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https://www.esaral.com/q/in-the-previous-problem-29214	1,713,291,354,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00461.warc.gz	703,420,048	10,944	# In the previous problem, Question: In the previous problem, suppose $m_{2}=2.0 \mathrm{~kg}$ and $m_{3}=3.0 \mathrm{~kg}$. What should be the mass $\mathrm{m}$ so that it remains at rest? Solution:	64	202	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2024-18	latest	en	0.782168
https://usacourseworkhelp.com/questions/in-reference-to-an-basic-atwood-pulley-physics-lab/	1,600,572,789,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00232.warc.gz	701,074,435	6,787	# In reference to an basic atwood pulley physics lab… ### Question Description: 20.50 In reference to an basic atwood pulley physics lab; 1. If we had used a heavier strong, the graphs (velocity vs. time) would not have been exactly linear, but would of curved slightly. Why? Which way would it curve? Hint: Think of acceleration at the beginning and acceleration at the end. 2. explain why a traction elevator uses a counterweight and discuss how they set the mass of the counterweight (reference to Newtons’ 2nd law of translation). 3. If we use the same setup as in lab with one simple change we can some up with a way to meaure the coeeficient of friction between a mass and the lab table. We use the same computer program (measures v vs. t) but plave the pulley system near the edge of the table with one mass hanging off the end (m1) and one mass resting on the table (m2). The mass m2 is able to slide towards the pulley system as m1 drops towards the floor (away from the pulleys). If the force of friction between m2 and the table is F=umg2, come up with a theoretical equation to figure out u (you can assume that you can get the same style of data that you did in this lab, that is, graphs, masses, etc.)	299	1,218	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.890625	3	CC-MAIN-2020-40	latest	en	0.931495
https://ilmihub.com/physics-quiz-physics-general-knowledge-mcqs.html	1,721,743,120,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00427.warc.gz	251,049,849	28,283	Home Physics MCQs Physics Quiz \| Physics General Knowledge MCQs # Physics Quiz \| Physics General Knowledge MCQs 975 ### Physics Quiz – Physics General Knowledge MCQs for NTS, PPSC, CSS, FPSC, PMS, OTS, PTS and Interviews This Physics Quiz contains Physics General Knowledge MCQs for NTS, PPSC, CSS, FPSC, PMS, OTS, PTS and Interviews. This post is the 10th and last post of this Physics Quiz Series. MCQs are taken from Caravan Comprehensive book. To visit all the Physics Quizzes you can Click Here. ## Physic Questions and Answers Quiz for Competitive Exams MCQs of this Quiz are starting from No.181. Click This for previous Quiz on Physics. 181. Which device is used to measure the specific gravity of a liquid? (a) Barometer (b) Ammeter (c) Manometer (d) Hydrometer (d) Hydrometer 182. Which device is used to measure humidity? (a) Tachometer (b) Hydrometer (c) Hygrometer (d) Galvanometer (c) Hygrometer 183. What splits light into its spectrum? (a) Ionometer (b) Spectrometer (c) Manometer (d) Stereoscope (b) Spectrometer 184. Angstrom is unit of which quantity? (a) Force (b) Heat (c) Resistance (d) Length (d) Length 185. A beam of which colour light passing through a prism scatters in seven colours? (a) Red (b) White (c) Black (d) Yellow (b) White 186. Dyne is unit of what quantity? (a) Heat (b) Length (c) Force (d) Resistance (c) Force 187. Ohm is the unit of which quantity? (a) Length (b) Heat (c) Force (d) Resistance (d) Resistance 188. The microphone is used to convert sound waves into _______. (a) Mechanical energy (b) Kinetic energy (c) Potential energy (d) Electrical energy (d) Electrical energy 189. If speed of rotation of earth increase what effect it would have on the weight of your body? (a) Weight decreases (b) Weight increases (c) Remains the same (a) Weight decreases 190. The art of designing a sequence of movements in which motion form or both are specified like dance motion is called _______. (a) Choreography (b) Dactylography (c) Plerytography (d) Metallography (a) Choreography 191. In which medium the sound travels fastest? (a) Steel (b) Water (c) Wood (d) Glass (d) Glass 192. Which instrument measure speed of ship? (a) Periscope (b) Angstrom (c) Sextant (d) Monometer (b) Angstrom 193. One horsepower is equal to how many Watt? (a) 820 Watt (b) 780 Watt (c) 760 Watt (d) 746 Watt (d) 746 Watt 194. The acceleration caused by gravity per second is_______. (a) 52 m (b) 48 m (c) 45 m (d) 40 m (a) 52 m 195. Which wave travel with almost the velocity of light? (a) Ultra-sonic wave (c) Blue wave (d) Solar wave 196. The frequency of which of the following is highest? (a) Lightwaves (b) Ultrasonic waves 197. Speed of sound in air is_______. (a) 480 m/s (b) 331 m/s (c) 300 m/s (d) 280 m/s (b) 331 m/s 198. LASER stands for______. (a) Light Against Skin by Emission of Radiation (b) Long Altitude Sight Emission Rays (c) Light Amplification by Stimulated Emission of Radiation (d) Left Align Sound Emission Rays (c) Light Amplification by Stimulated Emission of Radiation 199. FM stands for ________. (a) Frequency Module (b) Frequency Movement (c) Frequency Modulation (d) Ferentition Module (c) Frequency Modulation 200. One million cycle per second is called _______. (a) Megabyte (b) Megahertz (c) Gigabyte (d) Hedabyte (b) Megahertz To start from the first Quiz of Physics Visit This. If there is any mistake in the post, inform us in the comment section. • ### Physics Quiz Questions for Competitive Exams General Physics Questions and Answers \| Physics Questions for competitive exams Here is po… • ### Physics MCQs with Answers for Competitive Exams Physics MCQs with Answers for NTS, PPSC, CSS, FPSC, PMS, OTS, PTS Tests and Interviews Thi… • ### Physics GK Solved Multiple Choice Questions and Answers Physics GK Solved Multiple Choice Questions and Answers for Competitive Exams This post co…	1,068	3,920	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.828125	3	CC-MAIN-2024-30	latest	en	0.759776
https://machinesdontcare.wordpress.com/2009/06/01/object-orientated-coding-in-js/	1,531,869,442,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00159.warc.gz	721,111,784	18,897	01 Jun 09 ### Object-Orientated Coding In JS I’ve just (re)started leaning Objective-C, and thought I’d sabotage the learning-process (I always find a way to do this), by trying to do a bit of object-orientated coding in JavaScript, too. Of course, I’ve used objects in JS/QC before, but I’ve never used proper OO concepts when writing JS before. Here’s some example code: ```// Declare custom object to hold point data function point() {this.X = 0; this.Y = 0; this.Z = 0;} // Re-initialise point point.prototype.initpoint = function() {this.X = 0; this.Y = 0; this.Z = 0;} // Object method for point (creates new random point) point.prototype.newpoint = function(lastpoint, maxdist) { this.X = lastpoint.X + (Math.random() * (2 * maxdist) - maxdist); this.Y = lastpoint.Y + (Math.random() * (2 * maxdist) - maxdist); this.Z = lastpoint.Z + (Math.random() * (2 * maxdist) - maxdist); } // Calculates point on bezier-curve, taking 4 variables of type point and 1 value for time (in 0 > 1 range) point.prototype.bezierpoint = function(A, B, C, D, t) { var a = t; var b = 1 - t; this.X = A.X * Math.pow(b, 3) + 3 * B.X * Math.pow(b, 2) * a + 3 * C.X * b * Math.pow(a, 2) + D.X * Math.pow(a, 3); this.Y = A.Y * Math.pow(b, 3) + 3 * B.Y * Math.pow(b, 2) * a + 3 * C.Y * b * Math.pow(a, 2) + D.Y * Math.pow(a, 3); this.Z = A.Z * Math.pow(b, 3) + 3 * B.Z * Math.pow(b, 2) * a + 3 * C.Z * b * Math.pow(a, 2) + D.Z * Math.pow(a, 3); }``` The idea of OO coding is that you define objects, with particular properties, and then you can define methods, that act on those properties in some way. Once you’ve defined an object, its properties and its methods, the object becomes a kind of prototype, and you can then create lots of independent instances of that object, each with their own, independent properties. So, in the code above, I’m defining an object called ‘point’ to hold an X, Y and X position. ```function point() { this.X = 0; this.Y = 0; this.Z = 0; }``` Once I’ve defined the properties that every instance of the point object will have, I can define methods to do things with those properties. Here’s the simplest one: `point.prototype.initpoint = function() {this.X = 0; this.Y = 0; this.Z = 0;}` To create a new object of type ‘point’, I’d do this: `var p0 = new point();` I can then apply the methods I defined above to the X Y and Z values for p0, using dot-syntax: `p0.initpoint();` for example, would initialise all values for p0 to 0. The other methods I’ve defined are a bit more complicated, in that they require input variables to be passed to them when they are invoked, but the principle is the same. Of course, this kind of thing will be second-nature to anyone who has done work in C++, Java, or any of the many other OO programming and scripting languages (though the syntax may vary), but it’s quite novel to me. I’ve been messing around with bits of code for years, now, but, apart from the odd dabbling with ActionScript, I’ve never really got beyond relatively simple procedural, C-style coding. I think I’ll be using OO techniques like this a lot more in the future, though. It’s definitely something I’ll need to get my head around conceptually, when I learn to write Objective-C, so it makes sense to apply the same principles in JavaScript, too.	918	3,292	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2018-30	latest	en	0.712752
http://fungrim.org/entry/c92a6f/	1,582,295,045,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00129.warc.gz	61,990,245	4,087	Fungrim home page # Fungrim entry: c92a6f $\theta_{4}\!\left(z , \frac{\tau}{2}\right) = \frac{\theta_{4}^{2}\!\left(z, \tau\right) + \theta_{1}^{2}\!\left(z, \tau\right)}{\theta_{3}\!\left(0 , \frac{\tau}{2}\right)}$ Assumptions:$z \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \tau \in \mathbb{H}$ TeX: \theta_{4}\!\left(z , \frac{\tau}{2}\right) = \frac{\theta_{4}^{2}\!\left(z, \tau\right) + \theta_{1}^{2}\!\left(z, \tau\right)}{\theta_{3}\!\left(0 , \frac{\tau}{2}\right)} z \in \mathbb{C} \,\mathbin{\operatorname{and}}\, \tau \in \mathbb{H} Definitions: Fungrim symbol Notation Short description JacobiTheta$\theta_{j}\!\left(z , \tau\right)$ Jacobi theta function Pow${a}^{b}$ Power CC$\mathbb{C}$ Complex numbers HH$\mathbb{H}$ Upper complex half-plane Source code for this entry: Entry(ID("c92a6f"), Formula(Equal(JacobiTheta(4, z, Div(tau, 2)), Div(Add(Pow(JacobiTheta(4, z, tau), 2), Pow(JacobiTheta(1, z, tau), 2)), JacobiTheta(3, 0, Div(tau, 2))))), Variables(z, tau), Assumptions(And(Element(z, CC), Element(tau, HH)))) ## Topics using this entry Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub. 2020-01-31 18:09:28.494564 UTC	454	1,224	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.796875	3	CC-MAIN-2020-10	latest	en	0.351419
http://www.haverford.edu/physics-astro/course_materials/phys326/ps3.htm	1,369,058,539,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368699036375/warc/CC-MAIN-20130516101036-00094-ip-10-60-113-184.ec2.internal.warc.gz	490,076,305	6,270	Physics 326a-2012 Walter F. Smith Assignment 3 Due: Friday, Sept. 28 at the beginning of class Reading: Lab Manual J1 to J6, E1 to E4, C8 to C10, C13, H1 to H3 Assigned exercises: 3.1. The OP27 operational amplifier (op amp), made by Analog Devices, is one of the most commonly used for building scientific instruments. It’s an excellent, high-end op amp. Assume you’ve built an amplifier based around the OP27EP variant of this chip. (Most chips are available in several different grades, so as to suit a variatiey of customers.) a. Go on the internet and find the specs for the current noise density in (units of )and voltage noise density vn (units of ) at 1 kHz for this chip. Hint: Once you get to an appropriate site, you’ll want to look at the “datasheet” for this chip. b. Assume you use the amplifier you built to measure the Johnson noise of a 10 kW resistor at T = 290 K. Your measurement chain includes a high-pass filter with f3dB of 20 Hz and a low pass filter with f3dB of 2 kHz. What is the approximate total rms noise amplitude (referred to the amplifier input), including the amplifier noise? (Use the guide on pp. C11-C12 to calculate your effective noise bandwidth.) 3.2. For any op amp or amplifier, there is internal capacitance to ground from the output. Therefore, we can write . The stage which drives the output always has a maximum current Iout that it can deliver, so we see that this imposes a maximum value of , which is called the “slew rate”. In other words, the output voltage of the op amp or amplifier can’t change at a rate faster than the slew rate. Assume you’re using an amplifier with a slew rate of 20 V/ms, and the gain is set to 1000. You apply a sinusoidal input with a peak-to-peak amplitude of 0.01 V. What is the maximum frequency for this wave that you could apply without seeing distortion at the output due to the slew rate limit? Hint: the maximum frequency for a sinusoidal wave is different from that for a triangle wave. 3.3 A Swarthmore student draws your attention to the plot of Noise Figure for the SR560 amplifier. This is reproduced at right. Further details can be found in the manual for this amplifier, which is available online at: http://www.thinksrs.com/products/SR560.htm The student exclaims, “Look at that sweet spot at 1 kHz and about 200 kW ! The total noise at the output of the amplifier must be lower there than it would be, say, at 1 kHz and 10 W.” Explain why he’s wrong. (Note: a sweet spot is a particularly low noise region, or so thinks the Swat student…)	649	2,576	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2013-20	latest	en	0.846715
http://forums.xkcd.com/viewtopic.php?f=7&t=125039&view=print	1,561,019,155,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00386.warc.gz	71,271,510	5,677	Page 1 of 1 ### 2034: "Equations" Posted: Fri Aug 17, 2018 8:29 am UTC Title: All electromagnetic equations: The same as all fluid dynamics equations, but with the 8 and 23 replaced with the permittivity and permeability of free space, respectively. I feel like the fluid dynamics equations needs an outer product of the gradient operator or something. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 8:42 am UTC I'm not sure all the equations would fit on a T-shirt, but they would definitely fit as a full-body tattoo. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 10:10 am UTC Is the `truly deep physics equations` just because of the simple formulas that involve the planck constant or is there something else? ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 10:46 am UTC H_hat is a Hamiltonian (perhaps even a general differential operator), not Planck's constant. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 12:20 pm UTC (Missed out title text: All electromagnetic equations: The same as all fluid dynamics equations, but with the 8 and 23 replaced with the permittivity and permeability of free space, respectively.) IRT last one as: <something> without You is (nothingness/empty/pointless?) I'm fairly sure it is that, but it slightly out-geeks me. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 1:40 pm UTC I think I'm going to wait for the explainXKCD article to come up, then I'm going to re-enroll in college to see if Fluids, Thermodynamics, and Fields classes get any easier. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 2:11 pm UTC I didn't realize we knew enough about quantum gravity to actually have equations for it. ...or is that just ((representative of) \| (a parody of)) a string theory equation? ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 2:19 pm UTC richP wrote:I think I'm going to wait for the explainXKCD article to come up I'm still not much the wiser! ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 3:52 pm UTC SuperCow wrote:I didn't realize we knew enough about quantum gravity to actually have equations for it. ...or is that just ((representative of) \| (a parody of)) a string theory equation? We don't, at least not ones that belong to a generally accepted theory. SU(n) and U(n) each indicate a vector space (and certain operations) where each vector has n numbers. Nesting them doesn't make sense because 1) n needs to be an integer 2) n would typically be a specific fixed, small number, as these are less functions than a way to describe what kind of mathematical object we're using. Those numbers look like someone took the number space for Grand Unified Theory (electromagnetism + weak force + strong force) and made them more complicated, as if they were adding gravity to the mix. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 4:31 pm UTC Also, that "quantum gravity" expression... actually, several of the things on this... are not even equations. It's not an equation without an equals sign. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 4:51 pm UTC I'm pretty sure the joke about quantim gravity is that there are no equations at all, just some symmetry groups. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 5:29 pm UTC beojan wrote:H_hat is a Hamiltonian (perhaps even a general differential operator), not Planck's constant. Because as any fule kno, Alexander Hamilton was most famous for wearing hats. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 8:00 pm UTC The chemistry equations author was pre-med until he/she wrote the equation. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 9:36 pm UTC I'm highly suspicious that the "Deep Physics Equation" MUST be normalized to read "= 42", or it won't work. ### Re: 2034: "Equations" Posted: Fri Aug 17, 2018 10:43 pm UTC Number theory itself produces gravity waves, so there's a nesting that ought to be taken into account. ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 12:34 am UTC ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 3:27 am UTC Given the complexity of many formulae in modern physics, mathematics, etc. and the wide availability of Unicode I'm surprised we haven't gone beyond the standard greek/latin alphabet yet, instead choosing to use diacritics, typeface styles such as blackletter etc. There are plenty of characters available in other alphabets, Cyrillic, Japanese, Chinese, Egyptian Hieroglyphs and so on. Even emojis might start becoming a standard (why not use a sun for temperature or a car for velocity? just an idea) ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 4:09 am UTC gcgcgcgc wrote:Given the complexity of many formulae in modern physics, mathematics, etc. and the wide availability of Unicode I'm surprised we haven't gone beyond the standard greek/latin alphabet yet, instead choosing to use diacritics, typeface styles such as blackletter etc. There are plenty of characters available in other alphabets, Cyrillic, Japanese, Chinese, Egyptian Hieroglyphs and so on. Even emojis might start becoming a standard (why not use a sun for temperature or a car for velocity? just an idea) Mathematics has ventured into the Arabic alphabet in a few places - such as Alephs for cardinal infinities. ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 6:29 am UTC Aleph is Hebrew, not Arabic. So is bet. All the standard numerals are Arabic, though. (Unless you ask an actual Arab, then they're Indian). ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 8:14 am UTC gcgcgcgc wrote:Given the complexity of many formulae in modern physics, mathematics, etc. and the wide availability of Unicode I'm surprised we haven't gone beyond the standard greek/latin alphabet yet, instead choosing to use diacritics, typeface styles such as blackletter etc. There are plenty of characters available in other alphabets, Cyrillic, Japanese, Chinese, Egyptian Hieroglyphs and so on. Even emojis might start becoming a standard (why not use a sun for temperature or a car for velocity? just an idea) Well, I had friends with absolutely no aptitude for drawing and professor handwriting isn't known to produce easily legible greek letters. Based on those two experiences, I would not venture to read any handwritten calculations with tiny cars and chinese logograms in them, but have fun trying. ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 4:00 pm UTC Pfhorrest wrote:Aleph is Hebrew, not Arabic. So is bet. All the standard numerals are Arabic, though. (Unless you ask an actual Arab, then they're Indian). Good point. The first two letters of the Arabic alphabet are also called "aleph" and "beth" (both with a variety of transliterations), but look very different. ### Re: 2034: "Equations" Posted: Sat Aug 18, 2018 5:46 pm UTC I am not sure that we are supposed to take this one seriously, at least not literally. e to the (pi - infinity)? when pi is an index between something finite and infinity? this is probably a joke about using symbols with previous meaning inappropriately HEAT -> H2EAT is probably a joke.... the equations balances, sort of... but H2EAT? Would give you gas... Kinematics only for scalar potentials? Wasa matter, you don't like magnetism? And if we were really talking universal, we would have to do everything in at least 4 space... And a lot of glyphs that I have never seen before... Secrets of the Universe indeed... 42 belongs there somewhere. And, so damn it... where are the pyramids in all of this? ajohnso (not that I would suggest that RM would not have a secret meaning to his list... but perhaps the secret meaning is the lack of meaning?) ### Re: 2034: "Equations" Posted: Sun Aug 19, 2018 5:08 am UTC I suggest we translate the Hebrew letters into English before using them, based on their original Phoenician meanings. "Aleph" is "ox", "Bet" is "house", "Gimel" is "camel" (at least THAT one didn't change too much)... ### Re: 2034: "Equations" Posted: Wed Aug 22, 2018 8:15 am UTC I think we should start using digammas for functions. ### Re: 2034: "Equations" Posted: Wed Aug 22, 2018 10:19 am UTC rmsgrey wrote: Pfhorrest wrote:Aleph is Hebrew, not Arabic. So is bet. All the standard numerals are Arabic, though. (Unless you ask an actual Arab, then they're Indian). Good point. The first two letters of the Arabic alphabet are also called "aleph" and "beth" (both with a variety of transliterations), but look very different. Using the Arabic aleph would confuse people, because it's just I sloping the other way. I still hope Doctor Who runs out of letters at the ends of Greek and Roman alphabets and starts using the Arabic one, though, because watching the cast try to maintain straight faces while talking about how long the ha-shielded windows will protect them from the 100% wow-tonic environment while the engineers are repairing the yay-drive could make the show worth watching. ### Re: 2034: "Equations" Posted: Wed Aug 22, 2018 10:43 am UTC SpitValve wrote:I think we should start using digammas for functions. Well, in a way, we already do - though that particular digamma is actually coincidental! There is also one known use of Cyrillic: Ш is the Tate-Shafarevich group (and a few other things, apparently, but this particular use is definitely the Cyrillic letter). I'm surprised that runes and Japanese kana don't see more use, though... (Oh, and the one alphabet that I think would be awesome to see in math? Glagolitic.) ### Re: 2034: "Equations" Posted: Wed Aug 22, 2018 2:20 pm UTC + ( ^ ( - ) ) = / ( - ) + kthanxbye ### Re: 2034: "Equations" Posted: Wed Aug 22, 2018 6:21 pm UTC I think it's missing the most fundamental equation "f(x) = g(x)". Lothario O'Leary wrote:(Oh, and the one alphabet that I think would be awesome to see in math? Glagolitic.) ⰒⰎⰉⰈ ⰐⰑ! ### Re: 2034: "Equations" Posted: Fri Aug 24, 2018 5:36 am UTC most important equation? My vote is for: U+2205 U+2260 U+2205 If you want that in a prose form... "everything can be divided into two categories - those things that can be divided into two categories, and those that can't". watch for inappropriate binaries. aljohnso ### Re: 2034: "Equations" Posted: Fri Aug 24, 2018 9:33 pm UTC aljohnso wrote:most important equation? My vote is for: U+2205 U+2260 U+2205 If you want that in a prose form... "everything can be divided into two categories - those things that can be divided into two categories, and those that can't". watch for inappropriate binaries. aljohnso So, ∅ ≠ ∅ My prose interpretation of that is "an empty set is not equal to an empty set," which doesn't sound like an important equation, but rather, it sounds wrong. ### Re: 2034: "Equations" Posted: Mon Aug 27, 2018 9:38 pm UTC It depends on who you are defining things and which axioms you pick. Within ZFC that's wrong, but it's probably true in some formulations of set theory which don't assume an excluded middle (P doesn't necessarily equal ~~P).	2,926	10,962	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.65625	3	CC-MAIN-2019-26	latest	en	0.939351
http://m.csmonitor.com/Business/The-Circle-Bastiat/2011/0209/Cost-means-what-exactly	1,493,376,105,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917122933.39/warc/CC-MAIN-20170423031202-00412-ip-10-145-167-34.ec2.internal.warc.gz	239,793,449	11,137	Close X 'Cost' means what, exactly? View video Ted S. Warren / AP / File (Read caption) Working for wages is one thing that might have to be given up in order to do something else. So, assuming somebody can make at least \$8 per hour at a job, any other activity would cost at least \$8 per hour in foregone wages. View photo A couple of days ago, I wrote about prices, profits, and losses in the context of the Styrofoam-or-paper debate. Yesterday, we covered cost and supply in econ 100. Here is an email I sent to my students this morning that addresses the question of measuring costs using dollars and cents. One of the best treatments of the cost concept is James Buchanan’s Cost and Choice. Here’s the email: Here are a few quick thoughts on the use of dollars and cents to measure opportunity cost. They’re a little beyond what we’re doing this semester, they won’t be on any of the tests, and they will likely reappear in more formal fashion in intermediate microeconomics, but I thought that some of you might be interested: When we talk about “cost,” we’re talking about something subjective: a cost is the next-best opportunity that you have to forgo in order to do something. Since cost is subjective, how do we get away with expressing cost in terms of money? Using dollars and cents to approximate costs is actually more sound than it might appear to be at first. The market’s process of buying and selling generates market prices for things like corn, potatoes, textbooks—and labor of different kinds. We hypothesized yesterday that if the going you are each giving up the opportunity to work for wages in order to come to class. If you could have earned \$8/hour, then going to class for 75 minutes cost you \$10 in foregone wages. You might respond “but what if my next best opportunity is sleep, studying for another class, or playing video games rather than working? How do you get away with the ‘\$10 cost’ statement?” That’s a good question, and it illustrates that the cost we are estimating is at best a minimum cost: we know that the cost to you of going to class is at least \$10 even if there are a lot of activities between “going to class” and “working for wages” that you prefer to “working for wages.” Since we can observe market prices for different types of labor, and since most labor markets are extremely competitive, we can be pretty sure that going to class costs you at least \$10. Suppose you rank the different uses of the 75 minutes between 2 and 3:15 as follows: 1. Go to Econ 100 (I’m flattered) 2. Sleep 3. Play video games 4. Watch paint dry 5. Work for \$10 Value is subjective, so we can’t assign numbers that tell us anything about how much you value options 1-4; i.e., we can’t say that “Econ 100 gives you two gallons of happy while sleep only gives you 1.5 gallons of happy.” We can, however, observe the market price of labor and say that by going to class, you are giving up—at the very least—the opportunity to earn \$10. As this course is an introduction to economics, there are all sorts of interesting (?) details we will have to skip over. A few weeks ago, I sent you links to Ludwig von Mises’s Human Action and Murray Rothbard’s Man, Economy, and State, which provide unique and somewhat non-traditional approaches to economic thinking. Stephen Landsburg’s The Armchair Economist (which isn’t free online, unfortunately) provides a very good discussion of basic economics and might be a worthy addition to your nightstand.	781	3,502	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2017-17	latest	en	0.953225
https://www.physicsforums.com/threads/uniqueness-theorem.309845/	1,685,716,459,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648695.4/warc/CC-MAIN-20230602140602-20230602170602-00203.warc.gz	1,005,082,653	13,645	# Uniqueness theorem • captainjack2000 ## Homework Statement I have a situation with a charge distribution for a system of static charges in a vacuum. It then asks to state the uniqueness theorem for such a system. ## The Attempt at a Solution I know that the uniquessness theorem means that once you have found one solution to the system you have found THE solution. But how does this change for different systems? Is this a good enoug answer to what is the uniqueness theorem?	103	483	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2023-23	latest	en	0.949186
https://www.encyclopediaofmath.org/index.php/H%C3%B6lder_summation_methods	1,544,913,675,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00466.warc.gz	883,344,577	5,981	# Hölder summation methods A collection of methods for summing series of numbers, introduced by O. Hölder [1] as a generalization of the summation method of arithmetical averages (cf. Arithmetical averages, summation method of). The series $$\sum_{n=0}^\infty a_n$$ is summable by the Hölder method $(H,k)$ to sum $s$ if $$\lim_{n\to\infty}H_n^k=s,$$ where $$H_n^0=s_n=\sum_{k=0}^na_k,$$ $$H_n^k=\frac{H_0^{k-1}+\ldots+H_n^{k-1}}{n+1},$$ $k=1,2,\ldots$. In particular, $(H,0)$-summability of a series indicates that it converges in the ordinary sense; $(H,1)$ is the method of arithmetical averages. The $(H,k)$-methods are totally regular summation methods for any $k$ and are compatible for all $k$ (cf. Compatibility of summation methods). The power of the method increases with increasing $k$: If a series is summable to a sum $s$ by the method $(H,k)$, it will also be summable to that sum by the method $(H,k')$ for any $k'>k$. For any $k$ the method $(H,k)$ is equipotent and compatible with the Cesàro summation method of the same order $k$ (cf. Cesàro summation methods). If a series is summable by the method $(H,k)$, its terms $a_n$ necessarily satisfy the condition $a_n=o(n^k)$. #### References [1] O. Hölder, "Grenzwerthe von Reihen an der Konvergenzgrenze" Math. Ann. , 20 (1882) pp. 535–549 [2] G.H. Hardy, "Divergent series" , Oxford Univ. Press (1949) How to Cite This Entry: Hölder summation methods. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=H%C3%B6lder_summation_methods&oldid=32581 This article was adapted from an original article by I.I. Volkov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article	536	1,721	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2018-51	latest	en	0.75123
https://devsenv.com/example/codeforces-solution-beautiful-sets-of-points-solution-in-c,-c++,-java,-python	1,726,034,579,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00233.warc.gz	172,334,169	42,293	## Algorithm C. Beautiful Sets of Points time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Manao has invented a new mathematical term — a beautiful set of points. He calls a set of points on a plane beautiful if it meets the following conditions: 1. The coordinates of each point in the set are integers. 2. For any two points from the set, the distance between them is a non-integer. Consider all points (x, y) which satisfy the inequations: 0 ≤ x ≤ n0 ≤ y ≤ mx + y > 0. Choose their subset of maximum size such that it is also a beautiful set of points. Input The single line contains two space-separated integers n and m (1 ≤ n, m ≤ 100). Output In the first line print a single integer — the size k of the found beautiful set. In each of the next k lines print a pair of space-separated integers — the x- and y- coordinates, respectively, of a point from the set. If there are several optimal solutions, you may print any of them. Examples input Copy `2 2` output Copy `30 11 22 0` input Copy `4 3` output Copy `40 32 13 04 2` Note Consider the first sample. The distance between points (0, 1) and (1, 2) equals , between (0, 1) and (2, 0) — , between (1, 2) and (2, 0) — . Thus, these points form a beautiful set. You cannot form a beautiful set with more than three points out of the given points. Note that this is not the only solution. ## Code Examples ### #1 Code Example with C++ Programming ```Code - C++ Programming``` ``````#include <stdio.h> #include <algorithm> using namespace std; int main() { int n, m; scanf("%d%d", &n, &m); if(n == m) { printf("%d\n", n + 1); for(int i = 0, j = m; j >= 0; i++, j--) printf("%d %d\n", i, j); } else if(n > m) { printf("%d\n", min(n, m + 1)); for(int i = 0, j = m; j >= 0; i++, j--) printf("%d %d\n", i, j); } else if(n < m) { printf("%d\n", min(n + 1, m)); for(int i = 0, j = m; i <= n; i++, j--) printf("%d %d\n", i, j); } return 0; }`````` Copy The Code & Input cmd 2 2 Output cmd 3 0 1 1 2 2 0 ## Demonstration Codeforces Solution-Beautiful Sets of Points-Solution in C, C++, Java, Python	681	2,129	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2024-38	latest	en	0.648248
https://statanalytica.com/Your-program-should-read-the-array-to-be-sorted-from-unsorte	1,713,620,991,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00391.warc.gz	488,239,283	15,732	(5/5) Hire Me (/5) Hire Me (5/5) Hire Me (/5) Hire Me # Your program should read the array to be sorted from unsortedArrayXXX.dat that is in the same directory and save sorted arrays to files named mergeSortedXXX.dat INSTRUCTIONS TO CANDIDATES Department of Computer Engineering, Computer Programming I Term Project. Implement 4 sorting algorithms: 1. Merge Sort 2. Buble Sort 3. Your Algorithm 1 (Choose one known sorting algortihm) 4. Your Algorithm 2 (Choose another known sorting algorithm) Your program should read the array to be sorted from “unsortedArrayXXX.dat” that is in the same directory and save sorted arrays to files named “mergeSortedXXX.dat”. All 4 output files should have the same sorted array. “XXX” stands for different array sizes. For example, if the array size is 100 file name will be unsortedArray100.dat. Your program should also print the unsorted array and sorted arrays for each algorithm on computer screen. Your program should measure the sorting time used by each algorithm for the 4 array sizes 100, 1000, 10000 and 100000. The time should not include reading the array from the hard-drive, writing the sorted array to the hard-drive and printing arrays to the screen. The time measrument should only inlcude the time taken for sorting algorithms. Print the sort time on computer screen. And you can also dump all screen mesages to a file named as dump.txt by running your code under command promp adding “ > dump.txt” after your executable file. For example: C:MyCProjectsSortingcompareSortAlgorithms.exe > dump.txt In your report, compare the results on a chart and check it with their well-known performance metrics that are shown below: Name Average Complexity Merge Sort O(n2) Bubke Sort O(nlogn) Your Algorithm 1 Your Algorithm 2 Note 1: For a simple explanation of Big-O Notation pls check: https://justin.abrah.ms/computer-science/big-o-notation-explained.html Note 2: For an introduction to various sorting algorithms please check: https://en.wikipedia.org/wiki/Sorting_algorithm Chart below gives the input size vs. computer operations plots for mostly encountered Big-O complexities. Compare your charts for buble sort, merge sort and 2 other algorithms with the chart below and comment on results. You should fill in the following table with respected runtimes and construct a comparison chart for the 4 algorithms: Elapsed Time(ms) n=100 n=1000 n=10000 n=100000 Buble Sort Quick Sort Heap Sort Merge Sort Note: For each measurement run the algorithm for at least 10 times, and record the average score. Runtime will depend on many parameters such as the unrelated programs running under the same operating system. Make another comparison graph showing performances of algorithms for 4 different input size. Here performance is defined as 1/ElapsedTime. The shorter the elapsed time the higher the performance. Make a bar graph like shown below. On horizontal axis you will have 4 different array sizes. Performance of each algorithm will be plotted for each array size. Organize your main function and other funcitons for implementing different algorithms in seperate files. The project should be consisting of 3 source code files: main.c, sortingFuncitons.h, and sortingFunctions.c ## Related Questions ##### . The fundamental operations of create, read, update, and delete (CRUD) in either Python or Java CS 340 Milestone One Guidelines and Rubric Overview: For this assignment, you will implement the fundamental operations of create, read, update, ##### . Develop a program to emulate a purchase transaction at a retail store. This program will have two classes, a LineItem class and a Transaction class Retail Transaction Programming Project Project Requirements: Develop a program to emulate a purchase transaction at a retail store. This ##### . The following program contains five errors. Identify the errors and fix them 7COM1028 Secure Systems Programming Referral Coursework: Secure ##### . Accepts the following from a user: Item Name Item Quantity Item Price Allows the user to create a file to store the sales receipt contents Create a GUI program that:Accepts the following from a user:Item NameItem QuantityItem PriceAllows the user to create a file to store the sales receip ##### . The final project will encompass developing a web service using a software stack and implementing an industry-standard interface. Regardless of whether you choose to pursue application development goals as a pure developer or as a software engineer CS 340 Final Project Guidelines and Rubric Overview The final project will encompass developing a web service using a software stack and impleme Get Free Quote! 276 Experts Online	1,021	4,733	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2024-18	latest	en	0.771351
https://dokumen.pub/number-theory-and-its-applications-1nbsped-1032231432-9781032231433.html	1,719,168,225,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00301.warc.gz	190,688,594	200,960	# Number Theory and its Applications [1 ed.] 1032231432, 9781032231433 Number Theory and its Applications is a textbook for students pursuing mathematics as major in undergraduate and postgra 363 71 2MB English Pages 356 [366] Year 2022 Cover Title Page Dedication Preface 1 Prerequisites 2 Theory of Divisibility 2.1 Introduction 2.2 Divisibility 2.3 Worked out Exercises 2.4 Greatest Common Divisor 2.5 Least Common Multiple 2.6 Worked out Exercises 2.7 Linear Diophantine Equations 2.8 Worked out Exercises 2.9 Exercises: 3 Prime Numbers 3.1 Introduction 3.2 Primes & Fundamental Theorem of Arithmetic 3.3 Worked out Exercises 3.4 Exercises: 4 Theory of Congruences 4.1 Introduction 4.2 Congruences 4.3 Worked out Exercises 4.4 Linear Congruences 4.5 Worked out Exercises 4.6 System of Linear Congruences 4.7 Worked out Exercises 4.8 Exercises: 5 Fermat’s Little Theorem 5.1 Introduction 5.2 Fermat’s Little Theorem 5.3 Worked out Exercises 5.4 Wilson’s Theorem 5.5 Worked out Exercises 5.6 Exercises: 6 Arithmetic Functions 6.1 Introduction 6.2 The Sum and Number of Divisors 6.3 Worked out Exercises 6.4 Mobiüs μ-function 6.5 Worked out Exercises 6.6 Greatest Integer Function 6.7 Worked out Exercises 6.8 Exercises: 7 Euler’s Generalization and Ø–function 7.1 Introduction 7.2 Euler’s Ø–function 7.3 Worked out Exercises 7.4 Euler’s Theorem 7.5 Worked out Exercises 7.6 Properties of Ø–function 7.7 Worked out Exercises 7.8 Exercises: 8 Primitive Roots 8.1 Introduction 8.2 Multiplicative Order 8.3 Worked out Exercises 8.4 Primitive Roots for Primes 8.5 Worked out Exercises 8.6 Existence of Primitive Roots 8.7 Worked out Exercises 8.8 Index Arithmetic 8.9 Worked out Exercises 8.10 Exercises: 9 Theory of Quadratic Residues 9.1 Introduction 9.2 Quadratic Residues and Nonresidues 9.3 Worked out Exercises 9.4 Quadratic Reciprocity Law 9.5 Worked out Exercises 9.6 The Jacobi Symbol 9.7 Worked out Exercises 9.8 Exercises: 10 Integers of Special Forms 10.1 Introduction 10.2 Perfect Numbers 10.3 Worked out Exercises 10.4 Mersenne Primes 10.5 Worked out Exercises 10.6 Fermat Numbers 10.7 Worked out Exercises 10.8 Exercises: 11 Continued Fractions 11.1 Introduction 11.2 Finite Continued Fractions 11.3 Worked out Exercises 11.4 Infinite Continued Fractions 11.5 Worked out Exercises 11.6 Periodic Fractions 11.7 Worked out Exercises 11.8 Exercises: 12 Few Non-Linear Diophantine Equations 12.1 Introduction 12.2 Pythagorean Triples 12.3 Worked out Exercises 12.4 Fermat’s Last Theorem 12.5 Worked out Exercises 12.6 Exercises: 13 Integers as Sums of Squares 13.1 Introduction 13.2 Sum of Two Squares 13.3 Worked out Exercises 13.4 Sum of More than Two Squares 13.5 Worked out Exercises 13.6 Exercises: 14 Certain Applications on Number Theory 14.1 Fibonacci Numbers 14.2 Worked out Exercises 14.3 Pseudo-random Numbers 14.4 Worked out Exercises 14.5 Cryptology 14.6 Worked out Exercises 14.7 Exercises: Bibliography Index ##### Citation preview Number Theory and its Applications Satyabrota Kundu Supriyo Mazumder Levant Books India Dedicated to our Family Members. Preface From ancient times, number theory has always occupied the unquestioned historical importance of the subject. Number Theory is both pure and applied and at the same time both classical and modern. It has been the objective of the authors for quite some time to write an accessible and inviting textbook to number theory. Foremost, the present textbook will create an eﬀective instrument for both teaching and learning. The authors aim to integrate the richness and beauty of the subject and at the same time the book is full of unexpected usefulness. In the present text, the authors have worked hard to assemble many contrasting aspects of number theory into one standard textbook. This book is ideal for undergraduate and postgraduate level students. Certain levels of mathematical reasoning, basic algebra and real analysis are required as prerequisites to study the materials of our textbook. This textbook is designed in such a manner that it will serve the purpose for a wide range of readers. Some eﬀort has been devoted to make the ﬁrst few chapters less challenging, but gradually the chapters become more challenging. At each juncture, the instructor has to decide how deeply to pursue a particular topic before moving ahead to a new one. Chapter I introduces important topics like well-ordering property, principle of mathematical induction, Binomial theorem and many more in establishing results to study the materials of the textbook. Chapter II introduces the notion of divisibility, where with the help of the Euclidean algorithm, the greatest common divisors of a set of integers are introduced. Also the least common multiple and linear Diophantine equations are discussed. Chapter III deals with prime factorization, the fundamental theorem of arithmetic and factorization techniques are covered. Chapter IV introduces congruences and develops their fundamental properties. systems of linear congruences in one or more unknowns are discussed. The Chinese remainder theorem is also developed. Chapter V is concerned with Fermat’s Little theorem and Wilson’s theorem. Wilson’s theorem gives a congruence for factorials. Pseudo primes, strong pseudoprimes and absolute pseudoprimes(Carmichael numbers) are also introduced. Chapter VI gives a thorough discussion of the sum and number of divisors, M¨ obius function and greatest integer function. vi Number Theory and its Applications Chapter VII develops Euler’s φ-function, Euler’s theorem and properties of Euler’s φ-function. Explicit formulae are developed for these functions. Chapter VIII is devoted to the discussion of the order of an integer and of the primitive roots. Indices, which are similar to logarithms, are introduced. Primality testing based on primitive roots are described. Chapter IX includes discussions on quadratic residues and the famous law of quadratic reciprocity. The Legendre and Jacobi symbols are introduced and algorithms for evaluating them are described. Chapter X deals with Integers of Special forms where the special emphasis is given on Perfect numbers, Mersenne primes and Fermat’s numbers. Chapter XI treats simple ﬁnite and inﬁnite continued fractions. Special attention is paid to the Periodic fractions. Chapter XII covers the study on few non-linear Diophantine equations, where the various results on Pythagorean triples and Fermat’s Last theorem are done. Chapter XIII deals with the study of Integers as sum of squares, which includes sum of two squares and sum of more than two squares. Concluding chapter introduces the application part of number theory, where the Fibonacci numbers, Pseudo-random numbers and basic cryptology have been discussed. Here the Pseudo-random numbers and the techniques for generating them are discussed. In cryptology, Caeser cipher, Block cipher, Exponentiation cipher and RSA are based on modular arithmetic. After each section of the text there are worked-out exercises set containing problems of various levels of diﬃculty. Each set contains problems of a numerical nature, which should be done to develop computational skills. At the end of every chapter there is an exercise set to ﬁne-tune the practice for the students. Throughout the textbook, the readers will face some simple questions like Why! and Verify! to make the topics interesting for them to study. At the end of the textbook there is an extensive bibliography, where the lists of number theory texts and references are included for interested readers who would like to go for more details about some of the topics covered in the book. Finally, the authors would like to express their heartfelt gratitude to their respected teachers, friends and relatives, who provide constant support and inspiration in framing the book. Contents 1 Prerequisites 1 2 Theory of Divisibility 15 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2 Divisibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 20 2.4 Greatest Common Divisor . . . . . . . . . . . . . . . . . . . . . . 23 2.5 Least Common Multiple . . . . . . . . . . . . . . . . . . . . . . . 31 2.6 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 33 2.7 Linear Diophantine Equations . . . . . . . . . . . . . . . . . . . . 37 2.8 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 39 2.9 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 3 Prime Numbers 45 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 3.2 Primes & Fundamental Theorem of Arithmetic . . . . . . . . . . 46 3.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 56 3.4 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 4 Theory of Congruences 4.1 67 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 4.2 Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 73 4.4 Linear Congruences . . . . . . . . . . . . . . . . . . . . . . . . . . 77 4.5 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 81 4.6 System of Linear Congruences . . . . . . . . . . . . . . . . . . . . 87 4.7 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 91 4.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 viii 5 Fermat’s Little Theorem 5.1 Introduction . . . . . . . 5.2 Fermat’s Little Theorem 5.3 Worked out Exercises . 5.4 Wilson’s Theorem . . . 5.5 Worked out Exercises . 5.6 Exercises: . . . . . . . . Number Theory and its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Arithmetic Functions 6.1 Introduction . . . . . . . . . . . . . 6.2 The Sum and Number of Divisors . 6.3 Worked out Exercises . . . . . . . 6.4 Mobi¨ us μ-function . . . . . . . . . 6.5 Worked out Exercises . . . . . . . 6.6 Greatest Integer Function . . . . . 6.7 Worked out Exercises . . . . . . . 6.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 95 96 101 106 108 110 . . . . . . . . 113 113 114 118 126 129 132 137 143 7 Euler’s Generalization and φ–function 7.1 Introduction . . . . . . . . . . . . . . . 7.2 Euler’s φ–function . . . . . . . . . . . 7.3 Worked out Exercises . . . . . . . . . 7.4 Euler’s Theorem . . . . . . . . . . . . 7.5 Worked out Exercises . . . . . . . . . 7.6 Properties of φ–function . . . . . . . . 7.7 Worked out Exercises . . . . . . . . . 7.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 145 146 149 153 155 158 161 165 8 Primitive Roots 8.1 Introduction . . . . . . . . . . 8.2 Multiplicative Order . . . . . 8.3 Worked out Exercises . . . . 8.4 Primitive Roots for Primes . 8.5 Worked out Exercises . . . . 8.6 Existence of Primitive Roots 8.7 Worked out Exercises . . . . 8.8 Index Arithmetic . . . . . . . 8.9 Worked out Exercises . . . . 8.10 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 167 168 171 175 179 183 188 192 197 200 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contents ix 9 Theory of Quadratic Residues 203 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 9.2 Quadratic Residues and Nonresidues . . . . . . . . . . . . . . . . 203 9.3 Worked out Exercises 9.4 Quadratic Reciprocity Law . . . . . . . . . . . . . . . . . . . . . 217 . . . . . . . . . . . . . . . . . . . . . . . . 212 9.5 Worked out Exercises 9.6 The Jacobi Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . 225 9.7 Worked out Exercises 9.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 . . . . . . . . . . . . . . . . . . . . . . . . 221 . . . . . . . . . . . . . . . . . . . . . . . . 232 10 Integers of Special Forms 239 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 10.2 Perfect Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 10.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 244 10.4 Mersenne Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 10.5 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 254 10.6 Fermat Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 10.7 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 261 10.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 11 Continued Fractions 265 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 11.2 Finite Continued Fractions . . . . . . . . . . . . . . . . . . . . . 266 11.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 273 11.4 Inﬁnite Continued Fractions . . . . . . . . . . . . . . . . . . . . . 276 11.5 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 284 11.6 Periodic Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . 286 11.7 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 292 11.8 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 12 Few Non-Linear Diophantine Equations 295 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 12.2 Pythagorean Triples . . . . . . . . . . . . . . . . . . . . . . . . . 296 12.3 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 299 12.4 Fermat’s Last Theorem . . . . . . . . . . . . . . . . . . . . . . . 302 12.5 Worked out Exercises . . . . . . . . . . . . . . . . . . . . . . . . 303 12.6 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 x 13 Integers as Sums of Squares 13.1 Introduction . . . . . . . . . . . 13.2 Sum of Two Squares . . . . . . 13.3 Worked out Exercises . . . . . 13.4 Sum of More than Two Squares 13.5 Worked out Exercises . . . . . 13.6 Exercises: . . . . . . . . . . . . Number Theory and its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Certain Applications on Number Theory 14.1 Fibonacci Numbers . . . . . . . . . . . . . 14.2 Worked out Exercises . . . . . . . . . . . 14.3 Pseudo-random Numbers . . . . . . . . . 14.4 Worked out Exercises . . . . . . . . . . . 14.5 Cryptology . . . . . . . . . . . . . . . . . 14.6 Worked out Exercises . . . . . . . . . . . 14.7 Exercises: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 307 307 312 315 319 321 . . . . . . . 323 323 331 334 338 339 346 348 Bibliography 351 Index 353 1 Prerequisites Number theory is an important and signiﬁcant area of Mathematics. From ancient times, it has had a substantial impact on human civilization. For further development many great philosophers had contributed various aspects of number theory. Among those was Pythagoras (569 − 500 B.C). His famous Pythagorean theorem led many mathematicians to study squares and sum of squares. He has given us notable Pythagorean triplets. After 200 years, another Greek mathematician Euclid (350 B.C.) drew attention to the prime numbers. His investigations were based on the renowned Euclidean algorithm for ﬁnding greatest common divisors of two natural numbers. It plays a key role for evaluating prime factorisation of natural numbers. Few more studies on natural numbers had been done by Eratosthenes(276B.C.-196B.C.), Nicomachus(C.100) and Diophantus(C.250). The study of number theory started with the set of natural numbers, denoted by N = {1, 2, 3, 4, 5, . . .}. Two basic operations addition and multiplication are deﬁned on N. But in general, the subtraction is not deﬁned. The reason behind this is that the people will not be able to take Rs 5 from a person with Rs 4. Thus for u, v ∈ N, u − v is not always deﬁned. Here comes the notion of negative natural numbers that is −1, −2, −3, . . . so on. Therefore the set {1, 2, 3, 4, . . . , −1, −2, −3, −4, . . .} is an enlargement of N. But here 0 is missing. The concept of zero originated in ancient India, ancient Babylon and the Mayan civilisation. It is believed that it came into existence from 458 A.D. Actually the concept of zero was developed at diﬀerent times in each of these civilisations. But it was ﬁrst used in ancient India by Hindus and thereby Arabics use it as a number. Thus by inclusion of zero and the negative natural numbers create a new set known as set of integers, denoted by Z = {−1, −2, −3, −4, . . . ., 0, 1, 2, 3, 4, . . .}. Here the basic operations addition, 1 2 Number Theory and its Applications subtraction and multiplication is possible. But division, the inverse operation of multiplication is not deﬁned on Z. In other words, if u, v are elements of Z then uv is not always an element of Z. The set of integers plays an important role in the study of number theory. Next we proceed with our discussion by two important properties of natural numbers or sets of positive integers. The ﬁrst begins with a well ordering principle. The Well-Ordering Property: Every nonempty subset A of natural numbers has a least element. In other words, if A is a non-empty subset of N then ∃ m ∈ A such that m ≤ a, ∀a ∈ A. In particular, N itself has the least element 1. We now prove these two principles by taking help of each other. First we will prove the well-ordering principle using the principle of mathematical induction. (I)Deduction of mathematical induction by well ordering principle. Proof. Let A be the subset of natural numbers where 1 ∈ A and if k ∈ A then k + 1 ∈ A. Its suﬃces to show that A = N. To fulﬁll this, let B be the set of those natural numbers which does not belong to A. Then A ∪ B = N and A ∩ B = φ. To get the desired result we need to show that B = φ. If possible, let B = φ. Then by well ordering principle there exists a least element m of B such that m ≤ n, ∀ n ∈ B. Note that 1 = m as 1 ∈ A and hence 1 ∈ / B. Since m is the least element of B, therefore m − 1 ∈ / B and this gives m − 1 ∈ A. By hypothesis, we have (m − 1) + 1 ∈ A i.e. m ∈ A. This contradicts the fact that A ∩ B = φ. So our assumption that B = φ is wrong. This proves that A = N. (II) Deduction of well-ordering principle by mathematical induction. Proof. To the contrary, let A be the non empty subset of N having no least element. Let us construct a set B of natural numbers in such a way that any number x ∈ B is less than every member of A. Thus B = {x\|x ∈ N & x < a} for every a ∈ A. Then every member of A is either greater than or equal to each member of B. Therefore A ∩ B = φ. Note that 1 ∈ / A, otherwise 1 will become the least element of A. But by hypothesis A has no least element. Therefore 1 ∈ B and 1 < a for every a ∈ A. Assume that t ∈ B. Then t < a for every a ∈ A. We assert t + 1 ∈ / B. If so, then t + 1 be the next natural number larger than t. Then t + 1 will be the least element of A but A does not have any least element. Hence t + 1 ∈ B. Hence by principle of mathematical induction B ⊆ N. But A ∩ B = φ. Then N ∩ A = φ yields A = φ, a contradiction. Therefore A must have a least element. This completes the proof. Prerequisites 3 The above discussion leads to the following theorems related to mathematical induction. Theorem 1.0.1. Principle of induction: Let S be a subset of N such that (i) 1 ∈ S, (ii) if k ∈ S, then k + 1 ∈ S. Then S = N. Theorem 1.0.2. Let P (n) be u statement with n ∈ N satisfying (i) P (1) is true, (ii) P (k + 1) is true, whenever P (k) is true; then P (n) holds for all n. Example 1.0.1. Prove that for each n ≥ 2, (n + 1)! > 2n . The inequality holds for n = 2 since (2 + 1)! > 22 . Solution 1.0.1. Let the inequality hold for some natural number k ≥ 2. Then (k + 1)! > 2k . We are to prove that the inequality prevails for k + 1 ı.e. {(k + 1) + 1}! > 2k+1 . Now {(k + 1) + 1}! = (k + 2)! = (k + 2)(k + 1)! > 2 · 2k [∵ k + 2 > 2] = 2k+1 . Hence the inequality prevails for k + 1. Therefore by principle of induction, the inequality holds for all natural numbers ≥ 2. Remark 1.0.1. It is to be noted that the inequality fails to hold for n = 1. Example 1.0.2. Prove that for every n ≥ 1, 1 1 1 1 n + + + ··· + = . 1·2 2·3 3·4 n(n + 1) n+1 1 1 1 1 n + 2·3 + 3·4 + · · · + n(n+1) = n+1 be the statement Solution 1.0.2. Let P (n) : 1·2 for n ≥ 1. Note that P (1) is true(Verify!). Let k be the positive integer such that P (k) holds. We are to show P (k + 1) holds. Now 1 1 1 1 1 + + + ··· + + 1·2 2·3 3·4 k(k + 1) (k + 1)(k + 2) 1 k + , ∵ P (k) holds = k + 1 (k + 1)(k + 2) 1 (k + 1)2 = · k+1 k+2 k+1 . = (k + 1) + 1 Hence P (k + 1) holds. Hence the result follows. 4 Number Theory and its Applications m Example 1.0.3. Prove that 2m + 1 < 2 for all integers m ≥ 3. Solution 1.0.3. Let P (m) : 2m + 1 < 2m be the statement, for all integers m ≥ 3. Here for m = 3, we have 2 · 3 + 1 = 7 < 8 = 23 . Thus the statement is true for m = 3. Hence P (3) holds. Suppose k(≥ 3) is an integer for which the statement P (k) is true. To prove the validity of the argument of this statement we need to show P (m) is true for m = k + 1. Now 2(k + 1) + 1 = 2k + 1 + 2 < 2k + 2[Since P (m) is true for m = k] < 2k + 2k [since k ≥ 3] < 2k+1 . Thus the statement P (m) is true for m = k + 1. So by applying the principle of mathematical induction, we can say that P (m) is true for all m ≥ 3. Example 1.0.4. Prove that n2 < 2n for every integers n ≥ 5. Solution 1.0.4. Let P (n) : n2 < 2n be the given statement for all integers n ≥ 5. Here P (5) is true(verify!). Suppose k is a positive integer such that k ≤ 5 and P (k) is true. We are to show P (k + 1) is true. Now (k + 1)2 = k 2 + 2k + 1 < 2k + 2k + 1(by induction hypothesis) < 2k + 2k (see Example 1.0.3) < 2 · 2k = 2k+1 . So P (k + 1) holds. Hence this ﬁnishes the result. As we know counting is an important aspects of number system. There is a deep and signiﬁcant relation between counting techniques and number theory. We now continue our discussion with a signiﬁcant property of counting techniques, Pigeonhole principle, of the set of integers Z. It plays a key role in the study of numbers. The statement of this principle as follows. Theorem 1.0.3. Pigeonhole principle: The simplest form of the principle is that, if k + 1 or more objects are to be placed in k boxes then there is at least one box containing two or more of the objects. Proof. In anticipation of a contradiction, suppose that none of the k boxes has more than one object. Then the total number of objects will be k. This would lead to the contradiction as there are k + 1 or more objects. Prerequisites 5 In fact the ﬁrst formalisation of the notion of Pigeonhole principle is believed to be given by Dirichlet(1805 − 1859). He called it as drawer principle. But the term “Pigeonhole principle ”was ﬁrst coined by the mathematician Raphael M. Robinson in 1940. The next example illustrates this concept lucidly. Example 1.0.5. If there is a group of 368 people in a club, then there must be at least three people with the same birthday. The reason behind this is that the number of possibilities of birthdays in a year is 366. One can think that this principle is very much obvious. But this in fact would become an useful tool once we generalize it. The statement of this generalization is as follows: If n objects are placed in k boxes, then there is at least one box containing at least nk number of objects, where nk is the greatest integer not exceeding n k. Another important aspect of counting techniques which can be used frequently in the study of number theory is binomial theorem and it can be stated as follows. Theorem 1.0.4. For any two real numbers a, b and any positive integer n, (a + b)n = n n r=0 r an bn−r . One of the nice diagrammatic representation of binomial coeﬃcients is Pascal’s triangle, named after French mathematician Blaise Pascal. This looks like n=0 n=1 n=2 n=3 n=4 n=5 n=6 1 1 1 1 1 1 1 3 4 5 6 1 2 3 6 10 15 1 1 4 10 20 1 5 15 1 6 1 It shows that when two adjacent binomial coeﬃcients in the triangle are added, the binomial coeﬃcient in the next row between these two coeﬃcients are produced. All the above discussions are based on integers but in our next discussion we have highlighted those numbers which are of the form uv where u, v ∈ Z. Here v must be nonzero, otherwise the division will not be possible. From the division of integers, there comes the concept of decimal numbers. For example 45 = 0.8. 6 Number Theory and its Applications 2 4 Here we see that = 0.5 ı.e. we can cancel common multiple from both numerator and denominator till their greatest common divisor will be 1.This extension of set of integers leads to the notion of rational numbers, denoted by Q. Thus any number of the form uv where u, v ∈ Z with v = 0 and gcd(u, v) = 1, is called rational number. Actually the symbol Q comes from the German word ‘quotient ’which can be translated as ‘ratio ’, appeared in Bourbaki’s Algebra(1998). Classical greek and Indian mathematicians studied theory of national numbers, as part of the general study of number theory. The concept of rational numbers were from the ancient times. Those can be expressed as a ratio of two diﬀerent numbers that had no common divisors except 1. This prevalent concept which was there over a period of time was found to be inadequate, when in 500B.C. the Italian philosopher Hippasus proved the existence of irrational numbers. He was also a member of group of people who were called the Pythagorean mathematicians. His method of ﬁnding irrational numbers involve the technique of √ contradiction. There he had assumed that 2 is an irrational number. Then he moved on to show that no such rational number could exist. Therefore it should be something diﬀerent. However, Pythagorus believed in the absoluteness of number and could not accept the existence of irrational numbers. So he sentenced Hippasus to death by drowning. In 16th century, ﬁnally Europeans accepted the existence of negative integers and fractional numbers. Thereby in 17th century, many mathematicians used decimal fractions with modern notations. Subsequently from 19th century onwards, the irrationals were seperated into algebraic and transcendental part. It had remained almost dormant since Euclid. Actually irrationals are closely associated to continued fractions, which had received attention at the hands of Euler. Finally we end our discussions with the two fundamental theorems. Theorem 1.0.5. There any r ∈ Q satisfying r2 = 2. Proof. If possible, let ∃ r ∈ Q such that r2 = 2. Since r ∈ Q, therefore it is of the form xy with gcd(x, y) = 1 and y = 0. Hence x2 = 2y 2 . This proves x2 is even and consequently x is also so. Suppose x = 2k. Then 4k 2 = 2y 2 or 1, y 2 = 2k 2 . Hence y 2 is an even integer and so y is even. This shows gcd(x, y) = which ends up with contradiction. Thus for any rational quantity r, the equation √ x2 = 2 has no solutions. Furthermore, this theorem leads us to the fact that 2 is irrational. Finally our next theorem indicates on the existence of irrational numbers. Theorem 1.0.6. Let k be a non-square positive integer. Then any r ∈ Q such that r2 = k. Prerequisites 7 Proof. To the contrary, assume that ∃ a rational number r satisfying r = k. Then r can be written as ab where gcd(a, b) = 1. Then a2 = kb2 holds. For this positive integer k, there exists a positive integer l such that 2 l2 < k < (l + 1)2 ⇒ b2 l2 < a2 < (l + 1)2 b2 ⇒ bl < a < (l + 1)b ⇒ 0 < a − bl < b. (1.0.1) Further, (kb − al)2 = k 2 b2 − 2kbal + (al)2 = k[a2 − 2abl + (bl)2 ] = k(a − bl)2 2 ka − al (kb − al)2 a , So k = . Since a and b are prime to each other and (a − bl)2 b a − bl are two representation of k, therefore we have a − bl > b and this contradicts (1.0.1). So this completes the proof. Finally we are at the edge of culminating our discussions related to propagation of numbers. In the concluding part, we conduct the study of real numbers which is the union of rationals and irrationals. The set of real numbers is denoted by R. The creditability of uncovering R would not goes to a single person. The development took 2000 years and many renowned mathematicians had contributed with their important thoughts on R. The notion of R had started its development from ancient Greece. Euclid had developed a theory of proportions which is equivalent to modern theory of real numbers. In 17th century, John Napier and Simon Stevin introduced the concept of inﬁnite decimal expansion. Cantor and Dedekind had given their contributions to originate the modern theories of R. In the following sections we are going to discuss few important properties of R. Order Properties of R The linear relation ‘ 0, then xz < yz. 8 Number Theory and its Applications The set R together with the ordered relation deﬁned on it is said to be ordered set. For any x ∈ R, the concept of absolute property of R is deﬁned as \|x\| = x, if x > 0 = 0, if x = 0 = −x. if x < 0. Completeness Properties of R Next we are going to discuss the completeness property of R. The discussion starts with the following deﬁnition. Deﬁnition 1.0.1. Let K ⊂ R. Then a real number u is said to be an upper bound(lower bound) of K if x ∈ K and x ≤ u(x ≥ u). If K has an upper bound(lower bound), then it is called bounded above(bounded below). Example 1.0.6. Let K = {x ∈ R : 3 < x < 4}. Note that K is bounded above and 4 is the upper bound. Also K is bounded below and 3 is the lower bound. The last example raises the question, whether the upper bound 4 and the lower bound 3 is greatest or least respectively. The following deﬁnition of least upper bound and greatest lower bound will be the answer to the raised question. Deﬁnition 1.0.2. Let K ⊂ R. If K is bounded above(bounded below), then an upper bound (lower bound) is said to be the least upper bound(greatest lower bound) or supremum(inﬁmum) if it is less(greater) than every upper(lower) bounds of K. Actually it is a deeper property of R that for any non-empty bounded above(below) subset of K of R, the least upper bound(greatest lower bound) do exists. This property of R is called the supremum(inﬁmum) property. Note that we can establish these two properties are equivalent and one implies other. Furthermore, the supremum property of R can be treated as an axiom, known to be the completeness property of R. The statement is as follows: Statement 1.0.1. Axiom of least upper bound: If a set K is bounded above, then it has a least upper bound i.e. there exists a unique real number M satisfying 1. x ≤ M, ∀ x ∈ K. 2. for arbitrary (> 0), there exists an element α ∈ K such that M − < α ≤ M. Prerequisites 9 It is not necessary that every subset of R should have an upper bound. For instance it can be examined that N is not bounded above and thus it has no upper bound. Now taking help of the axiom of least upper bound, we are going to show the existence of greatest lower bound for a non–empty bounded below subset of R. Theorem 1.0.7. A non-empty subset K of R which is bounded below, has the greatest lower bound i.e.there exists a unique real number m such that 1. x ≥ m, ∀ x ∈ K. 2. For arbitrary (> 0), there exists an element β ∈ K such that m ≤ β < m + . Proof. Note that K is a non empty set, which is bounded below. Then ∃ k ∈ R such that x ≥ k, ∀ x ∈ K. Consider K = {−x : x ∈ K}. Let y ∈ K . Then −y ∈ K and −y ≥ K together implies y ≤ −k. This is true for all y ∈ K . This shows K is bounded above. Then from least upper bound axiom, we obtain M = sup K . We need to show −M = inf K. Let t ∈ K. Then −t ∈ K implies −t ≤ M . Thus we have t ≥ −M . Let (> 0) be chosen arbitrarily. Then α ∈ K such that M − < α ≤ M ⇒ −M ≤ −α < −M + . Since α ∈ K , then −α = β ∈ K and thus taking −M = m we have, 1. x ≥ m, ∀ x ∈ K. 2. for arbitrary (> 0), there exists an element β ∈ K such that m ≤ β < m + . Our next phase of discussion deals with the concept of functions between two arbitrary sets. The notion of function has an paramount importance not only in number theory. It helps to correlate between various algebraic structures. It is actually a rule of correspondence between the elements of two sets. The idea of function was developed in the seventeenth century. During this period mathematician Rene Descartes(1596 − 1650) used it to describe many mathematical relationship in his book Geometry. The term function was introduced by Gottfried Wilhelm Leibnitz(1646 − 1716), almost ﬁfty years after the publication of Geometry. The idea was further formalized by Leonhard Euler(1707−1783). However the present day conception of function is attributed to Dirichlet(1805 − 1859), who in 1837 proposed the deﬁnition of a function as 10 Number Theory and its Applications a rule of correspondence that assigns a unique value of an independent variable. Now a function from a set X to an another set Y can be deﬁned as follows: Deﬁnition 1.0.3. For two non-empty sets X and Y , a relation f from X into Y is called a function from X into Y if 1. domain of f is set X 2. f is well-deﬁned in the sense that for x1 , x2 ∈ X with x1 = x2 ⇒ f (x1 ) = f (x2 ). Actually f is a subset of X × Y , where X is referred as domain of f , denoted by Dom(f ), and the set Y is said to be co-domain of f . Here the set f (X) = {f (x) : x ∈ X} is a subset of Y . This f (X) is called range of f , denoted by Imf . For instance, consider X = Y = R. Then f (x) = 2x, ∀ x ∈ R is a function from R to R which is also called real valued function. Moreover, there are few types of functions such as f (x) = x ∀ x ∈ X from X to X, called identity function and f (x) = c for some constant c is known as constant function. Two functions f, g : X → Y are said to be equal if f (x) = g(x) ∀ x ∈ X. Sometimes functions can produce same values for diﬀerent elements of domain set such as if we consider f : R → R deﬁned by f (x) = \|x\|, then we can see that f (3) = f (−3) but 3 = −3. From this, we have the following deﬁnition. Deﬁnition 1.0.4. A function f : X → Y is said to be injective when for every x1 , x2 ∈ X if x2 ⇒ f (x1 ) = f (x2 ). x1 = Moreover, if for every y ∈ Y there exists at least one x ∈ x such that f (x) = y holds then f is said to be surjective. Intuitively, a surjective map covers the whole of the codomain set i.e. Im(f ) is the codomain of f . f is called bijective if it is both injective and surjective. Suppose A = {1, 2, 3} and B = {p, q, r, s, t} and f : A → B be deﬁned by f (1) = p, f (2) = r, f (3) = s. Since p, q, r have only one preimage and s, t have no preimage, therefore f is injective. On the other hand, suppose A = {1, 2, 3, 4, 5} and B = {p, q, r}. Let f : A → B and g : A → B be respectively deﬁned by f (1) = q, f (2) = p, f (3) = q, f (4) = r, f (5) = p; g(1) = r, g(2) = p, g(3) = p, g(4) = r, g(5) = r. Prerequisites 11 It is clear that f is surjective but g is not. Also, here f fails to be injective. Next let us consider the function f : R → R deﬁned by f (x) = 5x + 2. Here f is an example of a bijective function. Now consider a function f : R → R deﬁned by f (x) = sin x2 . Then we see that the function f is the amalgam of two functions f : R → R, where f (x) = sin x and g(x) = x2 . This originates the notion of composition of two functions. Deﬁnition 1.0.5. Consider two functions f : X → Y and g : Y → Z. Then the composition of f and g, denoted as g ◦ f , is deﬁned as (g ◦ f )(x) = g(f (x)) where g ◦ f : X → Z. Finally, we end this session of discussion by the concept of inverse of a function. Deﬁnition 1.0.6. Let f : X → Y be a bijective function. Then the function f −1 : Y → X is said to be inverse of f . The domain of the inverse function f −1 is the range of f and the range of f −1 is the domain of f . The concluding part of the chapter deals with the study of a special type of real-valued function whose domain set is, in particular, a set of naturals i.e.f : N → R, denoted by f (n) = xn where n ∈ N and xn ∈ R. In other words, it can be interpreted as 1 → x1 , 2 → x2 , . . . and so on. Therefore, the representation xn prevails. The set {xn : n ∈ N} is a subset of R which is said to be the range of the sequence. For example, let us consider a function f : N → R n n deﬁned as f (n) = n+1 , ∀ n ∈ N. Here the sequence is { n+1 }n , also denoted 1 2 3 n as { 2 , 3 , 4 , . . .}. The range set of this sequence is { n+1 : n ∈ N}. Now we are going to discuss the boundedness of sequence, whose deﬁnition is as follows: Deﬁnition 1.0.7. A sequence is said to be bounded above or bounded below according as the range set of the sequence is bounded above or below. It follows that the sequence {xn } is bounded above(below) if there exists a real quantity G(g) such that xn ≤ G(xn ≥ g) ∀ n ∈ N and G(g) is called the upper(lower) bound of the sequence. A sequence is said to be bounded if it is both bounded above and bounded below. 1 n =1− ≤ 1. Thus n+1 n+1 {xn } is bounded above and 1 is the upper bound. Consider { n1 } for all n ∈ N, where the values of the range set goes to zero if n → ∞. In other words, for all large values of n, the values of the range set cluster near zero. This generates the concept of convergence of a sequence. From the last example, we can see that xn = 12 Number Theory and its Applications Deﬁnition 1.0.8. A real number l is deﬁned as the limit of a sequence {xn } if for any arbitrary (> 0), ∃ m ∈ N such that \|xn − l\| < , ∀ n ≥ m. In this case, the sequence is said to be convergent and can be written as limn→∞ xn = l. The question arises: Is the limit l of the convergent sequence is unique? We are fortunate enough to have the following theorem as the answer to the question. Theorem 1.0.8. Limit of a convergent sequence is unique. Proof. Let the sequence {xn } converges to l and l . Then for a pre-assigned (> 0), ∃ m1 , m2 ∈ N such that \|xn − l\| < , ∀ n ≥ m1 , 2 and \|xn − l \| < , ∀ n ≥ m2 . 2 Let m = max{m1 , m2 }. Then, \|l − l \| = \|(xn − l) − (xn − l)\| ≤ \|(xn − l ) − (xn − l)\| < + = , ∀ n > m. 2 2 Since (> 0) is arbitrary, therefore the uniqueness follows. The discussion of the following theorem is based on the connection between convergence and boundedness. Theorem 1.0.9. A convergent sequence is bounded. Proof. Let the sequence {xn } converges to l. Then for a pre-assigned (> 0), ∃ m such that \|xn − l\| < , ∀ n ≥ m, i.e. l − < xn < l + , ∀ n ≥ m. If A and B be the greatest and least of the ﬁnite set {x1 , x2 , . . . , xn , . . . , l−, l+}, then B ≤ xn ≤ A ∀ n. Hence the sequence {xn } is bounded. Note that the converse of the above theorem is not true in general. For instance, consider the sequence {(−1)n } which is bounded but not convergent. So, if a sequence does not converge then two possibilities may arise. Either the sequence diverge or oscillates. Prerequisites 13 Deﬁnition 1.0.9. A sequence {xn } is said to be divergent and diverges to +∞, if for an arbitrary positive real number N , however large it is, ∃ m ∈ N such that xn > N, ∀ n ≥ m. Also, the sequence {xn } is said to be divergent and diverges to −∞, if for an arbitrary positive real number N , however large it is, ∃ m ∈ N such that xn < −N, ∀ n ≥ m. Finally, a sequence which is neither convergent nor divergent is said to be oscillatory sequence. For example, the sequence {2n } diverges to +∞ and the sequence {−n2 } diverges to −∞. The sequence {(−1)n } is an oscillatory one. Finally, we close this discussion with monotone sequence and few properties related to it. Deﬁnition 1.0.10. A sequence {xn } is said to be monotone increasing(decreasing) sequence if xn+1 ≥ (≤)n, ∀ n ∈ N. As an example, let us consider the sequence n2 i.e.{1, 4, 9, . . .}. This sequence is monotone increasing, however the sequence { 21n } i.e.{ 12 , 14 , 18 , . . .} is monotone decreasing. Theorem 1.0.10. A monotone increasing sequence, if bounded above, is convergent and converges to least upper bound. Proof. Let {xn } be the monotone increasing sequence, bounded above. Then for a pre-assigned (> 0), ∃ r ∈ N such that M − < xr ≤ M < M + . Here M is the least upper bound of {xn }. Therefore, M − < xr ≤ M + . Again, since {xn } is monotone increasing then for n ≥ r we ﬁnd xn ≥ xr . Thus, M − < xr < xn ≤ M < M + ⇒ \|xn − M \| < ∀ n ≥ r. This completes the proof. Remark 1.0.2. A monotone increasing sequence, if unbounded above, is divergent and diverges to +∞. Analogously, we can state that Theorem 1.0.11. A monotone decreasing sequence, if bounded below, is convergent and converges to greatest lower bound. Remark 1.0.3. A monotone decreasing sequence, if unbounded below, is divergent and diverges to −∞. 14 Number Theory and its Applications Exercises: n(n+1) 2 1. Using Mathematical induction show that 13 + 23 + · · · + n3 = for all positive integers n. 2 , 2. Using induction on n prove the identity n2 < n! for all integers n ≥ 4. 3. What is the coeﬃcient of x7 in (3x + 2)19 ? 4. Find the coeﬃcient of x5 y 7 in (x + y)10 . 5. Suppose S is a set of n + 1 integers. Prove that there exist distinct a, b ∈ S such that a − b is a multiple of n. 6. Show that in any group of n people, there are two who have an identical number of friends within the group. 7. Determine whether the function f : Z → Q deﬁned by f (x) = 2x , for all x ∈ R is bijective or not. 8. Let f, g : R → R be two functions, given by f (x) = \|x\| + x for all x ∈ R and g(x) = \|x\| − x for all x ∈ R. Evaluate f ◦ g and g ◦ f . √ √ 9. A sequence {xn } is deﬁned by xn = 5 and xn+1 = 5xn for n ≥ 1. Prove that lim xn = 5. 10. Show the convergence of the sequence {(1 + n1 )n }. 2 Theory of Divisibility “A marveilous newtrality have these things mathematicall, and also a strange participation between things supernaturall, immortall, intellectuall, simple and indivisible, and things naturall, mortall, sensible, componded and divisible.” – John Dee 2.1 Introduction Mathematics is the Universe’s natural tongue. From very beginning of our existence as a species, numbers have deeply mesmerised us. Due to Carl Friedrich Gauss “Number theory is one of the oldest branches of Mathematics which established a relationship between numbers belonging to the set of real numbers”. The pureness of Number Theory has charmed mathematicians generation after generation — each contributing to the branch that Carl Gauss described as the “Queen of Mathematics.” Today, however, a basic understanding of Number Theory is an absolute precursor to cutting-edge software engineering, speciﬁcally security-based software. Number Theory is at the heart of cryptography — which is itself experiencing a engrossing period of rapid evolution, ranging from the famous RSA algorithm to the wildly-popular blockchain world. Two clear-cut moments in history stand out as curvature points in the development of Number Theory. First, in archaic times, Euclid put forth his GCD (Greatest Common Divisor) algorithm — a splendid set of steps that simpliﬁes fractions to their simplest form using geometrical observations. Then, approximately two-thousand years later, Gauss formalized Euclid’s principles by com15 16 Number Theory and its Applications bining Euclid’s informal writings with his own extensive proofs in the timeless Disquistiones Arithmeticae. 2.2 Divisibility When an integer is divided by a second integer(= 0), the quotient may or may not be an integer. For instance, 36/6 = 6 is an integer, while 18/7 = 2.5 is not. This observation leads to the following deﬁnition. Deﬁnition 2.2.1. If a and b are integers, we say that b is divisible by a(= 0) if there exists an integer c such that b = ac. Also, we say that a is a divisor or factor of b, denoted by a\|b. If a does not divides b, then we write a b. Example 2.2.1. 10 is divisible by 5 because there exist an integer 2 such that 10 = 5 × 2. We say 5\|10. Proposition 2.2.1. For any integers a, b, c, d the following statements are true: 1. a\|0, 1\|a, a\|a. 2. a\|b ⇒ ca\|cb, ∀c ∈ Z. 3. a\|b and b\|c ⇒ a\|c. 4. a\|b and b\|a ⇒ a = ±b. 5. a\|b and a\|c ⇒ a\|(bx + cy) for arbitrary integers x and y. Proof. 1. Obvious. 2. Here, a\|b ⇒ b = da for some integer d, ⇒ cb = d(ca) ⇒ ca\|cb. 3. Here, a\|b ⇒ b = aq and c\|d ⇒ d = cp for some integers p and q. Therefore c = a(pq). Hence bd = ac(pq). Therefore ac\|bd, as pq is an integer. 4. Here, a\|b ⇒ b = ap for some integer p. Also, b\|c ⇒ c = bq for some integer q. Therefore c = bq = a(pq). Therefore a\|c. 5. Here, a\|b ⇒ b = ap for some integer p. Therefore b = bpq. Also, b\|a ⇒ a = bq for some integer q implies pq = 1. As p, q are integers either, p = q = 1 or p = q = −1. Therefore a = ±b. Theory of Divisibility 17 6. Here, a\|b ⇒ b = ap for some integer p and Here, a\|c ⇒ b = aq for some integer q. Therefore bx + cy = apx + aqy = a(px + qy). Now, px + qy ∈ Z as p, q, x, y ∈ Z. Therefore a\|(bx + cy). Theorem 2.2.1. The Division Algorithm: Given any two integers a and b, with b > 0 there exists unique integers q and r such that a = bq + r with 0 ≤ r < b. Proof. Let a and b be two ﬁxed integers with b = 0. Let A = {n ∈ N\|n = a − by, y ∈ Z}. Our claim is A = φ. For this there are two possibilities viz: 1. If a ≥ 0, then a − b(0) = a ≥ 0. So a − by is non–negative for y = 0(a ≥ 0). 2. If a < 0, then −a > 0. Since, b is a positive integer, we must have b ≥ 1. Multiplying the inequality by a positive quantity gives, (−a)b ≥ (−a) implies a − ab ≥ 0. So a − by is non–negative for y = a(< 0). Hence A = φ. Since, A ⊂ N, by well ordering principle A has a least element say r. Since, s ∈ A, therefore, r = a − by for some y = q. Thus we found integers r and q such that r = a − bq or a = r + bq. Since, r ∈ A, therefore r ≥ 0. Next our claim is r < b. On the contrary, if we assume r ≥ b, then 0 ≤ r − b = (a − bq) − b = a − b(q + 1) < r, which leads to a contradiction as r is the least in A. Hence r < b. Thus we found two integers q and r such that a = bq + r with 0 ≤ r < b. The last part of the proof deals with the uniqueness of q and r with the above properties. If possible, let there be two pair of integers r1 , q1 and r2 , q2 satisfying r1 + bq1 = a = r2 + bq2 (2.2.1) 0 ≤ r1 ≤ b and 0 ≤ r2 . (2.2.2) with We need to prove r1 = r2 and q1 = q2 . If r1 ≤ r2 , then 2.2.1 shows b(q1 − q2 ) = r2 − r1 . (2.2.3) Since by hypothesis, b > 0, r2 ≥ r1 , therefore q1 − q2 must be a non negative integer. Hence r2 − r1 must be one of 0, b, 2b, 3b, . . .. But 0 ≤ r2 ≤ r1 ≤ b implies r2 − r1 = 0. Hence by 2.2.3 and the preceeding equation together with the hypothesis b > 0, we have q1 = q2 . Similarly, taking r1 ≥ r2 , proves the uniqueness of q and r. 18 Number Theory and its Applications For another proof we give explicit formulae for the quotient and remainder in terms of the greatest integer function, which will be done in the consequent chapter of the book. Remark 2.2.1. 1. When b a, r satisﬁes strong inequality 0 < r < b. 2. Here q and r called quotient and remainder. 3. bq is largest multiple of b which does not exceed a. Example 2.2.2. Suppose we are dividing 51 by 5 then, 51 = 5 × 10 + 1. Comparing with the theorem we get, a = 51, b = 5, q = 10, r = 1.Here q = 10 is the quotient and r = 1 is remainder. Corollary 2.2.1. If a, b be two integers with b > 0, then there exists integers Q b and R such that a = bQ ± R, 0 ≤ R < . 2 Proof. From Division algorithm we have, for any two integers a and b with b > 0, there exists unique integers q and r such that a = bq + r, 0 ≤ r < b (2.2.4) We now consider three following cases: Case(i): Let r < and r = R in equation (2.2.4), we have a = bQ + R, 0 ≤ R 2b , then from equation(2.2.4) a = bq + r = b(q + 1) + (r − b) = b(q + 1) − (b − r). Taking q+1 = Q and b−r = R, we have a = bQ−R where R = b−r < b− 2b = 2b . Therefore a = bQ − R, 0 ≤ R < 2b . Now combining case (i) and (ii) we have, a = bQ ± R, 0 ≤ R 2b , the minimal remainder is R = r − b. Here we have done an illustration of the concept minimal remainder by an example. Let us choose a = 51 and b = 6 then 51 = 6 × 8 + 3(of the form a = bQ + R, Q = 8, R = 3). Also we can write 51 = 6 × 9 − 3(of the form a = bQ − R, Q = 9, R = 3). Thus Q and R are not unique as R = 2b = 3. Which is case (iii) of above corollary. Now if we choose a = 50 and b = 6 then 50 = 6 × 8 + 2. Thus in this case r = 2 < 2b = 3 and the minimal remainder is R = 2. Which is case (i) of above corollary. Now if we choose a = 52 and b = 6 then 52 = 6 × 8 + 4. Thus in this case r = 4 > 2b = 3 and the minimal remainder is R = r − b = 4 − 6 = −2. Which is case (ii) of above corollary. Theorem 2.2.2. Prove that every integer is of the form, 1. 3k or 3k ± 1. 2. 4k or 4k ± 1 or 4k ± 2. 3. 5k or 5k ± 1 or 5k ± 2. 4. 6k or 6k ± 1 or 6k ± 2 or 6k ± 3. Proof. From the above corollary any integer a is of the form a = bk ± r where b, k, r ∈ Z and 0 ≤ \|r\| ≤ b . 2 (2.2.5) 3 = 1.5. 1. When b = 3, we get from 2.2.5 a = 3k ± r where 0 ≤ \|r\| ≤ 2 Therefore r = 0, ±1. 4 = 2, ı.e. r = 2. When b = 4 we get from 2.2.5, a = 4k ± r, 0 ≤ \|r\| ≤ 2 0, ±1, ±2. Therefore a = 4k, 4k ± 1, 4k ± 2. 3. Rests treated as exercises. 20 Number Theory and its Applications 2.3 Worked out Exercises Problem 2.3.1. For any two integers a and b with b > 0, there exists unique b integers q1 and r1 such that a = bq1 + cr1 where 0 ≤ r1 < , c = ±1. 2 Solution 2.3.1. By division algorithm we have a = bq + cr, 0 ≤ r < b. b , take q1 = q, c = 1, r1 = r. Therefore a = bq1 + cr1 , 0 ≤ r1 < 2 b 2 , c = ±1. Case I r 2b , therefore 0 < b − r < 2b take q1 = q0 + 1, r1 = b − r and c2 = −1, therefore, a = bq1 + cr1 where 0 ≤ r1 < 2b , c = −1. b Case III r = 2b then q1 = q, c = 1, r1 = r. Therefore a = bq1 +cr1 , r1 = , c = 1 2 and if q1 = q + 1, r1 = b − r and c = −1. Therefore a = b(q + 1) − (b − b r) = bq1 + cr1 , = r, c = −1. In this case q1 and r1 is not unique, so 2 a = bq1 + cr1 , 0 ≤ r1 < 2b , c = ±1. Problem 2.3.2. Show that every square integer is of the form 5k or 5k ± 1 for some k ∈ Z. Solution 2.3.2. Note that every integer is of the form 5p, 5p±1, 5p±2 for some p ∈ Z. Square of these numbers are of the form: (5p)2 = 5 × 5p2 = 5k, where k = 5p2 is a positive integer (5p ± 1)2 = 25p2 ± 10p + 1 = 5(5p2 ± 2p) + 1 = 5k + 1, where k = 5p2 ± 2p + 1 ∈ Z (5p ± 2)2 = 25p2 ± 20p + 4 = 5(5p2 ± 4p + 1) − 1 = 5k − 1, where k = 5p2 ± 4p + 1 ∈ Z. Problem 2.3.3. Show that cube of any integer is of the form 9p, 9p + 1, 9p + 8(or 9p, 9p ± 1). Theory of Divisibility 21 Solution 2.3.3. Here, (3m)3 = 27m3 = 9p, where p = 3m3 ∈ Z (3m + 1)3 = 27m3 + 27m2 + 9m + 1 = 9(3m3 + 3m2 + m) + 1 = 9p + 1, where p = 3m3 + 3m2 + m ∈ Z (3m − 1)3 = 27m3 − 27m2 + 9m − 9 + 8 = 9(3m3 − 3m2 + m − 1) + 8 = 9p + 8, where p = 3m3 − 3m2 + m − 1 ∈ Z Also, (3m − 1)3 = 9(3m3 − 3m2 + m) − 1 = 9p − 1, where p = 3m3 − 3m2 + m ∈ Z. Problem 2.3.4. Prove that the expression a(a2 + 2) is an integer for a ≥ 1. 