Union_of_Local_Bases_is_Basis | https://proofwiki.org/wiki/Union_of_Local_Bases_is_Basis | [
"Local Bases",
"Topological Bases",
"Local Bases"
] | [
"Definition:Topological Space",
"Definition:Local Basis",
"Definition:Open Set/Topology",
"Definition:Basis (Topology)",
"Definition:Topology"
] | [
"Definition:Open Set/Topology",
"Definition:Local Basis",
"Definition:Local Basis",
"Set Union Preserves Subsets",
"Union is Smallest Superset/General Result",
"Definition:Set Equality/Definition 2",
"Definition:Basis (Topology)",
"Definition:Topology"
] |
proofwiki-4194 | Subtraction of Multiples of Divisors obeys Distributive Law | {{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$ | Let the (natural) number $AB$ be the same aliquant part of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted.
We need to show that $EB$ is also the same aliquant part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:400px
Let $GH = AB$.
Then whatever aliquant part $GH$ is of $CD$, the... | {{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$ | Let the [[Definition:Natural Number|(natural) number]] $AB$ be the same [[Definition:Aliquant Part|aliquant part]] of the [[Definition:Natural Number|(natural) number]] $CD$ that $AE$ [[Definition:Natural Number Subtraction|subtracted]] is of $CF$ subtracted.
We need to show that $EB$ is also the same [[Definition:Ali... | Subtraction of Multiples of Divisors obeys Distributive Law/Proof 1 | https://proofwiki.org/wiki/Subtraction_of_Multiples_of_Divisors_obeys_Distributive_Law | https://proofwiki.org/wiki/Subtraction_of_Multiples_of_Divisors_obeys_Distributive_Law/Proof_1 | [
"Algebra",
"Divisors",
"Subtraction",
"Subtraction of Multiples of Divisors obeys Distributive Law"
] | [] | [
"Definition:Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Natural Numbers",
"Definition:Subtraction/Natural Numbers",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"File:Euclid-VII-8.png",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:D... |
proofwiki-4195 | Subtraction of Multiples of Divisors obeys Distributive Law | {{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$ | A direct application of the Distributive Property:
{{begin-eqn}}
{{eqn | l = \frac m n b - \frac m n d
| r = \frac m n b + \frac m n \paren {-d}
| c =
}}
{{eqn | r = \frac m n \paren {b + \paren {-d} }
| c =
}}
{{eqn | r = \frac m n \paren {b - d}
| c =
}}
{{end-eqn}}
{{qed}} | {{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$ | A direct application of the [[Distributive Property]]:
{{begin-eqn}}
{{eqn | l = \frac m n b - \frac m n d
| r = \frac m n b + \frac m n \paren {-d}
| c =
}}
{{eqn | r = \frac m n \paren {b + \paren {-d} }
| c =
}}
{{eqn | r = \frac m n \paren {b - d}
| c =
}}
{{end-eqn}}
{{qed}} | Subtraction of Multiples of Divisors obeys Distributive Law/Proof 2 | https://proofwiki.org/wiki/Subtraction_of_Multiples_of_Divisors_obeys_Distributive_Law | https://proofwiki.org/wiki/Subtraction_of_Multiples_of_Divisors_obeys_Distributive_Law/Proof_2 | [
"Algebra",
"Divisors",
"Subtraction",
"Subtraction of Multiples of Divisors obeys Distributive Law"
] | [] | [
"Distributive Laws/Arithmetic"
] |
proofwiki-4196 | Proportional Numbers have Proportional Differences | {{:Euclid:Proposition/VII/11}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$. | As the whole $AB$ is to the whole $CD$, so let the $AE$ subtracted be to $CF$ subtracted.
We need to show that $EB : FD = AB : CD$.
:200px
We have that :$AB : CD = AE : CF$.
So from {{EuclidDefLink|VII|20|Proportional}} we have that whatever aliquot part or aliquant part $AB$ is of $CD$, the same aliquot part or aliqua... | {{:Euclid:Proposition/VII/11}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the [[Definition:Ratio|ratio]] of $a$ to $b$. | As the whole $AB$ is to the whole $CD$, so let the $AE$ subtracted be to $CF$ subtracted.
We need to show that $EB : FD = AB : CD$.
:[[File:Euclid-VII-11.png|200px]]
We have that :$AB : CD = AE : CF$.
So from {{EuclidDefLink|VII|20|Proportional}} we have that whatever [[Definition:Aliquot Part|aliquot part]] or [[D... | Proportional Numbers have Proportional Differences | https://proofwiki.org/wiki/Proportional_Numbers_have_Proportional_Differences | https://proofwiki.org/wiki/Proportional_Numbers_have_Proportional_Differences | [
"Ratios"
] | [
"Definition:Ratio"
] | [
"File:Euclid-VII-11.png",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Divisor (Algebra)/Integer/Aliquot Part",
"Definition... |
proofwiki-4197 | Galois Group is Group | Let $L / K$ be a normal extension.
Let $\Gal {L / K}$ be the Galois group of $L / K$.