3 Solution 2.3.4. Applying Division Algorithm, any integer a can be expressed 2 in the form 3q, 3q + 1, 3q + 2. Taking a = 3q we obtain a(a 3+2) = q(9q 2 + 2), an integer. Similarly putting a = 3q+1 and a = 3q+2 we obtain (3q+1)(3q 2 +2q+1) and (3q + 2)(3q 2 + 4q + 2) respectively, both of which are integers. Hence the result is proved. Problem 2.3.5. Show that one of every three consecutive integer is divisible by 3. Solution 2.3.5. Let a, a + 1, a + 2 be any three consecutive integers, then a is of the form 3p, 3p + 1, 3p − 1 where p ∈ Z. If a = 3p, then a is divisible by 3. If a = 3p + 1, then a + 2 = 3p + 3 = 3(p + 1) is divisible by 3. If a = 3p − 1, then a + 1 = 3p + 1 − 1 = 3p is divisible by 3. Problem 2.3.6. Find the minimal remainder of 416 with respect to (i) 37 (ii) 42. Solution 2.3.6. (i) Here a = 416, b = 37. Therefore 416 = 37 × 11 + 9(Why!). Therefore the minimal remainder is R = 9. (ii) Left to the reader. Problem 2.3.7. Show that an+1 − (a − 1)n − a is divisible by (a − 1)2 , a being an integer. 22 Number Theory and its Applications Solution 2.3.7. Since a ∈ Z, we have an+1 − (a − 1)n − a = a(an − 1) − (a − 1)n =a(a−1{a(an−1+· · · + 1 ) −n}) = ( a−1){(an+an−1+· · · +a)−n} = ( a−1)2{(an−1+an−2+· · · + 1 ) + ( an−2+· · · + 1 ) + · · · + 1 }. The given expression is divisible by (a − 1)2 . Problem 2.3.8. If both a and b are odd positive integers then a4 + b4 − 2 is divisible by 8. Solution 2.3.8. Let a = 2n1 + 1 and b = 2n2 + 1 be the odd positive integers, where n1 , n2 are positive integers. Thus we have, (2n1 + 1)4 + (2n2 + 1)4 − 2 = (2n1 )4 + 4 · (2n1 )3 + 6 · (2n1 )2 + 4 · (2n1 ) + 1 + (2n2 )4 + 4 · (2n2 )3 + 6 · (2n2 )2 + 4 · (2n2 ) + 1 − 2 = 16(n41 + n42 ) + 16(n31 + n32 ) + 24(n21 + n22 ) + 8(n1 + n2 ) = 8[2(n41 + n42 ) + 2(n31 + n32 ) + 3(n21 + n22 ) + (n1 + n2 )]. Problem 2.3.9. Show that the product of two integers of the form 4n + 1 is again of this form, while the product of two integers of the form 4n + 3 is of the form 4k + 1. Solution 2.3.9. Product of two integers of the form 4n + 1 gives us (4n + 1)(4m + 1) = 4(4mn + m + n) + 1 = 4k + 1, k ∈ Z. Similarly (4n + 3)(4m + 3) = 4(4mn + 3m + 3n + 2) + 1 = 4k + 1, k ∈ Z. Problem 2.3.10. Show that the square of every odd integer is of the form 8k+1. Solution 2.3.10. Let a be an odd integer. Then n = 2s + 1, s being an integer. Now, a2 = 4s(s + 1) + 1. If s is even, then s = 2m, m being an integer. Hence a2 = 8m(2m + 1) + 1 = 8k + 1, k = 2m + 1 ∈ Z. If s is odd, then s = 2m + 1. It follows, a2 = 8(2m + 1)(m + 1) + 1 = 8k + 1, k = (2m + 1)(m + 1) ∈ Z. Problem 2.3.11. Let m be a positive integer. We deﬁne m if m is even; 2, T (m) = 3m+1 , if m is odd. 2 Theory of Divisibility 23 We,thenformthesequenceobtainedbyiteratingT;m,T(m), T (T(m)), T (T(T (m))), . . . .Forinstance,startingwithm= 7 wehave7,11,17,26,13,20,10,5,8,4, 1,2, "well-knownconjecture,sometimescalledtheCollatz conjecture, BTTFSUTUIBUthesequenceobtainedbyiteratingT alwaysreachestheinteger1no NBUUFSXIJDIpositive integer m begins the sequence. 2k−1 , Show that the sequence obtained by iterating T starting with m = 3 where k is an even positive integer, k > 1, always reaches the integer 1. 2k Solution 2.3.11. If 3m is odd, then so is m. So T (m) = 3m+1 = 22 = 22k−1 . 2 Since T (m) is a power of 2, the exponent will decrease down to 1 with repeated iterations of T . Problem 2.3.12. Show that if a is an integer, then 3 divides a3 − a. Solution 2.3.12. Here a3 − a = a(a− 1)(a+ 1). Applying division Algorithm we have a = 3k, a = 3k +1 or a = 3k +2, k being an integer. If a = 3k and a = 3k + 1, then 3\|a and 3\|(a − 1) respectively. Finally, if a = 3k + 2 i.e. a + 1 = 3(k + 1), then 3\|(a + 1). Combining, it shows 3\|a(a − 1)(a + 1) = a3 − a. 2.4 Greatest Common Divisor If c and d be two arbitrary integers, not simultaneously zero, then the set of common divisors of c and d is a ﬁnite set of integers, always containing the integers +1 and −1(hence, their set of common divisors is non-null). Now every integer divides zero, so that if c = d = 0, then every integer serves as a common divisor of c and d. In this case, the set of common divisors of c and d turns to be inﬁnite. In this article, we are interested on the greatest integer among the common divisors of two integers. Deﬁnition 2.4.1. The greatest common divisor of two integers c and d, that are not both zero, is the greatest integer which divides both c and d. In other words, the above deﬁnition can be formulated as Deﬁnition 2.4.2. If c and d be two arbitrary integers, not simultaneously zero, the greatest common divisor of c and d is the common divisor e satisfying the following: 1. e\|a and e\|b. 2. If f \|a and f \|b then e ≥ f . The greatest common divisor of c and d is written as (c, d) or gcd(c, d). 24 Number Theory and its Applications Example 2.4.1. The common divisors of 20 and 80 are ±1, ±2, ±4, ±5, ±10and± 20. Hence gcd(20, 80) = 20. Similarly, looking at sets of common divisors, we ﬁnd that (12, 18) = 6, (50, 5) = 5, (19, 24) = 1, (0, 56) = 56, (−8, −16) = 8, and (−19, 361) = 19. We can also deﬁne the greatest common divisor of more than two integers. Deﬁnition 2.4.3. Let c1 , c2 , . . . , cn be integers, that are not all zero. The greatest common divisor of these integers is the greatest integer which is a common divisor of all of the integers in the set. The greatest common divisor of c1 , c2 , . . . , cn is denoted by (c1 , c2 , . . . , cn ) or gcd(c1 , c2 , . . . , cn ). Example 2.4.2. We see that (12, 18, 30) = 6 and (10, 15, 25) = 5. The following proposition can be used to ﬁnd the greatest common divisor of a set of more than two integers. Proposition 2.4.1. If c1 , c2 , . . . , cn are integers, not simultaneously zero, then gcd(c1 , c2 , . . . , cn ) = gcd c1 , c2 , . . . , (cn−1 , cn ) . Before proceeding for proof, let us explain the proposition with an example: To ﬁnd the greatest common divisor of the three integers 105, 140, and 350, we see that gcd(105, 140, 350) = gcd 105, (140, 350) = gcd(105, 70) = 35. Proof. In particular, a common divisor of the n integers c1 , c2 , . . . , cn is a divisor of cn−1 and cn and therefore, a divisor of (cn−1 , cn ). Also, any common divisor of the n − 2 integers c1 , c2 , . . . , cn−2 and (cn−1 , cn ), must be a common divisor of all n integers, for if it divides (cn−1 , cn ), it must divide both cn−1 and cn . Since the set of n integers and the set of the ﬁrst n − 2 integers together with the greatest common divisor of the last two integers have exactly the same divisors, their greatest common divisors are equal. Next we are particularly interested in pair of integers sharing no common divisors other than 1. Such pair of integers are said to be relatively prime or coprime. Deﬁnition 2.4.4. The integers c and d, not simultaneously zero, are said to be relatively prime(or coprime) if c and d have greatest common divisor (a, b) = 1. Example 2.4.3. Since, gcd(12, 13) = 1 therefore 12, 13 are relatively prime. We can also deﬁne the relatively prime of more than two integers. Theory of Divisibility 25 Deﬁnition 2.4.5. We say that the integers c1 , c2 , . . . , cn are mutually relatively prime(or coprime) if gcd(c1 , c2 , . . . , cn ) = 1. These integers are called pairwise relatively prime if for each pair of integers ci , cj from the set, gcd(ci , cj ) = 1, i.e., if each pair of integers from the set is relatively prime. If integers are pairwise relatively prime, they must be mutually relatively prime(Verify!). However, the converse fails is shown from the following example: Example 2.4.4. Consider the integers 15, 21, and 35. Since (35, 55, 77) = (35, (55, 77)) = (35, 11) = 1, we see that the three integers are mutually relatively prime. However, they are not pairwise relatively prime, because (35, 55) = 5, (35, 77) = 7 and (55, 77) = 11. Remark 2.4.1. Since the divisors of −a are the same as the divisors of a, it follows that gcd(a, b) = (\|a\|, \|b\|) (where \|a\| denotes the absolute value of a which equals a if a > 0, equals −a if a < 0) and equals 0 if a = 0. Hence we can restrict our attention to greatest common divisors of pairs of positive integers. We will show that the greatest common divisor of the integers c and d, not simultaneously zero, can be written as a sum of multiples of c and d. To phrase this more lucidly, we use the following deﬁnition: Deﬁnition 2.4.6. If c and d are integers, then a linear combination of c and d is a sum of the form mc + nd, where both m and n are integers. The following theorem relates deﬁnition 2.4.6 and greatest common divisors. Theorem 2.4.1. The greatest common divisor of the integers c and d, not simultaneously zero, is the least positive integer that is a linear combination of c and d.(In other words, given integers c and d, not both of which are zero, there exist integers m, n such that gcd(c, d) = mc + nd.) Before proceeding for the proof, let us illustrate the theorem succinctly with an example: Example 2.4.5. Consider the case in which c = 4 and d = 12. Here, the set S becomes S = {4(−2) + 12 · 1, 4(−1) + 12 · 1, 4 · 0 + 152 · 1, . . .} = {4, 8, 12, . . .}. Here 4 is the smallest integer in S, whence 4 = gcd(4, 12). Proof. Let e be the least positive integer such that e = ma + nb holds, m, n being integers.(Using the well-ordering property, there exist such least positive integer, also at least one of two linear combinations 1 · c + 0 · d and (−1) · c + 0 · b, 0 is positive, do exist). where c = 26 Number Theory and its Applications Claim(i) e\|c and e\|d. Claim(ii) e = gcd(c, d). To fulﬁll Claim(i), applying Division Algorithm, we have c = eq +r with 0 ≤ r < e. Now combining e = ma + nb and c = eq + r, we obtain r = (l − qm)c − qnd. This shows that the integer r is a linear combination of c and d. Since 0 ≤ r < e, and e is the least positive linear combination of c and d, we conclude that r = 0, and hence e\|a. In a similar manner, we can show that e\|d. For Claim(ii), all we need to show is that any common divisor f of c and d must divide e. Since e = ma + nb, if f \|c and f \|d, proves f \|e. This completes the proof. Remark 2.4.2. The foregoing argument is just an “existence” proof and does not provide a practical method for ﬁnding the values of m and n. The following theorem illustrates the relation between relatively prime integers and linear combinations(of relatively prime integers). Theorem 2.4.2. Let c and d integers, not simultaneously zero. Then c and d are relatively prime if and only if there exist integers m and n such that 1 = mc+nd. Proof. If c and d are relatively prime then gcd(c, d) = 1. By virtue of Theorem 2.4.1, there exist integers m and n satisfying 1 = mc+nd. In context of converse part, assume that 1 = mc+nd for some choice of m and n, and that e = gcd(c, d). Because e\|c and e\|d, Proposition 2.2.1 yields e\|(mc + nd), or e\|1 implies e = 1(Why!), and the desired conclusion follows. It is true, without adding an extra condition, that a\|c and b\|c together does not imply ab\|c. For instance, 6\|12 and 3\|12, but 6·3 12. Of course, if gcd(6, 3) = 1, then this situation would not arise. This brings us to Corollary the following corollary: Corollary 2.4.1. If c\|e and d\|e, with gcd(c, d) = 1, then cd\|e. Proof. As c\|e and d\|e, there exist integers m and n satisfying e = mc + nd. Now the relation gcd(c, d) = 1 implies 1 = ck + dl for some choice of integers k and l. Multiplying the last equation by e, we obtain e = e · 1 = e(ck + dl) = eck + edl. The appropriate substitutions on the right-hand side allows e = c(ds)k+d(cr)l = cd(sk + rl) implies cd\|e. The following few propositions address some properties of greatest common divisors. Theory of Divisibility 27 Proposition 2.4.2. Let a, b and c be integers with gcd(a, b) = d. Then a b 1. gcd , = 1. d d 2. gcd(a + cb, b) = gcd(a, b). 3. gcd(ma, mb) = md (m > 0). a b Proof. 1. Here a, b are integers with gcd(a, b) = d. Our claim is , have no d d common positive divisors other than 1. Assume that e is a positive integer a b a such that e\| and e\| . Then, there are integers k1 and k2 with = k1 e d d d b = k2 e, satisfying a = dek1 and b = dek2 . Hence de is a common and d a b , = 1. divisor of a and b. Hence e = 1(Why!). Consequently, gcd d d 2. Here a, b and c be integers with gcd(a, b) = d. Its suﬃces to show that the common divisors of a, b are exactly the same as the common divisors of a + cb, b ⇒ gcd(a + cb, b) = gcd(a, b). Let e be a common divisor of a, b. Then e\|(a + cb)(Why!), such that e is a common divisor of a + cb, b. If f is a common divisor of a + cb, b we see that f \| (a + cb) − cb = a(Why!), showing f is a common divisor of a, b. Hence gcd(a + cb, b) = gcd(a, b). 3. Since d = gcd(a, b) then ∃ integers x and y such that d = xa + yb(by Theorem 2.4.1). Then we have, m(xa + yb) = md ⇒ x(ma) + y(mb) = md As m > 0 then from the above equation we can assert that gcd(ma, mb) = m gcd(a, b) = md. Proposition 2.4.3. Prove that gcd(a, c) = 1 if and only if gcd(c − a, c). Proof. Every common divisor d of a and c is also a common divisor of c − a and a. Conversely, every common divisor d of c − a and a is also a common divisor of c − a + a = c and a. Therefore the greatest common divisor of a and c is the same as the greatest common divisor of c − a and a. So in general, gcd(a, c) = gcd(c − a, c) = 1. Proposition 2.4.4. Let a, b and c be integers with gcd(a, b) = 1. Then 1. If gcd(a, c) = 1, then gcd(a, bc) = 1. 28 Number Theory and its Applications 2. If gcd(a, b) = 1, and c\|a, then gcd(b, c) = 1. 3. If c (a + b), then gcd(a, c) = gcd(b, c) = 1. 4. If d\|ac, and d\|bc, then d\|c. Proof. 1. Since gcd(a, b) = 1, and gcd(a, c) = 1, therefore ∃ x, y, u, v ∈ Z such that 1 = ax + by = au + cv. ∴ 1 = (ax + by)(au + cv), = a(axy + byu + axu) + bcyu, = ak1 + bck2 , k1 = axy + byu + axu, k2 = yu. Hence gcd(a, bc) = 1. 2. left to the reader. 3. Since gcd(a, b) = 1, ∃ u, v ∈ Z such that au + bv = 1. Also, c (a + b) ⇒ ∃ m such that cn = a + b ⇒ cn − b = a. ∴ (cn − b)u + bv = 1, cnu − bu + bv = 1, cnu − b(u − v) = 1 ⇒ gcd(c, b) = 1. Similarly, gcd(c, a) = 1. 4. left to the reader. Our next theorem seems simple, but is of fundamental importance. Theorem 2.4.3. Euclid’s Lemma: If a\|bc, with gcd(a, b) = 1, then a\|c. Proof. By virtue of Theorem 2.4.2, writing 1 = am + bn, where m and n are integers. Multiplication of this equation by c produces c = 1 · c = (am + bn)c = acm + bcn. Because a\|ac and a\|bc, it follows that a\|(acm + bcn), which can be recast as a\|c. Remark 2.4.3. The condition gcd(a, b) = 1 is necessary is evident from the following example: 12\|9 · 8, but 12 9 and 12 8. Theorem 2.4.4. Let c, d be integers, not both zero. For a positive integer e, e = gcd(c, d) if and only if 1. e\|c and e\|d. Theory of Divisibility 29 2. Whenever f \|c and f \|d, then f \|e. Proof. Hint: Use Theorem 2.4.1. Simple application of the last theorem leads to the following proposition. Proposition 2.4.5. Let a, b and c be integers with gcd(a, b) = 1, then gcd(ac, b) = gcd(c, b). Proof. Let gcd(c, b) = d. Its suﬃces to show that d\|ab and secondly if k\|ac and k\|b then k\|d. Since d\|c, ∃ n such that dn = c so (dn)a = ca ⇒ d\|ca. Next, ∃ u, v ∈ Z such that d = au + cv. Since k\|b, then ∃ n such that kn = b. Hence d = cu + knv. Since gcd(a, b) = 1 ∃ p, q such that ap + bq = 1 ⇒ apc + bqc = c. ∴ d = (apc + bqc)x + kny, = axpc + bqcx + kny. But, k\|ac ⇒ ∃ r such that kr = ac. ∴ d = krpx + knqcx + kny, = k(rpx + nqcx + ny) ⇒ k\|d. Hence using Theorem 2.4.4 we obtain the desired result. Remark 2.4.4. The Theorem 2.4.4 sometimes serves as a deﬁnition of gcd(c, d). The advantage of using it as a deﬁnition is that order relationship is not involved. Thus, it may be used in algebraic systems having no order relation. Euclid’s Algorithm While ﬁnding the gcd of two integers (not both 0), we can of course list all the common divisors and pick the greatest one amongst those. However, if a and b are very large integers, the process is very much time consuming. However, there is a far more eﬃcient way of obtaining the gcd. That is known as the Euclid’s algorithm. This method essentially follows from the division algorithm for integers. To prove the Euclidean algorithm, the following lemma will be helpful. Lemma 2.4.1. If a = qb + r then the gcd(a, b) = gcd(b, r). Proof. Let d = gcd(a, b) and d1 = gcd(b, r). Then, d\|a, d\|b implies d\|(a − qb) ı.e, d\|r. Thus d is a common divisor of b and r, hence d\|d1 . Similarly, d1 \|b, d1 \|r implies d1 \|(bq + r) ı.e., d1 divides both a and b. Then, d1 \|d. Thus, d = d1 , as both d and d1 are positive by our deﬁnition of gcd. 30 Number Theory and its Applications Theorem 2.4.5. Euclid’s Algorithm: Let a and b (a > b) be any two integers . If r1 is the remainder when a is divided by b, r2 is the remainder when b is divided by r1 , r3 is the remainder when r1 is divided by r2 and so on. Thus rn+1 = 0, then the last non zero remainder rn is the gcd(a, b). Proof. Euclid’s algorithm is an eﬃcient way of computing the gcd of two integers by repeated application of the above lemma. At each step the size of the integers concerned gets reduced. Suppose we want to ﬁnd the gcd of two integers a and b, neither of them being 0. As gcd(a, b) = gcd(a, −b) = gcd(−a, b) = gcd(−a, −b), we may assume a > b > 0. By performing division algorithm repeatedly, we obtain a = bq1 + r1 , 0 ≤ r1 ≤ b. b = r1 q2 + r2 , 0 ≤ r2 ≤ r1 . r1 = r2 q2 + r3 , 0 ≤ r3 ≤ r3 . .. .. .=. rn−2 = rn−1 qn + rn , 0 ≤ rn ≤ rn−1 . rn−1 = rn qn+1 + rn+1 , 0 ≤ rn+1 ≤ rn . As we have a decreasing sequence of non-negative integers b > r1 > r2 > . . . > rn > rn+1 we must have rn+1 = 0 for some n. Then, by applying the previous lemma repeatedly, we ﬁnd that gcd(a, b) = gcd(r1 , b) = gcd(r2 , r1 ) = . . . = gcd(rn−1 , rn−2 ) = gcd(rn , rn−1 ) = rn . Thus, the last non-zero remainder rn in the above process gives us the gcd(a, b). Theorem 2.4.6. If k > 0, then gcd(ka, kb) = k gcd(a, b). Let us illustrate the statement of the above theorem with an example: gcd(12, 30) = gcd(3 · 4, 3 · 10) = 3 gcd(4, 10) = 3 gcd(2 · 2, 2 · 5) = 3 · 2 gcd(2, 5) = 6. Proof. If each of the equations appearing in the Euclidean Algorithm for a and b is multiplied by k, we obtain ak = (bk)q1 + r1 k, 0 ≤ r1 k ≤ bk. bk = (r1 k)q2 + r2 k, 0 ≤ r2 k ≤ r1 k. r1 k = (r2 k)q2 + r3 k, 0 ≤ r3 k ≤ r3 k. .. .. .=. rn−2 k = (rn−1 k)qn + rn k, rn−1 k = (rn k)qn+1 + 0. 0 ≤ rn k ≤ rn−1 k. Theory of Divisibility 31 But here the Euclidean Algorithm applied to the integers ak and bk, so that their greatest common divisor is the last nonzero remainder rn k; that is, gcd(ka, kb) = rn k = k gcd(a, b). Based on the above theorem, let us state and prove the following corollary: Corollary 2.4.2. Its suﬃces to consider the case k < 0. Then −k = \|k\| > 0 and gcd(ka, kb) = gcd(−ka, −kb) = gcd(\|k\|a, \|k\|b) = \|k\| gcd(a, b). 2.5 Least Common Multiple There is a concept parallel to that of the greatest common divisor of two integers, known as their least common multiple. Prime factorizations can also be used to ﬁnd the smallest integer that is a multiple of two positive integers(treated in later chapters). The problem of ﬁnding this integer arises when fractions are added. Deﬁnition 2.5.1. The least common multiple of two positive integers a and b is the smallest positive integer that is divisible by a and b, denoted by lcm(a, b) or [a, b]. The above deﬁnition can also be formulated as follows: Deﬁnition 2.5.2. The least common multiple of two nonzero integers a and b is the positive integer l satisfying the following: 1. a\|l and b\|l. 2. If a\|c and b\|c, with c > 0, then l ≤ c. Example 2.5.1. We have the following least common multiple: lcm(16, 20) = 80, lcm(24, 36) = 72, lcm(4, 20) = 20, and lcm(5, 13) = 65. Remark 2.5.1. Given nonzero integers a and b, lcm(a, b) always exists and lcm(a, b) < \|ab\|(Verify!). Proposition 2.5.1. For nonzero integers a and b, the following statements are equivalent(TFAE): 1. gcd(a, b) = \|a\|. 32 2. a b. Number Theory and its Applications 3. lcm(a, b) = \|b\|. Proof. (1)⇒(2): Let (1) holds. Then ∃ n ∈ Z such that b = a n. Now a > 0 ⇒ b = an ⇒ a\|b. Again, a < 0 ⇒ \|a\| = −1 ⇒ b = (−a)n ⇒ b = a(−n) ⇒ a\|b. a\|b. Hence (2) holds. (2)⇒(3): Let (2) holds. Then a \|b\| and clearly b \|b\|. Let c be another common multiple. Then a c and b c with c > 0. Now b c implies ∃ n ∈ Z such that c = bn and n ≥ 1. Thus, \|c\| = \|b\|\|n\| ≥ b which further gives \|c\| ≥ \|b\| and by deﬁnition \|b\| = lcm(a, b). (3)⇒(1): Let (3) holds. Therefore a \|b\| ⇒ \|a\| \|b\|. Let c be another common multiple. Then ∃ n ∈ Z such that a = cn ⇒ \|a\| = \|c\|\|n\|. But \|n\| ≥ 1 ⇒ \|c\|\|n\| ≥ \|c\| ⇒ \|a\| ≥ \|c\|.Therefore gcd(a, b) = \|a\|. The following theorem ﬁlled the gap between greatest common divisor and least common multiple. ab , where [a, b] (a, b) and (a, b) are the least common multiple and greatest common divisor of a and b, respectively. Theorem 2.5.1. If a and b are positive integers, then [a, b] = Proof. Let us begin with taking c = (a, b) and write a = cr, b = cs for integers r ab and s. If l = , then l = as = rb, making l a (positive) common multiple of a c and b. Now let d be any positive integer that is a common multiple of a and b, implies d = au = bv. As we know, there exist integers k and l such that c = ak + bl. As a result of which, d dc d(ak + bl) d d = = = k + l = vk + ul ⇒ l\|c ⇒ l ≤ c. l ab ab b a ab Hence l = lcm(a, b) and [a, b] = . (a, b) Remark 2.5.2. The alternate proof of the above theorem can be done using the prime factorizations of integers a and b (for further details refer to chapter Prime Numbers). Corollary 2.5.1. For any choice of positive integers a and b, [a, b] = ab if and only if (a, b) = 1. Proof. Obvious. Theory of Divisibility 33 We conclude this section with a simple but interesting proposition. Proposition 2.5.2. For a and b be two non zero integers. Then 1. gcd(a, b) = lcm(a, b) if and only if a = ±b. 2. If k > 0, then lcm(ka, kb) = k lcm(a, b). 3. If m is any common multiple of a and b, then lcm(a, b) m. Proof. 1. Let us consider gcd(a, b) = lcm(a, b) = d. Now by Theorem 2.5.1 we get d = ab. Since d a then ∃x ∈ Z such that dx = a. This implies d2 = dxb ⇒ d = xb ⇒ b d. Thus we have d b and d b which together implies d = ±b(by Proposition 2.2.1). By similar arguments we also have d = ±a. Therefore \|d\| = \|a\| = \|b\| implying a = ±b. Conversely let d = ±a holds. Then again by Proposition 2.2.1 we can assert that a b and b a. This claims that gcd(a, b) = lcm(a, b). 2. To prove this we are to start with gcd(ka, kb) · lcm(ka, kb) = k 2 \|ab\|. Then, k gcd(a, b) · lcm(ka, kb) = k 2 \|ab\| [ by Proposition 2.4.2] ⇒ gcd(a, b) · lcm(ka, kb) = k\|ab\| ⇒ gcd(a, b) · lcm(ka, kb) = k gcd(a, b) · lcm(a, b) ⇒ lcm(ka, kb) = k lcm(a, b). 3. Let us consider l = lcm(a, b) and by division algorithm ∃ integers q and r such that m = lq + r, 0 ≤ r < l. If r = 0 then obviously l m. If 0 < r < l then we can write r = m − lq. Since m and l are multiples of a and b then ∃ integers x, y, u, v such that r = ax − ayq = a(x − yq) and also r = bu − bvq = b(u − vq). This shows that r is a multiple of a, b and this contradicts the fact l = lcm(a, b). So r < l is not possible. This proves our assertion. 2.6 Worked out Exercises Problem 2.6.1. If a, b, c are integers, then gcd(a, bc) = 1 if and only if gcd(a, b) = gcd(a, c) = 1. 34 Number Theory and its Applications Solution 2.6.1. Let gcd(a, b) = gcd(a, c) = 1 holds. Then there exists integers m1 , n1 , m2 and n2 satisfying am1 + bn1 = 1 = am2 + cn2 . Therefore am1 cn2 + bn1 cn2 = cn2 = 1 − am2 , ⇒ a(m2 + cm1 n2 ) + bc(n1 n2 ) = 1. As m2 + cm1 n2 and n1 n2 are integers, therefore gcd(a, bc) = 1. Conversely, let gcd(a, bc) = 1 holds. We are to show gcd(a, b) = gcd(a, c) = 1. Let gcd(a, b) = 1. Then gcd(a, b) = d implies there exists m, n such that am + bn = d ⇒ acm + bcn = cd ⇒ a(cm) + b(cn) = cd. Therefore gcd(a, bc) = cd(= 1), a contradiction. Thus both a, b and a, c are coprime. Problem 2.6.2. Prove or disprove: If a\|(b + c), then either a\|b or a\|c. Solution 2.6.2. Hint: Take a = 3, b = 2, c = 7. Problem 2.6.3. If a\|bc, show that a\| gcd(a, b) gcd(a, c). Solution 2.6.3. Let gcd(a, b) = d1 and gcd(a, c) = d2 . Then ∃ x, y, u, v ∈ Z such that d1 = ax + by, & d2 = au + cv. Also, ∃ n ∈ Z satisfying an = bc. Now, d1 d2 = (ax + by)(au + cv), = a2 xu + acxv + abuy + bcyv, = a(axu + cxv + buy) + anyv, = a(axu + cxv + buy + nyv). ∴ a\| gcd(a, b) gcd(a, c). Problem 2.6.4. Prove that if d\|n, then (2d − 1)\|(2n − 1). Theory of Divisibility 35 Solution 2.6.4. We know that an − 1 = (a − 1)(an−1 + an−2 + . . . + a + 1), ∴ 2n − 1 = (2 − 1)(2n−1 + 2n−2 + . . . + 2 + 1), ∴ 2d − 1 = (2 − 1)(2d−1 + 2d−2 + . . . + 2 + 1). Since d\|n, ∃ x ∈ Z such that dx = n. Therefore 2n − 1 = 2dx − 1 = (2d )x − 1, ∴ (2d − 1) (2n − 1). = (2d − 1)(2d(x−1) + 2d(x−2) + . . . + 2d + 1). Problem 2.6.5. Prove that the product of any three consecutive integers is divisible by 6. Solution 2.6.5. Here we need to show 6\|a(a+1)(a+2), for any arbitrary a ∈ Z. Let S = a(a + 1)(a + 2). Here 6 = 3 · 2 and gcd(2, 3) = 1. If a is even, then 2\|a ⇒ 2\|S. And if odd, then 2 (a + 1) ⇒ 2\|S. Let a = 3q + r, q, r ∈ Z. Now r = 0, 1, 2. For all the values of r, 3\|S(verify!). Hence 2\|S, 3\|S together implies 6\|S. Problem 2.6.6. If a is an odd integer, then 24\|a(a2 − 1). Solution 2.6.6. Let us ﬁrst prove, a is of the form 8k + 1. Let a = 4q + r. Therefore r = 0 or 3(Why!). Therefore a2 = 16q 2 + 8q + 1 = 8k + 1, for r = 0 a2 = 16q 2 + 24q + 9 = 8k + 1, for r = 3. So a(a2 − 1) = a(8k), for some k. Hence 8 a(a2 − 1). Therefore 6 a(a2 − 1) ⇒ 3\|a(a2 − 1). As gcd(3, 8) = 1, hence 24\|a(a2 − 1)(Why!). Problem 2.6.7. If a is an integer not divisible by 2 or 3, then 24\|(a2 + 23). Solution 2.6.7. Let a = 12q + r, q, r ∈ Z with 0 ≤ r < 12. But here, r = 1, 5, 7, 11(Why!). Now a2 + 23 = (12q + r)2 + 23, = 144q 2 + 24qr + r2 + 23, = 24(6q 2 + qr) + r2 + 23. 36 Number Theory and its Applications Now, r = 1 gives r2 + 23 = 24 r = 5 gives r2 + 23 = 48 = 24 · 2 r = 7 gives r2 + 23 = 72 = 24 · 3 r = 11 gives r2 + 23 = 144 = 24 · 6 ∴ a2 + 23 = 24(6q 2 + qr) + 24 · k, for some k. Hence 24\|(a2 + 23). Problem 2.6.8. For n ≥ 1, and positive integers a, b, prove that gcd(an , bn ) = 1 where gcd(a, b) = 1. Solution 2.6.8. For n = 1, the statement is obvious. Let us assume the statement be true for n(> 1) = k i.e. gcd(ak , bk ) = 1. Now gcd(ak , bk+1 ) = gcd(ak , bk ) = 1[refer to the properties of GCD]. Since gcd(a, b) = gcd(b, a) = 1, then gcd(ak , bk+1 ) = 1 = gcd(ak+1 , bk+1 ). Problem 2.6.9. For n ≥ 1, and positive integers a, b, prove that the relation an \|bn implies a\|b. Solution 2.6.9. The relation is obvious for n = 1. If possible, let us assume the relation is true for n = k. Then an \|bn implies a\|b, which further implies ∃ x, y such that bk = xak & b = ay. b k bk+1 b = . ∴ xak+1 = abk = y y ∴ xyak+1 = bk+1 ⇒ ak+1 \|bk+1 . Problem 2.6.10. Prove that if gcd(a, b) = 1, then gcd(a + b, ab) = 1. Solution 2.6.10. Let c be the common divisor of a+b and ab. Then gcd(a, c) = gcd(b, c) = 1. Since c ab and gcd(c, a) = 1, then by Euclid’s Lemma we have c b. By similar reasoning, c a. As c ≤ gcd(a, b) = 1 ⇒ c = 1 ⇒ gcd(a + b, ab) = 1. Problem 2.6.11. Prove that the greatest common divisor of two positive integers divides their least common multiple. Solution 2.6.11. Let a, b > 0. We are to prove gcd(a, b) lcm(a, b). We know that gcd(a, b)lcm(a, b) = ab. Let d = gcd(a, b). Then ∃, m, n such that a = dn, b = dm. d · lcm(a, b) = (dn)(dm). ∴ lcm(a, b) = d(nm) ⇒ d lcm(a, b) ⇒ gcd(a, b) lcm(a, b). Theory of Divisibility 37 Problem 2.6.12. If a and b are prime to each other then prove that gcd(a + b, a2 + b2 ) = 1 or 2. Solution 2.6.12. Let gcd(a + b, a2 + b2 ) = d. Then d (a2 + b2 ) ⇐⇒ d (a + b)(a − b) + 2b2 . Since d (a + b), ∃ x such that dx = a + b. Let m ∈ Z be such that dm = (a+b)(a−b)+2b2 ⇒ dm = dx(a−b)+2b2 ⇒ d[m−x(a−b)] = 2b2 ⇒ d 2b2 . Now combining the facts d (a+b) and gcd(a, b) = 1, we ﬁnd gcd(b, d) = 1(Why!). Thus we get d b, which implies d\|2. Therefore d ≤ 2 implies d = 1 or 2. Problem 2.6.13. Let a, b, c be integers, no two of which are zero, and d = gcd(a, b, c). Show that d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b). Solution 2.6.13. Firstly, we will show d = gcd(gcd(a, b), c). Let f = gcd(a, b) and g = gcd(f, c). Now g\|f ⇒ g\|a, g\|b. Here g\|c ⇒ g ≤ d. Our next task is to show d\|f . Here for some x, y ∈ Z, f = ax + by[refer to Theorem 2.4.1]. Now a = du, b = dv for some u, v ∈ Z. Hence f = dux + dvy ⇒ d\|f . Now d\|c ⇒ d\|g ⇒ d ≤ g. Hence combining, d = g holds i.e. d = gcd(gcd(a, b), c). Proceeding as above, we can show that d = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b). Thus d = gcd(gcd(a, b), c) = gcd(a, gcd(b, c)) = gcd(gcd(a, c), b). 2.7 Linear Diophantine Equations Before delving deep into the topic, let us start with the following problem: A person wishes to buy ice cream bar for a get-together at home. After going to the ice cream parlour he came across with some ﬂavours: one is chocolate bar costing Rs.126 and another is strawberry bar costing Rs.99. He decided to buy both combinations with a budget of Rs.2000. Now the problem is; whether there exist any such combinations of these two ﬂavours? To answer this, let k denote the number of chocolate bars and l denote the number of strawberry bars, the person can purchase. Then we must have 126k + 99l = 2000, where both k and l are nonnegative integers. Now the need for Diophantine equation get along to ﬁnd the solutions of a particular equation, which follow from the set of integers. Diophantine equations get their name from the ancient Greek mathematician Diophantus, who wrote extensively on such equations. The type of diophantine equation ak + bl = c, where a, b and c are integers is called a linear diophantine equations in two variables. We now develop the theory for solving such equations. The following 38 Number Theory and its Applications theorem illustrates that when such an equation has solutions, and when there are solutions, explicitly describes them. Theorem 2.7.1. Let a, b be positive integers with d = gcd(a, b). If d c, the equation ax + by = c has no solutions(in integers). There are inﬁnitely many solutions(integers) if d\|c. Moveover in particular, if x = x0 , y = y0 is a solution of the equation, then all solutions are given by b a x = x0 + n, y = y0 − n, n being an integer. d d Before proceeding for the proof, ﬁrst we demonstrate the above theorem for ﬁnding all the integral solutions of the two diophantine equations described at the beginning of this section. We ﬁrst consider the equation 126x + 99y = 2000. The greatest common divisor of 126 and 99 is gcd(99, 126) = 9. Since 9 2000, we can say no integral solutions exist. Hence no combination of 126 and 99 rupees he can purchase. Proof. Assume that x and y are integers satisfying ax+by = c. Together d\|a and d\|b implies d\|c(Why!). Hence if d c there does not exists any integral solutions. So we assume that d\|c. Then from theorem (2.4.1), for some integers s, t d = as + bt. (2.7.1) Since, d\|c there exist some integer e such that de = c holds. Multiplying (2.7.1), we obtain c = a(se) + b(te). Hence one particular solution of the equation is given by x = x0 = se, y = y0 = te. b Now, to prove the remaining part of the theorem suppose x = x0 + n, y = d a y0 − n, n being an integer. Since, d b a ax + bl = a x0 + n + b y0 − n = ax0 + bl0 = c, d d we see that (x, y) is a solution. Next our claim is to show every solution of the equation ax + bl = c must be of the form described in the theorem. since, ax0 + bl0 = c, on subtraction we obtain a(x − x0 ) + b(y − y0 ) = 0 ⇒ a(x − x0 ) = b(−y + y0 ). (2.7.2) Theory of Divisibility 39 Dividing both sides by d, we see that a b (x − x0 ) = (−y + y0 ). d d a b = 1. Also, using Euclid’s By virtue of Proposition 2.4.2, we know gcd , d d a a Lemma it follows \|(y0 − y). Hence there exists an integer n with n = y0 − y d d a means y = y0 − n. Now putting this value of y into the (2.7.2), we ﬁnd d a b n implies x = x0 + n. a(x − x0 ) = b d d Example 2.7.1. A man wishes to purchase Rs 510 of travelers checks. The checks are available only in denominations of Rs 20 and Rs 50. How many of each denomination should he buy? Answer 2.7.1. Consider the equation 20k + 50l = 510. The greatest common divisor of 20 and 50 is (20, 50) = 10, and since 10\|510, there are inﬁnitely many integral solutions. Using the Euclidean algorithm, we ﬁnd that 20(−2)+50 = 10. Multiplying both sides by 51, we obtain 20(−102) + 50(51) = 510. Hence a particular solution is given by k0 = −102 and l0 = 51. Theorem 2.7.1 tells us that all integral solutions are of the form k = −102 + 5n and l = 51 − 2n. Since we want both k and l to be nonnegative, we must have −102 + 5n > 0 51 102 and n < . Since n is an integer, it follows and 51 − 2n > 0; thus, n > 5 2 that n = 21, 22, 23, 24, 25. Hence we have the following 5 solutions: (k, l) = (3, 9), (8, 7), (13, 5), (18, 3), (23, 1). 2.8 Worked out Exercises Problem 2.8.1. Examine the nature of the following Diophantine equations: 1. 14x + 35y = 93. 2. 33x + 14y = 115. 1. Here gcd(14, 35) = 7 and 7 93, hence not solvable. 2. Here gcd(33, 14) = 1 and 1 115, hence solvable. Solution 2.8.1. Problem 2.8.2. Determine all solutions, in positive integers, of the following Diophantine equations: 1. 158x − 57y = 7. 40 Number Theory and its Applications 2. 54x + 21y = 906. Solution 2.8.2. 1. To ﬁnd the solution of this equation we need to ﬁnd the gcd of 158, 57. Now applying Euclid’s Algorithm, we obtain 158 = 3 · 57 − 13 Again 1 = 3 − 2 = 3 − (5 − 3) 57 = 4 · 13 + 5 = 2 · 3 − 5 = 2(13 − 2 · 5) − 5 13 = 2 · 5 + 3 = 2 · 13 − 5 · 5 5=3·1+2 = 2 · 13 − 5(57 − 4 · 13) 3=2·1+1 = 22 · 13 − 5 · 57 = 22(3 · 57 − 158) − 5 · 57 = 61(57) + (−22) · 158. Thus, gcd(158, 57) = 1. ∴ 7 = (61 · 7)57 + (−22 · 7)158. Since gcd(158, 57)\|7 = 7 therefore, an integral solution do exist for the given equation. Hence (x0 , y0 ) = (−154, −427) is an integral solution. Hence all integral solutions of the given equation is of the form, −57 x = −154 + n = −154 − 57n > 0 ⇒ n < −2.7 ⇒ n ≤ −3 1 −158 y = −1510 + n = −1510 − 158n > 0 ⇒ n < −2.7 ⇒ n ≤ −3. 1 2. To ﬁnd the solution of this equation we need to ﬁnd the gcd of 54, 21. Now applying Euclid’s Algorithm, we obtain, 54 = 2 · 21 + 12 Again, 3 = 12 − 9 = 12 − (21 − 12) 21 = 12 · 1 + 9 = 2 · 12 − 21 12 = 9 · 1 + 3 = 2(54 − 2 · 21) − 21 9=3·3+0 Thus, gcd(54, 21) = 3 & 3 906. = 2 · 54 + (−5) · 21. ∴ 906 = (302 · 2)54 + (302 · (−5))21. Since gcd(54, 21)\|906 = 302 therefore, an integral solution do exist for the given equation. Hence (x0 , y0 ) = (604, −1510) is an integral solution. Hence all integral solutions of the given equation is of the form, 21 x = 604 + n = 604 + 7n > 0 ⇒ n > −86.3 3 −54 y = −1510 + n = −1510 − 18n > 0 ⇒ n < −83.9. 3 Theory of Divisibility 41 Thus, n = −84, −85, −86, which gives (x, y) = (16, 2), (9, 20), (2, 38). Problem 2.8.3. Determine all solutions of the Diophantine equation 24x + 138y = 18. Solution 2.8.3. First we need to calculate the gcd of 24 and 138 . Here, 138 = 5 · 24 + 18 Again, 6 = 24 − 18 = 24 − (138 − 5 · 24) 24 = 18 + 9 18 = 3 · 6 + 0 Thus, gcd(24, 138) = 6 & 6 18. = 6 · 24 − 138 ∴ 18 = (18)54 + (−3)138. So the integral solution is x0 = 18, y0 = −3. Thus the solution of this equation is, 138 n = 18 + 23n x = 18+ 6 24 n = −3 − 4n [n ∈ Z]. y = −3− 6 Problem 2.8.4. A farmer purchased 100 heads of livestock for a total cost of Rs.4000. Prices were as follow: sheep, Rs.120 each; hen, Rs.25 each; duck, Rs.50 each. If the farmer obtained at least one animal of each type how many had he bought? Solution 2.8.4. Let us consider the variables x, y and z for sheep, hen and duck respectively. Then from given hypothesis we have, x + y + z = 100 and 120x + 25y + 50z = 4000, where x, y, z ≥ 1. Then 24x + 5y + 10z = 800 and 24x + 10z + 5(100 − x − z) = 800 holds. Combining last two equations yield 19x + 5z = 300. Hence the solutions are x = 0, z = 60. Therefore x = 5k, z = 60 − 19k, y = 40 + 14k, k ∈ Z. Consequently, 5k ≥ 1 ⇒ k ≥ 1 60 − 19k ≥ 1 ⇒ k ≤ 3 40 + 14k ≥ 1 ⇒ k ≥ −2. Considering last three inequalities, we get 1 ≤ k ≤ 3 ⇒ k = 1, 2, 3. Therefore the possibilities are 5 sheep, 54 hens and 41 ducks or 10 sheep, 68 hens and 22 ducks or 15 sheep, 82 hens and 3 ducks. 42 Number Theory and its Applications 2.9 Exercises: 1. Let a, b and c be integers with gcd(a, b) = 1. Then gcd(a2 , b2 ) = 1. 2. Verify that 3a2 − 1 is never a perfect square. 3. For n ≥ 1, establish that the integer n(7n2 + 5) is of the form 6k. 4. For an odd integer n, show that n4 + 4n2 + 11 is of the form 16k. 5. Verify that if an integer is simultaneously a square and a cube (as is the case with 64 = 82 = 43 ), then it must be either of the form 7k or 7k + 1. 6. If a bc, show that a gcd(a, b) gcd(a, c). 7. Verify the followings: (a)the product of any four consecutive integers is divisible by 24; (b)the product of any ﬁve consecutive integers is divisible by 120. 8. Prove that the expression (3n)! (3!)n is an integer for all n ≥ 0. 9. Establish each of the statements below: (a)If a and b are odd integers, then 8 (a2 − b2 ). 2 (b)If a is an arbitrary integer, then 6 a(a + 11). 10. Assuming that gcd(a, b) = 1, prove that gcd(2a + b, a + 2b) = 1 or 3. 11. Prove that if gcd(a, b) = 1, then gcd(a + b, ab) = 1. 12. Find integers x, y, z satisfying gcd(198, 288, 512) = 198x + 288y + 512z. 13. Use the Euclidean Algorithm to obtain integers x and y satisfying gcd(1769, 2378) = 1769x + 2378y. 14. Examine the nature of the Diophantine equation 14x + 35y = 93. 15. Determine all solutions in the positive integers of 158x − 57y = 7. 16. Determine all solutions in the integers of 221x + 35y = 11. 17. Mr.Sen had gone to a medical shop to buy two medicines: medicine A and medicine B. By mistake, the chemist had given him the number of medicine A in place of medicine B and vice versa.Unaware of the fact, Mr.Sen received an extra amount Rs.68 from the shop keeper. Considering the price of each medicine A and medicine B to be Rs.10 and Rs.15 respectively, ﬁnd the least number of medicine A, Mr. Sen wanted to purchase. Theory of Divisibility 43 18. One hundred packets of dry food are distributed among 100 persons in such a way that every man, woman and child receives 3 packets, 2 packets, and half a packet respectively. Find the total number of persons over there? 3 Prime Numbers “God may not play dice with the universe, but something strange is going on with the prime numbers.” – Paul Erados, 3.1 Introduction A prime number is an integer or a whole number that has only two factors 1 and itself. In other words, a prime number can be divided only by 1 and itself. Also primes are greater than 1. For example, 3 is prime as it fails to be divided evenly by any number except for 1 and 3. However, 6 is not because it can be evenly divided by 2 and 3. The largest known prime number is 282,589,933 − 1, a number which has 24, 862, 048 digits when written in base 10. It was discovered by Patrick Laroche of the great internet Mersenne Prime search. Euclid recorded a proof that there does not exist any largest prime number and many mathematicians continue to search for large prime numbers. In 1978, few researchers used prime numbers to scramble and unscramble coded messages. This early form of encryption smoothen the way for Internet security, putting prime numbers at the heart of electronic commerce. Publickey cryptography, or RSA encryption, has simpliﬁed secure transactions of all times. The security of this type of cryptography depends on the diﬃculty of factoring large composite numbers, which is the product of two large prime numbers. Also, in modern banking security systems depend on the fact that large composite numbers cannot be factored in a short amount of time. Two 45 46 Number Theory and its Applications primes are considered secure if they are 2, 048 bits long, because the product of these two primes would be about 1, 234 decimal digits. Prime numbers have shown its existence in nature. Cicadas insect spend most of their time hiding, only reappearing to mate every 13 or 17 years. Why this particular number? Scientists invented that cicadas reproduce in cycles that minimize possible interactions with predators. Any predator reproductive cycle that divides the cicada’s cycle evenly means that the predator will hatch out the same time as the cicada at some point. For instance , if the cicada evolved towards a 12-year reproductive cycle, predators who reproduce at the 2, 3, 4 and 6 year intervals would ﬁnd themselves with plenty of cicadas to eat. By using a reproductive cycle with a prime number of years, cicadas would be able to minimize contact with predators. Simulation models of 1, 000 years of cicada evolution prove that there is a major advantage for reproductive cycle times based on primes. 3.2 Primes & Fundamental Theorem of Arithmetic Positive divisors of an integer have a great importance in the study of number theory. The integer 1 has only one positive divisor which is 1 itself. Any other integers has more than one divisor. At Least two divisors of them are 1 and the integer itself. There are integers which have divisors other than 1 and itself. The numbers which have only two divisors 1 and itself are called prime numbers. Deﬁnition 3.2.1. An integer p > 1 is said to be a prime number or prime if its only divisors are 1 and p itself. An integer which is not prime is known to be a composite number, having more than two(what are those?) divisors. Among the ﬁrst ten positive integers 2, 3, 5, 7 are prime numbers whereas 4, 6, 8, 9, 10 are examples of composite numbers. Here 1 is a special type of integer which is neither prime nor composite. Here the study of prime numbers starts with the study of prime divisors. Here 5 is prime where 5 3 but 5\|5 itself together implies 5\|15, leads us to the following theorem: Theorem 3.2.1. An integer p > 1 is prime if and only if p\|ab implies p\|a or p\|b. Proof. Let p be a prime number such that for any two integers a and b, p\|ab holds. If p\|a, then we are done. Let p a then the only divisors of p are 1 and p Prime Numbers 47 itself. As p is prime we have gcd(p, a) = 1 implies there exists integers r, t such that 1 = rp + at. Then b = brp + t(ab). Now p\|ab and p\|prb imply p\|b. Conversely, let p satisfy the condition and q, r be any integers such that p = qr where q < p. Thus p\|qr and by the condition we can say either p\|q or p\|r. But q\|p shows p\|r only. Therefore r = pt for some integer t. Hence p = qr = qpt implies qt = 1 implies q = 1. So 1 and p are only divisors of p. This shows p is prime. Let us now generalize the above theorem for more than two terms as follows: Theorem 3.2.2. If p is prime and p\|a1 a2 a3 · · · an , then p\|ai for some i = 1, 2, 3, . . . , n. Proof. We will prove this by mathematical induction. The statement is true for n = 1. With reference to theorem (3.2.1) the statement is true for n = 2. Let us assume the statement is true for n = k. Let n = k + 1 holds. Then p\|a1 a2 a3 · · · ak ak+1 . Also choose a1 a2 a3 · · · ak = b where b is an integer, thus p\|bak+1 . Now if p\|ak+1 we are done. If p ak+1 , then from n = 2 we have p\|b implies p\|a1 a2 a3 · · · ak which further implies p\|ai for some i by the induction hypothesis. Thus p\|ai for i = 1, 2, . . . , k+1. So the statement is true for n = k+1. Thus by principle of mathematical induction the theorem is proved. Corollary 3.2.1. If p, q1 , q2 , q3 , . . . , qn are all primes and p\|q1 q2 q3 · · · qn then p = qi for some i = 1, 2, . . . , n. Proof. By virtue of above theorem, we know that if p\|qi for some i = 1, 2, . . . , n. But qi being prime so qi is not divisible by any integer other than 1 and itself. Since, p > 1 then we have p = qi for some i = 1, 2, 3 . . . , n. Let us now consider few integers 35, 25, 10 and we see that 7\|35, 5\|25 and 2\|10. So the observation is that every integer has a prime factor. We now prove this result for any integer n ≥ 2. Theorem 3.2.3. Every integer n ≥ 2 has a prime factor. Proof. We prove the statement by mathematical induction method. Taking n = 2, the result is obvious as 2 itself is prime. Let us assume that each of the integers 2, 3, . . . , n − 1 has a prime factor. Now considering n > 2 we can say that the result is true if n is prime. If n is composite then n = rs for some integer r, s with 1 < r, s < n. Then by induction hypothesis r has a prime factor which is also a prime factor of n. So the theorem is proved. 48 Number Theory and its Applications The set of all positive integers is countably inﬁnite and the set of prime numbers is a subset of the set. So two possibilities to occur. One, the cardinality of the set is ﬁnite and the other which is countably inﬁnite. But the set of prime numbers that are countably inﬁnite is given in a theorem of Euclid (300 B.C.) and till the 21st century the proof is considered as an elegant proof of Mathematics. Theorem 3.2.4. Prime number set is countably inﬁnite. Proof. Let the number of primes be ﬁnite and we write them as p1 = 2, p2 = 3 · · · pn . Now let us consider a composite number m = p1 p2 · · · pn + 1 and m > 1. As m is composite it has a prime factor p(say). This p obviously one of p1 , p2 , · · · pn . Now p\|p1 p2 · · · pn + 1, p\|p1 p2 · · · pn together imply p\|1 [Applying x = −1, y = 1, b = p1 p2 · · · pn and c = p1 p2 · · · pn + 1 on a\|b, a\|c ⇒ a\|(bx + cy)]. This leads to a contradiction (Why!). So our assumption is wrong and the theorem is proved. All the above results lead us to the fact that any integer can be factorized if it is composite. The factorized integers can be prime or composite such as 20 = 4×5 where 4 is composite whereas 5 is prime. But the most interesting fact is that 20 = 22 × 5 where both 2 and 5 are prime. This factorization is known to be prime factorization. The following Fundamental Theorem of Arithmetic or the unique factorization theorem enlighten us about the fact: Theorem 3.2.5. Every positive integer n ≥ 2 can be expressed uniquely as product of primes, n = p1 p2 p3 · · · pr , where each pi is distinct for 1 ≤ i ≤ r. Proof. If n is prime then we are done. If n is composite then there exists an integer d such that d\|n with 1 < d < n. By well ordering principle, let p1 be the smallest of them. Here p1 must be prime otherwise t be any divisor of p1 such that 1 < t < p1 then t\|p1 and p1 \|n together imply t\|n which is a contradiction(Why!). So we have n = p1 n1 for some integer n1 where 1 < n1 < n. If n1 is prime then we are done. If n1 is composite then by the same argument we have another prime p2 and integer n2 where 1 < n2 < n1 such that n = p1 p2 n2 . Continuing this way we have a decreasing sequence of integers n > n1 > n2 > · · · > 1. This sequence is ﬁnite and after ﬁnite nn we will get a prime pr . This leads to prime factorization n = p1 p2 · · · pr . To prove the uniqueness let there be two distinct prime factorizations of n as n = p1 p2 · · · pr = q1 q2 · · · qs where r ≤ s and each of pi s and qj s are primes. These primes are in the ordering p1 ≤ p2 ≤ p3 ≤ · · · ≤ pr and q1 ≤ q2 ≤ q3 ≤ · · · ≤ qs . As p1 \|n this implies p1 \|q1 q2 · · · qs then by virtue of Corollary 3.2.1 p1 = qj for some j where 1 ≤ j ≤ n. This follows that p1 < q1 . Now cancelling the Prime Numbers 49 common factors from both the sides we have q2 q3 · · · qs = p2 p3 · · · pr . Continuing as above, up to r terms as r < s. After r-th step we have 1 = qr+1 qr+2 · · · qs which is absurd as qj s are prime. Hence r = s and p1 = q1 , p2 = q2 , . . . , pr = qs . So the factorization is unique. Let us consider an integer 15 which can be written as 5 × 3 where both 5 and 3 are distinct primes. But if we take 75 it can be expressed as 5 × 5 × 3 where we can see the representation of primes. By collecting those primes and replacing them by a single factor we can represent any integer by following corollary viz Corollary 3.2.2. Any positive integer can be uniquely written as p1n1 p2n2 · · · prnr where each ni is a positive integer and pi s are prime for i = 1, 2, 3, . . . , r with p1 < p2 < p3 < · · · < pr . From the above corollary we can assert that any arbitrary positive integer has an unique prime factorization. Now in the later part of this section we have given an alternative proof of the Theorem 2.5.1. For that we have to deﬁne the greatest common divisor and least common multiple of any two arbitrary integers in the light of prime factorization. Let us take two integers a and b with their unique prime factorizations a = pa1 1 p2a2 · · · pann , b = pb11 pb22 · · · pbnn with p1 < p2 < · · · < pn and ak , bk be non negative integers for k = 1, 2, · · · , n. Then gcd(a, b) = 1 pm2 · · · pmn and lcm(a, b) = pM1 pM2 · · · pMn where M = Max(ak , ak ) and pm k n n 1 2 1 2 mk = min(ak , bk ). Here to give alternative proof of the Theorem 2.5.1 we ﬁrst state and prove the lemma as follows: Lemma 3.2.1. If x and y are real numbers, then max(x, y) + min(x, y) = x + y. Proof. If x < y, then min(x, y) = x and max(x, y) = y, and again we ﬁnd that max(x, y) + min(x, y) = x + y. Similarly, If x > y, then min(x, y) = y and max(x, y) = x, and again we ﬁnd that max(x, y) + min(x, y) = x + y. Now using the above lemma, let us proceed for the alternate proof: Proof. Let a and b have prime-power factorizations a = p1a1 pa2 2 · · · pnan , b = pb11 pb22 · · · pbnn , where the powers are nonnegative integers and the primes pi s occurring in the prime-power factorizations of a and b. Let Mk = Max(ak , bk ) 50 Number Theory and its Applications and mk = min(ak , bk ). Then, we have 1 pM2 · · · pMn pm1 pm2 · · · pmn lcm(a, b) gcd(a, b) = pM n n 1 2 1 2 n +mn = p1M1 +m1 p2M2 +m2 · · · pM n = pa1 1 +b1 pa2 2 +b2 · · · pnan +bn = pa1 1 pb11 p2a2 pb22 · · · pann pbnn = ab. 2 3 The numbers 2, 3, 4 are integers and if we take , then this type are the 3 4 p where q = 0 and gcd(p, q) = 1. But there are rational numbers of the form q √ √ numbers of the form 2, 3 which can not be written as above. These are said to be irrational numbers. We are now going to introduce a famous result of Pythagoras on irrational numbers viz √ Theorem 3.2.6. The number 2 is irrational. √ √ a Proof. Let us suppose that 2 is a rational quantity. Then 2 = where a, b b a2 are integers relatively prime to each other. Squaring we have, 2 = 2 ⇒ a2 = 2b2 b implies b2 \|a2 . If b > 1, then by fundamental theorem of arithmetic we can say that there exists a prime p such that p\|b. Then it follows p\|a2 implies p\|a and hence gcd(a, b) ≥ p which is a contradiction unless b = 1. But if b = 1 holds then a2 = 2 which is impossible(Why!). Hence the proof. For further discussion of this chapter we will show our interest in ﬁnding extremely large primes. To do so our ﬁrst aim is to check whether a given integer is prime or not. We ﬁrst deal with this question by trial division of n using the following theorem viz Theorem 3.2.7. If n is a composite integer, then n has a prime factor not √ exceeding n. Proof. Since n is composite, we can write n = ab where a, b are integers with √ √ 1 < a ≤ b < n. There must be a ≤ n, if not then b ≥ a > n which leads to ab > n, which is not possible. Now from Theorem 3.2.3 the integer a must have √ a prime divisor p(say). Then p ≤ a ≤ n. Further if p\|a and a\|n implies p\|n. √ Then p is the required prime factor of n not exceeding n. We can use this theorem to ﬁnd all the primes less than or equal to a given positive integer n. This procedure is called Sieve of Eratosthenes. To illustrate Prime Numbers 51 the process, let us choose n = 81. Then by virtue of the above theorem, 81 has a √ prime factor less than or equal to 81 = 9. Since, the only prime less than 9 are 2, 3, 5, 7. We only have to ﬁnd those integers less than 81 which can be divisible by any one of those primes. In the below table we have shown a complete list of them. The multiples of any one or two or three of 2, 3, 5, 7 of the numbers in the table are cancelled by , C and respectively. C 2 10 A 18 A 2 6 34 42 A 5 0 58 66 A 7 4 3 11 19 2 7 35 A 43 5 1 59 67 7 5 4 12 A 20 A 2 A8 36 A 44 5 2 60 A 68 7 6 5 13 21 A 29 37 45 A 53 61 69 7 7 6 A 14 A 22 3 A0 38 46 5 A4 62 70 7 A8 7 15 A 23 31 39 47 5 5 63 71 79 8 16 A24 32 40 A48 56 64 72 80 9 17 25 33 41 49 57 65 73 81 The above table indicates that there exist many primes less than 81. In fact, from theorem (3.2.4), we have inﬁnitely many primes. A fairly natural question arises: Is it possible to estimate, how many primes are less than a positive real number x? We are fortunate enough to have the most renowned theorem of number theory, and of all mathematics, is the prime number theorem which answers this question. In 1793, Gauss speculated the theorem but it was an open problem until 1896, when a French mathematician J. Hadamard and a Belgian mathematician C. J. de la Vall´ ee-Poussin had proved it independently. So before going to state the theorem let us begin with a simple deﬁnition. Deﬁnition 3.2.2. The function π(x), where x is a positive real number, denotes the number of primes not exceeding x. We now state the prime number theorem, whose proof is beyond the scope of the book. π(x) Statement 3.2.1. In language of limits, the theorem can be stated as lim = x→∞ ln x 1. x The above stated theorem reﬂects the fact that for large values of x, is a ln x good approximation to π(x). Further, it is to be noted that it is not necessary to ﬁnd all primes not exceeding x in order to compute π(x). By virtue of counting 52 Number Theory and its Applications argument based on the Sieve of Eratosthenes,one can compute π(x) without ﬁnding all the primes less than x. Our next theorem addresses that the gap between consecutive primes is arbitrarily long. Theorem 3.2.8. For any positive integer n, there are at least n consecutive composite positive integers. Stated otherwise, there are arbitrarily large gaps in the series of primes. Proof. Consider the n consecutive positive integers (n + 1)! + 2, (n + 1)! + 3, . . . , (n + 1)! + n + 1. Now, 2 ≤ j ≤ n + 1 ⇒ j (n + 1)!. Finally, an appeal to Proposition 2.2.1 yields the desired result. The following example will exemplify our foregoing theorem. Example 3.2.1. For n = 5, the smallest 5 consecutive composite integers can be found by locating the ﬁrst pair of consecutive composite odd integers, 25 and 27. Hence the smallest 5 consecutive composite integers are 24, 25, 26, 27, and 28. These are considerably smaller than the integers (5+1)!+j = 6!+j = 720+j for j = 2, 3, 4, 5, 6. Also, the seven consecutive integers beginning with 8! + 2 = 40322 are all composite. However, these are much larger than the smallest seven consecutive composites 90, 91, 92, 93, 94, 95, and 96. Our next discussion is about the propagation of prime numbers of prime numbers. Let us choose p a prime and p˜ to be the product of all primes that are less than or equal to p. The numbers p˜ + 1 form are called “Euclidean numbers” as they appear in the proof of Theorem 3.2.4. For example, ˜ 2+1=2+1=3 ˜ 3+1=2·3+1=7 ˜ 5 + 1 = 2 · 3 · 5 + 1 = 31 = 59 · 509 is not prime. From these are all prime numbers but also we can see 13 two types of examples, we see that p˜ + 1 is not always a prime. Prime Numbers 53 If we consider a sequence of integers such as, n1 = 2 n2 = n1 + 1 n3 = n1 n2 + 1 .. . nk = n1 n2 · · · nk−1 + 1 where each nk > 1 and they are relatively prime to each other. If not, let gcd(ni , nj ) = d where i < j. Then d\|ni ⇒ d n1 n2 · · · ni · · · nj−1 . Since, d nj therefore d\|n1 n2 · · · nj−1 + 1 together imply d\|1, possible when d = 1. So our assertion, all nk s are pairwise relatively prime, is true. Now we can say that there are many distinct primes as there are integers nk . Let pn be n-th prime number. Then from Euclid’s proof we can estimate the rate of increase of pn . Here we have pn+1 ≤ p1 p2 · · · pn + 1 < pnn + 1. If n = 5 then 31 = p6 = p55 + 1 = 75 + 1 = 16808. Thus we have the following theorem viz Theorem 3.2.9. If pk be the k-th prime, then pk ≤ 22 k −1 . Proof. We will prove the theorem by Mathematical Induction on k. If k = 1, then the result is obvious. Let us assume k > 1. Then pk+1 ≤ p1 p2 · · · pk + 1 ≤ 2 · 22 · · · 2k−1 + 1 = 21+2+2 2 +···+2k−1 + 1 = 22 k −1 + 1. But 1 ≤ 22 −1 for all k. Therefore pk+1 ≤ 22 (How!). Thus the result is true for k + 1. Hence the proof. k k The last inequation of the above proof gives rise to an interesting corollary stated as follows: k Corollary 3.2.3. For k ≥ 1 there exists at least k + 1 primes less than 22 . Proof. left to the reader. Finally, we conclude this section with remarkable conjecture about primes, commonly known as Golbach’s Conjecture, stated by Christian Goldbach in a letter to Euler in 1742. Goldbach Conjecture: Every even positive integer greater than two can be written as the sum of two primes. 54 Number Theory and its Applications Let us exemplify the Conjecture with an example: 10 = 3 + 7 = 5 + 5 24 = 5 + 19 = 7 + 17 = 11 + 13 100 = 3 + 97 = 11 + 89 = 17 + 83 = 29 + 71 = 41 + 59 = 47 + 53 Next with the help of the following lemma, we are going to prove the fact that there exists inﬁnitely many primes of the form 4n + 3. Lemma 3.2.2. The product of two or more integers of the form 4n + 3 (n ∈ Z) results in the same form. Proof. It’s suﬃcient to prove the lemma with two integers of the form 4n + 1. Set k1 = 4n1 + 1, k2 = 4n2 + 1. Multiplying we obtain, k1 k2 = (4n1 + 1)(4n2 + 1) = 4(4n1 n2 + n1 + n2 ) + 1 = 4n + 1, [n = 4n1 n2 + n1 + n2 ∈ Z] which is the desired form. This facilitates the proof for the following theorem. Theorem 3.2.10. There exists inﬁnitely many primes of the form 4n + 3. Proof. Suppose there exists ﬁnitely many primes t1 , t2 , . . . , ts of the form 4n + 3. Also, consider N = 4t1 t2 . . . ts − 1 = 4(t1 t2 . . . ts − 1) + 3 to be a positive integer. Further, let N = k1 k2 . . . kn be the prime factorization of N . Since N is odd, then ki = 2, ∀ i. Thus ki is of the form, either 4n + 1 or 4n + 3. If ki is of the form 4n + 1, then using the lemma 3.2.2 we can say that N must be of the form 4n + 1. This is not the case here. Then N must contain one prime factor ki of the form 4n + 3. But, ki can not be found among t1 , t2 , . . . , ts . Otherwise this leads to ki \|1, which is not true. Thus our assumption of ﬁnitely many primes of the form 4n + 3 is wrong. The last theorem inspired us to ask a fairly question: Is the number of primes of the form (4n + 1) also inﬁnite? The following Dirichlet’s statement, whose proof is beyond the scope of the book, is the answer to the question. Prime Numbers 55 Theorem 3.2.11. If a and b are positive integers with gcd(a, b) = 1, then the arithmetic progression a, a + b, a + 2b, . . . contains inﬁnite number of primes. From Dirichlet’s statement it can be seen that there exists inﬁnitely many primes ending with 999, for instance 1999, 1000999, . . ., they appear in arithmetic progression given by 1000n + 999, with gcd(1000, 999) = 1. Theorem 3.2.12. There exists no arithmetic progression of the form a, a + b, a + 2b, . . . that consists of only primes. Proof. To begin with, consider a + nb = p, p being a prime. If nk = n + kp for k = 1, 2, 3 . . . then the nk -th term in the progression is a + nk b = a + (n + kp)b = (a + nb) + kbp = p + kbp = p(1 + kb) Since, p\|p(1 + kb), therefore p\|(a + nk b). Hence (a + nk b) can not be a prime, which is our desired result. Remark 3.2.1. From the above theorem, it’s quite clear that the progression contains inﬁnitely many composite numbers. Theorem 3.2.13. If all the n(> 2) terms of the arithmetic progression, p, p + d, p + 2d, . . . are primes, then q\|d where d being the common diﬀerence and q(< n) is a prime number. Proof. Consider a prime q < n. In anticipation of a contradiction, assume q d. Again, if possible let us assume that the ﬁrst q terms of the given progression will leave the same remainders when divided by q. Then ∃j, k ∈ Z with 0 ≤ j < k ≤ q − 1 or k − j ≤ q − 1 such that (p + jd) and (p + kd) generates same remainder when divided by q, which further implies q (k − j). But gcd(p, q) = 1 and by Euclid’s lemma q (k − j), which is impossible in the light of the inequality k − j ≤ q − 1. Hence the ﬁrst q terms of the given progression will leave q diﬀerent remainder upon division by q. Since they are extended from q integers 0, 1, 2, . . . , q − 1, one of them must be zero.This means for some t satisfying 0 ≤ t ≤ q − 1, q\|(p + td). Hence we conclude, p + td is composite because the inequality q < n ≤ p ≤ (p + td) holds (for if p ≤ n, then one of the term of the progression will be p(1 + d)). This leads to a contradiction and hence q d. 56 Number Theory and its Applications Remark 3.2.2. There is a conjecture that there exists arithmetic progression of ﬁnite length, consisting of consecutive prime numbers. For instance, 47, 53, 59 and 251, 257, 263, 269. Consider the function f : Z+ −→ Z deﬁned by f (n) = n2 + n + 41. There was a myth that the image set of the function was only primes. But in 1772, it was proved to be false by Leonhard Euler. Though the myth was true for n = 0, 1, 2 . . . , 39 but fails for n = 40, 41. Here f (40) = 40 · 41 + 41 = 412 , and f (41) = 41 · 42 + 41 = 41 · 43. Once again f (42) = 1847 turns out to be prime. The polynomial f (n) = (n2 + n + 41) is known as Euler polynomial. It is to be noted that no polynomial of the form n2 + n + q, q being prime, can perform better than Euler polynomial in giving primes for successive values of n. Theorem 3.2.14. There exists no non-constant polynomial f : Z+ −→ Z with integral coeﬃcients that generates solely prime numbers for n ∈ Z+ . Proof. To the contrary, assume that such a polynomial f does exists. Set f (n) = ak nk +ak−1 nk−1 +. . .+a2 n2 +a1 n+a0 where the coeﬃcients ai (i = 0, 1, 2 . . . , k) are integers with ak = 0. Let f (n0 ) = p, for some ﬁxed value n0 ∈ Z+ . Now, for any t ∈ Z, consider f(n0+tp) = ak(n0+tp)k+ak−1(n0+tp)k−1+. . . +a2(n0+tp)2+a1(n0+tp) + a0 = (ak nk + ak−1 nk−1 + . . . + a2 n2 + a1 n + a0 ) + pQ(t) = f (n0 ) + pQ(t) = p(1 + Q(t)), Q(t) being a polynomial in t with integral coeﬃcients. This shows p\|f (n0 + tp), which further implies f (n0 + tp) = p (t ∈ Z). This leads to a contradiction(Why!).Thus we have established the theorem. 3.3 Worked out Exercises Problem 3.3.1. The lucky numbers are generated by the screening process as follows: Let us begin with the set of positive integers. Starting the process by crossing out every second integer in the list, start the count with the integer 1. Other than 1 the smallest integer left is 3, continuing with the process every third integer left, beginning with the integer 1. The next integer left is 7, so we cross Prime Numbers 57 out every seventh integer left. Continuing as above, where at each stage we cross out every κth integer left where κ is the smallest integer left other than one. The integers that remain are the lucky numbers. Prove that the lucky number set is countably inﬁnite. Solution 3.3.1. At each stage of the procedure for generating the lucky numbers the smallest number left is κ, say, is designated to be a lucky number and inﬁnitely many primes are left after the deletion of every κ integer left. It follows that there are countably inﬁnite numbers of steps, and at every step a new lucky number is added to the sequence. Hence the proof. Problem 3.3.2. Show that the polynomial f (x) = x2 − x + 41 is prime for all integers x with 0 ≤ x ≤ 40. Furthermore, it is composite for x = 41. Solution 3.3.2. Hint: Find f (1), f (2), f (3), . . . , f (39), f (40). But f (41) is composite. Problem 3.3.3. Show that if g(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 where the coeﬃcients are integers, then ∃ y ∈ Z such that g(y) is composite. Solution 3.3.3. In anticipation to contradiction, suppose there any y ∈ Z such that g(y) is composite. Let y0 be a positive integer such that g(y0 ) = p, a prime. Let κ be any integer such that g(y0 + κp) = an (y0 + κp)n + an−1 (y0 + κp)n−1 +· · ·+a1 (y0 +κp)+a0 . By binomial expansion it follows that g(y0 +κp) = n aj xj0 + M p, M being an integer. Now p\|(g(y0 + M p) = g(y0 + κp) (Why!). j=0 Also g(y0 +κp) = p (Why!). This contradicts the fact that a polynomial of degree n takes on each value not more than n times. Hence there is an integer y such that g(y) is composite. Problem 3.3.4. Show that no integer of the form n3 + 1 is a prime, other than 2 = 13 + 1. Solution 3.3.4. Note that n must be positive. Otherwise no such integers are prime (Why!). Since n3 +1 = (n+1)(n2 −n+1), n3 +1 is not prime unless one of the two factors on the right hand side of the equation is 1 and the other is n3 +1. But (n + 1) > 1 for every positive integer n and the only way for n + 1 = n3 + 1 is when n = 1(Verify!). In this case, we have 13 + 1 = (1 + 1)(12 − 1 + 1) = 2. Hence 2 is the only prime of this form. Problem 3.3.5. Find all primes that are the diﬀerence of the fourth powers of two integers. 58 Number Theory and its Applications Solution 3.3.5. Suppose n = a − b = (a − b)(a + b)(a2 + b2 ), where a > b. The integer n cannot be prime because it is divisible by a + b which cannot be 1 or n. 4 4 Problem 3.3.6. Show that if a and n are positive integers such that an − 1 is prime, then a = 2 and n is prime. Solution 3.3.6. Let n be a composite number and k be any divisor of n. Then 1 < k < n and (ak − 1)\|(an − 1). As an − 1 is prime, so ak − 1 = 1(Why!). This is true, if a = 2 and k = 1. This leads to a contradiction as k > 1. Thus we have a = 2 and n is prime. Problem 3.3.7. Show that every integer greater than 11 is the sum of two composite integers. Solution 3.3.7. Let us assume that n be an integer greater than 11. Case I n is even: Then there exists an integer k such that n = 2k. Since n > 11, therefore n ≥ 12 and thus k ≥ 6. Now n − 4 = 2(k − 2) with k − 2 ≥ 4. By deﬁnition of divisibility, we have 2\|(n − 4) and (k − 2)\|(n − 4). By deﬁnition of compositeness, n − 4 is composite. Also n = (n − 4) + 4. As 4 is composite, therefore n is the sum of two composite numbers. Case II n is odd: Then there exists an integer k such that n = 2k + 1. Since n > 11, therefore n ≥ 13 and thus k ≥ 6. Now n − 9 = 2(k − 4) with k −4 ≥ 2. By deﬁnition of divisibility, we have 2\|(n−9) and (k −4)\|(n−9). Again by deﬁnition of compositeness, we have n − 9 is composite. Also n = (n − 9) + 9. As 9 is composite, therefore n is the sum of two composite numbers. Problem 3.3.8. If p ≥ 5 is a prime number, show that p2 + 2 is composite. Solution 3.3.8. By division algorithm, p = 6q + r where r = 0, 1, 2, 3, 4, 5. Case i: If r = 0, then p = 6q implies 6\|p, a contradiction(Why!). Hence r = 0 Case ii: If r = 2, then p = 6q + 2 implies 2\|p, a contradiction. Hence r = 2 Case iii: If r = 3, then p = 6q + 3 implies 3\|p, a contradiction. Hence r = 3 Case iv: If r = 4, then p = 6q + 4 implies 2\|p, a contradiction. Hence r = 4 Thus r = 1, 5 implies p = 6q + 1 or p = 6q + 5. Therefore 3\|(p2 + 2) in either case(Why!). Hence the proof. Prime Numbers 59 Problem 3.3.9. If p = 5 is an odd prime, prove that either p − 1 or p + 1 is divisible by 10. 2 2 Solution 3.3.9. Here p is of the form: 10q + 1, 10q + 3, 10q + 7, 10q + 9. But p= 10q + even since it can factor out 2, so fails to be prime. Now, (10q + 1)2 = 100q 2 + 20q + 1 ⇒ 10\|(p2 − 1) (10q + 3)2 = 100q 2 + 60q + 9 ⇒ 10\|(p2 + 1) (10q + 7)2 = 100q 2 + 140q + 49 ⇒ 10\|(p2 + 1) (10q + 9)2 = 100q 2 + 180q + 81 ⇒ 10\|(p2 − 1). Problem 3.3.10. If n > 1 is an integer not of the form 6k + 3, prove that n2 + 2n is composite. Solution 3.3.10. Here n is of the form 6q, 6q + 1, 6q + 2, 6q + 4, 6q + 5. Case (i): When n = 6q, then n2 + 2n = 36q 2 + 26q ⇒ 2\|(36q 2 + 26q ) as q > 0, hence a composite number. Case (ii): When n = 6q + 1, then n2 + 2n = 36q 2 + 12q + 26q+1 + 1 = 36q 2 + 12q +(2+1)(26q −· · ·+(−1)6q 16q )(W hy!) ⇒ 3\|(n2 +2n ), hence a composite number. Case (iii): When n = 6q+2, then n2 +2n = 36q 2 +24q+4+22 26q ⇒ 2\|(n2 +2n ), hence a composite number. case (iv): When n = 6q+4, then n2 +2n = 36q 2 +48q+16+24 26q ⇒ 2\|(n2 +2n ), hence a composite number. Case (v): Treated as an exercise.(Hint! Similar to 6q + 1 as above.) Problem 3.3.11. Prove that a positive integer a > 1 is a square if and only if in the prime factorizations of a all the exponents of the primes are even integers. Solution 3.3.11. Let a > 1 be square. Then a = n2 , for some integer n. Let 2 · · · p2ks shows all exponents are n = pk1 1 pk2 2 · · · pks s . Therefore n2 = p12k1 p2k s 2 even. Conversely, suppose all exponents of a = pk1 1 pk2 2 · · · pks s are even. Therefore 2 · · · p2ms = ki = 2mi for some mi and for every ki . Therefore a = p12m1 p2m s 2 m1 m2 ms 2 (p1 p2 · · · ps ) . Problem 3.3.12. An integer is said to be square-free if it is not divisible by the square of any integer greater than 1. Prove the following: 60 Number Theory and its Applications 1. An integer n > 1 is square-free if and only if n can be factored into a product of distinct primes. 2. Every integer n > 1 is the product of a square-free integer and a perfect square. Solution 3.3.12. 1. Let n > 1 be square free and n = pk1 1 pk2 2 · · · pks s be the prime factorization of it. Then ki ≥ 2 and therefore p2i \|n, a contradiction to the deﬁnition of square free. Therefore ki = 1. Hence n = p1 p2 p3 · · · ps with pi = pj . If possible, let n be not a square free and a2 \|n. Hence n = la2 , for some l ∈ Z. Let a = q1k1 q2k2 q3k3 · · · qrkr . Therefore p1 p2 p3 · · · ps = lq12k1 q22k2 q32k3 · · · qr2kr implies qj \|p1 p2 p3 · · · ps . By virtue of Corollary (3.2.1), qj = pi for some i = 1, 2, 3, . . . , s. After factoring out qj and pi , we still have p1 p2 p3 · · · ps = lq12k1 q22k2 q32k3 · · · qr2kr implies qj \|p1 p2 p3 · · · ps . But the original factorization p1 p2 p3 · · · ps was unique and qj was factored out. Hence qj fails to divide the remaining factorization, which shows n to be square free. 2. Let n = pk1 1 pk2 2 · · · pks s be the prime factorization of it. If ki is odd and ki > k k k 1, then ki − 1 is even. Let a = pr r1 pr r2 · · · pr rm , 1 ≤ ri ≤ s and kr is odd 1 2 m k with kr ≥ 1. Let b = pr1 pr2 · · · prm . Then a = bpr r1 1 i −1 kr −1 pr 2 kr −1 Also b is square free(Why!). Since kr − 1 is even, pr i 2 i i k · · · pr rm kl . 2li = pr . Let i i −1 m kl kl c = prl1 prl2 · · · prlm . Then, a = bc2 . Finally, suppose a\|n = pl1 1 pl2 2 · · · plj j m 1 2 where all kl are even as a\|n has factored out all of the odd exponents in j the canonical form of n. By previous problem above, a\|n = d2 ⇒ n = bc2 d2 = b(cd)2 , where b is square free. Problem 3.3.13. Find all prime numbers that divide 50!. Solution 3.3.13. All primes less than 50 will divide 50! because each is a term of 50!. By the fundamental theorem of Arithmetic, each term k of 50! that is non–prime has a unique prime factorization. And each term of the unique factorization of k is smaller than k, so is prime less than 50. There is no prime greater than 50 represented in this factorization of k. Hence all primes less than 50 will divide 50!. Problem 3.3.14. Show that any composite three-digit number must have a prime factor less than or equal to 31. Solution 3.3.14. We know 999 is the largest composite three digit number. Now √ 999 = 31.6. Here 31 is prime, so if a is composite, largest prime divisor is less √ than equal to a. Hence 31 is largest possible prime divisor. Prime Numbers 61 Problem 3.3.15. Prove that the prime number set is countably inﬁnite using the integer N = p! + 1. Solution 3.3.15. Let us assume there are ﬁnitely many primes, pn being the largest. Then N = pn ! + 1 = 1 · 2 · · · pn + 1. Now N must have a prime divisor pk with 1 ≤ k ≤ n(Why!). And pk \|1·2 · · · pn (Why!). Therefore pk \|(N −1·2 · · · pn ) ⇒ pk \|1 ⇒ pk = 1, a contradiction. Problem 3.3.16. Any integer n can be expressed as n = 2k m, where k ≥ 0 and m being an odd integer. Verify! Solution 3.3.16. With out any loss of generality, assume n > 0, for if n < 0 then −n = 2k m ⇒ n = 2k (−m). Now the following cases will arise: Case(i) If n is odd, then k = 0 and m = n. Case(ii) If n is even, then n = 2k1 , k1 < n. Case(iii) If k1 is odd, then we are done. Case(iv) If k1 is even, then k1 = 2k2 so n = 22 k2 where k2 < k1 < n. Continuing as above after i-th stage we have 2i ki , where ki < ki−1 . This is a ﬁnite process and after a certain stage we will reach at kt = 1 and there will be no odd integer after 1. In that stage, n = 2t kt = 2t · 1. Thus n can be expressed as n = 2k m, where k ≥ 0 and m being an odd integer. Problem 3.3.17. Prove or Disprove: Every positive integer can be written in the form p + a2 , where p is either a prime or 1, and a ≥ 0. Solution 3.3.17. Hint: 25 = p + a2 then consider a = 1, 2, 3, 4, 5. Problem 3.3.18. 6m + 1. 1. Prove: Any prime of the form 3n+1 is also of the form 2. The only prime of the form n3 − 1 is 7. Solution 3.3.18. 1. Here p = 3n + 1 is prime implies p is odd. Then p − 1 = 3n is even implies n is even. Hence n. = 2m, for integer m. Thus 3n + 1 = 6m + 1. 2. Here t = n3 − 1 = (n − 1)(n2 + n + 1). If n = 1, then t is prime. If n = 2, t = 7. If n > 2, then t will be a factor of two integers, neither of which is 1. Hence for n > 2, t can’t be prime. Problem 3.3.19. Find ﬁve primes of the form n2 − 2. 62 Number Theory and its Applications Solution 3.3.19. Hint: Consider n = 2, 3, 5, 7, 9. Problem 3.3.20. A positive integer n is said to be square-full, or powerful, if p2 \|n for every prime factor p of n. Prove that if n is square-full, then it can be written in the form n = a2 b3 , with a and b positive integers. Solution 3.3.20. Let n = pk1 1 pk2 2 · · · pkr r be the prime factorization of it. Since n is square-full, ki ≥ 2. Listing ﬁrst the odd exponents and then the even one, let us assume k k k k k k k k n = pk1 1 pk2 2 · · · pks s = qmm1 qmm2 qmm3 · · · qmms qnn1 qnn2 qnn3 · · · qnnt , 1 2 3 1 s 2 3 t where km are odd(so kms ≥ 3) and kn are even. Therefore for some vi , kn = i i i 2vi . Therefore k k k k n = qmm1 qmm2 qmm3 · · · qmms (qn2v1 qn2v2 qn2v3 · · · qn2vt ) 1 2 3 s km km km km = qm 1 1 2 2 km Hence, n = qm qm 1 1 2 2 3 3 km qm qm 3 3 1 km · · · qm 2 3 t (qn qn qn · · · qn ) . s s km qm · · · qm 1 v 1 v2 v 3 2 vt 2 3 t (Y ) , Y = qn qn qn · · · qnvt . v1 v2 v3 2 s 1 s 2 3 t Now km is odd and ≥ 3 together implies km − 3 is even. Thus i i m1 −3 m2 −3 m3 −3 n = qm qm qm · · · qm (qm 3 3 1 3 2 3 3 s 1 qm 2 qm 3 ms −3 · · · qm )(Y )2 . s Let mi − 3 = 2wi , qm1 qm2 qm3 · · · qms = b. Therefore 2w1 2w2 2w3 2ws n = b3 (qm qm qm · · · q m )(Y 2 ). 1 w1 w 2 3 Let X = qm qm qm 1 n = a2 b3 . 2 w3 2 2 s ws · · · qm . Then n = b3 X 2 Y 2 . Taking a = XY , we obtain s Problem 3.3.21. Given that p n for all primes p ≤ either a prime or the product of two primes. √ 3 n, prove that n > 1 is Solution 3.3.21. Assuming n to be composite and taking n = p1 p2 · · · pX with X ≥ 3, we know that √ √ 1 < 3 n < pi ≤ n. Therefore √ 3 √ n √ √ 3 n ≤ p2 ≤ n √ √ 3 n ≤ p3 ≤ n. n ≤ p1 ≤ √ √ √ Therefore n = ( 3 n)( 3 n)( 3 n) < p1 p1 p2 p3 = n ⇒ n < n. Hence X < 3 or = 2 or = 1. Thus n > 1 is either a prime(X = 1) or the product of two primes(X = 2). Prime Numbers 63 Problem 3.3.22. Prove that if n > 2, then there exists a prime p satisfying n < p < n!. Solution 3.3.22. For n > 2, n < n! − 1 < n!. If n! − 1 is prime, we are done. If n! − 1 is not prime, taking p to be a prime divisor, we have p < n! − 1. Suppose p ≤ n. Then p is one of the terms in 1, 2, 3, . . . , n. So p\|n!. Therefore p\|n! and p (n! − 1) together implies p n! − (n! − 1) = 1. Therefore p > n and hence the result. Problem 3.3.23. For n > 1, show that every prime divisor of n! + 1 is an odd integer that is greater than n. Solution 3.3.23. Because n! is even for n > 1, therefore n! + 1 is odd. Hence 2 (n! + 1), so every prime divisor of n! + 1 is odd. Suppose every prime divisor pi of n! + 1 is less than or equal to n. Since pi is one of the members of n!, therefore pi \|n!. Also pi \|(n! + 1) ⇒ pi \|(n! + 1) − n! = 1, a contradiction. Thus pi is greater than n. Problem 3.3.24. If a is a positive integer and integer. √ n a is rational, then √ n a is an √ r Solution 3.3.24. Let n a = , where r, s being integers and gcd(r, s) = 1 with s 0. Let r = p1 p2 · · · pX , s = q1 q2 · · · qY . Then pi = qj . Therefore s= (q1 q2 · · · qY )n a = (p1 p2 · · · pX )n . Therefore (p1 p2 · · · pX )n \|a. Let a = (p1 p2 · · · pX )n t, for some integer t. Therefore (q1 q2 · · · qY )n (p1 p2 · · · pX )n t = (p1 p2 · · · pX )n , implies qj = 1 for all j. Thus s = 1 and Problem 3.3.25. Prove for n ≥ 2, √ n √ r = n a, an integer. s n is irrational. √ Solution 3.3.25. Suppose, n ≥ 2, n n is rational. Then by Problem 3.3.24, it √ is an integer. Let n n = a. Then n = an . But n < 2n ⇒ an < 2n ⇒ either a < 2 or a = 1. Therefore n = 1n = 1, a contradiction. Problem 3.3.26. Prove that any odd prime p is of the form 4k + 1 or 4k + 3 for any non-negative integer k. 64 Number Theory and its Applications Solution 3.3.26. By Division Algorithm, any positive integer can be expressed in the form a = bq + r, 0 ≤ r < b or equivalently written as a = 4q + r, r = 0, 1, 2, 3. Now if; r = 0, a = 4q = 2(2q), an even integer. r = 1, a = 4q + 1 = 2(2q) + 1, an odd integer. r = 2, a = 4q + 2 = 2(2q) + 2 = 2(2q + 1) = 2m, an even integer. r = 3, a = 4q = 2(2q) + 3 = 2(2q + 1) + 1 = 2m + 1, an odd integer. Hence any odd prime p is of the form 4k + 1 or 4k + 3 for any non-negative integer k. Problem 3.3.27. If p and p2 + 8 are both prime numbers, prove that p3 + 4 is also prime. Solution 3.3.27. Referring to Problem 3.3.8, if p > 3 is prime, it is of the form (6k + 1) or (6k + 5). So for p = 6k + 1 or 6k + 5, we have p2 + 8 = 36k 2 + 12k + 9 or p2 + 8 = 36k 2 + 60k + 33 respectively. But 3 (36k 2 + 12k + 9) and 3 (36k 2 + 60k + 33). So p2 + 8 is not prime, provided p > 3. By the problem, both p and p2 + 8 are primes. Thus the only possibility is p = 3, which yields p2 + 8 = 17. Hence p3 + 4 = 31. Problem 3.3.28. Bertrand Conjecture: For any positive integer z, ∃ a prime p satisfying z ≤ p < 2z. Using this proves that for every n ≥ 2, ∃ a prime p with p < n < 2p. Solution 3.3.28. Case-I: n is odd: Since n ≥ 2 & k ≥ 1, ∃ k ∈ Z such that n = 2k + 1. Addressing to Bertrand’s Conjecture, ∃ a prime p satisfying k < p < 2k. Now p < (p + 1) < (2k + 1) = n ⇒ p < n. Further 2k < 2p ⇒ (2k +1) ≤ 2p ⇒ n ≤ 2p. But (2k +1) being odd and 2p is even, together conclude n < 2p. Thus ∃ a prime p such that p < n < 2p. Case-II: n is even: Since k ≥ 1, ∃ k ∈ Z such that n = 2k holds. An appeal to Bertrand’s Conjecture yields, k < p < 2k = n ⇒ p < n(p being a prime). Therefore n = 2k < 2p ⇒ n < 2p. Thus p < n < 2p. Problem 3.3.29. Let pn denote the n-th prime number. For n ≥ 3, prove that p2n+3 < pn pn+1 pn+2 . Solution 3.3.29. Note that pn+1 < 2pn . Therefore pn+3 < 2pn+2 . So p2n+3 < 4p2n+2 < 4pn+2 (2pn+1 ) = 8pn+2 pn+1 . Now p5 = 11 ⇒ 8pn+2 pn+1 < p5 pn+2 pn+1 . Therefore p2n+3 < pn pn+1 pn+2 , if n ≥ 5. For n = 4; p27 = 289 < p4 p5 p6 = 1001. For n = 3; p26 = 169 < p3 p4 p5 = 385. For n = 2; p25 = 121 < p2 p3 p4 = 105. Hence for n ≥ 3, p2n+3 < pn pn+1 pn+2 . Prime Numbers 65 Problem 3.3.30. There exist inﬁnitely many primes that do not belong to any pair of twin primes. Solution 3.3.30. Here gcd(5, 21) = 1. By Dirichlet’s theorem, the series 5 + 21k for k = 1, 2, 3, . . ., contains inﬁnitely many primes. Let p be one such prime. Then p = 5 + 21k(k ∈ Z) gives p + 2 = 7(1 + 3k) and p − 2 = 3(1 + 7k). Thus (p + 2) and (p − 2) fails to be prime. Hence all the primes contained in (5 + 21k) cannot be numbers of twin primes. Problem 3.3.31. Prove that there are inﬁnitely many primes of the form 6n+5. Solution 3.3.31. To the contrary, assume only a ﬁnite number of primes of the form (6n + 5). Let this be q1 , q2 , . . . , qs . Consider N = 6q1 q2 . . . qs − 1 = 6(q1 q2 . . . qs − 1) + 5. Let N = r1 r2 . . . rt be its prime factorization. Since N 2 for each i, so each ri can only be of the form 6n + 1, 6n + 3 or is odd, ri = 6n + 5. Since (6n + 1)(6m + 1) = 36mn + 6m + 6n + 1 = 6(6mn + m + n) + 1 = 6k + 1, where k = (6mn + m + n), this shows the product of two integers of the form (6n + 1) is of the same form. By similar reasoning, the product of two integers of the form (6n + 3) is also so. Furthermore, (6n + 1)(6m + 3) = 6(6mn + m + 3n) + 3 = 6k + 3, where k = (6mn + m + 3n). This implies, the product of two integers of the form (6n + 1) and (6n + 3) is of the form (6n + 3). So the only way for N to be of the form (6n + 5) is, N must contain at least one factor ri which is of the form (6n + 5). But any qi of the form 6n + 5. If such qi exists, then from construction of N we get N − 6q1 q2 . . . qs = −1. Furthermore N − 6q1 q2 . . . qs is divisible by a prime of the form (6n + 5), which contradicts our assumption(Why!). 3.4 Exercises: 1. Prove each of the assertions below: (a) The only prime of the form n3 − 1is 7. (b) The only prime p for which 3p + 1 is a perfect square is p = 5. (c) The only prime of the form n2 − 4 is 5. 66 Number Theory and its Applications 2. Given that p is a prime and p an , prove that pn an . 3. Establish each of the following statements: (a) If n > 4 is composite, then n divides (n − 1)!. (b) Any integer of the form 8n + 1, where n ≥ 1, is composite. 4. Prove that a positive integer a > 1 is a square if and only if in the canonical form of a all the exponents of the primes are even integers. 5. Verify that any integer n can be expressed as n = 2k m , where k ≥ 0 and m is an odd integer. 6. A positive integer n is called square-full, or powerful, if p2 n for every prime factor p of n (there are 992 square-full numbers less than 250, 000). If n is square-full, show that it can be written in the form n = a2 b3 , with a and b positive integers. 7. Given that p n for all primes p ≤ or the product of two primes. √ 3 n, show that n > 1 is either a prime 8. Show that any composite three-digit number must have a prime factor less than or equal to 31. 9. Let qn be the smallest prime that is strictly greater than Pn = p1 p2 . . . pn + 1. It has been conjectured that the diﬀerence qn − (p1 p2 . . . pn ) is always prime. Conﬁrm this for the ﬁrst ﬁve values of n. 10. Let pn denotes the n-th prime number and set dn = pn+1 − pn . Find ﬁve solutions of the equation dn = dn + 1. 11. For n > 3, show that the integers n, n + 2, n + 4 cannot all be prime. 12. A conjecture of Lagrange (1775) asserts that every odd integer greater than 5 can be written as a sum pl + 2p2 , where p1 , p2 are both primes. Conﬁrm this for all odd integers through 75. 13. Show that 13 is the largest prime that can divide two successive integers of the form n2 + 3. 14. Determine all twin primes p and q = p + 2 for which pq − 2 is also prime. 15. Let pn denote the n-th prime. For n > 3, show that pn < pl +p2 +. . .+pn−l . 4 Theory of Congruences “Number theorists are like lotus-eaters—having tasted this food they can never give it up.” – Leopold Kronecker 4.1 Introduction If two numbers a and b be such that the diﬀerence a − b is divisible by an integer n, then a and b are said to be “Congruent modulo n”. The number n is called the modulus and the statement “a is congruent to b(mod n)” is analytically written as a ≡ b(mod n). The quantity a is often said to be the base and the quantity b is called the residue or remainder. There are several types of residues. The common residue deﬁned to be the non-negative and smaller than n while the minimal residue is b or b − n, whichever is smaller than absolute terms. Perhaps, congruence arithmetic is mostly treated as a generalisation of the arithmetic of the clock. Since there are 60 minutes in an hour, “minute arithmetic” uses a modulus of n = 60. If a person starts at 30 minutes past the hour and then waits for another 55 minutes, 30+55 ≡ 25(mod 60), so the current time would be 25 minutes past the next hour. By similar way, “hour arithmetic” on a 12 hour clock uses a modulus of n = 12, so 9 O’ clock (a.m) plus four hours give 9 + 4 ≡ 1(mod 12) or 1 O’ clock (p.m). Theory of congruences are extremely useful in many areas of number theory. Test of divisibility is one of them.For instance, if the sum of the digits of an 67 68 Number Theory and its Applications integer is divisible by 3, then the original integer is divisible by 3. Also, congruences have their own restrictions. For instance, knowing the number of minutes past the hour is useful but knowing the hour the minutes are past is often more useful. So congruences discard absolute information. Also, if a ≡ b(mod n) and c ≡ d(mod n), then it follows that ax ≡ bx (mod n), but not usually xc ≡ xd (mod n) or ac ≡ bd (mod n). 4.2 Congruences The language of congruences was developed at the beginning of the nineteenth century by famous Mathematician Gauss. The language of congruence is extremely useful in number theory. Deﬁnition 4.2.1. If a and b are integers, we say that a is congruent to b modulo m if m\|(a−b), symbolically denoted by a ≡ b(mod m). If a and b are incongruent modulo m, then m (a − b) and is denoted by a ≡ b(mod m). Example 4.2.1. Since 6\|(20 − 2) = 18, therefore, 20 ≡ 2(mod 6). Similarly, 4 ≡ −5(mod 9) and 300 ≡ 6(mod 7). In working with congruences, the following proposition is needed. Proposition 4.2.1. If a and b are integers, then a ≡ b(mod m) if and only if there is an integer l such that a = b + lm. Proof. Let a ≡ b(mod m) hold. Then m\|(a − b) implies there exists an integer l such that a = b + lm. Conversely, if there exists an integer l such that a = b + lm holds, then a − b = lm implies l\|(a − b) implies a ≡ b(mod m). Here we have given an example to understand the above theorem lucidly. Example 4.2.2. Let us consider 16 ≡ 2(mod 7). Then 16 − 2 = 14 is divisible by 7 and also we can write 16 as 16 = 2 + 2 × 7. In the following theorem, we have shown some standard properties related to congruence relation which depicts how an algebraic operations(addition, subtraction, or multiplication) to both sides of a congruence preserves the congruence. Theorem 4.2.1. If a, b, c, d and m are integers with m > 0 satisfying a ≡ b( mod m) and c ≡ d(mod m), then 1. a ± c ≡ b ± d(mod m) 2. ac ≡ bd(mod m) Theory of Congruences 69 3. a ± c ≡ b ± c(mod m) 4. ac ≡ bc(mod m) 5. a(mod m) ≡ b(mod m). Proof. 1. Here a ≡ b(mod m) and c ≡ d(mod m) implies m\|(a − b) and m\|(c − d), which further implies there exists integers k and l satisfying a − b = km and c − d = lm. From the identity (a ± c) − (b ± d) = (a − b) ± (c − d) = km ± lm = m(k ± l), we see both m\| (a + c) − (b + d) and m\| (a − c) − (b − d) as k + l, k − l both are integers. Therefore a ± c ≡ b ± d(mod m). 2. Here a ≡ b(mod m) and c ≡ d(mod m) implies m\|(a − b) and m\|(c − d), this again implies that (c − d)b, (a − b)c both are divisible by m. Thus, (a−b)c+(c−d)b = (ac−bd) is divisible by m. Therefore ac ≡ bd(mod m). 3. Since a ≡ b(mod m), therefore m\|(a − b). Now (a ± c) − (b ± c) = a − b is divisible by m. Therefore a ± c ≡ b ± c(mod m). 4. Note that a ≡ b(mod m) ⇒ m\|(a − b). Now (a − b)c = ac − bc is divisible by m. Therefore ac ≡ bc(mod m). 5. As a ≡ b(mod m), then for some integer k we have a − b = km. Now k can be written as k = k1 − k2 where k1 , k2 are integers. Again, a − b = (k1 − k2 )m = k1 m − k2 m ⇒ a − k1 m = b − k2 m = r. Therefore r ≡ a( mod m), r ≡ b(mod m) ⇒ a(mod m) ≡ b(mod m). Example 4.2.3. Since 18 ≡ 3(mod 5) and 22 ≡ 2(mod 5), using Theorem (4.2.1) we see that 40 = 18 + 22 ≡ 3 + 2 ≡ 0(mod 5), −4 = 18 − 22 ≡ 3 − 2 ≡ 1( mod 5) and 396 = 18 · 22 ≡ 3 · 2 ≡ 6(mod 5). Example 4.2.4. Since 27 ≡ 3(mod 8), it follows 34 = 27 + 7 ≡ 3 + 7 ≡ 10( mod 8), 23 = 27 − 4 ≡ 3 − 4 ≡ −1(mod 8), and 25 = 27 − 2 ≡ 3 − 2 ≡ 1(mod 8). Next before proceeding further, the following example reﬂects the fact that a congruence is not necessarily retained when divided both sides by an integer. Example 4.2.5. We have 20 = 10 · 2 ≡ 4 · 2 = 8(mod 6). But 5 ≡ 2(mod 6). However, the next theorem provides us with a well-founded congruence when both sides of a congruence are divided by the same integer. 70 Number Theory and its Applications Theorem 4.2.2. If a, b, c and m are integers such that m > 0, d = gcd(c, m) m . and ac ≡ bc(mod m), then a ≡ b mod d Proof. Here ac ≡ bc(mod m) implies m\|(ac−bc) = c(a−b), which further implies there exists an integer k satisfying c(a − b) = km. Dividing both sides by d, c m km m c . Since gcd , = 1, it follows \|(a − b) ⇒ a ≡ b we have (a − b) = d d d d d m mod . d Theorem(4.2.2) has a corollary that is worth a separate statement. Corollary 4.2.1. For any arbitrary positive integers a and b, lcm(a, b) = ab if and only if gcd(a, b) = 1. Proof. Obvious. 5 15 ≡ Example 4.2.6. Since 15 = 5(mod 10) and gcd(5, 10) = 5, we see that 5 5 10 mod or 3 ≡ 1(mod 2). 5 Example 4.2.7. Since 42 ≡ 7(mod 5) and gcd(5, 7) = 1, we can conclude that 7 42 = mod 5 , or that 6 ≡ 1(mod 5). 7 7 In our next theorem using the principle of mathematical induction we have shown that if we increase the exponential power of elements of both sides of a congruence then the congruence relation is preserved. Proposition 4.2.2. Let a, b be any two integers. For some integer m > 0, if a ≡ b(mod m) holds then for any positive integer n, an ≡ bn (mod m) is also true. Proof. We are going to prove this theorem by the principle of mathematical induction. As a ≡ b(mod m) then the result is obviously true for n = 1. Let us assume the result is true for n = k. Then ak ≡ bk (mod m) holds. Now using the property(2) of Theorem (4.2.1)we have ak+1 ≡ bk+1 (mod m). Thus the result is true for n = k + 1. Therefore from the principle of mathematical induction the result is true for all n. Example 4.2.8. Here in this example we have tried to clarify the above proposition by an example. For that let us choose 8 ≡ 3(mod 5) then for n = 3 we see that 83 = 512 ≡ 27 = 33 (mod 5). In our following theorem we have shown the way to combine congruences of two same numbers with diﬀerent congruent moduli. To prove this theorem, ﬁrst we need to prove the following result. Theory of Congruences 71 Result 4.2.1. Let a be any integer and n1 , n2 be two positive integers with n1 \|a, n2 \|a. Then lcm(n1 , n2 )\|a. Proof. Let l be the least common multiple of n1 and n2 . If l a, then the division algorithm yields m = lq + r for some integers q and r where 0 ≤ r < l. Then r = m − lq. As l and m are multiples of a and b, then there exists integers t1 , t2 t1 , t2 such that m = at1 = bt1 , l = at2 = bt2 . Therefore r = at1 − at2 q = a(t1 − t2 q) and r = bt1 − bt2 q = b(t1 − t2 q). This shows that r is a multiple of both a and b. As l is least, then r ≥ l. This contradicts the fact 0 < r < l. Therefore m = lq ⇒ l\|m. Now the proof of the main theorem as follows. Theorem 4.2.3. For any integers a and b with positive integers t1, t 2,· · · tk if a≡b(modt1), a ≡b(modt2),· · · , a ≡b(modtk)thena≡b(modlcm(t1, t 2,· · · , t k)). Proof. Since a ≡ b(mod t1 ), a ≡ b(mod t2 ), · · · , a ≡ b(mod tk ) then we have, t1 \|(a − b), t2 \|(a − b), · · · tk \|(a − b). Now by above result we can say that lcm(t1 , t2 , · · · , tk )\|(a − b). This implies a ≡ b(mod lcm(t1 , t2 , · · · , tk )). In next corollary, we are going to describe an useful consequence of the above theorem. Corollary 4.2.2. For any integers a and b with positive relatively prime integers t1 , t2 , · · · tk if a ≡ b(mod t1 ), a ≡ b(mod t2 ), · · · , a ≡ b(mod tk ) then a ≡ b( mod (t1 t2 · · · tk )). Proof. Since a ≡ b(mod t1 ), a ≡ b(mod t2 ), · · · , a ≡ b(mod tk ), therefore t1 \|(a− b), t2 \|(a − b), · · · tk \|(a − b). As t1 , t2 , · · · tk are relatively prime integers, therefore lcm(t1 , t2 , · · · , tk ) = t1 t2 · · · tk . Then Theorem 4.2.3 gives a ≡ b(mod (t1 t2 · · · tk )). In the following proposition we have seen that the congruence relation is nothing but an equivalence relation. Proposition 4.2.3. Let m be any non-zero integer. Deﬁne a relation ‘ ≡ mod m’ on set of integers Z by a ≡ b(mod m) if and only if m\|(a − b). The relation ‘ ≡ mod m’ is an equivalence relation. Proof. A relation on a set is said to be equivalence if it is reﬂexive, symmetric and transitive. 1. Reﬂexivity: Since m\|(a − a), we see that a ≡ a(mod m). 72 Number Theory and its Applications 2. Symmetricity: If a ≡ b(mod m), then m\|(a − b). Hence there exists an integer l such that a − b = lm. This shows that (−l)m = b − a ⇒ m\|(b − a). Consequently, b ≡ a(mod m). 3. Transitivity: Let a ≡ b(mod m) and b ≡ c(mod m). Then m\|(a − b) and m\|(b − c) holds. Hence there exists integers k and l such that a − b = km and b − c = lm. Therefore a − c = (a − b) + (b − c) = km + lm = (k + l)m. Consequently, m\|(a − c) implies a ≡ c(mod m). Infact this equivalence relation is also called congruence relation. From the basic concept of algebra we can say that this equivalence relation always forms an equivalence class. In this case this is called congruence class. For example if we choose a positive integer 5 which leaves the remainder 0, 1, 2, 3, 4 when divides any integer. Here if we choose remainder as 1 then we have the set of integers {6, 11, 16, · · · } whose all the elements have remainder 1 when divided by 5. For that the above set can be written as [1] which is a congruence class modulo 5. Thus the deﬁnition of congruence class as follows. Deﬁnition 4.2.2. Let m be a positive integer and a be any integer then set of integers, {b : b ≡ a(mod m)} is called congruence class modulo m and denoted by [a]. From the above deﬁnition of congruence class another important fact we can discuss on integers. If we choose a set of integers say, {5, 21, −2, 62, 34} then for congruent modulo 5 we have, 5 ≡ 0(mod 5), 21 ≡ 1(mod 5), 62 ≡ 2( mod 5), −2 ≡ 3(mod 5), 34 ≡ 4(mod 5). Here we see that each of the elements of the above set are congruent modulo 5 with exactly one of the set {0, 1, 2, 3, 4}. Then this arbitrary set {5, 21, −2, 62, 34} is said to be a complete set of residue modulo 5. Now we are in the position to deﬁne that arbitrary set. Deﬁnition 4.2.3. An arbitrary set of m integers {a1 , a2 , · · · , am } is said to be a complete set of residue modulo m or CRS(mod m) if every integer of the set is congruent modulo m to exactly one of a1 , a2 , · · · , am . More speciﬁcally if, 1. ai ≡ aj (mod m), ∀i = j, i, j = 1, 2, · · · m 2. For each integer n, there exists a unique integer aj such that n ≡ aj ( mod m), j = 1, 2, · · · m. Obviously the set {0, 1, 2, · · · , m−1} forms a CRS(mod m). It is called trivial CRS(mod m). For an example if we choose m = 5 then the set {0, 1, 2, 3, 4} is the trivial CRS(mod 5). Theory of Congruences 73 Now in the following theorems we have shown here that addition and multiplication of any arbitrary element with all the elements of a complete residue system under some conditions preserves the properties of complete residue system. Theorem 4.2.4. If {a1 , a2 , · · · , am } is a set of complete residue system modulo m and c be any integer then {a1 + c, a2 + c, · · · , am + c} is also a set of complete residue system modulo m. Proof. It’s suﬃces to show that all the elements of {a1 +c, a2 +c, · · · , am +c} are distinct under congruent modulo m. Since {a1 , a2 , · · · , am } is a set of complete aj for i = j where i, j = 1, 2, · · · m. Thus residue system modulo m then ai ≡ ai − aj is not divisible by m. Also, (ai + c) − (aj + c) = ai − aj which follows that (ai + c) − (aj + c) is not divisible by m. Therefore ai + c ≡ aj + c(mod m) j where i, j = 1, 2, · · · m. This proves the theorem. for i = Theorem 4.2.5. If {a1 , a2 , · · · , am } is a set of complete residue system modulo m and c be any integer prime to m, then {ca1 , ca2 , · · · , cam } is also a set of complete residue system modulo m. Proof. Again here to prove this theorem we are to show all the elements of {ca1 , ca2 , · · · , cam } are distinct under congruent modulo m. Since {a1 , a2 , · · · , am } is a set of complete residue system modulo m then ai ≡ aj for i = j where i, j = 1, 2, · · · m. Thus ai −aj is not divisible by m. Also we have cai −caj = c(ai −aj ). Now c is prime to m implies gcd(c, m) = 1 and ai − aj is not divisible by m. Combining these two concepts we can conclude that cai − caj is not divisible by caj (mod m) for i = j where i, j = 1, 2, · · · m. This proves m. Therefore cai ≡ the assertion of this theorem. Combining the above two theorems, lead us to the following straightforward corollary: Corollary 4.2.3. If {a1 , a2 , · · · , am } is a set of complete residue system modulo m and c be any integer prime to m, then {ca1 + d, ca2 + d, · · · , cam + d} is also a set of complete residue system modulo m for any integer d. 4.3 Worked out Exercises Problem 4.3.1. Give an example to show that a2 = b2 (mod n) need not imply that a ≡ b(mod n). 74 Number Theory and its Applications Solution 4.3.1. Let us consider a = 5, b = 4, m = 3. Since 3 (25 − 16) = 9, therefore 52 ≡ 42 (mod 3). But 5 ≡ 4(mod 3). Problem 4.3.2. What is the remainder when the sum 15 + 25 + 35 + . . . + 995 + 1005 is divided by 4? Solution 4.3.2. Here 15 ≡ 1( mod 4) 1 ≡ 5 ≡ 9 . . . ( mod 4) 32 = 25 ≡ 0( mod 4) 2 ≡ 6 ≡ 10 . . . ( mod 4) 243 = 3 ≡ 3( mod 4) 3 ≡ 7 ≡ 11 . . . ( mod 4) 4 ≡ 0( mod 4) 4 ≡ 8 ≡ 12 . . . ( mod 4). 5 5 Each block of four numbers will have same remainder sum. Since 15 + 25 + 35 + 45 ≡ 1 + 0 + 3 + 0 ≡ 4 ≡ 0(mod 4), therefore 25 blocks will all have remainder 0 implies entire remainder is 0. Problem 4.3.3. For n ≥ 1, use congruence theory to establish 27 (25n+1 + 5n+2 ). Solution 4.3.3. Here 32 ≡ 5(mod 27) ⇒ 25 ≡ 5(mod 27). Now 25n ≡ 5n ( mod 27) 2 · 25n ≡ 2 · 5n ( mod 27). ∴ 25n+1 + 25n+2 ≡ 2 · 5n + 5n+2 ( mod 27) ≡ 5n (5 + 25)( mod 27) ≡ 5n · 27( mod 27) ≡ 0( mod 27). Problem 4.3.4. Find the remainder when the sum 1! + 2! + 3! + . . . + 100! is divided by 18. Solution 4.3.4. Note that 6! ≡ 0(mod 18) ⇒ (6 + n)! ≡ 0(mod 18) for n ∈ Z. Then 1! + 2! + 3! + · · · + 100! ≡ (1! + 2! + 3! + 4! + 5!)( mod 18) ≡ 153( mod 18) ≡ 9( mod 18). Therefore the remainder is 9. Problem 4.3.5. Prove for any integer a, a3 ≡ 0, 1, or 6(mod 7). Theory of Congruences 75 Solution 4.3.5. By division Algorithm, we have a = 7k + r, 0 ≤ r < 7. Now a = 7k : a3 = (7k)3 = 7 · 72 k 3 ⇒ a3 ≡ 0( mod 7). a = 7k + 1 : a3 = (7k + 1)3 = (7k)3 + 3 · (7k)2 + 3 · 7k + 1. ∴ a3 − 1 = 7[72 k 3 + 3 · 7k 2 + 3 · k] ⇒ a3 ≡ 1( mod 7) By similar way, a = 7k + 2 : a3 ≡ 1( mod 7) a = 7k + 3 : a3 ≡ 6( mod 7) a = 7k + 4 : a3 ≡ 1( mod 7) a = 7k + 5 : a3 ≡ 6( mod 7) a = 7k + 6 : a3 ≡ 1( mod 7). Problem 4.3.6. If {a1 , a2 , . . . , an } is a complete set of residues modulo n and gcd(a, n) = 1, prove that {aa1 , aa2 , . . . , aan } is also a complete set of residues modulo n. Solution 4.3.6. Consider aai and aaj with i = j, 1 ≤ i < j ≤ n. If aai and aaj are congruent moduli n, then aai − aaj = kn ⇒ a(ai − aj ) = kn for some k. Since gcd(a, n) = 1, Euclid’s Lemma gives n (ai − aj ), contradicting the aj . Therefore aai ≡ aaj . Hence by Proposition 4.2.1, the statement fact ai ≡ follows. Problem 4.3.7. Find the remainder when 10515 is divided by 7. Solution 4.3.7. Here 515 = 85 × 6 + 5 ⇒ 10515 = (106 )85 · 105 . Further, 102 ≡ 2( mod 7) ⇒ 106 ≡ 23 ≡ 1( mod 7) ⇒ (106 )85 ≡ 1( mod 7) ⇒ 10515 (106 )85 · 105 ≡ 1 · 5 ≡ 5( mod 7). So the desired remainder is 5. Problem 4.3.8. Verify that if a ≡ b(mod n1 ) and a ≡ b(mod n2 ), then a ≡ b(mod n), where the integer n = lcm(n1 , n2 ). Hence whenever n1 & n2 are relatively prime, a ≡ b(mod n1 n2 ). Solution 4.3.8. Let k1 , k2 ∈ Z be such that a − b = k1 n1 , & a − b = k2 n2 . Let d = gcd(n1 , n2 ). Then ∃ r ∈ Z such that n1 = dr. ∴ a − b = k2 n2 = k2 n2 n1 k n n = 2 1 2. dr r d 76 But Number Theory and its Applications n1 n2 d = lcm(n1 , n2 )[refer to Theorem 2.5.1]. ∴ a−b= k2 lcm(n1 , n2 ). r k Finally, our aim is to show 2 ∈ Z. Let s ∈ Z be such that n2 = ds. Since r a − b = k1 n1 = k2 n2 , then k1 dr = k2 ds ⇒ k1 r = k2 s. Since gcd(r, s) = 1, k therefore r divides k2 . This shows that r2 ∈ Z. Problem 4.3.9. Show that 41 divides 240 − 1. Solution 4.3.9. Here 220 = (25 )4 = (32)4 . This shows that 220 = (32)4 ≡ (−9)4 ≡ (81)2 (mod 41). However 81 ≡ −1(mod 41) ⇒ 220 ≡ 1(mod 41). Hence 41 (220 − 1). Problem 4.3.10. Justify, ak ≡ bk (mod n) and k ≡ j(mod n) need not imply that aj ≡ bj (mod n). Solution 4.3.10. Since 4 ≡ 9(mod 5), therefore 22 ≡ 32 ( mod 5), 2 ≡ 7( mod 5), 27 ≡ 37 ( mod 5)[Verify!]. Problem 4.3.11. If gcd(a, n) = 1, then prove that the integers c, c + a, c + 2a, c + 3a, . . . , c + (n − 1)a form a complete set of residues modulo n for any c. Solution 4.3.11. Consider c + ra & c + sa, r = s, 0 ≤ r, s ≤ n − 1. Suppose s > r. ∴ c + sa − (c + ra) = (s − r)a. Note that s ≤ n − 1, r ≤ n − 1 together implies s − r < n. Therefore n (s − r). Since gcd(a, n) = 1, therefore k ∈ Z such that (s − r)a = nk ⇒ c + sa ≡ c + ra. This completes the solution. Problem 4.3.12. Find all CRS modulo 6. Solution 4.3.12. Here the set {0, 1, 2, 3, 4, 5} forms trivial CRS modulo 6. By virtue of Theorem(4.2.4)and Theorem(4.2.5), we conclude that {k, k + a, k + 2a, k + 3a, k + 4a, k + 5a} forms a CRS modulo 6, where k is any arbitrary integer and a is an integer prime to 6. Problem 4.3.13. Prove that the integer 53103 + 10353 is divisible by 39. Theory of Congruences 77 Solution 4.3.13. Note that 39 = 3·13, 53 = 3·17+2 = 3·18−1, 103 = 34·3+1. Now 53 ≡ −1( mod 3) 53103 ≡ (−1)103 ( mod 3) 53 ≡ 1( mod 13) 53103 ≡ 1( mod 13). Furthermore, 103 ≡ 1( mod 3) 53 103 ≡ 1 ( mod 3) 53 103 ≡ −1( mod 13) 10353 ≡ −1( mod 13). Adding those congruences with respect to modulo 3 and modulo 13 we get, 53103 + 10353 ≡ 0(mod 3) and 53103 + 10353 ≡ 0(mod 13) respectively. This yields 3 (53103 + 10353 ), 13 (53103 + 10353 ). Since gcd(3, 13) = 1, therefore 3 · 13 = 103 39 (53 + 10353 ). 4.4 Linear Congruences The present section deals with the notion of linear equation in the sense of congruence relation. Consider a linear equation of the form 2x + 3y = 5 with 5 − 2x . If we two unknown integers x and y. Then it can be expressed as y = 3 5 − 2x consider as an integer then the above linear equation can be written as 3 2x ≡ 5(mod y). The foregoing congruence relation with unknown integer x is said to be linear congruence equation, whose deﬁnition is as follows: Deﬁnition 4.4.1. A congruence of the form ax ≡ b(mod m) where a, b, m are integers with m > 0 and x an unknown integer, is called linear congruence in one variable. Here we have dealt with the various aspects of linear congruences. In the beginning, we have tried to relate linear congruences with the linear Diophantine equation of two variables x and y. Our following theorem is based on that. Theorem 4.4.1. Let (x0 , y0 ) be an integral solution of ax + by = c for some integers a, b, c where a, b are non zero integers then x0 is the solution of ax ≡ c(mod m) considering m = \|b\|. Conversely, if x0 is a solution of the above congruence then there is an integer y0 for which (x0 , y0 ) is a solution of ax+by = c. 78 Number Theory and its Applications Proof. Since (x0 , y0 ) satisﬁes ax + by = c then we have by0 = c − ax0 . This shows that b divides ax0 − c. Therefore m = \|b\| divides ax0 − c and x0 becomes a solution of ax ≡ c(mod m). For the converse part we have x0 a solution of ax ≡ c(mod m). Since m = \|b\| divides ax0 − c then for some integer y0 we can write ax0 − c = by0 . This proves that (x0 , y0 ) satisﬁes ax + by = c. Now we are going to illustrate the above fact by following examples. Example 4.4.1. Here we have shown that a linear Diophantine equation 221x+ 35y = 11 can be solved using linear congruence. Firstly the equation 221x + 35y = 11 has been written as 221x ≡ 11(mod 35). Here the solution of this congruence equation is x ≡ 1(mod 35). Then we have x = 1 + 35t for some 1 [11 − 221 · 1] = −6 integer t. Here x0 = 1 is the particular value of x and y0 = 35 is particular value of y . Therefore y = −6 − 221t, x = 1 + 35t is the general solution. Example 4.4.2. Let us choose the congruence equation 5x ≡ 2(mod 26) and this has been written as 5x − 26y = 2 for some integer y. Here gcd(5, 26) = 1 can be written as 1 = 26 − 5 · 5. Thus here the particular value of x is x0 = −10. Then we have x = −10 − 26t for some integer t. Therefore x ≡ −10 ≡ 16( mod 26) is the solution of above congruence. Here in the Example 4.4.1 we have solved the linear Diophantine equation by converting it to linear congruence equation and also from the Example 4.4.2 we have solved the linear congruence equation by converting it to linear Diophantine equation. So from the above two examples we can say that the linear congruences and linear diophantine equations are relatable. In particular, we have seen that x = x0 is a solution of ax ≡ b(mod m) then any integer x1 ≡ x0 (mod m) is also a solution. Thus if we can ﬁnd a particular solution x0 of ax ≡ b(mod m), then all the elements belonging to the class of x0 , are the solutions of ax ≡ b(mod m). For instance, choose 4x ≡ ( mod 5) where x = 2 is a solution. Now it’s obvious that all the elements of [2] such as x = 7, 12 and so on are its solutions. Now the question arises, how many incongruent solutions modulo m do exist?. The following theorem reﬂects, under which condition it is possible to ﬁnd a solution of a linear congruence equation and if the solutions exist, how many of them are incongruent modulo m. Theorem 4.4.2. Let a, b, m are integers with m > 0 then the linear congruence ax ≡ b(mod m) has a solution if and only if d\|b, where d = gcd(a, m). If d\|b then it has exactly d numbers of incongruent solutions. Theory of Congruences 79 Proof. Theorem 4.4.1 asserts that any linear congruence ax ≡ b(mod m) is equivalent to linear Diophantine equation ax − my = b, for any integer y. So for particular integer x0 satisfying ax0 ≡ b(mod m) we get an integer y0 satisfying ax0 − my0 = b. Again by virtue of Theorem 2.7.1, if d b then any solutions. Also, if d\|b then the Diophantine equation ax + mk = b have inﬁnite number of m a n, k = k0 − n for some integer n. Here k0 is solutions given by x = x0 + d d a particular value for k. Then comparing both the diophantine equations, yields y = −k. Thus, the solutions of ax − my = b are given by m a a n, −y = −k0 − n ⇒ y = k0 + n. d d d x = x0 + Next, to determine the number of incongruent solutions of ax ≡ b(mod m), m m n and x2 = x0 + n as two solutions of ax ≡ b( consider x1 = x0 + d 1 d 2 mod m) for some integers n1 , n2 . If these two are congruent then, m m n1 ≡ x0 + n ( mod m). x0 + d d 2 m m ∴ n ≡ n ( mod m) d 1 d 2 m m m m. So using Theorem 4.2.2 we obtain n ≡ n ( = and Now, gcd m, 1 2 d d d m mod m). This proves that x = x0 + n has exactly d numbers of incongruent d solutions as n ranges through a complete residue system of residues modulo d. In the above theorem, taking a and m as relatively prime integers gives a straightforward corollary: Corollary 4.4.1. If a and m are relatively prime then the linear congruence ax ≡ b(mod m) has a unique solution modulo m. Next our aim is to exemplify the foregoing theorem and corresponding corollary by an example: Example 4.4.3. Consider the linear congruence 8x ≡ 16(mod 24). Since gcd(8, 24) = 8 and 8\|16, using the last theorem our aim is to show that ∃ 8 incongruent solutions modulo 24. Here x0 = 2 is a particular solution. Then 24 x ≡ 2+ t ≡ 2 + 3t(mod 24) are the incongruent solutions modulo 24 where 8 t = 0(1)7. Thus the solutions are x ≡ 2, 5, 8, 11, 14, 17, 20, 23(mod 24). Now, if we choose the congruence 8x ≡ 16(mod 23) then gcd(8, 23) = 1. Then by virtue of the last corollary, it has only one incongruent solution modulo 23 which is x ≡ 2(mod 23). 80 Number Theory and its Applications Example 4.4.4. Consider the linear congruence 8x ≡ 16(mod 24). Since gcd(8, 24) = 8 and 8\|16, using the last theorem our aim is to show that ∃, 8 incongruent solutions modulo 24. Here x0 = 2 is a particular solution. Then 24 x ≡ 2+ t ≡ 2 + 3t(mod 24) are the incongruent solutions modulo 24 where 8 t = 0(1)7. Thus the solutions are x ≡ 2, 5, 8, 11, 14, 17, 20, 23(mod 24). Now, if we choose the congruence 8x ≡ 16(mod 23) then gcd(8, 23) = 1. Then by virtue of the last corollary, it has only one incongruent solution modulo 23 which is x ≡ 2(mod 23). After solving a linear congruence equation, we are turning our discussion to solve a simultaneous system of linear congruences. This system actually came from Chinese puzzles as early as the ﬁrst century A.D. In number theory, the Chinese remainder theorem gives a unique solution to simultaneous linear congruences with coprime moduli. In its basic form, the Chinese remainder theorem will determine a number p that, when divided by some given divisors, leaves given remainders. The earliest known statement of the theorem is by the Chinese mathematician Sun-tzu Suan-ching in the 3rd century AD, whose original formulation was x ≡ 2(mod 3) ≡ 3(mod 5) ≡ 2(mod 7) with the solution x = 23 + 105k where k ∈ Z. The Chinese remainder theorem is widely used for computing with large integers, as it allows replacing a computation for which one knows a bound on the size of the result by several similar computations on small integers. Theorem 4.4.3. (Chinese Remainder Theorem): Let m1 , m2 , · · · mk be pairwise relatively prime integers. Then for k number of integers a1 , a2 , · · · ak the system of congruences x ≡ a1 (mod m1 ), x ≡ a2 (mod m2 ), · · · x ≡ ak (mod mk ) has a unique solution modulo M = m1 m2 · · · mk . M = m1 m2 · · · mr−1 mr+1 · · · mk is the product of all integers mr omitting mr , shows that gcd(Mr , mr ) = 1. Then from the Corollary 4.4.1 it is possible to ﬁnd a unique solution xr of the linear congruence Mr x ≡ 1(mod mr ). Our task is to show that the integer x ˜ = a1 M1 x1 + a2 M2 x2 + · · · ak Mk xk is a simultaneous solution of the above system. First our aim is to check, x ˜ satisﬁes x ≡ a1 (mod m1 ). Since all the integers M2 , M3 , · · · Mk contain m1 as a factor ˜ ≡ a1 M1 x1 (mod m1 ). As so ai Mi xi ≡ 0(mod m1 ) for all i = 2, 3, · · · k, then x M1 x1 ≡ 1(mod m1 ) it follows that x ˜ ≡ a1 (mod m1 ). This shows that x ˜ satisﬁes the linear congruence x ≡ a1 (mod m1 ). Proceeding as above, we can show that x ˜ also satisﬁes other congruences. Proof. Let Mr = Theory of Congruences 81 To proceed for the uniqueness part, let x and x ˜ be its two solutions. Then ˜(mod mr ) for r = 1, 2, · · · , k. Therefore mr divides (x − x ˜) we have x ≡ ar ≡ x for each r = 1, 2, · · · , k. Since all mr ’s are relatively prime then from Corollary 2.4.1 we have M = m1 m2 · · · mk \|(x − x ˜). This implies x ≡ x ˜(mod M = ˜ is the unique solution of the given system. m1 m2 · · · mk ). Therefore x In the following example, we have illustrated the preceding theorem lucidly. Example 4.4.5. Let us consider a system of simultaneous linear congruences as x ≡ 2(mod 3), x ≡ 4(mod 5), x ≡ 5(mod 7). Here M = 3 · 5 · 7 = 105 105 105 105 then we have M1 = = 35, M2 = = 21, M3 = = 15. As 3 5 7 Mr xr ≡ 1(mod mr ) so the linear congruences are 35x1 ≡ 1(mod 3), 21x2 ≡ 1( mod 5), 15x3 ≡ 1(mod 7). Those linear congruences are satisﬁed by x1 = 2, x2 = 1, x3 = 1 respectively. Thus a solution of the system is given by x ˜ = a1 M1 x1 + a2 M2 x2 + a3 M3 x3 = 140 + 84 + 75 ≡ 299(mod 105). Thus the unique solution of this system is x ˜ ≡ 89(mod 105). 4.5 Worked out Exercises Problem 4.5.1. Solve: (1) 36x ≡ 8(mod 102) (2) 140x ≡ 133(mod 301). Solution 4.5.1. 1. Since gcd(36, 102) = 6 8, therefore any solution. 2. Here 140 = 22 · 5 · 7, 301 = 7 × 43. Therefore gcd(140, 301) = 7 and 7\|133. Hence 7 incongruent solutions do exist. Dividing both sides of the congruence by 7 we have, 20x ≡ 19( mod 43) 40x ≡ 38 43x − 40x ≡ 43 − 38( mod 43) 3x ≡ 5( mod 43) 42x ≡ 70( mod 43) 43x − 42x ≡ 86 − 70( mod 43) x ≡ 16( mod 43). ∴ x ≡ 16 + 43t, for t = 0, 1, 2, 3, 4, 5, 6. ∴ x ≡ 16, 59, 102, 145, 188, 231, 274( mod 301). Problem 4.5.2. Using congruences, solve the Diophantine equations: 12x + 25y = 331. 82 Number Theory and its Applications Solution 4.5.2. Note that 12x ≡ 331( mod 25), or, 24x ≡ 662( mod 25), or, 25x − 24x ≡ 662 − 650( mod 25), or, x ≡ 12( mod 25). ∴ x = 12 + 25u, ∀u ∈ Z. Further, 25y ≡ 331( mod 12), or, 25y − 24y ≡ 331 − 324( mod 12), or, y ≡ 7( mod 12). ∴ y = 7 + 12v, ∀v ∈ Z. ∴ 12x + 25y = 12(12 + 25u) + 25(7 + 12v), or, 331 = 319 + 300u + 300v, or, 12 = 25u + 25v. ∴ x = 12 + 25u = 24 − 25v. Hence x = 24 − 25v, y = 7 + 12v for v ∈ Z. Problem 4.5.3. Solve: x ≡ 5(mod 6), x ≡ 4(mod 11), x ≡ 3(mod 17). Solution 4.5.3. Here x ≡ 5( mod 6) N = 6 · 11 · 17 = 1122. x ≡ 4( mod 11) N1 = 11 · 17 = 187. x ≡ 3( mod 17) N2 = 6 · 17 = 102. N3 = 6 · 11 = 66. Now 187x1 ≡ 1(mod 6) ⇒ 187x1 − 186x1 = x1 ≡ 1(mod 6). Again 102x2 ≡ 1( mod 11) 102x2 − 99x2 = 3x2 ≡ 1( mod 11) 21x2 ≡ 7( mod 11) 21x2 − 22x2 = −x2 ≡ 7( mod 11) 66x3 ≡ 1( mod 17) 66x3 − 68x3 = −2x3 ≡ 1( mod 17) 18x3 ≡ −9( mod 17) 18x3 − 17x3 = x3 ≡ −9( mod 17). ∴ x1 = 1, x2 = −7, x3 = −9. ∴ a1 N1 x1 = 5 · 187 · 1, a2 N2 x2 = 4 · (102) · (−7), a3 N3 x3 = 3 · (66) · (−9). ∴ a1 N1 x1 + a2 N2 x2 + a3 N3 x3 = −3703. ∴ x ≡ −3703 + 4 · 1122 = 785( mod 1122). Theory of Congruences 83 Problem 4.5.4. Obtain three consecutive integers, each having a square factor. Solution 4.5.4. Note that a ≡ 0(mod 22 ), a + 1 ≡ 0(mod 32 ), a + 2 ≡ 0( mod 52 ). Since 22 , 32 and 55 are relatively prime to each other, therefore by virtue of Chinese Remainder Theorem we ﬁnd a ≡ 0( mod 4) N = 4 · 9 · 25 = 900 a ≡ −1( mod 9) N1 = 9 · 25 = 225 a ≡ −2( mod 25) N2 = 4 · 25 = 100 N3 = 4 · 9 = 36. Now 225x1 ≡ 1(mod 4) ⇒ 225x1 − 224x1 = x1 ≡ 1(mod 4). Again 100x2 ≡ 1( mod 9) 36x3 ≡ 1( mod 25) 100x2 − 99x2 ≡ 1( mod 9) 72x3 ≡ 2( mod 25) x2 ≡ 1( mod 9) 72x3 − 75x3 ≡ −3( mod 25) 3x3 ≡ −2( mod 25) 24x3 ≡ −16( mod 25) 24x3 − 25x3 = −x3 ≡ −16( mod 25) x3 ≡ 16( mod 25). ∴ a1 N1 x1 + a2 N2 x2 + a3 N3 x3 = −1252. ∴ x ≡ 548( mod 900). Thus the desired three consecutive numbers are 548, 549, 550. Problem 4.5.5. Prove that the congruences x ≡ a(mod n) and x ≡ b(mod m) admit a simultaneous solution if and only if gcd(n, m) (a−b); if a solution exists, conﬁrm that it is unique modulo lcm(n, m). Solution 4.5.5. Suppose there exists a solution for x. Let d = gcd(n, m). This implies ∃ r, s ∈ Z such that n = dr, m = ds. Now x ≡ a( mod n) ⇒ x = a + nt, t ∈ Z, x ≡ b( mod m) ⇒ x = b + mk, k ∈ Z. ∴ a + nt = b + mk ⇒ nt − mk = b − a. Substituting for m, n we obtain d(sk − rt) = a − b ⇒ d = gcd(n, m) (a − b). 84 Number Theory and its Applications Next, let us assume d = gcd(m, n) and d\|(a − b). Then for some t ∈ Z, dt = a − b ⇒ ∃ x0 , y0 such that nx0 + my0 = d. Therefore dt = nx0 t + my0 t = a − b ⇒ my0 t + b = a − x0 tn. Let x ≡ a(mod n), x ≡ b(mod m). So ∃ a simultaneous solutions. Let y be any other solution. Then x ≡ a( mod n) y ≡ a( mod n). x ≡ b( mod m) y ≡ b( mod m). ∴ x ≡ y( mod n). x ≡ y( mod m). By virtue of worked out Problem 4.3.8, we obtain x ≡ y(mod lcm(m, n)). Problem 4.5.6. A certain integer between 1 and 1200 leaves the remainders 1, 2, 6 when divided by 9, 11, 13, respectively. What is the integer? Solution 4.5.6. From the given conditions, we have x ≡ 1( mod 9), 1 < x < 1200. x ≡ 2( mod 11), x ≡ 6( mod 13). Since 9, 13, 11 are relatively prime, therefore Chinese Remainder Theorem is applicable here. Rest proceeding similarly as in Problem 4.5.4, we obtain the integer 838. Problem 4.5.7. Obtain the two incongruent solutions modulo 210 of the system: 2x ≡ 3( mod 5) 4x ≡ 2( mod 6) 3x ≡ 2( mod 7). Solution 4.5.7. Here 2x ≡ 3( mod 5) (4.5.1) 4x ≡ 2( mod 6) (4.5.2) 3x ≡ 2( mod 7). (4.5.3) From(4.5.1), 4x ≡ 6( mod 5), 4x − 5x = x ≡ 1( mod 5), x ≡ −1 + 5( mod 5), x ≡ 4( mod 5). x 2 6 ≡ ( mod ), 2 2 2 2x ≡ 1( mod 3), From(4.5.2), 4 4x ≡ 2( mod 3), 4x − 3x = x ≡ 2( mod 3), x ≡ 2( mod 6). Theory of Congruences 85 Since gcd(4, 6) = 2, therefore from Theorem 4.4.2 we can say that there ∃ 2 incongruent solutions given by x0 , x0 + 62 , x0 being a solution. Here x0 = 2 is a solution, so 5 is the other. Therefore x ≡ 5(mod 6) is the other congruence equation. From (4.5.3), we obtain 6x ≡ 4( mod 7), 6x − 7x = −x ≡ −3( mod 7), −x ≡ −3( mod 7), ∴ x ≡ 3( mod 7). Therefore x ≡ 4(mod 5), x ≡ 2(mod 6) or x ≡ 5(mod 6), x ≡ 3(mod 7). Note that N = 5 · 6 · 7 = 210. Therefore N1 = 6 · 7 = 42, N2 = 5 · 7 = 35 and N3 = 5 · 6 = 30. Thus 42x1 ≡ 1( mod 5) 35x2 ≡ 1( mod 6) 42x1 − 40x1 = 2x1 ≡ 1( mod 5). 35x2 − 36x2 = −x2 ≡ 1( mod 6) 6x1 ≡ 3( mod 5) x2 ≡ 5( mod 6). 6x1 − 5x1 = x1 ≡ 3( mod 5) x1 ≡ 3( mod 5). 30x3 ≡ 1( mod 7) 30x3 − 28x3 = 2x3 ≡ 1( mod 7) 8x3 ≡ 4( mod 7) 8x3 − 7x3 = x3 ≡ 4( mod 7). Therefore a1 N1 x1 + a2 N2 x2 + a3 N3 x3 = 1214 or 1739(Verify!). Thus the solutions are x ≡ 164(mod 210) or x ≡ 59(mod 210). Problem 4.5.8. Obtain the eight incongruent solutions of the linear congruence 3x + 4y ≡ 5(mod 8). Solution 4.5.8. Set 3x ≡ 5 − 4y(mod 8). Since gcd(3, 8) = 1 and 1\|(5 − 4y), there exists one solution for any value of y. Because there are eight incongruent values of 5 − 4y(y = 0, 1, 2, 3, 4, 5, 6, 7), let us solve this for each values of y. First, let us take y ≡ 0(mod 7). Then 3x ≡ 5( mod 8) 15x ≡ 25( mod 8) 16x − 15x = x ≡ −1 ≡ 7( mod 8). 86 Number Theory and its Applications By similar reasoning, x ≡ 3(mod 8) for y ≡ 1(mod 8), x ≡ 7(mod 8) for y ≡ 2( mod 8), x ≡ 3(mod 8) for y ≡ 3(mod 8), x ≡ 7(mod 8) for y ≡ 4(mod 8), x ≡ 3(mod 8) for y ≡ 5(mod 8), x ≡ 7(mod 8) for y ≡ 6(mod 8) and x ≡ 3( mod 8) for y ≡ 7(mod 8). Problem 4.5.9. The basket-of-eggs problem is often phrased in the following form: One egg remains when the eggs are removed from the basket 2, 3, 4, 5, or 6 at a time; but, no eggs remain if they are removed 7 at a time. Find the smallest number of eggs that could have been in the basket. Solution 4.5.9. From the given conditions, we have x ≡ 1( mod 2) (4.5.4) x ≡ 1( mod 3) (4.5.5) x ≡ 1( mod 4) (4.5.6) x ≡ 1( mod 5) (4.5.7) x ≡ 1( mod 6) (4.5.8) x ≡ 0( mod 7). (4.5.9) If (4.5.6) is true, then x = 1 + 4n = 1 + 2(2n). Since gcd(2, 4) = 1, therefore we can eliminate (4.5.4). Moreover, if (4.5.8) is true, then x = 1 + 6n = 1 + 3(2n). Because gcd(3, 6) = 1, whence we can eliminate (4.5.5). Multiplying (4.5.6) by 3 and (4.5.8) by 2, we obtain 3x ≡ 3( mod 3 · 4) = 3( mod 12) (4.5.10) 2x ≡ 2( mod 2 · 6) = 2( mod 12). (4.5.11) ∴ 3x − 3 ≡ 2x − 2( mod 12), x ≡ 1( mod 12). (4.5.12) If (4.5.12) holds true, then (4.5.6) and (4.5.8) is also so. Now we have x ≡ 1( mod 5), x ≡ 0(mod 7) and x ≡ 1(mod 12). Note that 5, 7, 12 are relatively prime. Thus N = 5 · 7 · 12 = 420. Therefore N1 = 7 · 12 = 84, N2 = 5 · 12 = 60 and N3 = 5 · 7 = 35. Hence 84x1 ≡ 1( mod 5) 35x3 ≡ 1( mod 12) 84x1 − 85x1 = −1x1 ≡ 1( mod 5) 35x3 − 36x3 = −x3 ≡ 1( mod 12) x1 ≡ −1( mod 5). x3 ≡ 5 − 1( mod 12). Since a2 = 0, therefore 60x2 ≡ 1(mod 7). Thus a1 N1 x1 + a2 N2 x2 + a3 N3 x3 = −119(Verify!). Hence −119 + 420 = 301 eggs in basket. Theory of Congruences 4.6 87 System of Linear Congruences In this section, our discussion will be restricted to solve the system of linear congruence equations involving the same numbers of unknowns with the same modulus. Let us begin with an example. Consider the system of linear congruence equations: x + 2y ≡ 1(mod 5) (4.6.1) 2x + y ≡ 1(mod 5). (4.6.2) Now (4.6.1) × 2 − (4.6.2) yields 3y ≡ 1(mod 5). Note that 2 is the inverse of 3 modulo 5. So multiplying both sides of the foregoing equation by 2 we get, y ≡ 2(mod 5). Similarly, (4.6.2) × 2 − (4.6.1) we get, 3x ≡ 1(mod 5). Since 2 is the inverse of 3 modulo 5, therefore proceeding as above we get x ≡ 2(mod 5). Thus the solutions of the system of linear congruences are in pairs satisfying x ≡ 2(mod 5) and y ≡ 2(mod 5). This example motivates us to devise a general method for solving the system of linear congruences. Theorem 4.6.1. Let p, q, r, s, u, v and m be integers with m > 0, such that gcd(D, m) = 1 where D = ps − qr. Then the system of congruences px + qy ≡ u(mod m) (4.6.3) rx + sy ≡ v(mod m) (4.6.4) has a unique solution modulo m given by, ¯ x ≡ D(us − qv)(mod m) ¯ y ≡ D(pv − ur)(mod m) ¯ is the inverse of D modulo m. where D 88 Number Theory and its Applications Proof. Let us begin with a calculation. Here (4.6.3) × s − (4.6.4) × q yields Dx ≡ (us − qv)(mod m). ¯ is the inverse of D modulo m, therefore multiplying both sides by D ¯ we Since D get ¯ x ≡ D(us − qv)(mod m). ¯ on (4.6.3) × r − (4.6.4) × s gives Similarly, applying D ¯ y ≡ D(pv − ur)(mod m). Our claim is that any pair (x, y) is a solution. For this we have, ¯ px + qy ≡ D{p(us − qv) + q(pv − ur)}(mod m) ¯ ≡ Du(ps − qr)(mod m) ¯ ≡ DDu(mod m) ≡ u( mod m) ¯ and rx + sy ≡ D{r(us − qv) + s(pv − ur)}(mod m) ¯ ≡ Dv(ps − qr)(mod m) ¯ ≡ DDv(mod m) ≡ v(mod m). This proves the theorem. In the Theorem 4.6.1 we have discussed the solution for a system of two linear congruences with two unknowns. But the method fails for n linear congruences with n unknowns where n > 2. To overcome this, we require the algebra of matrices. The following deﬁnition on congruence relation between matrices will pave the way for our future discussions. Deﬁnition 4.6.1. For any two matrices S = (sij )n×k and T = (tij )n×k , S is said to be congruent to T modulo m(> 0) if sij ≡ tij (mod m) for every i and j with 1 ≤ i ≤ n, 1 ≤ j ≤ k. This is denoted as S ≡ T (mod m). 8 4 13 4 Example 4.6.1. Consider S = and T = . Then S ≡ T ( 9 7 14 12 mod 5). Proposition 4.6.1. For any two matrices [S]n×k and [T ]n×k with S ≡ T (mod m), ∃ matrices [U ]k×p and [V ]p×n respectively, with all integer entries, such that SU ≡ T U (mod m) and V S ≡ V T (mod m). Theory of Congruences 89 Proof. Let S = (sij )n×k , T = (tij )n×k and U = (uij )k×p be the matrices with n n integral entries. Now the entries of SU and T U are r=1 sir urj and r=1 tir urj respectively. Since S ≡ T (mod m), therefore we have sir ≡ tir (mod m) for all i and r. In view of Theorem 4.2.1 we get, n sir urj ≡ r=1 n tir urj (mod m). r=1 This proves SU ≡ T U (mod m). Similarly we can show that V S ≡ V T (mod m). We continue our development of the method for solving system of congruences, s11 x1 + s12 x2 + . . . s1n xn ≡ t1 (mod m) s21 x1 + s22 x2 + . . . s2n xn ≡ t2 (mod m) .... .. sn1 x1 + sn2 x2 + . . . snn xn ≡ tn (mod m). The system can be written as SX ≡ T (mod m), where ⎛ s1,1 ⎜ ⎜ s2,1 S=⎜ ⎜ .. ⎝ . sm,1 s1,2 s2,2 .. . sm,2 ⎞ ⎛ ⎞ ⎛ ⎞ x1 t1 s1,n ⎟ ⎜ ⎟ ⎜ ⎟ s2,n ⎟ ⎜ x2 ⎟ ⎜ t2 ⎟ ⎜ ⎟ ⎜ ⎟ .. ⎟ ⎟ X = ⎜ .. ⎟ and T = ⎜ .. ⎟ . ⎝ . ⎠ ⎝.⎠ . ⎠ · · · sm,n xn tn ··· ··· .. . This method is based on ﬁnding the inverse S¯ of S modulo m. Here S¯ is deﬁned ¯ ≡ S S¯ ≡ I(mod m), where In×n is the identity matrix. as SS 1 2 3 1 To illustrate this, let us choose S = . Then S¯ = where 3 4 4 2 11 5 1 0 6 10 1 0 ¯ = S S¯ = ≡ (mod 5) and SS ≡ (mod 5). 25 11 0 1 10 16 0 1 Next proposition describes a method for ﬁnding inverses of 2 × 2 matrices. a b Proposition 4.6.2. Let S = be a matrix with integer entries and m be c d a positive integer such that D = det S = ad − bc with gcd(D, m) = 1. Then the d −b ¯ ¯ is the inverse matrix S¯ = D is the inverse of S modulo m, where D −c a of D modulo m. 90 Number Theory and its Applications ¯ ≡ I( Proof. Whether S¯ is the inverse of S, it suﬃces to examine SS¯ ≡ SS mod m). For this, let us consider a b d −b ad − bc 0 1 0 ¯ ¯ ¯ S S¯ ≡ D ≡D ≡ DD ≡ I( mod m) c d −c a 0 ad − bc 0 1 Theory of Congruences 91 Example 4.6.2. Let us consider the system, x + 2y + 3z ≡ 1(mod 7) x + 3y + 5z ≡ 1(mod 7) x + 4y + 6z ≡ 1(mod 7). This can be written as SX ≡ T (mod m) where ⎛ 1 ⎜ S = ⎝1 1 2 3 4 ⎞ ⎛ ⎞ ⎛ ⎞ 3 x 1 ⎟ ⎜ ⎟ ⎜ ⎟ 5⎠ , X = ⎝y ⎠ and T = ⎝1⎠ . 6 z 1 ⎛ ⎞ −2 0 1 ⎟ ¯ = 6. Also adj S = ⎜ Here D = det S = −1. Then D ⎝−1 3 −2⎠ . Thus we 1 −2 1 have ⎛ ⎞ ⎛ ⎞ −2 0 1 −12 0 6 ⎜ ⎟ ⎜ ⎟ S¯ = 6 ⎝−1 3 −2⎠ = ⎝ −6 18 −12⎠ 1 −2 1 6 −12 6 ⎛ ⎞⎛ ⎞ ⎛ ⎞ −12 0 6 1 −6 ⎜ ⎟⎜ ⎟ ⎜ ⎟ X ≡ ⎝ −6 18 −12⎠ ⎝1⎠ (mod 7) ≡ ⎝ 0 ⎠ (mod 7). 6 −12 6 1 0 ∴ The solution is x ≡ 1(mod 7), y ≡ 0(mod 7), z ≡ 0(mod 7). 4.7 Worked out Exercises Problem 4.7.1. Find the solution of the system of linear congruences 2x + 3y ≡ 5(mod 7) x + 5y ≡ 6(mod 7). Solution 4.7.1. Multiplying second equation by 2 and then subtracting with ﬁrst one we get −7y ≡ −7( mod 7). This shows that y can take any residue modulo 7. If y = 0, then x ≡ 6(mod 7). So the ﬁrst solution is (6, 0). Continuing this manner we can ﬁnd other solutions too. 92 Number Theory and its Applications Problem 4.7.2. Find the inverse modulo 2 S= 1 5 for the matrix 2 2 ¯ ≡ 1(mod 5) ⇒ D ¯ = 3. Let S¯ Solution 4.7.2. Here D = 4 − 2 = 2. Then DD be the inverse of S. Then 2 −2 6 −6 1 4 S¯ = 3 = . ∴ S¯ ≡ ( mod 5). −1 2 −3 6 2 1 Problem 4.7.3. Find the inverse modulo 5 ⎛ 1 2 ⎜ S = ⎝1 2 1 4 for the matrix ⎞ 3 ⎟ 5⎠ 6 ¯ ≡ 1( Solution 4.7.3. Here D = 1(12 − 20) − 2(6 − 5) + 3(4 − 2) = −4. Then DD ¯ = 5. Now mod 7) ⇒ D ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ −1 0 4 −1 0 4 2 0 6 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ adj S = ⎝−1 3 −2⎠ shows S¯ = 5 ⎝−1 3 −2⎠ ≡ ⎝2 1 4⎠ ( mod 7). 2 −2 0 2 −2 0 3 4 0 4.8 Exercises: 1. Find the remainders when 250 and 4165 are divided by 7. 2. Establish the following divisibility statements by theory of congruence for integers n(≥ 1): (a) 7 52n + 3 · 25n−2 ; (b) 17 23n+1 + 3 · 52n+1 ; n+2 + 72n+1 . (c) 43 6 3. For n(≥ 1), show that (−13)n+1 ≡ (−13)n + (−13)n−1 (mod 181). 4. Find the remainder when 2340 is divided by 341. 5. Prove the assertions below: (a)If a is an odd integer, then a2 ≡ 1(mod 8). (b)For any integer a, a4 ≡ 0 or 1(mod 5). 6. Prove the following statements: (a) The product of any set of n consecutive integers is divisible by n. (b) Any n consecutive integers form a complete set of residues modulo n. Theory of Congruences 7. Using theory of congruence show that 89 244 − 1 and 97 248 − 1. 93 8. Solve the following linear congruences: (a) 5x ≡ 2(mod 26); (b) 34x ≡ 60(mod 98). 9. Using congruences, solve the Diophantine equations below: (a) 4x + 51y = 9; (b) 5x − 53y = 17. 10. Solve each of the following sets of simultaneous congruences: (a) x ≡ 5(mod 11), x ≡ 14(mod 29), x ≡ 15(mod 31); (b) 2x ≡ 1(mod 5), 3x ≡ 9(mod 6), 4x ≡ 1(mod 7), 5x ≡ 9(mod 11). 11. Obtain three consecutive integers, the ﬁrst of which is divisible by a square, the second by a cube, and the third by a fourth power. 12. Check that whether the system x ≡ 5(mod 6) and x ≡ 7(mod 15) has a solution or not. 13. Solve the system of congruences, 3x + 4y ≡ 5(mod 13) 2x + 5y ≡ 7(mod 13). 14. Find an integer having the remainders 2, 3, 4, 5 when divided by 3, 4, 5, 6 respectively. 15. Verify that 0, 1, 2, 22 , 23 , . . . , 29 form a complete set of residues modulo 11, but 0, 1, 22 , 32 , . . . , 102 does not. 16. Find the solution of the following system of linear congruences, 4x + y ≡ 5( mod 7) x + 2y ≡ 4( mod 7) 17. Find the solution of the following system of linear congruences, x + 3y ≡ 1( mod 5) 3x + 4y ≡ 2( mod 5). 18. Find the inverse modulo 5 for the matrix, 0 1 1 0 94 Number Theory and its Applications 19. Find the inverse modulo 7 for the matrix, ⎛ ⎞ 1 1 0 ⎜ ⎟ ⎝1 0 1⎠ 0 1 1 20. Find all solutions of the following system, x + y ≡ 1( mod 7) x + z ≡ 1( mod 7) y + z ≡ 1( mod 7). 5 Fermat’s Little Theorem “Perhaps, posterity will thank me for having shown that the ancients did not know anything.” – Pierre De Fermat 5.1 Introduction The famous French Mathematician Pierre de Fermat ﬁrst wrote what would become his “Little Theorem” in 1640. It states that for any prime number p and any integer a the expression (ap − a) is divisible by p as long as p does not divide a(the pair are relatively prime). Although a number n that does not divide an − a for some a must be a composite number but the converse is not necessarily true. An example, in the later section of the chapter, will justify the above argument. Thus, Fermat’s theorem gives a test that is necessary but not suﬃcient for primality. Also the theorem is applicable in public-key cryptography. As was typical of Fermat, no proof by him is known to exist. Although a proof in an unpublished manuscript dated around 1683 was given by German mathematician Gottfried Wilhelm Leibniz but in 1736 the proof, by Swiss Mathematician Leonhard Euler, had been known to be published. Some 2, 000 years old a special case of Fermat’s theorem known as the Chinese hypothesis which replaces a with 2, states that a number n is prime if and only if it divides 2n − 2. As proved later in the West, the Chinese hypothesis is only half right. Fermat’s little theorem concerns modular arithmetic. 95 96 5.2 Number Theory and its Applications Fermat’s Little Theorem Theorem 5.2.1. Fermat’s Little Theorem: Let p be a prime and p a then, ap−1 ≡ 1(mod p). Proof. Let us begin with the ﬁrst p − 1 positive multiples of a which are a, 2a, 3a, · · · , (p − 1)a. None of them are congruent modulo p to any other. Then ra ≡ sa(mod p) with 1 ≤ r < s ≤ p − 1 implies r ≡ s(mod p), which is not possible. Multiplying we get, a · 2a · 3a · · · · · · (p − 1)a ≡ 1 · 2 · 3 · · · · · · (p − 1)( mod p) ⇒ ap−1 ≡ 1( mod p). This is not the only way to prove the theorem. There are a lot more other interesting ways to prove this theorem. Mathematical induction is one among them. To begin with, we ﬁx a prime p. For this prime p it is obvious that 1p ≡ 1( mod p) i.e. 1p−1 ≡ 1(mod p), when a = 1. Suppose the statement prevails for a = k. Then k p−1 ≡ 1(mod p). Now we have to prove (k + 1)p−1 ≡ 1(mod p) for some base k + 1 ∈ Z and p (k + 1). Taking aid of binomial theorem we have, p p−1 p k k + 1. (k + 1) = k + + ··· + 1 l−1 p p p p! = Because for 1 ≤ l ≤ p − 1, it follows that p divides every l l!(p − l)! coeﬃcients of the terms of right hand side of the foregoing equation except k p and 1. Now taking modulo p we have (k + 1)p ≡ k p + 1(mod p). So by induction hypothesis we get (k + 1)p ≡ k + 1(mod p). Therefore the result holds for k + 1. Hence the principle of mathematical induction yields ap ≡ a(mod p) i.e.ap−1 ≡ 1( mod p) for all a ∈ Z such that p a. The above two proofs of Fermat’s Little theorem are mostly theoretic. Instead, we can provide some experimental ways by means of combinatorics to make the theorem more lively and natural. Choose p = 3, a = 2 where 3 2. Consider the following diagrams, Fermat’s Little Theorem 97 Figure 5.1: Fermat’s Little Theorem Here every angle of the triangles are associated with red and blue coloured balls. There are 23 = 8 ways to pick the colour of the balls. Also we see that 23 − 2 = 6 is divisible by 3. Therefore 23 ≡ 2(mod 3). Thus in general ap−1 ≡ 1( mod p) holds, where p a. Our next corollary investigates the question: Can we drop the condition gcd(a, p) = 1? Corollary 5.2.1. If p is prime, then ap ≡ a(mod p) for any integer a. Proof. When p \| a then ap ≡ 0 ≡ a(mod p) and if p a then by above theorem ap−1 ≡ 1(mod p) implies ap ≡ a(mod p). A simple but interesting question to ask: if an ≡ a(mod n) holds, then does it imply n is prime? The answer is in a negative sense. For instance, pick out n = 117. Then taking a = 2 we see that 2117 = (27 )16 · 25 where 27 = 128 ≡ 11( mod 117). Thus we ﬁnd 2117 ≡ 1116 · 25 (mod 117) ≡ 48 · 25 (mod 117) ≡ 221 ( mod 117). But 221 = (27 )3 . Hence 221 ≡ 113 (mod 117) ≡ 121 · 11(mod 117) ≡ 4 · 11(mod 117) ≡ 44(mod 117) ≡ 2(mod 117). Here we note 117 = 13 · 9. Hence, if an ≡ a(mod n) holds then n must be composite. Our future discussions will be based on some instances where those types of composite numbers even satisfy this congruence relation under some special 98 Number Theory and its Applications circumstances. Lemma 5.2.1. If p and q are distinct primes with ap ≡ a(mod q) and aq ≡ a( mod p) then, apq ≡ a(mod pq). Proof. It is very clear that (ap )q ≡ aq (mod p) ≡ a(mod p) and (aq )p ≡ ap ( mod q) ≡ a(mod q). So we have p\|apq and q\|apq . As gcd(p, q) = 1 then we can directly say that pq\|apq . Since gcd(a, b) = 1 and a\|c, b\|c together imply ab\|c, therefore apq ≡ a(mod pq). So the above lemma highlights the fact that the converse of Fermat’s theorem satisﬁes for some special type of composite numbers which can be expressed as the product of two distinct primes. These types of numbers are said to be pseudo-prime to the base a. Now we are in a position to deﬁne pseudoprime viz Deﬁnition 5.2.1. A composite integer n for which an ≡ a(mod n) is called a pseudoprime to the base a. If a = 2 then, it is called pseudo prime to the base 2 or simply pseudoprime. Let us take 341 = 11 · 31. So by Fermat’s Little Theorem we have 211 = 2·210 ≡ 2·1024(mod 31) ≡ 2·1(mod 31) ≡ 2(mod 31) and 231 = 2·(210 )3 ≡ 2·13 ( mod 11) ≡ 2(mod 11). Furthermore, gcd(11, 31) = 1. In view of Lemma(5.2.1) we can say that 211·31 = 2341 ≡ 2(mod 341) which further yields 341 as a pseudoprime. In fact, the ﬁrst ﬁve pseudoprimes are 341, 561, 645, 1105, 161038 and the ﬁrst four are odd. Finding pseudoprimes are diﬃcult as those are rarer than primes. There are only 245 pseudoprimes and 78498 primes less than 106 . We now try to construct an increasing sequence of pseudoprimes from the following theorem. Theorem 5.2.2. There are inﬁnitely many psuedo-primes to the base 2. Proof. Let n be a composite number. Then ∃ r, s ∈ Z such that n = rs where 1 < r ≤ s < n. Let Kn = 2n − 1 be any integer where (2r − 1)\|(2n − 1)[refer to Problem(2.6.4)] or (2r − 1)\|Kn , making Kn a composite quantity. As n is pseudo-prime then 2n ≡ 2(mod n). Hence 2n − 2 = kn for some k ∈ Z. Therefore 2Kn −1 = 2kn − 1 = (2n − 1)[2n(k−1) + · · · + 2n + 1] = Kn [2n(k−1) + · · · + 2n + 1] ≡ 0( mod Kn ). Hence 2Kn ≡ 2(mod Kn ). Therefore Kn is a pseudoprime. Fermat’s Little Theorem 99 Remark 5.2.1. The number Kn = 2n − 1 shown in the above theorem is said to be Mersenne number, which is named after Father Marin Mersenne (1588 − 1648)[for further studies refer to Chapter 10 Section 10.4 of this book]. The above discussion generates the fact that the pseudoprimes are the special type of composite numbers which satisﬁes the conditions of Fermat’s Little theorem. But in pseudoprime, we have a barrier of base element a i.e. for these types of numbers the condition of Fermat’s theorem does not satisfy for all base elements a. If we consider 561(= 3 × 11 × 17) with gcd(a, 561) = 1 for any a ∈ Z+ , we have gcd(a, 3) = 1 = gcd(a, 11) = gcd(a, 17). By virtue of Fermat’s theorem, we get a2 ≡ 1(mod 3), a10 ≡ 1(mod 11), a16 ≡ 1(mod 17) which imply a560 = (a2 )280 ≡ 1( mod 3) a560 = (a10 )56 ≡ 1( mod 11) a560 = (a16 )35 ≡ 1( mod 17). Since 3, 11, 17 are primes, the last three congruences together conclude a560 ≡ 1( mod 561). Therefore a561 ≡ a(mod 561) for all a ∈ Z+ with gcd(a, 561) = 1. The last example spotlights the fact that 561 is a special type of composite number which satisﬁes the condition of Fermat’s theorem for any integer. R.D.Carmichael ﬁrst noticed the existence of these types of numbers in the year 1910. Those numbers are called Carmichael numbers named after American Mathematician Carmichael. There are six Carmichael numbers 561, 1105, 1729, 2465, 2821, 6601 less than 10, 000. There are just 43 Carmichael numbers less than 106 and 1547 less than 1010 . Thus we are in a position to deﬁne the Carmichael number. Deﬁnition 5.2.2. The composite numbers n which satisfy the property an ≡ a( mod n) for all integers a are said to be absolute pseudoprime or Carmichael numbers. Next our aim is to establish the criterion for the existence of Carmichael numbers. Theorem 5.2.3 (Korselt’s Criterion). Let n be a composite square free integers; n = p1 p2 · · · pn where pi are distinct primes. If (pi − 1) (n − 1) for i = 1, 2, · · · , r then n is Carmichael number. Proof. Suppose that a is an integer satisfying gcd(a, pi ) = 1 for each i. Then, by Fermat’s theorem we have pi \|(api −1 − 1). As (pi − 1)\|(n − 1) so pi \|(an−1 − 1), as pi \|(an − a), for all a and i = 1, 2, 3, · · · , r. This implies n (an − a) for all a. Therefore n is Carmichael number. 100 Number Theory and its Applications The next theorem supplies the pertinent information about the prime factorizations of Carmichael numbers. Theorem 5.2.4. A Carmichael number must have at least three diﬀerent odd prime factors. Proof. Let n be a Carmichael number. Since n is composite and is the product of distinct primes so, it cannot have just one prime factor. Then assume, n = pq for some odd primes p, q with p > q. So n − 1 = pq − 1 = (p − 1)q + (q − 1) ≡ (q − 1) ≡ 0( mod p − 1), which render (p − 1) (n − 1). Since it has just two diﬀerent prime factors hence, n cannot be a Carmichael number. The development of primality of testing can be done further with the following: Deﬁnition 5.2.3. Let n be a positive integer with n − 1 = 2k t where k is a non– negative integer and t is an odd positive integer. We can say n passes Miller’s i test for the base a if either at ≡ 1(mod n) or a2 t ≡ −1(mod n) for some i with 0 ≤ i ≤ k − 1. The next theorem shows the idea of primality testing by means of Miller’s test. Theorem 5.2.5. If n is prime and a is a positive integer with n a, then n passes Miller’s test for the base a. Proof. Let n − 1 = 2k t where k is non–negative integer and t is an odd positive (n−1) k−ω integer. Let zω = a 2ω = a2 ·t for ω = 0, 1, 2, . . . , k. Since n is prime, by Fermat’s little theorem we have z0 = an−1 ≡ 1(mod n). Furthermore z12 = 2 n−1 2 a = an−1 = z0 ≡ 1(mod n) implies either z1 ≡ −1(mod n) or z1 ≡ 1( 2 n−1 n−1 2 2 2 = a 2 = z1 ≡ 1(mod n). mod n). If z1 ≡ 1(mod n) then z2 = a Thus either z2 ≡ 1(mod n) or z2 ≡ −1(mod n). Proceeding as above, z0 ≡ 2 = zω ≡ 1(mod n) or zω+1 ≡ 1( z1 ≡ z2 · · · zω ≡ 1(mod n) for ω < k. Also, zω+1 mod n). Thus continuing for ω = 1, 2, 3, . . . , k we ﬁnd that either zk ≡ 1(mod n) or zω ≡ −1(mod n) for some integer ω with 0 ≤ ω ≤ (k − 1). Hence n passes Miller’s test for the base a. Let us illustrate the above theorem by the following example. Choose n = 25 = 5 · 5. Then 724 = (74 )6 ≡ 1(mod 5) such that 25 is a pseudoprime to the Fermat’s Little Theorem 101 21 ·3 base 7. Also 24 = 2 · 3 then 7 ≡ −1(mod 25). Therefore 25 passes Miller’s test for base 7 as well as it is a pseudoprime. So getting motivated from the example we are going to deﬁne: 3 Deﬁnition 5.2.4. If n is composite and passes Miller’s test to the base a, then n is called strong pseudoprime to the base a. Let us illustrate the ideas behind the deﬁnition(5.2.4) with an example of strong pseudoprime which has passed Miller’s test. Consider n = 25326001. Then n−1 = 24 ×1582875. Here we can check that 21582875 ≡ −1(mod 25326001). This shows that 25326001 is a strong pseudoprime as it passes Miller’s test. Strong pseudoprimes are rare but there are still inﬁnitely many of them. We conclude this section with a theorem that reﬂects the existence of an inﬁnite number of strong pseudoprimes to the base 2. Theorem 5.2.6. There are inﬁnitely many strong pseudoprimes to the base 2. Proof. To begin with, suppose n to be an odd pseudoprime base 2. We claim that the composite number N = 2n − 1 is a strong pseudoprime to the base 2. Referring to Problem 2.6.4 we see that if n is composite then 2n − 1 is also so. Furthermore, if n is pseudoprime then we have 2n−1 ≡ 1(mod n). This implies that 2n−1 − 1 = nk for some odd integer k(> 0). We note that N − 1 = 2n − 2 = 2(2n−1 − 1) = 2nk. (N −1) As 2n = (2n − 1) + 1 = N + 1 ≡ 1(mod N ) then we can write 2 2 = 2nk ≡ 1( mod N ). The argument shows that N passes Miller’s test for base 2. Thus N becomes a strong pseudoprime base 2. An appeal to Theorem 5.2.2 concludes that there are inﬁnitely many strong pseudoprime to the base 2. This ﬁnishes the proof. 5.3 Worked out Exercises Problem 5.3.1. If gcd(a, 35) = 1, show that a12 ≡ 1(mod 35). Solution 5.3.1. As gcd(a, 35) = 1, therefore gcd(a, 7) = 1 = gcd(a, 5). An appeal to Fermat’s theorem indicates a6 ≡ 1(mod 7) ⇒ a12 = (a6 ) · (a6 ) ≡ 1( mod 7) and a4 ≡ 1(mod 5) ⇒ (a4 )3 ≡ 1(mod 5). Since gcd(5, 7) = 1, it follows that 35 (a12 − 1). Therefore a12 ≡ 1(mod 35). Problem 5.3.2. If gcd(a, 42) = 1 then 168 = 3 · 7 · 8 divides a6 − 1. 102 Number Theory and its Applications Solution 5.3.2. Because gcd(a, 42) = 1, therefore gcd(a, 7) = gcd(a, 3) = gcd(a, 2) = 1. By virtue of Fermat’s theorem, we ﬁnd a6 ≡ 1(mod 7), a2 ≡ 1( mod 3) and a ≡ 1(mod 2). Therefore a6 = (a2 )3 ≡ 1(mod 3). Moreover, a6 − 1 = (a3 − 1)(a3 + 1) = (a − 1)(a + 1)(a2 + a + 1)(a2 − a + 1). Because a is odd therefore, a > 0 ⇒ a ≥ 3. This yields 2 (a − 1), 4 (a + 1). Since 7, 3, 8 are 6 relatively prime to each other, therefore we get 168 (a − 1). Hence 8 (a6 − 1). Problem 5.3.3. If gcd(a, 133) = gcd(b, 133) = 1 then, show that 133\|(a18 −b18 ). Solution 5.3.3. We know that 133 = 7 · 19 and gcd(a, 19) = gcd(b, 19) = 1. Therefore in view of Fermat’s theorem we obtain a18 ≡ 1(mod 19) and b18 ≡ 1(mod 19). Hence a18 − b18 ≡ (1 − 1)(mod 19) ≡ 0(mod 19). Hence 19\|(a18 − b18 ). By similar reasoning, 7\|(a6 − b6 ). Since a18 − b18 = (a6 − b6 ) (a6 )2 + a6 b6 + (b6 )2 , therefore we have 7\|(a18 −b18 ). Thus 7·19 = 133\|(a18 − b18 ). Problem 5.3.4. Derive the following congruences: (a) a21 ≡ a(mod 15), ∀a. (b) a7 ≡ a(mod 42) ∀a. (c) a9 ≡ a(mod 30) ∀a. Solution 5.3.4. (a) Taking into consideration the corollary of Fermat’s theorem, we ﬁnd a21 ≡ a(mod 5) ⇒ (a5 )4 ≡ a4 (mod 5) ⇒ a21 ≡ a5 ≡ a( mod 15). Furthermore, a3 ≡ a(mod 3) ⇒ a21 ≡ a7 (mod 3). Again, (a3 )2 ≡ a2 (mod 3) ⇒ a7 ≡ a3 (mod 3) ≡ a(mod 3). Hence a21 ≡ a( mod 3). Thus, a21 ≡ a(mod 15). (b) As 42 = 7 · 3 · 2 by Fermat’s theorem we have a7 ≡ a(mod 7) and a3 ≡ a( mod 3). Therefore a6 ≡ a2 (mod 3) ⇒ a7 ≡ a3 (mod 3) ≡ a(mod 3). Also, a2 ≡ a(mod 2) ⇒ a6 ≡ a3 (mod 2) ≡ a(mod 2) ⇒ a7 ≡ a2 (mod 2) ≡ a( mod 2). Since 7, 3, 2 are prime to each other therefore, a7 ≡ a(mod (7 · 3 · 2)) ⇒ a7 ≡ a(mod 42). (c) Left to the reader. Problem 5.3.5. If gcd(a, 30) = 1, show that 60 (a4 + 59). Solution 5.3.5. Note that gcd(a, 30) = 1 implies gcd(a, 2) = gcd(a, 3) = gcd(a, 5) = 1. So gcd(a, 4) = gcd(a, 22 ) = 1. Now 60 = 22 · 3 · 5 and 60\|(a4 + 59) together implies a4 ≡ 1(mod 60). Here a2 ≡ 1(mod 3) implies a4 ≡ 1(mod 3) and a4 ≡ 1(mod 5). Further, a ≡ 1(mod 2) ⇒ a2 ≡ 1(mod 2) which leads to 2 (a2 − 1). Hence a2 ≡ (1 − 2)( mod 2) ≡ −1( mod 2). Fermat’s Little Theorem 103 Thus, combining the foregoing equation with 2 (a2 + 1) yields 2 (a4 − 1). Since 3, 4, 5 relatively prime to each other, therefore we can conclude 60\|(a4 −1). Hence a4 ≡ 1( mod 60) ≡ (1 − 60)( mod 60) ≡ −59( mod 60). This completes the solution. Problem 5.3.6. (a) Find the unit digit of 3100 using Fermat’s theorem. (b) For any integer verify that a5 and a have same unit digit. Solution 5.3.6. (a) Its suﬃces to consider modulo 10. Now we plan to use Fermat’s theorem to get 34 ≡ 1(mod 5). Therefore 3100 ≡ 1(mod 5). Moreover 3 ≡ 1(mod 2). Hence 3100 ≡ 1(mod 2). Further gcd(2, 5) = 1 ⇒ 10 3100 ≡ 1( mod 10). Hence the unit digit is 1. (b) By virtue of Fermat’s theorem, a5 ≡ a(mod 5) and a2 ≡ a(mod 2). Hence a4 ≡ a2 (mod 2) ≡ a(mod 2) implies a5 ≡ a2 (mod 2) ≡ a(mod 2). Thus a5 ≡ a(mod 10). Let 0 ≤ r < 10 holds. Then a5 − r ≡ a − r(mod 10). Therefore a5 − r ≡ 0(mod 10) ⇐⇒ a − r ≡ 0(mod 10). Therefore unit digit’s are same. Problem 5.3.7. If 7 a, then prove that either 7\|(a3 + 1) or 7\|(a3 − 1). Solution 5.3.7. By Fermat’s theorem, a6 ≡ 1(mod 7). Therefore 7\|(a6 − 1) but a6 − 1 = (a3 − 1)(a3 + 1). Therefore 7 (a3 + 1) implies 7\|(a3 − 1) and vice–versa. Problem 5.3.8. If p, q are distinct odd primes such that (p − 1)\|(q − 1) and gcd(a, pq) = 1, show that aq−1 ≡ 1(mod pq). Solution 5.3.8. Here gcd(a, pq) = 1 implies gcd(a, p) = 1 = gcd(a, q). Therefore with the help of Fermat’s Theorem we get ap−1 ≡ 1(mod p) and aq−1 ≡ 1( mod q). Since (p−1) (q −1), therefore q −1 = k(p−1)(k ∈ Z). Hence (ap−1 )k ≡ 1k (mod p) ≡ 1(mod p) ⇒ aq−1 ≡ 1(mod p). Thus pq\|(aq−1 − 1) ⇒ aq−1 ≡ 1( mod pq). Problem 5.3.9. If p, q are distinct primes then prove that pq−1 + pq−1 ≡ 1( mod pq). Solution 5.3.9. By virtue of Fermat’s theorem, pq−1 ≡ 1(mod q) implies q p−1 ≡ 0(mod q). Therefore pq−1 +pq−1 ≡ 1(mod q). Similarly, q p−1 +pq−1 ≡ 1( mod p). Further, gcd(p, q) = 1 yields pq−1 + pq−1 ≡ 1(mod pq). Problem 5.3.10. Establish the statement: If p is an odd prime, then 1p−1 + 2p−1 · · · (p − 1)p−1 ≡ (p − 1) ≡ −1(mod p). 104 Number Theory and its Applications Solution 5.3.10. Since p is odd prime, so p ≥ 3 and p a. if a < p then by Fermat’s theorem we have, ap−1 ≡ 1(mod p). For p − 1 terms we have, 1. 1p−1 ≡ 1(mod p) 2. 2p−1 ≡ 1(mod p) .. . 3. (p − 1)p−1 ≡ 1(mod p). Therefore 1p−1 + 2p−1 · · · (p − 1)p−1 ≡ (p − 1) ≡ −1(mod p). Problem 5.3.11. Conﬁrm 1105 = 5 · 13 · 17 is absolute pseudoprime. Solution 5.3.11. For any integer a, if 1105 a then 5 a, 13 a & 17 a. So by Fermat’s theorem, we have a4 ≡ 1(mod 5) ⇒ a1104 = (a4 )276 ≡ 1( mod 5). Also a12 ≡ 1(mod 13) ⇒ a1104 = (a12 )92 ≡ 1(mod 13). Moreover, a16 ≡ 1(mod 17) ⇒ a1104 = (a16 )69 ≡ 1(mod 17). As 5, 13, 17 are relatively prime to each other, therefore a1104 ≡ 1(mod 1105). Thus a1105 ≡ a(mod 1105) provided 1105 a. Clearly, a1105 ≡ a(mod 1105) prevails provided 1105 \| a. Hence 1105 is an absolute pseudo prime as it satisﬁes a1105 ≡ a(mod 1105) for any integer a. Problem 5.3.12. Prove that any integer of the form n = (6k+1)(12k+1)(18k+ 1) is an absolute pseudoprime if all three factors are prime; hence 1729 = 7·13·19 is also absolute pseudo–prime. Solution 5.3.12. Let p1 = 6k +1, p2 = 12k +1, p3 = 18k +1, be all primes. Now n = 36 · 36k 3 + 36 · 2k 2 + 36 · 9k 2 + 36k + 1. Therefore n − 1 = 36k[36 · k 2 + 11k + 1] gives p1 − 1 n − 1, p2 − 1 n − 1 and p3 − 1 n − 1. Since p1 , p2 , p3 are distinct primes and n is square free, therefore n is absolute pseudoprime. Problem 5.3.13. Show that 561 is the only Carmichael number of the form 3pq where p and q are primes. Solution 5.3.13. Let n = 3pq, with q > p odd primes, be a carmichael number. Then using Korselt’s criterion, we obtain (p − 1) (3pq − 1) = 3(p − 1)q + 3q − 1. So (p − 1) (3q − 1) ⇒ (p − 1)a = 3q − 1 for some a ∈ Z. Since q > p, we must have a ≥ 4. Similarly, ∃ b ∈ Z satisfying (q − 1)a = 3p − 1. Solving these two equations for p, q yields 2b + ab − 3 2b + 6 =1+ , ab − 9 ab − 9 2a + ab − 3 q= . ab − 9 p= (5.3.1) (5.3.2) Fermat’s Little Theorem 105 Since p > 3 being odd prime, therefore 4(ab−9) ≤ 2b+6 reduces to b(2a−1) ≤ 21. Now a ≥ 4 ⇒ b ≤ 3. Then, 4(ab − 9) ≤ 2b + 6 ≤ 12 ⇒ ab ≤ 21 ⇒ a ≤ 5. 4 Hence a = 4 or 5. If b = 3, then the denominator of (5.3.2) is multiple of 3. So the numerator must be multiple of 3, which is impossible as there any ‘a divisible by 3. Thus b = 1 or 2.The denominator of equation (5.3.2) must be positive, so ab > 9. Thus the only possible values for a and b is 5 and 2 respectively, which gives p = 11, q = 17. So 561 = 3 · 11 · 17 is the only Carmichael number of the form 3pq, where p and q are primes. Problem 5.3.14. Show that there are only a ﬁnite number of Carmichael numbers of the form n = pqr where p is a ﬁxed prime, and q and r are also primes. Solution 5.3.14. Assume r > q. Applying Korselt’s Criterion, we get (q − 1) (pqr − 1) = (q − 1)pr + pr − 1. Therefore (q − 1) (pr − 1) ⇒ pr − 1 = a(q − 1) for some a ∈ Z. Similarly, pq − 1 = b(r − 1) for some b ∈ Z. Since, r > q so a > b. Solving last two equations for q and r yields p(a − 1) + a(b − 1) , ab − p2 p(b − 1) + b(a − 1) . q= ab − p2 r= Because this last fraction must be an integer, we have ab − p2 ≤ p2 + pb − p − b, which further reduces to a(b − 1) ≤ 2p2 + p(b − 1), ⇒a−1≤ 2p2 p(b − 1) + ≤ 2p2 + p. b b So ∃ only ﬁnite values for a. Likewise, the same inequality gives b(a − 1) ≤ 2p2 + p(b − 1), ⇒ b(a − 1 − p) ≤ 2p2 − p. Since a > b and the denominator of the expression for q must be positive, we have a ≥ p + 1. Now, a = p + 1 gives (p + 1)(q − 1) = pq − p + q − 1 = pr − 1 ⇒ p\|q, a contradiction. 106 Number Theory and its Applications Therefore a > p + 1 ⇒ a − p − 1 > 0. The last inequality gives us b ≤ b(a − p − 1) ≤ 2p2 − p, which shows ∃ ﬁnitely many values of b. Because a, b determine q, r respectively, therefore there are only a ﬁnite number of Carmichael numbers of the form n = pqr. Problem 5.3.15. Show that 2047 is a strong pseudoprime base 2. Solution 5.3.15. Here n = 2047 yields n − 1 = 2046 = 2 × 1023. Now 21023 = (211 )93 = 204893 ≡ 1(mod 2047). So 2047 passes Miller’s test for base 2. Thus 2047 is a strong pseudoprime base 2. 5.4 Wilson’s Theorem Wilson’s theorem, in number theory, signiﬁes that any prime p divides (p−1)!+1, where n! is the factorial notation for 1×2×3×4×· · ·×n. For example, 7 divides (7 − 1)! + 1 = 6! + 1 = 721. The conjecture was ﬁrst published by the English mathematician Edward Waring in Meditationes Algebraicae (1770 ‘Thoughts on Algebra’), where he described it to the English mathematician John Wilson. After that it was proved by the French mathematician Joseph-Louis Lagrange in 1771. The converse of the theorem is also true; that is, (n − 1)! + 1 is not divisible by a composite number n. In theory, these theorems provide a test for primes; in practice, the calculations are impractical for large numbers. Theorem 5.4.1. Wilson’s Theorem: If p is a prime then (p − 1)! ≡ −1(mod p). Proof. Let us choose p > 3 and consider the linear congruence ax ≡ 1(mod p) where a is any one of 1, 2, 3, · · · , p − 1. Therefore gcd(a, p) = 1. Hence, it has an unique solution viz a˜ a ≡ (mod p) with 1 ≤ a ˜ ≤ p − 1. Because p is prime, a = a ˜ ⇔ a = 1 or a = p − 1 provided a2 ≡ 1(mod p) ⇒ (a − 1)(a + 1) ≡ 0( mod p). Therefore (a − 1) ≡ 0(mod p) or (a + 1) ≡ 0(mod p). Now if we delete 1 and p − 1, then the remaining 2, 3, . . . , p − 2 are set into pairs a and a ˜, where p−3 a a= ˜. So if these 2 congruences are multiplied, we obtain 2 · 3 · · · (p − 2) ≡ 1( mod p) ⇒ (p − 2)! ≡ 1(mod p) ⇒ (p − 1)! ≡ (p − 1) ≡ −1(mod p). Let us illustrate the use of the theorem by means of an example. Let us take Fermat’s Little Theorem 107 p = 11. Divide the integers 2, 3, 4, 5, 6, 7, 8, 9 into p−3 2 pairs such as 2 · 6 ≡ 1( mod 11) 3 · 4 ≡ 1( mod 11) 7 · 8 ≡ 1( mod 11) 5 · 9 ≡ 1( mod 11) Multiplying each pair together we obtain, 9! ≡ 1(mod 11). Hence 10! ≡ 1( mod 11), shows the result is true for p = 11. An interesting observation is that the converse is also true. Let n be a non–prime required integer. Then n must have a divisor d where 1 < d < n. As d ≤ n − 1, we have d\|(n − 1)!. Now from the condition we have, n\|((n − 1)! + 1). Hence combining the conditions, we have d\|((n − 1)! + 1). Thus d\|1 leads to contradiction, showing n is prime. Taking Wilson’s theorem and its converse together we can say that the condition is necessary and suﬃcient for an integer to be prime. Thus it gives us a condition of testing primality. Now we are at the end of this discussion with an application of Wilson’s theorem on quadratic congruences, where quadratic congruences assume the form Ax2 + Bx + C ≡ 0(mod m), where A ≡ 0(mod m) (otherwise the congruence would be a linear congruence). We will learn methods to evaluate these quadratic congruences. However, we will ﬁrst restrict our modulus m to being only an odd prime (3, 5, 7, 11, 13, . . .), or rather, any prime except 2. Now we are in a position to state the following theorem: Theorem 5.4.2. The quadratic congruence x2 + 1 ≡ 0(mod p) where p is an odd prime, has a solution if and only if p ≡ 1(mod 4). Proof. Let a be a solution of x2 +1 ≡ 0(mod p) then a2 ≡ −1(mod p). Since p a p−1 p−1 by Fermat’s theorem, we have 1 ≡ ap−1 (mod p) ≡ (a2 ) 2 (mod p) ≡ (−1) 2 ( mod p). The possibility that p = 4k + 3 for any integer k does not arise as p−1 (−1) 2 = (−1)2k+1 = −1. Therefore 1 ≡ (−1)(mod p) implies p\|2 which is p−1 a contradiction. So p is of the form 4k + 1. Now, (p − 1)! = 1 · 2 · · · · 2 p+1 · · · (p − 2)(p − 1) and 2 p − 1 ≡ −1( mod p) p − 2 ≡ −2( mod p) .. . p+1 ≡ 2 p−1 2 ( mod p). 108 Number Theory and its Applications 2 p−1 (mod p). If we assume p = 4k+1, 1 · 2··· 2 2 p−1 p−1 ! (mod p), by Wilson’s theorem. then (−1) 2 = 1. Therefore −1 ≡ 2 p−1 Therefore ! satisﬁes x2 + 1 ≡ 0(mod p). 2 Therefore (p−1)! ≡ (−1) 5.5 p−1 2 Worked out Exercises Problem 5.5.1. Find the remainder when 15! is divided by 17. Solution 5.5.1. Since (17 − 1)! = 16!, we have by virtue of Wilson’s theorem (17 − 1)! ≡ −1(mod 17). Therefore 16! ≡ −1(mod 17) ≡ 16(mod 17) ⇒ 15! ≡ 1( mod 17). Hence the remainder is 1. Problem 5.5.2. Find the remainder when 2(26)! is divided by 29. Solution 5.5.2. From Wilson’s theorem, we ﬁnd 28! ≡ −1(mod 29) ⇒ 28! ≡ 28( mod 29) ⇒ 27! ≡ 1(mod 29). Here we note that gcd(28, 29) = 1 ⇒ 27(26)! ≡ (1 + 29) = 30(mod 29) ⇒ 9(26)! ≡ 10(mod 29) ⇒ 9(26)! ≡ (10 + 29) = 9( mod 29) ⇒ 3(26)! ≡ 13(mod 29) ⇒ 3(26)! ≡ (13 + 29) = 42(mod 29) ⇒ (26)! ≡ 14(mod 29). Therefore 2(26)! ≡ 28(mod 29). Thus, 28 is the remainder. Problem 5.5.3. Show that 18! ≡ −1(mod 437). Solution 5.5.3. Note that 437 = 19 · 23, where both 19 and 23 are prime numbers. By Wilson’s theorem, we have 18! ≡ −1(mod 19) therefore 19\|(18! + 1) holds. So here the only thing we need to show is 23\|(18! + 1), because gcd(19, 23) = 1. Further by Wilson’s theorem, we obtain 22! ≡ −1(mod 23) ≡ 22( mod 23) ⇒ 21! ≡ 1(mod 23) ≡ 1 + 23 = 24(mod 23) ⇒ 7(20)! ≡ 8(mod 23) ⇒ 7 · 5 · 19! ≡ 2 ≡ 2 + 23 ≡ 25(mod 23) ⇒ 7 · 19 · 18! ≡ 5(mod 23) ≡ 5 + 23 = 28( mod 23) ⇒ 19 · 18! ≡ 4(mod 23) ⇒ 19 · 18! ≡ (4 − 23) = −19(mod 23) ⇒ 18! ≡ −1(mod 23). Therefore 23 (18! + 1) ⇒ 437 (18! + 1). Problem 5.5.4. Prove that for n(> 1) is prime if and only if (n − 2)! ≡ 1( mod n). Solution 5.5.4. By Wilson’s theorem and it’s converse we have, n is prime if and only if (n − 1)! ≡ −1(mod n). Hence (n − 1)! ≡ −1 + n = n − 1(mod n). Therefore (n − 2)! ≡ 1(mod n), as gcd(n, n − 1) = 1. Problem 5.5.5. If n is composite then show that (n − 1)! ≡ 0(mod n) except n = 4. Fermat’s Little Theorem 109 Solution 5.5.5. If n = 4, then (4 − 1)! = 3! = 6 ≡ 2(mod 4). Thus this equivalence is not true for n = 4. If n > 4 is a composite number, then n = r · s for some integers r and s. Since gcd(n, n − 1) = 1, therefore 1 < r < n − 1. So r must be the one of the factor of (n − 1)!. Similarly, for 1 < s < n − 1 the above argument is also true. If r = s, then r and s are diﬀerent factors of (n − 1)!. So n = r · s\|(n − 1)!. Therefore (n − 1)! ≡ 0(mod n). If r = s, then n = r 2 . Our claim is r < n2 . If not then, r ≥ n2 . Therefore 2 n = r2 ≥ n4 ⇒ 4 ≥ n. But this is not true because n > 4. Hence r < n 2 ⇒ 2r < n ⇒ 2r ≤ n − 1. Both r and 2r are factors of (n − 1)!, therefore r(2r) (n − 1)! ⇒ r2 (n − 1)!. Hence (n − 1)! ≡ 0(mod n). Problem 5.5.6. Given a prime p, establish (p − 1)! ≡ (p − 1)(mod 1 + 2 + 3 + · · · + (p − 1)). Solution 5.5.6. An appeal to Wilson’s theorem generates, (p − 1)! ≡ −1 = p − 1( mod p). Therefore p\|{(p − 1)! − (p − 1)}. We know that, 1 + 2 + 3 + · · · + (p − 1) = p(p − 1) . Since p − 1 is even, therefore (p−1) is an integer and (p−1) < (p − 1). 2 2 2 (p−1) Furthermore, (p − 1) {(p − 1)! − (p − 1)} ⇒ 2 {(p − 1)! − (p − 1)}. Because (p−1) (p−1) divide {(p − 1)! − (p − 1)}. Thus 2 , p = 1, therefore both p and 2 p(p−1) {(p − 1)! − (p − 1)} ⇒ (p − 1)! ≡ (p − 1)(mod 1 + 2 + 3 · · · + (p − 1)). 2 gcd Problem 5.5.7. If p is a prime prove that p (ap + (p − 1)! · a), for any integer a. Solution 5.5.7. Taking into consideration Euler’s generalisation theorem and Wilson’s theorem, we ﬁnd ap ≡ a(mod p) and −1 ≡ (p − 1)!(mod p) hold respectively. Multiplying last two congruences, we have −ap ≡ (p − 1)! · a(mod p). This proves, p (ap + (p − 1)! · a). Problem 5.5.8. If p is a prime prove that p\|((p − 1)! · ap + a), for any integer a. Solution 5.5.8. Hint: Same as Problem(5.5.7) Problem 5.5.9. Verify 4(29!) + 5! is divided by 31. Solution 5.5.9. An appeal to Wilson’s theorem gives, 30! ≡ −1(mod 31). Therefore 30 · 29! ≡ 31 − 1 = 30(mod 31) ⇒ 29! ≡ 1(mod 31). Hence 4(29)! ≡ 4( mod 31). Thus, we have 4(29!) + 5! ≡ 4 + 120 = 124(mod 31) ≡ 0(mod 31). Problem 5.5.10. Obtain the solution of x2 ≡ −1(mod 29). 110 Number Theory and its Applications Solution 5.5.10. As 29 ≡ 1(mod 4) so, ∃ a solution given by ! = ±14! . mod p)[refer to Theorem 5.4.2]. Therefore ± 29−1 2 p−1 2 ! ≡ −1( 2 Problem 5.5.11. Prove that the odd prime divisor of n2 + 1 is of the form 4k + 1. Solution 5.5.11. Let p be an odd prime divisor of n2 + 1. Therefore n2 + 1 ≡ 0( mod p). So n satisﬁes the quadratic congruence equation x2 ≡ −1(mod p). Hence p is of the form 4k + 1. Because p is of the form 4k + 3 it follows p−1 p−1 that, n2 ≡ −1(mod p) ⇒ 1 ≡ np−1 (mod p) ≡ (n2 ) 2 (mod p) ≡ (−1) 2 ( 4k+3−1 mod p) ⇒ 1 ≡ (−1) 2 (mod p) ≡ (−1)2k+1 (mod p) ≡ −1(mod p). This proves p 2, a contradiction. 5.6 Exercises: 1. Verify using Fermat’s theorem: 17 (11104 + 1). 2. Find the remainder of 97! when divided by 101. 3. Derive the congruence: a13 ≡ a(mod 3 · 7 · 13) for all integer a. 4. Find the remainder of 53! when divided by 61. 5. Prove 18351910 + 19862061 ≡ 0(mod 7). 6. Assume p a, p b, p is prime; (i) If ap ≡ bp (mod p) then, a ≡ b(mod p). (ii) If ap ≡ bp (mod p) then, ap ≡ bp (mod p2 ). 7. Using Fermat’s theorem, prove that for a odd prime p; (i) 1p−1 + 2p−1 + · · · + (p − 1)p−1 ≡ −1(mod p). (ii) 1p + 2p + · · · + (p − 1)p ≡ 0(mod p) 8. Conﬁrm that the followings are absolute prime: (a) 2821 = 7 · 13 · 31 (b) 2465 = 5 · 17 · 29. 9. Use Korselt’s criterion to determine which of them are Charmichael numbers: (a) 8911 (b) 10659 (c) 162401 (d) 126217. 10. Find the remainder when 3456 is divided by 7. 11. Find all positive integers n such that 22 n +1 is divided by 17. Fermat’s Little Theorem 111 12. Find 220 + 330 + 440 + 550 + 660 mod 17. 13. Determine whether 17 is a prime or not using Wilson’s theorem. 14. If p and p + 2 are a pair of primes then prove that 4((p − 1)! + 1) + p ≡ 0( mod p(p + 2)). 15. What is the remainder of 149! when divided by 139. 16. Find all Carmichael numbers of the form 5pq where p and q are primes. 17. Find a Carmichael number of the form 7 · 23 · q where q is an odd prime. 18. Show that 1373653 is a strong pseudoprime to base 2, 3. 19. Obtain the solution of x2 ≡ −1(mod 37). 6 Arithmetic Functions “Mathematics is the queen of sciences and number theory is the queen of mathematics.” – Carl Friedrich Gauss 6.1 Introduction There are few functions of special importance in connection with the study of the divisors of an integer. Any function whose domain of deﬁnition is the set of integers is known as the arithmetic function. Present chapter treats multiplicative functions deﬁned on integers, having the property that there is value at the product of two relatively prime integers equal to the product of the value of the function at these integers. In this chapter several results about multiplicative functions are done that will be used later on. We also deﬁne the sum of divisors and the number of divisors functions. Later on the Mo¨bius functions investigate integers in terms of their prime decomposition. The summatory function of a given function takes the sum of the values of f at the divisors of a given integer a. The Mo¨bius inversion formula is determined, which writes the values of f in terms of the values of its summatory function. We conclude the chapter by presenting the Greatest Integer function along with some interesting properties. 113 114 Number Theory and its Applications 6.2 The Sum and Number of Divisors In mathematics, function has an important role to play with diﬀerent topics. For advance study of number theory we have some important aspects related to functions. Any function whose domain is the set of integers are called the number theoretic function or arithmetic functions. The range set may be other than positive integers also. We will start our discussions with the sum and number of divisors. Deﬁnition 6.2.1. Given a positive integer n, τ (n) is deﬁned as total number of positive divisors of n. For an example if we choose n = 12, then τ (12) = 6 as the divisors are 1, 2, 3, 4, 5, 6, 12. In the following table we have shown few integers and their corresponding number of divisors. n τ (n) 2 2 3 2 4 3 5 2 6 4 7 2 8 4 9 3 10 4 Deﬁnition 6.2.2. Given a positive integer n, σ(n) is deﬁned as the sum of their divisors. For example if we choose n = 12, then σ(12) = 1 + 2 + 3 + 4 + 5 + 6 + 12 = 28. In the following table we have shown few integers and their corresponding sum of divisors. n σ(n) 2 3 3 4 4 7 5 6 6 12 7 8 8 15 9 13 10 18 Before going for further discussions we are going to interpret the symbol f (d) d\|n which means ‘Sum of values of f (d) as d runs over all positive divisors of n’. f (d). If n is prime This sum is denoted as F (n) and deﬁned as F (n) = d\|n then τ (n) = 2 and σ(n) = n + 1. The converse is also true is justiﬁed with f (d) = f (1) + f (2) + f (4) + f (5) + f (10) + f (20) ı.e. the given example: τ (n) = 1, σ(n) = d\|n σ(10) = d\|20 d\|n d, therefore, τ (10) = 1 = 1 + 1 + 1 + 1 + 1 = 4 and d\|10 d = 1 + 2 + 5 + 10 = 18. Those are already shown in the above two d\|10 given tables. The ﬁrst theorem of the chapter aims to ﬁnd the positive divisors of a positive integer where the prime factorisation of that positive integer is already known. Arithmetic Functions 115 Theorem 6.2.1. If n = pt11 pt22 · · · ptrr is the prime factorization of n > 1, then the positive divisors of n are precisely, those integers d of the form d = b b p11 p11 · · · pb1r , where 0 ≤ bi ≤ ti (i = 1, 2, 3, . . . r) and vice–versa. Proof. If d = 1, then b1 = b2 = · · · = br = 0 and d = n, then b1 = t1 , . . . , br = tr . Let n = dd where d, d > 1 holds. Then they can be expressed as product of primes where d = q1 q2 q3 . . . qs , d = r1 r2 r3 . . . ru considering qi , rj as primes. Hence pt11 pt22 · · · ptrr = q1 q2 q3 . . . qs r1 r2 r3 . . . ru . By uniqueness of primes some of qi is same as pj so collecting them we have, d = q1 q2 q3 . . . qs = pb11 · · · pb1r where bi = 0 is possible. Conversely, every number d = pb11 pb11 · · · pb1r turns out to be the divisor of n. Then we have, n = pt11 pt22 · · · ptrr = (pb11 pb22 · · · pbrr )(pt11 −b1 pt22 −b2 · · · ptrr −br ) = dd where d = pt11 −b1 pt22 −b2 · · · prtr −br and ti − bi ≥ 0 for all i, then d > 0 and d\|n. The next theorem deals with the formula for both the number theoretic functions τ (n) and σ(n). The previous two tables on these two functions illustrate, if the integer n is prime then τ (n) = 2 and σ(n) = n + 1. In particular if n = pα where p is prime then the divisors of pα are 1, p, p2 · · · pα thus τ (pα ) = α + 1 and pα+1 − 1 . Thus the general formula for these σ(pα ) = 1 + p + p2 + · · · + pα = p−1 two functions are as follows. Theorem 6.2.2. If n = pt11 pt22 · · · ptrr is prime factorization of r > 1 then the followings are true. 1. τ (n) = (t1 + 1)(t2 + 1)(t3 + 1) . . . (tr + 1). t +1 2. σ(n) = t +1 ptr +1 − 1 p11 − 1 p22 − 1 ··· r . p1 − 1 p2 − 1 pr − 1 Proof. 1. According to the above theorem, the positive divisors of n are precisely those integers d = pb11 pb11 · · · pb1r where 0 ≤ bi ≤ ti holds. So there are t1 + 1 choices for b1 , (t2 + 1) choices for b2 and continuing we have tr + 1 choices for br . Therefore total number of positive divisors are + 1) . . . (tr + 1). Hence τ (n) = (t1 + 1)(t2 + 1)(t3 + (t1 + 1)(t2 + 1)(t 3 1) · · · (tr + 1) = (tj + 1). 1≤j≤r 2. In order to evaluate σ(n) we consider the product, (1 + p1 + p21 + p31 + . . . pt11 )(1 + p22 + p32 + . . . pt22 ) · · · (1 + pr + p2r + . . . ptrr ) where each term in 116 Number Theory and its Applications brackets are positive divisor of each prime factorisation of n. Therefore σ(n) = (1 + p1 + p21 + p31 + . . . pt11 )(1 + p22 + p32 + . . . pt22 ) · · · (1 + pr + p2r + . . . ptrr ) t +1 t +1 ptr +1 − 1 p11 − 1 p22 − 1 ··· r p1 − 1 p2 − 1 pr − 1 t +1 pi i − 1 . = pi − 1 = 1≤i≤r We illustrate this with an example. Example 6.2.1. Let n = 150 = 2×3×52 then τ (150) = (1+1)(1+1)(2+1) = 12 22 − 1 3 2 − 1 5 3 − 1 = 372. and σ(150) = 2−1 3−1 5−1 Now the following deﬁnition deals with a special property of number theoretic functions known to be multiplicative property: Deﬁnition 6.2.3. A number theoretic function f is said to be multiplicative if, for positive integers m and n, f (mn) = f (m)f (n) where gcd(m, n) = 1. Remark 6.2.1. The function f (n) = 1, ∀ n ∈ Z is multiplicative because f (mn) = 1, f (m) = 1 & f (n) = 1, so that f (mn) = f (m)f (n). Similarly, the identity function g(n) = n, ∀ n ∈ Z is multiplicative, since g(mn) = mn = g(m)g(n). Observe that multiplicative functions f and g with the property f (mn) = f (m)f (n) and g(mn) = g(m)g(n) for all pairs of integers m and n, whether or not gcd(m, n) = 1, is said to be completely multiplicative functions. Now we are at the stage to discuss the multiplicative property of τ and σ. Theorem 6.2.3. The functions τ and σ are both multiplicative. Proof. Let m and n be two relatively prime integers both greater than 1, for if any one of them is 1 then the result is trivial. So our primal assumption is both m, n > 1. Let m = pt11 pt22 · · · ptrr and n = q1j1 q2j2 · · · qsjs be prime factorisation of m and n respectively where no pi = qr because gcd(m, n) = 1 and if any of pi ’s same as any of qr ’s then this leads to a contradiction(Why!). Therefore mn = pt11 pt11 pt22 · · · ptrr q1j1 q2j2 · · · qsjs and τ (mn) = [(t1 + 1)(t2 + 1) · · · (tr + 1)][(j1 + 1)(j2 + 1) · · · (js + 1)] = τ (m)τ (n) σ(mn) = pti i +1 − 1 q js +1 − 1 i pi − 1 qs − 1 1≤i≤r = σ(m)σ(n). 1≤j≤s Arithmetic Functions 117 We will continue our study on multiplicative functions of positive divisors for products of relatively prime integers. Next lemma is the ﬁrst step on this study. Lemma 6.2.1. If gcd(m, n) = 1, then the set of positive divisors of mn consists of all products d1 d2 where d1 \|m and d2 \|n and gcd(d1 , d2 ) = 1. Proof. Let us assume m, n > 1 and m = pt11 pt22 · · · ptrr and n = q1j1 q2j2 · · · qsjs be their respective prime factorization. Therefore mn = pt11 pt22 · · · ptrr q1j1 q1j1 q2j2 · · · qsjs . Hence any positive divisors d of mn represented in the form d = p1a1 pa2 2 · · · prar q1b1 q1b1 q2b2 · · · qsbs , where, 0 ≤ ai ≤ ti and 0 ≤ bi ≤ ji then d = d1 d2 where d1 = p1a1 pa2 2 · · · prar where d1 \|m and d2 = q1b1 q1b1 q2b2 · · · qsbs where d2 \|n and gcd(d1 , d2 ) = 1, as pi = qj because m and n are relatively prime. Before proceeding further we will illustrate the idea of a positive divisor f (d) is by means of an example. Let m = 4 and n = 5. Also F (n) = d\|n deﬁned earlier in this section. Here we choose f as an arithmetic function. In this example we will show F (20) = F (4)F (5). Now the divisors of 20 are 1, 2, 4, 5, 10, 20, therefore F (20) = f (1) + f (2) + f (4) + f (5) + f (10) + f (20). Also the divisors of 4 are 1, 2, 4 and of 5 are 1, 5. Thus we have, F (20) = f (1) + f (2) + f (4) + f (5) + f (10) + f (20) = f (1 · 1) + f (1 · 2) + f (1 · 4) + f (1 · 5) + f (2 · 5) + f (4 · 5) = f (1)f (1) + f (1)f (2) + f (1)f (4) + f (1)f (5) + f (2)f (5) + f (4)f (5) = (f (1) + f (2) + f (4))(f (1) + f (5)) = F (4)F (5) This shows the arithmetic function F is multiplicative. Now the theorem is as follows, Theorem 6.2.4. If f is a multiplicative function and F is deﬁned by F (n) = f (d) then F is also multiplicative. d\|n 118 Number Theory and its Applications Proof. Let m, n are relatively prime integer then F (mn) = f (d) = f (d1 d2 ), d1 \|m,d2 \|n d\|mn where gcd(d1 , d2 ) = 1 and f is multiplicative, then we have f (d1 d2 ) = f (d1 )f (d2 ). Therefore ⎛ ⎞⎛ ⎞ F (mn) = f (d1 )f (d2 ) = ⎝ f (d1 )⎠ ⎝ f (d2 )⎠ = F (m)F (n). d1 \|m,d2 \|n d1 \|m d2 \|n This proves the fact that F is multiplicative. 6.3 Worked out Exercises Problem 6.3.1. Prove that there are inﬁnitely many pairs of integers m and n with σ(m2 ) = σ(n2 ). Solution 6.3.1. There are inﬁnitely many integers k such that gcd(k, 10) = 1. Let us consider m = 5k, n = 4k. This implies there exist inﬁnitely many such m, n. Suppose k is prime with k = 2, 5. Now m2 = 52 k 2 and n2 = 42 k 2 = 24 k 2 . Theorem 6.2.2 yields 53 − 1 5−1 25 − 1 σ(n2 ) = 2−1 σ(m2 ) = 3 k −1 k3 − 1 = 31 . k−1 k−1 3 k −1 k3 − 1 · = 31 . k−1 k−1 · Thus there are inﬁnitely many pairs of integers m and n with σ(m2 ) = σ(n2 ). Problem 6.3.2. If n is a square-free integer, prove that τ (n) = 2s , where s is the number of prime divisors of n. Solution 6.3.2. Since n is square-free, therefore n = p1 p2 · · · pr where each pi = pj = for i = j. From Theorem 6.2.2, we obtain τ (n) = (k1 + 1)(k2 + 1) · · · (ks + 1), with ki = 1 for all i. Thus τ (n) = (1 + 1)(1 + 1) · · · (1 + 1) = 2 · 2 · · · 2 = 2s as there are s terms Problem 6.3.3. Prove that the following statements are equivalent: 1. τ (n) is an odd integer. Arithmetic Functions 119 2. n is a perfect square. k k Solution 6.3.3. 1⇒2:Suppose (1) holds. Let n = p1 1 p2 2 · · · pks s . Then using Theorem 6.2.2, we have τ (n) = (k1 + 1)(k2 + 1) · · · (ks + 1). Note that each ki + 1 is odd, so ki is even. Hence ki = 2ji implies 1 p2j2 · · · p2js = (pj1 pj2 · · · pjs )2 , n = p2j s s 1 2 1 1 which proves n is a perfect-square. k k 2⇒1:Suppose (2) holds. Then n = a2 for some a = p1 1 p2 2 · · · pks s , which implies 2 · · · p2ks . n = p12k1 p2k s 2 Therefore τ (n) = (2k1 + 1)(2k2 + 1) · · · (2ks + 1). Since each of 2ki + 1 is odd, therefore τ (n) is odd. Problem 6.3.4. For any positive integer z, prove 1 d\|z d = σ(z) . z z is a divisor of Solution 6.3.4. Note that d is a divisor of z if and only if d z(Why!). Therefore the set of divisors of z are given by {d1 , d2 , . . . , ds }, which z z z . Thus further can be expressed as , ,..., d1 d2 ds 1 z z z 1 1 , + + ... + =z + + ... + σ(z) = d1 + d2 + . . . + ds = d1 d2 ds d1 d2 ds implies 1 σ(z) 1 1 1 = . + + ... + = z d1 d2 ds d d\|z Problem 6.3.5. If z = q1t1 q2t2 · · · qsts is the prime factorization of z > 1, then prove that z 1 1 1 1> > 1− 1− ··· 1 − . σ(z) q1 q2 qs Solution 6.3.5. Since the divisors of z include 1 and z, therefore σ(z) ≥ z + 1 > z ⇒ z < 1. σ(z) By virtue of Theorem 6.2.2, we obtain t +1 σ(z) = t +1 q ts +1 − 1 q11 − 1 q22 − 1 ··· s . q1 − 1 q2 − 1 qs − 1 120 Number Theory and its Applications Therefore t z = σ(z) t +1 (q11 t q11 q22 · · · qsts t +1 −1)(q22 −1)···(qsts +1 −1) (q1 −1)(q2 −1)···(qs −1) t t (q11 q22 · · · qsts )(q1 − 1)(q2 − 1) · · · (qs − 1) = t +1 t +1 (q11 − 1)(q22 − 1) · · · (qsts +1 − 1) (q1 − 1)(q2 − 1) · · · (qs − 1) = t +1 (q11 t +1 −1)(q22 t q11 t q22 −1)···(qsts +1 −1) ···qsts (q1 − 1)(q2 − 1) · · · (qs − 1) q1 − 1t1 q2 − 1t2 · · · qs − = q1 q2 But qi > qi − 1 ti qi 1 qi − > 1 t qi i . 1 (6.3.1) qsts 1 . qi Thus (6.3.1) yields, (q1 − 1)(q2 − 1) · · · (qs − 1) q1 − 1t1 q2 − 1t2 · · · qs − z = σ(z) q1 q2 1 qsts (q1 − 1)(q2 − 1) · · · (qs − 1) q q · · · qs 12 1 1 1 1− ··· 1 − . = 1− q1 q2 qs > Hence z 1> > σ(z) 1 1− q1 1 1− q2 1 ··· 1 − qs . Problem 6.3.6. Prove if z > 1 is a composite number, then σ(z) > z + √ z. Solution 6.3.6. Since σ(z) = 1 + d1 + d2 + . . . + ds + z, its suﬃces to show that d1 + d2 + . . . + ds > √ z. Since z is composite, there exists di (for some i) such that 1 < di < z and di \|z. Therefore dz \|z and di < z together implies 1 < dz and 1 < di ⇒ d1 < 1 ⇒ i i i z di < z. Therefore z < z. 1< di Now two cases may arise: Arithmetic Functions √ √ Case(i) If di > z, then clearly 1 + di > z. So σ(z) = 1 + d1 + d2 + . . . + ds + z > z + Case(ii) If di ≤ 121 √ z. √ z, then √ 1 z 1 z √ ≤ ⇒ z=√ ≤ . d d z z i i √ √ z . Then dj \|z implies dj ≥ z. Therefore 1 + dj + z > z + z. di √ Hence from σ(z) = 1 + d1 + d2 + . . . + ds + z, it follows σ(z) > z + z. Let dj = Thus combining the above cases the assertion follows. Problem 6.3.7. For any integer k > 1, show that 1. there exist inﬁnitely many integers n for which τ (n) = k, 2. but at most ﬁnitely many n with σ(n) = k. Solution 6.3.7. 1. Let p be any prime and n = pk−1 . Then τ (n) = k(How!). Since there are inﬁnitely many primes, therefore there are inﬁnitely many n satisfying n = pk−1 and τ (n) = k. 2. Using Problem (6.3.5), we have σ(n) > n, ∀n. If σ(n) = k, for any k, then k serves as an upper bound to n. In fact, for any n ≥ k, σ(n) > k(How!). Hence there are at most k(> 1) integers such that σ(n) ≤ k. Problem 6.3.8. If pair of successive odd integers q and q + 2 that are both primes, called twin primes. For these q and q + 2 prove that σ(q + 2) = σ(q) + 2. Solution 6.3.8. The only divisors for any prime q are 1 and q itself. Therefore σ(q) = q + 1. Thus σ(q + 2) = q + 3 and σ(q) + 2 = q + 3. Thus q and q + 2 together implies σ(q + 2) = σ(q) + 2. Problem 6.3.9. Let f and g be multiplicative functions that are not identically zero and have the property that f (pk ) = g(pk ) for each prime p and k ≥ 1. Prove that f = g. Solution 6.3.9. Let n > 1 be an arbitrary integer with prime factorization given by n = q1t1 q2t2 · · · qsts . Then f (n) = f (q1t1 q2t2 · · · qsts ) = f (q1t1 )f (q2t2 ) · · · f (qsts ) = g(q1t1 )g(q2t2 ) · · · g(qsts ) = g(q1t1 q2t2 · · · qsts ) = g(n). 122 Number Theory and its Applications In particular, if n = 1 then f (1) = g(1) = 1. Hence combining all f = g holds. Problem 6.3.10. For any integer z > 1, prove that there exist integers z1 and z2 for which τ (z1 ) + τ (z2 ) = z. Solution 6.3.10. If z is prime, then τ (z) = 2. Since tau(1) = 1, taking z1 = z2 = 1 gives τ (z1 ) + τ (z2 ) = τ (z). t t If z is composite, let z = q11 q22 · · · qsts be its prime factorization. Then at tj least one of ti ≥ 2. Let qj be that factor. Therefore τ (z) = (t1 + 1)(t2 + 1) · · · (tj + 1) · · · (ts + 1) = tj (t1 + 1)(t2 + 1) · · · (ts + 1) + (t1 + 1)(t2 + 1) · · · (ts + 1). z t Let z1 = q1t1 q2t2 · · · qj j−1 · · · qsts and z2 = Hence tj qj = q1t1 q2t2 · · · qsts , where qi = qj . τ (z1 ) = (t1 + 1)(t2 + 1) · · · (tj − 1 + 1) · · · (ts + 1) = tj (t1 + 1)(t2 + 1) · · · (ts + 1). τ (z2 ) = (t1 + 1)(t2 + 1) · · · (ts + 1). Combining we obtain, τ (z1 ) + τ (z2 ) = τ (z). Problem 6.3.11. For any integer z ≥ 1, prove that d\|z τ (d)3 = d\|z 2 τ (d) . Solution 6.3.11. Since τ (z) is a multiplicative function, therefore [τ (mn)]3 = [τ (m)τ (z)]3 = [τ (m)]3 [τ (z)]3 . This shows τ (z)3 is a multiplicative function. τ (d)3 is multiplicative. MoreHence by virtue of Theorem 6.2.4, F (z) = over, the multiplicative property of G(z) = multiplicative(Why!). t t d\|z d\|z τ (d) implies H(z) = G(z)2 is Let z = q11 q22 · · · qsts be its prime factorization. Since F and H both are multiplicative, for z = q t , F (z) = H(z) holds. By similar reasoning, the relation F (z) = H(z) holds true for z = q1t1 q2t2 · · · qsts . Therefore considering z = q t and applying Theorem 6.2.1, all the divisors of z are given by 1, q, q 2 , . . . , q t . Thus Arithmetic Functions 123 τ (d)3 = τ (1)3 + τ (q)3 + . . . + τ (q t )3 d\|q t = 1 + (1 + 1)3 + (2 + 1)3 + . . . + (t + 1)3 2 τ (d) = 1 + 23 + 33 + . . . + (t + 1)3 2 (t + 1)(t + 2) = . 2 = [τ (1) + τ (q) + . . . + τ (q t )]2 d\|q t = [1 + (1 + 1) + (2 + 1) + . . . + (t + 1)]2 2 (t + 1)(t + 2) = . 2 2 τ (d)3 = τ (d) , so F (z) = H(z) for z = q t . Hence d\|q t d\|q t denote the sum of the sth powers of Problem 6.3.12. Given z ≥ 1, let σs (z) ds . Prove that the positive divisors of z; that is, σs (z) = d\|z σs (z) = s(t1 +1 ) q1 −1 q1s − 1 s(t2 +1 ) q2 −1 q2s − 1 ··· qrs(t1 +1 ) − 1 qrs − 1 , z = q1t1 q2t2 · · · qrtr being the prime factorization of z. Solution 6.3.12. By virtue of Theorem 6.2.1, all divisors of z are of the form q1a1 q2a2 . . . qrar , 0 ≤ ai ≤ ti . Therefore all the sth powers of the divisors of z are of the form q1a1 s q2a2 s . . . qrar s . Let us consider the product of sums (1 + q1s + q12s + q1t1 s ) . . . (1 + qrs + qr2s + qrtr s ). Each positive divisor to the sth power occurs only once as a term in the expansion of the product. Therefore σs (z) = (1 + q1s + q12s + q1t1 s ) . . . (1 + qrs + qr2s + qrtr s ). 