Then $\Gal {L / K}$ forms a group under the operation of composition of mappings. | The Galois group $\Gal {L / K}$ of $L / K$ is defined as:
:$\Gal {L / K} = \leftset{\sigma: L \to L: \sigma}$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\rightset{}$
We have that $\Gal {L / K}$ is a subset of the automorphism group on $L / K$.
We initially note that the Identity Mapping is Automo... | Let $L / K$ be a [[Definition:Normal Extension|normal extension]].
Let $\Gal {L / K}$ be the [[Definition:Galois Group of Field Extension|Galois group]] of $L / K$.
Then $\Gal {L / K}$ forms a [[Definition:Group|group]] under the operation of [[Definition:Composition of Mappings|composition of mappings]]. | The [[Definition:Galois Group of Field Extension|Galois group]] $\Gal {L / K}$ of $L / K$ is defined as:
:$\Gal {L / K} = \leftset{\sigma: L \to L: \sigma}$ is an [[Definition:Field Automorphism|automorphism]] of $L$ such that $\sigma$ [[Definition:Fixed Point|fixes]] $K$ point-wise$\rightset{}$
We have that $\Gal {... | Galois Group is Group | https://proofwiki.org/wiki/Galois_Group_is_Group | https://proofwiki.org/wiki/Galois_Group_is_Group | [
"Galois Theory"
] | [
"Definition:Normal Extension",
"Definition:Galois Group of Field Extension",
"Definition:Group",
"Definition:Composition of Mappings"
] | [
"Definition:Galois Group of Field Extension",
"Definition:Field Automorphism",
"Definition:Fixed Point",
"Definition:Subset",
"Definition:Automorphism Group/Group",
"Identity Mapping is Automorphism",
"Definition:Fixed Point",
"Definition:Empty Set",
"Definition:Closure (Abstract Algebra)/Algebraic ... |
proofwiki-4198 | Relation of Ratios to Products | {{:Euclid:Proposition/VII/19}}
That is:
:$a : b = c : d \iff ad = bc$ | Let $A, B, C, D$ be four (natural) numbers in proportion, so that $A : B = C : D$.
Let $A \times D = E$ and $B \times C = F$.
We need to show that $E = F$.
:300px
Let $A \times C = G$.
Then $A \times C = G$ and $A \times D = E$.
So from {{EuclidPropLink||book=VII|prop=17|title=Multiples of Ratios of Numbers}}:
: $C : D... | {{:Euclid:Proposition/VII/19}}
That is:
:$a : b = c : d \iff ad = bc$ | Let $A, B, C, D$ be four [[Definition:Natural Numbers|(natural) numbers]] in [[Definition:Proportion|proportion]], so that $A : B = C : D$.
Let $A \times D = E$ and $B \times C = F$.
We need to show that $E = F$.
:[[File:Euclid-VII-19.png|300px]]
Let $A \times C = G$.
Then $A \times C = G$ and $A \times D = E$.
S... | Relation of Ratios to Products | https://proofwiki.org/wiki/Relation_of_Ratios_to_Products | https://proofwiki.org/wiki/Relation_of_Ratios_to_Products | [
"Ratios"
] | [] | [
"Definition:Natural Numbers",
"Definition:Proportion",
"File:Euclid-VII-19.png"
] |
proofwiki-4199 | Ratios of Fractions in Lowest Terms | Let $a, b, c, d \in \Z_{>0}$ be positive integers.
Let $\dfrac a b$ be in canonical form.
Let $\dfrac a b = \dfrac c d$.
Then:
:$a \divides c$
and:
:$b \divides d$
where $\divides$ denotes divisibility.
{{:Euclid:Proposition/VII/20}} | Let $CD, EF$ be the least (natural) numbers of those which have the same ratio with $A, B$.
We need to show that $CD$ measures $A$ the same number of times that $EF$ measures $B$.
:250px
Suppose $CD$ is an aliquant part of $A$.
Then from {{EuclidPropLink|book = VII|prop = 13|title = Proportional Numbers are Proportiona... | Let $a, b, c, d \in \Z_{>0}$ be [[Definition:Positive Integer|positive integers]].
Let $\dfrac a b$ be in [[Definition:Canonical Form of Rational Number|canonical form]].
Let $\dfrac a b = \dfrac c d$.
Then:
:$a \divides c$
and:
:$b \divides d$
where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]... | Let $CD, EF$ be the least [[Definition:Natural Numbers|(natural) numbers]] of those which have the same [[Definition:Ratio|ratio]] with $A, B$.
We need to show that $CD$ [[Definition:Divisor of Integer|measures]] $A$ the same number of times that $EF$ [[Definition:Divisor of Integer|measures]] $B$.
:[[File:Euclid-VII... | Ratios of Fractions in Lowest Terms | https://proofwiki.org/wiki/Ratios_of_Fractions_in_Lowest_Terms | https://proofwiki.org/wiki/Ratios_of_Fractions_in_Lowest_Terms | [
"Ratios"
] | [
"Definition:Positive/Integer",
"Definition:Rational Number/Canonical Form",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Natural Numbers",
"Definition:Ratio",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"File:Euclid-VII-20.png",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Divisor (Algebra)/Integer/Aliquant Part",
"Definition:Divisor (Algebra)/Integer/... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.