124 Number Theory and its Applications Applying the formulae for the sum of the ﬁnite geometric series, s t1 s 2s (1 + qi + qi + qi ) = ⇒ σs (z) = s(t1 +1 ) q1 −1 q1s − 1 s(ti +1 ) −1 qis − 1 qi s(t2 +1 ) q2 −1 q2s − 1 , ··· qrs(t1 +1 ) − 1 qrs − 1 . Problem 6.3.13. For any positive integer z, show that σ(d) = d\|z Solution 6.3.13. Let H(n) = d\|z z d\|z G(mn) = = mn τ (d) d d\|mn mn d1 \|m,d2 \|n = d1 \|m,d2 \|n d1 d2 z d d τ (d). τ (d). Then τ (d1 d2 ) mn τ (d1 )τ (d2 ) d1 d2 m n τ (d1 ) τ (d2 ), since τ (d) is multiplicative d1 d2 d1 \|m,d2 \|n m n = τ (d1 ) τ (d2 ) d1 d2 = d1 \|m,d2 \|n d1 \|m,d2 \|n = G(m)G(n). Hence G is a multiplicative function. Using multiplicative property of σ(d) the function F (n) = d\|z σ(d) is multiplicative. t t Next let z = q11 q22 · · · qsts be its prime factorization. To prove F (z) = F (q1t1 q2t2 · · · qsts ) = F (q1t1 )F (q2t2 ) · · · F (qsts ) = G(q1t1 )G(q2t2 ) · · · G(qsts ) = G(q1t1 q2t2 · · · qsts ) = G(z), Arithmetic Functions 125 t t its suﬃces to show F (q ) = G(q ). Now F (q t ) = σ(d) d\|q t = q 0 + (q 0 + q 1 ) + . . . + (q 0 + q 1 + . . . + q s ) = (t + 1)q 0 + tq 1 + . . . + (1)q t = (1)q t + · · · + tq 1 + (t + 1). qt τ (d) G(q t ) = d t (6.3.2) d\|q = qt qt qt τ (q 0 ) + 1 τ (q 1 ) + . . . + t τ (q t ) 0 q q q = (1)q t + 2q t−1 + . . . + tq + (t + 1) = F (q t ), by (6.3.2). √ Problem 6.3.14. For any integer z ≥ 1, prove that τ (z) ≤ 2 z. √ √ √ z ≤ z. For if d > z or Solution 6.3.14. If d\|z, then either d ≤ z or d √ z > z, then d √ √ z d · = z > z z = z, d a contradiction. Let d1 , d2 , . . . , dt be the divisors of z where d1 < d2 < . . . < dt . Clearly, d1 = 1, dt = z. Now di \|z ⇒ z\|di and so z\|di must be one of di . pairing the divisors in such a way that di dj = z, where dj = z\|di . So either di ≤ dj or di ≥ dj . Case(i) t is even: Then we have 2t unique pairs {di , dj }(di = dj ) such that di dj = z. Let us arrange every pair in such a way that di < dj . Let dt be t t unique pairs, it must be ≤ dt . the largest of the di . Since there are 2 √ 2 But τ (z) = t and from above dt ≤ z. Thus, √ τ (z) √ ≤ z ⇒ τ (z) ≤ 2 z. 2 t−1 unique pairs {di , dj }(di = dj ) such that Case(ii) t is odd: Then we have 2 di dj = z and one pair {dr , dr } where dr dr = z. Let us arrange every unique pair in such a way that di < dj . Let dt be the largest of the di . Now if dr < dt , considering the pair {dj , dt } and applying the deﬁnition of dt , we obtain dt < dj , and dt dj = z. 126 Number Theory and its Applications Hence dr < dj ⇒ dr dr < dr dj ⇒ z < z, a contradiction. Hence dr > dt . √ But d2r = z ⇒ dr = z. As in Case(i), t−1 ≤ dt and t = τ (z). 2 Therefore √ τ (z) − 1 ≤ dt < dr = z, 2 which further implies √ √ τ (z) − 1 √ ≤ z, τ (z) − 1 < 2 z ⇒ τ (z) ≤ 2 z. 2 √ Hence τ (z) ≤ 2 z for both even and odd cases. Problem 6.3.15. Find the form of all positive integers n satisfying τ (n) = 10. What is the smallest positive integer for which this is true? t t Solution 6.3.15. Let n = q11 q22 · · · qsts be the prime factorization of n. Then τ (n) = (t1 + 1)(t2 + 1) . . . (ts + 1). If τ (n) = 10, then the possibilities are 10 and 5 × 2. This implies t1 + 1 = 10 or (t1 + 1)(t2 + 1) = 5 × 2. Thus n = q19 or n = q14 q2 where q1 , q2 are distinct primes. The smallest of such integers would be 29 or 24 × 3 or 34 × 2. Then the smallest among them is 24 × 3 = 48. 6.4 Mobi¨ us μ-function In this article, we will discuss an important arithmetic function called Mobi¨ us μfunction with some of its properties. Deﬁnition 6.4.1. For a positive integer n, deﬁne μ by the rules ⎧ ⎪ ⎨ μ(n) = ⎪ ⎩ 1 if n = 1; 0 if p2 \|n for some prime p; (−1)r if n = p1 p2 · · · pr , where pi ’s are distinct primes. For example we see that μ(2) = −1, μ(3) = −1, μ(4) = 0. Thus if we choose n = 15 = 3 × 5 then μ(15) = μ(3 · 5) = (−1)2 = 1. Now in the next theorem we are going to discuss the multiplicative property on Mobi¨ ous μ function. Theorem 6.4.1. The function μ is a multiplicative function. Arithmetic Functions 127 Proof. Its suﬃces to show that for any two relatively prime integers m and n, μ(mn) = μ(m)μ(n). This is trivial for m = n = 1. Now if we choose either p2 \|n or p2 \|m then p2 \|mn. Therefore μ(mn) = 0 = μ(m)μ(n). This case is also trivial. Now we assume m, n to be square-free integers. Then m = p1 p2 · · · pa , n = q1 q2 · · · qb where pi and qj are all distinct, then μ(mn) = μ(p1 p2 · · · pa q1 q2 · · · qb ) = (−1)a+b = (−1)a (−1)b = μ(m)μ(n). This proves that μ is a multiplicative function. Now from the above theorem we can see that both m and n are divisors of mn. A natural question arises how this function behaves with divisors of any integers. If n = 1 then the only divisor is d = 1 therefore μ(d) = μ(1) = 1. d\|1 So we have to discuss the divisors for those n > 1 and for that we need to apply μ(d) which has already been discussed in the ﬁrst section the formula F (n) = d\|n of this chapter. Our next theorem illustrates the clariﬁcation of this discussion. Theorem 6.4.2. For each positive integer 1, if n = 1; μ(d) = n ≥ 1, 0, if n > 1. d\|n where d is positive divisors of n. Proof. The assertion is obvious if n = 1, then μ(d) = μ(1) = 1. We proceed d\|n by mathematical induction on the number of diﬀerent prime factors of n when n > 1 and if n = pα , then μ(d) = μ(1) + μ(p) + . . . + μ(pα ) = 1 + (−1) = 0. d\|pα Since μ is multiplicative, using Theorem 6.2.4, F is also so. Thus if n = p1α1 p2α2 p3α3 · · · prαr , then d\|n μ(d) = α d\|p1 1 μ(d) α d\|p2 2 μ(d) · · · μ(d) = 0. r d\|pα r To illustrate the above theorem let us consider n = 12, the divisors of 12 are 1, 2, 4, 3, 6 and 12. Thus the required sum is, 128 Number Theory and its Applications μ(d) = μ(1) + μ(2) + μ(3) + μ(4) + μ(6) + μ(12) d\|12 = 1 + (−1) + (−1) + 0 + (−1)2 + 0 = 0. In mathematics, the classic Mobi¨ us inversion formula was introduced into number theory on 1832 by August Ferdinand Mobi¨ us, stated as follows: Theorem 6.4.3. Mobi¨ us Inversion Formulae: Let F and f be two number theo retic functions related by the formulae F (n) = f (d) and, f (n) = μ(d)F (n/d) d\|n d\|n for every n. If either of them is true then they satisfy both the formulae. f (d), then Proof. Let us ﬁrst choose that F (n) = d\|n μ(d)F (n\|d) = μ(d)F (d ), since integer d is the quotient when d\|n dd =n d\|n = μ(d) dd =n = e\|d f (e) ek =n μ(d), taking integer k = dh for some integer h. d\|k Now if k = 1, then using the Theorem 6.4.2 we have, f (d) for each n d\|n μ(d)f (e), as integer k is the quotient when e\|d dek=n = f (e), since F (n) = μ(d) = 1. Therefore d\|k μ(d)F (n\|d) = f (n). d\|n Conversely let, f (n) = μ(d)F (n/d) holds. Then, d\|n f (d) = d\|n d\|n = μ(d)F d \|d μ(d )F (p), d pq=n = ph =n F (p) d \|h d d Since d = d p, n = qd = d pq for some integer p and q μ(d ) where, h = d q for some integer h . Now again applying Theorem 6.4.2 we have, d \|h μ(d ) = 1 if h = 1 holds. f (d) = F (n). Therefore d\|n Arithmetic Functions 129 Before going to the last result of this section we see from Theorem 6.2.4 that if f (n) is multiplicative then F (n) is also multiplicative for each integer n. Now the question arises, is the converse assertion also true. The following theorem illustrates the answer of it. Theorem 6.4.4. If F is a multiplicative function and F (n) = f (s) then f s\|n is also multiplicative for any integer n and positive divisor s. Proof. Let m, n be relatively prime positive integers then any divisor s of mn can be uniquely written as s = s1 s2 where s1 \|m and s2 \|n where gcd(s1 , s2 ) = 1. Now by inversion formulae we have, f (mn) = μ(s)F s\|mn mn s = mn s1 s2 ⎞⎛ μ(s1 s2 )F s1 \|m,s2 \|n =⎝ μ(s1 )F s1 \|m m ⎠⎝ s1 μ(s2 )F s2 \|n n ⎠ s2 = f (m)f (n) (Why!). This proves the theorem. 6.5 Worked out Exercises Problem 6.5.1. Suppose a function Λ is deﬁned by ln p, if n = pk , where p is a prime and k ≥ 1; 0, otherwise. Prove that Λ(n) = d/n μ nd ln d = − d/n μ(d) ln d. Λ(n) = Solution 6.5.1. Let n = pk . Then n ln d = μ(pk ) ln 1+μ(pk−1 ) ln p+. . .+. . .+μ(pk−i ) ln pi +. . .+μ(p0 ) ln pk . μ d d/n Case(i) If k = 1, then d/n μ n d ln d = ln p(Verify!). Case(ii) If k > 1, then μ(pk−i ) = 0 except for i = 1, 2. Then the sum is same as k = 1. 130 Hence d/n Number Theory and its Applications μ n d ln d = ln p = Λ(n). Next μ(d) ln d = μ(p0 ) ln 1 + μ(p1 ) ln p1 + . . . + . . . + μ(pi ) ln pi + . . . + μ(pk ) ln pk . d/n For k > 1, μ(pk ) = 0 implies d/n μ(d) ln d = − ln p for all k. Hence μ(d) ln d = −Λ(n). d/n Remark 6.5.1. The function Λ in the Problem 6.5.1 is known as Mangoldt function. Problem 6.5.2. Let n = pk1 1 pk2 2 · · · pks s be the prime factorization of the integer n > 1. If f is a multiplicative function that is not identically zero, prove that μ(d)f (d) = (1 − f (p1 ))(1 − f (p2 )) · · · (1 − f (ps )). d/n Solution 6.5.2. Since μ and f is multiplicative, therefore μf is also so(Why!). By virtue of Theorem 6.2.4, F (n) = d/n μ(d)f (d) is multiplicative. Consider, F (pk ) = μ(d)f (d) d/pk = μ(1)f (1) + μ(p)f (p) + · · · + μ(pk )f (pk ) = μ(1)f (1) + μ(p)f (p)(W hy!) = 1f (1) + (−1)f (p) = f (1) − f (p). Since for a multiplicative function not identically zero, therefore f (1) = 1 implies F (pk ) = 1 − f (p). Thus μ(d)f (d) = (1 − f (p1 ))(1 − f (p2 )) · · · (1 − f (ps )). d/n Problem 6.5.3. Let S(n) denote the number of square-free divisors of n. Prove that S(n) = \|μ(d)\| = 2ω(n) , d/n where ω(n) is the number of distinct prime divisors of n. Solution 6.5.3. Consider, Arithmetic Functions 131 ⎧ ⎪ ⎨ 1, if n = 1; \|μ(n)\| = 0, if p2 \|n, p being prime ; ⎪ ⎩ 1, if n = p1 p2 · · · ps , pi being distinct. Let gcd(m, n) = 1. Then \|μ(1)\| = 1. If m = 1, then \|μ(mn)\| = \|μ(n)\| = \|μ(m)\|\|μ(n)\|. If p2 \|m, then p2 \|mn implies \|μ(mn)\| = 0 and \|μ(m)\| = 0. Hence \|μ(mn)\| = \|μ(m)μ(n)\|. Assume, both m, n are square-free. Let m = p1 p2 · · · ps , n = q1 q2 · · · qr with pi = qj as gcd(m, n) = 1. Clearly, \|μ(m)\| = \|μ(n)\| = \|μ(mn)\| = 1. Hence, \|μ(mn)\| = \|μ(m)\|\|μ(n)\|. This shows \|μ(n)\| is multiplicative. Using Theorem (6.2.4), S(n) = d/n \|μ(n)\| is also so. Now, consider n = pk . The divisors of n are 1, p, p2 , . . . , pk . Therefore \|μ(n)\| = 2. S(n) = d/n The number of square-free divisors of pk is 2 and is deﬁned by k1 k2 d/n \|μ(n)\|. Given that, n = p1 p2 · · · ps . From Theorem 6.2.1, all the square divisors of n are represented by n = pa1 1 p2a2 · · · pas s , 0 ≤ aj ≤ 1. Here the number of squarefree divisors of pi is 2, which are 1 & pi . It is true for all i = 1, 2, 3, . . . , s. Hence the total number of square-free integers is 2s = 2ω(n) , where ω(n) is the number of distinct prime divisors of n. Therefore ks S(n) = S(pk1 1 pk2 2 · · · pks s ) = S(pk1 1 )S(pk2 2 ) · · · S(pks s ) k1 ks = \|μ(p1 )\| · · · \|μ(ps )\| k d/p1 1 k d/p1 1 = 2s = 2ω(n) . Problem 6.5.4. The Liouville λ function deﬁned as (−1)t1 +t2 +...+ts , if z = pt11 pt22 · · · ptss , z > 1; λ(z) = 1, if z = 1. 1. Prove that λ is multiplicative. 2. For some positive integer z, prove that 1, if z = k2 for some integer k; λ(d) = 0, Otherwise. d/z Solution 6.5.4. 1. Let us consider two positive integers z and k with gcd(z, k) = t t v v 1, where z = p11 p22 · · · ptss , k = q1 1 q2 2 · · · qrvr . 132 Number Theory and its Applications t1 t2 v1 v2 Now zk = p1 p2 · · · ptss q1 q2 · · · qrvr , pi = qj . Hence λ(zk) = (−1)t1 +t2 +...+ts +v1 +v2 +...+vr = (−1)t1 +t2 +...+ts · (−1)v1 +v2 +...+vr = λ(z)λ(k). This shows that λ(z) is multiplicative function. 2. Let F (z) = d/z λ(d). Then by Theorem 6.2.4, F is multiplicative. Let z = pt . Then, F (z) = λ(1) + λ(p) + λ(p2 ) + . . . + λ(pt ) = 1 + (−1) + (−1)2 + (−1)3 + . . . + (−1)t−1 + (−1)t . Now, two cases may arise: Case(i) t is even: Then z = p2ω , t = 2ω for some positive integer ω. Therefore, taking m = pω , we obtain z = m2 . Also, F (z) = 1. t t Case(ii) t is odd: Then we have F (z) = F (pt ) = 0. Let n = p11 p22 · · · ptss . t t Since F is multiplicative, therefore F (n) = F (p11 )F (p22 ) · · · F (ptss ). t If z = k2 for some integer k, then all the ti ’s are even. So F (pi i ) = 1 and consequently, F (z) = 1. Again, if any of the ti is odd, then t F (pi i ) = 0. So F (z) = 0. Problem 6.5.5. For every integer z ≥ 3, prove that z t=1 μ(t!) = 1. Solution 6.5.5. Here μ(4) = 0(Why!). If n ≥ 4, then z! would contain 4 as a factor. Since μ is multiplicative, therefore for z ≥ 4, μ(z!) = 0. So, only case need to consider is z = 3. Now μ(1) = 1, μ(2) = −1 = μ(3) implies 3 t=1 μ(t!) = 1 + (−1) + 1 = 1. 6.6 Greatest Integer Function In this section we are going to discuss a special type of arithmetic function called greatest integer function. The domain of deﬁnition of this function is the set of real numbers and the range set is the set of integers. This function is very much useful for calculating continued fractions. The deﬁnition of the function as follows. Deﬁnition 6.6.1. For an arbitrary real number x, the largest integer less than or equal to x and denoted by [x] is called the greatest integer function. Arithmetic Functions 133 For an example we have [2.2] = 2 and [−2.2] = −3. Here for every real number x, there is a unique real number θ such that x = [x] + θ, 0 ≤ θ < 1, where θ is the fractional part of x. This θ sometimes denoted as {x} such that x = [x]+{x}, ∀x ∈ R. Actually the greatest integer function for any real number x follows the inequality x − 1 < [x] ≤ x. In our next proposition we have shown division algorithm using this inequality. Proposition 6.6.1. For any x ∈ R, prove division algorithm by the inequality x − 1 < [x] ≤ x. m Proof. Let q = m n and r = m − n n , clearly m = nq + r and we will show that m m the remainder satisﬁes the above inequality. As ∈ R then m −1 < m n n ≤ n. n Now multiplying by −n the above inequality and changing the order of inequality m we have, −m ≤ −n m n < n − m. Adding m we get, 0 ≤ m − n n < n ⇒ 0 ≤ r < n. We are to show this q and r are unique. Let us assume that they are not unique then m = nq1 + r1 = nq2 + r2 for q1 , q2 are quotients and 0 ≤ r1 , r2 < n where r1 , r2 are remainders. Now subtracting these two equations we have, 0 = n(q1 − q2 ) + (r1 − r2 ) thus (r2 − r1 ) = n((q1 − q2 )) which implies n\|(r1 − r2 ) but this is possible only if r1 − r2 = 0. Therefore r1 = r2 and q1 − q2 , which shows that q is unique quotient and r is unique remainder. Now we will discuss few properties related to this greatest integer function. Proposition 6.6.2. For any x, y ∈ R and m ∈ Z, the greatest integer function satisﬁes following properties: (i) [x + m] = [x] + m 0, (ii) [x] + [−x] = −1, if x ∈ Z; if x ∈ R \ Z. (iii) [x] + [y] ≤ [x + y] x! [x] (iv) = . m m Proof. Let x = n + θ for n ∈ Z, 0 ≤ θ < 1, (i) Here, x + m = (n + m) + θ where m + n ∈ Z, 0 ≤ θ < 1. Thus [x + m] = n + m = [x] + m. (ii) Here, −x = −n − θ, (−n − 1) + (1 − θ), 0 ≥ −θ > −1; 0 < 1 − θ ≤ 1. 134 Number Theory and its Applications Therefore [−x] = −(1 + n), if 1 − θ = 1; −n, if 1 − θ = 1. This proves that [x] + [−x] = −1, 0, if x ∈ / Z; if x ∈ Z. (iii) Let y = r + θ , r ∈ Z, 0 ≤ θ < 1. Therefore x + y = (n + r) + (θ + θ ), 0 ≤ (θ + θ ) < 2. Thus we have, [x + y] = n + r, n + r + 1, if 0 ≤ θ + θ < 1; if θ + θ ≥ 1. Hence [x] + [y] = n + r ≤ [x + y]. (iv) Now let x z ∈ Z and 0 ≤ θ < 1. Then we have = z + θ, m mz ∈ Z and 0 ≤ mθ < m. x = mz + mθ, Therefore Hence [x] = [z] = z as x ∈ Z. m x! [x] = . m m Now we are going to the application part of this greatest integer function. For that we choose an integer 7 whose factorial is 7! = 1·2·3·4·5·6·7 = 24 ·32 ·5·7. Here we can see that the highest power of 2 which divides 7! is 4. We can ﬁnd the exponent of any prime which occurs in prime factorization of any factorial of an integer by greatest integer function using the next theorem. ∞ n is the exponent of p appearing Theorem 6.6.1. If p is a prime then, pk k=1 in the prime factorization of n!. Arithmetic Functions 135 Proof. If p > n then p does not appear in the prime factorization of n!. Thus we have p ≤ n. Among who are divisible by the ﬁrst n positive integers those n n p are p, 2p, 3p · · · p. Thus there exists exactly multiples of p occuring p p n n in the product of n!. Among those integers p, 2p, 3p · · · p there are 2 p p n integers which are again divisible by p2 and they are p2 , 2p2 , 3p2 · · · 2 p2 . p After continuing these steps ﬁnitely many times we get the total number of ∞ n . times p divides n! is pk k=1 Now we will illustrate this theorem by means of an example. 10 = 5 integers Example 6.6.1. Let us take n = 10 and p = 2 then there are 2 which are divisible by p = 2 and they are 2, 4, 6, 8, 10. Among those integers 10 there are = 2 integers which are divisible by 4 and they are 4, 8. Now 22 10 among these two integers there are = 1 integers which are divisible by 23 10 10 10 + 2 + 3 = 8.Now 8 and it is 8 itself.Therefore the total number is 2 2 2 8 4 2 10! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 = 2 · 3 · 5 · 7. So the highest power of 2 is 8. In our next two results we are going to ﬁnd some common well known facts of mathematics by using the last theorem. Theorem 6.6.2. Ifn and r are positive integers with 1 ≤ r < n, then the n n! = binomial coeﬃcient is an integer. r r!(n − r)! n n! = Proof. For proving is an integer we are only to show n! is r r!(n − r)! divisible by r!(n − r)!. Now from the Theorem 6.6.1 we have the exponent of ∞ n highest power of prime p that divides n! is and the highest power of pi i=1 ∞ ∞ r (n − r) n! prime p that divides is + . Again from the r!(n − r)! pi pi i=1 i=1 Proposition 6.6.2(iii) we have [a + b] ≥ [a] + [b] for any two integers a, b. Then we have, n r (n − r) ≥ + . pi pi pi Taking the summation we get, ∞ ∞ ∞ n r (n − r) ≥ + . pi pi pi i=1 i=1 i=1 136 Number Theory and its Applications n! r!(n − r)! at least as many times in the denominator. As p is arbitrary so r!(n − r)! must n! divide n!. Thus is an integer. r!(n − r)! From the above inequality it follows that p occurs in the numerator of Corollary 6.6.1. For any positive integer r, the product of r consecutive integers is divisible by r!. Proof. Here we can assume the product of r consecutive integers as n(n − n! 1) · · · (n − r + 1) where n is largest. Here n(n − 1) · · · (n − r + 1) = = r!(n − r)! n! n! × r! and from the Theorem 6.6.2 we know that is an r!(n − r)! r!(n − r)! integer.This proves the assertion of this corollary. In our later part of discussion on greatest integer function we have shown some valuable relations between this function and other arithmetic functions. Their relationship comes out as, Theorem 6.6.3. Let f and F be two arithmetic functions such that F (n) = N f (d) where n is a positive integer. Then for any positive integer N , F (n) = d\|n N m=1 f (m) n=1 N . m Proof. We are going to start the theorem by the form of F (n). Taking the N N sum over this function we have F (n) = f (d). Here we are to collect n=1 n=1 d\|n the terms with equal values of f (d). Since each integer divides itself then the f (d) assertion for a ﬁxed positive integer m ≤ N , the term f (m) appears in d\|n if and only if m is a divisor of n is possible. Now to calculate the number of f (d) in which f (m) occurs as a term, it is suﬃcient to ﬁnd terms in the sum d\|n the number of integers from the set {1, 2,· · · N } which are divisible by m.From N of them. Thus for each m such that the Theorem 6.6.1 there are exactly n N diﬀerent positive 1 ≤ m ≤ N , f (m) is a term of the sum f (d) for n d\|n integers less than or equal to N . Therefore N n=1 d\|n f (d) = N m=1 f (m) N N = F (n) m n=1 Arithmetic Functions 137 This proves the theorem. Our next corollary is the immediate application of this theorem on two arithmetic functions τ (n) and σ(n). Corollary 6.6.2. If N is a positive integer then, N τ (n) = n=1 N N n=1 Proof. We know that τ (n) = n and N σ(n) = n=1 1 and σ(n) = d\|n N N n . n n=1 d. Now taking F (n) = τ (n) d\|n and f (n) = 1, for all n ∈ N we have from the Theorem 6.6.3 N τ (n) = N N . n Again taking F (n) = σ(n) and f (n) = n, for all n ∈ N we have from the Theorem N N N 6.6.3 . σ(n) = n n n=1 n=1 n=1 n=1 Now to visualize those two forms of τ (n) and σ(n) we will go through an example given below. Exercise 6.6.1. Let us consider N = 4 then, 4 τ (n) = τ (1) + τ (2) + τ (3) + n=1 τ (4) = 1 + 2 + 2 + 3 = 8. Now, 4 4 n=1 4 = [4] + [2]+ + [1] = 4 + 2 + 1 + 1 = 8. n 3 Also, 4 σ(n) = σ(1) + σ(2) + σ(3) + σ(4) = 1 + 3 + 4 + 7 = 15 n=1 and 4 4 4 = 1[4] + 2[2] + 3 + 4[1] = 4 + 4 + 3 + 4 = 15. n n 3 n=1 6.7 Worked out Exercises Problem 6.7.1. Find the highest power of 5 dividing 1000!. Solution 6.7.1. 1000 1000 1000 1000 + + = 249 ⇒ 5249 \|1000!. + 5 52 53 54 138 Number Theory and its Applications Problem 6.7.2. For an integer z ≥ 0, show that z z − − = z. 2 2 Solution 6.7.2. By deﬁnition, we have the following inequalities: z! z z −1< ≤ (6.7.1) 2 2! 2 z z z (6.7.2) − −1< − ≤− . 2 2 2 From equation (6.7.2), we have − − z2 < z2 + 1. Adding the last inequation with (6.7.1), we obtain z! z! z z − − < + + 1 = z + 1 ≤ z. (6.7.3) 2 2 2 2 Further, from inequation (6.7.2), we have z2 ≤ − − z2 . Adding the foregoing inequation with (6.7.1), we obtain z! z! − − (How!). (6.7.4) z≤ 2 2 z z Finally, (6.7.3) and (6.7.4) gives 2 − − 2 = z. Problem 6.7.3. If z ≥ 1 and q is a prime, then ﬁnd the exponent of the highest (2z)! . power of q that divides (z!)2 Solution 6.7.3. For any prime q, let s be the highest power of q that divides s (2z)!. If q\|z!, let k be the highest power of q such that q k \|z!. Thus qqk = q s−k . So s − k is the highest power of q satisfying q s−k (2z)! (z!) . Also, s − 2k is the (2z)! highest power of q satisfying q s−k (z!)2 . By virtue of Theorem 6.6.1, the highest ! ∞ power of q dividing (2z)! is k=1 q2zk and the highest power of q dividing z! is ! ∞ (2z)! z is given by, k=1 q k . Finally, the highest power of q dividing (z!)2 ∞ 2z k=1 qk ∞ ∞ z 2z z −2 = −2 k . qk qk q k=1 k=1 Problem 6.7.4. Let the positive integer z be written in terms of powers of the prime q so that we have z = ak q k + . . . + a2 q 2 + a1 q + a0 , where 0 ≤ ai < q. Find the exponent of the highest power of q appearing in the prime factorization of z!. Solution 6.7.4. Before ﬁnding the exponent of the highest power of q, let us state and prove the following lemma viz Arithmetic Functions 139 (q − 1) Lemma 6.7.1. For q > 1, z > 1; 1 q + 1 q2 + ... + 1 qz < 1. Proof. By principle of mathematical induction we are going to prove the above lemma. For k = 1, the lemma is trivial(Verify!). Suppose the lemma is true for z = k. Then 1 1 1 (q − 1) + 2 + . . . + k < 1. q q q Therefore 1 1 q−1 1 1 1 1 1 (q − 1) + 2 + . . . + k + k+1 = (q − 1) + 2 + . . . + k + k+1 q q q q q q q q 1 1 1 1 1 1 + . . . + k − − . . . − k + k − k+1 . =q q q q q q q By hypothesis q q 1 q + ... + 1 1 + ... + k q q − 1 qk 1 q − ... − 1 qk < 1, therefore 1 1 1 1 1 1 − . . . − k + k − k+1 < 1 + k − k+1 q q q q q q 1 < 1 − k+1 q < 1. Hence the lemma is true for z = k + 1. Using Theorem 6.6.1 the exponent of the highest power of q appearing in the prime factorization of z! is ∞ z i=1 qi a = ak q k−1 + . . . + a2 q + a1 + 0 q a0 a1 k−2 + 2 + . . . + a2 + + ak q q q + .. . + a0 + ak + . . . + . . . + k−1 + k q q ak a1 a0 + + . . . + . . . + k + k+1 . q q q In equation (6.7.5), all 0 ≤ ai ≤ q − 1. Note that a1 z qk ! q = ak q = (6.7.5) z q k−1 ! − ak−1 . 140 Number Theory and its Applications Therefore z q = z − a0 q1 z z q= − a1 q2 q .. . z z q = − ak−2 q k−1 q k−2 z z q = − ak−1 qk q k−1 z 0 = k − ak . q On adding the left and right column entries, we have z z z z + 2 + . . . + k−1 + k (q − 1) = z − (a0 + a1 + . . . + ak ). q1 q q q Hence ∞ z z − (a0 + a1 + . . . + ak ) = . qk q−1 k=1 Problem 6.7.5. Using Problem 6.7.4, ﬁnd the exponent of highest power of p dividing (pk − 1)!. Solution 6.7.5. Hint: Here, pk − 1 = (p − 1)(pk−1 + pk−2 + . . . + p + 1) = (p − 1)pk−1 + (p − 1)pk−2 + . . . + (p − 1)p + (p − 1). Since, p is prime, 0 ≤ p − 1 ≤ p so ak−1 = p − 1, ak−2 = p − 1, . . . , a1 = p − 1, a0 = p − 1. Take z = pk − 1 and apply the formulae in Problem 6.7.4. Problem 6.7.6. For any positive integer N , verify the formulae: N N λ(z) z z=1 Solution 6.7.6. Let F (z) = √ = [ N ]. λ(d), λ being the Liouville function deﬁned in d\|z Problem 6.5.4. Taking help of Theorem 6.6.3, we have N z=1 F (z) = N z=1 λ(z) N z . Arithmetic Functions 141 Moreover, by Problem (6.5.4) we have F (z) = Therefore N 1, if z = m2 for some integer m; 0, Otherwise. F (z) monitor the number of perfect squares less than or equal to z=1 N as F assigns a value of 1 to each z that can be expressed as a perfect square. Thus N N = Number of squares ≤ N. λ(z) z z=1 √ Next let us consider, [ N ] and perfect squares, which are 12 , 22 , 32 , and so on. For any N = m2 , there are exactly m perfect squares(positive integers) less √ than or equal to N . Suppose [ N ] is not an integer m be the largest integer √ satisfying m2 < N . Therefore N < (m + 1)2 ⇒ m < N < m + 1. Since √ √ m = N , therefore N is the number of perfect squares less than or equal to √ N N . Hence z=1 λ(z) Nz = [ N ]. Problem 6.7.7. If N is a positive integer, prove that τ (N ) = N N z z=1 N −1 − z Solution 6.7.7. Applying Corollary 6.6.2, yields fore . N N z=1 z = N z=1 τ (z). There- N N −1 N −1 N −1 N −1 = + z z N z=1 z=1 N −1 N −1 = . τ (z) + N z=1 N −1 N As Nz−1 < 1(∀N > 0), therefore NN−1 = 0. Hence z=1 Nz−1 = z=1 τ (z). N N −1 N Therefore z=1 Nz − Nz−1 = z=1 τ (z) − z=1 τ (z) = τ (N ). Problem 6.7.8. Given a positive integer N , prove: Solution 6.7.8. Let F (z) = d\|z z=1 μ(z) N z μ(d). By Theorem 6.4.2, we ﬁnd F (z) = N 1, if z = 1; 0, if z > 1. = 1. 142 Number Theory and its Applications By Theorem 6.6.1, we have N k=1 N N F (k) = μ(z) z z=1 = F (1) + F (2) + . . . + F (N ) = 1. N Hence z=1 μ(z) Nz = 1. Let us illustrate the problem taking N = 6. 6 6 6 6 6 6 6 = μ(1) + μ(2) + μ(3) + μ(4) + μ(5) + μ(6) μ(z) z 1 2 3 4 5 6 z=1 6 = 1 · 6 + (−1) · 3 + (−1) · 2 + 0 · 1 + (−1) · 1 + 1 · 1 =6−3−2+0−1+1 = 1. N Problem 6.7.9. Given a positive integer N , prove: z=1 μ(z) z ≤ 1. Solution 6.7.9. From Problem 6.7.8, we obtain N N μ(z) z z=1 =1⇒ N −1 z=1 N + μ(N ) = 1. μ(z) z Dividing by N , we obtain from foregoing equation N −1 N μ(N ) 1 1 = − . μ(z) N N N z=1 z (6.7.6) Again, N μ(z) z=1 z = N −1 z=1 = μ(z) μ(N ) + z N N −1 1 μ(N ) N . μ(z) + N z=1 z N (6.7.6) and (6.7.7) yields N μ(z) z=1 z N −1 N −1 N 1 1 1 N = − μ(z) + μ(z) N z=1 z N N z=1 z N −1 N N 1 1 = − + . μ(z) N z=1 z z N (6.7.7) Arithmetic Functions 143 Since N N < 1 and \|a + b\| ≤ \|a\| + \|b\|, \|a · b\| = \|a\| · \|b\|, 0 ≤ − z z 1 = 1, N N therefore N N −1 μ(z) N N 1 1 \|μ(z)\| − + N N z z z z=1 z=1 ≤ N −1 1 1 \|μ(z)\| + N z=1 N 1 1 (N − 1) + = 1 as \|μ(z)\| ≤ 1. N N Let us illustrate the problem taking N = 6. 6 μ(z) μ(1) μ(2) μ(3) μ(4) μ(5) μ(6) + + + + + = z 1 2 3 4 5 6 z=1 1 0 1 1 1 + − + + − + = 1 + − 2 3 4 5 6 5 1 1 = 1 + − + − + 6 5 6 13 2 = = 1 + − < 1. 15 15 6.8 Exercises: 1. Show that σ(n) = σ(n + 1) for n = 14, 206, 957. 2. For any positive integer n, prove that σ(n!) n! ≥1+ 1 2 + . . . + n1 . 3. Given a positive integer k > 1, show that there are inﬁnitely many integers n for which τ (n) = k, but at most ﬁnitely many n with σ(n) = k. 4. Prove that there are no positive integers n satisfying σ(n) = 10. 5. Show that for k ≥ 2, if 2k − 3 is prime, then n = 2k−1 (2k − 3) satisﬁes the equation σ(n) = 2n + 2. 6. Prove that if f and g are multiplicative functions, then so is their product f g and quotient fg (whenever the latter function is deﬁned). 7. For any positive integer n, show that z σ(d) = dτ (d). d d\|z d\|z 144 Number Theory and its Applications 8. Given z ≥ 1, let σs (z) denote the sum of the sth powers of the positive ds . Prove that σs is a multiplicative divisors of z; that is, σs (z) = d\|z function. 9. For each positive integer n, verify that μ(n)μ(n + 1)μ(n + 2)μ(n + 3) = 0. 10. If the integer n > 1 has a prime factorization n = pk11 pk22 . . . pkr r , prove the following (a) d\|n μ(d)σ(d) = (−1)r p1 p2 . . . pr ; 1 1 1 (b) d\|n μ(d) d = (1 − p1 )(1 − p2 ) . . . (1 − pr ). 11. If the integer n > 1 has a prime factorization n = pk11 pk22 . . . pkr r , then establish that d\|n μ(d)λ(d) = 2r . 12. Find the highest power of 7 dividing 2000!. 13. If n ≥ 1 and p is a prime, show that (2n)! (n!)2 is an even integer. 14. Find an integer n ≥ 1 such that the highest power of 5 contained in n! is 100. 15. Determine the highest power of 3 dividing 80! and the highest power of 7 dividing 2400!. 7 Euler’s Generalization and φ–function “Mathematicians have tried in vain to this day to discover some order in the sequence of prime numbers, and we have reasons to believe that it is a mystery into which the human mind will never penetrate.” – Leonhard Euler 7.1 Introduction Fermat’s work does not contribute much to the mathematics of his own day but creates a lasting impression on later generations of mathematics and Leonhard Euler (1707 − 1783) was the ﬁrst one to appreciate its signiﬁcance. Many theorems(without proof) were proclaimed by Fermat, and were later on proved by Euler with the similar arguments formulated by him. Present chapter deals with Euler’s generalization of Fermat’s Little Theorem, depending on a function which indeed was invented by Euler, but named by J. J. Sylvester(1814 − 1897) in 1883. Euler’s generalisations concerns congruences with prime moduli to arbitrary moduli. Euler’s theorem has great contributions in Economics. When F (L, K) is a production function then Euler’s theorem says that if factors of production are paid according to their marginal productivities, the total factor payment is equal to the degree of homogeneity of the production function times output. Euler has great contribution in other ﬁelds of science such as function theory and theory of music, as well as the relationship between music and mathematics. 145 146 Number Theory and its Applications We give an account of the most important results obtained by Euler in number theory. 7.2 Euler’s φ–function The chapter, Fermat’s little theorem addresses the congruence relation with a prime number. Now the question arises: Can we change the prime number by an arbitrary positive integer? The answer is in an aﬃrmative sense and Euler’s generalization is the important result which leads to that answer. Before going to this signiﬁcant result we need to introduce an important arithmetic function called Euler’s φ–function or Euler’s totient function. To meet the purpose, ﬁrst let us deﬁne this special type of function. Deﬁnition 7.2.1. For any positive integer n with n ≥ 1, Euler’s phi functionor Euler’s totient function denoted as φ(n) and deﬁned as the number of positive integers not exceeding n and relatively prime to n. Let us illustrate the above deﬁnition by some example, for which we displayed below a table of positive integers n and corresponding φ(n). n φ(n) 1 1 2 1 3 2 4 2 5 4 6 2 7 6 8 4 9 6 10 4 From the above table, it is clear that φ(1) = 1 and φ(p) = p − 1 for any prime p. Also, the converse with respect to second equality is true, i.e. if for any positive integer n, φ(n) = n − 1 holds then n is prime. Our next proposition directs us to ensure the proof of this statement. Proposition 7.2.1. If p is a prime then φ(p) = p − 1 holds and vice versa. Proof. If p is a prime, from the deﬁnition of φ-function, the number of integers which are less than p and prime to p is p − 1. Thus, φ(p) = p − 1 for every prime p. Conversely, let p be composite. Then it has a divisor q with 1 < q < p and 1. Now q belongs to the set {1, 2, 3, · · · p − 1} and q not relatively gcd(p, q) = prime to p implies φ(p) ≤ p − 2. Hence if φ(p) = p − 1 then p must be prime. The ﬁrst important agenda of this section is, for any arbitrary positive integer n what should be φ(n) when the prime factorisation of n is known. The next few results of this section helps us to reach that platform from where we can ﬁnd φ(n) for any arbitrary positive integer n. Euler’s Generalization and O\|—function Theorem 7.2.1. If p is prime and α > 0, then φ(pα ) = pα 1 − 147 1 . p Proof. Here we need to ﬁnd those positive integers for which gcd(n, pα ) = 1 that is p n. Now given below the arrangement of those positive integers not greater than pα . The arrangement is a rectangular array containing p columns and pα−1 rows: 1 2 ··· p p+1 p+2 · · · 2p .. .. .. .. . . . . α α p − p + 1 p − p + 2 · · · pα and there are pα−1 integers between 1 and pα which are divisible by p, namely p, 2p, 3p, . . . , pα−1 . p lies in rightmost column of the above array. Thus there are exactly pα − pα−1 α integers which are relatively to p and so by deﬁnition of the φ–function, prime 1 φ(pα ) = pα − pα−1 = pα 1 − . p To understand the above theorem lucidly by means of an example, let us choose p = 2 and α = 3. Now using the table we have: 1 3 5 7 2 4 6 8 Only the elements of the right sided column divides 23 . Thus φ(23 ) = φ(8) is the number of elements of the set {1, 3, 5, 7} which is 4 = 23−1 (2 − 1). We are now in the stage to ﬁnd the phi function for prime powers. But still a question arises, whether it is possible to ﬁnd the phi function of any positive integer directly whose prime factorization is given using the above theorem. The answer of this statement is in the aﬃrmative sense but for that we have to check the multiplicative property of this arithmetic function φ. The next part of the present section deals with this fact. Now we are in the position to state and prove the following theorem. Theorem 7.2.2. The function φ is a multiplicative function. Proof. It suﬃces to show that φ(mn) = φ(m)φ(n), where gcd(m, n) = 1. If any one of m, n is 1, the result is true(Why!). Thus we may assume m > 1, n > 1. We arrange the integers from 1 to mn into m × n order array as follows: 148 Number Theory and its Applications 1 m+1 2m + 1 .. . (n − 1)m + 1 2 ··· r m+2 ··· m+r 2m + 2 ··· 2m + r .. .. .. . . . (n − 1)m + 2 · · · (n − 1)m + r ··· m · · · 2m · · · 3m .. .. . . · · · mn We know that there are φ(mn) entries of the above array which are prime to mn(Why!) and this is same as the number of integers relatively prime to both m and n(refer to problem(2.6.1)). Now gcd(qm+r, m) = gcd(r, m), so the numbers in r−th column are relatively prime to m if and only if gcd(r, m) = 1. Thus there are only φ(m) columns containing integers relatively prime to m. Here every entry of that φ(m) columns are relatively prime to m. Now to show each of these φ(m) columns there are φ(n) integers which are relatively prime to n. In the entries of r-th column there are n integers r, m + r, . . . , (n − 1)m + r no two of which are congruent modulo n. If it happens, let (im + r) ≡ (jm + r)( mod n)(0 ≤ i < j < n). Therefore im ≡ jm(mod n) implies i ≡ j(mod n) as gcd(m, n) = 1, which leads to a contradiction. Thus the numbers in the r-th column are congruent modulo n to 0, 1, 2, · · · , n − 1, in some order. If s ≡ t( mod n) for some integer s and t then gcd(s, n) = 1 if and only if gcd(t, n) = 1. Thus r-th column contains as many integers, which are relatively prime to n, as does the set {0, 1, 2, . . . , n − 1}, namely φ(n). Therefore the total number of entries in the array that are relatively prime to both m, n is φ(m)φ(n). Finally, we are in the position to ﬁnd the phi-function for any arbitrary positive integer. Theorem 7.2.3. If the integer n > 1 has a prime factorization n = p1α1 p2α2 · · · prαr , 1 1 1 −1 ) · · · (pαr − pαr −1 ) = n then φ(n) = (p1α1 − pα ··· 1 − . 1− r 1 r p1 pr Proof. Since φ is multiplicative(Why!) and n has a prime factorization, n = p1α1 p2α2 · · · prαr then we have φ(n) = φ(p1α1 )φ(p2α2 ) · · · φ(prαr ). α α α −1 Again from the Theorem 7.2.1, we have φ(pj j ) = (pj j − pj j ) for each Euler’s Generalization and O\|—function 149 j = 1, 2, 3, · · · , r. Hence 1 −1 ) · · · (pαr − pαr −1 ) φ(n) = (p1α1 − pα r 1 r 1 1 α1 α2 αr 1− ··· 1 − = p1 p2 · · · pr p1 pr 1 1 =n 1− ··· 1 − . p1 pr The exempliﬁcation of the above theorem has been done from the following example: Example 7.2.1. Choose n = 720. Then the prime factorization of 720 is 24 · 32 · 5. Thus applying above theorem, we have 1 1 1 1− 1− , φ(360) = 720 1 − 2 3 5 1 2 4 = 720 · · · , 2 3 5 = 192. At the outset of this section, a table of positive integers and their corresponding phi-function, was displayed. There, φ(1) = φ(2) = 1 and the values of phi function for other integers are even. This is not a coincidence, as evident from our next theorem: Theorem 7.2.4. For n > 2, φ(n) is an even integer. Proof. Let us consider n = 2j with j ≥ 2. Then from Theorem 7.2.1, φ(n) = φ(2j ) = 2j−1 , an even integer. If n is not a power of 2 then it is divisible by some odd prime. Then n = pj m, where p being an odd integer and gcd(p, m) = 1. Therefore φ(n) = φ(pj )φ(m)(W hy!) = pj−1 φ(m)(p − 1), which is also even(Why!). 7.3 Worked out Exercises Problem 7.3.1. Verify that the equality φ(z) = φ(z + 1) = φ(z + 2) holds, when z = 5186. 150 Number Theory and its Applications Solution 7.3.1. Note that 2592 1 = 2592. 2 2593 2 6 12 18 5187 = 3 · 7 · 13 · 19, φ(5187) = 5187 = 2592. 3 7 13 19 1296 1 5188 = 22 · 1297, φ(5188) = 5188 = 2592. 2 1297 5186 = 2 · 2593, φ(5186) = 5186 z if and only if z = 2k . 2 1 k k k Solution 7.3.2. Let us consider z = 2 . Then φ(z) = φ(2 ) = 2 1 − = 2 z 2k−1 = . 2 z z Conversely, suppose φ(z) = . Then for to be an integer, z must be even. 2 2 k k Let z = 2k p2 2 · · · pkr r and assume ki = 0. Let q = p2 2 · · · pkr r . So q > 1 and gcd(2k , q) = 1. Problem 7.3.2. Prove: For some k ≥ 1, φ(z) = ∴ φ(z) = φ(2k q) = φ(2k )φ(q), 1 φ(q) = 2k−1 φ(q). = 2k 1 − 2 z Further = φ(z) = 2k−1 φ(q) ⇒ z = 2k φ(q). 2 k2 kr ∴ p2 · · · pr = φ(q) = φ(pk2 2 · · · pkr r ) 1 1 ··· 1 − . = pk2 2 · · · pkr r 1 − p2 pr ∴ p2 · · · pr = (p2 − 1) · · · (pr − 1). Therefore for each pi , pi = (pj − 1) for some j. This is impossible if ki = 0. Hence ki = 0. Thus the converse part follows. Problem 7.3.3. Prove that the equation φ(z) = φ(z + 2) is satisﬁed by z = 2(2p − 1) whenever p and 2p − 1 are both odd primes. Solution 7.3.3. Here 2p − 1 is an odd prime implies gcd(2, 2p − 1) = 1. 1 = 2p − 2. ∴ φ(z) = φ(2)φ(2p − 1) = (2p − 1) 1 − 2p − 1 Now z + 2 = 2(2p − 1) + 2 = 4p, p being an odd prime, yields gcd(4, p) = 1. 1 = 2p − 2. ∴ φ(z + 2) = φ(4)φ(p) = 2p 1 − p ∴ φ(z) = φ(z + 2). Euler’s Generalization and O\|—function 151 Problem 7.3.4. Show that there are inﬁnitely many integers n for which φ(n) is a perfect square. 1 Solution 7.3.4. For k ≥ 1, φ(2k ) = 2k 1 − = 2k−1 . If k is odd, then k − 1 2 is even. Let k = 2m + 1, for some m ∈ Z. ∴ φ(2k ) = φ(22m+1 ) = (2m )2 = a perfect square. Thus there are inﬁnitely many n = 2k , k being odd, and φ(n) is a perfect square. Problem 7.3.5. Prove that if the integer n has s distinct odd prime factors, then 2s φ(n). Solution 7.3.5. Let n = pk1 1 pk2 2 · · · pks s , pi > 2. ∴ φ(n) = pk1 1 −1 (p1 − 1)pk2 2 −1 (p2 − 1) · · · pks s −1 (ps − 1). As each pi is odd, so let pi = 2ri + 1 for some ri . ∴ φ(n) = pk1 1 −1 (p1 − 1)pk2 2 −1 (p2 − 1) · · · psks −1 (ps − 1)(2r1 )(2r2 ) · · · (2rs ), = 2s pk1 1 −1 (p1 − 1)pk2 2 −1 (p2 − 1) · · · pks s −1 (ps − 1)r1 r2 · · · rs . ∴ 2s φ(n). Problem 7.3.6. If every prime that divides n also divides m, prove that φ(nm) = nφ(m). Solution 7.3.6. Let p1 , p2 , . . . , ps be all those primes which divide both n and m. Suppose n = pk1 1 pk2 2 · · · pks s , m = p1j1 p2j2 · · · psjs q1m1 q2m2 · · · qrmr , q i being prime be such that qi=pj. ∴ nm = pk1 1 +j1 pk2 2 +j2 · · · pks s +js q1m1 q2m2 · · · qrmr . ∴ φ(nm) = pk1 1 +j1 pk2 2 +j2 · · · pks s +js q1m1 q2m2 · · · qrmr 1 1 1 1 1− ··· 1 − 1− ··· 1 − , p1 ps q1 qr 1 1 1 1 = p1j1 p2j2 · · · psjs 1 − ··· 1 − 1− ··· 1 − pk1 1 pk2 2 · · · pks s , p1 ps q1 qr = φ(m)pk1 1 pk2 2 · · · pks s , = n · φ(m). Problem 7.3.7. If φ(n)\|(n − 1), prove that n is a square-free integer. 152 Number Theory and its Applications Solution 7.3.7. Let n = pk1 1 pk2 2 · · · pks s and suppose n is not square-free such that ki ≥ 2 for some i. Now φ(n) = (pk1 1 − pk1 1 −1 ) · · · (pki i − pki i −1 ) · · · (pks s − pks s −1 ). k k −1 Since ki ≥ 2, ki − 1 ≥ 1, so pi (pi i − pi i ) ⇒ pi φ(n). By hypothesis φ(n) (n − 1) implies pi (n − 1). Again pi n yields pi n − (n − 1) ⇒ pi 1, which is a contradiction. Therefore for all i, ki = 1 implies n is square-free. Problem 7.3.8. Prove that there are no integers n for which φ(n) = n . 4 Solution 7.3.8. Here φ(1) = 1 = φ(2), φ(3) = 2 = φ(4). So the statement holds n true for n = 1, 2, 3, 4. Let n > 4. On the contrary, suppose φ(n) = . Let 4 n = pk1 1 pk2 2 · · · pks s , ki ≥ 1. 1 1 n ∴ φ(n) = n 1 − ··· 1 − = , p1 ps 4 1 (p − 1)(p2 − 1) · · · (ps − 1) = , ⇒ 1 p1 p2 · · · p s 4 ⇒ 4(p1 − 1)(p2 − 1) · · · (ps − 1) = p1 p2 · · · ps , ⇒ 2(p2 − 1) · · · (ps − 1) = p2 · · · ps , as p1 = 2. Since p2 , . . . , ps are all odd, therefore p2 · · · ps is odd. But 2(p2 − 1) · · · (ps − 1) is even. So p1 = 2 fails to work. Now if all p1 , p2 , . . . , ps are odd, then p1 p2 · · · ps is also so. Furthermore 4(p1 − 1)(p2 − 1) · · · (ps − 1) is even, which implies no such n exists. Problem 7.3.9. If p is a prime and k ≥ 2, show that φ(φ(pk )) = pk−2 φ((p − 1)2 ). Solution 7.3.9. Here φ(pk ) = pk−1 (p − 1). Since gcd(p, p − 1) = 1, therefore gcd(pk−1 , p − 1) = 1. Using the multiplicative property of φ, we obtain φ(φ(pk )) = φ(pk−1 (p − 1)) = φ(pk−1 )φ(p − 1) = pk−2 (p − 1)φ(p − 1). Now for every positive integer n, φ(n2 ) = nφ(n). Therefore (p − 1)φ(p − 1) = φ((p − 1)2 ). Hence φ(φ(pk )) = pk−2 (p − 1)φ(p − 1) = pk−2 φ((p − 1)2 ). Problem 7.3.10. If n = pk1 1 pk2 2 · · · pks s , then prove that 1 1 1 σ(n)φ(n) ≥ n2 1 − 2 1 − 2 ··· 1 − 2 . p1 p2 ps Euler’s Generalization and O\|—function 153 Solution 7.3.10. Note that σ(n) = k +1 p1 1 −1 p1 −1 s +1 −1 pk s ps −1 2k1 k1 −1 ··· k −1 and φ(n) = p1 1 (p1 − σ(n)φ(n) = (p 1 − p1 ) · · · (ps2ks − pks s −1 ). But 1) · · · pks s −1 (ps − 1). Therefore k −1 j k 2 2k k −1 2k p (pj j − pj j ) = pj j 1 − j2kj = pj j 1 − kj1+1 . For pj ≥ 1 we ﬁnd pj k +1 pj j ≥ p2j ⇒ pj 1 1 1 1 1 1 ≥ k +1 ⇒ − 2 ≤ − k +1 ⇒ 1 − 2 ≤ 1 − k +1 . 2 pj pj pj pj j pj j pj j 1 1− 2 . ∴ p j − pj pj 2 1 ∴ σ(n)φ(n) ≥ 1− 2 , pks s ps s 2 1 = pks s 1− 2 , ps s s 1 1 1 1 − 2 ··· 1 − 2 . = n2 1 − 2 p1 p2 ps 2kj 7.4 kj −1 k 2 ≥ pj j Euler’s Theorem The ﬁrst published proof of Fermat’s little theorem(stated in chapter 5 of this book) was given by Euler in 1736, where he had taken a prime p and an integer a. But later in the year, 1760 he succeeded in generalizing the result from prime p to an arbitrary integer n. This generalization is known as Euler’s generalization of Fermat’s theorem. The present section deals with the proof and related ideas associated with this remarkable theorem. Now, as a precursor to launch the proof of Euler’s generalization of Fermat’s theorem, we need the following lemma: Lemma 7.4.1. Let n > 1 and gcd(a, n) = 1. If k1 , k2 , · · · , kφ(n) are the positive integers less than and prime to n, then ak1 , ak2 , · · · , akφ(n) are congruent modulo n to k1 , k2 , · · · , kφ(n) in some order. Proof. Here we are going to show that no two of the integers ak1 , ak2 , · · · , akφ(n) are congruent modulo n. For if, aki ≡ akj (mod n) holds with 1 ≤ i < j ≤ φ(n) then ki ≡ kj (mod n), which is a contradiction since this two integers are less than n. Since, gcd(ki , n) = 1 ∀i and gcd(a, n) = 1 then from the worked out Problem 2.6.1) gcd(aki , n) = 1 ∀i. Let us ﬁx akj for some integer j, there exists unique integer b where 0 ≤ b < n for which akj ≡ b(mod n). Since, gcd(b, n) = gcd(akj , n) = 1, so b must be one of the integers k1 , k2 , · · · , kφ(n) . 154 Number Theory and its Applications This is true for all j. This proves that the numbers ak1 , ak2 , · · · , akφ(n) and the numbers k1 , k2 , · · · , kφ(n) are identical with respect to modulo n in a certain order. We now represent an example to make a lucid understanding of this lemma. For that let us take n = 9 and the set {1, 2, 4, 5, 7, 8} is a reduce system modulo 9. Since gcd(2, 9) = 1 then we have, 2 · 1 = 2, 2 · 2 = 4, 2 · 4 = 8, 2 · 5 = 10, 2 · 7 = 14, 2 · 8 = 16 is also a reduce system modulo 9. Theorem 7.4.1. (Euler): If n is a positive integer and gcd(a, n) = 1 then aφ(n) ≡ 1(mod n). Now before going to the proof, we illustrate the idea of it by an example. Example 7.4.1. From the last example, it is clear that both the sets {1, 2, 4, 5, 7, 8} and {2·1, 2·2, 2·4, 2·5, 2·7, 2·8} are reduced residue system of modulo 9. Therefore (2 · 1)(2 · 2)(2 · 4)(2 · 5)(2 · 7)(2 · 8) ≡ 1 · 2 · 4 · 5 · 7 · 8( mod 9), 26 · 1 · 2 · 4 · 5 · 7 · 8 ≡ 1 · 2 · 4 · 5 · 7 · 8( mod 9). Since we have gcd(1 · 2 · 4 · 5 · 7 · 8, 9) = 1, we conclude that 26 = 2φ(9) ≡ 1( mod 9). We now use the idea of this example to the following proof. Proof. Let us take n > 1 and k1 , k2 , · · · , kφ(n) be the positive integers less than n which are relatively prime to n. Since gcd(a, n) = 1, ak1 , ak2 , · · · , akφ(n) are congruent to k1 , k2 , · · · , kφ(n) (Why!). Then the least positive residue of ak1 , ak2 , · · · , akφ(n) are the integers k1 , k2 , · · · , kφ(n) in some order. Therefore (ak1 )(ak2 ) · · · (akφ(n) ) ≡ k1 k2 · · · kφ(n) ( mod n) and so aφ(n) k1 k2 · · · kφ(n) ≡ k1 k2 · · · kφ(n) ( mod n). Since gcd(ki , n) = 1 for each i so gcd(k1 k2 · · · kφ(n) , n) = 1[see Problem 2.6.1]. Thus the congruence becomes aφ(n) ≡ 1(mod n). Remark 7.4.1. If n = p is prime, then φ(p) = p − 1. Further if p a, then we have ap−1 ≡ 1(mod p), which is equivalent to Fermat’s Little theorem. Euler’s theorem has vast application in ﬁnding the modulo of a large number with respect to a positive integer. Applying Euler’s theorem, we can ﬁnd 301 congruent modulo to 99. Since gcd(4, 99) = 1 and φ(99) = of 4 with respect 1 2 10 1 φ(32 · 11) = 99 1 − 1− = 99 × × = 60, from Euler’s theorem 3 11 3 11 we have 460 ≡ 1(mod 99). Now 301 = 5 · 60 + 1, therefore 4301 ≡ (460 )5 · 41 ≡ 4( mod 99). Euler’s Generalization and O\|—function 7.5 155 Worked out Exercises Problem 7.5.1. Use Euler’s theorem to evaluate 2100000 modulo 77. Solution 7.5.1. Here gcd(2, 77) = 1, therefore 2φ(77) ≡ 1(mod 77). Now φ(77) = 6 · 10 = 60 ⇒ 260 ≡ 1( mod 77). Hence 260000 ≡ 1( mod 77), (260 )300 = 218000 ≡ 1( mod 77) ⇒ 236000 ≡ 1( mod 77). ∴ 296000 ≡ 1( mod 77), (260 )300 = 21800 ≡ 1( mod 77) ⇒ 23600 ≡ 1( mod 77). ∴ 299600 ≡ 1( mod 77), (260 )3 = 2180 ≡ 1( mod 77) ⇒ 2360 ≡ 1( mod 77). ∴ 299960 ≡ 1( mod 77). But 210 = 1024, 13 · 77 = 1001 ⇒ 210 ≡ 23(mod 77). Therefore 240 ≡ 234 ( mod 77) ⇒ 2100000 ≡ 234 (mod 77). Now 232 = 529 = 6 · 77 + 67 ⇒ 232 ≡ −10( mod 77) ⇒ 234 ≡ 100 ≡ 23(mod 77). Hence 2100000 ≡ 23( mod 77). Problem 7.5.2. For any prime p, prove that: τ (p!) = 2τ ((p − 1)!). Solution 7.5.2. Let p! = pk1 1 pk2 2 · · · pks s ·p = 1·2·3 · · · (p−1)·p and p1 , p2 , · · · ps be distinct primes. Here ki ≥ 0 are the integers for each i(= 1, 2, · · · s). Therefore k k k k (p − 1)! = p1 1 p2 2 · · · pks s . Since gcd(p, p1 1 p2 2 · · · pks s ) = 1, therefore τ (p!) = τ (p · pk1 1 pk2 2 · · · pks s ) = τ (p)τ (pk1 1 pk2 2 · · · pks s ) = τ (p)τ ((p − 1)!) = 2 · τ ((p − 1)!), ∵ τ (p) = 2. Problem 7.5.3. If gcd(a, n) = 1, show that the linear congruence ax ≡ b( mod n) has the solution x ≡ baφ(n)−1 (mod n). Solution 7.5.3. If x ≡ baφ(n)−1 (mod n), then ax = a(baφ(n)−1 ) = baφ(n) . Since gcd(a, n) = 1, by Euler’s theorem we have aφ(n) ≡ 1(mod n). ∴ ax = baφ(n) ≡ b · 1 ≡ b( mod n). Problem 7.5.4. Show that if gcd(a, n) = gcd(a − 1, n) = 1, then 1 + a + a2 + . . . + aφ(n)−1 ≡ 0( mod n). 156 Number Theory and its Applications Solution 7.5.4. By Euler’s theorem, we have gcd(a, n) = 1 ⇒ aφ(n) ≡ 1( mod n) ⇒ aφ(n) − 1 ≡ 0( mod n). But aφ(n) − 1 = (a − 1)(aφ(n)−1 + . . . + a2 + a + 1). Therefore (a − 1)(aφ(n)−1 + . . . + a2 + a + 1) ≡ 0(mod n). Since gcd(a − 1, n) = 1, therefore 1 + a + a2 + . . . + aφ(n)−1 ≡ 0(mod n). Problem 7.5.5. If m and n are relatively prime positive integers, prove that mφ(n) + nφ(m) ≡ 1(mod mn). Solution 7.5.5. Since gcd(m, n) = 1, therefore an appeal to Euler’s theorem produces mφ(n) ≡ 1( mod n) & nφ(m) ≡ 1( mod m). But nφ(m) ≡ 0(mod n) & mφ(n) ≡ 0(mod m). ∴ mφ(n) + nφ(m) ≡ 1 + 0 = 1( mod n), nφ(m) + mφ(n) ≡ 1 + 0 = 1( mod m). Since gcd(m, n) = 1, therefore combining we obtain mφ(n) +nφ(m) ≡ 1(mod mn). Problem 7.5.6. Find the units digit of 3100 by means of Euler’s theorem. Solution 7.5.6. Here gcd(10, 3) = 1. By Euler’s theorem, 3φ(10) ≡ 1(mod 10). Now, φ(10) = 4, therefore 34 ≡ 1(mod 10). Hence (34 )25 ≡ 1(mod 10). Therefore 3100 ≡ 1(mod 10). Thus, unit digit of 3100 is 1. Problem 7.5.7. Prove that a15 ≡ a3 (mod (215 − 23 )) for any integer a. Solution 7.5.7. Here, a15 − a3 = a3 (a12 − 1) = a3 (a6 + 1)(a6 − 1) = a3 (a6 + 1)((a3 + 1))(a3 − 1) = a3 (a6 + 1)((a3 + 1))(a2 + a + 1)(a − 1). 215 − 23 = 23 (26 + 1)((23 + 1))(22 + 2 + 1)(2 − 1) = 23 · 32 · 5 · 7 · 13. Applying the deﬁnition of Euler’s phi function we get, φ(8) = 4, φ(5) = 4, φ(13) = 12, φ(9) = 6, φ(7) = 6. Case(i) If gcd(a, 215 − 23 ) = 1, then this implies, gcd(a, 8) = 1, gcd(a, 13) = 1, gcd(a, 2) = 1, Euler’s Generalization and O\|—function 157 gcd(a, 5) = 1, gcd(a, 9) = 1. Now applying Euler’s theorem in all those above cases we can write, a4 ≡ 1( mod 8), a12 ≡ 1( mod 13), a4 ≡ 1( mod 5), a6 ≡ 1( mod 7), a6 ≡ 1( mod 9). Considering all the congruences together, we have a12 ≡ 1( mod 8 · 5 · 13 · 9 · 7). ∴ a15 ≡ a3 ( mod (215 − 23 )). Case(ii) If gcd(a, 215 − 23 ) = 1, then for some integer k, a = k(215 − 23 ), and a15 −a3 = (a14 −a2 )a = (a14 −a2 )k(215 −23 ) ⇒ a15 ≡ a3 ( mod (215 −23 )). Hence combining both the cases for any integer a, we get a15 ≡ a3 (mod (215 − 23 )). Problem 7.5.8. Use Euler’s theorem to conﬁrm that, for any integer z ≥ 0, 51\|1032z+9 − 7. Solution 7.5.8. Here, 51 = 17 · 3. Therefore φ(51) = 16 · 2 = 32. Also, gcd(10, 51) = 1 gives 10φ(51) = 1032 ≡ 1(mod 51). Thus, 1032z ≡ 1( mod 51). (7.5.1) Next, we are going to show 109 ≡ 7(mod 51). Now, 10 ≡ 7( mod 3), 10 ≡ 1( mod 3) ⇒ 1018 ≡ 1( mod 3). ∴ 109 = 108 · 10 ≡ 7 · 1( mod 3), or, 109 ≡ 7( mod 3). (7.5.2) −10 ≡ 7( mod 17), ∴ (−10)2 ≡ 72 = 49 ≡ −2( mod 17). ∴ (−10)8 = 108 ≡ (−2)4 = 16 ≡ −1( mod 17). ∴ (−10)9 ≡ −10 ≡ 7( mod 17). (7.5.3) (7.5.2) + (7.5.3) ⇒ 10 ≡ 7( mod 51). (7.5.4) 9 (7.5.1) + (7.5.4) ⇒ 10 32z 32z+9 10 · 10 ≡ 1 · 7( mod 51), 9 ≡ 7( mod 51). Thus, for any integer z ≥ 0, 51\|1032z+9 − 7. (7.5.5) 158 Number Theory and its Applications Problem 7.5.9. Prove that if a is an integer, then a7 ≡ a(mod 63). Solution 7.5.9. From Fermat’s little theorem we see that, a7 ≡ a(mod 7). So to prove this assertion we need to check a7 ≡ a(mod 9). If 9\|a then it is trivial. If 3 a then gcd(a, 9) = 1, so from Euler’s theorem it follows that aφ(9) = a6 ≡ 1( mod 9) or a7 ≡ a(mod 9). Thus together we have a7 ≡ a(mod 63). Problem 7.5.10. Solve the linear congruence 5x ≡ 3(mod 14) by Euler’s theorem. Solution 7.5.10. Here we multiply both sides of the congruence by 5φ(14)−1 = 55 . This gives 56 · x ≡ 3 · 55 (mod 14). Now by Euler’s theorem we have 5φ(14) = 56 ≡ 1(mod 14). This implies, x ≡ 3 · 55 ≡ 15 · 1111 ≡ 15 · 9 ≡ 9(mod 14). 7.6 Properties of φ–function Present section deals with some curious properties of Euler’s phi function related with some arithmetic functions. Discussion of this chapter commence with an important property of totient(φ) function, where the sum of values of φ(d) where d is the divisor of any positive integer n is always equal to n itself. Famous German mathematician Carl Friedrich Gauss was the ﬁrst person to notice that. φ(d) where d is posiTheorem 7.6.1. For each positive integer n ≥ 1, n = d\|n tive divisor of n. Proof. Let us choose n = 1 then, φ(d) = φ(1) = 1 = n. Thus the equality is d\|1 true in this case. Now we are only to prove the result for any positive integer n > 1. Let us choose a set Sn = {1, 2, 3, · · · , n} and \|Sn \| be the number of elements in Sn , then clearly \|Sn \| = n. For each divisor d of n we denote Sd be the set of all integers not exceeding n and gcd(m, n) = d for each m ∈ Sd . Now from m n the proposition (2.4.2) we have gcd(m, n) = d if and only if gcd , = 1. d d n We now have to show that each Sd has φ number of elements. Here for a d particular d all the elements of Sd are multiples d and less than or equal to of n n. Thus the elements of Sd are d, 2d, 3d, · · · , d. Now, let ad ∈ Sd be any d n element where gcd a, = e. Then clearly gcd(ad, n) = ed. Here ed = d if and d n only if e = 1 imply that only ad in Sd are those whose gcd a, = 1 that is the d Euler’s Generalization and O\|—function 159 n . Since each integers of the set {1, 2, 3, · · · , n} lies in exactly one number φ d n class Sd , we have the formula n = . But d runs through all positive φ d d\|n n n = φ φ(d). divisors of n so does . Thus ﬁnally we have, n = d d d\|n d\|n Here we have illustrated the above theorem by means of an example: Example 7.6.1. Let us choose a number n = 12 and the divisors of 12 are 1, 2, 3, 4, 6, 12. Thus the classes Sd are, S1 = {1, 5, 7, 11}, S2 = {2, 10}, S3 = {3, 9}, S4 = {4, 8}, S6 = {6}S12 = {12}. Now, φ(12) = 4, φ(6) = 2, φ(4) = 2, φ(3) = 2, φ(2) = 1, φ(1) = 1. Therefore φ(12) = φ(12) + φ(6) + φ(4) + φ(3) + φ(2) + φ(1) = 12 = n. d\|12 This shows the clariﬁcation of our above theorem. Also, the next part of our discussion is based on the last theorem. Here we illustrate the theorem with a suitable example, which totally depends on the multiplicative property of φ[for further details refer to theorem (7.2.2)]. Now for n = 1, the case is trivial. Let us choose n = 24 and apply the formula F (n) = d\|n φ(d) where F and φ are both number theoretic functions. Since φ is multiplicative, F is also so(Why!). Again n = 24 = 23 · 3 be the prime factorization of 24, which implies F (24) = F (23 )F (3). Now F (23 ) = φ(d) d\|23 = φ(1) + φ(2) + φ(4) + φ(8) = 1 + (2 − 1) + (22 − 2) + (23 − 22 )[∵ φ(pk ) = pk − pk−1 ] = 1 + 1 + 2 + 4 = 23 and F (3) = φ(d) d\|3 = φ(1) + φ(3) = 1 + 2 = 3. Therefore F (24) = 23 · 3 = 24 and thus we have n = 24 = F (24) = φ(d) d\|24 which is our desired result. Based on the last example, we are going to give the alternative proof of Theorem 7.6.1 as follows: 160 Number Theory and its Applications Proof. If n = 1, the case is trivial(Verify!). So we assume n > 1. Let us consider φ(d). Since φ is multiplicative, F is the number-theoretic function F (n) = d\|n also so. Let the prime factorization of n be given by n = p1d1 p2d2 · · · psds . Then F (n) = F (p1d1 )F (p2d2 ) · · · F (psds ). For each value of j, we obtain d F (pj j ) = φ(d) d d\|pj j d = φ(1) + φ(pj ) + φ(p2j ) + φ(p3j ) + . . . + φ(pj j ) d −1 d = 1 + (pj − 1) + (p2j − pj ) + (p3j − p2j ) + . . . + (pj j − pj j ) dj = pj . Hence F (n) = p1d1 p2d2 · · · psds = n ⇒ φ(d) = n. d\|n Now for the next part of discussion , let us choose a positive integer 20 and φ(20) = 8. Here the set of positive integers less than 20 and prime to 20 are {1, 3, 7, 9, 11, 13, 17, 19} and their sum is 1 + 3 + 7 + 9 + 11 + 13 + 17 + 19 = 80 = 1 2 × 20 × 8. This is not a coincidence, in fact our next theorem deals with it. Theorem 7.6.2. For n > 1, the sum of the positive integers less than n and 1 prime to n is nφ(n). 2 Proof. Let k1 , k2 , · · · , kφ(n) be the positive integers less than n and prime to n. Now using Proposition (2.4.3), we have from congruence relation, k1 + k2 + · · · + kφ(n) = (n − k1 ) + (n − k2 ) + · · · + (n − kφ(n) ) = nφ(n) − (k1 + k2 + · · · + kφ(n) ) Therefore, 2(k1 + k2 + · · · + aφ(n) ) = nφ(n) 1 ⇒ k1 + k2 + · · · + kφ(n) = nφ(n), 2 which proves the theorem. Finally at this point we can give an application of Mobi¨ ous Inversion formula, which leads us to the following theorem: Theorem 7.6.3. For any positive integer n, φ(n) = n μ(d) d\|n d . Before going to the proof, let us illustrate the theorem by means of an example: taking n = 14 we see that, Euler’s Generalization and O\|—function 161 μ(d) μ(2) μ(7) μ(14) = 14 μ(1) + + + 14 d 2 7 14 d\|14 −1 (−1) (−1)2 = 14 1 + + + 2 7 14 1 1 1 = 14 1 − − + 2 7 14 6 = 14 × = 6 = φ(14). 14 φ(d) = n and from Proof. From the Theorem (7.6.1) we know, F (n) = Mobi¨ ous inversion formulae we have, φ(n) = φ(n) = d\|n 7.7 μ(d) n . μ(d) = n d d d\|n d\|n n . Therefore we get, μ(d)F d d\|n Worked out Exercises Problem 7.7.1. For a square-free integer n > 1, show that τ (n2 ) = n if and only if n = 3. Solution 7.7.1. If n = 3, then τ (n2 ) = τ (32 ) = 2 + 1 = 3[refer to Theorem 6.2.2]. Next, suppose n is square-free with n > 1 and τ (n2 ) = n. Let n = p1 p2 · · · ps with pi = pj . Moreover, applying Theorem 6.2.2 we get τ (n2 ) = τ (p21 p22 · · · p2s ) = (2 + 1)(2 + 1) · · · (2 + 1) = 3s . ∴ τ (n2 ) = n = p1 p2 · · · ps = 3s , which implies all pi = 3. Hence n = 3 and s = 1. Problem 7.7.2. For n > 2, prove the inequality φ(n2 ) + φ((n + 1)2 ) ≤ 2n2 . √ Solution 7.7.2. If k is composite, then φ(k) ≤ k − k. As n2 is composite, √ so is (n + 1)2 . Therefore φ(n2 ) ≤ n2 − n2 = n2 − n. Again φ((n + 1)2 ) ≤ " (n + 1)2 − (n + 1)2 = n2 + n. Thus φ(n2 ) + φ((n + 1)2 ) ≤ 2n2 . Problem 7.7.3. Given an integer z, prove that there exists at least one k for which z\|φ(k). 162 Number Theory and its Applications Solution 7.7.3. Let k = p1 p2 · · · pr α1 α2 αr be such that 1 −1 · · · pαr −1 (p − 1) · · · (p − 1). φ(k) = pα 1 r 1 r 1 −1 · · · pαr −1 . So let z = q β1 · · · q βs . Choose k = q β1 +1 · · · q βs +1 . Our claim is z = pα s 1 1 r 1 s ∴ φ(k) = q1β1 · · · qsβs (q1 − 1) · · · (qs − 1), which implies z\|φ(k). Problem 7.7.4. Show that if z is a product of twin primes, say z = p(p + 2), then φ(z)σ(z) = (z + 1)(z + 3). Solution 7.7.4. Here gcd(p, p + 2) = 1, so φ(z) = φ(p)φ(p + 2) = (p − 1)(p + 1). But σ(z) = σ(p)σ(p + 2) = (p + 1)(p + 3). Therefore φ(z)σ(z) = (p − 1)(p + 1)2 (p + 3). Now (z + 1)(z − 3) = (p2 + 2p + 1)(p2 + 2p − 3) = (p + 1)2 (p + 3)(p − 1). Hence φ(z)σ(z) = (z + 1)(z + 3). Problem 7.7.5. Assuming d\|n, prove that φ(d)\|φ(n). Solution 7.7.5. Since d\|n, so assume n = pk11 pk22 . . . pkr r and d = pa1 1 pa2 2 . . . par r where 0 ≤ ai ≤ ki . Then φ(n) = n(1 − p11 )(1 − p12 ) . . . (1 − p1r ) and φ(d) = d(1 − p11 )(1 − p12 ) . . . (1 − p1r ). Since d\|n, then it follows that φ(d)\|φ(n). Problem 7.7.6. If z is a square-free integer, prove that for all integers k ≥ 2 σ(dk−1 )φ(d) = z k . d\|z Solution 7.7.6. Since φ and σ are multiplicative, F (z) = σ(dk−1 )φ(d) = d\|z d\|z σ(d)σ(d) · · · σ(d) φ(d), # \$% & (k−1)times is also so. Case(i) Let z be square-free and z = p. Then F (p) = σ(dk−1 )φ(d) d\|z = σ(1)φ(1) + σ(pk−1 )φ(p) =1+ pk−1+1 − 1 · (p − 1) = pk = z k . p−1 Euler’s Generalization and O\|—function 163 Case(ii) If z = p1 p2 · · · pr , then σ(dk−1 )φ(d) = F (z) = F (p1 )F (p2 ) · · · F (pr ) d\|z = pk1 pk2 · · · pkr = (p1 p2 · · · pr )k = z k . Problem 7.7.7. For any integer n, prove that 3\|σ(3n + 2). k k Solution 7.7.7. Let 3n + 2 = p1 1 p2 2 · · · psks . Since 3 ≡ 0(mod 3) and 3n + 2 ≡ 2( k k mod 3), therefore pi i ≡ 0(mod 3) for i = 1, 2, . . . , s. If all pi i ≡ 1(mod 3), then k k k k p1 1 p2 2 · · · pks s ≡ 1(mod 3). Since p1 1 p2 2 · · · pks s ≡ 2(mod 3), therefore ∃ one ki pi satisfying pi ≡ 2(mod 3). This implies pi ≡ 2(mod 3). Because if pi ≡ 0( k k mod 3) and pi ≡ 1(mod 3), then this yields pi i ≡ 0(mod 3) and pi i ≡ 1(mod 3) respectively. But this is not the case. Since pi ≡ 2(mod 3), then p2i ≡ 4 ≡ 1( mod 3) and p3i ≡ 2(mod 3). Therefore if pri ≡ 2(mod 3), then r is odd. Hence k pi i ≡ 2(mod 3), ki is odd. k ∴ σ(pki i ) = k k −1 (p − 1)(pi i + pi i + . . . + pi + 1) pi i+1 − 1 = i , pi − 1 pi − 1 = pki i + pki i −1 + . . . + pi + 1, and ki is odd. Since 2 ≡ (−1)(mod 3), therefore if r is odd then pri ≡ (−1)(mod 3) and if r is even then pri ≡ 1(mod 3). ∴ σ(pki i ) = pki i + pki i −1 + . . . + pi + 1 ≡ (−1) + 1 + . . . + (−1) + 1( mod 3) ≡ 0( mod 3.). ∴ 3\|σ(pi ) ⇒ 3\|(σ(pk1 1 )) · · · σ(pki i ) · · · σ(pks s ) ki ⇒ σ(pk1 1 pk2 2 · · · pks s ), ∵, σ is multiplicative 3\|σ(3n + 2). Problem 7.7.8. For any integer n > 1 has the form n = pk11 pk22 . . . pkr r , then show that d\|n μ(d)φ(d) = (2 − p1 )(2 − p2 ) . . . (2 − pr ). Solution 7.7.8. Since μ and φ are multiplicative then μ · φ is also multi plicative. Therefore F (n) = d\|n μ(d)φ(d) is also multiplicative. Note that k1 k 2 kr n = p1 p2 . . . pr is the prime factorization of n. Then μ(d)φ(d) F (pk ) = d\|pk = μ(1)φ(1) + μ(p)φ(p) + . . . + μ(pk )φ(pk ) = 1 + (−1)(p − 1) = 2 − p [∵ μ(pk ) = 0 for k ≥ 2]. 164 Number Theory and its Applications ∴ F (n) = (2 − p1 )(2 − p2 ) . . . (2 − pr ). Problem 7.7.9. If the integer z > 1 has the prime factorization z = q1k1 q2k2 · · · qsks , prove dφ(d) = d\|z 2k1 +1 q1 +1 q1 + 1 2k2 +1 q2 +1 q2 + 1 ··· qs2ks +1 + 1 qs + 1 . Solution 7.7.9. Since f (x) = x is multiplicative, therefore f · φ is also so. Hence F (z) = dφ(d), is multiplicative. d\|z Consider, F (q k ) = dφ(d) d\|q k = 1 · φ(1) + q · φ(q) + q 2 · φ(q 2 ) + . . . + q k · φ(q k ), = 1 + q(q − q 0 ) + q 2 (q 2 − q) + . . . + q k (q k − q k−1 ), = 1 + q 2 − q + q 4 − q 3 + q 6 − q 5 + . . . + q 2k − q 2k−1 , = 1 + (−1)1 q + (−1)2 q 2 + (−1)3 q 3 + . . . + (−1)2k q 2k . ∴ q 2k+1 + 1 = (q + 1)(q 2k − q 2k−1 + . . . + q 2 − q + 1)(W hy!). q 2k+1 + 1 = q 2k − q 2k−1 + . . . + q 2 − q + 1, q+1 q 2k+1 + 1 . ⇒ F (q k ) = q+1 ∴ dφ(d) = F (z) = F (q1k1 q2k2 · · · qsks ), d\|z = F (q1k1 )F (q2k2 ) · · · F (qsks ), 2k +1 2k +1 qs2ks +1 + 1 q2 2 + 1 q1 1 + 1 ··· . = q1 + 1 q2 + 1 qs + 1 Problem 7.7.10. Given k > 0, establish that there exists a sequence of k consecutive integers n + 1, n + 2, ..., n + k satisfying μ(n + 1) = μ(n + 2) = · · · = μ(n + k) = 0. Solution 7.7.10. Let pk be the kth prime. Then for i = j. gcd(p2i , p2j ) = 1. By Euler’s Generalization and O\|—function 165 virtue of Chinese Remainder theorem, ∃ a solution to: X ≡ −1( mod p21 ), X ≡ −2( mod p22 ), .. . X ≡ −k( mod p2k ), where p1 = 2, p2 = 3, . . . , pk = kth prime. If n = p1 p2 · · · pk and Ni = a simultaneous solution is φ(p21 ) X = (−1)N1 φ(22 ) ⇒ X = −N1 φ(p22 ) + (−2)N2 φ(32 ) − 2N2 φ(p2 ) + . . . + Nk φ(k2 ) − . . . − kNk k n pi , then . . ⇒ X + i ≡ 0( mod p2i ), for i = 1, 2, 3, . . . , k. ⇒ X + i = ap2i , for some integer a. Hence μ(X + i) = 0, i = 1, 2, 3, . . . , k. 7.8 Exercises: 1. Calculate φ(5040), φ(36000). 2. Prove the following assertions: (a) φ(3n) = 3φ(n) if and only if 3\|n. (b) φ(3n) = 2φ(n) if and only if 3 n. 3. If the integer n > 1 has r distinct prime factors, then show that φ(n) ≥ n 2r . 4. If n = pt11 pt22 . . . ptrr then prove the inequality τ (n)φ(n) ≥ n. 5. Prove that there are inﬁnitely many integers n satisfying φ(n) = n 3. 6. Show that Goldbach’s Conjecture implies that for each even integer 2n there exists integers n1 and n2 with φ(n1 ) + φ(n2 ) = 2n. 7. Use Euler’s theorem to establish the following: (a) For any integer a, a13 ≡ a(mod 2730). (b) For any odd integer a, a33 ≡ a(mod 4080). 8. For any prime p prove the following assertions: (a) σ(p!) = (p + 1)σ((p − 1)!); (b) φ(p!) = (p + 1)φ((p − 1)!). 166 Number Theory and its Applications 9. Prove that 4\|σ(4n + 3) for any positive integer n. 10. If the integer n > 1 has the prime factorization n = pt11 pt22 . . . ptrr then establish that: φ(d) k1 (p1 −1) )(1 + k2 (pp22−1) ) . . . (1 + kr (pprr−1) ). d\|n d = (1 + p1 11. Show that for any integer n, φ(n)\|n − 1 if and only if n is prime. 12. Prove that d\|n σ(d)φ( nd ) = nτ (n). 13. For a positive integer z, prove that μ2 (d) d\|z φ(d) = z . φ(z) 14. Show that if p and 2p + 1 are both odd primes, then n = 4p satisﬁes φ(n + 2) = φ(n) + 2. 15. For which positive integer n does φ(n) divides n? 8 Primitive Roots “The mathematician Pascal admires the beauty of a theorem in number theory; it’s as though he were admiring a beautiful natural phenomenon. Its marvellous, he says, what wonderful properties numbers have. It’s as though he were admiring the regularities in a kind of crystal.” – Ludwig Wittgenstein 8.1 Introduction In this chapter we have studied another important aspect of modular arithmetic called primitive root. To study primitive roots we have introduced the concept of order of an integer modulo k(∈ Z+ ). The order of an integer a modulo k is the least positive integer t for which at ≡ 1(mod k), where gcd(a, k) = 1. Basically for this value of k, a becomes the primitive root of k if t becomes φ(k). For instance 3 is a primitive root of 5, because 3φ(5) = 34 = 81 ≡ 1(mod 5) . When primitive root exists, it is very convenient to apply them in proofs and explicit constructions; for instance if p is an odd prime and a is a primitive root modulo p, the quadratic residues modulo p are exactly the even powers of the primitive root. Another important usefulness of primitive roots is to ﬁnd indices or discrete logarithms of integers. This discrete logarithm which follows similar properties as logarithm of positive real numbers, can be applied in simplifying the computations of modular arithmetic. The primitive root modulo k is mainly used in Cryptography, including the Diﬃ-Hellman key exchange. Also primitive roots are applicable in sound dif167 168 Number Theory and its Applications fusers. 8.2 Multiplicative Order In the last chapter we have seen that for any integer a and positive integer m with gcd(a, m) = 1, aφ(m) ≡ 1(mod m) where φ(m) is Euler’s phi function. Thus there exists at least one integer x which satisﬁes the equation ax ≡ 1( mod m). Then from well ordering principle we can assert that there exists a least positive integer n which satisﬁes the above congruence equation. This leads to the following deﬁnition. Deﬁnition 8.2.1. Let m > 1 and there exists an integer a such that gcd(a, m) = 1. Then the least positive integer x which satisﬁes the equation ax ≡ 1(mod m) is called order of integer a modulo m, denoted by ordm a. In 1801 Gauss introduced this notation in his text book Disquisitiones Arithmeticae. Example 8.2.1. Here, our goal is to ﬁnd an order of 3 modulo 11. We have 35 = 243 ≡ 1(mod 11). Therefore the integer 3 has order 5 modulo 11. In above example, if we consider 310 , 315 or powers in terms of multiple of 5 then those values will also satisfy the equation 3x ≡ 1(mod 11). Our next theorem deals with this fact. Theorem 8.2.1. If a(∈ Z) has order n modulo m i.e. ordm a = n then, ab ≡ 1( mod m) holds if and only if n divides b. Proof. Suppose n divides b. So b can be written as b = kn for some integer k. Since ordm a = n then, an ≡ 1( mod m) ⇒ (an )k ≡ 1k ( mod m) or ab ≡ 1( mod m). Conversely, let b be a positive integer satisfying ax ≡ 1(mod m). Then by the division algorithm ∃ q, r such that b = qn + r, 0 ≤ r < n. Therefore ab = (an )q .ar . Now ab ≡ 1(mod m) and an ≡ 1(mod m) together implies ar ≡ 1( mod m), which contradicts the fact that n is the least positive integer. Therefore we have b = qn and consequently n\|b. This proves the theorem. An immediate consequence of the last theorem leads to the following corollary. Corollary 8.2.1. If a and m are relatively prime with m > 0 then, order of the integer a modulo m must divides φ(m) i.e. ordm a φ(m). Primitive Roots 169 Proof. We know that gcd(a, m) = 1. Then by Euler’s theorem, a ≡ 1( x mod m) satisﬁes the equation a ≡ 1(mod m). If k be the order of a modulo m i.e. ordm a = k then, k must divide φ(m). φ(m) We now illustrate the fact by means of an example. For that, let us choose a = 5 and m = 12. Then the integers less than 12 and prime to 12 are 1, 5, 7, 11. Thus, φ(12) = 4. It is easy to see that 52 ≡ 1(mod 12) ⇒ ord12 5 = 2. Obviously 2 divides 4 = φ(12). Then we can write this as 54 ≡ 52 (mod 12) where 4 ≡ 2( mod 2). Our next result is based on the order of an integer. Theorem 8.2.2. If a has order n modulo m i.e. ordm a = n then, for some non negative integers i and j the congruence ai ≡ aj (mod m) holds if and only if i ≡ j(mod n) prevails. Proof. To start with, assume ai ≡ aj (mod m) where i ≥ j. Since gcd(a, m) = 1 then, ai−j ≡ 1(mod m)(Why?). As a has order n modulo m so, n (i − j)[refer to Theorem 8.2.1]. Therefore i ≡ j(mod n). For the converse part, let i ≡ j(mod n). Then i = qn + j for some integer q. Also, an ≡ 1(mod m). This shows that ai = anq+j = (an )q · aj ≡ aj (mod m). This ﬁnishes the proof. Theorem 8.2.2 gives an idea about the exponent x of a which satisﬁes ax ≡ 1( mod m). Now a fairly natural question presents itself: Is it possible to express the order of any exponent of a in terms of order a? The following theorem contains the answer. Theorem 8.2.3. If the integer a has order n modulo m, then at has order n modulo m for some t > 0. In other words, for some t > 0 if ordm a = n gcd(t, n) n . then ordm at = gcd(t, n) Proof. Let d = gcd(t, n). Then t = b1 d, n = b2 d with gcd(b1 , b2 ) = 1 for some n integers b1 and b2 . Now (at )b2 = (ab1 d ) d = (an )b1 ≡ 1(mod m). If at is assumed to have order r modulo m then by Theorem 8.2.1, we have r\|b2 . On the other hand, since a has order n modulo m, the congruence atr = (at )r ≡ 1(mod m) indicates n\|tr. This yields b2 d b1 dr ⇒ b2 b1 r. Now gcd(b1 , b2 ) = 1 shows b2 \|r. n n . Therefore at Also, r b2 and b2 r simultaneously implies r = b2 = = d gcd(t, n) n modulo m. has order gcd(t, n) We take immediate advantage of this theorem to prove the following corollary. 170 Number Theory and its Applications Corollary 8.2.2. Let a(∈ Z) has order n modulo m. Then at also has order n if and only if gcd(t, n) = 1 i.e. ordm at = n ⇐⇒ gcd(t, n) = 1. Proof. From Theorem 8.2.3, we have ordm at = only if gcd(t, n) = 1. n gcd(t,n) . Then ordm at = n if and Example 8.2.2. The following example will exemplify Theorem 8.2.3. Let us choose a = 4 and m = 13. Then 46 ≡ 1(mod 13). Thus ord13 4 = 6. Now 6 ord13 49 = = 2. In fact 418 = (46 )3 ≡ 1(mod 13). gcd(6, 9) In Corollary 8.2.1, we have seen that order of integer modulo m must divide φ(m), provided a and m are prime to each other. If the order of integer modulo m becomes exactly φ(m), then a is called primitive root of m. Deﬁnition 8.2.2. For an arbitrary integer a relatively prime to a positive integer m with ordm a = φ(m) then, a is called the primitive root of m. In fact, the above deﬁnition can be written as aφ(m) ≡ 1(mod m) where gcd(a, m) = 1. For instance if we choose a = 5 and m = 7 then, φ(7) = 6 as 7 is prime and we see that 56 ≡ 1(mod 7) verifying 5 as a primitive root of 7. Also there are some cases where we cannot ﬁnd primitive roots. For example if we choose a = 7 and m = 6 then, the integers less than 6 and prime to 6 are 1, 5. Thus, φ(6) = 2 and also 7 ≡ 1(mod 6) but 72 ≡ 1(mod 6). So in this case 7 is not a primitive root modulo 6. Our next theorem is going to exhibit an important property of primitive roots. Theorem 8.2.4. If a and m(> 0) are relatively prime and if a is primitive root of modulo m, then the integers a1 , a2 , · · · aφ(m) forms a reduced residue system modulo m. Proof. Its suﬃces to prove, all of the integers a1 , a2 , · · · aφ(m) are incongruent modulo m. If not, let ai ≡ aj (mod m) for 1 ≤ i < j ≤ φ(m). Thereby using Theorem 8.2.2, we can say that i ≡ j(mod φ(m)). But this is possible only if i = j, which proves that a1 , a2 , · · · aφ(m) are incongruent modulo m. Hence a1 , a2 , · · · aφ(m) forms a reduced residue system of modulo m. The following corollary deals with ﬁnding the number of primitive roots: Corollary 8.2.3. If an integer m(> 0) has primitive root then, the number is exactly φ(φ(m)). Proof. Let a be a primitive root modulo m. Then from Theorem 8.2.4, all the integers a1 , a2 , · · · aφ(m) forms a reduced residue system modulo m. Again using Primitive Roots 171 Corollary 8.2.2, we can assert that at is a primitive root modulo m if and only if gcd(t, φ(m)) = 1. As there are φ(φ(m)) such integers then the number of primitive roots are exactly φ(φ(m)). Example 8.2.3. Let us choose a = 2 and m = 11. Then the integers less than 11 and prime to 11 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. In fact φ(11) = 10. Its obvious that 210 ≡ 1(mod 11). Thus 2 is a primitive root modulo 11. Now from the above corollary, we can say there are exactly φ(φ(10)) = 4 primitive roots as the integers less than 10 and prime to 10 are 1, 3, 7, 9. Now 23 = 8 ≡ 8( mod 11), 27 = 128 ≡ 7(mod 11), 29 = 512 ≡ 6(mod 11). So here the primitive roots of 11 are 2, 6, 7, 8. 8.3 Worked out Exercises Problem 8.3.1. Find the order of the integers 2, 3, & 5 modulo 19. Solution 8.3.1. Here φ(19) = 18. Therefore the divisors of 18 are 1, 2, 3, 6, 9, 18. Under modulo 19, 22 ≡ 4, 23 ≡ 8, 26 ≡ 7, 29 ≡ 18, 218 ≡ 1 imply order of 2 is 18. By similar reasoning, 32 ≡ 9, 33 ≡ 8, 36 ≡ 7, 39 ≡ 18, 318 ≡ 1 generates order of 3 to be 18 and 52 ≡ 6, 53 ≡ 11, 56 ≡ 7, 59 ≡ 1 produces order of 5 to be 9. Problem 8.3.2. If k is a positive integer and a is an integer relatively prime to k such that order of a modulo k is k − 1 then, show that k is prime. Solution 8.3.2. An appeal to Euler’s theorem yields ak−1 ≡ 1(mod k) and aφ(k) ≡ 1(mod k). If φ(k) < k − 1, then this leads to a contradiction. Hence φ(k) = k − 1. If k were composite, then it would have a divisor d with 1 < d < k. Furthermore, k is a divisor of itself and this implies φ(k) ≤ (k − 2). But φ(k) = k − 1 shows that k is not composite and hence a prime. Problem 8.3.3. Assume that the order of a modulo n is h and the order of b modulo n is k. Show that the order of ab modulo n divides hk; in particular, if gcd(h, k) = 1 then ab has order hk. 172 Number Theory and its Applications Solution 8.3.3. We note that ah ≡ 1( mod n) ⇒ ahk ≡ 1( mod n), bk ≡ 1( mod n) ⇒ bkh ≡ 1( mod n). ∴ (ab)hk = ahk bhk ≡ 1( mod n). Thus by virtue of Theorem 8.2.1, order of ab divides hk. Next suppose gcd(h, k) = h h k k 1. Let h = p1 1 p2 2 · · · phr r & k = q1 1 q2 2 · · · qsks . Here qj = pj as gcd(h, k) = 1. Let ω be the order of ab. Then ω hk. So ω = p1l1 p2l2 · · · prlr q1m1 q2m2 · · · qsms , l l where 0 ≤ li ≤ hi , 0 ≤ mi ≤ ki . Let ω = hx ky , hx = p11 p22 · · · prlr , ky = m m ˜ & k = kk ˜ . q1 1 q2 2 · · · qsms . Thus hx \|h, ky \|k. Let h = hh x y ∴ (ab)hx ky = ahx ky bhx ky ≡ 1( mod n). ˜ ∴ (ahx ky bhx ky )h ≡ 1( mod n). ˜ hx ky hx ky h But (a b (8.3.1) ˜ k hh ˜ k hh x y x y ) =a b = (a ) (bh )ky ≡ (bh )ky ( mod n), as ah ≡ 1( mod n). (8.3.2) h ky (8.3.1) and (8.3.2) together imply (bh )ky ≡ 1(mod n). Since order of b is k, therefore we get k\|hky (refer to Theorem 8.2.1). Since gcd(h, k) = 1, therefore k\|ky . Hence k\|ky , ky \|k ⇒ k = ky . Similarly, h = hx . Finally, ω = hk. So gcd(h, k) = 1 ⇒ order of ab = hk. Problem 8.3.4. Prove that the odd prime divisors of the integer n4 + 1 are of the form 8k + 1. Solution 8.3.4. Assume p (n4 + 1). Therefore n4 ≡ −1(mod p) ⇒ n8 ≡ 1( mod p). Here gcd(n, p) = 1, for if n = kp (k ∈ Z) ⇒ n4 = (kp)4 ⇒ p\|n4 . This combines with p\|(n4 + 1) gives p\|1, which is impossible. Let r be the order of n modulo p. Therefore r\|8 [refer to Theorem 8.2.1]. Thus, r = 1, 2, 4, 8. Here order of n can not be 1, for n ≡ 1 ⇒ n4 ≡ 1. Similarly, order of n can not be 2 and 4(Why!). Hence order of n modulo p must be 8. Therefore 8\|φ(p) ⇒ 8\|(p − 1) ⇒ p = 8k + 1 where k ∈ Z. Problem 8.3.5. Prove that if p and q are odd primes and q (ap −1), then either q\|(a − 1) or else q = 2kp + 1 for some integer k. Solution 8.3.5. Our ﬁrst claim is gcd(a, q) = 1. For if, let gcd(a, q) = d(> 1). Then d\|q & q (ap − 1) together yields d (ap − 1). Also d\|a implies d\|1, a Primitive Roots 173 contradiction. Hence our ﬁrst claim is fulﬁlled. Now q (ap − 1) ⇒ ap ≡ 1( mod q). Let r be the order of a modulo q. Then r\|p [refer to Theorem 8.2.1]. Moreover p being prime shows r = 1 or p. If r = 1, then a ≡ 1(mod q) ⇒ q\|(a − 1). If r = p, then p\|φ(q)(Explain Why!). But φ(q) = q − 1. Therefore p\|(q − 1) ⇒ pk˜ = q − 1, k˜ ∈ Z. But q being odd implies q − 1 is even. Since p is odd, k˜ must be even, so k˜ = 2k for some k ∈ Z. Therefore p(2k) = q − 1, q = 2pk + 1 for some k. Problem 8.3.6. Verify that 2 is a primitive root of 19, but not of 17. Solution 8.3.6. Here, φ(19) = 18, 26 = 64 ≡ 7( mod 9) 72 = 49 ≡ 11( mod 19) 73 ≡ 77 = 4 · 19 + 1 ≡ 1( mod 19) 218 = (26 )3 ≡ 73 ≡ 1( mod 19). Therefore 218 = 2φ(19) ≡ 1(mod 19). Suppose order of 2 modulo 19 = r(< 18). This implies r\|18. So r = {1, 2, 3, 6, 9}. Now, 21 ≡ 1( mod 19) ⇒ r = 1. 22 ≡ 1( mod 19) ⇒ r = 2. 23 ≡ 1( mod 19) ⇒ r = 3. 26 ≡ 7 ≡ 1( mod 19) ⇒ r = 6. 29 ≡ 8 · 7 = 56 ≡ 18( mod 19) ⇒ r = 9. So φ(19) = 18 is the smallest integer r for which 2r ≡ 1(mod 19). Hence 2 is a primitive root of 19. Next for 17, φ(17) = 16. Let r be the order of 2. Therefore r ∈ {1, 2, 8, 16}. Here r = 1, 2, 4(Verify!). Now 28 = 15 · 17 + 1 ≡ 1( mod 17). ∴ 28 ≡ 1( mod 17). So order of 2mod 17 is 8 and not 16. Hence 2 is not a primitive root of 17. Problem 8.3.7. Let r be a primitive root of the integer n. Prove that rk is a primitive root of n if and only if gcd(k, φ(n)) = 1. Solution 8.3.7. Since r has order φ(n) modulo n, by virtue of Theorem 8.2.3, φ(n) rk has order modulo n. Now if gcd(k, φ(n)) = 1, then rk has order gcd(k, φ(n)) φ(n) implies rk is a primitive root of n. 174 Number Theory and its Applications k Conversely, suppose r is a primitive root of n. Then, rk has order φ(n) φ(n) ⇒ gcd(k, φ(n)) = 1. modulo n. From above, φ(n) = gcd(k,φ(n)) Problem 8.3.8. Prove that if p and q > 3 are both odd primes and q\|Rp , then q = 2kp + 1 for some integer k. 10p − 1 10p − 1 . If q\|Rp then for some r, qr = ⇒ Solution 8.3.8. Here, Rp = q q q(qr) = 10p −1. This shows q\|10p −1 implying q\|(10−1) or q = 2kp+1 for some k[refer to worked out Problem 8.3.5]. Since q > 3, then q (10 − 1). Therefore q = 2kp + 1 for some integer k. Problem 8.3.9. Prove that a is a primitive root modulo odd prime p if and only if a is an integer with gcd(a, p) = 1 such that a (p−1) q ≡ 1( mod p) for all prime divisors q of p − 1. Solution 8.3.9. Let a be a primitive root modulo p. Then the order of a modulo p is φ(p) = p − 1. Thus for any prime divisor q of p − 1 we can assert that a (p−1) q ≡ 1( mod p). In anticipation of a contrapositive argument, let a be not a primitive root of p. Then there exists an integer t < p − 1 with at ≡ 1(mod p). Now from Corollary 8.2.1, we can say that t divides p − 1. This implies that p − 1 = st for some integer s(> 1). Then we have p−1 = t. Let q be a prime divisor of s. This s s = t · ( ). From this we get, shows that p−1 q q a (p−1) q s s = at( q ) = (at ) q ≡ 1( mod p). This ﬁnishes the proof of the converse part. Problem 8.3.10. Prove that if a is a primitive root modulo m, then a is also primitive root modulo m, where a is an inverse of a modulo m. Solution 8.3.10. Let a be a primitive root modulo m, then we have aφ(m) ≡ 1( mod m). Let the order of a modulo m be k. Then by virtue of Corollary 8.2.1 we get k\|φ(m). Since a is inverse of a modulo m then, aa ≡ 1( mod m) ⇒ (aa)k ≡ 1( mod m) ⇒ ak (a)k ≡ 1( mod m) ⇒ ak ≡ 1( mod m). This shows that φ(m)\|k. Combining k\|φ(m) and φ(m)\|k yields k = φ(m). This proves that a is also primitive root modulo m. Primitive Roots 8.4 175 Primitive Roots for Primes In the foregoing section we have dealt with an important concept, known as primitive roots. In the present section we will study whether every prime has a primitive root or not. To proceed further, the concept of polynomial congruence will act as an important tool. The deﬁnition of polynomial congruence is as follows: Deﬁnition 8.4.1. Let f (x) be a polynomial with integral coeﬃcients, then the expression f (x) ≡ 0(mod m) is known to be polynomial congruence modulo m. Here in polynomial congruence an integer a is a root of f (x) modulo m if f (a) ≡ 0(mod m). Also if a is a root of f (x) modulo m then it is obvious that every integer which are congruent to a modulo m is also a root of f (x) ≡ 0( mod m). For instance, let f (x) = x2 + 2x + 4 has two incongruent roots modulo 6 and they are x ≡ 2(mod 6), x ≡ −4(mod 6). If we choose g(x) = x2 + 2 then it has no roots modulo 6. The next theorem deals with roots of a polynomial modulo p, where p is prime. Theorem 8.4.1. (Lagranges): Let f (x) = an xn + an−1 xn−1 + · · · + a0 and p an (n ≥ 1), where p is prime then the polynomial congruence f (x) ≡ 0( mod p) has at most n incongruent solutions modulo p. Proof. Let us apply the principle of Mathematical induction on n, degree of f (x). If n = 1 then f (x) = a1 x + a0 . Since p a1 then gcd(a1 , p) = 1, thus the congruence equation a1 x ≡ a0 (mod p) has unique solution[refer to Corollary 4.4.1]. Thus the result is true for n = 1. Suppose the statement is true for all polynomials whose degree is k(> 1). Let the degree of f (x) be k + 1, whose one of the roots being chosen as α. Then we have, f (x) = (x − α)g(x) + r, where the degree of g(x) is k with integral coeﬃcients. Also, r is constant. Since α is a root of f (x) then, f (α) = (α − α)g(α) + r ≡ 0(mod p). So we can write, f (x) = (x − α)g(x)(mod p). Now as the degree of g(x) is k then p does not divide the leading coeﬃcient of g(x), otherwise p will divide the leading coeﬃcient of f (x) whose degree is k + 1. If β be another root of f (x) then, f (β) ≡ 0(mod p) ⇒ (β − α)g(β) ≡ 0(mod p). As α, β are diﬀerent roots under congruence modulo p, then we have g(β) ≡ 0( mod p). As the degree of g(x) is k, by induction hypothesis, at most k numbers of β are possible. Therefore f (x) ≡ 0(mod p) has at most k + 1 incongruent solutions modulo p. Thus the statement is true for n = k + 1 and hence it is true for all integers n(≥ 1). 176 Number Theory and its Applications This result leads to an observation that is useful in certain situations; namely, Corollary 8.4.1. If p is prime and q divides p−1, then the congruence equation, xq − 1 ≡ 0(mod p) has exactly q incongruent solutions. Proof. Here q\|(p − 1) implies p − 1 = qk for some integer k. Then, xp−1 − 1 = (xq − 1)(xq(k−1) + xq(k−2) + · · · + xq + 1) = (xq − 1)R(x). Here R(x) = xq(k−1) + xq(k−2) + · · · + xq + 1 whose degree is q(k − 1) = qk − q = (p − 1) − q. Applying Theorem 8.4.1 we can say that R(x) has at most (p − 1) − q incongruent roots. Also from Fermat’s little theorem, xp−1 − 1 ≡ 0( mod p) has p − 1 incongruent roots under modulo p. Now let x = c be a solution of xp−1 −1 ≡ 0(mod p) but not a solution of R(x) ≡ 0(mod p). Then x = c must be the solution of xq − 1 ≡ 0(mod p). So Theorem 8.4.1 yields the number of incongruent solutions of xq −1 ≡ 0(mod p) is exactly (p−1)−(p−1)+q = q. Now we are in a stage to assert that for any prime p and any divisor q of p − 1, the congruence equation whose degree is q has exactly q diﬀerent solutions under modulo p. Our next theorem focuses on the number of integers having a given order under modulo p. Theorem 8.4.2. Let p be prime and d a positive divisor of p − 1. Then the number of incongruent integers of order d modulo p is φ(d). Proof. Let d\|(p − 1) and χ(d) be the number of positive integers having order d modulo p which are less than p. Since every integer lying between 1 and p − 1 has order d for some d\|(p − 1), it follows p−1= χ(d). d\|(p−1) Again from Theorem 7.6.1 we observe that, p−1= φ(d). d\|p−1 Combining the last two equations, we obtain d\|(p−1) χ(d) = d\|(p−1) φ(d). (8.4.1) Primitive Roots 177 Claim: d\|(p − 1) ⇒ χ(d) ≤ φ(d). From (8.4.1) it follows that χ(d) = φ(d). If χ(d) = 0 then we are done. Let us choose χ(d) > 0, which shows that there exists an integer n of order d. Then there exists d number of integers n, n2 , n3 , · · · , nd which are incongruent modulo p. Also, these integers satisfy the equation xd ≡ 1( mod p) because, (nk )d = (nd )k ≡ 1(mod p) where k ∈ {1, 2, · · · , d}. However the Corollary 8.2.2 signiﬁes that any power k of a is of the order d modulo p provided gcd(k, p) = 1. Finally, we conclude that k is no one but φ(d) and also χ(d) = φ(d). So the number of incongruent integers having order d modulo p is φ(d). This proves the theorem. If we substitute d = p − 1 then we arrive at the following corollary: Corollary 8.4.2. Every prime has a primitive root. Proof. In the above theorem, taking d = p − 1, p being a prime, we can say that there are φ(p − 1) incongruent integers of order p − 1 modulo p. Using the deﬁnition of primitive root, for any integer a relatively prime to p, the order of a becomes d = p − 1 = φ(d) modulo p implies a is a primitive root. Remark 8.4.1. Using Corollary 8.2.3, for any prime p which has a primitive root, the number of primitive roots are φ(φ(p)) = φ(p − 1). Finally, the following example illustrates the above results: Example 8.4.1. Choose p = 29. Then d = 4 is a divisor of p − 1 = 28. So the number of integers incongruent modulo 31 of order 4 is φ(4) = 2. Also, there are φ(φ(29)) = φ(28) = 14 number of primitive roots. Taking n = 12, shows 124 ≡ 1( mod 29)(Verify!), satisﬁes x4 − 1 ≡ 0(mod 29). So the number of integers of the form 12k where 1 ≤ k ≤ 28, are relatively prime to and less than 29. Now 28 = 4, if k = 7, 21. the order of 12k is gcd(k, 28) Finally we need to evaluate 127 and 1221 modulo 29. In fact, 127 ≡ 17( mod 29) 1221 ≡ 12( mod 29). Thus 12 and 17 are the only integers of order 4 modulo 29. Again if we choose n = 2 and d = p − 1 = 28 then we have 228 ≡ 1(mod 29). This shows that 2 is the primitive root modulo 29. The concluding part of the section deals with the following table which lists the smallest positive primitive root for each prime below 200. 178 Prime 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 Number Theory and its Applications Least Positive Primitive Root 1 2 2 3 2 2 3 2 5 2 3 2 6 3 5 2 2 2 2 7 5 3 2 Prime 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 Least Positive Primitive Root 3 5 2 5 2 6 3 3 2 3 2 2 6 5 2 5 2 2 2 19 5 2 3 Table 8.1: Primitive root for Primes Primitive Roots 8.5 179 Worked out Exercises Problem 8.5.1. Prove for an odd prime p, the only incongruent solutions of x2 ≡ 1(mod p) are 1 and p − 1. Solution 8.5.1. Since p is odd prime therefore, 2 (p − 1). Hence with reference to Corollary 8.4.1, the congruence equation has exactly 2 incongruent solutions. Clearly, 1 is a solution as 1 ≡ 1(mod p). Now p − 1 is also a solution as (p−1)2 ≡ 1(mod p). Therefore 1 and p−1 are solutions and they are incongruent modulo p. Problem 8.5.2. Find all the positive integers less than 43 having order 6 modulo 43, provided 3 is a primitive root of 43. Solution 8.5.2. By virtue of Theorem 8.2.2, 3k (1 ≤ k ≤ 42) are incongruent. 42 Thus, all integers less than 43 are congruent to 3k having the order gcd(k, 42) modulo 43[refer to Theorem 8.2.3]. Now 42 = 6 ⇒ gcd(k, 42) = 7 ⇒ k = 7, 35. gcd(k, 42) Thus, 37 , 335 have order 6 modulo 43. Here, 33 = 27, 34 = 81 ≡ −5( mod 43), ∴ 37 ≡ −135 ≡ −135 + 3 · 43 ≡ 37 ≡ −6( mod 43). 37 · 37 ≡ (−6)(−6) ≡ 36 ≡ (−7)( mod 43), 318 ≡ (−7)(−5) ≡ 35 ≡ (−8)( mod 43). 332 ≡ 314 · 318 ≡ (−7)(−8) ≡ 56 ≡ 13( mod 43). 333 ≡ 39 ≡ −4( mod 43), ∴ 335 ≡ 9(−4) = −36 ≡ 7( mod 43). Hence 7(≡ 335 ), 37(≡ 37 ) have order 6 modulo 43. Problem 8.5.3. Considering r as a primitive root of the odd prime p, prove the following: 1. The congruence r p−1 2 ≡ −1(mod p) holds. 2. If r1 is any other primitive root of p, then rr1 is not a primitive root of p. 3. If for the integer r , rr ≡ 1(mod p) holds then r is also a primitive root of p. 180 Number Theory and its Applications Solution 8.5.3. 1. From Fermat’s theorem, we have rp−1 ≡ 1(mod p). As p−1 p−1 2 exists. 2 ∈ Z, therefore r ∴ rp−1 − 1 ≡ 0( mod p) ⇒ (r If (r p−1 2 p−1 2 − 1)(r p−1 2 + 1) ≡ 0( mod p). − 1) ≡ 0(mod p), then r fails to have order p − 1(Why!). ∴ (r p−1 2 + 1) ≡ 0( mod p) ⇒ r p−1 2 ≡ −1( mod p). 2. On the contrary, if rr1 would have been a primitive root of p then its order would be p − 1. By virtue of (1), r ⇒r p−1 2 p−1 2 ≡ −1( mod p), p−1 & r1 2 ≡ −1( mod p), p−1 r1 2 ≡ (−1)(−1)( mod p), ⇒ (rr1 ) p−1 2 ≡ 1( mod p). A contradiction to our hypothesis(Why!). Hence the statement follows. 3. Let us assume 1 ≤ r ≤ p − 1. For if r = p then, r ≡ 0(mod p). If r > p then, by division algorithm we can write r = qp + s for some integers q and s with 0 ≤ s < p − 1. This implies r ≡ s(mod p) and both r and s have same order. This follows that gcd(r , p) = 1. Now consider (r )k with 1 ≤ k ≤ p − 1. If k < p − 1 and (r )k ≡ 1(mod p) then, 1 = 1k ≡ (rr )k = rk (r )k ≡ rk (mod p). This contradicts the fact that r is a primitive root modulo p. Therefore 1 = 1p−1 ≡ (rr )p−1 = rp−1 (r )p−1 ≡ rk (mod p). Now (r )k ≡ 1(mod p)(1 ≤ k < p − 1) and gcd(r , p) = 1 together implies that r is a primitive root of p. Problem 8.5.4. For a prime p > 3, prove that the primitive roots of p occur in incongruent pairs r, r˜ where r˜ r = 1(mod p). Solution 8.5.4. Referring to Theorem 7.2.4, for n > 2, φ(n) is an even integer so that φ(n) ≥ 2. Let r be one primitive root of p. Then r, r2 , . . . , rp−1 are congruent to 1, 2, 3, . . . , p − 1 in some order and so r, r2 , . . . , rp−1 are incongruent[refer to Theorem 8.2.4]. Since p > 3, there are at least three members in the r = rp−1 ≡ 1(mod p). list. Let r˜ = rp−2 . Then r and r˜ are incongruent and r˜ So r˜ is another primitive root(refer last problem), thereby it follows that if r is a primitive root of p(> 3) ∃ another primitive root r˜ incongruent to r such that r˜ r = 1(mod p). Problem 8.5.5. Use the fact that each prime p has a primitive root to give a diﬀerent proof of Wilson’s theorem. Primitive Roots 181 Solution 8.5.5. Let r be a primitive root of p. Now, 1, 2, 3, . . . , p − 1 are the positive integers relatively prime to p. Also, φ(p) = p − 1. Then, by virtue of Theorem 8.2.4, r, r 2 , . . . , rp−1 are congruent modulo p to 1, 2, 3, . . . , p−1 in same order. ∴ r · r 2 · · · rp−1 ≡ 1 · 2 · 3 · · · · (p − 1)( mod p) ⇒ r1+2+3+...+(p−1) ≡ (p − 1)!( mod p) ⇒r p(p−1) 2 ⇒ (r ≡ (p − 1)!( mod p) ) ≡ [(p − 1)!]2 ( mod p). p−1 p (8.5.1) Since r is a primitive root of p, therefore rp−1 ≡ 1(mod p). So (rp−1 )p ≡ 1( mod p). Thus from equation(8.5.1), we obtain [(p − 1)!]2 ≡ 1( mod p) ⇒ [(p − 1)!]2 − 1 ≡ 0( mod p) ⇒[(p − 1)! + 1][(p − 1)! − 1] ≡ 0( mod p). (p−1)p If (p − 1)! − 1 ≡ 0(mod p), then r 2 ≡ (p − 1)! ≡ 1(mod p). But rp−1 ≡ 1( (p−1)p p−1 p−1 mod p) ⇒ rp ≡ r(mod p). Therefore r 2 = (rp ) 2 ≡ r 2 ≡ 1(mod p). p−1 Thereby contradicts the fact that the order of r = p − 1, as < (p − 1). 2 Hence (p − 1)! + 1 ≡ 0(mod p) ⇒ (p − 1)! ≡ (−1)(mod p). Problem 8.5.6. If p is a prime, show that the product of the φ(p − 1) primitive roots of p is congruent modulo p to (−1)φ(p−1) . Solution 8.5.6. Since r is a primitive root of p, therefore r, r2 , . . . , rp−1 are congruent modulo p to 1, 2, 3, . . . , p − 1 in some order[refer to Theorem 8.2.4]. If s be any other primitive root of p, it must be congruent to one of 1, 2, 3, . . . , p−1. Therefore s is congruent to one of r, r2 , . . . , rp−1 . Hence all primitive roots of p are of the form rk , where 1 ≤ k ≤ p − 1. Since gcd(k, p − 1) = 1, therefore we have k = p − 1. So k must be of the form 1 ≤ k < p − 1. Let us denote these φ(p − 1) integers as k1 , k2 , . . . , kφ(p−1) , 1 ≤ ki < p − 1. Thus, the product of these primitive roots is r k1 r k 2 · · · r kφ(p−1) =r k1 +k2 +...+kφ(p−1) . Now Theorem 7.6.2 yields k1 + k2 + . . . + kφ(p−1) = ∴ r k1 +k2 +...+kφ(p−1) 1 (p − 1)φ(p − 1). 2 1 = r 2 (p−1)φ(p−1) . 182 Number Theory and its Applications For p > 2, φ(p − 1) is even[refer to Theorem 7.2.4]. So 2 φ(p − 1). 1 1 ∴ r 2 (p−1)φ(p−1) = (r(p−1) ) 2 φ(p−1) , 1 ≡ (1) 2 φ(p−1) ≡ 1( mod p), ∵, 1 φ(p − 1) ≥ 1, for p > 2. 2 Since φ(p − 1) is even, (−1)φ(p−1) = 1. This shows that, r k1 +k2 +...+kφ(p−1) ≡ (−1)φ(p−1) ( mod p). For p = 2, the only primitive root is 1, φ(2) = 1 and so, r k1 +k2 +...+kφ(p−1) = 1 ≡ (−1)( mod 2), ∵, 1 ≡ (−1)( mod 2). Hence the formula holds for p = 2. Problem 8.5.7. Show that if p is a prime and p = 2q + 1, where q is an odd prime and a is a positive integer with 1 < a < p − 1, then p − a2 is a primitive root modulo p. Solution 8.5.7. Since p is a prime, therefore φ(p) = p − 1 = 2q. Now the possible orders of p − a2 are 1, 2, q and 2q. If p − a2 has the order 2, then we get (p − a2 )2 = p2 − 2pa2 + a4 ≡ a4 ( mod p). Here a2 ≡ 1(mod p), as a can not be of order 4. Thus we have a ≡ ±1(mod p). But this is not possible, because 1 < a < p − 1. This shows p − a2 does not posses order 2. If p − a2 has order q, then using binomial theorem we get (p − a2 )q ≡ −(a2q ) ≡ −1( mod p). Again, this proves p − a2 does not posses order q. Hence (p − a2 ) has the order 2q and (p − a2 ) becomes a primitive root modulo p. Problem 8.5.8. Let r be a primitive root of the prime p with p ≡ 1( mod 4). Show that −r is also a primitive root. Solution 8.5.8. Let r be a primitive root modulo p where p ≡ 1(mod 4). Let t be the order of −r modulo p. Thus, from Corollary 8.2.1 we have t\|(p − 1). This shows p − 1 = kt for some integer k. Here we need to show k = 1. For that, let us suppose k > 1. Two cases arise: Primitive Roots 183 Case-I: If k is odd, then t = is even. This gives (−r) = r ≡ 1(mod p), which contradicts the fact that r is a primitive root modulo p. p−1 Case-II: If k is even, then (−r)t = (−r) k ≡ 1(mod p). As k is even, p−1 p−1 2 then p−1 ≡ 1(mod p). Here p−1 k 2 gives (−r) 2 is even as, p ≡ 1(mod 4). Hence p−1 k (−r) p−1 2 = (−1) p−1 2 (r) t p−1 2 =r p−1 2 t ≡ 1( mod p). Again we arrive at a contradiction(Why!). Thus the only possibility remains k = 1. This shows −r is a primitive root modulo p. 8.6 Existence of Primitive Roots A recent section of this chapter deals with the existence of primitive roots for primes. In the present section, our main aim is to ﬁnd the existence of primitive roots for composite numbers. We discourse this fact by square of a prime number, which are speculated in the following theorems. Lemma 8.6.1. If p is an odd prime having a primitive root r modulo p, then rp−1 ≡ 1(mod p2 ). Proof. Since r is a primitive root modulo p, then the order of r is φ(p) = p − 1. Also we have, rp−1 ≡ 1(mod p). Let n be the order of r modulo p2 , then rn ≡ 1( mod p2 ). This shows that p2 (rn − 1) ⇒ p rn − 1. Therefore rn ≡ 1(mod p). Again Theorem 8.2.3 gives, φ(p) = (p − 1) n and Corollary 8.2.1 generates 2 n φ(p ) = p(p − 1). Now combining them, we can say that either n = p − 1 or n = p(p − 1). Finally, our task is to show n = p − 1 is not possible. Let us choose s = r + p where s becomes a primitive root modulo p [∵ s ≡ r(mod p)]. Now applying Binomial Theorem, sp−1 = (r + p)p−1 , p − 1 p−3 2 r p + · · · + pp−1 , 2 = rp−1 + (p − 1)rp−2 p + ≡ rp−1 + (p − 1)rp−2 p( mod p2 ), ≡ rp−1 − prp−2 ( mod p2 ). From this congruence equation, we can assert that sp−1 ≡ 1(mod p2 ). Otherwise, if sp−1 ≡ 1(mod p2 ) holds then prp−2 ≡ 0(mod p2 ) ⇒ p2 \|rp−2 p. Thus p\|rp−2 (∵ gcd(p, r) = 1), which is not possible(Why!). This shows that for any 1(mod p2 ). primitive root r modulo p, rp−1 ≡ 184 Number Theory and its Applications In the preceding lemma, we have not considered the case n = p(p − 1). The following theorem shines on this case. Theorem 8.6.1. If p is an odd prime with primitive root r modulo p then either r or r + p is a primitive root modulo p2 . Proof. Let n be the order of r modulo p2 . Then Lemma 8.6.1 yields, either n = p − 1 or n = p(p − 1). Consider n = p(p − 1) = φ(p2 ). Then r becomes primitive root modulo p2 . Now, if s = r + p then s becomes primitive root modulo p. Again, from the last Lemma we have rp−1 ≡ 1(mod p2 ). Hence the order of s modulo p2 is p(p − 1) = φ(p2 ). Consequently s = r + p is a primitive root modulo p2 . To illustrate the above theorem, let us pick-out p = 11. Then we get 210 ≡ 1( mod 11). This asserts, 2 is a primitive root modulo 11. Now, 211(11−1) = 2110 ≡ 1( mod 121)(Verify!). Concluding that 2 is the primitive root of 121. Now if we choose, p = 487 then 10 will be the primitive root modulo 487. Here we can see that 497 = 487 + 10 becomes the primitive root modulo 4872 . In fact, p = 487 is the smallest prime for which there exists a primitive root that is not also a primitive root modulo 4872 . Till now, our discussions are based on primitive root modulo p2 for an arbitrary odd prime p. Now a fairly natural question presents itself: what will happen, if we take any integral exponents of p greater than 2. As a prelude, we need a technical lemma. Lemma 8.6.2. Let p be an odd prime and r be the primitive root of p such that rp−1 ≡ 1(mod p2 ). Then for each positive integer n ≥ 2, we also have pn−2 (p−1) r ≡ 1(mod pn ). Proof. Let us apply the principle of mathematical induction on n ≥ 2. From Theorem 8.6.1, we know that if p has a primitive root r then r is also primitive root modulo p2 satisfying rp−1 ≡ 1(mod p2 ). So the lemma holds true for n = 2. Let us assume that the lemma is true for all n ≥ 2. This shows that pn−2 (p−1) 1(mod pn ). Now we are to show the lemma is true for n + 1. r ≡ As p being an odd prime yields gcd(r, p) = gcd(r, pn−1 ) = gcd(r, pn ) = 1. Using Euler’s theorem, we obtain n−2 rp (p−1) = rφ(p n−1 ) ≡ 1( mod pn−1 ). This shows that for any integer k, we have n−2 rp (p−1) = (1 + kpn−1 ). (8.6.1) Primitive Roots 185 Here p k, otherwise k = tp [for some integer t], contradicts our hypothesis n−2 rp (p−1) ≡ 1(mod pn ). Now taking p-th power on both sides of (8.6.1), we ﬁnd n−1 rp (p−1) = (1 + kpn−1 )p n−1 = 1 + p(kp p (kpn−1 )2 + · · · + (kpn−1 )p )+ 2 ≡ 1 + kpn ( mod pn+1 ) Since p k, then we get n−1 rp (p−1) ≡ 1( mod pn+1 ). This establishes the validity of the incongruence for n + 1. So by principle of mathematical induction the incongruence of the statement holds true for every n ≥ 2. Thus, the last lemma allows us to state and prove the following theorem. Theorem 8.6.2. Let p be an odd prime and r be a primitive root modulo p2 . Then, for all positive integers n, r is a primitive root modulo pn . Proof. Here, for any odd prime p, r is a primitive root modulo p2 . Then, for n−2 1(mod pn ). Let d be the order any n > 2, Lemma 8.6.2 indicates rp (p−1) ≡ of r modulo pn . Using, Corollary 8.2.1 we have d\|φ(pn ) = pn−1 (p − 1). Again, we see that rd ≡ 1(mod pn ) ⇒ rd ≡ 1(mod p)(How!). Also from Theorem 8.2.1 we have, p − 1 = φ(p) d. Considering d\|pn−1 (p − 1) and (p − 1)\|d simultaneously, we obtain d = pt (p − 1), where t is an integer such that 0 ≤ t ≤ n − 1. If we take t < n − 1, then n−2 rp (p−1) t = (rp (p−1) pn−2−t ) ≡ 1( mod pn ), n−2 1( as d being the order of r modulo pn . This contradicts the fact rp (p−1) ≡ n n−1 n (p − 1) = φ(p ), consequently r becomes a primitive mod p ) . Hence d = p n root modulo p . The following example will lucidly explain the foregoing theorem. Example 8.6.1. Let us choose p = 5 then φ(5) = 4. Now we show that 34 ≡ 1( mod 5). Here 320 = (34 )5 ≡ 1(mod 25). Thus 3 is a primitive root modulo 5 and also for modulo 25. Then using Theorem 8.6.2 we can say that 3 is also a primitive root modulo 5n , for all positive integers n. In fact, here we have shown it for n = 3. For that φ(p3 ) = p2 (p − 1) = 25 × 4 = 100. Now 35 ≡ (−7)( mod 125) ⇒ 3100 ≡ 720 ≡ 1(mod 125), as 710 ≡ (−1)(mod 125). 186 Number Theory and its Applications Till now, our discussion was based on primitive root modulo exponents of odd primes. Proceeding further, on primitive roots modulo powers of 2, we ﬁgure out 1 and 3 to be primitive roots modulo 2 and 22 respectively. Now, the following theorem furnishes a means for the situation modulo 2n for n > 2. So let us begin with a technical lemma. Lemma 8.6.3. If a be an odd integer then a integer n ≥ 3. φ(2n ) 2 = a2 n−2 ≡ 1(mod 2n ) for any Proof. Considering principle of mathematical induction on n, let n = 3. Then a2 ≡ 1(mod 8). Since a is an odd integer, let us take a = 2m + 1 where m ≥ 0. ∴ (2m + 1)2 ≡ 1( mod 8) ⇒ 4(m + 1)m ≡ 0( mod 8). Obviously 8 4m(m + 1) as m being a non-negative integer and m(m + 1) is a product of consecutive integers. So the lemma is true for n = 3. Let us assume the lemma is true for some integer k > 3. Then for any odd k−2 ≡ 1(mod 2k ) holds, which implies ∃ an integer a the equivalence relation a2 2k−2 k integer d such that a = 1 + d · 2 . Squaring both sides, we obtain a2 k−1 = (1 + d2k )2 , = 1 + d2k+1 + d2 22k , ≡ 1( mod 2k+1 ). This asserts that the lemma is true for n = k + 1. So the equivalence relation n−2 ≡ 1(mod 2n ) is true for all n ≥ 3. a2 Consequence of the last lemma leads to the next theorem. Theorem 8.6.3. The integer 2n has no primitive root for n ≥ 3. Proof. From the deﬁnition of primitive roots, it follows that if 2n possesses a n primitive root then it satisﬁes aφ(2 ) ≡ 1(mod 2n ) for any odd integer a and n ≥ 3. But Lemma 8.6.3 establish, a2 n−2 =a φ(2n ) 2 ≡ 1( mod 2n ). So there are no primitive roots for n ≥ 3. For further clariﬁcation, let us pick out n = 4 and a = 5. Then φ(24 ) = 8. φ(24 ) Here 5 2 = 54 ≡ 1(mod 24 ). So 5 is not a primitive root modulo 24 . Now, instead of taking exponents of 2 or exponents of odd prime if we take a composite number which can be factored into a product of two relatively primes, then any primitive roots modulo that composite number. Next theorem is based on that intent. Primitive Roots 187 Theorem 8.6.4. If for any two integers m, n > 2, gcd(m, n) = 1 then the integer mn has no primitive root. Proof. Let us start with an integer a such that gcd(a, mn) = 1. Then gcd(a, m) = 1 and gcd(a, n) = 1. Choose b = lcm(φ(m), φ(n)) and d = gcd(φ(m), φ(n)). Theorem 7.2.4 guarantees that both φ(m) and φ(n) are even numbers, which implies d ≥ 2. Also, Theorem 2.5.1 yields b= φ(mn) φ(m)φ(n) ≤ . [∵ φ is multiplicative] d 2 From Euler’s theorem we have, aφ(m) ≡ 1(mod m) and aφ((n) ≡ 1(mod n). φ(n) φ(m) Therefore ab = (aφ(m) ) d ≡ 1(mod m) and ab = (aφ(n) ) d ≡ 1(mod n). Since gcd(m, n) = 1, combining we get ab ≡ 1(mod mn). But here b = φ(mn) which shows a is not a primitive root modulo mn. For further illustration of this theorem, choose m = 3, n = 4 and a = 5. Then φ(3 × 4) = φ(3) × φ(4) = 2 × 2 = 4. Now 52 = 25 ≡ 1(mod 12) shows 5 is not a primitive root modulo 12. We have demonstrated that all exponents of odd primes possess primitive roots. Also, the only powers of 2 which have primitive roots are 2 and 4. Consider the form m = 2pn where p is an odd prime and n is a positive integer. The following theorem deals with those forms of integers having primitive roots. Theorem 8.6.5. Let p be an odd prime and n ≥ 1, then the form 2pn possesses primitive roots. n Proof. Let a be a primitive root modulo pn . Then aφ(p ) ≡ 1(mod pn ). As φ n is multiplicative so, φ(2pn ) = φ(2)φ(pn ) = φ(pn ). Thus we have aφ(2p ) ≡ 1( mod pn ). Two cases may arise: n Case I a is odd: Then aφ(2p ) ≡ 1(mod 2). Now gcd(2, p) = 1 ⇒ gcd(2, pn ) = n 1 ⇒ aφ(2p ) ≡ 1(mod 2pn ). This proves, a is a primitive root modulo 2pn . n Case II a is even: Then a+pn is odd. So (a+pn )φ(2p ) ≡ 1(mod 2). As a+pn ≡ n n a(mod pn ), so (a + pn )φ(2p ) ≡ 1(mod pn ). Therefore (a + pn )φ(2p ) ≡ 1( mod 2pn ), by a similar argument as above. This shows a+pn is a primitive root modulo 2pn . Referring to Example 8.6.1, 3 is a primitive root modulo 5n (n ≥ 1). Then from the above theorem, 3 also becomes a primitive root modulo 2·5n . In fact, if we choose n = 3 then m = 2×53 = 250 and φ(2×53 ) = 52 ×4 = 100 holds. Here 188 Number Theory and its Applications 3 = (3 ) ≡ (−7) ≡ 151(mod 250). Finally, 3100 ≡ (151)4 ≡ 1(mod 250). This proves, 3 is a primitive root of 2 × 53 . Again, if a = 2 then 2 + 5n is odd for n ≥ 1. From preceding theorem, 2 + 5n becomes a primitive root modulo 2 × 5n . In particular, if we prefer n = 2 then 2 + 52 = 27 becomes a primitive root modulo 2 × 52 = 50, as 2720 = (275 )4 ≡ 74 ≡ 1(mod 50). 20 5 4 4 Remark 8.6.1. After discussing all the theorems and results in this section we are in this point to say that an integer k > 1 possesses a primitive root if and only if k = 2, 4, pn , 2pn , where p is an prime. 8.7 Worked out Exercises Problem 8.7.1. Determine all the primitive roots of 32 , 33 , & 34 . Solution 8.7.1. Remark (8.6.1) guarantees the existence of primitive roots for 3k . Here, 2 is a primitive root of 3. Therefore either 2 or 2 + 3 will be primitive roots of 3k (k ≥ 2). If r be a primitive root of p then, order of r(mod p2 ) is p − 1 or p(p − 1). Hence order of 2(mod 32 ) is 3 − 1 or φ(32 ). But 22 = 4 ≡ 1( 2 mod 32 ), so 2φ(3 ) ≡ 1(mod 32 ). Thus, 2 is a primitive root of 32 , 33 , & 34 . Next, let us calculate the other primitive roots of 32 , 33 , & 34 . (a) 32 : There are φ(φ(32 )) = φ(6) = 2 primitive roots. Since φ(32 ) = 6 therefore, 2h will have order 6(Why!) ⇐⇒ gcd(h, 6) = 1 or h = 1, 5. Thus, 25 = 32 ≡ 5( mod 32 ). Hence the primitive roots of 32 are 2, 5. (b) 33 : There are φ(φ(33 )) = φ(18) = 6 primitive roots and all are of the form 2k such that gcd(k, 18) = 1 or k = 1, 5, 7, 11, 13, 17. Thus, 25 = 32 ≡ 5( mod 27), 27 ≡ 5 · 22 ≡ 20( mod 27), 211 ≡ 52 · 2 = 50 ≡ 23( mod 27), 213 ≡ 23 · 22 = 92 ≡ 11( mod 27), 217 ≡ 211 · 25 · 22 ≡ 23 · 5 · 2 = 230 ≡ −40 ≡ 14( mod 27). Hence the primitive roots of 33 are 2, 5, 11, 14, 20, 23 (c) 34 : Left to the reader. Problem 8.7.2. Find the primitive root for all positive integers k modulo 13k . Primitive Roots 189 Solution 8.7.2. As 2 = 4096 ≡ 1(mod 13), so 2 is a primitive root of 13. Here Theorem 8.6.1 can be brought into work to assert that either 2 or 2+13 = 15 is a primitive root of 132 . Now we have, 212 ≡ 40(mod 132 ) and (212 )13 = 2156 ≡ 4013 ≡ 1(mod 132 ). Therefore 2 is a primitive root of 132 . So Theorem 8.6.2 asserts that 2 is a primitive root of 13k . 12 Problem 8.7.3. Obtain a primitive root of 34. Solution 8.7.3. Here we note that 34 ≡ −4(mod 17) ⇒ 316 ≡ 1(mod 17). Hence 3 is a primitive root of 17. Appealing Theorem 8.6.5 we can conclude that 3 is a primitive root of 2 × 17 = 34. Problem 8.7.4. For any odd prime p, prove that any primitive rootr of pn is also a primitive root of p. Solution 8.7.4. Here, gcd(r, pn ) = 1 ⇒ gcd(r, p) = 1. Let k be the order of r(mod p). ∴ rk ≡ 1( mod p). ∴ k ⇒ \|φ(p) ⇒ k (p − 1). (8.7.1) Also, rk = 1 + sp, s ∈ Z. So for n > 1 we obtain rkp (n−1) (n−1) = (1 + sp)p , n−1 n−1 n−1 p p sp + (sp)2 + . . . + (sp)p . =1+ 1 2 n−1 p , for 1 ≤ k ≤ pn−1 & But, pn−1 k p (sp)k , for 1 ≤ k ≤ pn−1 . n−1 pn−1 p n 2 pn−1 ∴ p sp + . (sp) + . . . + (sp) 1 2 ∴ rkp n−1 ≡ 1( mod pn ). Since r is a primitive root of pn , therefore φ(pn ) kpn−1 [Why!], φ(pn ) = pn−1 (p − 1). ∴ (p − 1) k. (8.7.2) Hence combining (8.7.1) and (8.7.2) yields k = p − 1, so order of r mod p is p − 1. 190 Number Theory and its Applications Problem 8.7.5. Prove that 3 is a primitive root of all integers of the form 7k and 2 · 7k . Solution 8.7.5. Here 2( mod 7), 33 ≡ 6( mod 7), 31 ≡ 1( mod 7), 32 ≡ 34 ≡ 4( mod 7), 35 ≡ 5( mod 7), 36 ≡ 1( mod 7). ∴ 3φ(7) = 36 ≡ 1( mod 7). So 3 is a primitive root of 7. Therefore order of 3mod 72 is 7 − 1 or 7(7 − 1). But, 34 = 81 ≡ 32( mod 72 ). ∴ 36 ≡ 9 · 32 ≡ 43( mod 72 ). 1( mod 72 ). ∴ 36 ≡ Thus, order of 3(mod 72 ) is φ(72 ) = 7(7 − 1). Now, Lemma 8.6.2 displays that for k ≥ 2, (k−2) (7−1) 1( mod 7k ). ≡ 37 And Theorem 8.6.2 asserts 3 is a primitive root of 7k , k ≥ 1. Finally, 3 being an odd primitive root for 7k , Theorem 8.6.5 establishes that 3 is a primitive root of 2 · 7k . Problem 8.7.6. Assume that r is a primitive root of the odd prime p and (r + tp)p−1 ≡ 1(mod p2 ). Show that r + tp is a primitive root of pk for each k ≥ 1. Solution 8.7.6. Since r ≡ (r + tp)(mod p), so r and r + tp have same order. Therefore r + tp is also a primitive root of p. Since any primitive root of has order mod p2 of p − 1 or p(p − 1), then r + tp has order modulo p2 of p − 1 or p(p − 1). By the given condition, (r + tp)p−1 ≡ 1(mod p2 ), order of r + tp is not p − 1 so must be φ(p2 ) = p(p − 1). Therefore r + tp is a primitive root of p, is also a primitive root of p2 . Finally, r + tp is a primitive root of pk for each k ≥ 1[refer to Lemma 8.6.2 and Theorem 8.6.2]. Problem 8.7.7. Obtain all the primitive roots of 41. Solution 8.7.7. With reference to Table 8.1 we see that 6 is a primitive root of 41. Also by Theorem 8.2.4 we can say that all other primitive roots are congruent to one of 6, 62 , . . . , 640 . Again, 41 has φ(φ(41)) = φ(40) = (23 −22 )×(5−1) = 16 incongruent primitive roots. Now by Theorem 8.2.3 we can assert, if 6 has order 40 then 6b has order 40 . Here 6b will be a primitive root if gcd(b, 40) = 1. This implies, gcd(b, 40) Primitive Roots 191 b = 1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39. Finally taking the above b’s as exponent of 6, we can ﬁnd other primitive roots modulo 41(Do it!). Problem 8.7.8. A Carmichael function, after the famous American Mathematician Robert Carmichael, associates to every positive integer k there exists a positive integer v such that {av ≡ 1(mod k)}, for all integers a with gcd(a, k) = 1. Moreover, the least universal exponent of k is known to be the minimal universal exponent of k, denoted by λ(k). Minimal universal exponent of k is often termed as reduced totient function or the least universal exponent. In general, if k = 2m pt11 pt22 . . . ptrr be the prime factorization of n(> 1) then λ(k) = lcm(λ(2m ), φ(pt11 ), φ(pt22 ), . . . , φ(ptrr )) where λ(2) = 1, λ(22 ) = 2 and λ(2m ) = 2m−2 for m ≥ 3. Using the deﬁnition, prove the following statements: 1. For k = 2, 4, pt , 2pt , where p is an odd prime, λ(k) = φ(k). 2. If gcd(a, 2k ) = 1, then aλ(2 k ) ≡ 1(mod 2k ). 3. If gcd(a, k) = 1 then aλ(k) ≡ 1(mod k). Solution 8.7.8. 1. Pertaining the above deﬁnition, we ﬁnd λ(2) = 1. Now φ(2) = 2 − 20 = 1 ⇒ λ(2) = φ(2). Again, by virtue of the deﬁnition, λ(22 ) = 2 = 22 − 21 = φ(22 ). Now for k = 2pt , we see that λ(k) = lcm(λ(2), φ(pt )) = lcm(1, φ(pt )) = φ(pt ) & φ(2φ(pt )) = φ(2)φ(φ(pt )) = φ(φ(pt )). These two together gives λ(2φ(pt )) = φ(2φ(pt )). Similarly, for k = pt we get λ(φ(pt )) = φ(φ(pt )). k 2. Since gcd(a, 2k ) = 1 then, by Euler’s theorem we can say that aφ(2 ) ≡ 1( mod 2k ). Taking mathematical induction into consideration, the problem prevails for k = 1, 2[Verify!]. Consider the problem to be true for k = m ≥ m 3. Then we can assert that aλ(2 ) ≡ 1(mod 2m ). This shows that aλ(2 m ⇒ a2λ(2 ) m = 1 + 2m r ) = (1 + 2m r)2 ⇒ aλ(2 m+1 ) = 1 + 2 · 2m r + 22m r2 ⇒ aλ(2 m+1 ) = 1 + 2m+1 r + 2m−1 2m+1 r2 ⇒ aλ(2 m+1 ) = 1 + (r + r2 2m−1 )2m+1 . This shows that aλ(2 is true for all k. m+1 ) ≡ 1(mod 2m+1 ). Thus we can say that the result 192 Number Theory and its Applications t 3. Let k = p , p being an odd prime. Then by (1) we have λ(k) = φ(k). Since gcd(a, k) = 1 then, by Euler’s theorem aλ(k) = aφ(k) ≡ 1(mod k). If m k = 2m pt11 pt22 . . . ptrr then, by (2) we see that aλ(2 ) ≡ 1(mod 2m ). Since λ(2m ) lcm[λ(2m ), φ(pt11 ), φ(pt22 ), . . . , φ(ptrr )] it follows aλ(k) ≡ 1(mod 2m ). ti Again by Euler’s theorem we have, aφ(pi ) ≡ 1(mod ptii ) which further ast1 t2 m tr serts that alcm[λ(2 ),φ(p1 ),φ(p2 ),...,φ(pr )] ≡ 1(mod ptii ). Combining λ(ptii ) = φ(ptii ) [by (1)] and λ(ptii ) lcm[λ(2m ), φ(pt11 ), φ(pt22 ), . . . , φ(ptrr )], it follows aλ(k) ≡ 1(mod ptii ). This is true for all i = 1, 2, . . . , r. Finally, applying corollary(4.2.2) we can conclude that aλ(k) ≡ 1(mod k). Problem 8.7.9. Prove that if gcd(a, k) = 1, then the linear congruence ax ≡ b( mod k) has the solution x ≡ baλ(k)−1 (mod k). Solution 8.7.9. By virtue of the Problem 8.7.8(3), we get aλ(k) ≡ 1(mod k). Then, multiplying both sides of the congruence by b we get baλ(k) ≡ b( mod k) b · a · aλ(k)−1 ≡ b( mod k) a baλ(k)−1 ≡ b( mod k). This shows that x ≡ baλ(k)−1 (mod k) is a solution of ax ≡ b(mod k). 8.8 Index Arithmetic The present section deals with an important aspect of modular arithmetic, introduced by Gauss in his Disquistiones Arithmeticae. Theorem(8.2.4) highlights the idea that if a is a primitive root modulo m then the integers a1 , a2 , · · · , aφ(m) forms a reduced system of modulo m. For any arbitrary integer r relatively prime to m, ∃ unique x with 1 ≤ x ≤ φ(m), satisﬁes ax ≡ r(mod m). This motivates us to give the following deﬁnition. Deﬁnition 8.8.1. Let m be a positive integer with primitive root a. If r be another integer relatively prime to m, then ∃ unique integer x, with x(1 ≤ x ≤ φ(m)), such that ax ≡ r(mod m) holds is said to be index of r base a modulo m. ind r Also, this index is denoted as ind ra . Thus we can write a a ≡ r(mod m). In fact, if x is an index of r base a then x = ind ra , where we do not need to indicate modulo m which is already prime to a. Choose any integer ŕ, which is also congruent modulo r. Then r ≡ ŕ(mod m) where ŕ is also relatively r ŕ prime to m. Hence aind a ≡ r(mod m) and aind a ≡ ŕ(mod m) together imply r ŕ ind a a ≡ aind a (mod m). Finally an appeal to Theorem 8.2.2 concludes that Primitive Roots 193 ind a ≡ ind notion of index. r ŕ a (mod φ(m)). The upcoming example will help to understand the Example 8.8.1. Pick out m = 5. Then φ(5) = 4 ⇒ 24 ≡ 1(mod 5). This shows that 2 is primitive root modulo 5. Now, 24 ≡ 1( mod 5), 23 ≡ 3( mod 5), 22 ≡ 4( mod 5), 21 ≡ 2( mod 5) and it follows that ind 12 = 4, ind 22 = 1, ind 32 = 3, ind 42 = 2. The following theorem depicts some properties of indices similar to logarithms of real numbers. Here the equality sign in logarithms is replaced with congruence relation modulo m. Sometimes indices are called discrete logarithms of integers. Theorem 8.8.1. If m has a primitive root a and ind relative to a then, r a denotes the index of a 1. ind 1a ≡ 0(mod φ(m)). 2. ind ar1 r2 ≡ ind ar1 + ind ar2 (mod φ(m)). k 3. k · ind ra = ind ar (mod φ(m)), for k > 0. Proof. 1. As a is a primitive root modulo m, from Euler’s theorem we have aφ(m)≡1( mod m) , where φ(m) is the smallest power of a which is congruent modulo m. ∴ ind 1a ≡ 0( mod m). 2. From the deﬁnition of indices we have, r aind a1 ≡ r1 ( mod m), r and aind a2 ≡ r2 ( mod m). Multiplying the last two congruence gives, r r aind a1 +ind a2 ≡ r1 r2 ( mod m). Again, r r2 aind a1 ≡ r1 r2 ( mod m). Now, the property of congruence yields, r r2 aind a1 r r ≡ aind a1 +ind a2 ( mod m). Finally, by virtue of Theorem 8.2.2, we obtain the desired property. 194 Number Theory and its Applications 3. Using deﬁnition of indices, we have rk aind a ≡ rk ( mod m) r r and ak·ind a ≡ (aind a )k ≡ rk ( mod m). Then from the property of congruences, we obtain r rk ak·ind a ≡ aind a ( mod m). Finally, applying Theorem 8.2.2 we get k k · ind ra = ind ra ( mod φ(m)). For an illustration of the above properties, let us consider the following example. Example 8.8.2. Let us choose, m = 7 then φ(7) = 6 and a = 5. Now, 51 ≡ 5( mod 7), 52 ≡ 4( mod 7), 53 ≡ 6( mod 7), 54 ≡ 2( mod 7), 55 ≡ 3( mod 7), 56 ≡ 1( mod 7). Here we note that ind 65 = 3, ind 35 = 5, ind 45 = 2, ind 55 = 1, ind 25 = 4. Let r1 = 5 and r2 = 4 with gcd(5, 4) = 1 and m = 7. Now, ind 20 ≡ ind 55 + ind 45 ≡ 1 + 2 ≡ 3( mod 6). 5 Again, ≡ ind 65 ( mod 6). 20 ≡ 6( mod 7) ⇒ ind 20 5 ∴ ind 4×5 ≡ ind 45 + ind 55 ( mod φ(7)). 5 Take k = 2, then 2 ind 35 ≡ 2ind 35 ≡ 2 × 5 ≡ 4( mod 6). Here 9 ≡ 2( mod 7). ∴ ind 95 ≡ ind 25 ( mod 6). 2 ∴ ind 35 ≡ 2ind 35 ( mod φ(7)). Actually, this concept of indices has important applications in solving binomial congruences of the type xk ≡ r(mod m), k ≥ 2 where m is a positive integer having primitive root and gcd(r, m) = 1. The following example will reﬂect on solving an equation of this type. Primitive Roots 195 Example 8.8.3. Solve the congruence 3x ≡ 5(mod 11) using indices. As m = 11, so φ(11) = 10. Also, 210 ≡ 1(mod 11). Thus 2 is the primitive root modulo 11. Now the table of indices as follows: 4 a ind a2 1 10 2 1 3 8 4 2 5 4 6 9 7 7 8 3 9 6 10 5 4 ind 3x ≡ ind 52 ≡ 4( mod 10). 2 4 ≡ ind 32 + 4ind x2 ( mod 10), Also, ind 3x 2 ≡ 8 + 4ind x2 ( mod 10). ∴ 8 + 4ind x2 ≡ 4( mod 10). ∴ 4ind x2 ≡ −4( mod 10), ≡ 6( mod 10). Since gcd(2, 10) = 2 from Theorem(4.2.2), we obtain 2ind x2 ≡ 3( mod 5). ∴ ind x2 ≡ 4, 9( mod 5). Thus we have, x ≡ 24 , 29 ( mod 11). So x ≡ 5, 6( mod 11) are the solutions here. Now in this particular example, the number of primitive roots are given by φ(φ(11)) = 4. In this solution, we have taken 2 as a primitive root. The other primitive roots are 8, 7, 10. Now consider any one of them say 7, then the above table may diﬀer but the ﬁnal solution remains unaltered. Let’s justify this with an illustration: a ind a7 1 10 2 3 3 4 4 6 5 2 6 7 7 1 8 9 9 8 4 ind 3x ≡ ind 57 ≡ 2( mod 10). 7 4 ≡ ind 37 + 4ind x7 ( mod 10), Also, ind 3x 2 ≡ 4 + 4ind x7 ( mod 10). ∴ 4 + 4ind x2 ≡ 2( mod 10). ∴ 4ind x2 ≡ −2( mod 10), ≡ 8( mod 10). 10 5 196 Number Theory and its Applications Since, gcd(4, 10) = 2, 2ind x7 ≡ 4( mod 5). ∴ ind x7 ≡ 7, 2( mod 5). Thus we have, x ≡ 77 , 720 ( mod 11). So x ≡ 5, 6( mod 11) are the solutions here. Next, our focus will be on the congruence of the form xk ≡ r(mod m), k ≥ 2 where m is a positive integer having primitive root and gcd(r, m) = 1. To discuss further, let us have the following deﬁnition: Deﬁnition 8.8.2. If m and k are positive integer with gcd(r, m) = 1 for any integer r, then we say that r is a kth power residue m provided the congruence xk ≡ r(mod m), k ≥ 2 has a solution. Now our next theorem deals with the case where m possesses a primitive root of this type of congruence equations which are solvable. Theorem 8.8.2. Let m be a positive integer having a primitive root and gcd(r, m) = 1. Then the congruence xk ≡ r(mod m), k ≥ 2 has a solution if and only if φ(m) a d ≡ 1(mod m), where gcd(k, φ(m)) = d; if it has a solution then there are exactly d number of incongruent solutions modulo m. Proof. Let a be a primitive root modulo m. We know that xk ≡ r(mod m), k ≥ 2 holds if and only if k ind xa = tφ(m) + ind ra for some integer t. Now, k ind xa − tφ(m) = ind ra becomes a linear diophantine equation. From Theorem 2.7.1, we can say that this equation possess a solution if d\|ind ra . Also, there are d number of incongruent solutions. As d\|ind ra ⇒ ind ra = dn for some integer n. Thus, φ(m) φ(m) ind ra = × (dn), d d φ(m) ⇒ ind ra ≡ 0( mod φ(m)). d ∴ r φ(m) d ≡ 1( mod m). Remark 8.8.1. In particular,taking m as prime say p with gcd(r, p) = 1, the p−1 congruence xk ≡ r(mod m), k ≥ 2 has a solution if and only if a d ≡ 1( mod m) where d = gcd(k, p − 1). Finally to ﬁne tune the above theorem, let us illustrate it by an example: Primitive Roots 197 Example 8.8.4. Let us consider a congruence equation by x ≡ 4(mod 5), taking m = 5. Then 2 d = gcd(2, φ(5)) = gcd(2, 4) = 2. Here φ(m) φ(m) 4 = = 2 & 4 d = 42 ≡ 1( mod 5), d 2 holds. So using the foregoing theorem, the given congruence equation is solvable. 8.9 Worked out Exercises Problem 8.9.1. The following is a table of indices for the prime 17 relative to the primitive root 3: a ind a3 1 16 2 14 3 1 4 12 5 5 6 15 7 11 8 10 9 2 10 3 11 7 12 13 13 4 14 9 15 6 16 8 With the aid of this table, solve the following congruences: 1. x12 ≡ 13(mod 17). 2. 8x5 ≡ 10(mod 17). Solution 8.9.1. 1. Here, x12 ≡ 13(mod 17), gcd(13, 17) = 1. ∴ 12ind x3 ≡ ind 13 ( mod 16), ind 13 = 4. 3 3 ∴ 12ind x3 ≡ ind 13 ≡ 4( mod 16), gcd(12, 16) = 4. 3 Dividing by 4, we get 3ind x3 ≡ 1(mod 4). Hence ind x3 = 3, 7, 11, 15. Thus, x = 6, 7, 10, 11(refer to the table). Therefore x ≡ 6, 7, 10, 11(mod 17). 2. Here 8x5 ≡ 10(mod 17), gcd(10, 17) = 1. ∴ ind 83 + 5ind x3 ≡ ind 10 ( mod 16). 3 ⇒ 10 + 5ind x3 ≡ 3( mod 16). ⇒ 5ind x3 ≡ −7( mod 16). Since gcd(5, 16) = 1, therefore one solution do exists. This implies, 15ind x3 ≡ −21( mod 16), ⇒ − ind x3 ≡ −21( mod 16), ⇒ ind x3 ≡ 21( mod 16) ≡ 5( mod 16). Thus from table, x = 5. Hence x ≡ 5(mod 17). 198 Number Theory and its Applications Problem 8.9.2. Find the remainder when 324 · 513 is divided by 17. Solution 8.9.2. Here 324 · 513 ≡ x(mod 17) & gcd(1, 17) = 1, so just one solution exists. ∴ 24ind 33 + 13ind 53 ≡ ind x3 ( mod 16) ⇒ 24(1) + 13(5) ≡ ind x3 ( mod 16) [by Problem 8.9.1]. ⇒ 89 ≡ 9 ≡ ind x3 ( mod 16), ∴ x ≡ 14( mod 17). Hence remainder is 14. Problem 8.9.3. Show that the congruence x3 = 3(mod 19) has no solutions, whereas x3 = 11(mod 19) has three incongruent solutions. Solution 8.9.3. 1. Here x3 = 3(mod 19) & gcd(3, 19) = 1. Since gcd(3, φ(19)) = 18 gcd(3, 18) = 3 and 3 3 = 36 = 33 − 33 ≡ 8 · 8 ≡ 7 ≡ 1(mod 19), therefore x3 = 3(mod 19) has no solutions[refer to Theorem 8.8.2]. 2. Here x3 = 11(mod 19) & gcd(11, 19) = 1. Since gcd(3, φ(19)) = gcd(3, 18) = 18 3 and 11 3 = 116 ≡ (−8)6 ≡ (64)3 ≡ (7)3 ≡ 49 · 7 ≡ 11 · 7 ≡ 1(mod 19), therefore there are three incongruent solutions[refer to Theorem 8.8.2]. Problem 8.9.4. If r is a primitive root of the odd prime p, prove that = ind p−1 = ind −1 r r Solution 8.9.4. 1 (p − 1). 2 1. Since −1 ≡ (p − 1)(mod p), therefore ind −1 = ind p−1 . r r . Then rx ≡ (p − 1)(mod p). As p is odd, p − 1 is even. 2. Let x = ind p−1 r So p−1 2 exists. ∴ rp−1 ≡ 1 ≡ (p − 1)2 ( mod p) ⇒ rp−1 ≡ (p − 1)2 ( mod p) ⇒ r p−1 2 ≡ (p − 1)( mod p) or r p−1 2 ≡ −(p − 1)( mod p). p−1 If r 2 ≡ −(p − 1)(mod p) ≡ 1(mod p), then r would not have order p−1 p−1 p−1(Why!). Hence r 2 ≡ −(p−1)(mod p) ≡ 1(mod p). So r 2 ≡ (p−1)( = 12 (p − 1). mod p) holds. Thus by deﬁnition, ind p−1 r Problem 8.9.5. For which values of b is the exponential congruence 9x ≡ b( mod 13) solvable? Primitive Roots 199 Solution 8.9.5. Here 2 is a primitive root of 13(check!). To solve, we need to construct the following table: a ind a2 1 12 2 1 3 4 4 2 5 9 6 5 7 11 8 3 9 8 10 10 11 7 12 6 Therefore xind 92 ≡ ind b2 (mod 12), because ind 92 = 8. Since gcd(8, 12) = 4, therefore the foregoing congruence yields 4 ind b2 . This implies ind b2 = 4, 8, 12. Thus, from the above table we ﬁnd b = 3, 9, 1. Problem 8.9.6. Determine the integers a(1 ≤ a ≤ p − 1) such that the congruence x4 ≡ a(mod p) has a solution for p = 7, 11, & 13. Solution 8.9.6. Since 36 ≡ 1(mod 7) and 210 ≡ 1(mod 11) therefore, 3, 2 is the primitive root of 7, 11 respectively. To construct the table of indices for 7, the following congruences will be particularly helpful: 31 ≡ 3( mod 7), 32 ≡ 2( mod 7), 33 ≡ 6( mod 7), 34 ≡ 4( mod 7), 35 ≡ 5( mod 7), 36 ≡ 1( mod 7). Thus the corresponding table is as follows: a ind a3 1 6 2 2 3 1 4 4 5 5 6 3 Again, the following congruences will provide assistance in constructing the table of indices for 11: 21 ≡ 2( mod 11), 22 ≡ 4( mod 11), 23 ≡ 8( mod 11), 24 ≡ 5( mod 11), 25 ≡ 10( mod 11), 26 ≡ 9( mod 11), 27 ≡ 7( mod 11), 28 ≡ 3( mod 11), 29 ≡ 6( mod 11), 210 ≡ 1( mod 11). So the required table is: a ind 11 2 1 10 2 1 3 8 4 2 5 4 6 9 7 7 8 3 9 6 10 5 1. p = 7: x4 ≡ a( mod 7), 4ind x ≡ ind a ( mod 6), gcd(4, 6) = 2. ∴ 2 a ⇒ ind a = 2, 4, 6. Hence a = 2, 4, 1 200 Number Theory and its Applications 2. p = 11: left to the reader. 3. p = 13: left to the reader. Problem 8.9.7. If r and r are both primitive roots of the odd prime p, show that for gcd(a, p) = 1 ind ar ≡ ind ar ind rr ( mod p − 1). Solution 8.9.7. Let x ≡ ind mod p). a r (mod p), y ≡ ind a r (mod p) & z ≡ ind r r ( ∴ by deﬁnition (r )x ≡ a( mod p), (r)y ≡ a( mod p) & (r )z ≡ r( mod p) ⇒ (r )zy ≡ (r)y ( mod p). ∴ (r )x ≡ (r)y ( mod p) ≡ (r )zy ( mod p). By Theorem 8.2.2, x ≡ zy(mod p−1). Therefore ind ar ≡ ind ar ind rr (mod p−1). 8.10 Exercises: 1. Find the order of the integers 2, 3 and 5 modulo 23. 2. If a has order 2k modulo odd prime p, then show that ak ≡ −1(mod p). 3. Prove that φ(2n − 1) is a multiple of n for any n > 1. 4. Verify: The odd prime divisors of the integer n2 + n + 1 which are diﬀerent from 3 are of the form 6k + 1. 5. Find two primitive roots of 10. 6. Prove that for any odd prime p, the prime divisors of 2p − 1 are of the form 2kp + 1. 7. If p be an odd prime then show that the congruence xp−2 +. . .+x2 +x+1 ≡ 0( mod p) has exactly p − 2 incongruent solutions and they are the integers 2, 3, . . . , p − 1. 8. If 3 is a primitive root of 43 then prove that all the positive integers less than 43 has the order 21 modulo 43. 9. Find all positive integers less than 61 having order 4 modulo 61. Primitive Roots 201 10. Let r be a primitive root of the odd prime p. Show that if p ≡ 3(mod 4) holds then −r has order p−1 2 modulo p. 11. Find all primitive roots of 82 and 38. 12. Prove that the integer 20 has no primitive roots. 13. Obtain a primitive root for any integer k of the form 11k and 17k . 14. For any odd prime p show that there are as many primitive roots of 2pn as of pn . 15. Solve the following congruences by using a table of indices for a primitive root of 11: (a) 7x3 ≡ 3(mod 11). (b) x8 ≡ 10(mod 11). 16. Constructing a table of indices for the prime 17 with respect to the primitive root 5, solve 7x ≡ 7(mod 17). 17. Determine the integers a(1 ≤ a ≤ 12) such that the congruence ax4 ≡ b( mod 13) has a solution for b = 2, 5, and 6. 18. Determine whether the two congruences x5 ≡ 13(mod 23) and x7 ≡ 15( mod 29) are solvable. 19. Find the solutions of the exponential congruences 5x ≡ 4(mod 19) and 4x ≡ 13(mod 17). 9 Theory of Quadratic Residues “Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and forthwith it is something entirely diﬀerent.” – Johann Wolfgang von Goethe 9.1 Introduction In the chapter Fermat’s little theorem, we deﬁned quadratic congruence equation of the form ax2 + bx + c ≡ 0(mod p), p being an odd prime and a ≡ 0( mod p). There we have discoursed the solution of the quadratic congruence of the type x2 + 1 ≡ (mod p). Also, how to solve a quadratic congruence of that form, was being treated there. Following the path of solving a quadratic congruence, we need an important fact of modern number theory, known as quadratic reciprocity law, a major contribution of Gauss in 1796. After that many eminent mathematicians of 19th century had given their important contributions on this aspect. In the present chapter we have illustrated various important facts and results based on quadratic reciprocity law. 9.2 Quadratic Residues and Nonresidues As quadratic reciprocity law deals with solving a quadratic congruences of the type ax2 + bx + c ≡ 0(mod p), p being an odd prime with gcd(a, p) = 1. Also, 203 204 Number Theory and its Applications Theory of Quadratic Residues 205 Proof. Here, the quadratic residue is of the form x ≡ r(mod p) where p is the odd prime and r be an integer with gcd(r, p) = 1. To compute quadratic residue of p among the integers 1, 2, · · · , p − 1, we ﬁrst need to establish that the number of incongruent solutions modulo p be either zero or two. Set x = b as a solution and (−b)2 = b2 ≡ r(mod p). Then x = −b is also a solution of x2 ≡ r(mod p). −b(mod p) otherwise b ≡ −b(mod p) implies 2b ≡ 0(mod p), which Here b ≡ is not possible as b2 ≡ r(mod p) and p r. There exist no other solutions of the given equation as x2 ≡ b2 (mod p) implies p (x − b)(x + b). Then either p\|(x − b) ⇒ x ≡ b(mod p) or p\|(x + b) ⇒ x ≡ −b(mod p). Because the number p−1 of incongruent solutions are two, there must be exactly quadratic residues 2 and non quadratic residues modulo p taken from 1, 2, · · · , p − 1. 2 The last theorem enables us to determine the number of quadratic residues taken from 1, 2, · · · , p − 1. Now we pose the following question: Under what circumstances an integer r becomes a quadratic residue of odd prime p? In order to answer this question, we need the Euler’s Criterion. Theorem 9.2.2. (Euler’s Criterion): Let p be an odd prime with gcd(r, p) = 1. p−1 Then r is a quadratic residue modulo p if and only if r 2 ≡ 1(mod p). Proof. Let r be a quadratic residue modulo p. Then ∃ b ∈ {1, 2, · · · , p − 1} such p−1 that b2 ≡ r(mod p). This implies, bp−1 ≡ r 2 (mod p). Then with the use of p−1 Fermat’s theorem, we have r 2 ≡ bp−1 ≡ 1(mod p) as gcd(b, p) = 1. p−1 p−1 Conversely if r 2 ≡ 1(mod p), then r 2 ≡ 1(mod p). Let b be a primitive root modulo p. Then r ≡ bi (mod p) for some integer i where 1 ≤ i ≤ p − 1. This i(p−1) p−1 shows, b 2 ≡ r 2 ≡ 1(mod p). By Theorem 8.2.1 the order of b must divide i(p − 1) . As b is a primitive root, so its order is φ(p) = p − 1. Thus to divide 2 i(p − 1) , i must be even, say 2j. It follows that, (bj )2 = b2j = bi ≡ r(mod p). 2 Hence bj is a solution of x2 ≡ r(mod p). This proves that r is quadratic residue of p. As p is odd prime and gcd(r, p) = 1, an appeal to Fermat’s little theorem p−1 p−1 p−1 yields, (r 2 − 1)(r 2 + 1) ≡ rp−1 − 1 ≡ 0(mod p). Thus, either r 2 ≡ 1( p−1 mod p) or r 2 ≡ −1(mod p) holds. Here both the conditions fail to agree together, otherwise 1 ≡ −1(mod p) implies p\|2 leads to a contradiction. As p−1 p−1 r 2 ≡ 1(mod p) prevails for quadratic residue modulo p, then obviously r 2 ≡ −1(mod p) holds for quadratic non residue modulo p. The last part of the theorem generates an obvious corollary. 206 Number Theory and its Applications Corollary 9.2.1. Let p be an odd prime with gcd(r, p) = 1. Then r is a quadratic p−1 residue modulo p or quadratic non residue modulo p according as r 2 ≡ 1( p−1 mod p) or r 2 ≡ −1(mod p) prevails. The next example illustrates Euler’s Criterion. 11−1 Example 9.2.2. If p = 11, we have 3 2 = 35 ≡ 1(mod 11). Also 4 45 ≡ 1(mod 11). So 3 and 4 are quadratic residue modulo 11. 11−1 2 = Unlike our earlier discussions, from now on our eﬀorts will be simpliﬁed by the symbol ( pr ). As the notation ( pr ) was introduced by Legendre, it is called Legendre symbol. Deﬁnition 9.2.2. Let p be an odd prime and r be an integer with gcd(r, p) = 1, then the Legendre symbol ( pr ) is deﬁned as, 1, if r, is a quadratic residue modulo p; r = p −1, if r, is a quadratic non residue modulo p For example, Legendre ifwe take p= 7, the symbol can be expressed as 1 2 4 3 5 6 = = = 1 and = = = −1 as 1, 2, 4 are quadratic 7 7 7 7 7 7 residue modulo 7 and 3, 5, 6 are quadratic non residue modulo 7. By virtue of Legendre symbol, Euler’s criterion can be portrayed as p−1 r ≡ r 2 ( mod p). p The following theorem discourse few properties related to Legendre symbol. Theorem 9.2.3. Let p be an odd prime and r, s be integers not divisible by p. Then 1. If r ≡ s(mod p) then ( pr ) = ( ps ). 2. r s rs = . p p p 3. r2 = 1. p Proof. 1. As r ≡ s(mod p) then x2 ≡ r(mod p) has a solution if and only if x2 ≡ s(mod p) has a solution. Hence with respect to modulo p we have ( pr ) = ( ps ). Theory of Quadratic Residues 207 2. If p\|rs then the proof is obvious. Consider the case when p rs and here gcd(p, rs) = 1. By virtue of Euler’s Criterion, rs p−1 ≡ (rs) 2 ( mod p) p rs p−1 p−1 ≡ (r) 2 (s) 2 ( mod p) p rs r s ≡ ( mod p). p p p Here Legendre symbol assume the values 1 and −1. If possible, let ( rs p ) = r s r ( p )( p ). Then 1 ≡ −1(mod p) ⇒ p\|2, which is impossible. Thus ( p )( ps ) = ( rs p ). 2 3. As ( pr ) is either 1 or −1, it follows ( rp ) ≡ ( pr )( pr ) ≡ 1(mod p). Hence we are done. We now consider the problem of ﬁnding Legendre symbol of those prime p whose quadratic residue is −1. This can be achieved by Euler’s Criterion. Theorem 9.2.4. If p is an odd prime, then 1, if p ≡ 1( mod 4) −1 = p −1, if p ≡ 3 ≡ −1( mod 4) Proof. For any integer k, if p ≡ 1(mod 4) then p is of the form p = 4k + 1. By Euler’s criterion we have, p−1 −1 ≡ (−1) 2 ≡ (−1)2k ≡ 1( mod p). p Also, for some integer k , if p ≡ 3(mod 4) then p is of the form p = 4k + 3. By similar reasoning as above we get, p−1 −1 ≡ (−1) 2 ≡ (−1)2k +1 ≡ −1( mod p). p This proves, −1 p = 1, if p ≡ 1( mod 4) −1, if p ≡ 3 ≡ −1( mod 4). The last theorem depicts that the congruence equation of the form x2 ≡ −1( mod p) has at least one solution if and only if p is of the form p = 4k + 1 for any integer k. Now we must dispose of an example supporting the fact. 208 Number Theory and its Applications Example 9.2.3. Let us consider the congruence x2 ≡ 50(mod 17). We have to examine the solvability of this equation. This can be accomplished by evaluating 50 −1 ( 50 17 ). Now 50 ≡ −1(mod 17) yields ( 17 ) = ( 17 ). But from Euler’s Criterion it 17−1 −1 2 follows, ( −1 ≡ 1(mod 17), which asserts ( 50 17 ) ≡ (−1) 17 ) = ( 17 ) = 1. This testify the solvability of x2 ≡ 50(mod 17). The following theorem dispense an important application regarding the number of odd primes. Theorem 9.2.5. For any positive integer k, There are inﬁnitely many primes of the form 4k + 1. Proof. In anticipation of a contradiction, assume that there are ﬁnitely many primes p1 , p2 , · · · pm and consider N = (2p1 p2 · · · pm )2 + 1. Here N is odd so there exists some odd prime p which divides N . Thus (2p1 p2 · · · pm )2 ≡ −1( mod p). Now an appeal to Theorem 9.2.4 renders that the Legendre symbol some positive integer k. This ( −1 p ) = 1 prevails if and only if p = 4k + 1 for conﬁrms that p is one of p1 , p2 , · · · pm . Hence pi [N − (2p1 p2 · · · pm )2 ] or pi 1 for any i = 1, 2, · · · m. This contradicts the fact that there are ﬁnitely many primes of the form 4k + 1. Hence the proof. p−1 Theorem 9.2.6. If p is an odd prime then r=1 ( pr ) = 0, where both the number of quadratic residue and non residue modulo p are exactly p−1 2 . Proof. Let a be a primitive root of p. Then by Theorem 8.2.4 the integers a1 , a2 , · · · ap−1 forms a reduced residue system modulo p, where φ(p) = p − 1. Let r be any integer between 1 and p − 1. Then for a unique positive integer k(1 ≤ k ≤ p − 1), r ≡ ak (mod p) holds. Now with the aid of Euler’s Criterion, we ﬁnd k p−1 p−1 a r = ≡ (ak ) 2 ≡ (a 2 )k ≡ (−1)k ( mod p). p p p−1 As a is a primitive root of p and gcd(a, p) = 1, then a 2 ≡ −1(mod p). But p−1 p−1 both ( pr ) and (−1)k are equal to either 1 or −1. Hence r=1 ( pr ) = k=1 (−1)k = 0. It follows both the number of quadratic and quadratic nonresidues are exactly p−1 2 . The proof of the last theorem generates an important fact, which we have in the form of following corollary. Corollary 9.2.2. Let p be an odd prime. Then the quadratic residues of p are congruent modulo p to the even powers of a primitive root a of p; the quadratic nonresidues are congruent to the odd powers of a. Theory of Quadratic Residues Proof. Follows from the fact rem. 209 ( pr ) = k ( ap ) ≡ (−1) (mod p) of the foregoing theok For an exempliﬁcation of the fact, set p = 7. Here 3 is a primitive root modulo 7. By the even powers of 3, the quadratic residues modulo 7 are given by 32 ≡ 2(mod 7), 34 ≡ 4(mod 7) and 36 ≡ 1(mod 7); by the odd powers of 3, the quadratic nonresidues modulo 7 are 31 ≡ 3(mod 7), 33 ≡ 6(mod 7) and 35 ≡ 5(mod 7). We are in a position to state and prove an elegant result due to Gauss. The result provides another criterion to ﬁnd out whether an integer a relatively prime to p is a quadratic residue modulo p. Theorem 9.2.7. (Gauss Lemma): Let p be an odd prime and r be an integer with gcd(r, p) = 1. If i denotes the number of integers among r, 2r, · · · ( p−1 2 )r where remainders exceed by p2 when divided by p, then ( pr ) = (−1)i . Proof. Here none of the integers among r, 2r, · · · ( p−1 2 )r are congruent to zero modulo p as gcd(r, p) = 1. Let r1 , r2 , · · · rj be the remainders, when divided by p and 0 < rk < p2 (k = 1, 2, · · · j). Also, s1 , s2 , · · · si are the remainders when divided by p and p > st > p2 (t = 1, 2, · · · i). Then i + j = ( p−1 2 ) and the integers, r1 , r2 , · · · rj , p − s1 , p − s2 , · · · p − si are all positive and less than p2 . Claim: p − st = rk . If not, let us choose p − st = rk for some t and k. Then ∃ u, v ∈ Z with 1 ≤ u, v ≤ p−1 2 satisfying st ≡ ur(mod p) and rk ≡ vr( mod p). Then, (u + v)r ≡ st + rk ≡ p ≡ 0(mod p). As gcd(r, p) = 1 so (u + v)r ≡ 0(mod p). But this is impossible as 1 < u + v ≤ p − 1. Now the integers r1 , r2 , · · · rj , p − s1 , p − s2 , · · · p − si are among 1, 2, · · · p−1 2 in some order. So their product gives, p−1 ! = r1 r2 · · · rj (p − s1 (p − s2 ) · · · (p − si ) 2 p−1 ! ≡ r1 r2 · · · rj (−s1 )(−s2 ) · · · (−si )( mod p) 2 p−1 ! ≡ (−1)i r1 r2 · · · rj s1 s2 · · · si ( mod p). 2 Also, the integers r1 , r2 , · · · rj , p − s1 , p − s2 , · · · p − si are congruent modulo p among r, 2r, · · · ( p−1 2 )r in some order. Hence p−1 p−1 i ! ≡ (−1) r(2r) · · · r ( mod p) 2 2 p−1 p−1 p−1 ! ≡ (−1)i r 2 !( mod p). 2 2 210 Number Theory and its Applications p−1 2 i As gcd(p, ( p−1 (mod p). Now by Euler’s Criterion, 2 )!) = 1 so 1 ≡ (−1) r p−1 r i 2 ≡ (−1) (mod p). This implies that, ( pr ) = (−1)i . (p) ≡ r Example 9.2.4. Let r = 4 and p = 7. To ﬁnd ( 47 ) compute the positive residues of 1 · 4, 2 · 4, 3 · 4. They are 4, 1, 5 modulo 7 respectively. Since exactly two of the integers are greater than 72 , then from Gauss Lemma we can say that ( 47 ) = (−1)2 = 1. This shows that 4 is a quadratic residue modulo 7. Now in later part of the present section, by Gauss Lemma, we have characterized all primes p that have 2 as a quadratic residue modulo p. p2 −1) 2 Theorem 9.2.8. If p is an odd prime, then = (−1)( 8 . Hence 2 is a p quadratic residue of all primes p = ±1(mod 8) and a quadratic non residue of all primes p = ±3(mod 8). 2 Proof. From Gauss Lemma we have = (−1)i where i is the number of p integers among 2 · 1, 2 · 2, 3 · 2, · · · ( p−1 2 ) · 2 which have remainders greater than p p p−1 2 when divided by p. Here the integers 2j < 2 for 1 ≤ i ≤ 2 if and only if p j < p4 . Hence there are [ p4 ] integers which are less than p2 . Thus i = p−1 2 − [4] integers are greater than p2 . So by Gauss Lemma we have, p−1 p 2 = (−1) 2 −[ 4 ] . p To prove this we are to show, p − 1 p (p2 − 1) − ≡ ( mod 2). 2 4 8 Consider congruence class modulo 8 as both sides of the above congruence depends only on the congruence class of p modulo 8. As p is an odd prime so p is of the forms 8j + 1, 8j + 3, 8j + 5, 8j + 7. Now, 1 = 4j − 2j = 2j if, p = 8j + 1 then i = 4j− 2j + 4 3 if, p = 8j + 3 then i = 4j + 1− 2j + = 4j + 1 − 2j = 2j + 1 4 5 = 4j + 2 − 2j − 1 = 2j + 1 if, p = 8j + 5 then i = 4j + 2− 2j + 4 7 if, p = 8j + 7 then i = 4j + 3− 2j + = 4j + 3 − 2j − 1 = 2j + 2. 4 Theory of Quadratic Residues 211 Thus, when p = 8j + 1 or p = 8j + 7 then i is even and ( p2 ) = 1; when p = 8j + 3 or p = 8j + 5 then i is odd and ( p2 ) = −1. So when p ≡ ±1(mod 8) then 2 is a quadratic residue of all primes; when p ≡ ±3(mod 8) then also 2 is a quadratic nonresidue of all primes. Now if p ≡ ±1(mod 8) then p = 8j ± 1 where j is an integer. So p2 − 1 (8j ± 1)2 − 1 = 8 8 64j 2 ± 16j = 8 = 8j 2 ± 2j. p2 −1 2 Thu, in this case p 8−1 is even and (−1) 8 = 1 = ( p2 ). Also, if p ≡ ±3(mod 8) then p = 8j ± 3 where j is an integer such that, p2 − 1 (8j ± 3)2 − 1 = 8 8 64j 2 ± 48j + 8 = 8 2 = 8j ± 6j + 1. Thus, in this case p2 −1 8 prime p we ﬁnd (−1) is odd and (−1) p2 −1 8 p2 −1 8 = −1 = ( p2 ). Finally, for any odd = ( p2 ). Example 9.2.5. To exemplify the last theorem, pick out p = 23 for which 232 −1 2 2 ( 23 ) = (−1) 8 = (−1)66 = 1 holds. Again, consider p = 11 such that ( 11 )= (−1) 112 −1 8 = (−1)15 = −1. Now, our next theorem directs another way of ﬁnding a primitive root. Because the proof of this theorem reﬂects some crucial applications of Legendre symbol, we include it in the present chapter instead of previous one. Theorem 9.2.9. Let p and 2p+1 be two odd primes. Then the integer (−1) is a primitive root of 2p + 1. p−1 2 ·2 Proof. We begin the proof by taking q = 2p + 1. As p is an odd prime, then p is of the form 1 + 4k or 3 + 4k for any positive integer k. Hence p−1 Case(i) p = 1 + 4k: Then p ≡ 1(mod 4) ⇒ (−1) 2 · 2 = 2. Since q is an odd prime and φ(q) = q − 1 = 2p imply the order of 2 modulo p is one of 1, 2, p, 2p. With the help of Euler’s Criterion and Legendre symbol together q−1 we have, ( 2q ) ≡ 2 2 = 2p (mod q). Now p = 1 + 4k ⇒ q = 8k + 3 ⇒ q ≡ 3( mod 8). In view of Theorem(9.2.8) we get, ( 2q ) = −1 which shows 2p ≡ −1( 212 Number Theory and its Applications mod q). This concludes that p is not the order of 2 modulo q. Because 22 ≡ 1(mod q) ⇒ q\|3, is clearly impossible, the order of 2 is neither 1 nor 2. So the order of 2 modulo q is 2p. This leads us to the fact that 2 becomes a primitive root modulo q = 2p + 1. p−1 Case(ii) p = 3 + 4k: Then p ≡ 3(mod 4) ⇒ (−1) 2 · 2 = −2. Again, p = 4k + 3 ⇒ q = 8k + 7 ⇒ q ≡ −1(mod 8). From Theorem 9.2.8 we have, −1 2 ( 2q ) = 1 ⇒ (−2)p ≡ −1(mod p), since (−2)p ≡ ( −2 q ) = ( q )( q )(mod q) −1 and ( q ) = −1(refer to Theorem 9.2.4). Now, applying similar arguments as case(i), we can say that −2 is a primitive root of q = 2p + 1. For further illustration of this theorem, if we consider p = 5 then we can examine that 2 is a primitive root of 2p + 1 = 11. Also for p = 11, −2 becomes the primitive root modulo 2p + 1 = 23. Before retiring from the ﬁeld, we should mention another application of Legendre symbol on primes. The theorem here as follows. Theorem 9.2.10. There are inﬁnitely many primes of the form 8k − 1. Proof. To the contrary, suppose there are ﬁnitely many primes p1 , p2 , · · · pn . Let T = (4p1 p2 · · · pn )2 − 2 and p be an odd prime divisor of T . This shows 4p1 p2 · · · pn is the solution of the equation (4p1 p2 · · · pn )2 ≡ 2(mod p). Thus ( p2 ) = 1 holds. Now, by Theorem 9.2.8 we have p ≡ ±1(mod 8). If p ≡ 1( mod 8) then p is of the form 8k + 1 which implies T of the form 16k + 2. This is impossible as T is of the form 16k − 2. Thus the prime divisor must be of the form p = 8k − 1. Combining p \|T and p \|(4p1 p2 · · · pn )2 yields p \|2, a contradiction. So the number of primes of the form 8k − 1 is inﬁnite. 9.3 Worked out Exercises Problem 9.3.1. For an odd prime p, prove that the quadratic residues modulo p are congruent modulo p to the integers 2 12 , 22 , 32 , · · · , ( p−1 2 ) 2 Solution 9.3.1. Let us take a = 12 , 22 , 32 , · · · , ( p−1 2 ) . p−1 p−1 . Now if we take b = 1, 2, 3, · · · , p−1 Then a 2 = 1p−1 , 2p−1 , · · · ( p−1 2 ) 2 , then gcd(b, p) = 1 as 1 ≤ b < p − 1 and p is a prime. By Fermat’s Theorem, we have p−1 bp−1 ≡ 1(mod p). This implies a 2 ≡ 1(mod p). Applying Euler’s Criterion, 2 we can say that 12 , 22 , 32 , · · · , ( p−1 2 ) are quadratic residue of p. Theory of Quadratic Residues 213 Next we are to show 1 , 2 , 3 , · · · incongruent modulo p. Let us p−1 2 2 b. Then we have (a+b)(a−b) ≡ choose a ≡ b (mod p), 1 ≤ a, b ≤ 2 and a = p−1 p−1 0(mod p). Now a + b ≤ 2 + 2 = p − 1, implies gcd(a + b, p) = 1. So we can divide both sides of the congruence by a + b. This shows that a − b ≡ 0(mod p). Thus we have a ≡ b(mod p), implies a = b. This is a contradiction.This shows 2 that 12 , 22 , 32 , · · · , ( p−1 2 ) are incongruent modulo p. Finally we have to show that any quadratic residue a modulo p must be 2 2 congruent to 12 , 22 , 32 , · · · , ( p−1 2 ) . Let x0 be the solution of x ≡ a(mod p) where 1 ≤ x0 ≤ p − 1. Here p − x0 is also a solution. Then one of x0 and p − x0 must be less than or equal to p − 1. Therefore one of x20 or (p − x0 )2 is 2 2 2 equal to 12 , 22 , 32 , · · · , ( p−1 2 ) . Since x0 ≡ a(mod p) then we have, (p − x0 ) = 2 p2 − 2px0 + x20 ≡ a(mod p). Thus a must be congruent to 12 , 22 , 32 , · · · , ( p−1 2 ) . 2 2 2 2 , ( p−1 2 ) Problem 9.3.2. Solve the following quadratic congruence: 3x2 + 9x + 7 ≡ 0( mod 13). Solution 9.3.2. To solve this, choose y = 2ax + b = 6x + 9 where a = 3 and b = 9. We transfer this equation in terms of y 2 ≡ d(mod 13), where d = b2 − 4ac = −3. This implies that y 2 ≡ −3 ≡ 10(mod 13). Thus y ≡ 6, 7( mod 13). Using y = 6x + 9 we see that, 6x + 9 ≡ 6( mod 13) or, 6x + 9 ≡ 7( mod 13) 6x ≡ −3( mod 13) 6x ≡ −2( mod 13) 6x ≡ 36( mod 13) 12x ≡ −4( mod 13) x ≡ 6( mod 13) x ≡ 4( mod 13). Problem 9.3.3. Prove that 3 is a quadratic residue of 23. Solution 9.3.3. Here we have, 3 23−1 2 = 311 = 9(27)3 . Then we can see that, 9(27)3 ≡ 9(4)3 ( mod 23) ≡ 9 · 64( mod 23) ≡ 9 · (−5)( mod 23) ≡ 1( mod 23). By Euler’s Criterion, we cay say that 3 is a quadratic residue of 23. Problem 9.3.4. Given that a is a quadratic residue of the odd prime p, prove that p − a is a quadratic residue or nonresidue of p according as p ≡ 1(mod 4) or p ≡ 3(mod 4). 214 Number Theory and its Applications k−1 ≡ −1( mod p) (9.3.1) k−1 Note that (a2 )2 = a2k ≡ 1(mod p) and φ(p) = p − 1 = 2k . If n be the order r of modulo p, then n must divide 2k . Let n = 2r with a2 ≡ 1(mod p) for some r < k. If r = k − 1, then (9.3.1) does not hold. This is a contradiction. If r r r+1 r < k − 1, then (a2 )2 = a2·2 = a2 ≡ 1(mod p) and continuing this way after 2r (k − 1) − r times we get, a ≡ 1(mod p) which again contradicts (9.3.1). This shows that n = 2k = φ(2k + 1), which implies a is a primitive root modulo 2k + 1. Problem 9.3.7. Find the value of the following Legendre symbol ( −72 131 ). 2 2 −3 ·2 Solution 9.3.7. Note that ( −72 131 ) = ( 131 ·2 2 −1 2 ) = ( 131 )( 131 ) [∵ ( ap ) = 1]. How- −1 ) = (−1) ever, by Euler’s Criterion we have ( 131 131−1 2 = −1. Since 131 = 16·8+3, Theory of Quadratic Residues 215 therefore by Theorem 9.2.8 we see that ( −72 131 ) = (−1)(−1) = 1. 2 ( 131 ) = −1. These two together yields 5 ) Problem 9.3.8. Use Gauss’s lemma to compute the Legendre symbol ( 19 p Solution 9.3.8. Note that p = 19. Then we can see that p−1 2 = 9 and 2 = 9.5, which further yields S = {5, 10, 15, 20, 25, 30, 35, 40, 45} ≡ {5, 10, 15, 1, 6, 11, 16, 2, 7}( mod 19). Thus the numbers 10, 15, 16, 11 are greater than 9.5. So by Gauss 5 ) = (−1)4 = 1. Lemma we have for n = 4, ( 19 Problem 9.3.9. Prove that 2 is not a primitive root of any prime of the form p = 3 · 2n + 1, except when p = 13. Solution 9.3.9. To verify 2 is not a primitive root, its suﬃces to show that 2 is a quadratic residue of p. Now p − 1 = 3 · 2n and by Theorem 9.2.8 we have, p − 1 ≡ 0(mod 8) for n ≥ 3. This shows that ( p2 ) = 1 and 2 is quadratic residue of p.Hence 2 is not a primitive root of p. Consider n = 1. Then p = 3 · 2 + 1 = 7 implies p ≡ 7(mod 8). Moreover, by Theorem 9.2.8 we have ( p2 ) = 1. This shows that 2 is not a primitive root. If we consider n = 2, then p = 13 implies p ≡ 5(mod 8). Again by virtue of Theorem 9.2.8, we have ( p2 ) = −1. So in this case, 2 is a quadratic non residue and hence a primitive root of 13 also. Problem 9.3.10. For an odd prime p, show that there are quadratic nonresidues of p that are not primitive roots of p. p−1 2 − φ(p − 1) Solution 9.3.10. In view of Theorem 9.2.6, we can say that there exist p−1 2 number of quadratic residues and non residues of p. Let a be a quadratic residue of p. Then it can not be a primitive root because p−1 a 2 ≡ 1(mod p)[by Corollary 9.2.1]. But φ(p) = p − 1 implies a is not a primitive root. Thus if r is a primitive root of p, then it must be congruent to quadratic non residue of p. Let S be the set of quadratic non residue of p. Then S contains p−1 2 elements. From Corollary 8.2.3, there are φ(p − 1) numbers of elements of S which are primitive roots of p. Thus [ p−1 2 − φ(p − 1)] elements of S are not primitive roots of p. Problem 9.3.11. If p is an odd prime, prove that p−2 a(a + 1) a=1 p = −1. Solution 9.3.11. Let us choose an integer a such that gcd(a, p) = 1. Then from linear congruence we can say that there exists an integer a satisfying aa ≡ 1( mod p) for 1 ≤ a ≤ p − 2. 216 Number Theory and its Applications Here both a and a runs from 1 to p − 2. However if a = p − 1, then a(p − 1) ≡ 1(mod p). This implies −a ≡ 1(mod p) ⇒ a + 1 ≡ 0(mod p), which shows that p − 1 = a, a contradiction. Let a1 a ≡ 1(mod p) and a2 a ≡ 1(mod p). Then a1 a ≡ a2 a (mod p) implies a1 = a2 . This shows that if a runs from 1 to p − 2, then each a from 1 to p − 2 is represented only once. As a runs from 1 to p − 2, so 1 + a runs from 2 to p − 1. ∴ aa ≡ 1( mod p) ⇒ a + aa ≡ a + 1( mod p) ⇒ a(1 + a ) ≡ a + 1( mod p) ⇒ a2 (a + 1) ≡ a(a + 1)( mod p). Thus we have, a(a+1) p 2 = a (ap +1) = 1+a . Now we can see that, p p−2 a(a + 1) a=1 p = p−2 1 + a p 1+a =2 = p−1 a a =2 By Theorem 9.2.6we have, −1. p−1 a a=1 p p p−1 = a p a=2 = p−1 a 1 − . p p a=1 = 0 and 1 p = 1. This proves Problem 9.3.12. If p ≡ 7(mod 8), then prove that p\|2 that 2n − 1 is composite for n = 239 p−1 2 p−2 a(a+1) a=1 p − 1. Hence show p−1 Solution 9.3.12. From Euler’s Criterion we have, ( p2 ) ≡ 2 2 (mod p). Moreover, since p ≡ 7(mod 8) therefore by Theorem 9.2.8 we have ( p2 ) = 1. Thus p−1 p−1 together we say that 2 2 − 1 ≡ 0(mod p). Hence p 2 2 − 1 . For the remaining part, to verify 2239 − 1 is composite, we need to express for some p of the form 7 + 8k(k ∈ Z). Hence 239 = p−1 239 as p−1 2 2 ⇒ p = 503 = 7 + 62 · 8. Problem 9.3.13. If the prime p > 3, prove that p divides the sum of its quadratic residues. = Theory of Quadratic Residues 217 Solution 9.3.13. Let r be a primitive root modulo p and a1 , a2 , · · · , a p−1 be the 2 quadratic residue of p. By Corollary 9.2.2, r2 , r4 , · · · rp−1 are congruent to the quadratic residue of p. Therefore a1 + a2 + · · · + a p−1 ≡ r2 + r4 + · · · rp−1 ( mod p) 2 ≡ r2 (1 + r2 + r4 + · · · rp−3 )( mod p) [∵ p > 3]. Since r is a primitive root, so rp−1 ≡ 1(mod p). Therefore, a1 + a2 + · · · + a p−1 ≡ r2 + r4 + · · · rp−3 + rp−1 ( mod p) 2 ⇒ a1 + a2 + · · · + a p−1 ≡ (1 + r2 + r4 + · · · rp−3 )( mod p). 2 Now from the above two congruence equation for p > 3 we have, r2 (1 + r2 + r4 + · · · rp−3 ) ≡ (1 + r2 + r4 + · · · rp−3 )( mod p). Here p must divide (1 + r2 + r4 + · · · rp−3 ), otherwise r2 ≡ 1(mod p) which contradicts the fact that r is primitive root as p > 3. Thus we have, p\|(a1 + a2 + · · · + a p−1 ). 2 Problem 9.3.14. Show that the odd prime divisor p of the integers 9n + 1 are of the form p ≡ 1(mod 4). Solution 9.3.14. Here p is an odd prime divisor of 9n + 1. Then 9n + 1 ≡ 0( mod p) or (3n )2 ≡ −1(mod p). Since p is odd, therefore either p ≡ 1(mod 4) or p ≡ 3(mod 4). Now by Theorem 9.2.4, if p ≡ 3(mod 4) we can say that there does not exists any solution of x2 ≡ −1(mod p). So there is no solution of (3n )2 ≡ −1(mod p). Thus p is the divisor of 9n + 1 if it must be of the form p ≡ 1(mod 4). 9.4 Here in this section we have discussed another important of modular arithmetic which gives us the path to solve quadratic congruences modulo prime numbers. p If we choose two odd primes p and q then both the Legendre symbols q q and are deﬁned. A fairly question presents itself: Is it possible to ﬁnd p any one of them with the helpof other. The answer is contained in the notion p q p−1 q−1 ‘Quadratic Reciprocity Law’: · = (−1) · . Also, this leads q p 2 2 2 2 to the fact that if x ≡ p(mod q) is solvable then x ≡ q(mod p) is also so and 218 Number Theory and its Applications vice versa. Now before going to the proof of above quadratic reciprocity law we have to prove a signiﬁcant lemma ﬁrst. This lemma further leads us to prove this important law of modular arithmetic. Lemma p is an odd prime and r is an odd integer with gcd(r, p) = 1 9.4.1. If p−1 r 2 then = (−1) t=1 [tr/p] . p p − 1 Proof. To begin with, consider the integers r, 2r, · · · r among which some 2 are greater than p2 and the rest are less than p2 when divided by p. Let r1 , r2 , · · · rn be the remainders less than p2 and s1 , s2 , · · · sm be the remainders greater than p − 1 p r by p we get, tr = qt p + wt where 2 . Dividing every element of r, 2r, · · · 2 wt tr 1 ≤ wt ≤ p − 1. Then p = qt + p ⇒ qt = [tr/p] for 1 ≤ t ≤ p−1 2 . Hence tr = [tr/p]p + wt . If the remainder wt < p2 , it is one of r1 , r2 , · · · rn and if wt > p2 then it is one of s1 , s2 , · · · sm . Adding both sides of the last equation we have, p−1 2 p−1 tr = t=1 2 [tr/p] · p + t=1 n ri + i=1 m sj (9.4.1) j=1 Taking the proof of Gauss Lemma into consideration, the integers r1 , r2 , · · · rn , p− s1 , p − s2 , · · · p − sm are among 1, 2, · · · p−1 2 in some order. Therefore p−1 2 t= t=1 n ri + i=1 m (p − sj ) = pm + j=1 n ri − i=1 m sj (9.4.2) j=1 Subtracting (9.4.2) from (9.4.1)we get, p−1 (r − 1) 2 t=1 p−1 t = p( 2 [tr/p] − m) + 2 t=1 m sj (9.4.3) j=1 As p and r are both odd, the application of p ≡ r ≡ 1(mod 2) on (9.4.3) tells us, p−1 2 t=1 p−1 t≡1·( 2 t=1 p−1 [tr/p] − m)( mod 2) ⇒ m ≡ 2 [tr/p]( mod 2). (9.4.4) t=1 Finally, Gauss Lemma leads us to conclude that ( pr ) = (−1)m = (−1) p−1 2 t=1 [tr/p] . Theory of Quadratic Residues 219 The above lemma is not only a tool to prove quadratic reciprocity law but 5 ), also used to evaluate Legendre symbol. For instance, if we wish to prove ( 11 11−1 2 by above lemma we only need to evaluate t=1 [t · 5/11]. Here, 5 [t·5/11] = [5/11]+[10/11]+[15/11]+[20/11]+[25/11] = 0+0+1+1+2 = 4. t=1 5 Hence ( 11 ) = (−1)4 = 1. Now, the actual details of the Gauss Quadratic Reciprocity Law appear below. Theorem 9.4.1. If p and q are distinct odd primes, then ( pq )( pq ) = (−1)( p−1 q−1 2 )( 2 ) Proof. To start with, consider a pair of integers (x, y) with 1 ≤ x ≤ p−1 2 and p−1 q−1 . Now we divide the total number of pair ( )( ) into two 1 ≤ y ≤ q−1 2 2 2 groups, which depends on the relative sizes of qx and py. Geometrically it can be interpreted by a diagonal of a rectangle from (0, 0) to ( p2 , 2q ) as shown in the ﬁgure below: Figure 9.1: Here the equation of the diagonal is y = ( pq )x, with slope pq . First, we wish to ensure that the pair do not coincide with the diagonal. If so, then py = qx ⇒ q\|py. Then either q\|p or q\|y holds. Since gcd(p, q) = 1 so q p. Also 1 ≤ y ≤ q−1 2 implies q y. Thus, py = qx is not possible. . 220 Number Theory and its Applications 1 qx Let qx > py . Then we have, 1 ≤ x ≤ p− 2 and 1 ≤ y ≤ p . So for each x lying p−1 between 1 and 2 , there are [qx/p] number of integers satisfying 1 ≤ y ≤ qx p . p−1 2 [qi/p]. So the total number of such pair are i=1 q−1 Again, if qx < py then 1 ≤ y ≤ 2 and 1 ≤ x ≤ py q . Proceeding as above, q−1 2 the total number of such pair are j=1 [pj/q]. As the total number of coordinates p−1 p−1 q−1 p−1 q−1 2 2 = in x and y are p−1 i=1 [qi/p] + i=1 [qi/p]. 2 · 2 , therefore 2 · 2 Appealing to Lemma 9.4.1 we get, q−1 p−1 2 [pj/q] q p 2 = (−1) j=1 × (−1) i=1 [qi/p] q p q−1 p−1 2 [pj/q]+ 2 [qi/p] j=1 i=1 = (−1) = (−1)( p−1 q−1 2 )·( 2 ) . This proves the law of quadratic reciprocity. Corollary 9.4.1. If p and q are distinct odd primes, then 1, if p ≡ 1( mod 4) or q ≡ 1( mod 4); q p = q p −1, if p ≡ q ≡ 3( mod 4). p−1 q−1 q−1 Proof. As ( pq )( pq ) = (−1)( 2 )·( 2 ) , then the number ( p−1 2 ) · ( 2 ) is even if and only if at least one of them is of the form 4j + 1 for some integer j. This shows q−1 that p ≡ 1(mod 4) or q ≡ 1(mod 4). Also ( p−1 2 ) · ( 2 ) is odd if and only if both p and q are of the form 4j + 3 for some integer j, which further implies p ≡ q ≡ 3(mod 4). Our task is complete. The above corollary gives birth to the following obvious corollary. Corollary 9.4.2. If p and q are distinct odd primes, then q ( p ), if p ≡ 1( mod 4) or q ≡ 1( mod 4); p = q q −( p ), if p ≡ q ≡ 3( mod 4). 5 For example, if we consider the Legendre symbols ( 11 ) and ( 11 5 ) then Theorem (9.4.1) yields, 5−1 11−1 5 11 = (−1) 2 · 2 11 5 = (−1)10 = 1 5 Thus, according to the last corollary we have ( 11 5 ) = ( 11 ) = 1. Finally, we conclude this section with the following theorem. Theory of Quadratic Residues 221 Theorem 9.4.2. If p = 3 is an odd prime, then 1, if p ≡ ±1( mod 12) 3 = p −1, if p ≡ ±5( mod 12). Proof. Appealing to Corollary(9.4.2) we get, p ( 3 ), 3 = p −( p3 ), if p ≡ 1( mod 4) if p ≡ −1( mod 4). Set p ≡ 1(mod 3). Recalling Theorem 9.2.3 we get ( p3 ) = ( 13 ) = 1. Also, if p ≡ −1( 3−1 −1 2 ( mod 3) then by similar reasoning we have, ( p3 ) = ( −1 3 ) and ( 3 ) ≡ (−1) p mod 3). Combining we get ( 3 ) = −1. Thus, 1, if p ≡ 1( mod 3) p = 3 −1, if p ≡ −1( mod 3). Combining p ≡ 1(mod 4) and p ≡ 1(mod 3) yield p ≡ 1(mod 12)(Why!). Also, p ≡ −1(mod 4) and p ≡ −1(mod 3) together gives p ≡ −1(mod 12). Suppose p ≡ 1(mod 4) and p ≡ −1(mod 3). Then for z, z ∈ Z we have p − 1 = 4z, p + 1 = 3z . On solving, we get p + 7 = 12(z − z) which implies p ≡ −7 ≡ 5(mod 12). Finally, for p ≡ −1(mod 4) and p ≡ 1(mod 3) there exist integers t, t such that p + 1 = 4t and p − 1 = 3t . Proceeding as above, we get p ≡ 7 ≡ −5(mod 12). This leads us to, 1, if p ≡ ±1( mod 12) 3 = p −1, if p ≡ ±5( mod 12). 9.5 Worked out Exercises Problem 9.5.1. Evaluate the following Legendre symbol: ( 1234 4567 ). Solution 9.5.1. Here 4567 ≡ 3(mod 4) and 1234 = 2 · 617. Also 617 ≡ 1( 222 Number Theory and its Applications mod 4). An appeal to Theorem 9.2.8 and Corollary 9.4.2) yields 2 617 4567 1234 = = [∵, 4567 ≡ 7( mod 8)] 4567 4567 4567 617 248 2 31 = = [∵, 248 = 23 · 31] 617 617 617 617 28 31 = = = 617 31 31 7 31 3 7 = =− =− = 31 7 7 3 1 = = 1. 3 Problem 9.5.2. Show that if p is an odd prime, then 1, if p ≡ 1( mod 8) or p ≡ 3( mod 8) −2 = p −1, if p ≡ 5( mod 8) or p ≡ 7( mod 8). −1 2 Solution 9.5.2. Note that ( −2 p ) = ( p )( p ). By Theorem 9.2.4, we can say that 2 ( −1 p ) = 1 if p ≡ 1(mod 4) and by Theorem 9.2.8 we have ( p ) = 1 if p ≡ ±1( −1 mod 8). By similar arguments, ( p ) = −1 if p ≡ 3(mod 4) and ( p2 ) = −1 if p ≡ ±3(mod 8). However if p ≡ 1(mod 8), then p ≡ 1(mod 4) which gives −2 ( −2 p ) = 1. Furthermore if p ≡ 3(mod 8), then p ≡ 3(mod 4). Thus ( p ) = (−1) · (−1) = 1. Moreover if p ≡ −3 ≡ 5(mod 8), then p ≡ 5 ≡ 1(mod 4), which shows ( −2 p ) = (−1) · 1 = −1. Also if p ≡ 7(mod 8), then p ≡ 7 ≡ 3( mod 4) shows ( −2 p ) = (1) · (−1) = −1. Combining all those results we obtain, 1, if p ≡ 1( mod 8) or p ≡ 3( mod 8) −2 = p −1, if p ≡ 5( mod 8) or p ≡ 7( mod 8) Problem 9.5.3. Determine whether the following quadratic congruence is solvable or not: 3x2 + 6x + 5 ≡ 0(mod 89). Solution 9.5.3. Let us consider the equation 3x2 + 6x + 5 ≡ 0(mod 89). So 3x2 + 6x + 5 ≡ 0( mod 89) ⇒ (6x + 6)2 + 24 ≡ 0( mod 89) ⇒ (6x + 6)2 ≡ 65( mod 89). Let y = 6x + 6. Then y 2 ≡ 65(mod 89). As 13 ≡ 1(mod 4), then by Corollary 9.4.2) we can write, 89 11 13 2 13 = = = = = −1[Since 11 ≡ 3( mod 8)] 89 13 13 11 11 Theory of Quadratic Residues 223 Moreover 5 ≡ 1(mod 4). Then by Corollary 9.4.2 we can say that, 2 5 89 4 2 = = = =1 89 5 5 5 Together we have, 65 89 = −1. This shows that the given equation is not solvable. Problem 9.5.4. Verify that if p > 3 is an odd prime, then 1, if p ≡ 1( mod 6) −3 = p −1, if p ≡ 5( mod 6) −1 3 Solution 9.5.4. Let us take ( −3 p ) = ( p )( p ). To prove the assertion we are to consider two cases: Case(i): Let p ≡ 1(mod 6). Then p − 1 = 6t for some integer t(> 0). 1. If t is even, then t = 2t for some integer t (> 0). This shows that p − p−1 2 = (−1)6t = 1 1 = 12t or p ≡ 1(mod 12). Therefore ( −1 p ) = (−1) and ( p3 ) = 1[by Theorem 9.4.2]. Therefore ( −3 p ) = 1. 2. If t is odd, then t = 2t + 1. This shows that p − 1 = 6(2t + 1) or p−1 2 p ≡ 7 ≡ −5(mod 12). Therefore ( −1 = (−1)3+6t = −1. p ) = (−1) Then by Theorem 9.4.2 we have ( −3 p ) = 1. Hence p ≡ 1(mod 6) ⇒ −3 ( p ) = 1. Case(ii): Let p ≡ 5(mod 6). Then p − 5 = 6k for some integer k(> 0). 1. If k is even, then t = 2k for some integer k (> 0). This shows p−1 2 = that p − k = 12k or p ≡ 5(mod 12). Therefore ( −1 p ) = (−1) 3 −3 2+6k = 1 and ( p ) = −1[by Theorem 9.4.2]. Therefore ( p ) = (−1) −1. 2. If k is odd, then t = 2k + 1. This shows that p − 5 = 6(2k + 1) p−1 2 = (−1)5+6t = or p ≡ −1(mod 12). Therefore ( −1 p ) = (−1) −1. Then by Theorem 9.4.2 we have ( −3 p ) = −1. Therefore p ≡ 5( −3 mod 6) ⇒ ( p ) = −1. Problem 9.5.5. Prove that there exist inﬁnitely many primes of the form 8k+3. Solution 9.5.5. To the contrary, let us assume that there exists ﬁnitely many primes of the form 8k + 3 say, p1 , p2 , · · · , pn . Consider M = (p1 p2 · · · pn )2 + 2. pi . For if p = pi , Then M is odd and it has an odd prime divisor p such that p = then p\|M and p\|(p1 p2 · · · pn )2 together imply p\|2. Now M ≡ 0(mod p) imply (p1 p2 · · · pn )2 ≡ −2(mod p). So either p ≡ 1( mod 8) or p ≡ 3(mod 8)[refer to Problem 9.5.2]. Let M = q1k1 q2k2 · · · qsks and 224 Number Theory and its Applications qi ≡ 1(mod 8) for all i. But pi ≡ 3(mod 8) imply p2i ≡ 9 ≡ 1(mod 8). This shows that (p1 p2 · · · pn )2 + 2 ≡ 3(mod 8) i.e. M ≡ 3(mod 8), a contradiction to the fact that M = q1k1 q2k2 · · · qsks ≡ 1(mod 8). Therefore all qi ’s can’t be of the form qi ≡ 1(mod 8). So there must be some odd prime divisor qi = p of M such that p ≡ 3(mod 8). This contradicts the fact that pi ’s are ﬁnite. Problem 9.5.6. Show that ( p5 ) = 1 if and only if p ≡ 1, 9, 11, 19(mod 20). Solution 9.5.6. By deﬁnition of Legendre symbol, we have p is an odd prime. If p ≡ 1, 9, 11, 19(mod 20), then we have p ≡ 1, 9, 11, 19(mod 5). This shows that p ≡ 1, 4(mod 5). As 5 ≡ 1(mod 4), by Corollary 9.4.2 we ﬁnd ( p5 ) = ( p5 ) = ( 15 ) or( 45 ). In any case ( p5 ) = 1. For the converse part, let ( p5 ) = 1 holds. Since 5 ≡ 1(mod 4), therefore 5−1 ( p5 ) = ( p5 ) = 1. So by Euler’s Criterion, p 2 ≡ 1(mod 5) yields p ≡ 1, 4( mod 5). Furthermore, for an odd prime we have p ≡ 1, 3(mod 4). Those above congruences imply that 4p ≡ 4(mod 20) or 4p ≡ 16(mod 20) and also 5p ≡ 5( mod 20) or 5p ≡ 15(mod 20). Subtracting them we get, p ≡ −1or 11(mod 20). This follows that p ≡ 1, 9, 11, 19(mod 20). Problem 9.5.7. If p and q are odd primes satisfying p = q +4a for some integer a, prove that ( ap ) = ( aq ). Solution 9.5.7. Since p = q + 4a, therefore ( pq ) = ( q+4a q ). But q + 4a ≡ 4a( p 4a a mod q). Thus ( q ) = ( q ) = ( q ). Similarly, q = p − 4a implies ( pq ) = ( −a p ) = −1 a ( p )( p ). As p is odd, two cases arise: q a Case(i): If p ≡ 1(mod 4), then Theorem 9.2.4 yields ( −1 p ) = 1. So ( p ) = ( p ). By Corollary 9.4.2, we ﬁnd ( pq ) = ( pq ) as p ≡ 1(mod 4). Then applying all above notions we get, ( ap ) = ( aq ). Case(ii): If p ≡ 3(mod 4), then by Theorem 9.2.4 we have ( −1 p ) = −1. Thus q a we get ( p ) = −( p ). As p = q + 4a, then p ≡ q(mod 4) implies q ≡ 3(mod 4). Moreover, by Corollary 9.4.2 we have ( pq ) = −( pq ). In this case we can say that ( aq ) = −( pq ) = ( ap ). Problem 9.5.8. Find a prime number p that is simultaneously expressible in the forms x2 + y 2 , u2 + 2v 2 and r2 + 3s2 . Solution 9.5.8. Let us take x2 + y 2 = p. Then u2 v2 2 ≡ −2(mod p) and rs2 ≡ −3(mod p). −2 −3 the condition ( −1 p ) = ( p ) = ( p ) = 1. x2 y2 ≡ −1(mod p). Similarly To ﬁnd the value of p we need to apply Theory of Quadratic Residues 225 −2 Therefore for ( −1 p ) = 1, we get p ≡ 1(mod 4). Similarly, we get ( p ) = 1 provided p ≡ 1(mod 8) or p ≡ 3(mod 8). However, if p ≡ 1(mod 6) then ( −3 p ) = 1. If p ≡ 1(mod 24), then p ≡ 1(mod 4), p ≡ 1(mod 8) and p ≡ 1(mod 6) holds together. Let us consider p = 1 + 24k for k = 1, 2, 3 · · · . Then p = 25, 49, 73, · · · . In particular, 73 is a prime and it can be written as 82 + 32 = 73 = 12 + 2(6)2 = 52 + 3(4)2 . Problem 9.5.9. Show that the prime divisors p(= 3) of the integer n2 − n + 1 are of the form 6k + 1. Solution 9.5.9. Note that n2 − n + 1 is odd for all n ≥ 1(Verify!). If p is a prime divisor of n2 − n + 1, then p > 3 as n2 − n + 1 = 2 and p = 3. If p\|n2 − n + 1, then p also divides (2n − 1)2 + 3. Therefore (2n − 1)2 ≡ −3(mod p) follows that ( −3 p ) = 1. This shows that p ≡ 0, 1, 2, 3, 4, 5(mod 6). If p = 0, 2, 4 ≡ ( mod 6), then we have p ≡ 0, 2, 4(mod 2), which is not possible as p > 3. If p ≡ 3( mod 6), then p ≡ 3(mod 6) implies 3\|p, a contradiction as p > 3. Thus for p ≡ 5( mod 6) we have ( −3 p ) = −1 [by Problem 9.5.4]. This shows that p ≡ 1(mod 6) holds and p is of the form 1 + 6k for some integer k. 9.6 The Jacobi Symbol In this section, we conduct a study on another important symbol of number theory known as Jacobi symbol. This is the generalization of Legendre symbol studied in previous section. The symbol acts as an utilitarian in evaluating Legendre symbol. Deﬁnition 9.6.1. Let n be a positive integer and n = pn1 1 pn2 2 · · · pnmm be its prime factorization where each pi ’s are distinct. Then the Jacobi symbol nr is n 1 r n2 nm deﬁned by nr = pn1 pn2r···pnm = pr1 · · · prn , r being any positive p2 m 1 2 r integer relatively prime to n. Here the symbols pi for each i = 1, 2, · · · m are known to be Legendre symbol. 2 2 2 2 2 To illustrate, let us choose a = 2, b = 147. Then 147 = 72 ·3 = 7 3 = 2 (−1)(−1) = −1. This demonstrates, in particular, Jacobi symbol is Legendre symbol if n is prime. However for any composite number n, the value of nr fails to assure the solvability of the congruence x2 ≡ r(mod p). Since x2 ≡ 3(mod 5) and x2 ≡ (mod 7) have no solutions, therefore the equation x2 ≡ 3(mod 5) has no 3 3 3 = 5 7 = (−1)(−1) = 1. Thus for solution. Although we can see that 35 226 Number Theory and its Applications any composite integer n and any prime divisor p of n, x2 ≡ r(mod n) has a solution if x2 ≡ r(mod p) is solvable. Jacobi symbol also relish certain interchangeable properties to those of the Legendre symbol. Theorem 9.6.1. Let n be a composite number and r, s be integers such that gcd(r, n) = 1 and gcd(s, n) = 1 respectively. Then the following properties are true: 1. If r ≡ s(mod n) then ( nr ) = ( ns ). s rs r = . 2. n n n n−1 −1 = (−1) 2 . 3. n n2 −1 2 = (−1) 8 4. n Proof. To prove these four assertions, consider n = pn1 1 pn2 2 · · · pnmm to be its prime factorization. 1. Let p be the prime divisor of n. Then r ≡ s(mod p) holds. Taking Theorem 9.2.3 into consideration, we obtain pr = ps . This implies, n 1 n 2 nm r r r r = ··· n p1 p2 pn n 1 n 2 nm s s s = ··· p1 p2 pn s . = n r s 2. Again, by Theorem 9.2.3 we ﬁnd rs pi = pi pi for each i. Thus we can deduce that, nm n1 n2 rs rs rs rs = ··· n p1 p2 pn n1 n2 n m n 1 n 2 nm r s s r r s = ··· ··· p1 p2 pn p1 p2 pn s r . = n n r ≡ r p−1 2 (mod p) for any −1 p−1 integer r with gcd(r, p) = 1, p being any odd prime. Then p = (−1) 2 . 3. With the help of Euler’s Criterion we ﬁnd p Theory of Quadratic Residues 227 Consequently, −1 n −1 = p1 n1 = (−1) −1 p2 n2 ··· −1 pn n m n1 (p1 −1) n (p −1) n (p −1) + 2 22 +···+ m 2m 2 . Now, n can be expressed as n = pn1 1 pn2 2 · · · pnmm = {1 + (p1 − 1)}n1 {1 + (p2 − 1)}n2 · · · {1 + (pm − 1)}nm . Here for each i, the term pi − 1 is even. Then we have {1 + (pi − 1)}ni ≡ 1 + ni (pi − 1)(mod 4). Also, {1 + ni (pi − 1)}{1 + nj (pj − 1)} ≡ 1 + ni (pi − 1) + nj (pj − 1)(mod 4) because ni nj (pi − 1)(pj − 1) is multiple of 4. Hence n ≡ 1 + n1 (p1 − 1) + n2 (p2 − 1) + · · · + nm (pm − 1)( mod 4) n−1 n1 (p1 − 1) n2 (p2 − 1) nm (pm − 1) ≡ + + ··· + ( mod 2) 2 2 2 2 This shows that ( −1 n ) = (−1) n−1 2 . p2 −1 4. In light of Theorem 9.2.8, for any odd prime p if ( p2 ) = (−1) 2 holds then 2 becomes the quadratic residue provided p ≡ ±1(mod 8). Thus, we ﬁnd n m n1 n 2 2 2 2 2 = ··· n p1 p2 pm = (−1) n1 (p2 −1) n2 (p2 −1) n (p2 −1) 1 2 + +···+ m 8m 8 8 . Again, n can be expressed as n2 = {1 + (p21 − 1)}n1 {1 + (p22 − 1)}n2 · · · {1 + (p2m − 1)}nm . Since pi ≡ ±1(mod 8) then p2i ≡ 1(mod 8) or p2i − 1 ≡ 0( mod 8). Thus {1 + (p2i − 1)}ni ≡ 1 + ni (p2i − 1)(mod 64) and {1 + ni (p2i − 1)}{1 + nj (p2j − 1)} ≡ 1 + ni (p2i − 1) + nj (p2j − 1)(mod 64). Hence n2 ≡ 1 + n1 (p21 − 1) + n2 (p22 − 1) + · · · + nm (p2m − 1)(mod 64), which further 2 n (p2 −1) n (p2 −1) n (p2 −1) + 2 82 + · · · + m 8m (mod 8). This yields implies n 8−1 ≡ 1 81 ( n2 ) = (−1) n2 −1 8 . Now the following theorem highlights the fact that the reciprocity law agrees for both Jacobi symbol and the Legendre symbol. Theorem 9.6.2. Let m and n relatively prime odd positive integers. Then n−1 n ( m−1 2 )( 2 ) (m n )( m ) = (−1) 228 Number Theory and its Applications · · · qsms be prime factorizations Proof. Let n = and m = of n and m respectively. Then, n r m i m = n pi i=1 r s qj ni mj = and pi i=1 j=1 mj s n n = m qj j=1 n m r s pi i j = . qj i=1 j=1 pn1 1 pn2 2 · · · pnr r q1m1 q2m2 r s m n pi qj ni mj = . From the quadratic reciprocity n m q j pi i=1 j=1 qj −1 pi qj pi −1 = (−1)( 2 )( 2 ) . Hence law, we achieve q j pi Thus we get, m n n m = r s (−1)( m (q −1) ni (pi −1) )( j 2j ) 2 i=1 j=1 s r ni (pi −1) mj (qj −1) ( )( ) 2 2 = (−1) j=1 i=1 r ni (pi −1) s mj (qj −1) ( ) ( ) 2 2 j=1 = (−1) i=1 = (−1)( n−1 m−1 2 )( 2 ) . This completes the proof. Our future eﬀort is develop an elegant algorithm for evaluating Jacobi symbol. Let m and n be two relatively prime positive integers with m > n. Assume J0 = m and J1 = n. Applying division algorithm on J0 and J1 we have, J0 = J1 · q1 + 2t1 J2 where q1 is quotient and 2t1 J2 is the remainder. Here, 2t1 J2 is constructed in such a way that J2 is an odd positive integer less than J1 ; and t1 be a non negative integer. Now the repeated use of division algorithm gives, J1 = J2 · q2 + 2t2 J3 J2 = J3 · q3 + 2t3 J4 . = .. Jn−3 = Jn−2 · qn−2 + 2tn−2 Jn−1 Jn−2 = Jn−1 · qn−1 + 2tn−1 · 1, Theory of Quadratic Residues 229 where each tj is non-negative integer and Ji is an odd positive integer less than Ji−1 for i = 2, 3, · · · n − 1. Here the sequence of equations get discontinued, where Jk = 1 for some positive integer k as gcd(m, n) = 1. To illustrate the algorithm, let us take m = J0 = 225 and n = J1 = 29. Then, 225 = 29 · 7 + 21 · 11 29 = 11 · 2 + 20 · 7 11 = 7 · 1 + 22 · 1 Formatting all our discussions in the following theorem, which demonstrates, how this algorithm evaluates Jacobi symbols. Theorem 9.6.3. Let m and n be positive integers with m > n. Then 2 tn−1 (J −1) t1 (J 2 −1) t2 (J 2 −1) J −1 J −1 J −1 J −1 n−1 m 1 2 +( 12 )( 22 )+···+( n−2 )( n−1 ) 8 2 2 ) = (−1) 8 + 8 +···+ , n where Ji and ti are the integers for i = 1, 2, · · · n − 1 as described above. ( Proof. By virtue of Theorem 9.6.1, we see that t1 t1 t1 (J 2 −1) 2 J2 2 J2 J2 J0 m 1 = = = (−1) 8 . = n J1 J1 J1 J1 J1 Again in the light of Theorem 9.6.2, we see that J2 J1 −1 J2 −1 J 1 2 = (−1) 2 . J1 J2 Combining them, gives t1 (J 2 −1) J m J1 −1 J2 −1 1 1 + 2 8 = (−1) 2 . n J2 Continuing with these steps we get, ti (J 2 −1) Ji−1 Ji −1 Ji+1 −1 Ji i + 2 8 = (−1) 2 , Ji Ji+1 for i = 2, 3, · · · n − 1. Finally combining all those inequalities we obtain, J −1 Jn−1 −1 tn−1 (J 2 −1) J −1 J −1 t1 (J 2 −1) t2 (J 2 −1) m n−1 2 1 2 + 12 +···+ n−2 8 2 2 2 = (−1) 8 + 8 +···+ . n Example 9.6.1. Let m = 225 and n = 29. Then we get, 2(292 −1) 0(112 −1) 2(72 −1) 29−1 11−1 11−1 9−1 225 = (−1) 8 + 8 + 8 +( 2 )( 2 )+( 2 )( 2 ) , 29 = (−1)210+12+70+20 = (−1)312 = 1. 230 Number Theory and its Applications Our previous discussion tells us about the solvability of quadratic congruence modulo odd prime p. Consider two composite numbers 441 and 1764, then factorize them as 441 = 32 × 72 , 1764 = 22 × 32 × 72 . Thus an odd composite number can be represented as product of distinct odd primes and if the composite number is even, then in the factorization of it some powers of 2 occurs along with the product of distinct odd primes. If we consider an odd composite number mn k m1 m2 mn 1 m2 m = pm 1 p2 · · · pn and an even composite number n = 2 p1 p2 · · · pn , then k i for it is necessary to ﬁnd solvability of quadratic congruences modulo 2 and pm i 2 each prime pi and positive integers k, mi . Solving the equations x ≡ r(mod m) and x2 ≡ r(mod n) is equivalent to solve the systems x2 ≡ r(mod 2k ), x2 ≡ 2 mn 1 r(mod pm 1 ) · · · x ≡ r(mod pn ). This motivates us to state and prove the following theorems, which deals with the solvability of quadratic congruences 2 k i x2 ≡ r(mod pm i ) and x ≡ r(mod 2 ) for any odd prime pi and k > 0. Theorem 9.6.4. If p is an odd prime and gcd(r, p) = 1, then the congruence x2 ≡ r(mod pn ) (n ≥ 1) has a solution if and only if ( pr ) = 1. Proof. For the ‘if’ part, let the congruence equation x2 ≡ r(mod pn ) has a solution. Then x2 ≡ r(mod p) is solvable. Hence ( pr ) = 1. For the ‘only if’ part, let ( pr ) holds. We intend to use the principle of mathematical induction on n to prove the solvability of x2 ≡ r(mod pn ). The result is trivial for n = 1. Let us consider the statement prevails for n = k. Then x2 ≡ r(mod pk ) is solvable. For n = k + 1 its suﬃces to show that x2 ≡ r(mod pk+1 ) is solvable. As x2 ≡ r(mod pk ) is solvable, let x = x0 be its solution. Then x20 = r + bpk holds, for some integer b. To examine whether x2 ≡ r(mod pk ) is solvable, we need to ﬁnd the solution of 2x0 y ≡ −b(mod p). As gcd(2x0 , p) = 1, therefore this linear congruence is solvable with its solution y = y0 . Let us consider the integer x1 = x0 + y0 pk . Squaring we get, (x0 + y0 pk )2 = x20 + 2x0 y0 pk + y02 p2k = r + (b + 2x0 y0 )pk + y02 p2k . Here p (b + 2x0 y0 ). Then applying congruence modulo pk+1 we get, x2 ≡ r( mod pk+1 ). Thus the statement is true for n = k + 1, so the induction step is complete. Let us illustrate Theorem 9.6.4 with an example, where we can ﬁnd a solution of x2 ≡ 14(mod 52 ). To accomplish this, we see that x2 ≡ 14(mod 5) has a solution x0 = 7 as 45 = 1. Now 72 = 49 = 14 + 7 · 5, where b = 7 and the linear equation 2 × 7 × y ≡ −7(mod 5) has a solution y0 = 2. Thus Theory of Quadratic Residues 231 x1 = x0 + y0 p = 7 + 2 × 5 = 17. Let us examine, whether x1 = 17 is the solution of x2 ≡ 14(mod 52 ). Now consider even composite number, where we can compute the case for modulo 2k (k ≥ 1). The following is the theorem based on that. Theorem 9.6.5. Let r be an odd integer. Then 1. x2 ≡ r(mod 2) always solvable. 2. x2 ≡ r(mod 4) has a solution if and only if r ≡ 1(mod 4). 3. x2 ≡ r(mod 2n ), for n ≥ 3 has a solution if and only if r ≡ 1(mod 8). Proof. 1. Obvious. 2. Since the square of any odd integer congruent to 1 modulo 4, therefore r must be of the form 4k + 1 to solve x2 ≡ r(mod 4). Thus the equation x2 ≡ r(mod 4) has a solution if and only if r ≡ 1(mod 4). In this case x = 1 and x = 3 are two solutions modulo 4. 3. Note that the square of any odd integer is congruent to 1 modulo 8. Then x2 ≡ r(mod 2n ) is solvable implies r is of the form 8m + 1 for any integer m. This proves r ≡ 1(mod 8). Conversely, let r ≡ 1(mod 8) holds. With the help of mathematical induction, we will prove x2 ≡ r(mod 2n ) is solvable. Take n = 3. Then the congruence becomes x2 ≡ r(mod 8). Thus x2 ≡ 1(mod 8) is certainly solvable. Here x = 1, 3, 5, 7 satisﬁes x2 ≡ 1(mod 8). Thus the result is true for n = 3. Next, let us assume the result be true for n = k(> 3). Then the congruence x2 ≡ r(mod 2k ) admits a solution x0 , which implies ∃ b ∈ Z such that x20 = r + b · 2k . As r is odd, so is the integer x0 . Thus gcd(x0 , 2) = 1 leads to the fact that x0 y ≡ −b(mod 2) has a solution y = y0 . Choose an integer x1 = x0 + y0 2k−1 . Then squaring we get, (x0 + y0 2k−1 )2 = x20 + x0 y0 2k + y02 22k−2 = r + (b + x0 y0 )2k + y02 22k−2 Here 2\|(x0 y0 + 2) yields x2 ≡ r(mod 2k+1 ). So the result becomes true for n = k + 1. Hence the converse part is established. The ﬁnal theorem of this section is the combined eﬀect of the proof of last two theorems. 232 Number Theory and its Applications Theorem 9.6.6. Let n(> 1) be a composite number with the prime factorization n = 2k pn1 1 pn2 2 · · · pnt t and r be any positive integer with gcd(r, n) = 1. Then the congruence equation x2 ≡ r(mod n) is solvable if and only if 1. r pi = 1 for i = 1, 2, · · · t; 2. r ≡ 1(mod 4), if 4\|n but 8 n; r ≡ 1(mod 8), if 8\|n. 9.7 Worked out Exercises 27 Problem 9.7.1. Evaluate the following Jacobi symbol ( 101 ). Solution 9.7.1. Here we have, 27 101 = 3 101 3 3 101 = [by Corollary 9.4.2 since 101 ≡ 1( mod 4)] 3 3 2 = −1. = 3 Problem 9.7.2. Let a and b are relatively prime integers such that b is odd and positive and a = (−1)s 2t q where q is an odd integer. Prove that, 2 a ( b−1 )s+( b 8−1 )t q 2 = (−1) . b b Solution 9.7.2. Note that gcd(a, b) = 1 and b is an odd positive integer. As a = (−1)s 2t q for any odd integer q, then it follows that a (−1)s 2t q = b b s t −1 2 q = b b b q b b−1 b2 −1 q . = (−1)( 2 )s+( 8 )t b = (−1)( b−1 2 )s (−1)( b2 −1 8 )t Problem 9.7.3. For which positive integers n that are relatively prime to 15 does the Jacobi symbol ( 15 n ) equal to 1. Theory of Quadratic Residues 233 Solution 9.7.3. An appeal to the reciprocity law of Jacobi symbol, for any integer n with gcd(n, 15) = 1 it follows that Here n 15 15 n = n ( 15 ), n ), −( 15 if n ≡ 1( mod 4) if n ≡ 3( mod 4). n n and 5 3 = 1, if n ≡ 1( mod 3) n = 3 −1, if n ≡ 2( mod 3) 1, if n ≡ 1 or 4( mod 5) n = 5 −1, if n ≡ 2 or 3( mod 5). Combining we get, n 15 = 1, if n ≡ 1, 2, 4, 8( mod 15) −1, if n ≡ 1, 7, 13, 14( mod 15). Applying Chinese Remainder Theorem, ( 15 n ) = 1 holds if and only if n ≡ 1, 7, 11, 17, 43, 53 or 59(mod 60)(Verify!). Problem 9.7.4. Let n be an odd square-free positive integer. Prove that there is an integer a such that gcd(a, n) = 1 and ( na ) = −1. Solution 9.7.4. As n is odd and square-free, it is of the form n = p1 p2 · · · pt where each pi ’s are primes. Then ( na ) can be written as ( na ) = ( pa1 )( pa2 ) · · · ( pat ). Let us choose b as one of quadratic non residue among ( p12−1 ) numbers of quadratic non residues of p1 . This implies ( pb1 ) = −1. As gcd(a, n) = 1 and n = p1 p2 · · · pt , then applying Chinese Remainder Theorem we can assume a is a solution of the system x ≡ b(mod p1 ), x ≡ 1(mod p2 ), · · · x ≡ 1(mod pt ). Then we have, ( pa1 ) = ( pb1 ) = −1, ( pa2 ) = ( p12 ) = 1, · · · ( pat ) = ( p1t ) = 1. This proves that ( na ) = ( pa1 )( pa2 ) · · · ( pat ) = −1. Problem 9.7.5. Using the Generalized Quadratic Reciprocity Law, determine whether the congruence x2 ≡ 231(mod 1105) is solvable. Solution 9.7.5. First note that 231 = 3 · 7 · 11 and 1105 = 5 · 13 · 17 with 231 231 ) = 1 or ( 1105 ) = −1. gcd(231, 1105) = 1. For this we need to show either ( 1105 234 Number Theory and its Applications ( 231−1 )( 1105−1 ) 2 2 231 Now ( 1105 )( 1105 231 ) = (−1) 231 1105 = 1. Thus we have, 1105 181 = = 231 231 181 181 181 = 3 7 11 6 5 1 = 3 7 11 2 3 11 = = −1 7 7 5 So x2 ≡ 231(mod 1105) is not solvable. Problem 9.7.6. Let us consider a positive integer r such that r is not a perfect square and satisﬁes r ≡ 0(mod 4) or r ≡ 1(mod 4). Now for this r we deﬁne Kronecker symbol as follows, 1, if r ≡ 1( mod 8) r = 2 −1, if r ≡ 5( mod 8). 2 ) if 2 r, where the symbol of the right hand Now for this r prove that, ( 2r ) = ( \|r\| side is a Jacobi symbol. 2 Solution 9.7.6. If r > 0, then clearly ( \|r\| ) = ( 2r ) and by Theorem 9.6.1 we have r 2 −1 (−r)2 −1 r 2 −1 2 2 ( 2r ) = (−1) 8 . Also if r < 0, then ( \|r\| ) = ( −r ) = (−1) 8 = (−1) 8 . But here 2 r means r is odd and r ≡ 1(mod 4) together gives r ≡ 1(mod 8) and r ≡ 5(mod 8). Now for r ≡ 1(mod 8) and r ≡ 5(mod 8) we see that (−1) 1 and (−1) 2 ). ( 2r ) = ( \|r\| r 2 −1 8 r 2 −1 8 = = −1 holds respectively. Together these two assertions imply Problem 9.7.7. Find the solution of x2 + 5x + 6 ≡ 0(mod 53 ). Solution 9.7.7. Here x2 + 5x + 6 ≡ 0(mod 53 ) ⇒ (x + 3)(x + 2) ≡ 0(mod 53 ). Therefore x ≡ −3, −2 ≡ 122, 123(mod 53 ). Problem 9.7.8. Solve the following quadratic congruence: x2 ≡ 2(mod 73 ). Solution 9.7.8. To solve this equation, we start with the congruence equation x2 ≡ 2(mod 7). The solution of this equation is x0 = 3. Now x20 = 32 = 2 + b · 7 ⇒ b = 1. Now the linear equation 2 · 3 · y ≡ −1(mod 7) has the solution y0 = 1. So x1 = x0 + y0 × 7 = 10 is the solution of x2 ≡ 2(mod 72 ). Further, x2 ≡ 2(mod 72 ) has the solution x0 = 10. Now x20 = 102 = 2 + b · 49 ⇒ b = 2. The linear equation 2 · 10 · y ≡ −2(mod 7) has the solution y0 = 2. So x1 = x0 + y0 × 49 = 108 is the solution of x2 ≡ 2(mod 73 ). Thus the ﬁnal solutions are x ≡ 108, −108 ≡ 108, 235(mod 73 ). Theory of Quadratic Residues 235 Problem 9.7.9. First determine the values of a for which the congruence is solvable and then ﬁnd the solution of that: x2 ≡ a(mod 25 ). Solution 9.7.9. Note that x2 ≡ a(mod 25 ) is solvable, if a ≡ 1(mod 8)(Why?). Therefore a = 1, 9, 17 or 25 as 25 = 32. For a = 1: x ≡ ±1, ±1 + 24 ≡ 1, 15, 17, 31(mod 25 ). For a = 9: x ≡ ±3, ±3 + 24 ≡ 3, 13, 19, 29(mod 25 ). For a = 17: x2 ≡ 17+32(mod 25 ) ∴ x ≡ ±7, ±7+24 ≡ 7, 9, 23, 25(mod 25 ). For a = 25: x ≡ ±5, ±5 + 24 ≡ 5, 11, 21, 27(mod 25 ). Problem 9.7.10. Prove that if the congruence x2 ≡ a(mod 2n ), where a is odd and n ≥ 3, has a solution, then it has exactly four incongruent solutions. Solution 9.7.10. From Theorem 8.6.3, it follows that 2n has no primitive root for n ≥ 3. Thus x2 ≡ a(mod 2n ) has a solution. Since a is odd and x0 is a solution, then x0 must be odd. Also −x0 is a solution. Let x1 be any other solution. Then x21 ≡ a(mod 2n ) implies x21 ≡ x20 ( 0 0 )( x1 −x )≡ mod 2n ) for n ≥ 3. Therefore (x1 +x0 )(x1 −x0 ) ≡ 0(mod 2n ) ⇒ ( x1 +x 2 2 x1 +x0 x1 −x0 n−2 ), n ≥ 3. But here 2 + 2 = x1 is odd, so any one of them 0(mod 2 is even. 0 0 0 If x1 −x is even and x1 +x is odd, then (x1 − x0 )( x1 +x ) ≡ 0(mod 2n−1 ) ⇒ 2 2 2 x1 ≡ x0 (mod 2n−1 ). 0 0 0 On the other hand if x1 −x is odd and x1 +x is even, then (x1 +x0 )( x1 −x )≡ 2 2 2 n−1 n−1 ) ⇒ x1 ≡ −x0 (mod 2 ). Thus together we have x1 = ±x0 + 0(mod 2 k2n−1 . If k is odd then for k = 2r + 1 we have, x1 = ±x0 + 2n−1 + r2n ≡ ±x0 + 2n−1 ( mod 2n ). If k is even then for k = 2r we have, x1 = ±x0 + r2n ≡ ±x0 (mod 2n ). Therefore the only incongruent solutions are ±x0 , ±x0 + 2n−1 . 9.8 Exercises: 1. Find the solutions of the following quadratic congruences: (a) x2 + 5x + 1 ≡ 0(mod 7) (b) x2 + 7x + 10 ≡ 0(mod 11) (c) 5x2 + 6x + 1 ≡ 0(mod 23). 2. Given that a is a quadratic residue of the odd prime p, prove that a is not a primitive root of p. 236 Number Theory and its Applications 3. Let p be an odd prime and gcd(a, p) = 1. Establish that the quadratic congruence ax2 + bx + c ≡ 0(mod p) is solvable if and only if b2 − 4ac is either zero or a quadratic residue of p. 4. Find all the quadratic residues of 19, where 2 is a primitive root of 19. 5. Evaluate the following Legendre symbols: −23 18 461 −219 111 (a) ( 19 23 ) (b)( 59 ) (c)( 43 ) (d)( 773 ) (e)( 383 ) (f)( 991 ). 6. Use Gauss’s lemma to compute each of the Legendre symbols below : 8 5 (a) ( 11 ) (b)( 19 ) (c)( 11 23 ). 7. Let a and b be integers not divisible by prime p. Show that there is either one or three quadratic residues among the integers a, b and ab. 8. Given a prime p, show that, for some choice of n > 0, p divides (n2 − 2)(n2 − 3)(n2 − 6) 9. If the prime p > 5, show that p divides the sum of the squares of its quadratic nonresidues. 10. Given that p and q = 4p+1 are both primes, then prove that any quadratic nonresidue of q is either a primitive root of q or has order 4 modulo q. 11. Determine whether the given quadratic congruence is solvable 2x2 + 5x − 9 ≡ 0(mod 101). 12. Find a congruence describing all primes for which 7 is a quadratic residue. 13. Prove that there are inﬁnitely many primes of the form 5k − 1. 14. Verify that the prime divisors p of the integer 2n(n + 1) + 1 are of the form p ≡ 1(mod 4). 15. Solve the quadratic congruence x2 ≡ 11(mod 35). 16. Evaluate the following Jacobi symbols: 2663 10001 (a) ( 1009 2307 ) (b)( 3299 ) (c)( 20003 ). 17. For which positive integers n that are relatively prime to 30 does the Jacobi symbol ( 30 n ) equal to 1. 18. Evaluate the following Kronecker symbol: 5 ) (b)( 101 (a) ( 12 200 ). Theory of Quadratic Residues 237 19. Show that if n1 and n2 be positive integers with gcd(a, n1 n2 ) = 1, then ( n1an2 ) = ( na1 )( na2 ). 20. Show that if n1 and n2 are positive integers relatively prime to a and n1 ≡ n2 (mod \|a\|), then ( na1 ) = ( na2 ). 21. Solve the congruence x2 ≡ 31(mod 114 ). 22. Solve each of the following quadratic congruences: (a) x2 ≡ 7(mod 33 ); (b) x2 ≡ 14(mod 53 ). 23. Solve the congruence x2 ≡ 9(mod 23 · 3 · 52 ). 24. Show that 7 and 18 are the only incongruent solutions of x2 ≡ −1(mod 52 ). 25. Solve the congruence 2x2 + 1 ≡ 0(mod 112 ). 10 Integers of Special Forms “An equation means nothing to me unless it expresses a thought of God.” – Srinivasa Ramanujan 10.1 Introduction Over the centuries, integers with certain properties were studied extensively by several mathematicians. Present chapter portrayed few integers viz Perfect number, Mersenne Primes and Fermat number, where various theorems together with few properties are proved. We begin with the notion of perfect number. 10.2 Perfect Numbers In ancient times, Greek Mathematicians have found some interesting integers which can be written as sum of their divisors. For example 6 = 1 + 2 + 3 where 1, 2, 3 are divisors of 6. They have named those type of integers as perfect integers or perfect numbers. Thus the deﬁnition of perfect number are as follows: Deﬁnition 10.2.1. If n is a positive integer with σ(n) = 2n, then n is said to be a perfect number. For example we have σ(28) = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 · 28. σ(6) = 1 + 2 + 3 + 6 = 12 = 2 · 6. 239 240 Number Theory and its Applications Here 6, 28 are perfect numbers. Also, the Greek Mathematicians discovered the method of ﬁnding an even perfect number. Here the perfect number 6, 28 can be written as 6 = 2 · 3 = 22−1 (22 − 1), 28 = 23−1 (23 − 1) = 4 · 7 respectively. Thus the numbers are in the multiplication of even and prime numbers. Euclid ﬁrst tried to ﬁnd the form of a perfect number. After 2000 years Euler gave us a concrete proof about the form of a perfect number. Now the following theorem deals with the form of a perfect number. Theorem 10.2.1. For any positive integer n is an even perfect number if and only if it is of the form n = 2k−1 (2k − 1)(k > 1), 2k − 1 being prime. Proof. Let p = 2k − 1. Then n = 2k−1 p. Since, k > 1 then 2k−1 is not a prime and gcd(2k−1 , p) = 1. From the theorem, the sum of divisor function σ is multiplicative. Then σ(n) = σ(2k−1 p) = σ(2k−1 )σ(p) 2(k+1)−1 − 1 × (p + 1) [from Theorem 6.2.2] 2−1 = (2k − 1)(p + 1) = = 2k (2k − 1) = 2n, which shows n is a perfect number. Conversely, let n be an even perfect number. Then for some positive integer s and positive odd integer t, n = 2s t. Since, gcd(2s , t) = 1, from Theorem 6.2.3 applying multiplicative property of σ we have σ(n) = σ(2s t) = σ(2s )σ(t) = (2s+1 − 1)σ(t). Since, n is perfect, thus we have σ(n) = 2n = 2s+1 t. Combining we get, 2s+1 t = (2s+1 − 1)σ(t). As gcd(2s+1 , 2s+1 − 1) = 1, then 2s+1 \|σ(t). Thus there exists an integer q satisfying σ(t) = 2s+1 · q. Thus we have, (2s+1 − 1) · 2s+1 · q = 2s+1 · t. Therefore (2s+1 − 1)q = t, which shows q\|t and q = t, for if q = t then 2s+1 − 1 = 1 ⇒ s = 0, which is not possible. Now t + q = (2s+1 − 1)q + q = 2s+1 · q = σ(t). Next our claim is q = 1 otherwise t will be a composite number. If q = 1 then the divisors Integers of Special Forms 241 of t are 1, q and t implies σ(t) ≥ 1 + q + t, contradicts the fact σ(t) = q + t. Hence q = 1 and t = 2s+1 − 1, σ(t) = t + 1 which together shows t is prime. Thus, n = 2s (2s+1 − 1), where 2s+1 − 1 is prime. Taking s = k − 1, we have s > 0 ⇒ k > 1 and n = 2(k − 1)(2k − 1). Hence the proof. From the above theorem, it is clear that if our tasks is to ﬁnd even perfect number then we only need to check the primes which is of the form 2n − 1. In search of that we ﬁrst show that the exponent n must be prime. Our next theorem deals with it. Theorem 10.2.2. If n is a positive integer and 2n − 1 is prime then n must be prime. Proof. Let us assume, n be not prime. Then n is of the form n = rs, 1 < r, s < n. Then 2n − 1 = (2r )s − 1. Therefore 2n − 1 = (2r − 1)(2r(s−1) + 2r(s−2) + . . . + 1). Here in the right hand side, both the factors are greater than 1 which shows 2n − 1 is composite if n is so. Thus if 2n − 1 is prime then n must be prime. In particular, if we take n = 2, 3, 5 then 2n − 1 = 3, 7, 31 are all primes, which shows that 2(22 − 1) = 6, 22 (23 − 1) = 28, 24 (25 − 1) = 496 are all perfect numbers. It is to be noted that all the even perfect number ends with 6m or 8. Since, the prime number set is countably inﬁnite, therefore the numbers of the form 2p − 1 is also inﬁnite, for some prime p. As a consequence, we conclude that the perfect number set is countably inﬁnite. Now the question arises: Does all the even perfect number ends with 6 or 8? The following theorem deals with this fact: Theorem 10.2.3. Every even perfect number ends with either 6 or 8 i.e.either n ≡ 6(mod 10) or n ≡ 8(mod 10). Proof. Let n be an even perfect number. Then n is of the form n = 2k−1 (2k − 1)(k > 1), where2k−1 is a prime number. Now since 2k−1 is prime, k is also so(Why!). The case is trivial if we choose k = 2, then n = 6. Now our claim is to prove the statement for k > 2. Since, k is prime and greater than 2, it must be odd. So k must be of the form 4t + 1 or 4t + 3 for any positive integer t. Now, in the rest part of the theorem our claim is to show 1. the last digit 6 corresponds to 4t + 1 or 242 Number Theory and its Applications 2. the last digit 8 corresponds to 4t + 3. To fulﬁll the above claims, further we need to prove 16m ≡ 6(mod 10). Applying principle of mathematical induction we see that the case is trivial for m = 1. Let for m = r, 16r ≡ 6(mod 10), holds true. Now for m = r + 1, we have 16r+1 = 16r · 16 ≡ 6 · 16( mod 10). Therefore 16r+1 ≡ 36( mod 10) ≡ 6( mod 10). Next to fulﬁll claim(1), let us take k = 4t + 1. Then, n = 24t+1−1 (24t+1 − 1) = 28t+1 − 24t = 2 · 162t − 16t. Hence n ≡ 2 · 6 − 6(mod 10) ≡ 6(mod 10)(How!). For claim(2), taking k = 4t + 3 we obtain n = 24t+2 (24t+3 − 1) = 28t+5 − 24t+2 = 2 · 162t+1 − 4 · 16t . Hence n ≡ 2 · 6 − 4 · 6 ≡ −12(mod 10) ⇒ n = 8(mod 10). Consequently, every even perfect number has a last digit ends with 6 or 8. Let us continue our discussion on the last digit of an even perfect number more deeply with some examples. In our foregoing examples instead of taking k = 3 which gives n = 28, if we take k = 7, we have n = 26 (27 − 1) = 8128. Here the last two digits are 2, 8. This is not a coincidence, in fact in our next corollary we have shown how the last two digits always become 28. Corollary 10.2.1. An even perfect number n ends with either 6 or 28. Proof. The case of last digit 6 follows form the above theorem. For the remaining part, let us take n = 2k−1 (2k − 1)(k > 1), 2k−1 being prime. Since n is an even perfect number and k = 4t + 3 for a positive integer t, we have 2k−1 = 24t+2 = 16t · 4 ≡ 6 · 4( mod 10) ≡ 4( mod 10). As k > 2 then 4\|2k−1 and the number formed by last two digits of 2k−1 is divisible by 4. Then under congruent modulo 100, the possibilities are 2k−1 ≡ 4, 24, 44, 64 or 84(mod 100). Since the last digit is 4, therefore 2k−1 ≡ 4(mod 10). Now, let us take 2k−1 ≡ 4(mod 100) then 2 · 2k−1 ≡ 2 · 4 − 10(mod 100) ≡ 7(mod 100) and so on. Then we have, 2k − 1 = 2 · 2k−1 − 1 ≡ 7, 47, 87, 27 or 67( mod 100) Integers of Special Forms 243 Thus multiplying, we have n = 2k−1 (2k − 1) ≡ (4 · 7, 24 · 47, 44 · 87, 64 · 27) or 84 · 67( mod 100), ⇒ n ≡ 28( mod 100)(Verify!). The above discussions help us to recognize the form of an even perfect number. Here we conclude our discussions with some properties to elaborate even perfect number related to triangular number, a number which can be arranged in triangular form. For instance, let us consider 3 = 1 + 2; 6 = 1 + 2 + 3; 10 = 1 + 2 + 3 + 4; 15 = 1 + 2 + 3 + 4 + 5; 21 = 1 + 2 + 3 + 4 + 5 + 6; 28 = 1 + 2 + 3 + 4 + 5 + 6 + 7; 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8; 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, . . ., which can be arranged in triangular form as shown in the diagram below. Figure 10.1: Triangular Diagram Our next result focusses on the relation between them. Proposition 10.2.1. If N is an even perfect number then N is triangular. Proof. Let us choose a triangular number n. Then n can be written as n = 1 1 + 2 + 3 + · · · + r = r(r − 1) for some integer r. Now, since N is a perfect 2 number N is of the form, N = 2k−1 (2k − 1) = 1 k k−1 (2 )(2 ), 2 which shows N is of the triangular form. This proves N is triangular. We can also write perfect number as a sum of cubes by our next proposition. Proposition 10.2.2. If N is an even perfect number then N = 13 + 23 + . . . + n+1 (2 2 − 1)3 , n > 1. 244 Number Theory and its Applications Proof. Let us consider m = 2 n−1 2 . Then, 13 + 23 + . . . + (2m − 1)3 = (13 + 23 + . . . + (2m)3 ) − (23 + 43 + . . . + (2m)3 ) (m)2 (m + 1)2 (2m)2 (2m + 1)2 − 23 × 4 4 2 2 = m (2m − 1). = On substituting the value of m, we obtain 13 + 33 + . . . + (2 10.3 n+1 2 − 1)3 = 2n−1 (2n − 1) = N (W hy!). Worked out Exercises Problem 10.3.1. Show that the integer n = 210 (211 −1) is not a perfect number. Solution 10.3.1. Note that n = 210 (211 − 1). To show n is a perfect number, it’s suﬃces to show that σ(n) = 2n. Since gcd(210 , 211 − 1) = 1, therefore σ(n) = σ(210 )σ(211 − 1) as σ is multiplicative. Furthermore, 211 − 1 has a prime factorization given by 211 − 1 = 23 · 89. For any prime p, σ(p) = p + 1 holds. Then σ(211 − 1) = σ(23)σ(89) = 24 × 90 yields, 1 σ(n) = 211 × (1 − ) × 24 × 90 2 = 210 × 24 × 90 211 × (211 − 1) = = 2n. Therefore n is not a perfect number. Problem 10.3.2. Verify that no power of a prime can be a perfect number. Solution 10.3.2. Let p be a prime and n be any positive integer. We claim that n+1 2pn . But σ(pn ) = 1 + p + p2 + · · · + pn = p p−1−1 . Consider p = 2. σ(pn ) = 2n+1 . Let the assertion be true for p > 2. Then Then σ(2n ) = 2n+1 − 1 = n+1 n − 1 = 2p (p − 1) holds. It follows that, p pn+1 − 1 = 2pn+1 − 2pn ⇒ pn+1 − 2pn + 1 = 0 ⇒ pn (p − 2) = −1. Integers of Special Forms 245 This leads to a contradiction, as pn and p − 2 are greater than or equal to 1. This proves pn is not a perfect number. Problem 10.3.3. For any perfect number n show that 1 d\|n ( d ) = 2. Solution 10.3.3. As n is a perfect number, so σ(n) = 2n. Further we have σ(n) = d\|n d. Combining the pieces, we get d\|n d = 2n. Multiplying both sides by n1 gives, 1 1 σ(n) = d = 2. n n d\|n Referring to Problem 6.3.4, we have 1 d\|n ( d ) = 2. Problem 10.3.4. Prove that no divisor of a perfect number can be perfect. Solution 10.3.4. Let m be a proper divisor of a perfect number n. As n is perfect, so the last problem generates d\|n d1 = 2. Thus d\|m d1 < d\|n d1 = 2. This proves m is not perfect. Problem 10.3.5. Find the last two digits of the perfect number n = 219936 (219937 − 1). Solution 10.3.5. Since gcd(2, 25) = 1, therefore using the Euler’s Theorem we get 2φ(25) ≡ 1(mod 25) i.e. 220 ≡ 1(mod 25). This implies 219334 ≡ 214 ( mod 25). Further 214 ≡ 9(mod 25). Stitching the facts, we get 219336 ≡ 4 · 9( mod 100). Therefore 219337 ≡ 72(mod 100), which shows that, 219936 (219937 − 1) ≡ 36 × 71( mod 100) ≡ 56( mod 100). Therefore the last two digits of the perfect number is 56. Problem 10.3.6. Prove that there are the only two 3-perfect number of the form n = 2k · 3 · p for any odd prime p. Solution 10.3.6. If for any positive integer n, σ(n) = kn for k ≥ 3 holds then n is called k-perfect number. Thus for 3-perfect number we have σ(n) = 3n. We claim that, for any odd prime p the integers 120 and 672 are the only 3-perfect numbers of the form n = 2k · 3 · p. Assume p = 3. Then n = 2k · 32 . Thus σ(n) = σ(2k )σ(32 ) = 13(2k+1 − 1). This yields σ(n) = 3n, which is not possible as 13 3n. 246 Number Theory and its Applications Now consider p > 3. Then 3n = σ(n) = σ(2k )σ(3)σ(p) = (2k+1 − 1) × 4 × (p + 1). If k = 0, then σ(n) = 4×(p+1) = 9p and if k = 1 then σ(n) = 12×(p+1) = 6p. This shows that k is always greater than 1. Taking advantage of this form of n and dividing by 4 we ﬁnd, 2k−2 · 32 · p = (2k+1 − 1)(p + 1). (10.3.1) But gcd(p, p + 1) = gcd(2k−2 , 2k+1 − 1) = 1, which implies p\|2k+1 − 1 and 2k−2 \|p + 1. Thus from the (10.3.1) we obtain, 2k−2 · 32 · p = p · t1 · 2k−2 · t2 for some integer t1 , t2 . This gives t1 t2 = 9. So three cases may arise: 1. If t1 = 1, t2 = 9 holds, then we have p + 1 = 9 · 2k−2 and 2k+1 − 1 = p. The last two equations together gives 9 · 2k+2 = 2k+1 , which is not possible. 2. If t1 = 9, t2 = 1 holds, then p + 1 = 2k−2 and 2k+1 − 1 = 9 · p implies 9 · 2k−2 − 9 = 2k+1 − 1. On solving we get p = 7, k = 5. 3. If t1 = 3, t2 = 3 holds. Then we have p + 1 = 3 · 2k−2 and 2k+1 − 1 = 3 · p .Then solving we get p = 5, k = 3. Thus the possible values are n = 25 × 3 × 7 = 672 and n = 23 × 5 × 3 = 120. Problem 10.3.7. If n > 6 is an even perfect number, prove that n ≡ 4(mod 6). Solution 10.3.7. Here n > 6 is even perfect number, so it is of the form 2k−1 (2k − 1) where k(> 2) and 2k − 1 are odd primes. Since gcd(2, 3) = 1, therefore by Fermat’s theorem 2k−2 (2k −1) ≡ 2(2−1)(mod 3) ⇒ 2k−1 (2k −1) ≡ 4( mod 6). Hence n ≡ 4(mod 6). Problem 10.3.8. Let n be a positive integer. Deﬁne the sequence n1 , n2 , n3 , · · · recursively by n1 = σ(n) − n and nk+1 = σ(nk ) − nk for k = 1, 2, 3, · · · . Prove that if n is perfect, then n = n1 = n2 = · · · . Solution 10.3.8. Let n = n1 be perfect. Then n2 = σ(n)−n = 2n−n = n = n1 . Continuing this way at j-th stage we get, nj = σ(nj−1 ) − nj−1 = σ(n) − n = n, ∀j ≥ 1. This proves that n = n1 = n2 = · · · . Integers of Special Forms Problem 10.3.9. For any even perfect number n = 2 2k \|(σ(n2 ) + 1). 247 k−1 (2 − 1), prove that k Solution 10.3.9. As n is perfect, so it is of the form n = 2k−1 (2k − 1) where 2k − 1 is prime. As gcd(2k − 1, 2k−1 ) = 1, then σ(n2 ) = σ(22k−2 )σ((2k − 1)2 ) = (1 + 2 + · · · + 22k−2 )(1 + 2k − 1 + (2k − 1)2 ) = (22k−1 − 1)(22k − 2k + 1) = 22k−1 (22k − 2k + 1) − 22k + 2k − 1. From this it is clear that 2k (σ(n2 ) + 1). Problem 10.3.10. A number n is said to be super perfect number if σ(σ(n)) = 2n. Now using this concept of superperfect number, show that if n = 2k with 2k+1 − 1 a prime, then n becomes a super perfect number. Solution 10.3.10. Here n = 2k . Then σ(n) = σ(2k ) = 2k+1 − 1. As 2k+1 − 1 is prime, so σ(σ(n)) = σ(2k+1 − 1) = 2k+1 = 2n. Thus 2k is super perfect number. Problem 10.3.11. Let n be a non perfect positive integer. Then n is called deﬁcient if σ(n) < 2n and is called abundant if σ(n) > 2n. Every non perfect positive integer is either deﬁcient or abundant. If n does not satisfy both the conditions then it is called perfect number. Now using this fact prove that every prime power is deﬁcient. Solution 10.3.11. Pick n = pk for some prime p and positive integer k. Then pk+1 − 1 . As p ≥ 2, then 2pk − 1 < pk+1 is always true. It we have, σ(pk ) = p−1 pk+1 − 1 < 2pk = 2n follows that pk+1 − 1 < 2(pk+1 − pk ) = 2pk (p − 1). Thus p−1 justiﬁes n as deﬁcient. 10.4 Mersenne Primes Our earlier discussions on pseudoprime[refer to chapter-5] based on Fermat’s Little theorem, where for any composite integer n, 2n − 1 is a pseudoprime. Any number of the form 2n − 1 is known to be n-th Mersenne number, after French Mathematician Father Marin Mersenne. Now, the present section deals with seeking some prime numbers of that form. From previous section, we have seen that for any positive integer n, if 2n − 1 is prime then n must be prime. But in general the converse of the last statement is not true, which gives birth to the 248 Number Theory and its Applications notion of Mersenne prime i.e. for any prime p, if the integer 2p − 1 is also prime then it is said to be Mersenne prime. Using the results on Mersenne Prime, we will discuss various methods of primality testing. Mersenne primes are used in the Mersenne twister PRNG (pseudo-random number generator), these are used extensively in simulations, Montecarlo methods, etc. The CWC mode for block ciphers can uses M127 as a prime number because x(mod 2127 − 1) is very easy to compute. Deﬁnition 10.4.1. If p is a prime and Mp = 2p − 1 is also so, then Mp is called Mersenne prime. Example 10.4.1. The Mersenne number M7 = 27 −1 is prime but the Mersenne number M11 = 211 − 1 = 2047 = 23 × 89 is not so. Here M7 is Mersenne prime but M11 is not. Here in the above example we have seen that M7 is prime but M11 is composite. Thus the formula that 2p − 1 is prime for any prime p fails to be true in general. So the hope for constructing prime numbers using this formula goes in vain. Now our next discussions focusses on constructing various methods for determining whether this special types of Mersenne numbers are prime or not. In next theorem, we have presented one particular type of test based on that. Theorem 10.4.1. If p and p = 2p + 1 are primes then p does not divide both Mp and Mp + 2 together. Proof. Since p is prime then by Fermat’s Little Theorem we have,2p −1 ≡ 1( mod p ) ∴ 2p −1 − 1 ≡ 0( mod p ) ⇒ (2 p −1 2 − 1)(2 p −1 2 + 1) ≡ 0( mod p ) ⇒ (2p − 1)(2p + 1) ≡ 0( mod p ) ⇒ Mp (Mp + 2) ≡ 0( mod p ) This follows that p \|Mp or p \|Mp + 2 but not together. If p divides both of them then this implies p \|2 which is impossible. This proves our theorem. To illustrate the above theorem lucidly let us choose p = 5 then p = 2 × 5 + 1 = 11. Here M5 = 25 − 1 = 31 clearly 11 31 but 11\|(31 + 2) = 33. This shows the signiﬁcance of the above theorem. Now if we choose p = 11 then p = 2 × 11 + 1 = 23. Here M11 = 211 − 1 = 2047 and clearly 23\|2047 as 2047 = 23 × 89 but 23 (2047 + 2) = 2049. In this two examples we have faced Integers of Special Forms 249 two cases where in the ﬁrst p \|Mp + 2 and in the second p \|Mp . Now the question arises: under what condition p will divide Mp . We have discussed about that circumstances in our following theorem. Theorem 10.4.2. If p = 2p + 1 is prime for any prime p then, 1. p \|Mp implies p ≡ 1(mod 8) or p ≡ 7(mod 8) 2. p \|Mp + 2 implies p ≡ 3(mod 8) or p ≡ 5(mod 8) p −1 Proof. 1. Here p = 2p + 1 and p \|Mp together implies 2 2 = 2p ≡ 1( mod p ). As p \|Mp then from the Theorem 9.2.8 we can say that the Legendre symbol ( p2 ) = 1 whenever p ≡ ±1(mod 8). Thus p \|Mp implies p ≡ 1(mod 8) or p ≡ 7(mod 8). p −1 2. Similarly if p \|Mp +2 then we have, 2 2 = 2p ≡ −1(mod p ), As p \|Mp +2 then again from the Theorem 9.2.8 we can say that the Legendre symbol ( p2 ) = −1 whenever p ≡ ±3(mod 8). Thus p \|Mp + 2 implies p ≡ 3( mod 8) or p ≡ 5(mod 8). Our next corollary is an immediate consequence of above theorem. Corollary 10.4.1. If p and p = 2p+1 both are odd primes, then p \|Mp provided p ≡ 3(mod 4). Proof. As p is odd then it is of the form 4n + 1 or 4n + 3 for any non negative integer n. Now if p = 4n + 3 then p becomes 8n + 7. Now from the above theorem we can say that p \|Mp . Also if p = 4n + 1 then p becomes 8n + 3. So again by the above theorem we have p Mp . Here for an example if we choose p = 3 then p = 2 × 3 + 1 = 7 both are odd. Here M3 = 23 − 1 = 7 and obviously 7\|M3 . Now we are going to discuss few more results on primality testing of Mp . Here we have seen under which circumstances Mp has a divisor or not. The following theorem is ﬁrst among them. Theorem 10.4.3. If p is an odd prime then any prime divisor of Merenne number Mp is of the form 2kp + 1 where k is a positive integer. Proof. Let us choose p be any prime divisor of Mp = 2p − 1 then we have, 2p ≡ 1( mod p ). If 2 has the order n modulo p then by the Theorem (8.2.1) we can say that n\|p. Here if n = 1 then p \|1 since 2n ≡ 1(mod p ) which is impossible. Thus we have n = p as p is odd prime. Again from Fermat’s little theorem we have, 250 Number Theory and its Applications p −1 2 ≡ 1(mod p ) as gcd(2, p ) = 1. Now again applying Theorem 8.2.1 we have seen that n\|p − 1. But n = p which shows p\|p − 1. This implies p − 1 = pm for some integer m. Therefore p is of the form 1 + pm. Now if m is odd then p is even which is a contradiction as p is prime divisor of Mp . Thus we have m = 2k an even number and p is of the form 2kp + 1. Now we have the second following theorem to be discussed here. Theorem 10.4.4. If p is an odd prime then any prime divisor p of Mp is of the form p ≡ ±1(mod 8). p+1 Proof. Let p = 2m + 1 is a prime divisor of Mp . If we choose b = 2 2 then b2 − 2 = 2p+1 − 2 = 2Mp ≡ 0(mod p ). This implies b2 ≡ 2(mod p ) and taking m-th power both sides of congruence we arrive at, b2m = bp −1 ≡ 2m (mod p ). Here gcd(b, p ) = 1 as p is prime and so by Fermat’s theorem bp −1 ≡ 1(mod p ). Now combining these two congruences we have, 2m ≡ 1(mod p ) which shows that p \|Mm . Finally by Theorem 10.4.2 we have the conclusion that p ≡ ±1( mod 8). From this above two theorem we can arrive at a conclusion that these two theorems can be used to decide whether a Mersenne number prime or not. Following examples are illustration of this fact. Example 10.4.2. Let us consider M11 = 211 − 1 = 2047 and the prime factors √ √ of M11 are less than M11 = 2047 = 45.24. But the prime divisors here of the form 22k + 1 for any positive integer k. So the integers are 23 and 45 less √ than M11 . We can check that 23\|2047 asserting that M11 is composite number. Again if we choose M13 = 213 − 1 = 8191 and the prime factors of M13 are less √ √ than M15 = 8191 = 90.504. They are of the form 26k + 1 for k > 0. So the numbers are 27, 53 and 79. Here 53 and 79 are the primes but they does not divide M13 . So M13 is a prime. It’s deserve mentioning, that since 1914 ﬁrst 12 Mersenne primes (hence, 12 perfect numbers) had been known. The last Mersenne prime(11th one), M89 was discovered independently in 1911 by Powers and Cunningham. In 1876 Lucas founded the prime M127 , the largest prime for the next 75 years. Because there are special primality tests for Mersenne numbers, it has been possible to determine whether extremely large Mersenne numbers are prime. Following is one such primality test. This test, commonly known as The LucasLehmer Test, has been extensively used to ﬁnd the largest known Mersenne primes, which are the largest known primes. The proof of this test is beyond the scope of this book but may be found in Lenstra [] and Sierpinski []. Integers of Special Forms 251 Statement 10.4.1. Lucas-Lehmer Test: Let p be a prime and let Mp = 2p−1 denote the p-th Mersenne number. Deﬁne a sequence of integers recursively by setting r1 = 4, and for k ≥ 2, 2 − 2( mod Mp ), 0 ≤ rk < Mp . rk ≡ rk−1 Then, Mp is prime if and only if rp−1 ≡ 0(mod Mp ). The following example will illustrate the application of the Lucas-Lehmer Test: Example 10.4.3. Let us consider the Mersenne number M7 = 27 − 1 = 127. Then r1 = 4 imply r2 ≡ r12 − 2 ≡ 14(mod 127). Now r3 ≡ r22 − 2 = 194 ≡ 67( mod 127), r4 ≡ r32 − 2 = 4487 ≡ 42(mod 127),r5 ≡ r42 − 2 ≡ 111(mod 127) and r6 ≡ r52 − 2 ≡ 0(mod 127). Thus we can conclude that M7 is prime. Various activities has been directed towards the discovery of Mersenne primes, especially since each new Mersenne prime discovered has become the largest prime known, and for each new Mersenne prime, there exist a new perfect number. At the present time, a total of 29 Mersenne primes are known and these include all Mersenne primes Mp with p < 62981 and with 75000 < p < 100000. The following table illustrates the list of known Mersenne Primes. 252 Number Theory and its Applications p 2 3 5 7 13 17 19 31 61 89 107 127 521 607 1279 2203 2281 3217 4253 4423 9689 9941 11213 19937 21701 23209 44497 86243 132049 216091 Number of decimal digits in Mp 1 1 2 3 4 6 6 10 19 27 33 39 157 183 386 664 687 969 1281 1332 2917 2993 3376 6002 6533 6987 13395 25962 39751 65050 Date of Discovery Ancient Times Ancient Times Ancient Times Ancient Times Mid 15th century 1603 1603 1772 1883 1911 1914 1876 1952 1952 1952 1956 1952 1957 1961 1961 1963 1963 1963 1971 1978 1979 1979 1983 1983 1985. Many mathematicians believe that there are inﬁnitely many Mersenne primes, but a proof of this seems to be an open problem. As p increases, known Mersenne primes Mp clearly become more scarce. It has been hypothecated that about two primes Mp should be expected for all primes p in an interval x < p < 2x. One of the renowned problems of number theory is whether there exist any odd perfect numbers. Although no odd perfect number has been produced so Integers of Special Forms 253 far. But, it is possible to ﬁnd certain conditions for the existence of odd perfect numbers. The oldest of these was due to Euler, is reﬂected in the form of the following theorem: 2 2jr Theorem 10.4.5. If n is an odd perfect number, then n = pk11 p2j where 2 · · · pr pi ’s are distinct odd primes and p1 ≡ k1 ≡ 1(mod 4). Proof. Let n = pk11 pk22 · · · pkr r be the prime factorization of n. As n is perfect so we have, 2n = σ(n) = σ(pk11 )σ(pk22 ) · · · σ(pk11 ). Now n is odd imply n ≡ ±1(mod 4) then σ(n) = 2n ≡ 2(mod 4). This shows that σ(n) is divisible by 2 but not by 4. Thus one of σ(pki i ) is even and the rest of are odd. Let us choose σ(pk11 ) is even. As pi ’s are odd primes then pi ≡ ±1(mod 4). Then for pi ≡ −1(mod 4) we have, σ(pki i ) = 1 + pi + p2i + · · · + pki i ≡ 1 + (−1) + · · · + (−1)ki ( mod 4) Therefore σ(pki i ) ≡ 0(mod 4) if ki is odd and σ(pki i ) ≡ 1(mod 4) if ki is even. Here σ(n) ≡ 2(mod 4) implies σ(pk11 ) ≡ 2(mod 4) if σ(pki i ) ≡ 1(mod 4) (i = 2, 3, · · · r). Otherwise σ(pki i ) ≡ 0(mod 4) signiﬁes that 4 divides σ(pki i ) which is not possible. Now here σ(pk11 ) ≡ 2(mod 4) gives us k1 ≡ 1(mod 4). Also for other values of i as σ(pki i ) ’s are odd the σ(pki i ) ≡ ±1(mod 4). This also implies ki ≡ 0(mod 4) or ki ≡ 2(mod 4). Thus in any cases ki ’s are even integer for i = 2, 3, · · · r. This proves the theorem. Now in the following corollary we have given the form of an odd perfect number. This is also an immediate consequence of above theorem. Corollary 10.4.2. If n is an odd perfect number then it is of the form n = pt k 2 , where p is an odd prime with p k and p ≡ t ≡ 1(mod 4). Then we have n ≡ 1( mod 4) also. Proof. From the above theorem it follows that, n = pk11 pk22 · · · pkr r = pk11 (pj22 · · · pjrr )2 = pt k 2 , where p1 = p, k1 = t & k = pj22 · · · pjrr . As p ≡ 1(mod 4) then we have pt ≡ 1(mod 4). Also k is odd imply k ≡ 1( mod 4) or k ≡ 3(mod 4). Thus in any case we have k 2 ≡ 1(mod 4). Then it follows that n = pt k 2 ≡ 1 · 1 ≡ 1(mod 4). 254 10.5 Number Theory and its Applications Worked out Exercises Problem 10.5.1. Any two distinct Mersenne numbers are coprime. Solution 10.5.1. Let p, q be two distinct primes where q > p. Note that gcd(p, q) = 1. It can also be obtained by Euclid’s algorithm. Then for any integer k we have q = kp + r1 , p = k1 r1 + r2 , r1 = k2 r2 + r3 , .... .. rn−2 = rn−1 kn−1 + rn , = rn−1 kn−1 + 1. Let l be the common divisor of Mp = 2p − 1 and Mq = 2q − 1. Then we have, ∴ l [(2q − 1) − 2q−p (2p − 1)], ⇒l (2q−p − 1), ⇒l [(2q−p − 1) − 2q−2p (2p − 1)], ⇒l (2q−2p − 1), .... .. ⇒l (2q−kp − 1), ⇒l (2r1 − 1), .... .. ⇒l (2rn − 1), ⇒l = 1. ∴ gcd(Mp , Mq ) = 1. Problem 10.5.2. Show that the Mersenne number M13 = 213 − 1 is prime and hence the integer 212 (213 − 1) is perfect. Solution 10.5.2. Note that the Mersenne number M13 = 213 − 1 has prime divisor of the form 2k · 13 + 1[see Theorem 10.4.3]. Applying Example 10.4.2, Integers of Special Forms 255 we can say that M13 is prime. Claim: n = 212 (213 − 1) is perfect. Thus, σ(n) = σ(212 M13 ) = σ(212 )σ(M13 ) = (213 − 1)(213 − 1 + 1) = 213 (213 − 1) = 2n. Hence the integer 212 (213 − 1) is perfect. Problem 10.5.3. Prove that the Mersenne number M29 is composite. Solution 10.5.3. The given Mersenne number M29 = 229 − 1 has the prime √ divisor of the form 2k · 29 + 1, for any integer k. From M29 < 23171, we obtain few possible divisors viz 59, 117, 233. Amongst them we can verify that 233\|M29 , which implies M29 is a composite number. Problem 10.5.4. A pair of integers m and n is said to be amicable numbers if σ(m) = m + n = σ(n) holds. Now for such pair m and n prove that, −1 −1 1 1 + = 1. d d d\|m Solution 10.5.4. Note that bining we get, d\|n 1 d\|m d = d d\|m m and 1 d\|n d = d d\|n n . Com- −1 −1 1 1 m n + + = d d σ(n) σ(m) d\|m d\|n m+n [∵ σ(m) = σ(n) = m + n] m+n = 1. = Problem 10.5.5. Show that for any odd prime p neither p nor p2 can be one of an amicable pair. Solution 10.5.5. Let p be an arbitrary prime. Then σ(p) = p + 1. Let p and 3 −1 . Thus p2 − 1 = p2 be the amicable pair. Then σ(p) = p + 1 = σ(p2 ) = pp−1 3 2 p − 1 ⇒ p = 1. This is not possible. So neither p nor p can be one of the amicable pair. Problem 10.5.6. Prove that for any odd prime p if n = pa2 is an odd perfect number, then n ≡ p(mod 8). 256 Number Theory and its Applications 2 Solution 10.5.6. Assume that n = pa is an odd perfect number. Then a is an odd integer. Thus a is of the form 4k ±1. Now a2 = (4k ±1)2 = 16k 2 ±8k +1 ≡ 1( mod 8). ∴ n = pa2 ≡ p(mod 8). Problem 10.5.7. Use the Lucas-Lehmer test to determine whether the Mersenne number M11 is prime or not. Solution 10.5.7. Here the Mersenne number is M11 = 211 − 1 = 2047. Then r1 = 4 ⇒ r2 ≡ r12 − 2 = 14(mod 2047). Now r3 ≡ r22 − 2 = 195(mod 2047) and r4 ≡ r32 − 2 = 1177(mod 2047). So continuing this way(Do it!) we ﬁnd r10 ≡ 0( mod M11 ). So M11 is not a prime. Problem 10.5.8. If the integer n > 1 is a product of distinct Mersenne primes, prove that σ(n) = 2k for some integer k. Solution 10.5.8. Let Mp = 2p − 1 and Mq = 2q − 1 be the distinct Mersenne primes and n = Mp Mq . Then σ(n) = σ(Mp )σ(Mq ) = (Mp + 1)(Mq + 1) = 2p · 2q = 2p+q = 2k . ∴ σ(n) is of the form 2k for any integer k. Problem 10.5.9. If p is an odd prime of the form 4k + 3 for any integer k and q = 2p + 1 be another prime then prove that q divides the Mersenne number Mp = 2p − 1. Solution 10.5.9. Here p is of the form 4k+3. Then q = 2p+1 = 2(4k+3)+1 = 8k + 7. By Theorem 9.2.8 we have, ( 2q ) = 1 as q ≡ 7(mod 8). Moreover, by q−1 Euler’s Criterion we see that 2 2 ≡ 2p ≡ 1(mod q). This shows that q (2p − 1). Problem 10.5.10. Prove that if n is a positive integer and 2n + 1 is prime, and if n ≡ 0(mod4) or n ≡ 3(mod 4) then 2n + 1 divides the Mersenne number Mn = 2n − 1, while if n ≡ 1(mod 4) or n ≡ 2(mod 4), then 2n + 1 divides Mn + 2 = 2n + 1. Solution 10.5.10. If n ≡ 0(mod4) or n ≡ 3(mod 4), then 2n + 1 ≡ 1(mod 8) 2 ) = 1(refer or 2n + 1 ≡ 7(mod 8) respectively. In both these cases, we have ( 2n+1 2n+1−1 to Theorem 9.2.8). By Euler’s Criterion, we get 2 2 = 2n ≡ 1(mod 2n + 1). n This proves that 2n + 1 (2 − 1). Furthermore, if n ≡ 1(mod4) or n ≡ 2(mod 4), then 2n + 1 ≡ 3(mod 8) or 2 2n + 1 ≡ 5(mod 8) respectively. In both these cases, we obtain 2n+1 = −1(refer to Theorem 9.2.8). By Euler’s Criterion, 2 yields 2n + 1 (2n + 1). 2n+1−1 2 = 2n ≡ −1(mod 2n+1). This Integers of Special Forms 10.6 257 Fermat Numbers In this section our discussion is based on particular numbers of the form Fn = n 22 + 1 where n is a non negative integer. They are called Fermat numbers, named after the French mathematician Pierre de Fermat who ﬁrst studied numbers of this form. Here we have also discussed few basic properties and primality of Fermat numbers. Fermat’s ﬁrst conjecture was that all the numbers of these 5 type were prime. But in 1732 Euler showed that the number F5 = 22 + 1 is composite. Then it become a question whether there are inﬁnitely many prime of this form or not. Thus to start our discussion we have given the following deﬁnition of Fermat numbers as well as Fermat primes. n Deﬁnition 10.6.1. An integer is of the form Fn = 22 + 1, n ≥ 0 is called Fermat number. If Fn is prime then it is called Fermat prime. For example we can see that the ﬁrst ﬁve Fermat numbers F0 = 3, F1 = 5, F2 = 17, F3 = 257 and F4 = 65, 537 all are primes. But the conjecture that 5 Fermat numbers are primes fails to F5 . Here F5 = 22 + 1 = 641 × 6700417 is composite. In fact we have an elementary proof that 641\|F5 due to G.Bennet. Theorem 10.6.1. The Fermat number F5 is divisible by 641. Proof. Let us choose u = 27 and v = 5 then we have, 1 + uv = 1 + 27 · 5 = 641. It is seen that , 1 + uv − v 4 = 1 + v(u − v 3 ) = 1 + (27 − 53 )v = 1 + 3v = 24 . Now 5 F5 = 22 + 1 = 24 · 228 + 1 = 24 · u4 + 1 = (1 + uv − v 4 )u4 + 1 = (1 + uv)[u4 + (1 − uv)(1 + u2 v 2 )] This shows that 641 divides F5 as 1 + uv = 641. We are now going to prove some basic properties of Fermat numbers in our next theorem. Theorem 10.6.2. 1. For n ≥ 1, Fn = (Fn−1 − 1)2 + 1 2. For n ≥ 1, Fn − 2 = F0 F1 · · · Fn−1 2 3. For n ≥ 2, Fn = Fn−1 − 2(Fn−2 − 1)2 4. For n ≥ 2, Fn − Fn−1 = 22 n−1 F0 F1 · · · Fn−2 258 Number Theory and its Applications Proof. 1. (Fn−1 − 1)2 + 1 = (22 n−1 n + 1 − 1)2 + 1 = 22 + 1 = Fn . 2. We have to prove this result by principle of mathematical induction. When n = 1, we have F1 − 2 = 5 − 2 = 3 = F0 . Let the result is true for n = k then, Fk − 2 = F0 F1 · · · Fk−1 . Now, F0 F1 · · · Fk + 2 = (Fk − 2)Fk + 2 k k = (22 − 1)(22 + 1) + 2 = 22 k+1 +1 = Fk+1 ∴, Fk+1 − 2 = F0 F1 · · · Fk . Thus the result is true for n = k + 1. So by the principle of mathematical induction the result is true. 3. 2 Fn−1 − 2(Fn−2 − 1)2 = (22 n−1 + 1)2 − 2(22 n = 22 + 2 · 22 n−1 n−2 )2 + 1 − 2 · 22 n−1 n = 22 + 1 = Fn , n ≥ 2 4. We have to prove this result by principle of mathematical induction. When n = 2, we have F1 + 22 · F0 = 5 + 22 · 3 = 17 = F2 . Now let us assume the k−1 result is true for n = k. Then we have Fk = Fk−1 + 22 F0 F1 · · · Fk−2 . Now, Fk + 22 F0 F1 · · · Fk−1 = Fk + 22 k k−1 (22 = Fk + 22 k−1 (Fk − Fk−1 )Fk−1 2k = (2 k−1 + 1) + 22 k k−1 k F0 F1 · · · Fk−2 )Fk−1 k (22 − 22 k−1 (22 k−1 + 1) k = (22 + 1) + 22 (22 − 1) = 1 + 22 k+1 = Fk+1 Thus the result is true for n = k + 1. So by the principle of mathematical induction the result is true. We are now going to exhibit an important property of Fermat number. Integers of Special Forms 259 Theorem 10.6.3. Let m and n be distinct non negative integers with m < n. Then the Fermat numbers Fn and Fm are relatively prime. Proof. From the Theorem 10.6.2 we have, Fn − 2 = F0 F1 · · · Fm · · · Fn−1 (m < n). Let us assume d is a common divisor of Fm and Fn then d\|Fm and d\|Fn − 2. Now combining together we have d\|2. This is possible only when d = 1 or d = 2. As Fm and Fn are odd then d = 2 is not possible. Thus we have d = 1. This proves the theorem. This theorem leads us to the fact that there are inﬁnitely many primes. Since gcd(Fm , Fn ) = 1 then every Fermat number Fn has a prime factor pn (say) n. Now the following theorem we have which imply pm = pn whenever m = given another way on primality testing of Fn . This theorem is called Pepin Test due to T.Pepin. Theorem 10.6.4. (Pepin’s Test):For any m ≥ 1 the Fermat number Fm = Fm −1 m 22 + 1 is prime if and only if 3 2 ≡ −1(mod Fm ) holds. Fm −1 Proof. Let us choose that the congruence, 3 2 ≡ −1(mod Fm ) holds. We are to show that Fm is prime. Now squaring both sides of the congruence, we have 3Fm −1 ≡ 1(mod Fm ) . Let us choose Fm is composite and p be any prime factor of Fm . Then we can say that, 3Fm −1 ≡ 1(mod p). Let k be the order of 3 modulo p then by the Theorem 8.2.1 we have k\|(Fm − 1). This implies that m k\|22 and k must be a power of 2. Let k = 2r for any r ≤ 2m−1 then squaring 2m−1 ≡ 1(mod p). This both sides of 3k ≡ 1(mod p) repeatedly yields that, 32 Fm −1 implies that 3 2 ≡ 1(mod p). Then from the given condition it follows that 1 ≡ −1(mod p). This shows that p = 2 which is a contradiction. Thus the m only possiblity is k = 22 = Fm − 1. Since k = Fm − 1 ≤ p − 1 and p\|Fm then we have p = Fm and consequently Fm becomes a prime. Conversely if m Fm = 22 + 1 is prime for m ≥ 1, then by the law of quadratic reciprocity we m have, ( F3m ) = ( F3m ) = ( 23 ) = −1, since Fm ≡ (−1)2 + 1 ≡ 2(mod 3). Now by Euler’s Criterion we have, ( F3m ) ≡ 3 3 Fm −1 2 Fm −1 2 (mod Fm ). Now together gives us ≡ −1(mod Fm ). This proves the theorem. 1 F1 −1 Example 10.6.1. Let us choose m = 1, then F1 = 22 + 1 = 5 and 3 2 = 32 = 9 ≡ −1(mod 5). So by Pepin’s Test we have F1 is prime. Similarly taking m = 2 by Pepin’s test we can see that F2 = 17 is prime. The following is our ﬁnal theorem of this section where we have shown an important aspect on ﬁnding the divisor of a perfect number. In the year 1747, Euler established that every prime factor of Fm is of the form k · 2n+1 + 1 then 260 Number Theory and its Applications after 100 years later in 1879 Edouard Lucas improved this result. So from there we have this following theorem. Theorem 10.6.5. For m ≥ 2, any prime divisor p of a Fermat number Fm is of the form p = n · 2m+2 + 1. m Proof. Let p be a prime divisor of Fm then, 22 ≡ −1(mod p). Now squaring m+1 both sides of the congruence we have 22 ≡ 1(mod p). Let k be the order of 2 modulo p then we have from the Theorem 8.2.1, k\|2m+1 . Here k is not of m the form 2r where 1 ≤ r ≤ m, otherwise this leads to 22 ≡ 1(mod p). This is a contradiction as 1 ≡ −1(mod p) imply p = 2. So k is of the form 2m+1 . Since order of 2 modulo p divides φ(p) = p − 1, then 2m+1 \|p − 1. Thus m ≥ 2 and 2m+1 \|p − 1 together shows that p ≡ 1(mod 8). So by Theorem(9.2.8) we p−1 have the Legendre symbol ( p2 ) = 1. Also by Euler’s Criterion, 2 2 ≡ ( p2 ) = 1( mod p). Thus from the Theorem 8.2.1 we have k = 2m+1 \| (p−1) 2 . So we obtain p = n · 2m+2 + 1 for some integer n. 3 To illustrate this theorem let us choose F3 = 22 + 1 = 257 then any prime √ factor of F3 is less than F3 . Now by last theorem any prime factor of F3 is of the form k · 25 + 1 = 32k + 1 for any integer k. As p > 1 then there is no such integer k and hence F3 is prime. Finally we conclude this section with a geometrical aspects of Fermat numm bers. A Fermat number Fm = 22 + 1 for m ≥ 1 is called a Fermat prime if m is prime. Now if we think each value of a Fermat prime as a unit square then those m−1 and unit squares can be arranged as a square whose length of each side is 22 2 an extra unit square. For example if we choose m = 2 then F2 = 22 + 1 = 17. Now here we have 17 unit squares and they can be arranged as a square of side 2−1 = 4 and an extra unit square. This has been shown in below ﬁgure. length 22 Figure 10.2: Integers of Special Forms 261 3 Similarly for F3 = 22 + 1 = 257 = 162 + 1 we have a square whose side length is 16 and an extra unit square. Here determining a Fermat number is composite or not is equivalent to arrange all the unit squares as a rectangle or not. If we can’t arrange them as a rectangle then it is prime otherwise composite number. See the ﬁgure below for F2 = 17. Figure 10.3: 10.7 Worked out Exercises n Problem 10.7.1. Show that 22 + 5 is composite for each integer n(> 0). n Solution 10.7.1. For any positive integer n, 2n is even. Then we have, 22 ≡ 1( n n mod 3). This shows that, 22 + 5 ≡ 1 + 5 ≡ 0(mod 3). Since 3 22 + 5, therefore n 22 + 5 is composite. Problem 10.7.2. Prove that every Fermat number Fn is either a prime or a pseudoprime. Solution 10.7.2. We need to verify, if Fn is composite then it is pseudoprime. n n For that, we need to prove (22 )2 +1 ≡ 2(mod Fn ). We know that, n 22 ≡ −1( mod Fn ) n ⇒ (22 )2 ⇒2 2 2n n 2n −n ≡ (−1)2 2n −n ( mod Fn ) ≡ 1( mod Fn ) ⇒ (22 )2 n +1 ≡ 2( mod Fn ). Problem 10.7.3. Taking fourth powers of congruences on 5·27 ≡ −1(mod 641) deduce that 232 + 1 ≡ 0(mod 641) and hence 641\|F5 . 262 Number Theory and its Applications Solution 10.7.3. Note that 5 · 27 ≡ −1(mod 641). Then 54 · (27 )4 ≡ (−1)4 ( mod 641) ⇒ 625 · 228 ≡ 1( mod 641) ⇒ − 16 · 228 ≡ 1( mod 641) ⇒ − 232 ≡ 1( mod 641). 5 This implies that 641\|232 + 1 = 22 + 1 = F5 . Problem 10.7.4. For n ≥ 2, show that the last digit of the Fermat number n Fn = 22 + 1 is 7. n Solution 10.7.4. First, we prove 22 ≡ 6(mod 10) for n ≥ 2. For this, apply 2 principle of mathematical induction. For n = 2, we have 22 = 24 = 16 ≡ 6( mod 10). Thus the statement is true for n = 2. Let the result be true for n = k. k+1 k To prove the result for n = k+1, let us take 22 = 24 ·22 ≡ 16·6 ≡ 6(mod 10). n So the result is true for n = k +1. Thus we have 22 +1 ≡ 7(mod 10). Therefore the unit digit is 7. Problem 10.7.5. For n ≥ 1, prove that gcd(Fn , n) = 1. Solution 10.7.5. Let us assume that d = gcd(Fn , n) and p be any prime divisor of d. Then p\|Fn and p\|n. Taking into consideration Theorem 10.6.5, p is of the form and p = k · 2n+2 + 1. As p\|n, then we have p = k · 2n+2 + 1 ≤ n. This leads to a contradiction. Thus gcd(Fn , n) = 1. Problem 10.7.6. For any odd integer n, show that 3\|2n + 1. Solution 10.7.6. Let us assume n is odd. Then it is of the form 2k + 1 for any positive integer k. This follows that 2n = 22k+1 = 2 · 4k . Claim: 2 · 4k ≡ 2( mod 3). To fulﬁll the claim, apply principle of mathematical induction. Let us take k = 1. Then 2 · 4 = 8 ≡ 2(mod 3). So the result is true for k = 1. Let the result be true for k = t. This follows 2 · 4t ≡ 2(mod 3). Let k = t + 1. Then we have 2 · 4t+1 = 4 · 2 · 4t ≡ 4 · 2 ≡ 2( mod 3). So the result is true for k = t + 1. Thus by principle of mathematical induction, the result is true for all integers n. Therefore 2 · 4k ≡ 2(mod 3) ⇒ 2 · 4k ≡ −1( mod 3). This proves 3\|(2n + 1). n Problem 10.7.7. For any Fermat number Fn = 22 + 1, establish that Fn ≡ 5( mod 9) or Fn ≡ 8(mod 9) according as n is odd or even. Integers of Special Forms 263 n Solution 10.7.7. Let Fn = 22 + 1 be the arbitrary Fermat number for n > 0. Now 2n ≡ (−1)n (mod 3) implies that if n is odd then 2n is of the form 3k + 2 where k is even and if n is even then 2n is of the form 3k + 1 for some odd integer k . If n is odd then we have, n 22 + 1 = 23k+2 + 1 ≡ 4 + 1 ≡ 5( mod 9). If n is even then we have, n 22 + 1 = 23k +1 + 1 ≡ (−1) · 2 + 1 ≡ −1 ≡ 8( mod 9). Problem 10.7.8. Use Pepin’s test to show the Fermat number 257 is prime. 3 F3 −1 Solution 10.7.8. Note that F3 = 22 + 1 = 257. So 3 2 = 3128 . Moreover, 3128 = (38 )16 ≡ 13616 ≡ 644 ≡ 2412 ≡ 256 ≡ −1(mod 257). Then by Pepin’s Test F3 = 257 is prime. Problem 10.7.9. From Pepin’s test, conclude that 3 is a primitive root of every Fermat prime. p p Solution 10.7.9. Let Fp = 22 + 1 is prime. Then φ(Fp ) = 22 . So order of 3 p k p−k ≡ 1(mod Fp ) if p ≥ k. modulo Fp is a power of 2, say 2k . Then 32 = 32 ·2 Fp −1 Fp −1 2 ≡ −1(mod Fp ). Then 3 ≡ 1(mod Fp ). So By Pepin’s Test, we have 3 order of 3 modulo Fp is φ(Fp ). This proves that 3 is the primitive root for every prime as p is arbitrary. 10.8 Exercises: 1. Verify that a perfect square cannot be a perfect number. 2. Establish the following assertions concerning k-perfect number: (a)If n is a 3-perfect number and 3 n, then 3n is 4-perfect. (b)If n is a 5-perfect number and 5 n, then 5n is 6-perfect. 3. Prove that if n = 28 is an even perfect number, then n ≡ 1 or − 1(mod 7). 4. Prove that if 2k − 1 is prime, then the sum 2k−l + 2k + 2k+l + · · · + 22k−2 will yield a perfect number. For instance, 23 −1 is prime and 22 +23 +24 = 28, which is perfect. 264 Number Theory and its Applications 5. (a) Show that any divisor of a deﬁcient or perfect number is deﬁcient. (b) Show that any multiple of an abundant or perfect number is abundant. 6. (a) Show that 16 is a superperfect number. (b) Show that for an odd prime p if n = p2 , then n is not super perfect. 7. Prove that the Mersenne number M19 is a prime; hence, the integer n = 218 (219 − 1) is perfect. 8. If m and n are an amicable pair, with m even and n odd, then n is a perfect square. 9. Show that Any odd perfect number n can be represented in the form n = pa2 , where p is a prime. 10. If n is an odd perfect number, prove that n has at least three distinct prime factors. 11. Use the Lucas-Lehmer test to determine whether the Mersenne numbers M3 and M13 are prime or not. 12. Find the smallest prime divisor q > 3 of each of the integers 229 + 1 and 241 + 1. 13. For any prime p > 3, prove that 13 (2p + 1) is not divisible by 3. 14. Establish that any Fermat prime Fn can be written as the diﬀerence of two squares, but not of two cubes. 15. Prove that if n is odd pseudoprime then the Mersenne number Mn is also pseudoprime. 11 Continued Fractions “If equations are trains threading the landscape of numbers, then no train stops at pi.” – Richard Preston 11.1 Introduction Being a natural object, continued fractions appear in many areas of Mathematics, sometimes in an unexpected way. The old name of continued fraction was “anthyphaeiretic ratios”, which the Dutch mathematician and astronomer in 1687 made the ﬁrst practical application explaining how to use convergents to ﬁnd the best rational approximations for ratios. To build a mechanical planetarium, motivated him to do so. Later on, many renowned mathematicians including Euler, Jacobi, Gauss and Cauchy get attracted to continued fractions. Continued fractions ﬁnd its applications in some areas of contemporary Mathematics. The application of continued fraction lies in the cryptography, to explain a kind of attack on the RSA system. Nowadays, there are mathematicians who continue to develop the theory on continued fractions. The sections of the chapter starts with ﬁnite continued fractions and then subsequently we move to inﬁnite and periodic continued fractions. 265 266 11.2 Number Theory and its Applications Finite Continued Fractions 137 . Applying Euclidean We commence this section with an example, by taking 33 137 Algorithm on yields 33 137 = 4 · 33 + 5 33 = 6 · 5 + 3 5=1·3+2 3 = 1 · 2 + 1. Dividing both sides of every equations by the divisor of that equation, we obtain 137 33 33 5 5 3 3 2 5 1 = 4 + 33 33 5 3 1 =6+ =6+ 5 5 3 2 1 =1+ =1+ 3 3 2 1 =1+ . 2 =4+ Combining we get, 137 5 1 =4+ =4+ . 33 33 6 + 1+ 1 1 1+ 1 2 The foregoing expression in this sequence of equations is a simple ﬁnite continued 137 . This motivates us to start with the following deﬁnition. fraction of 33 Deﬁnition 11.2.1. A ﬁnite continued fraction is an expression of the form 1 b0 + , 1 b1 + .. . + bn−1 + 1 bn where b0 , b1 , b3 , . . . , bn are real numbers with b1 , b3 , . . . , bn being positive. The real quantities b1 , b3 , . . . , bn are known to be partial quotients (also known as partial denominators) of the continued fraction. The continued fraction is simple (canonical) if the real numbers b0 , b1 , b3 , . . . , bn are all integers. Symbolically, continued fractions are denoted by [b0 ; b1 , b3 , . . . , bn ] to avoid writing out continued fractions. Continued Fractions 267 The ﬁrst theorem of the section reﬂects on the representation of every ﬁnite simple continued fractions. Later on using Euclidean Algorithm, we will work on the representation of every rational number as a ﬁnite continued fraction. Theorem 11.2.1. Every ﬁnite simple continued fraction represents a rational number. Proof. Let [b0 ; b1 , b2 , b3 , . . . , bk ] with 1 ≤ k ≤ n be b ﬁnite simple continued fraction. We proceed by mathematical induction on k. For k = 1, we have b b +1 [b0 ; b1 ] = b0 + b1 = 0 b1 , a rational number. Assume the theorem is true for 1 1 k = m. Then for 1 ≤ m ≤ n, we have [b0 ; b1 , b2 , b3 , . . . , bm ] is a rational number. Now for k = m + 1, we obtain [b0 ; b1 , b2 , b3 , . . . , bm , bm+1 ] = b0 + 1 . [b1 ; b2 , b3 , b4 , . . . , bm , bm+1 ] By induction hypothesis, [b1 ; b2 , b3 , b4 , . . . , bm , bm+1 ] is rational which implies it can be expressed in the form rs with r, s ∈ Z, s = 0 and gcd(r, s) = 1. Thus, [b0 ; b1 , b2 , b3 , . . . , bm , bm+1 ] = b0 + 1 r s = b0 + b r+s s = 0 , r r a rational number. Hence by induction method we are done. Theorem 11.2.2. Any rational number can be expressed as a ﬁnite simple continued fraction. Proof. Let x be a rational quantity. Then ∃ u, v ∈ Z with v = 0 and gcd(u, v) = 1 such that x = uv . Suppose r0 = u and r1 = v. Then an appeal to Euclidean algorithm yields the following sequence of equations viz r0 = r1 q1 + r2 , 0 < r2 < r1 . r1 = r2 q2 + r3 , 0 < r3 < r2 . r2 = r3 q2 + r4 , 0 < r4 < r3 . .. .. .=. rn−3 = rn−2 qn−2 + rn−1 , 0 < rn−1 < rn−2 . rn−2 = rn−1 qn−1 + rn , 0 < rn < rn−1 . rn−1 = rn qn . In the above equations, q2 , q3 , . . . , qn are all positive integers. Rewriting the algorithm in fractional form we get, 268 Number Theory and its Applications u r r 1 = 0 = q1 + 2 = q1 + r1 v r1 r1 r 2 r1 r 1 = q2 + 3 = q2 + r2 r2 r2 r 3 r2 r 1 = q3 + 4 = q1 + r3 r3 r3 r 4 .. .. .=. rn−3 r = qn−2 + n−1 = qn−2 + rn−2 rn−2 rn−2 r = qn−1 + n = q1 + rn−1 rn−1 rn−1 = qn . rn 1 rn−2 rn−1 1 rn−1 rn r Upon substitution, the value of r1 from the second equation into the ﬁrst of the 2 algorithm, we get u 1 = q1 + . 1 v q2 + r /r 2 By similar manner, substituting the value of previous one, we obtain 3 r2 r3 u 1 = q1 + v q2 + q + 1 3 from the third equation into the . 1 r3 /r4 Continuing this manner, we obtain 1 u = q1 + v . 1 q2 + .. . + qn−1 + 1 qn Therefore [q1 ; q2 , q3 , . . . , qn ]. This proves any rational number can be expressed as a ﬁnite simple continued fraction. In view of foregoing theorem, an interesting question to ask about uniqueness of the representation of a rational number as a ﬁnite simple continued fraction. The next theorem will answer the question. Continued Fractions 269 Theorem 11.2.3. Every rational number can be expressed exactly by two ﬁnite simple continued fraction expansions. Proof. Let η be a rational number. Then ∃ p, q ∈ Z with q = 0 and gcd(p, q) = 1 such that η = pq . In view of the Theorem 11.2.2 we can say that η can be expressed as [b1 ; b2 , · · · bn ]. Then from Euclidean algorithm taking r0 = p and r1 = q we have, r0 = r1 b1 + r2 , 0 < r2 < r1 . r1 = r2 b2 + r3 , 0 < r3 < r2 . r2 = r3 b2 + r4 , 0 < r4 < r3 . .. .. .=. rn−2 = rn−1 bn−1 + rn , 0 < rn < rn−1 . rn−1 = rn bn . The last step of Euclidean algorithm can be written as, rn−1 = (bn − 1)rn + rn . Then in this case last step will be rn = 1 · rn . Thus the expression of simple continued fraction will be [b1 ; b2 , · · · bn − 1, 1]. This proves that every rational number can be expressed exactly by two ﬁnite simple continued fraction expansions. Next, we will talk about the numbers obtained from a ﬁnite continued fraction by removing the expression at diﬀerent stages. Deﬁnition 11.2.2. The continued fraction [b0 ; b1 , b3 , . . . , bk ] where k(∈ Z+ ) ≤ n, is said to be the k–th convergent of the continued fraction [b0 ; b1 , b3 , . . . , bn ]. The k–th convergent of [b0 ; b1 , b3 , . . . , bn ] is denoted by Ck . Let b0 , b1 , . . . , bn be real numbers, with b1 , . . . , bn be positive. Let us deﬁne simple fundamental recurrence relations pk and qk (k = 0, 1, 2, 3, . . . , n) as follows: p0 = b0 q0 = 1 p1 = b1 b0 + 1 q1 = b1 pk = bk pk−1 + pk−2 qk = bk qk−1 + qk−2 where k = 2, 3, 4, . . . , n. We use this to start with a formulae for the convergents. Theorem 11.2.4. The k-th convergent Ck of the ﬁnite continued fraction p [b0 ; b1 , b2 , b3 , . . . , bn ] has the value Ck = qk , 0 ≤ k ≤ n. k 270 Number Theory and its Applications Proof. We proceed by principles of mathematical induction on k. For k = 0, 1, 2, we ﬁnd p b0 = 0, 1 q0 1 b1 b0 + 1 p = = 1, C1 = b0 + b1 b1 q1 p 1 b2 (b1 b0 + 1) + b0 C2 = b0 + = 2. 1 = b b + 1 q2 b1 + b 2 1 C0 = b0 = 2 Hence the theorem is true for k = 0, 1, 2. If possible, let the theorem be true for k = m where 2 ≤ m ≤ n. Then Cm = + pm−2 b p pm = m m−1 . qm bm qm−1 + qm−2 (11.2.1) Dependence of the real numbers pm−1 , pm−2 , qm−1 , qm−2 on the partial quotients b1 , b2 , b3 , . . . , bn allow us to replace the real number bm by bm + b 1 in the m+1 (11.2.1), to get Cm+1 = [b0 ; b1 , b2 , b3 , . . . , bm , bm+1 ] 1 = [b0 ; b1 , b2 , b3 , . . . , bm , bm + ] bm+1 bm + b 1 pm−1 + pm−2 m+1 = bm + b 1 qm−1 + qm−2 m+1 bm+1 (bm pm−1 + pm−2 ) + pm−1 bm+1 (bm qm−1 + qm−2 ) + qm−1 b p p + pm−1 = m+1 m = m+1 , bm+1 qm + qm−1 qm+1 = which is true for k = m + 1. Hence we are done. Theorem 11.2.5. If Ck is the k-th convergent of the ﬁnite continued fraction [b0 ; b1 , b2 , b3 , . . . , bn ], then for 1 ≤ k ≤ n; pk qk−1 − qk pk−1 = (−1)k−1 . Proof. We apply mathematical induction on k to prove the theorem. For k = 1, we have p1 q0 − q1 p0 = (b1 b0 + 1) · 1 − b1 b0 = 1 = (−1)1−1 . Assume the theorem is true for k = m. Then for 1 ≤ m ≤ n, we have pm qm−1 − qm pm−1 = (−1)m−1 . Continued Fractions 271 Now, pm+1 qm − qm+1 pm = (bm+1 pm + pm−1 )qm − (bm+1 qm + qm−1 )pm = −(pm qm−1 − qm pm−1 ) = −(−1)m−1 = (−1)m , so the theorem prevails for k = m+1. This completes the proof by induction. An immediate consequence of the foregoing theorem are the following corollaries. Corollary 11.2.1. For 1 ≤ k ≤ n, gcd(pk , qk ) = 1. Proof. If d = gcd(pk , qk ); then by virtue of above theorem, we obtain d (−1)k−1 . Because d > 0, this enables us to conclude d = 1. Corollary 11.2.2. For the k-th convergent of the ﬁnite simple continued fraction [b0 ; b1 , b2 , b3 , . . . , bn ], the following identities are true. 1. for 1 ≤ k ≤ n; Ck − Ck−1 = (−1)k−1 qk qk−1 . 2. for 2 ≤ k ≤ n; Ck − Ck−2 = bk (−1)k−2 qk qk−2 . Proof. 1. An immediate consequence of Theorem 11.2.5, yields Ck − Ck−1 = p pk (−1)k−1 − k−1 = . qk qk−1 qk qk−1 2. With the aid of pk = bk pk−1 + pk−2 & qk = bk qk−1 + qk−2 , we note that p pk − k−2 qk qk−2 pk qk−2 − pk−2 qk = qk qk−2 + pk−2 )qk−2 + pk−2 (bk qk−1 + qk−2 ) (b p = k k−1 qk qk−2 − pk−2 qk−1 ) b (p q = k k−1 k−2 qk qk−2 Ck − Ck−2 = = bk (−1)k−2 [by Theorem(11.2.4)]. qk qk−2 This ﬁnishes the second identity. 272 Number Theory and its Applications With the aid of last two corollaries, we have the following theorem. Theorem 11.2.6. Let Ck be the k-th convergent of the ﬁnite simple continued fraction [b0 ; b1 , b2 , b3 , . . . , bn ]. Then C1 > C3 > C5 · · · > · · · ; C0 < C2 < C4 · · · < · · · . Also for all i = 0, 1, 2, 3, 4, . . .; every even-numbered convergent C2i is less than every odd-numbered convergent C2i+1 . Proof. As the partial quotients are positive real numbers and qj > 0 ∀ j, replacing k by k + 2 in Corollary 11.2.2(2), we ﬁnd that Ck+2 − Ck = bk+2 (−1)k qk+2 qk > 0, if k is even say k = 2i; < 0, if k is odd say k = 2i − 1. Hence C1 > C3 > C5 · · · > · · · ; (11.2.2) C0 < C2 < C4 · · · < · · · . (11.2.3) For the ﬁnal part, its suﬃces to show that C2r−1 > C2s . In view of Theorem 11.2.4, we ﬁnd that Ck − Ck−1 = p pk (−1)k−1 − k−1 = . qk qk−1 qk qk−1 For k = 2i − 1, we get C2i−1 − C2i = (−1)2i−2 > 0 ⇒ C2i−1 > C2i . qk qk−1 Hence using (11.2.2) and (11.2.3), we get blending all inequalities together C2r−1 > C2s+2r−1 > C2s+2r > C2s . We conclude this section with the theorem concerning the relation associated with the denominator of the k-th convergent of the ﬁnite continued fraction. Theorem 11.2.7. If qk is the denominator of the k-th convergent Ck of the ﬁnite continued fraction [b0 ; b1 , b2 , b3 , . . . , bn ], then qk−1 ≤ qk for 1 ≤ k ≤ n; the strict inequality holds when k > 1. Continued Fractions 273 Proof. We will take the help of mathematical induction on k to establish the theorem. For k = 0, q0 = 1; k = 1, q1 = b1 and b1 ≥ 1. Thus q1 ≥ q0 . Assume, the theorem persist for k = m with 1 ≤ m < n. Then, qm−1 ≤ qm . Now qm+1 = bm+1 qm + qm−1 > bm+1 qm ≥ 1 · qm = qm . So the inequality prevails for k = m + 1. Hence by induction the theorem is established. 11.3 Worked out Exercises Problem 11.3.1. Find the rational number, expressed in lowest terms, represented by each of the following simple continued fractions: (i)[3; 7, 15, 1] (ii)[2; 1, 2, 1, 1, 4] Solution 11.3.1. (i) By virtue of Deﬁnition(11.2.1), we obtain 1 [3; 7, 15, 1] = 3 + 7+ 1 =3+ 1 7+ 15 + 1 = 1 355 . 113 16 Remark 11.3.1. [3; 7, 15, 1] is a good approximation for π. (ii) 1 [2; 1, 2, 1, 1, 4] = 2 + = 1 1+ 87 . 32 1 2+ 1+ 1 1+ 1 4 Problem 11.3.2. Find the simple continued fraction, not terminating with the 17 partial quotient of 1, of each of the following rational numbers: (i) (ii) 9 746 − . 830 Solution 11.3.2. Here we need to apply the proof of Theorem(11.2.2) to ﬁnd the simple continued fraction expansion. The tool required here is Euclidean algorithm. (i) Applying Euclidean algorithm on 17 and 9 we have, 17 = 1 · 9 + 8 9=1·8+1 8 = 8 · 1. Thus the sequence of quotients give the continued fraction expansion [1; 1, 8]. 274 Number Theory and its Applications (ii) Again applying Euclidean algorithm on −746 and 830 we have, −746 = −1 · 830 + 84 830 = 9 · 84 + 74 84 = 1 · 74 + 10 74 = 7 · 10 + 4 10 = 2 · 4 + 2 4 = 2 · 2. Thus the sequence of quotients give the continued fraction expansion [−1; 9, 1, 7, 2, 2]. Problem 11.3.3. Find the convergence of the continued fraction expansion of 75 19 . Solution 11.3.3. We need to ﬁnd the continued fraction expansion of plying Euclidean algorithm on 75 and 19 we have, 75 19 . Ap- 75 = 3 · 19 + 18 19 = 1 · 18 + 1 18 = 18 · 1. Thus the sequence of quotients generate the continued fraction [3; 1, 18]. Using the fundamental recurrence relations pk and qk (k = 0, 1, 2), we compute p0 = 3 q0 = 1 p1 = 1 · 3 + 1 = 4 q1 = 1 p2 = 18 · 4 + 3 = 75 q2 = 18 · 1 + 1 = 19. Thus the convergents are C0 = 3 1 = 3, C1 = 4 1 = 4 and C2 = 75 19 . pk = [bk ; bk−1 , · · · , b1 , b0 ] where Problem 11.3.4. Prove that if b0 > 0 then pk−1 pk−1 pk Ck−1 = qk−1 and Ck = qk , k ≥ 1 are successive convergents of the continued fraction [b0 ; b1 , · · · bn ]. Solution 11.3.4. Since Ck−1 and Ck (k ≥ 1) are successive convergents of [b0 ; b1 , · · · bn ], therefore fundamental recurrence relations yields pk = bk pk−1 + pk−2 . Using the relation successively gives, pk pk−1 Again, = bk + pk−2 = bk + pk−1 pk−1 = bk−1 + pk−2 1 pk−2 pk−3 1 pk−1 pk−2 . . Continued Fractions 275 Thus, pk pk−1 = bk + 1 bk−1 + 1 . pk−2 pk−3 Proceeding this way, we get p2 1 = b2 + p1 . p1 p0 Since p0 = b0 and p1 = b1 b0 + 1 holds, therefore all the steps we get, pk pk−1 . 1 .. pk pk−1 = b1 + b10 prevails. Piling up 1 = bk + bk−1 + Therefore p1 p0 . + b1 + 1 b0 = [bk ; bk−1 , · · · , b1 , b0 ]. Problem 11.3.5. Prove that if the simple continued fraction expression of the rational number β, β > 1, is [b0 ; b1 · · · , bn ], then the simple continued fraction expression of β1 is [0; b0 · · · , bn ]. Solution 11.3.5. Here β is a rational number then for any r, s ∈ Z with s = 0 and gcd(r, s) = 1 we can write β = rs . Since β > 1 then we have r > s. Also [b0 ; b1 , · · · , bn ] is the expression of simple continued fraction rs . Again β1 = rs < 1 implies s < r. Applying Euclidean algorithm on s and r we get s = 0 · r + s, r = b0 · s + b1 and rest of the steps are same as rs . Thus the simple continued fraction expression of β1 is [0; b0 · · · , bn ]. The following problem deals with an alternating way of ﬁnding the general solution of linear Diophantine equations using simple ﬁnite continued fractions. Problem 11.3.6. Determine the general solutions of the Diophantine equation 18x + 5y = 24. Solution 11.3.6. Note that gcd(18, 5) = 1. The given equation can be re-written as 5 24 18 x+ y = ⇒ 18x + 5y = 24. (11.3.1) 1 1 1 We proceed to solve using two steps. In ﬁrst step, we will ﬁnd the particular solution of (11.3.1) when equated to 1 i.e. 18x + 5y = 1 and thereby in the second step ﬁnd the general solution of the given equation. 276 Number Theory and its Applications 18 5 (or if one prefers ) as a simple continued fracStep I We begin by writing 5 18 tion. An appeal to Euclidean algorithm yields 18 = 3 · 5 + 3 5=1·3+2 3=1·2+1 2 = 2 · 1, 18 1 = [b0 ; b1 , b2 , b3 ] = [3; 1, 1, 2] = 3 + . Here using simple 5 1 + 1+1 1 2 fundamental recurrence relations pk , qk for k = 0, 1, 2, 3 we ﬁnd p0 = b0 = 3, p1 = b1 b0 + 1 = 4, p2 = b2 p1 + p0 = 7, p3 = b3 p2 + p1 = 18; q0 = 1, q1 = b1 = 1, q2 = b2 q1 + q0 = 2, q3 = b3 q2 + q1 = 5. The convergents of this simple ﬁnite continued fraction are given by, so that C0 = p0 p p p 18 . = 3, C1 = 1 = 4, C2 = 2 = 7, C3 = 3 = q0 q1 q2 q3 5 In view of Theorem 11.2.5(taking k = 3) we obtain p3 q2 −q3 p2 = (−1)3−1 = 1, which further yields 18(2) − 5(7) = 1. (11.3.2) Taking into consideration (11.3.1), when relation (11.3.2) is multiplied by 24 gives, 18(48) + 5(−168) = 24. Thus, a particular solution of the Diophantine (11.3.1) is x0 = 48, y0 = −168. Step II Now by virtue of Theorem 2.7.1, the general solution is given by the equations x = x0 + bt = 48 + 5t, y = y0 − at = −168 − 18t; t = 0, ±1, ±2, . . . . This completes the solution. 11.4 Inﬁnite Continued Fractions In the present section, we will study inﬁnite continued fractions and establish how to express a real quantity with the help of an inﬁnite continued fraction. We will depict the utilization of the continued fraction representation of a real number to generate rational numbers that are approximations of this real number. Continued Fractions 277 In the subsequent system, we will study inﬁnite continued fractions of quadratic irrationalities. Let us begin with an inﬁnite sequence of positive integers a0 , a1 , a2 , . . . , . . . ,. To deﬁne and to study inﬁnite continued fraction, we need few results from mathematical analysis which we have covered in prerequisites. The following theorem is of fundamental importance, based on the relation between Fibonacci sequence and inﬁnite continued fraction. To know Fibonacci sequence in details, refer to Chapter14(sec 14.1). Theorem 11.4.1. For the simple inﬁnite continued fraction [b0 ; b1 , b2 , b3 , . . . , bn ], p the inequality qk ≥ Uk (k = 1, 2, . . .) holds where Ck = k is the k-th convergent qk of the fraction and Uk denotes the k-th Fibonacci number(See Chapter 14). Proof. We establish by mathematical induction. For k = 1, 2 we have q1 = b1 ≥ 1 = U1 and q2 = b2 q1 + q0 = b2 b1 + b0 ≥ U2 respectively. So the result is true for k = 1, 2. Let us assume the result be true for k = m. Then qm ≥ Um . We are to prove the result for k = m + 1. Now qm+1 = bm+1 qm + qm−1 , ≥ bm+1 Um + Um−1 , ≥ Um + Um−1 , = Um+1 , from the deﬁnition of Fibonacci sequence. Hence we are done. The following theorem deﬁne inﬁnite continued fractions as limits of ﬁnite continued fractions. The limit η described in the statement of the theorem is called the value of the inﬁnite simple continued fraction [b0 ; b1 , b2 , b3 , . . . , bk , . . .]. Theorem 11.4.2. Let {b0 , b1 , b2 , . . .} be an inﬁnite sequence of integers with bi , i = 1, 2, 3, . . . being positive. Let Ck = [b0 ; b1 , b2 , b3 , . . . , bk ]. Then limk→∞ Ck = η. Proof. Let m ∈ Z+ . From Theorem 11.2.5, we see that C1 > C3 > C5 > · · · > Cm−1 , C0 < C2 < C4 · · · < Cm and C2j > C2k+1 whenever 2j ≤ m and 2k + 1 < m. Taking into consideration all possible values of m, we get C1 > C3 > C5 > · · · > C2n−1 > C2n+1 > · · · , C0 < C2 < C4 · · · < C2n−2 < C2n < · · · 278 Number Theory and its Applications andC2j> C 2k+1∀j,k∈Z .Sinceboththesequences {C1, C 3, C 5,· · · , C 2n−1, C 2n+1,· · · ,}and{C0, C 2, C 4· · · , C 2n−2, C 2n,· · · } are NPOPtoneandbounded,thereforetheywillconvergetoadeﬁnitelimitsaythe TFquence{C1, C 3, C 5,· · · , C 2n−1, C 2n+1,· · · ,}convergestoη1and{C0, C 2, C 4· · · , C 2n−2, C 2n,^·convergestoη2i.e. + lim C2n+1 = η1 , lim C2n = η2 . k→∞ k→∞ Claim: η1 = η2 . Using Corollary 11.2.2(2), we ﬁnd that C2n+1 − C2n = p2n+1 p (−1)(2n+1)−1 1 − 2n = = . q2n+1 q2n q2n+1 q2n q2n+1 q2n Also qk ≥ k ∀ k ∈ Z+ [for details refer to Theorem 11.4.1], yields 1 q2n+1 q2n Hence C2n+1 − C2n = s ⇒ < 1, q2n0 +1 this leads to the contradiction as (rq2n −sp2n ) cannot lie between 0 and 1. Hence with this ﬁnishes the proof. The following theorem shows the uniqueness of the inﬁnite simple continued fraction that represent the same irrational number. Theorem 11.4.4. If the two inﬁnite continued fractions [b0 ; b1 , b2 , b3 , . . . , bk , . . .] and [c0 ; c1 , c2 , c3 , . . . , ck , . . .] are equal(or represent the same irrational number), then bn = cn ∀ n ≥ 0. Proof. Let x = limn→∞ [b0 ; b1 , b2 , b3 , . . . , bk , . . .]. Then, C0 < x < C1 ⇒ b0 < x < b0 + 1 ⇒ b0 < x < (b0 + 1) ∵ b1 ≥ 1. b1 Hence [x] = b0 . Now, assume [b0 ; b1 , b2 , b3 , . . . , bk , . . .] = x = [c0 ; c1 , c2 , c3 , . . . , ck , . . .]. Then b0 + 1 1 = x = c0 + . [b1 ; b2 , b3 , b4 , . . . , bk , . . .] [c1 ; c2 , c3 , c4 , . . . , ck , . . .] (11.4.5) With the help of the reason stated in ﬁrst paragraph of the proof, we have from (11.4.1) b0 = [x] = c0 ⇒ [b1 ; b2 , b3 , b4 , . . . , bk , . . .] = [c1 ; c2 , c3 , c4 , . . . , ck , . . .]. By similar reasoning as stated, we next obtain b1 = [x] = c1 ⇒ [b2 ; b3 , b4 , b5 , . . . , bk , . . .] = [c2 ; c3 , c4 , c5 , . . . , ck , . . .]. Hence the theorem is true for k = 0, 1. Now assume bk = ck which implies [bk+1 ; bk+2 , bk+3 , bk+4 , . . . , . . .] = [ck+1 ; ck+2 , ck+3 , ck+4 , . . . , . . .]. 280 Number Theory and its Applications Applying the same argument, we see that bk+1 = ck+1 ⇒ [bk+2 ; bk+3 , bk+4 , bk+5 , . . . , . . .] = [ck+2 ; ck+3 , ck+4 , ck+5 , . . . , . . .]. Thus by mathematical induction, our task is complete. We will now show that every irrational number can be uniquely expressed by an inﬁnite simple continued fraction. Theorem 11.4.5. Let η = η0 be irrational and deﬁne the sequence {b0 , b1 , b2 , . . .} recursively by bk = [ηk ], ηk+1 = 1 , k = 0, 1, 2, 3, . . . . . . . ηk − bk Then η = limk→∞ Ck , where Ck = [b0 ; b1 , b2 , b3 , . . . , bk ]. Proof. From the given recursive deﬁnition, we ﬁnd bk is an integer for every k. We are to prove ηk is irrational for every k. We will take the help of mathematical induction on k to fulﬁll our claim. For k = 0, η0 = η is irrational. Let for k = m, ηm be irrational number. Now, ηm+1 = 1 1 ⇒ ηm = bm + . ηm − bm ηm+1 (11.4.6) If ηm+1 is rational, then ηm is also so, which contradicts the induction hypothesis. Hence ηm+1 is irrational number. Now as ηm is irrational and bk is an integer, we know that ηm = bm and bm < ηm < bm + 1(Why!), ⇒ 0 < ηm − bm < 1, 1 ⇒ ηm+1 = > 1, ηm + bm ⇒ bm+1 = [ηm+1 ] ≥ 1, for k = 0, 1, 2, 3, . . . . This shows bi ; i = 1, 2, . . . are all positive integers. On Continued Fractions 281 repetitive use of (11.4.2), we ﬁnd 1 = [b0 ; η1 ] η1 1 = b0 + 1 = [b0 ; b1 , η2 ] η = η0 = b0 + b1 + η1 2 . = .. . = .. 1 = b0 + = [b0 ; b1 , b2 , . . . , bk , ηk+1 ]. 1 b1 + 1 b2 + .. . + bk + 1 ηk+1 Now our next claim is [b0 ; b1 , b2 , . . . , bk , ηk+1 ] → η as k → ∞. By virtue of Theorem 11.2.3, we have η = [b0 ; b1 , b2 , . . . , bk , ηk+1 ] η p + pk+1 , = k+1 k ηk+1 qk + qk−1 where Cj = pj is the jth convergent of [b0 ; b1 , b2 , . . . , bk , . . .]. Hence qj ηk+1 pk + pk+1 p − k, ηk+1 qk + qk−1 qk −(pk qk−1 − pk−1 qk ) = (ηk+1 qk + qk−1 )qk η − Ck = = (−1)k [By Theorem(11.2.5)]. (ηk+1 qk + qk−1 )qk Since, ηk+1 qk + qk−1 > bk+1 qk + qk−1 = qk+1 , therefore, \|η − Ck \| 0 such that \|sη − r\| < \|qk η − pk \|, then s ≥ qk+1 . Proof. Given that \|sη −r\| < \|qk η −pk \|; to the contrary assume that 1 ≤ s ≤ qk+1 . To begin with consider the system of simultaneous equations pk x + pk+1 y = r (11.4.7) qk x + qk+1 y = s. (11.4.8) Now, (11.4.7) × qk − (11.4.8) × pk gives (pk+1 qk − pk qk+1 )y = rqk − spk . Taking into consideration Theorem(11.2.5), we ﬁnd y = (−1)k (rqk − spk ). By similar reasoning; (11.4.8) × pk+1 − (11.4.7) × qk+1 yields x = (−1)k (spk+1 − rqk+1 ). Claim: x = 0, y = 0. If x = 0, then spk+1 = rqk+1 . Since gcd(pk+1 , qk+1 ) = 1, therefore qk+1 \|s ⇒ qk+1 ≤ s, contradicts our assumption. If y = 0, then r = pk x, s = qk x[by (11.4.7) and (11.4.8)] implies \|sη − r\| = \|x\|\|qk η − pk \| ≥ \|qk η − pk \|, [∵ \|x\| ≥ 1]. This steers contradiction to the stated condition. Hence x = 0, y = 0. Finally we note that x and y are of opposite signs. Consider y < 0. Since qk x = s − qk+1 y [refer to (11.4.7)], we have x > 0 because qk x > 0 and qk > 0. When p p y > 0 we see that qk x < 0 as s < qk+1 . So x < 0. Since qk < η < qk+1 , we ﬁnd k k+1 qk η − pk and qk+1 η − pk+1 are of opposite sign. From (11.4.7) and (11.4.8), we obtain \|sη − r\| = \|(qk x + qk+1 y)η − (pk x + pk+1 y)\| = \|x(qk η − pk ) + y(qk+1 η − pk+1 )\|. Continued Fractions 283 Combining the conclusions of the previous two paragraphs, we get x(qk η − pk ) and y(qk+1 η − pk+1 ) have the same sign, so \|sη − r\| = \|x\|\|\|x(qk η − pk )\| + \|y\|\|\|x(qk+1 η − pk+1 )\| ≥ \|x\|\|x(qk η − pk )\| ≥ \|x(qk η − pk )\|, ∵ \|x\| ≥ 1. Finally, this contradicts our given fact. Hence 1 ≤ s ≤ qk+1 fails. Consequently, the proof is complete. The following well-known theorem leads to the fact that the convergents of the inﬁnite simple continued fraction of an irrational number are good approximations to η. Theorem 11.4.7. Dirichlet’s Theorem on Diophantine Approximation: If η is r an irrational number, then there are inﬁnitely many rational numbers such s r 1 that η − < 2 . s 2s r Proof. Let k be the k-th convergent of the continued fraction of η. Following sk the proof of Theorem 11.4.5, we get 1 1 η − rk < < 2 [∵, sk < sk+1 ]. sk sk sk+1 sk rk (convergents of η) k = 1, 2, 3, . . . are inﬁnitely many rational sk numbers satisfying the conditions stated in the theorem. Consequently, The concluding theorem of this section focusses on the fact that any rational number which approximates very closely an irrational number must be a convergent of the inﬁnite simple continued fraction of this number. u Theorem 11.4.8. If η is an irrational number and is a rational number with v u 1 u v > 0 such that η − < 2 , then is a convergent of the simple continued v 2v v fraction representation of η. u Proof. Assume that is not a convergent of the simple inﬁnite continued fracv p p tion expansion of η. Then ∃ successive convergents k and k+1 such that qk qk+1 qk ≤ s < qk+1 . From Theorem 11.4.6, we see that 1 u \|qk η − pk \| ≤ \|vη − u\| = v η − < . v 2v 1 p η − k < , [Dividing by qk ]. qk 2vqk 284 Number Theory and its Applications Since \|vpk − uqk \| ≥ 1(we know that vpk − uqk is non-zero integer as follows that pk u = ), it v qk 1 \|vpk − uqk \| pk u ≤ = − vqk vqk q v k p u k ≤ η − + η − qk v 1 1 < + 2. 2vqk 2v (where we have used the triangle inequality to obtain the second inequality above). Hence 1 1 < 2 ⇒ 2vqk > 2v 2 ⇒ qk > v, 2vqk 2v contradicting our assumption. 11.5 Worked out Exercises Problem 11.5.1. Solution 11.5.1. √ 1+ 5 . Find the simple inﬁnite continued fraction of 2 √ √ 1+ 5 1+ 5 . Then a0 = = 1. Let us consider η0 = 2 √2 5+1 1 2 11.4.5 we obtain η1 = = = η0 . = √ η0 − a0 2 5−1 Applying Theorem √ 1+ 5 This gives = [1; 1, 1, 1, 1, . . . , . . .]. 2 Problem 11.5.2. Prove that the ﬁrst four partial quotients of the simple inﬁnite e−1 continued fraction of are 0, 2, 6, 10. e+1 e−1 Solution 11.5.2. Suppose η0 = e−1 = 0. In view of . Then b = 0 e+1 e+1 1 e+1 e+1 Theorem(11.4.5), we get η1 = . Then, b1 = [η1 ] = e−1 = 2. = η0 − b0 e−1 e−1 1 1 = − . Consequently, b2 = [η2 ] = − Again, η2 = = e+1 η1 − b1 e−3 e−1 − 2 e−1 e−3 1 1 = 6. By similar manner, η3 = = − , b = = e−3 η2 − b2 η2 − 6 7e − 19 3 [η3 = 10]. Thus the ﬁrst four partial quotients of the simple inﬁnite continued e−1 fractions of are b0 = 0, b1 = 2, b2 = 6, b3 = 10. e+1 Problem 11.5.3. Let γ and δ be two real numbers. Then γ is equivalent to δ if ∃ integers a, b, c and d satisfying ad − bc = ±1 and γ = aδ+b cδ+d . Prove that if γ is equivalent to δ, then δ is also equivalent to γ. Continued Fractions 285 Solution 11.5.3. Since γ is equivalent to δ then we can write γ = some integers a, b, c and d with ad − bc = ±1. Now, aδ+b cδ+d for γcδ + γd = aδ + b γcδ − aδ = b − γd δ(γc − a) = b − γd δ= γd − b . γc − a Here we have (−a)d − (−b)c = bc − ad = ±1. This results in δ is equivalent to γ. Problem 11.5.4. Prove that the i-th convergent of the simple inﬁnite continued fraction of η1 is the reciprocal of the (i − 1)-th convergent of the simple inﬁnite continued fraction of η, where η(> 1) is an irrational quantity. Solution 11.5.4. Let the simple inﬁnite continued fraction of η be [b0 ; b1 , b2 , . . .]. 1 1 Then = = [0; b0 , b1 , b2 , . . .][refer to Problem(11.3.5)]. Now the η [b0 ; b1 , b2 , . . .] 1 i-th convergent of is η 1 where [0; b0 , b1 , b2 , . . . , bi−1 ]. Therefore [0; b0 , b1 , b2 , . . . , bi−1 ] = 0 + a0 + B1 1 1 . Here [b0 ; b1 , b2 , . . . bi−1 ] B = [b1 ; b2 , b3 , b4 , . . . , bi−1 ] = = [b0 ; b1 , b2 , . . . bi−1 ] b0 + B1 is the (i − 1)-th convergent of η. This proves the statement. Problem 11.5.5. Let η be an irrational number with simple inﬁnite continued fraction expansion [b0 ; b1 , b2 , . . .]. Prove that the simple continued fraction of −η is [−b0 − 1; 1, b1 − 1, b2 , b3 , . . .] if b1 > 1 and [−b0 − 1; b2 + 1, b3 , . . .] if b1 = 1. Solution 11.5.5. Here η is an irrational number with simple inﬁnite continued fraction expansion [b0 ; b1 , b2 , . . .]. Let us take B = [b2 ; b3 , b3 , b4 , . . .]. If b1 > 1, then we have [b0 ; b1 , b2 , . . .] + [−b0 − 1; 1, b1 − 1, b2 , b3 , . . .] 1 1 = b0 + + (−b0 − 1) + b1 + B1 1 + (b 1−1) + B1 1 B − b0 − 1 + = b0 + b1 B + 1 1+ = b B−B+1 B −1+ 1 b1 B + 1 b1 B + 1 = 0. 1 B b1 B−B+1 286 Number Theory and its Applications If b1 = 1, then we need to consider B = [b3 ; b4 , b5 , b6 , . . . , ]. Thus we have, [b0 ; b1 , b2 , . . .] + [−b0 − 1; b2 + 1, b2 , b3 , . . .] 1 1 = b0 + + (−b0 − 1) + 1 b1 + b + 1 (b2 + 1) + B1 2 = b0 + = B 1 1+ B b2 B+1 −1+ B b2 B + B + 1 B b2 B + 1 −1+ b2 B + 1 + B b2 B + B + 1 = 0. 11.6 Periodic Fractions In continuation with example(11.4.1), we ﬁnd b5 = b1 , b6 = b2 and so on. The process iterates repeatedly, which makes the sequence of partial quotients periodic. So this example motivates us to study the notion of periodic inﬁnite simple continued fractions. In the present section, we will show that the necessary and suﬃcient condition for an inﬁnite continued fraction to be periodic is that the real number represented by the continued fraction is a quadratic irrational. So let us begin with a deﬁnition. Deﬁnition 11.6.1. An inﬁnite simple continued fraction [b0 ; b1 , b2 , . . . , . . .] is said to be periodic if ∃ positive integers N and k such that bn = bn+k for all positive integers n with n ≥ N . Symbolically, periodic fraction are represented by [b0 ; b1 , b2 , bN −1 , bN , bN +1 , . . . , bN +k−1 ]. It means periodic inﬁnite simple continued fraction are expressed as [b0 ; b1 , b2 , bN −1 , bN , bN +1 , . . . , bN +k−1 , bN , bN +1 , . . .]. To characterize irrational numbers with periodic inﬁnite simple continued fractions, we need the following deﬁnition. Deﬁnition 11.6.2. Quadratic Irrationalities: A real number η is said to be a quadratic irrational if η is irrational and if η is a root of a quadratic equation with integral coeﬃcients given by Ax2 + Bx + C = 0, where A, B, C are integers. Before we proceed with important results, let us start with an important theorem about quadratic irrationalities. Theorem 11.6.1. A real number η is said to be a quadratic irrational if and only √ a+ b if ∃ integers a, b, c such that b is not a perfect square and satisﬁes η = c where b > 0 and c = 0. Continued Fractions 287 Proof. Let us assume the real number η to be a quadratic irrational. Then η is irrational and also ∃ integers A, B, C such that η√is a root of the quadratic −B ± B 2 − 4AC equation Ax2 + Bx + C = 0. Hence η = with A = 0. 2A 2 2 Since η is real, therefore B − 4AC > 0. Also B − 4AC can not be a perfect square[Why!]. Considering the values of a, b, c as follows: a = b, b = B 2 − 4AC, C = −2A or a = −B, b = B 2 − 4AC, c = 2A, √ a+ b we obtain η = where b > 0 and c = 0. c For the converse part, assume that for some integers √ a, b, c with b > 0 and a+ b c= 0 such that b is not a perfect square and η = . Then η is irrational. c Also, √ √ a+ b η= ⇒ b = cη − a. c Squaring, we get b = (cη − a)2 ⇒ c2 η 2 − 2acη + (a2 − b) = 0. This implies A = c2 , B = −2ac, C = a2 − b. Hence η is a quadratic irrational. This completes the proof. An immediate consequence of Theorem 11.6.1 leads to a corollary of importance. Corollary 11.6.1. Let η be a quadratic irrational. Then ∃ integers a, b and c 0, b(> 0) is not a perfect square and c (b − a2 ). with c = √ a+ b Proof. We have η = . Then multiplying numerator and denominator by c√ 2 a\|c\| + bc \|c\| we obtain η = 0 . Here a\|c\|, c\|c\| & bc2 are all integers with c\|c\| = c\|c\| 2 2 0. Also bc > 0 as b > as c = √ 0 and bc is not a√perfect square as b is not so. a+ b a\|c\| + bc2 with η = gives Hence comparing η = c\|c\| c b − a2 = bc2 − a2 c2 = c2 (b − a2 ) ⇒ c (b − a2 ). This completes the proof. The following theorem can be applied when periodic simple continued fractions represent quadratic irrationalities. Theorem 11.6.2. If η be a quadratic irrational and if r, s, t, u are integers, rη + s then is either rational or a quadratic irrational. tη + u 288 Number Theory and its Applications Proof. Since η is a quadratic irrational, the last lemma ensures the existence of √ a+ b integers a, b, c such that b is not a perfect square and satisﬁes η = with c 0. On simpliﬁcation, we get b > 0 and c = √ rη + s (ar + cs) + r b √ = tη + u (at + cu) + t b √ [(ar + cs)(at + cu) − rtb] + [r(at + cu) − t(ar + cs)] b . = (at + cu)2 − t2 b Hence if the coeﬃcients of b vanishes, then rη + s becomes rational, otherwise tη + u quadratic irrational. For our future discussions of simple continued fractions of quadratic irrationalities, we will need the notion of the conjugate of a quadratic irrationality, whose deﬁnition is as follows. √ a+ b Deﬁnition 11.6.3. Let η = , where b > 0 and c = 0, be a quadratic c √ a− b . irrational. Then the conjugate of η, denoted by, η¯ is deﬁned by η¯ = c Theorem 11.6.3. If the quadratic irrational η is a root of the polynomial Ax2 + Bx + C = 0, where A, B, C are integers, then the other root of the polynomial is η¯, the conjugate of η. Proof. Obvious. The following theorem deals with few properties involving quadratic irrationalities. √ √ a + b1 d a + b2 d Theorem 11.6.4. If η1 = 1 and η2 = 2 are quadratic irrac1 c2 tionals, then the following statements are true: 1. (η1 + η2 ) = η¯1 + η¯2 . 2. (η1 − η2 ) = η¯1 − η¯2 . 3. η1 η2 = η1 η2 . 4. η1 η2 = η1 . η2 Proof. Left to the reader. Continued Fractions 289 The central theorem about periodic simple continued fractions is commonly known as Lagrange’s theorem. Although, Euler proved a part of the theorem. In 1737 Euler proved that a periodic inﬁnite simple continued fraction represents a quadratic irrationality. Later in 1770, Joseph-Louis Lagrange showed that a quadratic irrationality has a periodic continued fraction. To prove the converse part of Lagrange’s theorem, that the simple continued fraction of a quadratic irrationality is periodic, we need the following theorem for obtaining the continued fraction of a quadratic irrational. Theorem 11.6.5. Let √ η be a quadratic irrational. Then ∃ integers A0 , B0 & d A0 + d with B0 = 0, d > 0 and d is not a perfect square with such that η = B0 √ Ak + d 2 , bk = [ηk ], bk+1 = bk Bk − B0 (d − A0 ). Deﬁne recursively, ηk = Bk 2 d − Ak+1 , k = 0, 1, 2, 3, . . .. Then η = [b0 ; b1 , b2 , b3 , . . .]. Ak & Bk+1 = Bk Proof. Taking aid of mathematical induction, we will prove that Ak , Bk ∈ Z with Bk = 0 and Bk (d − A2k ) for k = 0, 1, 2, 3, 4, . . .. For k = 0, the assertion follows from the given condition of the theorem. Assume the statement is true for k = m. Then Am , Bm ∈ Z with Bm = 0 and Bm (d − A2m ) holds. We are to show that the statement is true for k = m + 1. Now Am+1 = bm Bm − Am ∈ Z. Further, Bm+1 = d − A2m+1 = 1 {d − (bm Bm − Am )2 } Bm Bm 1 2 = {d − b2m Bm − A2m + 2bm Am Bm } Bm d − A2m = + (2bm Am − b2m Bm ) Bm d − A2m = + bm (2Am − bm Bm ). Bm (11.6.1) By induction hypothesis, Bm (d − A2m ). Hence by (11.6.1) Bm+1 ∈ Z. Since d d − A2m+1 is not a perfect square, we get d = A2m ⇒ Bm+1 = 0. Since Bm = Bm+1 therefore, Bm+1 (d − A2m+1 ). For the concluding part of the theorem, its suﬃces 290 Number Theory and its Applications 1 for k = 0, 1, 2, 3, . . .. Now ηk − bk √ Ak + d η k − bk = − bk B √ k d − (bk Bk − Ak ) = Bk √ d − A2k+1 B d − Ak+1 1 √ = = = . = √ k+1 Bk ηk+1 Bk ( d + Ak+1 ) d + Ak+1 to show that ηk+1 = Hence from the foregoing equation the desired result follows. Theorem 11.6.6. Lagrange’s Theorem: Any periodic simple inﬁnite continued fraction of an irrational number is a quadratic irrational, and vice-versa. Remark 11.6.1. It is to be noted that this theorem is diﬀerent from Lagrange’s theorem on polynomial congruences discussed in Chapter 8. Proof. Let η be the irrational number and the simple inﬁnite continued fraction of it be periodic. We are to prove η is quadratic irrational. Then η = [b0 ; b1 , b2 , b3 , . . . , bN −1 , bN , bN +1 , . . . , bN +k ]. Let β = [bN ; bN +1 , . . . , bN +k−1 , bN +k ] = [bN ; bN +1 , . . . , bN +k−1 , bN +k , β]. Then from Theorem 11.2.3 it follows that β= βpk + pk−1 , βqk + qk−1 where pk = lim [bN ; bN +1 , . . . , bN +k−1 , bN +k ] qk k→∞ pk−1 = lim [bN ; bN +1 , . . . , bN +k−1 , bN +k ]. qk−1 k→∞ Here we note that β is irrational[Why!] and in view of (11.6.2), we ﬁnd qk β 2 + (qk−1 − pk )β − pk−1 = 0. Consequently, β is quadratic irrational. Further, η = [b0 ; b1 , b2 , b3 , . . . , bN −1 , β]. (11.6.2) Continued Fractions 291 βpN −1 + pN −2 p p where N −1 and N −2 are convergent of βqN −1 + qN −2 qN −1 qN −2 [b0 ; b1 , b2 , b3 , . . . , bN −1 ]. Taking into consideration Theorem 11.6.2, η is quadratic irrational as β is so. Hence we are done. For the converse part, √ assume η to be a quadratic irrational. Then by TheA0 + d orem 11.6.5 η = . Furthermore, we have η = [b0 ; b1 , b2 , b3 , . . . , ] where B0 √ d − A2k+1 A + d , bk = [ηk ], Ak+1 = bk Bk − Ak+1 , Bk+1 = , k = ηk = k Bk Bk+1 0, 1, 2, 3, 4, . . .. Since η = [b0 ; b1 , b2 , b3 , . . . , ηk ] therefore, by virtue of Theorem 11.2.3 we ﬁnd A η + Ak−2 . Taking conjugates on both sides, we obtain η = k−1 k Bk−1 ηk + Bk−2 Then η = η= Solving for ηk yields, Ak−1 ηk + Ak−2 [refer to Theorem 11.6.4]. Bk−1 ηk + Bk−2 ⎡ Bk−2 η − ⎣ ηk = − Bk−1 η − Since η = lim k→∞ Ak−2 Bk−2 Ak−1 Bk−1 (11.6.3) ⎤ ⎦. Ak−2 A = lim k−1 Bk−2 k→∞ Bk−1 therefore, lim k→∞ η− Ak−2 Bk−2 η− Ak−1 Bk−1 = 1. Hence by deﬁnition of convergence ∃ N ∈ Z such that ηk < 0 ∀ k ≥ N . Since ηk > 0 for k ≥ 1, we get √ √ √ Ak + d Ak − d 2 d ηk − ηk = − = > 0. Bk Bk Bk So Bk > 0 ∀ k ≥ N . Since Bk Bk+1 = d − A2k+1 , we see that for k ≥ N 0 ≤ Bk ≤ Bk Bk+1 = d − A2k+1 ≤ d. (11.6.4) Also k ≥ n yields A2k+1 ≤ d = A2k+1 − Bk Bk+1 , which gives − d < Ak+1 0, then p qk = [bk ; bk−1 , · · · , b2 , b1 ] where Ck−1 = k−1 qk−1 qk−1 pk , k ≥ 1 are successive convergents of the continued fraction qk [b0 ; b1 , · · · bn ]. and Ck = 5. Using continued fraction, determine the general solutions of the Diophantine equation 364x + 227y = 1. √ √ 6. Find the simple inﬁnite continued fractions of the followings: (i) 2 (ii) 5. 7. Determine the ﬁrst four partial quotients of the following simple inﬁnite e2 − 1 . continued fractions : (i) 2π (ii) 2 e +1 8. Prove that a real number r is equivalent to itself. 9. Prove that if r1 , r2 and r3 are real numbers such that r1 and r2 are equivalent and r2 and r3 are equivalent, then r1 and r3 are equivalent. √ √ 10. Find the simple continued fractions of the followings: (i) 37 (ii) 209 √ √ 13 − 2 (iii)1 + 2 (iv) . 7 11. Determine the quadratic irrationality of the followings: (i) [5; 10] (ii)[1; 2, 3] (iii)[2; 1, 5]. 12. For an integer m with m ≥ 2, prove that the simple continued fraction of √ m2 − m is [m − 1; 2, 2m − 2]. 13. Prove that for any positive integer m the simple continued fraction of √ √ m2 + 2 is [m; m, 2m]. Hence ﬁnd 51. 14. Show that for any positive odd integer m(> 3) the simple continued frac√ m−3 tion of m2 − 4 is [m − 1; 1, m−3 2 , 2, 2 , 1, 2m − 2]. √ 15. Find the period length of the simple continued fraction of m2 + n, where m and n are integers, n > 1 and n\|2m. 12 Few Non-Linear Diophantine Equations “I count Maxwell and Einstein, Eddington and Dirac, among “real” mathematicians. The great modern achievements of applied mathematics have been in relativity and quantum mechanics, and these subjects are at present at any rate, almost as “useless” as the theory of numbers.” – Godfrey Harold Hardy 12.1 Introduction Diophantine equations are those equations restricted to only integral solutions. In chapter2 (Section 2.7), we have studied only linear Diophantine equations. There we discussed how the integral solutions can be calculated. But what about non-linear Diophantine equations? Unlike linear Diophantine equations, in general there is no method for solving all non-linear Diophantine equations. However, few non-linear Diophantine equations as well as their certain families can be solved using many results. The present chapter addresses few non-linear Diophantine equations which are solvable. In the ﬁrst section, we consider the Diophantine equations of the form x2 + y 2 = z 2 , where x, y, z forms the length of the side of the right triangle. A triple (x, y, z) of integers that solve this equation is called Pythagorean Triples. Subsequently, we will focus on the famous Diophantine equation xn + y n = z n where n > 2. Fermat stated that there any solutions of this Diophantine equation, a statement commonly known as 295 296 Number Theory and its Applications Fermat’s Last theorem. The proof of the statement was one of the greatest challenges of mathematics. After a long span of time, in 1995 English mathematician Sir Andrew John Wiles gave the ﬁrst proof of the statement. The proof is beyond the scope of the book. 12.2 Pythagorean Triples Deﬁnition 12.2.1. A Pythagorean triple(also known as Pythagorean triangle) is a set of three integers a, b, c satisfying a2 + b2 = c2 and is said to be primitive if gcd(a, b, c) = 1. For instance if we take 5, 12, 13 where 52 + 122 = 132 , then 5, 12, 13 are said to form Pythagorean triple. Similarly, 3, 4, 5 and 6, 8, 10 are also so. Here the ﬁrst two triples 5, 12, 13 and 3, 4, 5 are primitive but the last one is not as gcd(6, 8, 10) = 2. For a Pythagorean triple 6, 8, 10, if we divide each element of the triple by 2, then the triple becomes 3, 4, 5 which is both primitive and Pythagorean. In general, for a Pythagorean triple a, b, c where gcd(a, b, c) = d, ∃ integers a , b , c where a = da , b = db , c = dc with gcd(a , b , c ) = 1. This triple a , b , c also form a Pythagorean triple as a2 + b2 = c2 a2 + b2 = = c2 . d2 d2 Hence a , b , c forms a primitive Pythagorean triple. Here the key idea is very simple, all Pythagorean triples can be found by forming integral multiples of primitive Pythagorean triples. Next, the following two lemmas will pave the way for ﬁnding all primitive Pythagorean triples. Lemma 12.2.1. If a, b, c is a primitive Pythagorean triple, then they are mutually prime to each other. Proof. Its suﬃces to show that, for any primitive Pythagorean triple a, b, c; gcd(a, b) = gcd(a, c) = gcd(b, c) = 1. To begin with, consider gcd(a, b) > 1. Then ∃ a prime p such that p gcd(a, b). This implies p\|a and p\|b. Hence p\|(a2 + b2 ). As p\|c2 then p\|c, otherwise for some integer k and r with 0 < r < k; c = pk + r ⇒ c2 = p(p2 k 2 + 2kr) + r2 . This contradicts p c2 . So p\|a, p\|b, p\|c altogether gives gcd(a, b, c) > 1. This is impossible as a, b, c is a primitive Pythagorean triple. Therefore gcd(a, b) = 1. Similarly, gcd(a, c) = gcd(b, c) = 1. Few Non-Linear Diophantine Equations 297 Taking advantage of the last lemma, we will proceed for the next lemma. But before that, let us state and prove a simple but signiﬁcant result related to congruence. Result 12.2.1. Square of any integer is either congruent to 0 or 1 modulo 4. Proof. Consider n ∈ Z. If n is even, then n = 2k for some integer k. On squaring we obtain n2 = 4k 2 ≡ 0( mod 4). If n is odd, then n = 2k + 1 for some integer k . Proceeding as above, we obtain n2 = 4k 2 + 4k + 1 ≡ 1( mod 4). Lemma 12.2.2. Let a, b, c be three positive integers with gcd(a, b) = 1 and ab = c2 . Then ∃ m, n ∈ Z such that a = m2 and b = n2 . Proof. For a = 1 or b = 1, the lemma is obvious. So let us begin by taking a > 1 and b > 1. Let a = pk1 1 pk2 2 · · · pkr r , b = q1j1 q2j2 · · · qsjs , be the prime factorizations of a and b respectively. Since gcd(a, b) = 1, therefore px s are same as qy for x = 1, 2, 3, . . . , r and y = 1, 2, 3, . . . , s. Again 2t ab = c2 ⇒ pk1 1 pk2 2 · · · pkr r q1j1 q2j2 · · · qsjs = z12t1 z22t2 · · · zl l , 2t 2t 2t where c = z1 1 z2 2 · · · zl l be the prime factorization of c. Now we appeal to Fundamental theorem of Arithmetic to obtain the fact that the prime powers on both sides of the equation will be same. So each px is same as zω for some ω = 1, 2, 3, . . . , l with coordinated exponents kx = 2tω . In similar manner, each qy is same as zω with coordinated exponents jy = 2tω . Blending them, yields k1 k2 kr j1 j2 js a = m2 and b = n2 where m = p12 p22 · · · pr 2 and n = q12 q22 · · · qs2 . Hence we are done. Making use of Lemma 12.2.2, we come up to the characterization of Pythagorean triple. Theorem 12.2.1. The positive integers a, b, c with 2 b form a primitive Pythagorean triple if and only if there exist integers x and y such that a = x2 − y 2 , b = 2xy, c = x2 + y 2 where x > y and both cannot be even or odd simultaneously. 298 Number Theory and its Applications Proof. Let a, b, c be a Pythagorean triple. Because b is even, both a and c are odd. Then a + c and a − c are even, which implies ∃ integers r, s(> 0) such that a−c a+c and s = . Also, r= 2 2 a2 + b2 = c2 ⇒ b2 = c2 − a2 ⇒ 2 c−a b c+a . = 2 2 2 Here we note that gcd(r, s) = 1. If not, let gcd(r, s) = d. Then d\|r, d\|s together yields d (r + s) = a and d (r − s) = c. Thus d gcd(a, c) = 1, which is impossible. In view of Lemma 12.2.2, ∃ integers x, y such that r = x2 and s = y 2 . Hence a, b, c can be written in terms of x, y as a = r − s = x2 − y 2 √ b = 4rs = 2xy (12.2.2) c = r + s = x2 + y 2 . (12.2.3) (12.2.1) Claim: gcd(x, y) = 1. To the contrary, assume d = gcd(x, y). Then d \|x and d \|y. This renders d a = x2 − y 2 , d b = 2xy, d x2 + y 2 ; which contradicts gcd(a, b, c) = 1. Putting all discussions into together, we conclude gcd(x, y) = 1. The ﬁnal touch is to consider both x, y to be odd integers. An appeal to (12.2.1) bring forth all the integers a, b, c to be even. This results in contradiction to gcd(a, b, c) = 1. Since gcd(x, y) = 1, so x, y cannot be even or odd simultaneously. For the converse part, let us begin with a2 + b2 = (x2 − y 2 )2 + 4x2 y 2 = (x2 + y 2 )2 = c2 , which shows a, b, c form a Pythagorean triple. For the primitive part, assume gcd(a, b, c) = d(> 1). Then ∃ a prime p such that p gcd(a, b, c). Here p = 2, otherwise a becomes even which is not possible as a = x2 − y 2 and both x, y cannot be even or odd simultaneously. Also, p\|a and p\|c together gives p (a + c) = 2x2 and p (a − c) = 2y 2 . Hence p\|x, p\|y contradicts gcd(x, y) = 1. Thus our assumption, gcd(a, b, c) = d fails. Hence gcd(a, b, c) = 1 and a, b, c forms a Pythagorean triple. Let us run through a speciﬁc example to exemplify the theorem lucidly. Example 12.2.1. Consider x = 7 and y = 4. Here gcd(7, 4) = 1. Following Few Non-Linear Diophantine Equations 299 the above theorem, we get a = x2 − y 2 = 33 b = 2xy = 56 c = x2 + y 2 = 65, is a primitive Pythagorean triple. Taking advantage of the previous theorem, our last theorem leads to an interesting geometric fact related to Pythagorean triple (Pythagorean triangle). Theorem 12.2.2. The radius of the inscribed circle of a Pythagorean triple (Pythagorean triangle) is always an integer. Proof. Geometrically, Pythagorean triple is a right angle triangle whose sides are of integral length. Let r denotes the radius of the inscribed circle. Here the area of the triangle is ab = r(a + b + c). Now, a2 + b2 = c2 and as the triple is not primitive then for any integer k > 0 a = k(x2 − y 2 ), b = 2kxy, c = k(x2 + y 2 ), where x, y are positive integers. On substitution, we obtain r = k(x−y)y. Hence the proof is established. 12.3 Worked out Exercises Problem 12.3.1. Find all primitive Pythagorean triples a, b, c with b = 40. Solution 12.3.1. From Theorem 12.2.1 the form of Pythagorean triple we get, a = x2 − y 2 , b = 2xy, c = x2 + y 2 . Since b = 40 and gcd(x, y) = 1 with 2 (x + y) (Why!), therefore we have either x = 5, y = 4 or x = 20, y = 1. So the triples are 9, 40, 41 and 399, 40, 401. Problem 12.3.2. Prove that if a, b, c is a Pythagorean triple, then exactly one of a, b and c is divisible by 4. Solution 12.3.2. If a, b, c is a Pythagorean triple, then a = x2 − y 2 , b = 2xy, c = x2 + y 2 . where either x is even or y. Then xy must be even. This shows that 2\|xy ⇒ 4\|2xy = b. This proves our assertion. 300 Number Theory and its Applications Problem 12.3.3. Let x1 = 3, y1 = 4, z1 = 5 and xn , yn , zn for n = 2, 3, 4, . . . be deﬁned recursively by, xn+1 = 3xn + 2zn + 1 yn+1 = 3xn + 2zn + 2 zn+1 = 4xn + 3zn + 2. Show that xn , yn , zn is a Pythagorean triple. Solution 12.3.3. In anticipation of mathematical induction, the basis step is x21 + y12 = 32 + 42 = 52 = z12 . This shows that x1 , y1 , z1 is a Pythagorean triple. Let us assume that xn , yn , zn is a Pythagorean triple. Now, 2 x2n+1 + yn+1 = (3xn + 2zn + 1)2 + (3xn + 2zn + 2)2 = 18x2n + 8zn2 + 24xn zn + 18xn + 12zn + 5 = (16x2n + 9zn2 + 24xn zn + 16xn + 12zn + 4) + (2x2n − zn2 + 2xn + 1) 2 = zn+1 + {(x2n + 2xn + 1) + (x2n − zn2 )} 2 = zn+1 + {(xn + 1)2 − yn2 } 2 = zn+1 [∵ xn + 1 = yn ]. So the result is true for all integer n ≥ 1. Problem 12.3.4. Find all solutions in positive integers of the Diophantine equation x2 + 2y 2 = z 2 . Solution 12.3.4. Here 2y 2 = z 2 − x2 = (z + x)(z − x). If we consider x as even z−x number, then z is also even. This implies x+z 2 and 2 both are integers. If gcd( z−x 2 , z + x) = 1, then by Lemma 12.2.2 there exists integers m and n 2 2 such that z−x 2 = m and z + x = n . Thus solving we get, x= m2 − 2n2 n2 + 2m2 , y = mn, z = . 2 2 z+x , z − x) = 1 then by the same argument we Similarly, if we consider gcd( 2 get 2m2 − n2 2m2 + n2 x= , y = mn, z = . 2 2 Problem 12.3.5. Prove that if a, b, c is a primitive Pythagorean triple in which b and c are consecutive positive integers, then a = 2y + 1, b = 2y(y + 1), c = 2y(y + 1) + 1 for some y > 0. Few Non-Linear Diophantine Equations 301 Solution 12.3.5. Here the Pythagorean triple are such that b + 1 = c. Now a = x2 − y 2 , b = 2xy, c = x2 + y 2 . This implies 2xy + 1 = x2 + y 2 (x − y)2 = 1 x = y + 1. This shows that the Pythagorean triple is of the form a = (y + 1)2 − y 2 = 2y + 1, b = 2y(y + 1), c = 2y(y + 1) + 1. Problem 12.3.6. Show that there exists inﬁnitely many primitive Pythagorean triple a, b, c whose even member b is a perfect square. Solution 12.3.6. Let us consider a triple as a = n4 − 4, b = 4n2 , c = n4 + 4. It is clear that gcd(a, b, c) = 1. Now we are to verify these triples a, b, c as a Pythagorean triple. Here a2 + b2 = (n4 − 4)2 + 16n4 = n8 − 8n4 + 16 + 16n4 = (n4 + 4)2 = c2 . As n is arbitrary integer, then there are inﬁnite number of Pythagorean triple of these type. Problem 12.3.7. Find all Pythagorean triples containing the integer 12. Solution 12.3.7. Note that we need to ﬁnd all primitive triples containing a divisor of 12; 2, 3, 4, 6, 12. The triple is of the form a = x2 − y 2 , b = 2xy, c = x2 + y 2 , and gcd(x, y) = 1. Now here b is even. Thus if b = 2xy = 2, then x = y = 1 and a = 0 which is not possible. If b = 2xy = 4, then x = 2, y = 1 which shows a = 3, c = 5. If b = 2xy = 6, then x = 3, y = 1 and they are not of opposite parity. If b = 2xy = 12, then either x = 6, y = 1 which shows a = 35, c = 37 or x = 3, y = 2 showing a = 5, c = 13. 3 as 9 can not be written as sum of two squares. If a = 3 = x2 −y 2 = Here c = (x + y)(x − y), then x = 2, y = 1, b = 4, and c = 5. Therefore the Pythagorean triples containing 12 are; (9, 12, 15), (35, 12, 37), (5, 12, 13) and (16, 12, 20). Problem 12.3.8. For an arbitrary positive integer n, show that there exists a Pythagorean triangle the radius of whose inscribed circle is n. 302 Number Theory and its Applications Solution 12.3.8. Let n be an arbitrary positive integer and r be the radius of m + n − p . the circle inscribed in the triangle having sides m , n , p . Then r = 2 2 2 2 2 2 Here if m = 2n + 1, n = 2n + 2n, p = 2n + 2n + 1, then m + n = p forms a Pythagorean triangle and r = n. 12.4 Fermat’s Last Theorem In the foregoing section, we dealt with the solvability of Diophantine equations of the form a2 + b2 = c2 where a, b, c form a primitive Pythagorean triple. Consider the question, whether the solvability of the last equation can be preserved if we change the exponents of a, b, c by an integer greater than 2? Fermat has stated a conjecture, commonly known to be Fermat’s Last theorem, that it is impossible to write a cube as a sum of two cubes and in general any exponent as a sum of two similar exponents. Many Mathematicians had given futile eﬀorts to prove the assertion. Statement 12.4.1. Fermat’s Last Theorem: The Diophantine equation an + bn = cn has no solution in non-zero integers a, b, c where n is an integer ≥ 3. Our future discussions will be based on a special case when n = 4, although it is the time for all positive integers m with 3 ≤ m ≤ 12500. The technique used for the proof is known as ‘method of inﬁnite descent’ introduced by Fermat. By virtue of the method we can establish that the Diophantine equation a4 +b4 = c2 has no solution. This leads to the stronger aspect than showing Fermat’s Last theorem for n = 4, since any solution of a4 + b4 = c4 = (c2 )2 gives a solution of a4 + b4 = c2 . Theorem 12.4.1. The Diophantine equation a4 + b4 = c2 has no non-zero solutions for the positive integers a, b, c. Proof. To the contrary, suppose a0 , b0 , c0 be a positive solutions of a4 + b4 = c2 . Here we note that gcd(a0 , b0 ) = 1, for if gcd(a0 , b0 ) = 1 then gcd(a0 , b0 ) = d(say). This implies a0 = da1 , b0 = db2 for some positive integers a1 and b1 with gcd(a1 , b1 ) = 1. As a0 , b0 , c0 satisﬁes a4 + b4 = c2 , we can write a40 + b40 = c20 or (a20 )2 + (b20 )2 = (c0 )2 so that a20 , b20 , c0 is a primitive Pythagorean triple. In view of Theorem 12.2.1, we ﬁnd a20 = s2 − t2 b20 = 2st c0 = s2 + t2 . Few Non-Linear Diophantine Equations 303 where s, t are positive integer with s > t and both s and t can not be even or odd simultaneously. Then a20 + t2 = s2 . Consider t be even. Then s is odd. It follows for some positive integer r, t = 2r. Now t20 = 2st = 4sr 2 b or 20 = sr . Again, since gcd(s, t) = 1, then gcd(s, r) = 1. taking into consideration Lemma 12.2.2, we ﬁnd s = c21 , r = y12 for some positive integer c1 and y1 . Now gcd(s, t) = 1 ⇒ gcd(s, a0 , t) = 1. This yields s, a0 , t to be Pythagorean triple. Now, by virtue of Theorem 12.2.1 we get a20 = u2 − v 2 , t = 2uv, s = u2 + v 2 , where gcd(u, v) = 1, u > v > 0. Then uv = 2t = r = y12 . Employing Lemma 12.2.2, produces u = a21 and v = b21 . Substitute these values in u2 + v 2 = s to obtain a41 + b41 = c21 . Here, c1 < c41 = s2 < s2 + t2 = c0 . Continuing this manner inﬁnite number of times, we obtain monotone decreasing sequence {c0 , c1 , c2 , c3 , . . . , . . . , . . .}. Since ∃ only a ﬁnite number of positive integers less than c0 , then this leads to a contradiction, which makes our assumption wrong. Thus the Diophantine equation a4 + b4 = c2 has no non-zero solutions for the positive integers a, b, c. Taking advantage of the theorem, we have the following corollary. Corollary 12.4.1. The equation a4 +b4 = c4 has no solution in positive integers. Theorem 12.4.2. The area of a Pythagorean triangle can never be equal to a perfect square. Proof. Consider a Pythagorean triangle with sides a, b, c satisﬁes a2 + b2 = c2 . The area of the triangle is 12 ab. Let the area be a perfect square. Then for some integer x, we have 12 ab = x2 . It follows 2ab = 4x2 . On addition and subtraction with a2 + b2 = c2 , we obtain (a + b)2 = c2 + 4x2 , (a − b)2 = c2 − 4x2 . Therefore (a2 − b2 )2 = c4 − (4x2 )2 = c4 − (2x)4 . In view of last remark, it follows that any solution of the last equation, which makes our task complete. 12.5 Worked out Exercises Problem 12.5.1. Show that the equation xp + y p = z p has no solutions in non-zero integers, p being an odd prime. 304 Number Theory and its Applications Solution 12.5.1. Let n(≥ 3) ∈ Z. By Fermat’s theorem, xn + y n = z n has no solution for integers x, y, z. If n has an odd prime factor then n = pk for some k ∈ Z+ . Then we have, (xk )p + (y k )p = (z k )p . Consequently, xp + y p = z p has no solution in non-zero integers. If n has no odd factor then n is some power of 2. Since n > 2, 4\|n ⇒ n = 4k for some k ∈ Z+ . Thus (xk )4 + (y k )4 = (z k )4 has no solution. Problem 12.5.2. Show that the Diophantine equation x4 − y 4 = z 2 has no solutions in non-zero integers. Solution 12.5.2. Rewriting the Diophantine equation x4 − y 4 = z 2 as x4 = z 2 + y 4 , we get by Theorem 12.2.1 z = 2st, y 2 = s2 − t2 , x2 = s2 + t2 . Therefore s4 − t4 = x2 y 2 = (xy)2 is a smaller solution as 0 < s < Now, y 2 = 2st, z = s2 − t2 , x2 = s2 + t2 , √ s2 + t2 = x. where s is even and t being odd. Since gcd(s, t) = gcd(2s, t) = 1, in view of Lemma 12.2.2 we obtain 2s = ω 2 , t = v 2 . As ω is even so it takes the form ω = 2u, s = 2u2 . Hence x2 = 4u4 + v 4 . Also, this implies that 2u2 , v 2 , x forms a primitive Pythagorean triple. Again an appeal to Theorem 12.2.1 for a > b > 0, we ﬁnd 2u2 = 2ab, v 2 = a2 − b2 , x = a2 + b2 where gcd(a, b) = 1. Thus u2 = ab ensures a = c2 , b = d2 [refer to Lemma 12.2.2]. Hence v 2 = a2 − b2 = c4 − d4 , which is also a smaller solution [Why!]. In consequence, we arrive at a contradiction in both the cases. Hence we are done. Problem 12.5.3. Use Fermat’s Little theorem to verify that if p is prime and xp + y p = z p holds, then p (x + y − z). Solution 12.5.3. By Fermat’s Little Theorem, for any two integers x, y we have xp ≡ x(mod p) and y p ≡ y(mod p). Then from equation xp + y p = z p we have xp + y p ≡ (x + y)( mod p) ⇒ z p ≡ (x + y)( mod p) ⇒ z ≡ (x + y)( mod p), as z p ≡ z( mod p) ⇒ p (x + y − z). Problem 12.5.4. Show that the Diophantine equation x2 +y 2 = z 3 has inﬁnitely integral solutions by showing that for each positive integer k, x = 3k 2 − 1, y = k(k 2 − 3), z = k 2 + 1, form a solution. Few Non-Linear Diophantine Equations 305 Solution 12.5.4. Here x2 + y 2 = (3k 2 − 1)2 [k(k 2 − 3)]2 = 9k 4 − 6k 2 + 1 + k 2 (k 4 − 6k 2 + 9) = (k 2 + 1)3 = z 3 . Thus the Diophantine equation x2 + y 2 = z 3 has inﬁnitely integral solutions. Problem 12.5.5. In a Pythagorean triple x, y, z prove that not more than one of x, y or z can be a perfect square. Solution 12.5.5. To the contrary, let us begin with the Pythagorean triple x, y, z such that more than one of x, y or z can be a perfect square. As x, y, z forms Pythagorean triple, it follows x2 + y 2 = z 2 . Now, two cases may arise: Case I If x = m2 , y = n2 for ∀ m, n ∈ Z, then x4 + y 4 = z 2 has a solution. But from Theorem 12.4.1, no solution exists of this form. Hence the assumption is not possible. Case II If x = m2 , y = z 2 for ∀ m, n ∈ Z, then x4 + y 4 = z 2 has a solution, which is not possible by Problem 12.5.2. So our assumption fails for both the cases. Problem 12.5.6. Verify that the only solution in relatively prime positive integers of the equation x4 − y 4 = 2z 2 is x = y = z = 1. Solution 12.5.6. The given Diophantine equation is x4 − y 4 = 2z 2 . (12.5.1) Squaring both sides of (12.5.1), we get (x4 − y 4 )2 = 4z 4 or (x4 − y 4 )2 + 4x4 y 4 = 4z 4 or 4z 4 − 4x4 y 4 = (x4 − y 4 )2 2 4 x − y4 or z 4 − x4 y 4 = . 2 (12.5.2) From Problem 12.5.2, we can assert that (12.5.2) has no non-zero solution. Thus at least one of z, xy, x ± y is non-zero. Moreover, (12.5.1) and (12.5.2) have the same solutions. If x = y = 0, then x4 = y 4 . Now from (12.5.1) we get, y 4 = x4 = z 2 . Thus the only possibility is x = y = z = 1. Proceeding as above, in other cases too we can ﬁnd x = y = z = 1. 306 Number Theory and its Applications Problem 12.5.7. Show that the equation positive integers. Solution 12.5.7. Let the equation sides by (xyz)4 we obtain 1 x4 + 1 y4 1 1 1 + 4 = 2 has no solution in 4 x y z = 1 z2 be solvable. Multiplying both (yz)4 + (xz)4 = (x2 y 2 z)2 . (12.5.3) It is of the form a4 + b4 = c2 . In view of Theorem 12.4.1, we can say that the (12.5.3) has no non-zero solution for positive integers yz, xz, x2 y 2 z. Consequently, our assumption proved to be wrong. This ﬁnishes our result. 12.6 Exercises: 1. Find all Pythagorean triples (x, y, z) with z ≤ 40. 2. Obtain all primitive Pythagorean triples x, y, z in which x = 60. 3. Show that if (x, y, z) is a primitive Pythagorean triple then, exactly one of x, y and z is divisible by 5. 4. Prove that in a primitive Pythagorean triple x, y, z the product xy is divisible by 12, hence 60 \| xyz. 5. Find all solutions in positive integers of the diophantine equation x2 +3y 2 = z2. 6. Prove that if x, y, z is a primitive Pythagorean triple in which the diﬀerence z − y = 2, then x = 2t, y = t2 − 1, z = t2 + 1 for some t > 1. 7. Show that there exist inﬁnitely many Pythagorean triples x, y, z in which x and y are consecutive triangular numbers. Exhibit three of these. 8. Find formulas for the integers of all Pythagorean triples (x, y, z) with z = y + 1. 13 Integers as Sums of Squares “Arithmetic is numbers you squeeze from your head to your hand to your pencil to your paper till you get the answer.” – Carl Sandburg 13.1 Introduction Sum of squares of integers is the sum of two squares, three squares, four squares or n squares. In arithmetic operations we have sum of n terms. But there are many techniques for the calculation of sum of squares. In statistics and basic algebra, we have various results using sum of squares. From beginning many famous mathematicians like Euler, Lagrange and Gauss have contributed a lot to develop the concept of sum of squares. The basic problem they have encountered with is that, whether any integer can be expressed as sum of squares or not. Obviously, we have shown that some numbers cannot be represented as sums of two squares; some numbers cannot be represented as sums of three squares, and all numbers can be represented as sums of four squares. Along the way, we also prove that numbers satisfying certain conditions can be represented as sums of two squares. 13.2 Sum of Two Squares Throughout history, Mathematicians have been interested to ﬁnd nature of numbers in many ways; mainly by addition and multiplication. Then, is it possible 307 308 Number Theory and its Applications that these two operations be used together in the form of sum of squares? For instance, 25 can be expressed as 32 + 42 but if we consider 11 then it cannot be written as the sum of two squares. Our future discussions will be based on the nature and properties of those integers which can be written as sum of two squares. To begin with, we will go for necessary and suﬃcient condition for which a positive integer can be expressed as sum of two squares. Our ﬁrst theorem is the leading step towards this journey. Theorem 13.2.1. If p and q are two numbers such that each can be represented as the sum of two squares, then so is their product pq. Proof. Let p = a2 + b2 and q = c2 + d2 be the representation of p and q respectively, where a, b, c, d are integers. Now we have pq = (a2 + b2 )(c2 + d2 ) = (ac + bd)2 + (ac − bd)2 , where ac + bd and ac − bd are integers. Hence the proof. For further illustration if we choose p = 5 = 22 + 12 and q = 10 = 32 + 12 then we get, pq = 50 = 49 + 1 = 72 + 12 . In this example we have seen that the prime number 5 has the sum of two squares representation. But this is not true for all primes. In fact 3 = c2 + d2 has no integral solution. This leads us to our next theorem. Theorem 13.2.2. Any prime p satisfying p ≡ 3(mod 4) can not be represented as a sum of two squares. Proof. Suppose to the contrary, p can be written as p = a2 + b2 for some integers a and b. Now for a ≡ 0, 1, 2 or 3(mod 4) we have a2 ≡ 0 or 1(mod 4). The same congruence relation holds for b. Adding both we get, a2 + b2 ≡ 0, 1 or 2(mod 4). This contradicts the fact that p ≡ 3(mod 4). Thus p can not be represented as sum of two squares and we are done. The proof of the above theorem employs the fact that any prime which is congruent to 1 modulo 4 can be expressed as sum of two squares. We are now going to prove this statement. But to prove this we need to prove a wonderful lemma of congruence due to Norwegian Mathematician Alex Thue. In this lemma we have applied an important principle of mathematics namely Pigeon-hole principle. Lemma 13.2.1. (Thue) Let p be any prime and a be an integer with gcd(a, p) = √ √ 1. Then there exists integers x and y with 0 < \|x\| < p and 0 < \|y\| < p so that x ≡ ay(mod p). Integers as Sums of Squares 309 Proof. Let k = [ p] be an integer such that k ≤ p < (k + 1) . The number of pairs for which 0 ≤ x ≤ k and 0 ≤ y ≤ k is (k + 1)2 . Here (k + 1)2 is greater than p so by Pigeon-hole principle there exists two pairs for which x1 − ay1 ≡ x2 − ay2 ( mod p). This implies x1 − x2 ≡ ay1 − ay2 (mod p). Let x = x1 − x2 and y = y1 − y2 , then we get x ≡ ay(mod p). Here we need to show that 0 < \|x\| < k and 0 < \|y\| < k for x, y = 0. Let us consider one of them is zero then the other is also zero. If both x and y are zero then the pairs (x1 , y1 ) and (x2 , y2 ) are same. But this is not the case. Thus none of x or y is zero. This proves the lemma. 2 2 Now we are in a position to prove the following theorem. Theorem 13.2.3. (Fermat) An odd prime p is expressible as a sum of two squares if and only if p ≡ 1(mod 4). Proof. Let us suppose that p can be written as sum of two squares, p = a2 + b2 . As p is prime so p a and p b. Now from theory of linear congruence there exists an integer c such that bc ≡ 1(mod p). Now multiplying both sides of p = a2 + b2 by c2 and taking modulo p we have, (ac)2 ≡ −1( mod p). Here −1 is the quadratic residue of p. So by Legendre symbol we have, ( −1 p )=1 and applying Theorem 9.2.4 we get, p ≡ 1(mod 4). For the converse part, let us assume that p ≡ 1(mod 4). An appeal to Theorem 9.2.4 gives x2 ≡ −1(mod p) for an integer x. In view of Lemma 13.2.1 we can assert that for such an integer x there are integers a and b such that √ √ 0 < \|a\| < p, 0 < \|b\| < p so that, a ≡ xb( mod p) ⇒ a2 ≡ x2 b2 ( mod p) ⇒ a2 ≡ −b2 ( mod p) ⇒ a2 + b2 ≡ 0( mod p). This shows that p (a2 + b2 ) ⇒ p ≤ a2 + b2 . But a2 + b2 < p + p = 2p. Together we have a2 + b2 = p. From the last theorem it is clear that any prime p which is of the form 4k + 1, k being any integer, can be expressed as sum of two squares. For instance, if p = 13 then it can be written as 13 = 22 + 32 . Is the representation for p = 13 unique? We answer this below. 310 Number Theory and its Applications Integers as Sums of Squares 311 2 2 of two squares. Thus k is of the form, k = r + s where r and s are integers. Therefore n = m2 k = (mr)2 + (ms)2 . Conversely, let n be represented as sum of two squares n = m2 k = a2 + b2 . Also let p be a prime factor of k. We are to show p is not of the form 4t + 3. Let d = gcd(a, b) then a = dk1 , b = dk2 where gcd(k1 , k2 ) = 1. Thus we have d2 (k12 + k22 ) = m2 k. 2 As k is square-free so d2 \|m2 , then we can write, k12 + k22 = ( m d ) k = tk for some integer t. This shows that k12 + k22 ≡ 0( mod p). As gcd(k1 , k2 ) = 1, then one of k1 and k2 is prime to p. Let k1 be prime to p, then there exists k1 such that k1 k1 ≡ 1(mod p). Therefore (k1 k1 )2 + (k2 k1 )2 ≡ 0( mod p) ⇒ (k2 k1 )2 ≡ −1( mod p). This shows −1 is quadratic residue of p and (− p1 ) = 1. Now applying Theorem 9.2.4 we get p ≡ 1(mod 4). As p is arbitrary, this proves that k has no prime factor of the form 4t + 3 for any integer t(≥ 0). Example 13.2.1. We are now going to illustrate the above theorem lucidly by some examples. Let us choose n = 245 = 72 · 5 = 72 (22 + 12 ) = 142 + 72 , which can be expressed as sum of two squares. Here the prime factor 7 of the form 4t + 3 is of even power. But if we consider this prime factor 7 as of odd power then it can not be expressed as sum of two squares. For example let n = 1715 = 73 · 5 and this can not be written as sum of two squares. From this above discussion yields the following corollary. Corollary 13.2.2. Any positive integer n can be represented as sum of two squares if and only if each of its prime factors of the form 4t + 3 are of even powers. Finally we conclude the present section with the following theorem on representing an integer as diﬀerence of two squares. Theorem 13.2.5. Any positive integer n can be represented as diﬀerence of two squares if and only if n is not of the form 4t + 2 for some integer t. 312 Number Theory and its Applications Proof. Suppose n be represented as diﬀerence of two squares. Then n = a2 − b2 for some integers a and b. Here for any integer a we have, a2 ≡ 0 or 1(mod 4). This follows that, n = a2 − b2 ≡ 0, 1 or 3( mod 4). So n can not be of the form 4t + 2 for any integer t. Conversely, let n be not of the form 4t+2, t ∈ Z. Then n ≡ 0, 1 or 3(mod 4). If n ≡ 1 or 3(mod 4), then n + 1 and n − 1 are both even integers. Thus n can be expressed as 2 2 n+1 n−1 n= − . 2 2 Also if n ≡ 0(mod 4) then we have n= n 2 n 2 +1 − −1 . 4 4 This shows that n can be expressed as diﬀerence of two squares. To illustrate this theorem, let us take n = 15 and it can be expressed as 15 = 82 − 72 . But n = 14 can not be expressed as the diﬀerence of two squares. Taking advantage of the treatment of Theorem 13.2.5 yields an obvious corollary. Corollary 13.2.3. An odd prime is the diﬀerence of squares of two successive numbers. 13.3 Worked out Exercises Problem 13.3.1. Represent each of the primes 113, 229 and 373 as a sum of two squares. Solution 13.3.1. Note that 113 ≡ 1(mod 4), 229 ≡ 1(mod 4) and 373 ≡ 1( mod 4). Thus the given odd primes are expressible in the form 4t + 1. Applying Theorem 13.2.3 we see that these numbers can be expressed as sum of two squares. The expressions are, 113 = 82 + 72 , 229 = 152 + 22 and 373 = 182 + 72 . Problem 13.3.2. Show that a prime p can be written as sum of two squares if and only if the congruence x2 + 1 ≡ 0(mod p) admits a solution. Solution 13.3.2. Let p = a2 + b2 be the prime number where a, b are positive integers. Then a2 + b2 ≡ 0(mod p) with b ≡ 0(mod p). On dividing both sides of the congruence by b2 yields, a 2 + 1 ≡ 0( mod p). b Integers as Sums of Squares This shows that x Conversely, let x is a quadratic residue of Theorem 9.2.4, we sum of two squares. 313 is the solution of x + 1 ≡ 0(mod p). + 1 ≡ 0(mod p) admits a solution. This implies that −1 modulo p. By Legendre symbol we have ( −1 p ) = 1. In view get p ≡ 1(mod 4). This proves that p can be expressed as = ab 2 2 Problem 13.3.3. Prove that the positive integer n as many representations as the sum of two squares as does the integer 2n. Solution 13.3.3. Let n be a positive integer which can be represented as n = a2 + b2 for any integer a and b. For this representation of n we get, 2n = (a + b)2 + (a − b)2 where both a + b and a − b are integers. From these two representations of n and 2n, we can say that n has as many as representations as the sum of two squares as does 2n and vice versa. Problem 13.3.4. Show that of any four consecutive integers, at least one of them is not represented as a sum of two squares. Solution 13.3.4. Let n be an integer which can be written as sum of two squares. Then n is of the form n = a2 + b2 for any integers a and b. Here both a2 and b2 are congruent to 0 or 1 modulo 4. This shows that a2 + b2 is congruent to 0, 1 or 2 modulo 4. If n ≡ 0(mod 4), then n + 3 ≡ 3(mod 4). If n ≡ 1(mod 4), then n + 2 ≡ 3(mod 4). If n ≡ 2(mod 4), then n + 1 ≡ 3(mod 4). Thus for all these cases there is at least one integer among n, n+1, n+2, n+3 which is congruent to 3 modulo 4 and that integer can not be written as sum of two squares. Problem 13.3.5. If a positive integer n is not the sum of squares of two integers then prove that n can not be represented as sum of two squares of rational numbers. Solution 13.3.5. As n can not be written as sum of two squares, it has a prime factor p of the form 4t + 3 for any integer t(≥ 0). Further, referring to Corollary 13.2.2, we can say that p occurs in odd powers. This shows pk \|n for some odd a 2 c 2 integer k. If n can be expressed as n = + for some integers a, b, c, d b d 2 2 2 with b, d = 0, we get n(bd) = (ad) + (cd) . Here in the left hand side, p occurs to be an odd power but in the right hand side it happens to be an even power [by Corollary 13.2.2]. This is not possible. Thus n can not be written as sum of two squares of rational numbers. 314 Number Theory and its Applications Problem 13.3.6. A positive integer is representable as the diﬀerence of two squares if and only if it is the product of two factors that are both even or both odd. Solution 13.3.6. Let n be the positive integer, which can be represented as diﬀerence of two squares. Thus we have, n = a2 − b2 = (a + b)(a − b). Let a + b = 2m + 1 and a − b = 2m . This implies a = b= 2(m − m ) + 1 . Therefore 2 2(m + m ) + 1 and 2 n = a2 − b2 1 = [4(m + m )2 + 4(m + m ) + 1 − 4(m − m )2 − 4(m − m ) − 1] 4 = 8m (2m + 1). Comparing we get a − b = 8m , which is not true. This asserts that both a + b and a − b are either even or odd. Conversely, let n = m1 m2 where both m1 and m2 are even say m1 = 2k1 and m2 = 2k2 . Then we get 1 [(m1 + m2 )2 − (m1 − m2 )2 ] 4 = [(k1 + k2 )2 − (k1 − k2 )2 ]. n= If both m1 and m2 are odd, say m1 = 2k1 + 1 and m2 = 2k2 + 1, then we get 1 [(m1 + m2 )2 − (m1 − m2 )2 ] 4 = [{2(k1 + k2 ) + 2}2 − 4(k1 − k2 )2 ] n= = [(k1 + k2 + 1)2 − (k1 − k2 )2 ]. This shows that in both the cases the integer n can be expressed as diﬀerence of two squares. Problem 13.3.7. Verify that 45 is the smallest positive integer admitting three distinct representations as the diﬀerence of two squares. Solution 13.3.7. Considering the Problem(13.3.6), we can say that both the factors a + b and a − b are either even or odd, for any integer n = a2 − b2 . Here n = 45 can be written as 1 × 45 or 3 × 15 or 5 × 9. Thus the distinct representations of 45, as the diﬀerence of two squares are n = 232 − 222 or 92 − 62 or 72 − 22 . Integers as Sums of Squares 315 Problem 13.3.8. Express the integers 650 and 1450 as sum of two squares. Solution 13.3.8. Note that 13 = 32 + 22 , 29 = 52 + 22 and 50 = 72 + 12 . Now 650 = 13 · 50 = (32 + 22 )(72 + 12 ) = (3 · 1 + 2 · 7)2 + (3 · 7 − 2 · 1)2 [Why!] = 172 + 192 . 1450 = 29 · 50 = (52 + 22 )(72 + 12 ) = (5 · 1 + 2 · 7)2 + (5 · 7 − 2 · 1)2 [Why!] = 192 + 332 . 13.4 Sum of More than Two Squares The foregoing section describes the aspects of those integers which can be written as sum of two squares. But there exist few integers which can not be expressed as sum of two squares. For instance, if we consider 26 then 26 = 12 + 32 + 42 . The present section deals with those types of integers which can be expressed as sum of three squares or more than that. We started with the study of the existence of integers which can be expressed as sum of three squares of integers. We pave the way with a signiﬁcant lemma. Lemma 13.4.1. An integer congruent to 7 modulo 8 cannot be represented as sum of three squares of integers. Proof. Suppose that an integer N of the form N ≡ 7(mod 8) can be expressed as N = x2 + y 2 + z 2 where x, y, z ∈ Z. We note that square of any integer is congruent to 0, 1 or 4 modulo 8. Applying these two assertions we can say that N is congruent to 0, 1, 2, 3, 4, 5 or 6 modulo 8. This contradicts our assumption and hence we are done. Taking the lemma into consideration, we can prove the following theorem. Theorem 13.4.1. The positive integers of the form 4a (8b+7), where a and b are non negative integers, cannot be expressed as sum of three squares of integers. Proof. Here the positive integer N is of the form 4a (8b + 7), where a and b are non negative integers. Let us consider a ≥ 1 and N = x2 + y 2 + z 2 . Then N ≡ 0( 316 Number Theory and its Applications mod 4) implies x, y and z must be even. Putting x = 2x1 , y = 2y1 and z = 2z1 we get, 4a−1 (8b + 7) = x21 + y12 + z12 for some integers x1 , y1 and z1 . Again, consider a − 1 ≥ 1. Then proceeding as above and reducing the exponent of 4 we obtain a = 0. This concludes that N ≡ 7(mod 8). Thus N can be expressed as sum of three squares of integers. But this contradicts the Lemma 13.4.1. This completes the proof. The foregoing theorem highlights the fact that there are integers which can not be written as sum of three squares. For instance, if we choose N = 28 then we see that it can not be written as sum of three squares. But 28 can be written as 28 = 52 + 12 + 12 + 12 . Here 28 has been written as sum of four squares. A natural question is to ask whether every integer can be written as sum of four squares or not. We settle this below. Theorem 13.4.2. If p and q be two positive integers which can be expressed as sum of four squares, then their product is also so. Proof. Let p = p21 + p22 + p23 + p24 and q = q12 + q22 + q32 + q42 . Now we can have pq = (p21 + p22 + p23 + p24 )(q12 + q22 + q32 + q42 ) = (p1 q1 + p2 q2 + p3 q3 + p4 q4 )2 + (p1 q2 − q1 p2 + p3 q4 − p4 q3 )2 + (p1 q3 − q1 p3 + p4 q2 − p2 q4 )2 + (p1 q4 − p4 q1 + p2 q3 − p3 q2 )2 . Here the above identity can be veriﬁed by simple calculation. This shows pq can also be written as sum of four squares. Consider p = 12 and q = 13. Then 12 = 32 + 12 + 12 + 12 and 13 = 32 + 22 + 02 + 02 . Here 12 · 13 = (32 + 12 + 12 + 12 )(32 + 22 + 02 + 02 ) = (9 + 2)2 + (6 − 3)2 + (2 − 3)2 + (−3 − 2)2 = 121 + 9 + 1 + 25 = 156. This establishes the fact behind the above theorem. We are now going to conclude this section by proving the fact that any integer can be expressed as sum of four squares of integers. To prove this assertion, we need some lemmas.The fundamental theorem of arithmetic says that every integer greater than 1 can be expressed as a product of powers of distinct primes. Combining this with the assertion of Theorem 13.4.2, if we can show that any Integers as Sums of Squares 317 prime can be written as sum of four squares, then we are done. Here for the prime p = 2, we have 2 = 12 + 12 + 02 + 02 . So the last assertion is trivially true. To fulﬁll the purpose for odd primes, let us state and prove the following lemmas. Lemma 13.4.2. If p is an odd prime, then there exists an integer m < p such that 1 + x2 + y 2 = mp, 0 < m < p, for some integers x and y with 0 ≤ x ≤ p 2 and 0 ≤ y ≤ p respectively. 2 2 Proof. Let us consider x from the set {0, 1, 2, · · · , p−1 2 }. Then x have all of 2 2 diﬀerent congruences modulo p. If not, let us consider x1 ≡ x2 (mod p), then p\|(x1 − x2 )(x1 + x2 ) ⇒ x1 ≡ ±x2 (mod p). This is not possible. p−1 , the numbers −1 − y 2 are all Applying the same idea for y = 0, 1, 2, · · · 2 p+1 incongruent modulo p. Thus we have two sets of numbers of incongruent 2 modulo p for x and y respectively. But in totality there are p + 1 number of elements in these two sets where only p number of elements are residue modulo + p − 1, for p. This shows that there exists at least one element from 0, 1, 2, · · · 2 2 2 which x is congruent to −1 − y modulo p. Hence x2 ≡ −1 − y 2 ( mod p) ⇒ 1 + x2 + y 2 ≡ mp. p 2 p 2 and y 2 < together implies 2 2 Now x2 1. If m is even then three possible cases may arise: all the integers x, y, z, w are even; all are odd; any two of them are even and other two are odd. For all those possibilities we can arrange them as x ≡ y(mod 2) and z ≡ w(mod 2). It x−y x+y z−w z+w , , , are all integers and follows that 2 2 2 2 2 2 2 2 m x+y x−y z+w z−w p. + + + = 2 2 2 2 2 This contradicts the minimality of m. If m is odd then there exists integers α, β, γ, δ such that x ≡ α(mod m), y ≡ m β(mod m), z ≡ γ(mod m) and w ≡ δ(mod m) with −( m 2 ) < α, β, γ, δ < 2 . Now we have α2 + β 2 + γ 2 + δ 2 ≡ x2 + y 2 + z 2 + w2 ( mod m) and this implies, α2 + β 2 + γ 2 + δ 2 = tm for some integer t and 0 ≤ α2 + β 2 + 2 γ 2 + δ2 < 4 m = m2 ⇒ 0 ≤ t < m. 2 Here t > 0, since for t = 0 we have α = β = γ = δ = 0 and consequently x ≡ y ≡ z ≡ w ≡ 0(mod m). This shows that m2 \| mp which is not possible as 1 < m < p. Again by Theorem 13.4.2 we have m2 tp = (tm)(mp) = (α2 + β 2 + γ 2 + δ 2 )(x2 + y 2 + z 2 + w2 ). This shows tp = A2 + B 2 + C 2 + D2 where (xα + yβ + zγ + wδ) m (xβ − yα + zδ − wγ) B= m (xγ − yδ − zα + wβ) C= m (xδ + yγ − zβ − wα) D= . m A= Integers as Sums of Squares 319 Again this contradicts the choice of m. Thus m must be 1. This proves the lemma. We are now in a position to state the fundamental theorem about representation of any integer as sum of four squares. The proof follows from the Theorem 13.4.2, Lemma 13.4.3 and Lemma 13.4.4 together with fundamental theorem of arithmetic. Theorem 13.4.3. Every positive integer is the sum of squares of four integers. 13.5 Worked out Exercises Problem 13.5.1. If p = q12 + q22 + q32 where p, q12 , q22 , q32 are all primes, show that some qi = 3 for i = 1, 2, 3. Solution 13.5.1. Observe that p, q1 , q2 , q3 are primes. Then each qi is either congruent to 0 or 1 modulo 3. If all qi congruent 1 modulo 3, then p ≡ 1 + 1 + 1( mod 3). Thus p ≡ 0(mod 3) and since p is prime, p = 3 is the only possibility. Note that 3 = 12 + 12 + 12 implies qi = 1 for i = 1, 2, 3. This is not possible as qi ’s are primes. So atleast one of qi ’s is congruent to 0 modulo 3. Since qi ’s are prime numbers, hence the result follows. Problem 13.5.2. For a given positive integer show that n or 2n is a sum of three squares. Solution 13.5.2. If n is sum of three squares, then there is nothing to prove. Let n cannot be expressed as sum of three squares. Then it is of the form 4a (8b + 7) for some non negative integers a and b. Now 2n = 2 · 4a (8b + 7) and if 2n cannot be written as sum of three squares, then it is of the form 4e (8d + 7) for some non negative integers c and d. Comparing we get 2 · 4a = 4e , which is impossible. Thus 2n is not of the form 4e (8d + 7) and hence it can be expressed as sum of three squares. Problem 13.5.3. Represent the integers 90 and 23 as the sum of three squares. Solution 13.5.3. Since 90 ≡ 2(mod 8), therefore from Lemma 13.4.1 it follows that 90 can be expressed as sum of three squares. In fact we have 90 = 82 +52 +12 . Again, 23 ≡ 7(mod 8). In view of Lemma 13.4.1, we can say that 23 cannot be expressed as sum of three squares of integers. Problem 13.5.4. Verify that every positive odd integer is of the form a2 + b2 + 2c2 , where a, b, c are integers. 320 Number Theory and its Applications Solution 13.5.4. For any positive integer n, let 2n + 1 be the required positive integer. Now 4n + 2 is not of the form 4a (8b + 7) for some positive integers a and b. So by virtue of Theorem 13.4.1, we can say that 4n + 2 can be expressed as sum of three squares. Let 4n + 2 = x2 + y 2 + z 2 , for some integers x, y and z. Here exactly two amongst x2 , y 2 , z 2 are congruent to 1 modulo 4. Let us consider x, y be two odd integers and z be even. Now x + y 2 x − y 2 z 2 2n + 1 = + +2 . 2 2 2 Thus 2n + 1 = a2 + b2 + 2c2 , where a = x+y 2 , b= x−y 2 and c = z 2 are integers. Problem 13.5.5. Prove or disprove that the sum of two integers each representable as the sum of three squares of integers is also representable as sum of three squares. Solution 13.5.5. In general, the statement is not true. Choose 4 = 22 + 02 + 02 and 3 = 12 + 12 + 12 . Then their sum 3 + 4 = 7 can be written as 7 ≡ 7(mod 8). From Lemma 13.4.1, we can say that 7 cannot be expressed as sum of three squares. Problem 13.5.6. Express 21 and 89 as the sum of four squares. Solution 13.5.6. Here 21 = 3×7 and both 3 and 7 are odd primes. By virtue of Lemma 13.4.4, both 3 and 7 can be expressed as sum of four squares. Moreover, from Theorem 13.4.2 we have 21 = 3 × 7 and it can be expressed as sum of four squares. The expression is 21 = 42 + 22 + 12 + 02 . Furthermore, 89 is an odd prime. Then from Lemma 13.4.4, we can assert that 89 can also expressed as sum of four squares and it is of the form, 89 = 92 + 22 + 22 + 02 . Problem 13.5.7. Write 105 = 7 × 15 and 510 = 15 × 34 as the sum of four squares. Solution 13.5.7. Here 15 = 32 + 22 + 12 + 12 , 7 = 22 + 12 + 12 + 12 and 34 = 42 + 42 + 12 + 12 . Taking help of proof of the Theorem 13.4.2 we get, 105 = 15 × 7 = (32 + 22 + 12 + 12 )(22 + 12 + 12 + 12 ) = (2 · 3 + 2 · 1 + 1 · 1 + 1 · 1)2 + (3 · 1 − 2 · 2 + 1 · 1 − 1 · 1)2 + (3 · 1 − 2 · 1 + 1 · 1 − 2 · 1)2 + (3 · 1 − 2 · 1 + 2 · 1 − 1 · 1)2 = 102 + 22 + 12 + 02 . Similarly, we can ﬁnd the expression for 510 = 15 × 34. Integers as Sums of Squares 321 Problem 13.5.8. Show that every integer n, n ≥ 170, is the sum of the squares of ﬁve positive integers. Solution 13.5.8. Note that any integer can be expressed as sum of four squares. Then choose m = n − 169, which can be expressed as sum of four squares. Let m = x2 + y 2 + z 2 + w2 for some integers x, y, z and w. If x, y, z are 0, then n = w2 + 169 = w2 + 102 + 82 + 22 + 12 . If x, y are 0, then n = z 2 + w2 + 169 = z 2 + w2 + 122 + 42 + 32 . If x is 0, then n = y 2 + z 2 + w2 + 169 = y 2 + z 2 + w2 + 122 + 52 . If none are 0, then n = x2 + y 2 + z 2 + w2 + 169 = x2 + y 2 + z 2 + w2 + 132 . Problem 13.5.9. A prime p can be represented as the diﬀerence of two cubes if and only if it is of the form p = 3k(k + 1) + 1, for some k. Hence express 7 as the diﬀerence of two cubes. Solution 13.5.9. Let p be a prime which can be expressed as diﬀerence of two cubes. Then we have p = a3 − b3 = (a − b)(a2 + ab + b2 ). Since p is prime, therefore either a − b = 1 or a2 + ab + b2 = 1. If a2 + ab + b2 = 1, then a = 1 and b = −1 gives p = 2 = 13 − (−1)3 which is not of the form 3k(k + 1) + 1. If a − b = 1, then a = 1 + b which gives p = (1 + b)3 − b3 = 3b(b + 1) + 1 for any integer b. Thus p is of the form 3k(k + 1) + 1 for any integer k. Conversely, if p = 3k(k + 1) + 1, then p = (k + 1)3 − k 3 for any integer k. Thus p can be expressed as diﬀerence of two cubes. For the second part, pick k = 1. Then p = 3·1(1+1)+1 = 7 can be expressed as 7 = 23 − 13 . 13.6 Exercises: 1. Show that each of the integers 2n , where n = 1, 2, 3, · · · , can be expressed as sum of two squares. n 2. Prove that every Fermat number Fn = 22 + 1, where n ≥ 1, can be expressed as the sum of two squares. 3. Express 101 and 490 as the sum of two squares. 4. If n is a triangular number, show that each of the three successive integers 8n2 , 8n2 + 1, 8n2 + 2 can be written as a sum of two squares. 322 Number Theory and its Applications 5. Prove that a positive even integer can be written as the diﬀerence of two squares if and only if it is divisible by 4. 6. Express 11 and 28 as the sum of three squares if possible. 7. Show that every positive integer is either of the form a2 + b2 + c2 or a2 + b2 + 2c2 , where a, b, c are integers. 8. Show that the only prime p that is representable as the sum of two positive cubes is p = 2. 9. Express 12 and 99 as the sum of four squares. 10. Prove that no positive integer n, n = ±4(mod 9), is the sum of three cubes. 14 Certain Applications on Number Theory “Mathematics is the language of nature.” – Leonardo Fibonacci 14.1 Fibonacci Numbers The word Fibonacci is coined after the mathematician Leonardo Fibonacci, who in the 12th century, get obstructed while studying a curious problem. Fibonacci started with a pair of ﬁctional and slightly unbelievable baby rabbits, a baby boy rabbit and a baby girl rabbit. After one month, they were fully grown and in the next month two more baby rabbits (again a boy and a girl) were born. The succeeding month these babies were fully grown and the ﬁrst pair had two more baby rabbits (again, a boy and a girl). By similar manner, the next month the two adult pairs each have a pair of baby rabbits and the babies from last the month mature. Fibonacci asked how many rabbits a single pair can produce after a year (every month each adult pair produces a mixed pair of baby rabbits who mature the next month). He realised that the number of adult pairs in a given month is the total number of rabbits (both adults and babies) in the previous month. Writing An for the number of adult pairs in the nth month and Rn for the total number of pairs in the nth month, this gives An = Rn−1 . Fibonacci also realised that the number of baby pairs in a given month is the number of adult pairs in the previous month. Writing Fibonacci also realised that the number of baby pairs in a given month is the number of adult pairs 323 324 Number Theory and its Applications in the previous month. Writing Bn for the number of baby pairs in the nth month, this gives for the number of baby pairs in the nth month, this gives Bn = An−1 Rn−2 . Therefore the total number of pairs of rabbits (adult+baby) in a particular month is the sum of the total pairs of rabbits in the previous two months: Rn = An + Bn = Rn−1 + Rn−2 . Starting with one pair, the sequence we generate is exactly the sequence at the start of this article. And from that we can see that after twelve months there will be 144 pairs of rabbits. Theorem 14.1.1. In a Fibonacci sequence, the successive terms are relatively prime. Proof. Its suﬃces to show that for a Fibonacci sequence, gcd(Vn , Vn+1 ) = 1 ∀n ≥ 1. On the contrary, let d = gcd(Vn , Vn+1 ), d > 1. Then d Vn and d Vn+1 . Also, d (Vn+1 − Vn ) ⇒ d Vn−1 . Now, Vn − Vn−1 = Vn−2 . Combining d Vn and d Vn−1 yield d Vn−2 . Similar argument generates d Vn−3 , d Vn−4 , . . . , d V3 , d V2 , d V1 . But, V1 = 1 and d V1 leads to contradiction. Hence the proof. The identity Vm+n = Vm−1 Vn + Vm Vn+1 (14.1.1) is primal to bring out many spectacular features of the Fibonacci sequence. For ﬁxed m ≥ 2, we proceed by mathematical induction on n. For n = 1, the identity takes the form Vm+1 = Vm−1 V1 + Vm V2 = Vm−1 + Vm , which is trivially true. Assume that the identity is true for n = k. Then Vm+k = Vm−1 Vk + Vm Vk+1 & Vm+(k−1) = Vm−1 Vk−1 + Vm Vk . (14.1.2) (14.1.3) Combining (14.1.2) and (14.1.3) gives Vm+k + Vm+(k−1) = Vm−1 (Vk + Vk−1 ) + Vm (Vk+1 + Vk ). Following the deﬁnition of Fibonacci sequence, the foregoing equation is the same as Vm+k+1 = Vm−1 Vk+1 + Vm Vk+2 , which is the desired identity with n = k + 1. Hence by induction method the identity is true for m ≥ 2 and n ≥ 1. Certain Applications on Number Theory Theorem 14.1.2. For m ≥ 1, n ≥ 1, Vm Vmn . 325 Proof. Here we argue by principles of mathematical induction on n. For n = 1, the theorem is trivially true. Let Vm Vmn be true for n = 1, 2, 3, . . . , k. An appeal to the identity (14.1.1) yields Vm(k+1) = Vmk+m = Vmk−1 Vm + Vmn Vm+1 . By induction hypothesis, Vm Vmk ⇒ Vm Vmk−1 Vm + Vmn Vm+1 ⇒ Vm Vm(k+1) . Hence the proof. Lemma 14.1.1. If m = qn + r, then gcd(Vm , Vn ) = gcd(Vr , Vn ). Proof. Here, we have gcd(Vm , Vn ) = gcd(Vqn+r , Vn ), = gcd(Vqn−1 Vr + Vqn Vr+1 , Vn ) by identity(14.1.1). Combining the fact that gcd(a + c, b) = gcd(a, b) whenever b c and the foregoing theorem yields, gcd(Vqn−1 Vr + Vqn Vr+1 , Vn ) = gcd(Vqn−1 Vr , Vn ). Claim: gcd(Vqn−1 , Vn ) = 1. To fulﬁll this, consider gcd(Vqn−1 , Vn ) = d. Com bining d Vn and Vn Vqn imply d Vqn and therefore, d is a common divisor of the Fibonacci numbers Vqn−1 and Vqn . Since successive Fibonacci numbers are relatively prime, this results in d = 1. Finally, with the help of the fact gcd(a, c) = 1 ⇒ gcd(a, bc) = gcd(a, b), ∀a, b, c ∈ Z, we obtain the desired lemma. Theorem 14.1.3. The greatest common divisor of two Fibonacci numbers is again a Fibonacci number. Proof. Its suﬃces to show that gcd(Vm , Vn ) = Vd where gcd(m, n) = d and Vm , Vn and Vd are Fibonacci numbers. If possible, set m ≥ n. An appeal to Euclidean algorithm to m and n gives m = q1 n + r1 , 0 < r1 < n n = q2 r1 + r2 , 0 < r2 < r1 r 1 = q3 r 2 + r 3 , 0 < r3 < r2 .. . rn−2 = qn rn−1 + rn , rn−1 = qn+1 rn + 0. 0 < rn < rn−1 326 Number Theory and its Applications By the last lemma, we obtain gcd(Vm , Vn ) = gcd(Vr1 , Vn ) = gcd(Vr1 , Vr2 ) = · · · = gcd(Vr n−1 Now addressing Theorem 14.1.2, we have rn rn−1 ⇒ Vrn Vr n−1 , Vrn ). (How!). But rn = gcd(m, n)(Why!). Blending all the facts generate the desired one. Theorem 14.1.4. Let n ∈ Z+ and α = √ 1+ 5 2 . Then Vn > αn−2 , ∀ n ≥ 3. Proof. We proceed by mathematical induction on n to obtain the desired inequality. For n = 3, V3 = 2 > α. So the inequality prevails for n = 3. Assume the inequality holds for all integers k with k ≤ n. Then αk−2 < Vk . Since √ 1+ 5 α = 2 is one of the roots of x2 − x − 1 = 0, therefore α2 − α − 1 = 0. Hence α2 = k + 1. So αn−1 = α2 · αn−3 = (α + 1) · αn−3 = αn−2 + αn−3 . By induction, we have the following inequalities αn−2 < Vn , αn−3 < Vn−1 . Thus, αn−1 < Vn + Vn−1 = Vn+1 ⇒ Vn+1 > αn−1 . Hence the inequality is true for n = k + 1. So our task is complete. Theorem 14.1.5. Let Vk+1 and Vk+1 be successive terms of the Fibonacci sequence. Then the Euclidean algorithm takes exactly k divisions to show that gcd(Vk+1 , Vk+2 ) = 1. Proof. Taking aid of Euclidean algorithm and the deﬁning relation for Fibonacci sequence Vn = Vn−1 + Vn−2 at every step, we ﬁnd that Vn+2 = Vn+1 · 1 + Vn Vn+1 = Vn · 1 + Vn−1 .. .. .=. V4 = V3 · 1 + V2 V3 = V2 · 2 + 0 V2 = 0 · 2 + V1 = 2. Certain Applications on Number Theory 327 Hence the Euclidean algorithm takes exactly n-divisions to show gcd(Vk+1 , Vk+2 ) = 1. Our next discussions will be based on Lam´ e s theorem, ﬁrst proved by a French Mathematician Gabriel Lam´ e of 19th century, which calculates the number the number of division required to ﬁnd greatest common divisor by means of Euclidean Algorithm. Theorem 14.1.6. Lam´ e’s Theorem: The number of divisions needed to ﬁnd the greatest common divisor of two positive integers using the Euclidean Algorithm, does not exceed ﬁve times the number of digits in the smaller of the two digits. Proof. Taking into consideration, the Euclidean algorithm to ﬁnd the greatest common divisor of a = r0 and b = r1 with a > b, we get the following set of equations; r0 = r1 q1 + r2 , 0 ≤ r2 ≤ r1 . r1 = r2 q2 + r3 , 0 ≤ r3 ≤ r2 . r2 = r3 q2 + r4 , 0 ≤ r4 ≤ r3 . .. .. .=. rn−3 = rn−2 qn−2 + rn−1 , 0 ≤ rn−1 ≤ rn−2 . rn−2 = rn−1 qn−1 + rn , 0 ≤ rn ≤ rn−1 . rn−1 = rn qn . We have used n-divisions. We note that each of the quotients q1 , q2 , . . . , qn−1 is greater than or equal to 1 and qn ≥ 2, since rn < rn−1 . Hence r n ≥ 1 = V2 , rn−1 ≥ 2rn ≥ 2V2 = V3 , rn−2 ≥ rn−1 + rn ≥ V3 + V3 = V4 , rn−3 ≥ rn−2 + rn−1 ≥ V4 + V3 = V5 , .. . .. . .. . r2 ≥ r3 + r4 ≥ Vn−1 + Vn−2 = Vn , b = r1 ≥ r2 + r3 ≥ Vn + Vn−1 = Vn+1 . Thus after using n-divisions in Euclidean algorithm, we have b ≥n+1 . Taking √ advantage of Theorem 2.4.5, we obtain Vn+1 > αn−1 for n > 2 where α = 1+2 5 . Hence b > αn−1 . Here, we note that log b > (n − 1) log α > n−1 1 ,[ ∵ < log α] 5 5 328 Number Theory and its Applications It gives, n − 1 < 5 log b. Consider b to have k decimal digits. Then b < 10k and log b < k, which further implies (n − 1) < 5k. Theorem 14.1.7. Zeckendorf representation: Any positive integer can be expressed as a sum of distinct Fibonacci numbers, no two of which are consecutive. Proof. Because two consecutive numbers of the Fibonacci sequence may be combined to give the next one, it is pointless to have consecutive Fibonacci numbers in the representation of an integer. Thus, whenever possible, Vk +Vk−1 is replaced by Vk+1 in the following proof. We proceed by principles of mathematical induction on n > 2. Its suﬃces to show that the integers from the set {1, 2, 3, 4, . . . , Vn−1 } is a sum of numbers from the set {Vi i = 1, 2, 3 . . . , n − 2}, repetition being not allowed. Consider the ﬁrst 10 positive integers. 1 = V1 6 = V1 + V3 + V4 = V2 + V3 + V4 2 = V3 7 = V3 + V5 = V1 + V2 + V3 + V4 3 = V4 8 = V 4 + V5 4 = V 1 + V4 9 = V 1 + V 4 + V5 5 = V 3 + V4 10 = V3 + V4 + V5 . So the theorem prevails for ﬁrst few positive integers. Suppose the statement is true for n = k. Select N in such a way that Vk − 1 < N < Vk+1 . Since, N − Vk−1 < Vk+1 − Vk−1 = Vk it implies N − Vk−1 is representable as a sum of distinct numbers from the set {Vi i = 1, 2, 3 . . . , k − 2}. Consequently, N and the integers from the set {1, 2, 3, 4, . . . , Vk+1 − 1} can be written, without repetition, as a sum of numbers from the set {Vi i = 1, 2, 3 . . . , k − 2, k − 1}. Hence we are done. Theorem 14.1.8. Binet formula: For any integer n ≥ 1, - √ n √ n . 1+ 5 1− 5 1 . − Vn = √ 2 2 5 Proof. To begin with, consider the quadratic equation x2 − x − 1 = 0. (14.1.4) Certain Applications on Number Theory 329 √ √ 1− 5 1+ 5 and δ = . Since Let γ and δ be the roots of (14.1.4). Then γ = 2 2 γ and δ are the roots of (14.1.4), therefore γ2 − γ − 1 = 0 (14.1.5) δ − δ − 1 = 0. (14.1.6) 2 Multiplying (14.1.5) by γ n and (14.1.6) by δ n , we obtain γ n+2 − γ n+1 − γ n = 0 (14.1.7) δ n+2 − δ n+1 − δ n = 0. (14.1.8) Now, {(14.1.7) − (14.1.8)}/(γ − δ) gives γ n+1 − δ n+1 γ n − δn γ n+2 − δ n+2 = + . γ−δ γ−δ γ−δ Set Vn = γ n − δn . Then the foregoing equation can be simpliﬁed as, γ−δ Vn+2 = Vn+1 + Vn , n ≥ 1. Since γ and δ are the roots of the (14.1.4) therefore from the relation between roots and coeﬃcients we have √ γ + δ = 1, γδ = −1, γ − δ = 5. Thus, V1 = 1, V2 = 1, V3 = 2, V4 = 3 and so on. Hence {V1 , V2 , V3 , . . .} forms a Fibonacci sequence, where the nth term of the sequence is given by γ n − δn Vn = . γ−δ An immediate consequence of Binet formula is the following two identities: 2 − Vn2 = V2(n+1) . Theorem 14.1.9. For n ≥ 1, Vn+2 Proof. Since γδ = −1 ⇒ (γδ)2k = 1, for k ≥ 1. Also, γ + δ = 1. Then, 2 n 2 n+2 γ γ − δn − δ n+2 2 2 Vn+2 − Vn = − γ−δ γ−δ = γ 2(n+2) + δ 2(n+2) − γ 2n − δ 2n (γ − δ)2 = (γ 2 − δ 2 )(γ 2(n+1) − δ 2(n−1) ) (γ − δ)2 = (γ + δ)V2(n+1) = V2(n+1) . Hence proved. 330 Number Theory and its Applications 2 Theorem 14.1.10. For any integer n ≥ 1, V2n+1 V2n−1 − 1 = V2n . Proof. Since γδ = −1 ⇒ (γδ)2k = 1, for k ≥ 1. Also, γ 2 +δ 2 = (γ+δ)2 −2γδ = 3. Then, 2n−1 2n+1 γ γ − δ 2n+1 − δ 2n−1 V2n+1 V2n−1 − 1 = −1 γ−δ γ−δ 2n−1 2n+1 γ γ − δ 2n+1 − δ 2n−1 √ √ −1 = 5 5 γ 4n + δ 4n + (γ 2 + δ 2 ) − 5 = 5 γ 4n + δ 4n − 2 = 5 γ 4n + δ 4n − 2(γδ)2n = 5 2n 2n 2 γ −δ 2 √ = = V2n . 5 Before retiring from the ﬁeld, the remaining theorems of this section deals with the prime factors of Fibonacci numbers. Theorem 14.1.11. For a prime p > 5, either p\|Vp−1 or p\|Vp+1 but not both. (αp −αp ) Proof. Using Binet’s formula we get, Vp = √5 2 . If we expand p-th power of α and α2 by the binomial theorem then we obtain, p p−1 √ p √ p 1 5 2 5 5 + ··· + Vp = √ 1 + 5+ p p 1 2 2 5 p p−1 √ p √ p 1 5 2 5 5 − ··· − 5+ − √ 1− p 1 2 2p 5 p 2 p p p−1 1 p 5 + ··· + 5+ + 5 2 . = p−1 5 3 1 p 2 Now using the facts that kp ≡ 0(mod p) for 1 ≤ k ≤ p−1 and 2p−1 ≡ 1(mod p) we get, p−1 p p−1 5 2 ≡ 5 2 ( mod p). Vp ≡ 2p−1 Vp ≡ p Now since gcd(5, p) = 1, then by Fermat’s little theorem 5p−1 ≡ 1(mod p). This shows Vp2 ≡ 1(mod p). Now if we apply an well known identity Vp2 = Vp−1 Vp+1 + (−1)p−1 modulo p then we have, Vp−1 Vp+1 ≡ 0(mod p). As p > 5 then gcd(p − 1, p + 1) = 2 and applying Theorem 14.1.3 together we have found that gcd(Vp−1 , Vp+1 ) = V2 = 1. Finally we can conclude that either p\|Vp−1 or p\|Vp+1 but not both. Certain Applications on Number Theory 331 Theorem 14.1.12. Let p ≥ 7 be a prime satisfying p ≡ 2(mod 5) or p ≡ 4( mod 5) then 2p − 1 divides Vp provided 2p − 1 is prime. Proof. Let p is of the form 5k + 2p forp some integer k. Now from Binet’s formula p p (α −α ) ) . Vp = √5 2 . Squaring both sides and expanding α2p we have, Vp = (α √−β 5 and α22p by 5Vp2 = 22p−1 2p 2 2p 2p p 5 + ··· + 5+ 5 + 2. 4 2 2p 1 1+ Since 2p − 1 is a prime then by Fermat’s little theorem we have 2(2p−1)−1 ≡ 1( mod 2p − 1) or 22p−1 ≡ 2(mod 2p − 1). Also 2p k ≡ 0(mod 2p − 1) for 2 ≤ k < 2p − 1. Applying these two facts together the above expression reduces to, 2(5Vp2 ) ≡ (1 + 5p ) + 4( mod 2p − 1) or 2Vp2 ≡ 1 + 5p−1 ( mod 2p − 1). 5 (mod 2p − 1)[by Euler’s criterion]. Again from 2p − 1 Corollary 9.4.2 it is clear that, 5 2p − 1 10k + 3 3 = = = = −1. 2p − 1 5 5 5 Now, 5p−1 = 5 2p−2 2 Thus we have, 2Vp2 ≡ 1 + (−1)(mod 2p − 1). This shows that 2p − 1 divides Vp . If p ≡ 4(mod 5) then p = 5k + 4 for some integer k . Then applying the Corollary 9.4.2 again we see that, 5 2p − 1 10k + 7 2 = = = = −1. 2p − 1 5 5 5 Then we have, 2Vp2 ≡ 1 + (−1)(mod 2p − 1). This also shows that 2p − 1 divides Vp . 14.2 Worked out Exercises Problem 14.2.1. Show that the sum of squares of the ﬁrst n Fibonacci numbers is given by, V12 + V22 + · · · + Vn2 = Vn Vn+1 . Solution 14.2.1. Let us consider, Vt Vt+1 − Vt−1 Vt = Vt (Vt+1 − Vt−1 ) = Vt2 for t ≥ 2. 332 Number Theory and its Applications Now taking t = 1, 2, 3, · · · , n we have, V12 = V1 V2 V22 = V2 V3 − V1 V2 V32 = V3 V4 − V2 V3 .. . . = .. Vn2 = Vn Vn+1 − Vn−1 Vn . Adding those equations we get, V12 + V22 + · · · + Vn2 = Vn Vn+1 . Problem 14.2.2. Evaluate gcd(V15 , V20 ). Solution 14.2.2. From Theorem 14.1.3 it is clear that gcd(Vm , Vn ) = Vd where d = gcd(m, n). Now here gcd(15, 20) = 5 then gcd(V15 , V20 ) = V5 = 5. Problem 14.2.3. If gcd(m, n) = 1, prove that Vm Vn Vmn ∀ m, n ≥ 1. Solution 14.2.3. As m, n ≥ 1 then applying Theorem 14.1.2 we can say that Vm \|Vmn and Vn \|Vmn . Since gcd(m, n) = 1, then we have gcd(Vm , Vn ) = V1 = 1. Thus applying Corollary 2.4.1 we can conclude that Vm Vn \|Vmn for all m, n ≥ 1. 2 Problem 14.2.4. Show that V2n+1 = Vn+1 +Vn2 whenever n is a positive integer. Solution 14.2.4. We will solve this problem by the principle of mathematical induction. For this let us assume n = 1. Then we have V3 = 2 = V22 +V12 = 1+1. If we choose n = 2 then we get, V5 = 5 = 22 + 12 = V32 + V22 . Thus the result is true for n = 1, 2. Now we have to assume that the identity is true for all n 2 2 less than or equal to k. Then we can assume that V2k−3 = Vk−1 + Vk−2 and 2 . Finally we need to calculate V2k+1 . Thus, V2k−1 = Vk2 + Vk−1 V2k+1 = V2k + V2k−1 = V2k−1 + V2k−2 + V2k−1 = 2V2k−1 + (V2k−1 − V2k−3 ) = 3V2k−1 − V2k−3 . Now substituting all the assumptions in above equation we get, 2 2 2 V2k+1 = 3(Vk2 + Vk−1 ) − (Vk−1 + Vk−2 ) 2 = 3Vk2 + 2Vk−1 − (Vk − Vk−1 )2 2 = 2Vk2 + Vk−1 + 2Vk Vk−1 = 2Vk2 + (Vk+1 − Vk )2 + 2Vk (Vk+1 − Vk ) 2 = Vk+1 + Vk2 . Certain Applications on Number Theory 333 Now by applying principle of mathematical induction we can say that the result is true for all positive integer n. Problem 14.2.5. Show that the simple continued fraction, terminating with V the partial quotient of 1, of k+1 (k ∈ Z+ ) is [1; 1, 1, 1 . . . , 1](k-times) where Vk Vk denote the k-th Fibonacci number. Solution 14.2.5. In view of recursion formulae for the Fibonacci sequence, the V Euclidean algorithm for k+1 yields; Vk Vk+1 = 1(Vk ) + Vk−1 Vk = 1(Vk−1 ) + Vk−2 .. .. .=. V2 = 1(V1 ). So Vk+1 = [1; 1, 1, 1 . . . , 1](k-times). Hence the proof. Vk Problem 14.2.6. Prove that the sum of the ﬁrst n Fibonacci numbers with odd indices is given by the formula V1 + V3 + V5 + · · · + V2n−1 = V2n . Solution 14.2.6. We will prove this formula by principle of mathematical induction. Here we can see that, V1 = V2 = 1, V1 + V3 = 3 = V4 . So the identity is true for n = 1 and n = 2. Let us assume that the identity is true for n = k. We now show that the identity is true for n = k + 1.Here, V1 + V3 + V5 + · · · + V2n−1 = V2k + V2k+1 [by induction hypothesis] = V2k+2 . Thus the result is true for n = k + 1. So by principle of mathematical induction we can say that the result is true for all positive odd integer n. Problem 14.2.7. Find the Zeckendorf representation of 50 and 110. Solution 14.2.7. Zeckendorf representation we can assert that any positive integer can be expressed as a sum of distinct Fibonacci numbers, no two of which are consecutive. Here, 50 = 34 + 13 + 3 = V9 + V7 + V4 and 110 = 89 + 21 = V11 + V8 . Problem 14.2.8. From the Binet’s formula of Fibonacci numbers, derive the relation V2n+2 V2n−1 − V2n V2n+1 = 1, n ≥ 1. 334 Number Theory and its Applications Solution 14.2.8. From Binet’s formula we have, Vn = √ 1, α1 β1 = −1, α1 − β1 = 5. Now αn −β n 1√ 1 5 where α1 + β1 = α2n−1 − β12n−1 α12n+2 − β12n+2 √ √ × 1 5 5 α12n − β12n α12n+1 − β12n+1 √ √ − × 5 5 α14n+1 − α12n+2 β12n−1 − α12n−1 β12n+2 + β14n+1 = 5 α14n+1 − α12n β12n+1 − α12n+1 β12n + β14n+1 − 5 α12n β12n+1 + α12n+1 β12n − α12n+2 β12n−1 − α12n−1 β12n+2 = 5 V2n+2 V2n−1 − V2n V2n+1 = α2 = α12n β12n [(α1 + β1 ) − ( β 1 + 1 β12 α1 )] 5 α1 β1 − {(α1 + β1 )3 − 3α1 β1 (α1 + β1 )} = 5α1 β1 −1 − (1 + 3) = 1. = −5 14.3 Pseudo-random Numbers Randomly chosen numbers are very much useful for performing computer simulations of some complicated statistical or mathematical phenomena. But the methods commonly used to generate random numbers are not a random process. Random numbers are called pseudo-random numbers when they are generated by some deterministic process but they qualify the predetermined statistical test for randomness. The ﬁrst input will determine the sequence of Pseudo random numbers generated by any method of determining them. One such method is called mid-square method. This method was ﬁrst introduced by Von Neumann in the year 1949. Mid-square method Mid-Square method was the ﬁrst method used to generate Pseudo random numbers. In order to generate a sequence of Pseudo random numbers, we need to select a number with n digits. Then squaring that n-digit number, the next number will be the middle part of that squared number. For better explanation, let us consider a four digit number. Subsequently, we are to square this four digit number. From this squared digit number we need to take out the middle Certain Applications on Number Theory 335 four digit. This will give us the second random number. After that we are to iterate this method to obtain a sequence of random numbers. Here we need to remember certain things. We know that the square of a four digit number has eight digits or fewer than this. Those fewer than eight digits has to be considered as eight digit number by adding zero as initial digit or digits. However the sequence of numbers produced in this method appears to be random, and they are useful for computer simulations. This sequence of random numbers are called pseudo random numbers. Let us illustrate the above discussion with an example. Take a two digit number 69 as the seed. So 69 is our initial approximation. Squaring 69 we get 692 = 4761. On applying mid-square method we obtain 76 as second term. The third term is 77 as 762 = 5776. Continuing this way we will get thirteenth term 05 as 842 = 7056. Then the fourteenth term will be 02 as 52 = 25 = 0025. This process will be terminated at sixteenth term as 22 = 4 = 0004 and the middle two digits will become 00. Some discrepancies lies with the Mid-Square method. The most undesirable feature is that this process produces small set of random numbers. For instance, the last example shows that after sixteenth stage the random numbers will occur only zero. So to overcome this drawback we need an algorithm, known as Linear Congruential Method, for ﬁnding pseudo-random numbers. Linear Congruential Method To begin with, consider some integers a, c, m with the constraints m > 0, 2 ≤ a < m, 0 ≤ c ≤ m. Select the seed x0 such that 0 ≤ x0 ≤ m. To obtain the sequence of pseudo random numbers, we need to construct the recursive congruence relation xn+1 ≡ axn + c( mod m); 0 ≤ xn+1 < m, for n = 0, 1, 2, . . . . Note that m is modulus, a is multiplier and c is the increment. The following example will lucidly illustrate the method. Example 14.3.1. Consider m = 19, a = 5 and c = 2. If we take the seed 336 Number Theory and its Applications x0 = 6, then from xn+1 ≡ axn + c(mod m) we get x1 ≡ 5 × 6 + 2 ≡ 32( mod 19) ⇒ x1 ≡ 13( mod 19) x2 ≡ 5 × 13 + 2 ≡ 67( mod 19) ⇒ x2 ≡ 10( mod 19) x3 ≡ 5 × 10 + 2 ≡ 52( mod 19) ⇒ x3 ≡ 14( mod 19) x4 ≡ 5 × 14 + 2 ≡ 72( mod 19) ⇒ x4 ≡ 15( mod 19) x5 ≡ 5 × 15 + 2 ≡ 77( mod 19) ⇒ x5 ≡ 1( mod 19) x6 ≡ 5 × 1 + 2 ≡ 7( mod 19) x7 ≡ 5 × 7 + 2 ≡ 7( mod 19) ⇒ x7 ≡ 18( mod 19) x8 ≡ 5 × 18 + 2 ≡ 92( mod 19) ⇒ x8 ≡ 16( mod 19) x9 ≡ 5 × 16 + 2 ≡ 82( mod 19) ⇒ x9 ≡ 6( mod 19) x10 ≡ 5 × 6 + 2 ≡ 32( mod 19) ⇒ x10 ≡ 13( mod 19). So we see that x1 ≡ x10 ≡ 13(mod 19). Hence it follows that xk = xk−9 for k ≥ 10. Thus the sequence is 13, 10, 14, 15, 1, 7, 18, 16, 6, 13, . . . and it contains 9 diﬀerent numbers. The following theorem suggest us to ﬁnd the terms of the sequence of pseudo random numbers generated by the linear congruential method directly from seed, multiplier and increment. Theorem 14.3.1. The terms of the sequence generated by linear congruential method are given by, xn ≡ an x0 + c(an − 1) mod m , 0 ≤ xn < m. (a − 1) Proof. We can prove the result by mathematical induction on n. For n = 1 we get, x1 ≡ ax0 + c(mod m), 0 ≤ x1 < m. This is true from the formula of linear congruential method. Let us assume that the result be true for n = k. Then the k −1) congruence relation xk ≡ ak x0 + c(a mod m , holds for 0 ≤ xk < m. Since (a−1) xk+1 ≡ axk + c(mod m), 0 ≤ xk+1 < m holds, then we have c(ak − 1) + c( mod m) xk+1 ≡ a ak x0 + (a − 1) a(ak − 1) + 1 ( mod m) ≡ ak+1 x0 + c (a − 1) (ak+1 − 1) ≡ ak+1 x0 + c ( mod m). (a − 1) This shows that the result is true for n = k + 1. So by principle of mathematical induction we can say that the result is true for all integers n. Certain Applications on Number Theory 337 In the Example 14.3.1 we have seen that the sequence of pseudo random numbers contain 9 diﬀerent numbers for the sequence 13, 10, 14, 15, 1, 7, 18, 16, 6, 13, . . .. Here the total number of diﬀerent elements in the sequence is called period length and is deﬁned as follows. Deﬁnition 14.3.1. The maximum length of the sequence obtained without repetition in a linear congruential pseudo random number generator is called period length. We can see that the longest possible length for a linear congruential generator is the modulus m. In the next theorem, we have given light on this fact. This is also a characterisation theorem of a good generator. Theorem 14.3.2. The linear congruential pseudo random number generator produces a sequence of period length m if and only if gcd(c, m) = 1, a ≡ 1( mod p) for all primes p dividing m, and a ≡ 1(mod 4) provided 4 m. The proof of this theorem beyond the scope of the book. For further reference see Knuth[8]. In particular if c = 0 in the recursive relation xn+1 ≡ axn + c(mod m), then we get the relation as xn+1 ≡ axn (mod m), 0 < xn+1 < m. Here m is the modulus and a is the multiplier. This method is called the pure multiplicative congruential method. Here the terms of the sequence of pseudo random numbers are obtained by xn ≡ an x0 (mod m) with 0 < xn+1 < m, where x0 is the seed. If L is the period length of the sequence of pseudo random numbers obtained recursively by pure multiplicative congruential method, then L is the least universal exponent. The following proposition indicates this fact. Proposition 14.3.1. The largest possible period length of the sequence of pseudo random numbers, generated recursively using pure multiplicative generator is λ(m), where λ(m) is the least universal exponent modulo m. Proof. Let L be the period length of the sequence of pseudo random numbers generated recursively using pure multiplicative generator. Then x0 ≡ aL x0 ( mod m), where a is the multiplier and x0 is the seed. Now gcd(x0 , m) = 1 yields aL ≡ 1(mod m). An appeal to the deﬁnition of least universal exponent, the forgoing congruence relation implies that the largest possible period length is λ(m). We conclude this section with another method of pseudo random generator. In this method, we need to consider a positive integer m as modulus and the 338 Number Theory and its Applications initial term x0 as seed of the generator. Here the sequence of pseudo random numbers has been generated by the congruence relation, xn+1 ≡ x2n ( mod m), 0 < xn+1 < m. 14.4 Worked out Exercises Problem 14.4.1. Apply the mid square method for generating pseudo random numbers for 6139. Solution 14.4.1. Here the initial seed is 6139. Squaring 6139 gives (6139)2 = 37687321. Applying mid square method, we get x2 = 6873. Furthermore, squaring and thereby applying this method we ﬁnd x3 = 2381. Similarly we get x4 = 6691, x5 = 7694, x6 = 1976, x7 = 9045 and so on. Problem 14.4.2. Find the period length of the sequence of pseudo random numbers generated by linear congruential method with x0 = 2 and xn+1 ≡ 4xn +7( mod 25). Solution 14.4.2. Here the initial seed x0 = 2. Then applying linear congruential method we have, x1 ≡ 4 × 2 + 7 ≡ 15( mod 25) ⇒ x2 ≡ 4 × 15 + 7 ≡ 17( mod 25) ⇒ x3 ≡ 4 × 17 + 7 ≡ 0( mod 25) ⇒ x4 ≡ 4 × 0 + 7 ≡ 7( mod 25) ⇒ x5 ≡ 4 × 7 + 7 ≡ 10( mod 25) ⇒ x6 ≡ 4 × 10 + 7 ≡ 22( mod 25) ⇒ x7 ≡ 4 × 22 + 7 ≡ 20( mod 25) ⇒ x8 ≡ 4 × 20 + 7 ≡ 12( mod 25) ⇒ x9 ≡ 4 × 12 + 7 ≡ 5( mod 25) ⇒ x10 ≡ 4 × 5 + 7 ≡ 2( mod 25). Thus the period length is 10. Problem 14.4.3. Find the multiplier a of a linear congruential generator xn+1 ≡ axn + c(mod m) for which the period length is m = 225 − 1, where gcd(c, m) = 1. Solution 14.4.3. We have 225 −1 = 31·601·1801. Taking into account Theorem 14.3.2, we can say that a ≡ 1(mod p) for p = 31, 601, 1801. Finally using Chinese Remainder Theorem we get a ≡ 1(mod 225 − 1). Problem 14.4.4. Using linear congruential method if either a = 0 or a = 1 is applied for multiplier in the generation of pseudo random numbers, then prove that the resulting sequence fails to an eﬀective option for the same. Certain Applications on Number Theory 339 Solution 14.4.4. If a = 0 we have xn+1 ≡ c(mod m) which means that the sequence is constant for n ≥ 1. This is not an eﬀective choice for the sequence of pseudo random numbers. If a = 1 we have xn+1 ≡ xn + c(mod m), shows that the terms of the sequence diﬀer by a constant modulo m. This is also not an eﬀective choice for the sequence of pseudo random numbers. Problem 14.4.5. Find the sequence of numbers generated by the square pseudo random number generator with modulus 77 and seed 8. Solution 14.4.5. Here the initial seed is x0 = 8 and the modulus is m = 77. Then we have, x1 ≡ 82 ≡ 64( mod 77) x2 ≡ 642 ≡ 15( mod 77) x3 ≡ 152 ≡ 71( mod 77) x4 ≡ 712 ≡ 36( mod 77) x5 ≡ 362 ≡ 64 ≡ x1 ( mod 77). Thus the sequence of pseudo random number is 8, 64, 15, 71, 36, 64, · · · . 14.5 Cryptology From ancient times coding and decoding of messages are two very familiar words. With the passage of time, the signiﬁcance of coding and decoding increases. In defence, the secret communication plays a key role. Now a days the privacy of data is needed for bank and any other ﬁnancial transactions. In twenty ﬁrst century, communications through social media creates a huge impact in our daily life. Essence of secured and encrypted medium is a vital issue for these types of communications. Hence there is a great deal of interest in the techniques of making messages unintelligible to everyone except the intended receiver. Many mathematicians and computer scientists have given their important contribution towards the growth of the discipline devoted to secrecy systems called cryptography. Cryptography is the part of cryptology that deals with design and application of secrecy system and cryptoanalysis is aimed at breaking these systems. Actually the word cryptography is the combination of two Greek words ‘Krypto’means ‘hidden’and ‘Graphein’means ‘to write’. In simple words, a message that is to be altered into a secret form is called plaintext while after transformation to a secret form, these messages is said to be ciphertext. The process of converting a plaintext to a ciphertext is called 340 Number Theory and its Applications encrypting and the reverse process of converting from ciphertext to plaintext is said to be decrypting or deciphering. A cipher is a method for altering a plaintext message to ciphertext by changing the letters of plaintext using a transformation. This section is devoted with few types of methods of encryption and decryption. One of the most ancient Cryptographic method, ﬁrst introduced by great Roman emperor Julius Caesar, is called Character or Monographic cipher. Here we will discuss the newest form of Caesar cipher which was invented in late 1970’s. The method starts with translating English alphabetical letters to integers from 0 to 25, as shown in the table below. Letters Integers Letters Integers A 0 N 13 B 1 O 14 C 2 P 15 D 3 Q 16 E 4 R 17 F 5 S 18 G 6 T 19 H 7 U 20 I 8 V 21 J 9 W 22 K 10 X 23 L 11 Y 24 M 12 Z 25 The Caesar cipher is categorised as a substitution cipher in which each the alphabet in the plaintext is shifted by a ﬁxed number down the alphabet. To explain this by modular arithmetic, let us consider P to be the numeric equivalent of plaintext letter and C be the numerical equivalent of ciphertext letter. Then we have, C ≡ P + 3(mod 26) where 0 ≤ C ≤ 25 and shifting done by three down the alphabet. According to this cipher, if we want to send a message: HOW ARE YOU then we can write this as, 7 14 22 0 17 4 24 14 20. Taking help of the congruence relation we have, 10 17 25 3 20 7 1 17 23. Now translating back to letters, we get KRZ DUH BRX. This is the message we will send. Now the receiver will decipher it by the congruence relation, P ≡ C − 3 ≡ C + 23(mod 26). The Caesar cipher is one of the family of similar ciphers described by a shift transformation, C ≡ P + k(mod 26), 0 ≤ C ≤ 25. Certain Applications on Number Theory 341 Here k is the number of shifting in alphabetical order. More generally we will consider the transformation of the type, C ≡ aP + b(mod 26), 0 ≤ C ≤ 25. Here a, b are integers with gcd(a, 26) = 1. These are called aﬃne transformations. Shift transformations are aﬃne transformations with a = 1. Here to conduct the cipher, we need gcd(a, 26) = 1 so that P and C runs through complete system of residues modulo 26. There are φ(26) = 12 possibilities of a and 26 possibilities of b, giving total 12 × 26 = 312 types of transformations. Here the relationship of ciphertext to plaintext is given by, P ≡ a(C − b)(mod 26), 0 ≤ P ≤ 25. Here a is the inverse of a modulo 26. The monographic cipher is very simple and insecured. To avoid the vulnerability of monographic cipher, we are going to present block or polygraphic cipher, introduced by Lester S. Hill in 1929. This cipher is also called Hill cipher. Hill cipher uses matrix algebra to encrypt blocks of any desire length, where each block contains plaintext letters of speciﬁed length. After the encryption, each block of plaintext letters encrypted to ciphertext letters of same length. Here our discussion starts with digraphic cipher, where each block contains two letters of plaintext and it is to be encrypted to a block of two letters of ciphertext. First, we are to split the message into two blocks of two letters. If the number of letter is odd, then we are to add a dummy letter ‘X’ so that the ﬁnal block contains two letters. Consider a message, BEWARE OF THE MESSENGER. We now split it into blocks of two letters as, BE WA RE OF TH EM ES SE NG ER. Secondly, we need to translate these blocks into its numerical equivalents. So here we translate them to obtain 1 4 22 0 17 4 14 5 19 7 4 12 4 18 18 4 13 6 4 17. We are to convert each block of plaintext numbers P1 P2 into ciphertext numbers C1 C2 by system of linear congruence modulo 26. The formula is given by, C1 ≡ aP1 + bP2 ( mod 26) C2 ≡ cP1 + dP2 ( mod 26) 342 Number Theory and its Applications where a, b, c and d are integers. Here in this example we are going to apply the system, C1 ≡ 3P1 + 10P2 ( mod 26) C2 ≡ 9P1 + 7P2 ( mod 26). So by this formula for P1 = 1 and P2 = 4 the ciphertext numbers, C1 ≡ 3 · 1 + 10 · 4 ≡ 43 ≡ 17( mod 26) C2 ≡ 9 · 1 + 7 · 4 ≡ 37 ≡ 11( mod 26). Thus the ciphertext numbers obtained for entire message is, 17 11 14 16 13 25 14 5 23 12 2 16 10 6 16 8 21 3 10 25. Finally we need to translate these blocks into letters, RL OQ NZ OF XM CQ KG QI VD KZ. The decrypting process in this cipher system has to be done with the help of Theorem(4.6.1). Here in our example the relationship becomes, P1 ≡ 21C1 + 22C2 ( mod 26) P2 ≡ 25C1 + 9C2 ( mod 26). Actually the matrix representation of digraphic cipher is 3 10 P1 C1 ≡ ( mod 26) 9 7 C2 P2 Taking help of proposition(4.6.2) we ﬁnd the inverse of 21 22 26 is and the relationship becomes, 25 9 21 P1 ≡ 25 P2 3 9 10 7 under modulo 22 C1 ( mod 26). 9 C2 In general, the polygraphic cipher where the block length is n(> 2) we need to apply congruences of matrices. Here the matrix relationship becomes, C ≡ AP ( mod 26) Certain Applications on Number Theory 343 where A is an n × n matrix with gcd(det A, 26) = 1, ⎛ ⎞ ⎛ ⎞ C1 P1 ⎜ ⎟ ⎜ ⎟ C P ⎜ 2⎟ ⎜ 2⎟ ⎜ . ⎟ and ⎜ . ⎟ ⎜ . ⎟ ⎜ . ⎟ ⎝ . ⎠ ⎝ . ⎠ Cn Pn For deciphering we need matrix A which is inverse of A modulo 26. This A can be obtained by Proposition(4.6.3). Hence to determine the plaintext from ciphertext, we use the relationship, P ≡ AC( mod 26). This polygraphic cipher can only be used for small values of n. For large values of n this method becomes extremely infeasible. In such situation we can carry out our study of Cryptology to another cipher, called exponentiation cipher. This cipher is based on modular exponentiation. This was ﬁrst introduced by Pohig and Hellman in the year 1978. This is less vulnerable to frequency analysis than block cipher. To start this procedure, let us consider a prime p and an exponent e ﬁrst such that gcd(e, p − 1) = 1. Next we have to translate the letters of the message into numerical equivalents. For that we use the relationship given below. Letters Integers Letters Integers A 00 N 13 B 01 O 14 C 02 P 15 D 03 Q 16 E 04 R 17 F 05 S 18 G 06 T 19 H 07 U 20 I 08 V 21 J 09 W 22 K 10 X 23 L 11 Y 24 M 12 Z 25 After the numeric conversion we need to group the letters in the message in · · · 25&. For blocks of k letters, where k is chosen so that 2525 · · · 25& < p < 2525 # \$% # \$% k−times k+1−times example if we choose p = 1423 then we should use the blocks of k = 2 letters, because 2525 < 1423 < 252525. Finally we form ciphertext block C from plaintext block P by the relation, C ≡ P e ( mod p), 0 ≤ C < p. We now illustrate this by an example. For that let us choose p = 101 and e = 3. We want to encipher the message G 06 O 14 O 14 D 03 M 12 O 14 R 17 N 13 I 08 N 13 G 06 344 Number Theory and its Applications Since 25 < p < 2525 then we have to use block of k = 1 letters. Now we form ciphertext block from plaintext block by the relation, C ≡ P 3 ( mod 101), 0 ≤ C < 101. This gives us the series, 14 17 17 27 11 17 65 76 07 76 14. Now to decipher the message that has been encoded using exponential cipher, we need to know the deciphering key an integer d such that de ≡ 1(mod p − 1), so that d is the inverse of e modulo p −1. The existence of d has been ensured by the fact that gcd(e, p − 1) = 1. Note that de can be written as de = 1 + n(p − 1) for some integer n. Now we have, C ≡ P e ( mod p) ⇒ C d ≡ P ed ( mod p) ≡ P · P n(p−1) ( mod p) ≡ P · (P p−1 )n ( mod p). Using Fermat’s little theorem we get, P p−1 ≡ 1( mod p) as p P. Thus we obtain, P ≡ C d ( mod p). This is the relation for converting encrypted text to plaintext. The exponentiation cipher has major drawbacks when employed in a network with many users. Here each pair of communicants must employ an enciphering key which is kept secret from the other communicants of that network. The reason is that, if the enciphering key is known then the deciphering key can be obtained by simple computations of number theory. There is another disadvantage of changing encryption key frequently. To avoid those diﬃculties, we need to assign a key to each pair of individuals which must be kept secret from other individuals of the network. Public Key cipher system has to be introduced to overcome those diﬃculties. The aim of public key cipher is to establish secret communications in a network of n individuals. Here each individual produce a key k which is not known to other individuals. We need to pass the private information through the enciphering transformation E(k) for a key k to obtain ciphertext. Now for k1 , k2 , · · · kn keys, if the individual i wishes to send a message to individual j Certain Applications on Number Theory 345 then for plaintext P the ciphertext C is computed by transformation Ekj where C = Ekj (P ). To decipher the message, j applies the deciphering transformation Dkj on C to obtain P satisfying Dkj (C) = Dkj (Ekj (P )) = P . Today the public key cipher, known as RSA, is the most widely used cipher. In the year 1978, RSA was proposed by Ronald Rivest, Adi Shamir and Leonard Adleman. This public key cipher is based on modular exponentiation. To construct RSA algorithm, we need to use the keys in pairs (e, n) where e is exponent and n is modulus. Here n must be the product of large primes p and q i.e. n = pq, so that gcd(e, φ(n)) = 1. Here consider the integer e satisfying 1 < e < φ(n). Now to encipher the message, we ﬁrst translate the letters into numerical using the table described in exponentiation cipher. Finally to encipher a plaintext block P , apply the relation E(P ) = C ≡ P e ( mod n), 0 < C < n where C is ciphertext block. To decipher the ciphertext block C, we need to obtain the inverse exponent d of e modulo φ(n). The existence of d is ensured by the fact gcd(e, φ(n)) = 1. Here d and e follow the congruence relation de ≡ 1(mod φ(n)) i.e. de = k φ(n) + 1 for some integer k . Finally we have, D(C) ≡ C d ≡ P ed ( mod n) ≡ P kφ(n)+1 ( mod n) ≡ (P φ(n) )k · P ≡ P ( mod n). (14.5.1) As the probability that P and n not relatively prime is very small, so gcd(P, n) = 1. In view of Euler’s theorem, we obtain P φ(n) ≡ 1(mod n) in congruence (14.5.1). The pair (d, e) is said to be the deciphering key. Let us exemplify this algorithm with an example. Here we want to encipher the message: BEST WISHES, where e = 3 and n = 2669 = 17 × 157 = pq. Now the numeric equivalence of letters of the message is, 01 04 18 19 22 08 18 07 04 18. After applying the congruence C ≡ P 3 (mod 2669) for P = 0104 we have, C ≡ (104)3 ≡ 1215( mod 2669). So here the ciphertext is 1215. Similarly, by means of the last congruence the message becomes, 12 15 12 24 14 71 00 23 01 16. 346 14.6 Number Theory and its Applications Worked out Exercises Problem 14.6.1. Encipher the message: SURRENDER, using the aﬃne transformation C ≡ 11P + 18(mod 26). Solution 14.6.1. First translate the message into its numerical equivalents, S 18 U 20 R 17 R 17 E 4 N 13 D 3 E 4 R 17 Then encipher each numerical equivalence using the relation C ≡ 11P + 18( mod 26). For P = 18, C ≡ 11 × 18 + 18 ≡ 8(mod 26) and continuing we get, 8 4 23 23 10 5 25 10 23. Now translating to letters we have, I E X X K F Z K X. Problem 14.6.2. Decipher the message: RTOLKTOIK, which was encripted by the transformation, C ≡ 3P + 24(mod 26). Solution 14.6.2. To begin with, convert the transformation by inverse of 3 modulo 26. Note that 3 · 9 ≡ 1(mod 26). Thus 9 is inverse of 3 modulo 26. Now we have, 9(C − 24) ≡ 3 · 9P ( mod 26) ∴ P ≡ 9C + 18( mod 26). Translating the alphabets in numeric we get, R 17 T 19 O 14 L 11 K 10 T 19 O 14 I 8 K 10 If C = 17 then P ≡ 9 × 17 + 18 ≡ 15(mod 26). This gives us, 15 7 14 13 4 7 14 12 4. and converting back to letters we get, PHONE HOME. Problem 14.6.3. Decipher the ciphertext message: UW DM NK QB EK, which was encripted using digraphic cipher which sends the plaintext P1 P2 into ciphertext block C1 C2 with, C1 ≡ 23P1 + 3P2 ( mod 26) C2 ≡ 10P1 + 25P2 ( mod 26). Certain Applications on Number Theory 347 Solution 14.6.3. First we translate the message into numeric equivalents, U W D M N K Q B E K 20 22 3 12 13 10 16 1 4 10. 23 3 1 3 Now the matrix has the inverse under modulo 26. Now for 10 25 10 3 U W the plaintext block is, 1 3 20 8 ≡ ( mod 26). 10 3 22 6 Thus U W has been transferred to IG. Similarly transferring other ciphertexts we get, IG NO RE TH IS. Problem 14.6.4. What is the plaintext message that corresponds to the ciphertext 12 13 09 02 05 39 12 08 12 34 11 03 13 74 . produced using modular exponentiation with modulus p = 2591 and enciphering key e = 13? Solution 14.6.4. Here p−1 = 2590 and 13·797 ≡ 1(mod 2590). Then d = 797. Applying the plaintext formula P ≡ C 797 ( mod 2591) we ﬁnd, 03 14 13 14 19 17 04 00 03 19 07 08 18 23 . Converting this to letters we get, DO NO TR EA DT HI SX where X is dummy alphabet. Problem 14.6.5. If the ciphertext message: 05041874034705152088235607360468 is produced by RSA cipher with e = 5 and n = 2881, ﬁnd the plaintext message. Solution 14.6.5. Since n = 2881 = 43 · 67 and φ(n) = 42 · 66 = 2772, the inverse of 5 is 1109 modulo 2772(∵ 5 · 1109 ≡ 1(mod 2772)). Now we need to take each 4-digit block as ciphertext to convert into plaintext using the formula P ≡ C 1109 (mod 2881). Calculating congruences for each 4-digit block ciphertext we get plaintext block as, 348 Number Theory and its Applications 04 00 19 02 07 14 02 14 11 00 19 04 02 00 10 04 . Converting the letters we get, EA TC HO CO LA TE CA KE. 14.7 Exercises: 1. Prove that Un+3 − Un = 2Un+1 whenever n is a positive integer. 2. Evaluate gcd(U24 , U36 ). 3. If 3 divides n + m, prove that Un−m−1 Un + Un−m Un+1 is an even integer. 4. Show that the sum of the ﬁrst n Fibonacci numbers with even indices is given by the formula U2 + U4 + U6 + · · · + U2n = U2n+1 − 1. 5. Find the Zeckendorf representation of 85 and 200. 6. Find the sequence of pseudo random numbers generated using the middlesquare method, taking 369 as the seed. 7. Find the period length of the pure multiplicative pseudo random number generator xn ≡ axn−1 (mod 231 − 1) for the multiplier a = 17. 8. Find the sequence of numbers generated by the square pseudo random number generator with modulus 1001 and seed 5. 9. Find a good choice for the multiplier a in the pure multiplicative pseudo random number generator xn+1 ≡ axn (mod 101). 10. Considering m = 100, a = 17 and c = 43 where the initial seed is x0 = 27, ﬁnd the sequence of pseudo random numbers using linear congruential method. 11. If the Caesar cipher produced KDSSB ELUWKGDB, what is the plaintext message? 12. Using the linear cipher C ≡ 5P + 11(mod 26), encrypt the message NUMBER THEORY IS EASY. 13. Decrypt the message YLF QX PCRIT, which was encrypted using the aﬃne transformation C ≡ 21P + 5(mod 26). Certain Applications on Number Theory 349 14. Using the digraphic cipher that sends the plaintext block P1 P2 to the ciphertext block C1 C2 , with C1 ≡ 8P1 + 9P2 ( mod 26) C2 ≡ 3P1 + 11P2 ( mod 26), encrypt the message DO NOT SHOOT THE MESSENGER. 15. What is the ciphertext that is produced when RSA encryption with key e = 7 and n = 2627 is used to encrypt the message LIFE IS A DREAM? Bibliography [1] Burton,D., Elementary Number Theory, 6th Edition, McGraw-Hill Science/Engineering/Math; 6th edition. [2] Silverman, J.H., A Friendly Introduction to Number Theory, Pearson; 4th edition(January 28, 2012). [3] Apostle, Tom, M., Introduction to Analytic Number Theory, Springer. [4] Rosen, R.K, Elementary Number Theory and its Applications(6th Edition), Pearson Addision Wesely. [5] Niven, I., Zuckerman, H., An Introduction to the Theory of Numbers(4th Edition), John Wiley & Sons. [6] Baker, A., A concise Introduction to the Theory of Numbers, Cambridge University Press. [7] Koblitz, N., A course in number theory and cryptography, 2nd Edition, Springer-Verlag. [8] Knuth, D.E., Art of Computer Programming: Semi Numerical Algorithms(Vol.2)(2nd Edition), Addision-Wesley; Reading Massachusetts; 1980, [9] Hill, L.S., Concerning certain linear transformation apparatus of cryptography(Vol.38), American Mathematical Monthly(1931), [10] CW Coppel, W.A., Number Theory An Introduction to Mathematics: Part A, Springer. 351 Index abundant, 247 amicable numbers, 255 amicable pair, 255 arithmetic function, 117 arithmetic progression, 55, 56 Bertrand Conjecture, 64 Binet formula, 328 boundedness, 12 Caesar cipher, 340 Carmichael function, 191 Carmichael numbers, 99 Chinese Remainder Theorem, 84 Chinese remainder theorem, 80 ciphertext, 339 Collatz conjecture, 23 complete residue system, 73 composite, 46, 48, 98, 120 composite numbers, 99 congruence, 68 congruence class, 72 congruence relation, 70 congruent, 68 congruent modulo, 72 consecutive, 136 continued fraction, 266 convergence, 12 coprime, 24, 25 countably inﬁnite, 48 cryptoanalysis, 339 cryptography, 95, 339 Cryptology, 343 deciphering key, 345 decryption, 340 deﬁcient, 247 diﬀerence of two squares, 311 Diophantine equation, 37, 275, 300 Diophantine equations, 39 Dirichilet’s Theorem, 283 divisible, 16 division algorithm, 17, 133 encryption, 340 equivalence class, 72 equivalence relation, 71 Euclid’s Algorithm, 30, 40 Euclid’s Lemma, 28 Euclidean Algorithm, 266 Euler polynomial, 56 Euler’s Criterion, 205 Euler’s generalization, 153 Euler’s phi function, 146, 156, 158 exponent, 138 exponential congruence, 198 exponentiation cipher, 343 factorial, 134 factorisation, 114 Fermat number, 257, 260 353 354 Fermat prime, 257 Fermat’s Last Theorem, 302 Fermat’s Little Theorem, 96 Fibonacci number, 325 Fibonacci sequence, 324 ﬁnite continued fraction, 266 ﬁnite simple continued fraction, 267 Gauss Lemma, 209 geometric series, 124 Golbach’s Conjecture, 53 greatest common divisor, 23, 24, 27 greatest integer function, 132 greatest lower bound, 9 Hill cipher, 341 index, 192, 193 inﬁnite continued fractions, 276 integer, 98 irrational, 6, 50, 63 Number Theory and its Applications lucky numbers, 56 M¨ obius inversion, 113 mathematical induction, 2 Mersenne number, 99, 247, 248 Mersenne Prime, 248 Mid-Square method, 334 Miller’s test, 100 minimal remaider, 21 minimal remainder, 19 Mobio¨us Inversion, 160 monographic cipher, 341 monotone decreasing sequence, 13 monotone increasing sequence, 13 multiplicative function, 117, 147 multiplicative property, 116 Pascal’s triangle, 5 Pepin’s Test, 259 perfect number, 239, 241 period length, 337 periodic fraction, 286 Jacobi symbol, 225 Pigeonhole principle, 5 Korselt’s Criterion, 99 plaintext, 339 Kronecker symbol, 234 polygraphic cipher, 341 polynomial congruence, 175 Lame´ s theorem, 327 polynomial congruences, 290 least common multiple, 31 prime, 46, 114, 134 least universal exponent, 337 primitive Pythagorean triple, 296, 297, least upper bound, 8 302 Legebdre symbol, 219 primitive root, 167, 170, 171, 173, 174 Legendre symbol, 206, 249 product, 123 linear combination, 25, 26 pseudo-random numbers, 334 linear congruence, 77, 78, 80, 85, 87, pseudoprime, 98 106, 192 public-key, 95 linear congruential generator, 337 pure multiplicative generator, 337 linear congruential method, 336 Pythagorean triple, 296 linear Diophantine equation, 77, 78 Liouville function, 140 quadratic congruence, 107, 203, 222 Lucas-Lehmer Test, 251 quadratic irrational, 286 Index quadratic non residue, 210 quadratic nonresidue, 204 quadratic reciprocity, 220 Quadratic Reciprocity Law, 217 quadratic reciprocity law, 203 quadratic residue, 204, 205, 208 rational, 50 rational numbers, 6 recursion formulae, 333 reﬂexive, 71 relatively prime, 117 remainder, 75 residue, 67 RSA algorithm, 345 Sieve of Eratosthenes, 50 simple continued fraction, 333 simple fundamental recurrence relations, 276 simple inﬁnite continued fraction, 277 square free, 99 successive convergents, 274 sum of four squares, 318 sum of three squares, 315 sum of two squares, 308 super perfect number, 247 symmetric, 71 Thue, 308 transitive, 71 triangular number, 243 twin primes, 121 well-ordering, 2 Wilson’s theorem, 106 Zeckendorf representation, 328 355	224,754	595,037	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.890625	4	CC-MAIN-2024-26	latest	en	0.750133
https://www.coursehero.com/file/6473068/17-graphs/	1,513,193,270,000,000,000	text/html	crawl-data/CC-MAIN-2017-51/segments/1512948530668.28/warc/CC-MAIN-20171213182224-20171213202224-00333.warc.gz	734,636,752	24,103	17-graphs - Algorithms Lecture 17 Basic Graph Properties... This preview shows pages 1–2. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Algorithms Lecture 17: Basic Graph Properties [ Sp’10 ] Obie looked at the seein’ eye dog. Then at the twenty-seven 8 by 10 color glossy pictures with the circles and arrows and a paragraph on the back of each one...and then he looked at the seein’ eye dog. And then at the twenty-seven 8 by 10 color glossy pictures with the circles and arrows and a paragraph on the back of each one and began to cry. Because Obie came to the realization that it was a typical case of American blind justice, and there wasn’t nothin’ he could do about it, and the judge wasn’t gonna look at the twenty- seven 8 by 10 color glossy pictures with the circles and arrows and a paragraph on the back of each one explainin’ what each one was, to be used as evidence against us. And we was fined fifty dollars and had to pick up the garbage. In the snow. But that’s not what I’m here to tell you about. — Arlo Guthrie, “Alice’s Restaurant” (1966) I study my Bible as I gather apples. First I shake the whole tree, that the ripest might fall. Then I climb the tree and shake each limb, and then each branch and then each twig, and then I look under each leaf. — Martin Luther 17 Basic Graph Properties 17.1 Definitions A graph G is a pair of sets ( V , E ) . V is a set of arbitrary objects that we call vertices 1 or nodes . E is a set of vertex pairs, which we call edges or occasionally arcs . In an undirected graph, the edges are unordered pairs, or just sets of two vertices; I will usually write uv instead of { u . v } to denote the undirected edge between u and v . In a directed graph, the edges are ordered pairs of vertices; I will usually write u v instead of ( u . v ) to denote the directed edge from u to v . We will usually be concerned only with simple graphs, where there is no edge from a vertex to itself and there is at most one edge from any vertex to any other. Following standard (but admittedly confusing) practice, I’ll also use V to denote the number of vertices in a graph, and E to denote the number of edges. Thus, in an undirected graph, we have ≤ E ≤ V 2 , and in a directed graph, 0 ≤ E ≤ V ( V- 1 ) . If ( u , v ) is an edge in an undirected graph, then u is a neighbor or v and vice versa. The degree of a node is the number of neighbors. In directed graphs, we have two kinds of neighbors. If u v is a directed edge, then u is a predecessor of v and v is a successor of u . The in-degree of a node is the number of predecessors, which is the same as the number of edges going into the node. The out-degree is the number of successors, or the number of edges going out of the node. A graph G = ( V , E ) is a subgraph of G = ( V , E ) if V ⊆ V and E ⊆ E .... View Full Document {[ snackBarMessage ]} Page1 / 11 17-graphs - Algorithms Lecture 17 Basic Graph Properties... This preview shows document pages 1 - 2. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	796	3,259	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2017-51	latest	en	0.933183
https://www.jiskha.com/members/profile/posts.cgi?name=Kimber	1,516,104,476,000,000,000	text/html	crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00060.warc.gz	944,943,303	3,863	# Posts by Kimber Total # Posts: 22 Math My teacher said the answer is 6.7% but I don't know how to get that Math The population in a 55 square mile city, City A, is 44,000. A neighboring city, City B, measures 65 square miles and has 48,750 citizens. Determine which city has a greater population density and by what percentage its population density is greater than the other city's. I... Math ho. Explain why 6.7320508075689 is a irrational. Math ho. A piece of ribbon 25 meters long is cut into pieces of equal length. It is possible to get a piece with irrational length? Science Why are Canadians air masses considered to be dry air masses? MATH HI. THE REGION INCLUDING THE CIRCLE AT THE FREE-THROW LINE TO THE BASELINE IS GIVEN. FIND THE AREA OF THE REGION. USE PI AS 3.14. IN THE PICTURE GIVEN, ITS 12 FEET BY 19 FEET RECTANGLE WITH A CIRCLE INTERSECTION AS A FREE-THROWN LINE. English He is Dieing ,And not being able to do anything ,but knowing that his family is with him . He passes away. MATH I DON'T KNOW ??? HELP Chemistry What molarity of KNO3 should be used in order for dilution of 63.2 mL to yield 253 mL of 0.319 M KNO3? Chemistry to how many milliliters must 29.6 mL of 4.25 M HCL be diluted to make 0.654 M HCL? Chemistry How many milliliters of 2.50 M KI must be diluted to make 365 mL of 0.432 M KI? Chemistry If 25.9 mL of 300 M NaCl is diluted to 136 mL, what is the molarity of the dilute solution? Chemistry and with this one I did 0.1667(750.0)=0.225(v) 125.025/0.225=555.7?? Chemistry Describe how to make 750.0 mL of 0.1667 M NH4C2H3O2 from 0.225 M NH4C2H3O2? Chemistry ok so did I do it right?? 0.225(750.0)=0.1667(v) 168.75/0.1667=1012.3?? Chemistry How many milliliters of 0.225 M NH4C2H3O2 are needed to make 750.0 mL of 0.1667 M NH4C2H3O2? pre algebra what is the length of a banana-14m,14cm or 14mm? physics A 30g bullet with a speed of 400m=s strikes a glancing blow to a target brick of mass 1:0kg. The Brick breaks into two fragments. The bullet deflects at a angle of 30 above the +x-axis and has a reduced speed of 100m=s. One piece of the brick with mass (0.75kg) goes off ... physics A 30g bullet with a speed of 400m=s strikes a glancing blow to a target brick of mass 1:0kg. The Brick breaks into two fragments. The bullet deflects at a angle of 30 above the +x-axis and has a reduced speed of 100m=s. One piece of the brick with mass (0.75kg) goes off ... math The shortest side of a polygon of area 196 square inches is 4 inches long. What is the area of a similar polygon whose shortest side is 8 inches long? mathematics if a circle has a radius of 6 inches,what will the area of the circumscribed equilateral triangle be? orientalism of muslim and arab american According Pyne, how have post-9-11 government responses affected prejudice and discrimination against Muslims, Arabs, and related groups? o List two to three characteristics of Orientalism. How may Orientalism and prejudice contribute to hate crimes against these groups? o In...	893	3,018	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2018-05	latest	en	0.890837
https://www.pw.live/chapter-is-matter-around-us-pure-class-9/true-solution	1,670,230,336,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00723.warc.gz	979,940,808	19,050	# TRUE SOLUTION ## Is matter around us pure of Class 9 A solution in which particles of the solute are broken down to such a fine state, that they cannot be seen under a powerful microscope is called a true solution. ### Characteristics of a True Solution : • A true solution is always clear and transparent, i.e., light can easily pass through it without scattering. • The particles of a solute break down to almost molecular size and their diameter is of the order of 1 nm (10-9 m) or less. • A true solution can completely pass through a filter paper as particle size of solute is far smaller than the size of pores of filter paper. • A true solution is homogeneous in nature. • In a true solution, the particles of solute do not settle down, provided temperature is constant. • From a true solution, the solute can easily be recovered by evaporation or crystallisation. ### Concentration of a Solution : It is defined as the amount of solute present in a given quantity of the solution. The most common method for expressing the concentration of a solution is called percentage method. The concentration of solution refers to the percentage of solute present in solution. Furthermore, the percentage of solute can be expressed in terms of : • mass of the solute • volume of the solute. • Concentration of a solution in terms of mass percentage of solute : If a solution is formed by dissolving a solid solute in a liquid solvent then the concentration of solution is expressed in terms of mass percentage of solute and is defined as under : The concentration of solution is the mass of the solute is grams, which is present in 100 g of a solution. The percentage concentration of a solution refers to mass of solute in 100 g of solution and not 100g of solvent, i.e., water. The concentration of a solution in terms of mass percentage of solute is calculated by the formula given below : Concentration of solution (ii) Concentration of a solution in terms of volume percentage of solute : If a solution is formed by dissolving a liquid solute in a liquid solvent, then the concentration of the solution is expressed in terms of volume percentage of solute. The concentration of a solution is the volume of the solute in milliners, which is present in 100 milliliters of a solution. It is very important to keep in mind that the percentage concentration of solution refers to volume of solute in 100 ml of solution and not 100 ml of solvent, i.e., water. The concentration of a solution in terms of volume percentage of the solute is calculated by the formula given below : Concentration of solution ### Saturated, Unsaturated and Supersaturated Solutions: A solution, in which no more solute can be dissolved at that temperature, is called a saturated solution. A solution, in which more quantity of solute can be dissolved without raising its temperature, is called an unsaturated solution. A solution which temporarily contains more solute than the saturation level (i.e. maximum solute) at a particular temperature, is called a supersaturated solution. ### Test for saturated, unsaturated and supersaturated solutions: In order to test, whether a given solution is saturated or unsaturated, add some more solute to this solution and try to dissolve by stirring with glass rod keeping temperature constant. If more solute does not dissolve in the given solution, then it must be a saturated solution and if more solute dissolves, it must be an unsaturated solution. On the other hand, supersaturated solution can be easily distinguished from the saturated solution simply by adding a few crystals of solute dissolved. If the precipitation of some additional solute occurs, it is a supersaturated solution; otherwise it is a saturated solution. Ex.1 What is the meaning of 15% solution of NaCl ? Solution: 15% solution of NaCl is a solution 100 g of which contains 15 g of NaCl and 85 g of water. Ex.2 Calculate the amount of glucose required to prepare 250 g of 5% solution of glucose by mass. Solution: % of solute Mass of solute = Ex.3 A solution contains 50 mL of alcohol mixed with 150 mL of water. Calculate concentration of this solution. Solution: This solution contains a liquid solute (alcohol) mixed with a liquid solvent (water), so we have to calculate the concentration of this solution in terms of volume percentage of solute (alcohol). Now, we know that: Concentration of solution Here, Volume of solute (alcohol) = 50 mL And. Volume of solvent (water) = 150 mL So, Volume of solution = Volume of solute + Volume of solvent = 50 + 150 = 200 mL Now, putting these values of ‘volume of solute’ and ‘volume of solution’ in the above formula we get : Concentration of solution percent (by volume) Thus, the concentration of this alcohol solution is 25 per cent or that it is a 25%. Ex.4 How much water should be added to 16 ml acetone to make its concentration 48%? Solution: Concentration of solution x = × 100 = 33.33 ml Volume of solvent = 33.33 - 16 = 17.33 ml. ### SOLUBILITY: The maximum amount of solute which can be dissolved in 100 gms of a solvent at a specified temperature is known as the solubility of that solute in that solvent (at that temperature). e.g. A maximum of 36 gms of common salt (NaCl) can be dissolved in 100 g. of water at 20°C (or 293 K). Therefore, the solubility of common salt in water at 20°C is 36 g. Different substances have different solubilities in the same solvent. Let us understand with the help of an experiment : • Take approximately 50 mL of water each in two separate beakers. • Add common salt in one beaker and sugar or barium chloride in the second beaker with continuous stirring. • When no more solute can be dissolved, heat the contents of the beaker. • Start adding the solute again. • Is the amount of common salt and sugar or barium chloride, that can be dissolved in water at a given temperature, the same? • What would happen if you were to take a saturated solution at a certain temperature and cool it slowly? ### Discussion: • The amounts of common salt, sugar and barium chloride that can be dissolved in water (50 mL) at room temperature are different. • When a saturated solution at a certain temperature is cooled, the solubility decreases and the amount of the solute which exceeds the solubility at lower temperature crystallizes out of the solution. Conclusion: Different substances have different solubilities in a given solvent at the same temperature and, in general, the solubility decreases as the solution is cooled and the extra amount of solute crystallizes out. Ex. 2.5 g of a solute is dissolved in 25 g of water to form a saturated solution at 298 K. Find out the solubility of the solute at this temperature. Solution: Mass of the solute = 2.5 g Mass of the solvent = 25 g ∴ Solubility of the solute =	1,547	6,858	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.34375	3	CC-MAIN-2022-49	latest	en	0.937673
https://istopdeath.com/find-the-roots-zeros-using-the-rational-roots-test-fxx34x2-140x-143/	1,670,387,859,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00549.warc.gz	358,963,528	18,353	Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+4x^2-140x-143 If a polynomial function has integer coefficients, then every rational zero will have the form where is a factor of the constant and is a factor of the leading coefficient. Find every combination of . These are the possible roots of the polynomial function. Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is , which means it is a root. Simplify the expression. In this case, the expression is equal to so is a root of the polynomial. Simplify each term. Raise to the power of . Raise to the power of . Multiply by . Multiply by . Subtract from . Since is a known root, divide the polynomial by to find the quotient polynomial. This polynomial can then be used to find the remaining roots. Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by . Place the numbers representing the divisor and the dividend into a division-like configuration. The first number in the dividend is put into the first position of the result area (below the horizontal line). Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend . Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line. Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend . Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line. Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend . Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line. All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder. Simplify the quotient polynomial. Solve the equation to find any remaining roots. Use the quadratic formula to find the solutions. Substitute the values , , and into the quadratic formula and solve for . Simplify. Simplify the numerator. Raise to the power of . Multiply by . Multiply by . Multiply by . Simplify the expression to solve for the portion of the . Simplify the numerator. Raise to the power of . Multiply by . Multiply by . Multiply by . Change the to . Rewrite as . Factor out of . Factor out of . Move the negative in front of the fraction. Simplify the expression to solve for the portion of the . Simplify the numerator. Raise to the power of . Multiply by . Multiply by . Multiply by . Change the to . Rewrite as . Factor out of . Factor out of . Move the negative in front of the fraction. The final answer is the combination of both solutions. The polynomial can be written as a set of linear factors. These are the roots (zeros) of the polynomial . The result can be shown in multiple forms. Exact Form: Decimal Form: Find the Roots/Zeros Using the Rational Roots Test f(x)=x^3+4x^2-140x-143	661	3,132	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5625	5	CC-MAIN-2022-49	latest	en	0.899025
https://quizizz.com/en-us/triangle-theorems-worksheets-grade-8	1,716,340,272,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00860.warc.gz	416,606,624	27,996	## Recommended Topics for you Triangle Theorems 21 Q 6th - 8th Triangle Theorems 20 Q 8th Triangle Theorems 10 Q 8th Triangle Theorems 20 Q 8th Triangle Theorems - Solve 20 Q 8th Investigating Triangle Theorems Practice 17 Q 8th - 10th Triangle Theorems Review 18 Q 7th - 8th Triangle Theorems 26 Q 8th Triangle Theorems 15 Q 8th Parallel Lines and Triangle Theorems Quick Check 10 Q 8th Triangle Theorems 17 Q 8th Angle Relationships and Triangle Theorems 25 Q 8th External/internal triangle theorems 6 Q 8th 8B Triangle Theorems Quiz 25 Q 6th - 8th 8th Final - Transversal Angles And Triangle Theorems 21 Q 8th Triangle Theorems Review 20 Q 8th - 10th Triangle Theorems 30 Q 8th Triangle Theorems 22 Q 8th Topic 23 Triangle Theorems 12 Q 8th Triangle Theorems 21 Q 8th Triangle Angle Theorems 16 Q 8th - 11th Triangle Congruence Theorems 14 Q 8th - 10th Angles of a Triangle Theorems 32 Q 8th ## Explore printable Triangle Theorems worksheets for 8th Grade Triangle Theorems worksheets for Grade 8 are an essential resource for teachers looking to enhance their students' understanding of math and geometry concepts. These worksheets are specifically designed to cater to the learning needs of Grade 8 students, focusing on the various triangle theorems that form the basis of their geometry curriculum. Teachers can utilize these worksheets to provide ample practice opportunities for their students, reinforcing their understanding of the subject matter. The worksheets cover a wide range of topics, including the Pythagorean theorem, triangle congruence, and similarity, as well as angle and side relationships within triangles. By incorporating these Triangle Theorems worksheets for Grade 8 into their lesson plans, teachers can ensure that their students develop a strong foundation in math and geometry. Quizizz is an excellent platform for teachers to supplement their Triangle Theorems worksheets for Grade 8, as it offers a variety of interactive quizzes and activities that can be used to assess students' understanding of math and geometry concepts. In addition to the worksheets, Quizizz provides teachers with a plethora of resources, such as customizable quizzes, engaging game-based activities, and real-time feedback on student performance. This allows teachers to identify areas where their Grade 8 students may need additional support or practice. Furthermore, Quizizz integrates seamlessly with popular learning management systems, making it easy for teachers to incorporate these resources into their existing lesson plans. By utilizing Quizizz alongside Triangle Theorems worksheets for Grade 8, teachers can create a comprehensive and engaging learning experience for their students, ensuring their success in math and geometry.	655	2,825	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2024-22	latest	en	0.926642
https://pybenchmarks.org/u64q/program.php?test=fannkuchredux&lang=python3&id=3	1,579,958,603,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00336.warc.gz	602,791,562	3,951	## performance measurements Each table row shows performance measurements for this Python 3 program with a particular command-line input value N. N CPU secs Elapsed secs Memory KB Code B ≈ CPU Load 1013.173.4747,392894 97% 99% 91% 99% 1013.093.4848,124894 100% 92% 96% 95% 1013.033.4448,036894 97% 99% 91% 99% Read the ↓ make, command line, and program output logs to see how this program was run. Read fannkuch-redux benchmark to see what this program should do. ### notes Python 3.3.1 (default, Apr 11 2013, 12:45:45) [GCC 4.7.2] on linux ## fannkuch-redux Python 3 #3 program source code ```# The Computer Language Benchmarks Game # http://benchmarksgame.alioth.debian.org/ # # contributed by Joerg Baumann # many thanks to Oleg Mazurov for his helpful description from sys import argv from math import factorial from multiprocessing import cpu_count, Pool from itertools import islice, starmap def permutations(n, start, size): p = bytearray(range(n)) remainder = start for v in range(n - 1, 0, -1): rotation_count, remainder = divmod(remainder, factorial(v)) for _ in range(rotation_count): p[:v], p[v] = p[1:v + 1], p[0] if size < 2: yield p[:] else: rotations = [(v, factorial(v)) for v in range(3, n)] for i in range(start + 2, start + size + 2, 2): yield p[:] p[0], p[1] = p[1], p[0] yield p[:] p[1], p[2] = p[2], p[1] for v, modulo in rotations: if i % modulo != 0: break p[:v], p[v] = p[1:v + 1], p[0] def alternating_flips_generator(n, start, size): maximum_flips = 0 alternating_factor = 1 for permutation in permutations(n, start, size): first = permutation[0] if first: flips_count = 1 while True: permutation[:first + 1] = permutation[first::-1] first = permutation[0] if not first: break flips_count += 1 if maximum_flips < flips_count: maximum_flips = flips_count yield flips_count * alternating_factor else: yield 0 alternating_factor = -alternating_factor yield maximum_flips alternating_flips = alternating_flips_generator(n, start, size) return sum(islice(alternating_flips, size)), next(alternating_flips) def fannkuch(n): assert(n > 0) total = factorial(n) with Pool() as pool: else: checksum, maximum = sum(checksums), max(maximums) print("{0}\nPfannkuchen({1}) = {2}".format(checksum, n, maximum)) if __name__ == "__main__": fannkuch(int(argv[1])) ``` ### make, command-line, and program output logs ```Wed, 22 Jan 2020 18:09:21 GMT COMMAND LINE: /usr/bin/python3 fannkuchredux.python3-3.python3 10 PROGRAM OUTPUT: 73196 Pfannkuchen(10) = 38 ```	791	2,498	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.859375	3	CC-MAIN-2020-05	longest	en	0.548737
https://nl.mathworks.com/matlabcentral/answers/360158-is-subset-s-a-subspace-of-r3	1,669,906,706,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710813.48/warc/CC-MAIN-20221201121601-20221201151601-00273.warc.gz	475,468,357	27,596	# is subset S a subspace of R3? 12 views (last 30 days) Hannah Blythe on 7 Oct 2017 Edited: Cedric Wannaz on 8 Oct 2017 S - {(2x-y, xy, 7x+2y): x,y is in R} of R3 Three requirements I am using are i. S is nonempty ii. For any u,v in S, u+v is in S iii. For any c in R and u in S, cu is in S So far I have proved the first condition. For the second condition I said a,b,c,d are in R and that u=(2a-b, ab, 7a+2b) and v=(2c-d, cd, 7c+2d). When I added them I got {2(a+c)-(b+d), (ab)+(cd), 7(a+c)+2(b+d)}. What should I do from here? Cedric Wannaz on 7 Oct 2017 Edited: Cedric Wannaz on 7 Oct 2017 Good start. As (a+c) in R and so is (b+d), you demonstrated that the first component of u+v is compatible with the definition of S. What about the second component? In other words, can you develop/expand/factor ab+cd into a product of (a+c) and (b+d)? If you can, check the third component. If you cannot, well unless it's obvious it's not a proof (maybe it's just that you are unable to do it, but it could be possible). Yet you may be able to use what you understand about the mismatch for building a counter-example, and you only need one to show that it doesn't work. Cedric Wannaz on 8 Oct 2017 Edited: Cedric Wannaz on 8 Oct 2017 Now that you have probably finished your exercise, here is S with the origin displayed as a red dot: Looking at this, you understand why, when you pick two vectors defined by points that are part of S, their sum is likely not part of S. And here is the little bit of code that produced this figure (minus the rotations that I did by hand), in case your exercise was related to MATLAB: [U,V] = meshgrid( -10:10, -10:10 ) ; X = 2U-V ; Y = U.V ; Z = 7U+2V ; for k = 1 : 4 subplot( 2, 2, k ) ; hold on ; grid on ; surf( X, Y, Z ) ; plot3( 0, 0, 0, 'r.', 'MarkerSize', 20 ) ; end	590	1,810	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2022-49	latest	en	0.94409
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-9-inequalities-and-problem-solving-9-3-absolute-value-equations-and-inequalities-9-3-exercise-set-page-599/64	1,534,905,651,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221219242.93/warc/CC-MAIN-20180822010128-20180822030128-00328.warc.gz	893,416,315	14,983	## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $a=-4$ The absolute value of $x,$ written as $\|x\|,$ is the distance of $x$ from $0,$ and hence, is always a nonnegative number. Therefore, the left side of the given inequality, $\|a+4\|\le0 ,$ is always nonnegative. This means that the inequality is only satisfied when $\|a+4\|=0$ or when $a+4=0.$ Using the properties of equality to isolate the variable, the expression $a+4=0$ is equivalent to \begin{array}{l}\require{cancel} a+4=0 \\\\ a=-4 .\end{array} In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$	184	631	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.375	4	CC-MAIN-2018-34	longest	en	0.82776
https://www.autoitscript.com/forum/topic/125098-performance-optimization-syntax-answers-and-restructuring-ideas-welcome/	1,579,606,522,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250603761.28/warc/CC-MAIN-20200121103642-20200121132642-00087.warc.gz	771,004,988	43,727	# Performance optimization - syntax answers and restructuring ideas welcome ## Recommended Posts My program is very dependent on running fast. I currently have this code running about 8 times each second (at the end of an endless loop until exiting): ```If NOT (StringLen(\$System_Log) == 0) PrintLogToTxt() \$System_Log = "" EndIf``` Note: \$System_Log is a global string containing between 0 to 150,000 characters each loop (varies). This should be faster than using `If NOT (\$System_Log == "")` or `If \$System_Log <> ""` ..but I would like confirmation on this and/or tips on further optimization. I'm thinking about running the first mentioned code every 2, or 4 loops instead of every. Would this speed up things? (the \$System_Log string would be 4x larger) Question 2) I also have gazillions of Case, Switch, and If-Else statements. After some research into this - "If" is obviously the slowest. But which is the fastest in general - case or switch? ##### Share on other sites ```If \$System_Log then PrintLogToTxt() \$System_Log = "" EndIf``` ```\$b = "1" For \$i = 2 To 150000 \$b &= "1" Next \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b Then \$y +=1 Next ConsoleWrite(TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if NOT (StringLen(\$b) == 0) Then \$y +=1 Next ConsoleWrite(TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 If NOT (\$b == "") Then \$y +=1 Next ConsoleWrite(TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b <> "" Then \$y +=1 Next ConsoleWrite(TimerDiff(\$timer) & @tab & \$y & @CRLF)``` ```\$a = 1 \$b = 2 \$timer = TimerInit() For \$i = 0 To 1000000 If \$a = \$b Then Else EndIf Next ConsoleWrite(TimerDiff(\$timer) & @CRLF) \$timer = TimerInit() For \$i = 0 To 1000000 Select Case \$a = \$b Case Else EndSelect Next ConsoleWrite(TimerDiff(\$timer) & @CRLF) \$timer = TimerInit() For \$i = 0 To 1000000 Switch \$a Case \$b Case Else EndSwitch Next ConsoleWrite(TimerDiff(\$timer) & @CRLF)``` Edited by KaFu ##### Share on other sites Thanks for some input. (New) Question 3) If I have a lot of boolean variables which i check and/or set, which is faster: ```If \$myBool Then \$myBool = False EndIf``` or just simply set it every time (at this specific line): `\$myBool = False` This code runs a few hundred times per second, so it will matter in the long run. ##### Share on other sites I've thought about optimization in the past with AutoIt, and basically came to the conclusion that the less text AutoIt has to parse, the better. The fastest - set \$myBool to 0 each time. 'False' isn't really needed other than for text output. A setting of any value other than 0 can be interpreted as 'True'. ##### Share on other sites I think the same might be said for the size of the AU3 file (or I could be wrong) If I have a single If...EndIf statement, I tend to have it on the one line. `If \$iInt = 0 Then ConsoleWrite("0" & @CRLF)` ```If \$iInt = 0 Then ConsoleWrite("0" & @CRLF) EndIf``` ##### Share on other sites I've thought about optimization in the past with AutoIt, and basically came to the conclusion that the less text AutoIt has to parse, the better. I've always found multiline If's to be faster than single line If's though. Doesn't stop me from using single line If's. To answer Q3: From what I can test it seems that evaluating this: ```If False Then ;this never gets executed \$myBool = False EndIf``` Takes about twice as long as setting this, when \$myBool is always false: `\$myBool = False` In fact this: ```\$myBool = False \$myBool = True``` Is only marginally slower than evaluating the if statement above, even if the if statement never has to actually change any variables. So in no circumstance could it be faster to use the if statement. This changes when you are setting a lot of variables, or more complex variables, but for a bool, just set it every time. One interesting thing I read on the topic a while ago is that autoit takes longer to execute code with longer variable and function names. So you might want to replace it all for names with 1 to 3 characters depending on how many names you need. I think there might be an obfuscator option to do so. The ultimate performance optimization would probably to learn something besides AutoIt for these applications. (that's not going to make me popular now is it?) ##### Share on other sites I think there might be an obfuscator option to do so. This is my standard option to boost performance ... #Obfuscator_Parameters=/sf /sv /om /cs=0 /cn=0 ##### Share on other sites @Xibalba You might also like Bowmore's AU3Profiler, search the Examples forum ##### Share on other sites whatever Edited by MvGulik "Straight_and_Crooked_Thinking" : A "classic guide to ferreting out untruths, half-truths, and other distortions of facts in political and social discussions." "The Secrets of Quantum Physics" : New and excellent 2 part documentary on Quantum Physics by Jim Al-Khalili. (Dec 2014) "Believing what you know ain't so" ... Knock Knock ... ##### Share on other sites I've always found multi line IFs to be faster than single line IFs though. Doesn't stop me from using single line IFs. I'm of the opinion that the first priority should be that the code should be clear and the logic used easy to follow. Only when that has been achieved and the application is working as expected should we worry about speed and then concentrate on those areas which are the real bottle necks. Tvern I was interested in your comment regarding the speed of multi line IFs verses single line IFs, so I did a bit of testing. It appears that there is a significant difference in speed between the two forms. Strangely though the difference is reversed depending on whether the test expression evaluates true or false as shown in the tests below. In the tests labeled AT BT CT DT the expression evaluates to true. In the tests labeled AF BF CF DF the expression evaluates to false. ```\$b = "qwertyuiytrew" ;evaluates to true \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b <> "" Then \$y +=1 Next ConsoleWrite("AT " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i=1 To 1000000 If \$b<>"" Then \$y+=1 EndIf Next ConsoleWrite("BT " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b Then \$y+=1 Next ConsoleWrite("CT " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b Then \$y+=1 EndIf Next ConsoleWrite("DT " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$b="" ;evaluates to false \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b <> "" Then \$y +=1 Next ConsoleWrite("AF " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i=1 To 1000000 If \$b<>"" Then \$y+=1 EndIf Next ConsoleWrite("BF " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b Then \$y+=1 Next ConsoleWrite("CF " & TimerDiff(\$timer) & @tab & \$y & @CRLF) \$y = 0 \$timer = TimerInit() For \$i = 1 To 1000000 if \$b Then \$y+=1 EndIf Next ConsoleWrite("DF " & TimerDiff(\$timer) & @tab & \$y & @CRLF)``` Edit: Spelling Edited by Bowmore "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to build bigger and better idiots. So far, the universe is winning."- Rick Cook ##### Share on other sites whatever Edited by MvGulik "Straight_and_Crooked_Thinking" : A "classic guide to ferreting out untruths, half-truths, and other distortions of facts in political and social discussions." "The Secrets of Quantum Physics" : New and excellent 2 part documentary on Quantum Physics by Jim Al-Khalili. (Dec 2014) "Believing what you know ain't so" ... Knock Knock ... ##### Share on other sites Tvern & Bowmore, that's interesting regarding the multi--line If/EndIf's. I've tested Bowmore's code, removing all unneeded spaces and adding a sleep before execution of the main script (sometimes AutoIt needs 'initialization' time with a script), and this is the result: ```AT 2680.35570615504 1000000 BT 1900.54062567844 1000000 CT 1917.61630901653 1000000 DT 1120.32454568287 1000000 AF 1306.79400458893 0 BF 1542.82777028091 0 CF 547.604857968695 0 DF 803.950824347737 0``` So on my PC, ~800 milliseconds extra for a matched single-line 'If..Then'. And ~250 ms. extra for non-matching multiline 'If..Then.EndIf' statements. Very interesting. I guess the scripter could determine which case is most likely to happen and then tweak the code based on that. Or just stick with 'If..Then..Endif', as that has the least performance impact either way. *edit: Thinking on this.. most of my code within loops where there is a single-line 'If..Then' usually results in an ExitLoop or Return, so those are best left alone.. Hmm.. and most people don't have anything running 1000000 times, so the scores above need to be taken with that in mind. Edited by Ascend4nt ##### Share on other sites Hmm.. and most people don't have anything running 1000000 times, so the scores above need to be taken with that in mind. Yeah I doubt this will have much practical use in most applications, but it's nice to have an idea of how these things behave. I was pretty surprised with the results I gor from Bowmore's script myself. ##### Share on other sites @Ascend4nt I agree it is easy to get carried away with small performance differences that don't have much impact in the real world. However personally I frequently use AutoIt at work to manipulate files with 25 to 50 million lines so small differences can somtimes make a significant time the application takes to run. Compared to using AutoIt in the first place is the biggest productivity boost. For many jobs, even if it is a one off the time taken to write a script and run it is far quicker than the alternative especially when you have a number of preprepared template scripts for different types of similar jobs. If you know something about the data you are processing significant performance increases can be achieved by ensuring the the most frequently occurring scenarios are checked for first in IF...Then; Switch or Select blocks. "Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to build bigger and better idiots. So far, the universe is winning."- Rick Cook ##### Share on other sites whatever Edited by MvGulik "Straight_and_Crooked_Thinking" : A "classic guide to ferreting out untruths, half-truths, and other distortions of facts in political and social discussions." "The Secrets of Quantum Physics" : New and excellent 2 part documentary on Quantum Physics by Jim Al-Khalili. (Dec 2014) "Believing what you know ain't so" ... Knock Knock ... ## Create an account or sign in to comment You need to be a member in order to leave a comment ## Create an account Sign up for a new account in our community. It's easy! Register a new account × • Wiki • Back • #### Beta • Git • FAQ • Our Picks × • Create New...	3,098	11,293	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.515625	3	CC-MAIN-2020-05	longest	en	0.7547
https://www.w3spoint.com/armstrong-number-in-c	1,685,667,715,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648245.63/warc/CC-MAIN-20230602003804-20230602033804-00361.warc.gz	1,149,173,304	14,407	# Armstrong Number in C Sure! An Armstrong number, also known as a narcissistic number, is a number that is equal to the sum of its own digits each raised to the power of the number of digits. For example, 153 is an Armstrong number because 1^3 + 5^3 + 3^3 = 153. Here’s an example of how you can check for Armstrong numbers in C: ```#include <stdio.h> #include <math.h> int isArmstrong(int number); int main() { int number; printf("Enter a number: "); scanf("%d", &number); if (isArmstrong(number)) { printf("%d is an Armstrong number.\n", number); } else { printf("%d is not an Armstrong number.\n", number); } return 0; } int isArmstrong(int number) { int originalNumber, remainder, result = 0, n = 0; originalNumber = number; // count the number of digits while (originalNumber != 0) { originalNumber /= 10; ++n; } originalNumber = number; // calculate the result while (originalNumber != 0) { remainder = originalNumber % 10; result += pow(remainder, n); originalNumber /= 10; } // check if the number is Armstrong if (result == number) { return 1; } else { return 0; } } ``` In this program, we define the `isArmstrong` function to check whether a given number is an Armstrong number. It calculates the sum of each digit raised to the power of the number of digits. If the resulting sum is equal to the original number, it returns 1; otherwise, it returns 0. In the `main` function, we take input from the user and call the `isArmstrong` function to determine whether the input number is an Armstrong number. Finally, we display the appropriate message based on the result. Please note that this code assumes that the input number is a non-negative integer. Categories C	424	1,694	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.15625	3	CC-MAIN-2023-23	longest	en	0.612105
https://fr.mathworks.com/matlabcentral/cody/problems/1952-count-decimal-digits-of-a-number/solutions/364877	1,611,037,996,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703517966.39/warc/CC-MAIN-20210119042046-20210119072046-00365.warc.gz	349,531,483	16,823	Cody # Problem 1952. Count decimal digits of a number Solution 364877 Submitted on 5 Dec 2013 by Tobias Schäfer This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass %% x = 248; y_correct = 3; assert(isequal(ndigits(x),y_correct)) 2 Pass %% x=1589 y_correct=4; assert(isequal(ndigits(x),y_correct)); x = 1589 3 Pass %% x=100 y_correct=3; assert(isequal(ndigits(x),y_correct)); x = 100 4 Pass %% x=-10; y_correct=2; assert(isequal(ndigits(x),y_correct)); ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting!	205	709	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.890625	3	CC-MAIN-2021-04	latest	en	0.523843
https://stackoverflow.com/questions/21462605/what-happens-when-loop-gets-to-the-end/21462722	1,624,360,844,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488517048.78/warc/CC-MAIN-20210622093910-20210622123910-00210.warc.gz	488,189,510	38,957	# What happens when loop gets to the end? I am python beginner, with no previous programming knowledge. I apologize for the name of the topic, but I simply could not make a better one. Here is what I want: ``````letter = "w" # search for this letter in the list bellow listL = ["z","b","y","a","c"] for let in listL: if let == letter: print "found it" break else: if let == listL[-1]: print "nope can't find it" continue `````` I have a list of letters, and I want to search for a particular letter in that list. If I find the letter, then everything is ok, and the for loop should stop. If I do not find it, I would like the loop to stop that current iteration, and try with the next letter in the list. If not a single letter in the list has that particular letter, then it should print "nope can't find it". The upper code works. But I was wondering if this could be wrote a bit clearly? And by clearly I do not mean "advanced", but scholar way, an-example-from-the-book way. Thank you. Python offers an `else` statement for the for loop that is executed if the loop ends without being broken: ``````for let in llistL: if let == letter: print("Found it!") break else: print("nope could'nt find it") `````` That would be the "scholar way" for a `for` loop, however if you just test the presence of an element in a list, Arkady's answer is the one to follow. ``````if letter in listL: print "found it" else: print "nope..." `````` Simply use in ``````if let in listL: print("Found it") else: `````` edit : you were faster by 30 s, congrats ;) Your loop will keep looping until it either breaks (found it!) or the list is exhausted. You do not need to do anything special to "stop that current iteration, and try with the next letter in the list". We don't need a `continue` when a letter doesn't match, this will happen automatically as long as there are more letters to check. We only want to display "nope can't find it" after we've searched through the entire list, so we don't need to check until the end. This `else` statement corresponds to the `for` loop, instead of the `if` in your previous code. ``````letter = "w" # search for this letter in the list bellow listL = ["z","b","y","a","c"] for let in listL: if let == letter: print "found it" break #found letter stop search else: #loop is done, didn't find matching letter in all of list print "nope can't find it" `````` There is actually a `for: else:` construct in Python, where the `else` runs if the `for` loop doesn't `break`: ``````for let in listL: if let == letter: print("Found it") break else: `````` Alternatively, you can use `list.index`, which will give the index of an item if found in a list and raise a `ValueError` if it isn't found: ``````try: index = listL.index(letter) except ValueError:	745	2,797	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2021-25	latest	en	0.94407
https://xorshammer.com/2021/04/08/but-why-is-proof-by-contradiction-non-constructive/	1,686,270,975,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224655244.74/warc/CC-MAIN-20230609000217-20230609030217-00247.warc.gz	1,158,652,564	23,114	# But Why Is Proof by Contradiction Non-Constructive? We think of a proof as being non-constructive if it proves “There exists an $x$ such that $P(x)$ without ever actually exhibiting such an $x$. If you want to form a system of mathematics where all proofs are constructive, one thing you can do is remove the principle of proof by contradiction: the principle that you can prove a statement $P$ by showing that $\neg P$ is false. (Let’s leave aside set-theoretical considerations for the moment.) But one thing you can ask is: exactly why is the principle of proof by contradiction non-constructive? In the paper Linear logic for constructive mathematics, Mike Shulman gave an answer which I found quite mind-blowing: Imagine you’re proving $\exists x\,P(x)$ by contradiction. This means that you allow yourself to assume $\neg\exists x\,P(x)$ and show a contradiction from there. The assumption is equivalent to $\forall x\,\neg P(x)$, but in order to use such an assumption, you actually have to produce an $x$, so shouldn’t that be constructive? The answer is that yes it will be, unless you use the hypothesis more than once! So (the paper reasons), you can form a constructive system of mathematics not by removing the law of proof by contradiction, but by requiring you to only use a hypothesis $P$ once when proving a statement $P\rightarrow Q$. Absolutely amazing! It gets even more amazing: Once you’ve committed to doing that, there’s a question: Does proving $(A\wedge B)\rightarrow C$ mean you’re allowed to use both $A$ and $B$ in your proof of $C$, or that you could use either to prove $C$? Both are reasonable interpretations, so conjunction splits into two separate connectives. Dually, disjunction also splits into two connectives, and these two connectives can be given the interpretations of “constructive-or” and “non-constructive-or”. Fantastic!	426	1,875	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.828125	4	CC-MAIN-2023-23	latest	en	0.924804
myessaymaster.com	1,606,268,625,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00352.warc.gz	402,317,022	7,398	# Total number of people unemployed Essay As total population increases, the number of persons employed may possibly increase but so would the number of persons without a job hence accounting for the increase in the above formula’s numerator and then the unemployment rate. 2 . Why frictional lack of employment is deemed desirable? Frictional unemployment arises from immobility in labor force which is not always occupationally mobile or perhaps geographically perfect. It has two categories namely search unemployment (when people are searching for better jobs than the ones they may be offered) and casual employment (when folks are between jobs example actors, roof repairers). It is appealing because it suggests growth and choice within an economy. Likewise, it is unconscious unemployment that is people are not forced out of work due to awful economic circumstances or a negative business ambiance (Sean, 2005). 3. The manuscript with this text was typed with a friend. A new secretary been hired to complete the same work, GDP might have been higher, even though the volume of output would have been identical. Exactly why is this? Does this make sense? GROSS DOMESTIC PRODUCT stands for Major Domestic Outcome and is thought as the value of result produced in the country on the 12 moth period. It could be measured in three fundamental ways specifically the income method, the output method plus the expenditure approach. According to the income method, GROSS DOMESTIC PRODUCT is tested in terms of wages, rent and profits. Therefore , a secretary would be paid out wages to type this manuscript which will would put towards the GROSS DOMESTIC PRODUCT total when a friend would do the be employed by free. four. GDP in 1981 was \$2. 96 trillion. It grew to \$3. '07 trillion more than 20 years ago, yet the quantity of output in fact decreased. How is this possible? The above trend is possible in the event that output was being measured in current prices. Though the volume of goods and services continued to be the same, all their prices may possibly have gone up during the 12 months period as a result of inflation which caused the GDP to grow to \$3. 07 trillion. Another reason can be the fact that production of tangible products decreased however the provision of intangible products (such because banking, insurance and teaching services) in fact went up. 5. If perhaps gross expense is not really large enough to switch the capital that depreciates within a particular year, is net investment better or lower than zero? What happens to our production possibilities? Net investment would be lesser than zero. The production opportunities would go down because the capital being used up is if she is not replaced quickly enough. 6th. If you anticipated prices to rise, would it become advantageous to take out a loan with a fixed interest rate mortgage? In terms of the importance of money after some time, explain for what reason or obtain. Yes, it might be advantageous. This is due to as rates would increase, the value of money would decrease. We because borrowers can be paying the same amount but it would now be worth significantly less to the loan provider because funds has misplaced its worth and \$1 can no longer nevertheless the same items it used to before. 7. Essay Extra Credit (5 points). Discover two organizations that reap the benefits of deflation and two that lose. Those who would gain would consist of lenders lending on a fixed interest rate and pensioners receiving a fixed quantity. Those who could lose will be borrowers credit on a fixed interest rate and folks giving out real estate on lease. References Publication Sean, Meters. F. (2005). Economics fundamentals. For Dummies.	744	3,730	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2020-50	latest	en	0.964802
https://clay6.com/qa/39373/find-the-ratio-in-which-yz-plane-divides-the-segment-formed-by-joining-the-	1,601,183,116,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400250241.72/warc/CC-MAIN-20200927023329-20200927053329-00079.warc.gz	324,443,024	6,145	# Find the ratio in which $YZ$ plane divides the segment formed by joining the points $(-2,4,7)$ and $(3,-5,8)$ $\begin{array}{1 1}3:2 \\ 2:3 \\ 3:4 \\ 4:3\end{array}$	68	167	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2020-40	latest	en	0.768471
https://www.coursehero.com/file/8650531/Homework-4-Solution/	1,529,497,133,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267863518.39/warc/CC-MAIN-20180620104904-20180620124904-00456.warc.gz	784,741,099	87,264	Homework 4 Solution # Homework 4 Solution - MAD 5305 Graph Theory Spring 2012... This preview shows pages 1–3. Sign up to view the full content. MAD 5305 Graph Theory, Spring 2012 CRN: 21139 Classes: PHY 109, MWF 10:45-11:35am Assignment 4 – solution Question 1 . (2 points) (a) Describe the breadth first search (BFS) algorithm when applied to a connected graph. (b) Explain how BFS can be used to find the components of a disconnected graph. (c) The time-complexity of BFS when applied to a connected graph with n vertices and m edges is O ( m + n ), where O ( x ) denotes a function f ( x ) satisfying f ( x ) Cx for some constant C > 0 and for all large enough x . Give an explanation of the term time-complexity , and explain why the time-complexity of BFS is O ( m + n ). Solution . (a) Here is a description of BFS using a queue Q (which is a data-structure): The Breadth First Search (BFS) algorithm Input . A connected graph G = ( V, E ), and an initial vertex u V . Output . A BFS tree ( V, E ( T )) with root u so that for every vertex v , the unique path from u to v is a shortest path between u, v in G . 1. (Initialization) E ( T ) := , first := 1, last := 1, Q [ first ] := u . Set visited [ u ] := true , and visited [ v ] := false for all v negationslash = u . 2. (Iteration) while ( first last ) do (that is, while queue is not empty) 3. x := Q [ first ], first := first + 1; 4. for all neighbors y of x do 5. if ( visited [ y ] = false ) then 6. last := last + 1, Q [ last ] := y , add { x, y } to E ( T ), visited [ y ] := true ; 7. end if; 8. end do; 9. end do; (b) Given a disconnected graph, apply BFS to an aribtrarily chosen vertex u , and obtain the set of vertices (from the spanning tree) of the component containing u . Repeat BFS for any un-visited vertex, and obtain (the vertices from) another component. Oviously this process can be repeated until all components are found. (c) The time-complexity of an algorithm is the “number of steps” required by the algorithm in the worst case among a set of instances (inputs) with given parameter(s). For example, in BFS, the set of instances is the set of connected graphs with n vertices and m edges. In order to find the time-complexity of BFS, we see that for each execution of the while loop in line 2, the operations in line 3 are executed once. Thus line 3 is executed n times (since graph is connected, and there are n vertices), and hence at most C n opertations, for some constant C , are incurred by line 3. The for loop in line 4 is executed summationdisplay x V deg( x ) = 2 m times, and each time that the for loop is iterated, a constant number of operations are performed. Thus, the number of steps incurred by the for loop in line 4 is at most C ′′ m for some constant C ′′ . This shows that the total number of operations required by BFS is at most C ( n + m ) for some constant C . In time-complexity, we use O ( m + n ), denoting that the time-complexity of BFS is 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document linear in m and n (as opposed to square, cubic, exp, or log). (For example, a linear-time algorithm is obviously better than an exponential-time algorithm, and the difference is huge when the problem size is big.) This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern	1,099	4,318	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.890625	4	CC-MAIN-2018-26	latest	en	0.864817
http://www.physicsforums.com/tags.php?tag=work	1,394,338,461,000,000,000	text/html	crawl-data/CC-MAIN-2014-10/segments/1393999671637/warc/CC-MAIN-20140305060751-00041-ip-10-183-142-35.ec2.internal.warc.gz	494,600,153	10,161	Threads Tagged with work Thread / Thread Starter Last Post Replies Views 1. The problem statement, all variables and given/known data Nitrogen was compressed from initial pressure of 101... T 10:13 PM Sunfire 1 101 Introductory Physics Homework 1. The problem statement, all variables and given/known data In a video game, a 70kg villain making an elusive... T 09:10 PM Simon Bridge 4 191 Introductory Physics Homework 1. The problem statement, all variables and given/known data A small steel ball of mass .0283kg is placed on the... T 05:05 PM haruspex 3 118 Introductory Physics Homework Hi! I wonder how a refridgerator work. I understand that the Ideal Gas Law pV=N_{mol}RT is used somehow. 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Feb28-14 02:36 AM yazz912 10 178 Calculus & Beyond Homework I have gathered experimental data regarding this heat transfer problem. Unfortunately right now there is no simple way... Feb27-14 10:27 AM capterdi 3 489 General Engineering When earth pulls a mass with gravitational force why does the energy of the earth-mass system decrease? Isn't work... Feb27-14 09:12 AM Drakkith 6 352 General Physics 1. The problem statement, all variables and given/known data W = f x D W = work f = force D = distance What... Feb26-14 04:14 PM haruspex 3 162 Introductory Physics Homework If i have a condition like : Adding (1100 10002)10] + (0011 11002)10] = (1 0000 0100)26010 that is how i do the... Feb26-14 04:00 PM meBigGuy 1 169 Electrical Engineering 1. The problem statement, all variables and given/known data A bucket of water with mass 100kg is on the ground... Feb26-14 03:35 PM haruspex 5 151 Calculus & Beyond Homework 1. 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Feb24-14 12:19 PM TSny 7 108 Introductory Physics Homework 1. The problem statement, all variables and given/known data http://i62.tinypic.com/b7f66x.png 2. Relevant... Feb24-14 01:33 AM haruspex 6 145 Introductory Physics Homework Say for example I have a hollow cube with equal length/width on all sides. Let's also assume that the outside walls of... Feb23-14 04:49 AM 256bits 3 361 General Engineering 1. The problem statement, all variables and given/known data An ideal gas at initial temperature T1 and pressure P1... Feb22-14 03:19 PM Emspak 2 121 Introductory Physics Homework 1. The problem statement, all variables and given/known data... Feb22-14 09:05 AM Pranav-Arora 46 741 Introductory Physics Homework I don't know if analog computers exist or not but if they were to be made, how would they work? I have idea of how... Feb21-14 07:32 PM FactChecker 11 1,521 Computing & Technology In First law of thermodynamics , work done by the gas is given by pΔV . I have few doubts regarding this- 1)Is... Feb21-14 10:34 AM Chestermiller 3 216 General Physics 1. The problem statement, all variables and given/known data A system can be taken from state A to state B where SA =... Feb21-14 12:00 AM Simon Bridge 1 110 Advanced Physics Homework Hello all, I was wondering if it was possible to have energy without an associated force. More specifically, can... Feb20-14 11:27 PM ahyaa 3 206 Classical Physics	1,893	6,958	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.625	3	CC-MAIN-2014-10	latest	en	0.836662
https://www.hackmath.net/en/example/4147	1,561,392,381,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627999615.68/warc/CC-MAIN-20190624150939-20190624172939-00171.warc.gz	766,219,903	7,262	# The swing To swing the two girls. Aneta weight 45 kg and Simon 35 kg weight. How far should sit Simon from the middle of swing so it is balanced, if we know that Aneta is sitting at distance 1,5m? How far are girls sitting apart? Result r2 = 1.929 m x = 3.429 m #### Solution: Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! #### To solve this example are needed these knowledge from mathematics: Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? ## Next similar examples: 1. Meat loses Meat loses 18% of its weight by smoking. How much raw meat butcher used to manufacture 35 kilos of smoked? 2. Drying herbs Herbs lose 75% weight By drying. How many fresh herbs do you have to collect if you want to make 1.5 kg of dried herbs? 3. Empty canister The canister is filled with oil weight 17 kg. If filled only half weight 9 kg. What weight does my empty canister have? 4. Crystal water The chemist wanted to check the content of water of crystallization of chromic potassium alum K2SO4 * Cr2 (SO4) 3 * 24 H2O, which was a long time in the laboratory. From 96.8 g of K2SO4 * Cr2 (SO4) 3 * 24 H2O prepared 979 cm3 solution of base. S 5. Holidays - on pool Children's tickets to the swimming pool stands x € for an adult is € 2 more expensive. There was m children in the swimming pool and adults three times less. How many euros make treasurer for pool entry? 6. Theorem prove We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started? 7. Glass Trader ordered from the manufacturer 200 cut glass. The manufacturer confirmed the order that the glass in boxes sent a kit containing either four or six glasses. Total sent 41 boxes. a) How many boxes will contain only 4 glasses? b) How many boxes will co 8. Cows Agricultural cooperative has increased the number of housed cows by 14% to 285 units. By how many cows increased agricultural cooperative the number of cows? 9. The price The price of the land increased by 17%. What was the original price of the land if it now costs 46800 €? 10. Sale off Product cost 95 euros before sale off. After sale off cost 74 euros and 10 cents. About what percentage of produt became cheaper? 11. Find the 7 Find the number that is smaller than 5 5/12 by as much as 2 2/13 is smaller than 6 1/6 12. Fifth of the number The fifth of the number is by 24 less than that number. What is the number? 13. Baskets of apples Grandfather gathered apples into baskets. He filled one large and three small baskets. Small baskets are identical and each can hold 3 kg of apples less than in a large basket. In small baskets together 3 kg of apples over a large basket. How many kilogram 14. Pine How much of a mixed forest rangers want to cut down, if their head said: "We only cut down pine trees, which in our mixed forest 98%. After cut pine trees constitute 94% of all trees left." 15. Boxes 200 boxes have been straightened in three rows. The first was 13 more than in the second, and in the second was one fifth more than in the third one. How many boxes are in each row? 16. Heating plant Workers dump imported coke at heating plant. On the first day used half of the amount, the second day three quarters the rest and on the third day left 120 tons. How much coke they dump at heating plant? 17. Mean if the mean of the set of data 5, 17, 19, 14, 15, 17, 7, 11, 16, 19, 5, 5, 10, 8, 13, 14, 4, 2, 17, 11, x is -91.74, what is the value of x?	960	3,598	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.265625	3	CC-MAIN-2019-26	latest	en	0.937531
https://johnwbarrett.wordpress.com/category/mathematics/state-sum-models/	1,685,268,227,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00580.warc.gz	391,015,905	22,454	# Gravity matters John Barrett's research website ## Linearised regge calculus Here is a paper that I wrote a draft of back in 1987 but never made it to publication: Linearised Regge Calculus. A long long time ago, I worked on a discrete version of general relativity called Regge Calculus. This is a rather elegant theory invented by the late great Tullio Regge using simplicial manifolds with Riemannian (or Lorentzian) metrics that are flat on each simplex (“piecewise flat”). I was particularly interested in how this could mimic general relativity on large scales, i.e., I wanted to show that if you worked to a finite resolution so that the (smaller) simplicial structure wouldn’t be apparent, then the metrics approximated nice smooth ones obeying Einstein’s equations. This is a hard problem, but can be simplified by looking a weak gravitational fields, those that are close to flat space. By linearising the equations, one gets something that can be analysed. I got my result in a paper I was rather proud of, A convergence result for linearised Regge calculus. From the abstract: “…solutions of the linearised Regge equations converge to analytic solutions of the linearised Einstein equations.” However it needed a technical result that is sort of “obvious” but actually highly technical, that the calculation of the curvature from the metric has the same properties in the linearised Regge case as it does for linearised Einstein. I worked it all out, with heavy use of the simplicial approach to algebraic topology which I had learnt by giving supervisions (tutorials to students) on C.R.F Maunder’s course in Cambridge, following his book Algebraic Topology. The results are summarised very briefly in the paper The fundamental theorem of Regge calculus. I called it that because it seemed pretty fundamental, and probably at that age I didn’t care whether anyone else agreed or not (but obviously the editor of the journal didn’t object). The actual proof is in the preprint posted above. I don’t know where I sent it, but it seems it didn’t make the CERN or the KEK preprint libraries (and the arXiv didn’t exist then, of course). I always meant to revise the presentation a little before publishing. But I never got around to it; there must have been something more important to do. And now I no longer have the tex file I can’t upload it to arXiv (can I?) People keep asking me for it, so here it is! Written by johnwbarrett 18 October 2018 at 19:26 ## Dichromatic state sum models Manuel Bärenz and I have just finished a paper on 4-dimensional topological state sum models. It is posted here (since 12 Jan 2016), and on arXiv. The idea of the paper is to squeeze more out of the Crane-Yetter state sum model and perhaps indicate how to get a viable quantum gravity model (or condensed matter model) from it. The original CY was defined just using the quantum group version of SU(2), and turns out to be rather too simple be an interesting physics model. The reason is that the group SU(2) is used to “colour” both one- and two-dimensional dual edges. This has the effect that the quantum theory can’t “see” the difference between the one- and two-dimensional stuff. All this is best seen using a handle decomposition rather than a triangulation. There’s an operation that changes 1-handles into 2-handles (thus changing the topology of the manifold) and the problem with the original CY is that it is invariant under this operation – which is a property that isn’t wanted. As a consequence, CY is the same on lots of different manfolds, which is why it is “too simple”. One of the things that Manuel and I have is an efficient translation between the triangulation picture and the handle picture. In the handle picture, Jerome Petit had the idea that the 1- and 2-handles can be coloured differently, to give a new set of models that he called “dichromatic”. We have understood that in the triangulation picture, this corresponds to the CY invariant being “nonmodular” (the original CY is “modular”). Interestingly, there are also models in the handle picture that don’t have a CY description at all. We calculated a few simple examples and found that one of them has configurations that are a plausible analogue of “teleparallel gravity” in the formulation given by Baez and Wise. This doesn’t yet mean we have a new quantum gravity model because, firstly, we only used finite groups instead of Lie groups (to keep things simple) and secondly, it isn’t clear that the action will be the gravity action. Still, it is an interesting direction. What it needs next is to do a lot more examples. Probably any really interesting examples will involve representations of a group or quantum group that are non-unitary. New territory indeed! Written by johnwbarrett 12 January 2016 at 04:37 ## Como lectures with one comment I’m giving some lectures at the SIGRAV school in Como on “Non-commutative geometry and quantum gravity”. I’ll put the slides up here as they happen. Lecture 1: The Planck scale. This lecture reviews the experimental evidence for a fundamental Planck scale. Various theoretical approaches to modelling the Planck scale are mentioned very briefly, mainly concentrating on the predictions of the spectral action suggested by non-commutative geometry. Lecture 2: Non-commutative geometry: axioms, examples. The axioms of non-commutative geometry are given in this lecture alongside an explanation of how they are satisfied by an ordinary (commutative) manifold. The example given by the standard model internal space is also summarised. Lecture 3: Matrix geometries and fuzzy spheres. The lecture explains my project to replace the usual space-time manifold with a non-commutative geometry based on finite-dimensional matrices. Examples with spherical symmetry are considered. Lecture 4: State sum models, using 2d examples. A different topic is introduced in this lecture: 2d topological state sum models defined using a diagrammatic calculus. Some recent progress in defining models that are sensitive to the spin structure of the surface (and are thus fermionic) is explained. Lecture 5: Non-commutative geometry in 2d state sum models. The final lecture connects the topic of state sum models and non-commutative geometry. It shows how defects in a 2d state sum model can carry a non-commutative geometry. ——— Giving the lectures turned out to be enormously useful. I’m writing a paper called “matrix geometries and fuzzy spaces as spectral triples“, which I hope will appear soon is out now. The paper will be much better for having had to explain the ideas in the lectures. Written by johnwbarrett 2 June 2014 at 09:39	1,466	6,687	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2023-23	latest	en	0.949274
http://forums.wolfram.com/mathgroup/archive/2005/Jul/msg00047.html	1,585,846,850,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00093.warc.gz	72,724,434	7,814	Re: Explicit solution to Root[] • To: mathgroup at smc.vnet.net • Subject: [mg58449] Re: Explicit solution to Root[] • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> • Date: Sat, 2 Jul 2005 04:07:13 -0400 (EDT) • Organization: The Open University, Milton Keynes, England • References: <da2mmv\$932\$1@smc.vnet.net> • Sender: owner-wri-mathgroup at wolfram.com Mukhtar Bekkali wrote: > Here is the code: > > \!$\(Root[\(-2$\ #1\^3 + 2\ #1\^4 - #1\ Root[$-4$ - 3\ #1 + 66\ > #1\^2 + > 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 &, 1] - 6\ #1\^2\ > Root[$-4$ - \ > 3\ #1 + 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 &, > 1] + 6\ #1\^3\ Root[$-4$ - 3\ #1 + > 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 &, 1] - 5\ \ > Root[$-4$ - 3\ #1 + 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 > &, \ > 1]\^2 - 6\ #1\ Root[$-4$ - 3\ #1 + 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 > + > 216\ #1\^5 &, 1]\^2 + 6\ #1\^2\ Root[$-4$ - 3\ #1 + > 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 &, > 1]\^2 - 2\ Root[$-4$ - > 3\ #1 + 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + 216\ #1\^5 > &, 1]\ > \^3 + 2\ #1\ Root[$-4$ - 3\ #1 + 66\ #1\^2 + 80\ #1\^3 - 108\ #1\^4 + > 216\ \ > #1\^5 &, 1]\^3 &, 2];\)\) > > I would guess it is a number. I applied RootReduce, ToRadicals, N or > combinations of thereof, however, nothing seem to convert the above > expression into an explicit number. What command or sequence of > > Mukhtar Bekkali > Hi Mukhtar, Is this what you are looking for (N seems to work pretty well)? In[1]:= Root[-2#1^3 + 2#1^4 - #1Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1] - 6#1^2Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1] + 6#1^3Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1] - 5Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1]^2 - 6#1Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1]^2 + 6#1^2Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1]^2 - 2Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1]^3 + 2#1Root[-4 - 3#1 + 66#1^2 + 80#1^3 - 108#1^4 + 216#1^5 & , 1]^3 & , 2] Out[1]= Root[-24 + 6#1 + 51#1^2 - 40#1^3 - 54#1^4 + 54#1^5 & , 1] In[2]:= N[%] Out[2]= 1.1122081402235109 Best regards, /J.M. • Prev by Date: Re: Explicit solution to Root[] • Next by Date: Re: Can't assign value to symbols • Previous by thread: Re: Explicit solution to Root[] • Next by thread: Re: Explicit solution to Root[]	1,326	2,572	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2020-16	longest	en	0.378078
https://manycoders.com/excel/formulae/cubemember-excel/	1,713,518,585,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00045.warc.gz	334,878,948	23,568	# Cubemember: Excel Formulae Explained ## Key Takeaway: • CUBEMEMBER is a powerful Excel formula that allows you to extract data from multidimensional data sources. With CUBEMEMBER, you can easily filter and manipulate data from multiple sources, making it an ideal tool for data analysts and business professionals. • While CUBEMEMBER may seem intimidating at first, mastering the basics of Excel formulae and syntax is essential. By understanding the fundamentals, you’ll be able to create complex formulae and manipulate data more efficiently. • Essential Excel formulae like SUM, AVERAGE, and COUNT are crucial for data analysis. Likewise, logical formulae like IF, AND, and OR can be used for decision-making. Incorporating these formulae into your workflow can save you time and effort in manipulating data. Are you struggling to understand Excel Formulae? CUBEMEMBER provides an easy and comprehensive way to get an overview of each formula and their underlying concepts. Get all your Excel queries answered in one place! ## CUBEMEMBER: Beginner’s Guide to Excel Formulae Ever been overwhelmed by the vast number of Excel formulae? As an Excel user, I know it well. So I’ve put together this beginner’s guide. It focuses on one particular function: CUBEMEMBER. We’ll start by understanding the basics, so even rookies can follow. Next, we’ll explore the commonly used formulae and how to use them in CUBEMEMBER. By the end, you’ll have a strong foundation in Excel formulae and know CUBEMEMBER’s power. ### Understanding the basics of Excel formulae What is an Excel formula? It’s an expression that uses math to calculate one or more cells in a spreadsheet. To write basic formulae, type “=” followed by a math expression in the designated cell. Functions are useful for forming complex formulae, as they do specific calculations and return a value. Excel formulae help people analyze and process data, making it easier to get insightful results quickly. Because of this, most employers now consider Excel proficiency an essential skill. To learn more, let’s review the types of formulae and their applications. ### Different types of formulae and their applications Understanding different types of formulae and their applications is key to analyzing data with Excel. Here’s a snapshot of some common formula types and what they could be used for: Formula Type Use/Application SUM Adding up values in a range to get a total. AVERAGE Figuring out the average value in a range of cells. MAX/MIN Finding the highest or lowest value in a series. IF Testing if a condition is true or false. Allows making logical statements. For example, the SUM function can add up columns or rows of numbers quickly. This is helpful when dealing with large datasets. It can also be used to total the number of products sold. The AVERAGE formula is handy to work out the average value in a range. Say you want to know the average customer satisfaction rating from your survey data. You could use MAX/MIN to work out things like maximum revenue earned by a company or minimum temperature recorded at a particular location. IF can be manipulated in many different ways. It could be used to determine if certain employees meet criteria such as years of service or attendance records. To illustrate, an accountant I once saw managed payroll reporting hours worked by members on shift work scheduling using Excel’s IF formulae. This enabled them to efficiently manage over 100 employees. So, to help readers, here’s a step-by-step guide that breaks down Excel formula syntax. ## Excel Formulae Syntax: A Step-by-Step Guide Let’s dive into the world of Excel formulae syntax! Learn the step-by-step process to create complex formulas. Mastering syntax is the foundation of successful data analysis, so it’s important to understand how to make and use these formulas. In the first sub-section, we’ll explain the basics of writing formulae. Transform data into usable and meaningful insights quickly. In the second sub-section, we’ll take you on a journey to create complex formulae with syntax. Become a data wizard with the tools we provide! ### Mastering the syntax of Excel formulae Learn how to utilize Excel formulae syntax with ease! Familiarize yourself with the formula bar and understand the functions in the library. Know the basics of a formula – operators, arguments, and how they are arranged. Utilize references like cell referencing, range references, and relative vs absolute cell references. Test and troubleshoot your formulas by checking for errors. Increase your spreadsheet power and accuracy! Understand the purpose of each element in a formula. Create various calculations using different data types and manipulate data sets across worksheets. Avoid errors and redundancies in computations. Mastering Excel formulae syntax is essential for advanced users. Don’t miss out on developing these skills! Create complex formulae using syntax for efficient and accurate data processing. ### Creating complex formulae using syntax Open Excel and click a cell for the formula. Start with an equal sign (=) followed by the name of the function, like CUBEMEMBER. Put cell references or ranges in parentheses as arguments. Add arithmetic or comparison operators if needed. Close parentheses and press Enter. Nested functions have multiple functions combined with semicolons within parentheses. They use Cell References like \$A\$1 or B2. Syntax rules include using capital letters for function names. Syntax can be confusing at first. Mistakes lead to wrong results. Mastering syntax unlocks advanced data analysis capabilities. Add it to your learning list! It gives access to deep insights that are otherwise missed. In our next section, we will discuss essential Excel Formulae for data manipulation. Techniques for easy data arrangement for analysis will be included. ## Essential Excel Formulae for Data Manipulation Excel lovers, I know the feeling of dread when you gaze at a data set that looks impossible. Raw data is not always ready for analysis, and without the proper tools, it can seem like a huge challenge to turn it into insights. That’s where Excel formulas come in! In this CUBEMEMBER series, we’ll go into some important Excel formulas for data manipulation. These will make it easier to sum, average, and count your data. We’ll also look into logical formulas like IF, AND, OR, for decision-making, and text formulas (LEFT, RIGHT, MID) for text processing. Let’s begin! ### Sum, Average, and Count formulae for data analysis The SUM formula adds up the cells within a range. AVERAGE computes the average of the cells within a range. COUNT counts the number of cells with numbers in the given range. Using these formulae correctly, large datasets can be manipulated quickly and accurately. The SUM function can uncover trends by adding values in rows and columns. The AVERAGE function shows central tendencies or patterns in data sets. To improve data analysis, combine SUM and COUNT functions with IF or AND to make decisions based on criteria. Or use pivot tables for more intricate analysis. Pivot tables make reports from large datasets with lots of variables, producing concise summaries. Next, we will cover logical formulae such as IF, AND, OR for decision making. ### Logical formulae like IF, AND, OR for decision making The IF function is used to set a logical test. It then specifies what should happen if the condition is met, and what should happen if it is not met. This helps you manipulate your data effectively. The AND function is used when multiple conditions must be met to get a result. It only returns “True” if all conditions are met. With the OR function, multiple criteria can be true at once. It returns “True” as long as one or more of the criteria are met. Using IF, AND, OR formulae saves time. It also removes human errors from manual data manipulations. An example of this is the 2016 US Presidential Election projections. CNN used modified Excel formulas to project voting counts live. The formulas served as a base for custom algorithms that collected county-by-county voting results. Finally, Text Formulae (LEFT, RIGHT, MID) are great for text processing. They help with large amounts of text data in Excel. ### Text Formulae (LEFT, RIGHT, MID) for text processing Make text processing a breeze with LEFT, RIGHT and MID Formulae! Here’s a quick guide to get you up to speed: • LEFT(text,num_chars): Extracts the leftmost characters from a string. • RIGHT(text,num_chars): Extracts the rightmost characters from a string. • MID(text,start_num,num_chars): Extracts characters from the middle of a text string, based on start position and length. These formulas can be combined with others, like CONCATENATE or SUBSTITUTE, for dealing with large amounts of data. Pro Tip: Excel is not case-sensitive, so using UPPER, LOWER or PROPER before manipulating your text can help you avoid errors! Ready to take things to the next level? Advanced Excel Formulae are here for you! ## Advanced Excel Formulae: The Next Level I’m an Excel user and I’m ready to kick it up a notch. This article will help me do just that. We’ll learn some powerful formulae that can seriously up my spreadsheet game. Specifically, VLOOKUP, HLOOKUP, and INDEX/MATCH. Plus, date & time formulae like NOW, TODAY, and DATE. And financial formulae such as PMT, PV, and FV for financial modelling. Get ready to take my Excel skills to the next level with these amazing formulae! ### Lookup and Reference Formulae (VLOOKUP, HLOOKUP, INDEX/MATCH) Lookup and Reference Formulae are powerful Excel formulae used for finding and fetching particular information from tables or ranges. These formulae are key for managing large datasets easily. For instance, VLOOKUP vertically searches the left-most column of a table, and returns the corresponding data from the same row. HLOOKUP is almost the same but it works in reverse. INDEX/MATCH matches two criteria: MATCH locates the row and column with the required data, then INDEX returns the value from the worksheet. These formulae help you to automate data retrieval and avoid mistakes when manually seeking data from big spreadsheets. Research by PwC shows that 41% of time spent on financial reporting can be automated using excel-based solutions such as Lookup and Reference Formulae. After that we will talk about Date and Time Formulae (NOW, TODAY, DATE) in more detail. ### Date and Time Formulae (NOW, TODAY, DATE) The Date and Time Formulae (NOW, TODAY, DATE) are a great tool for Excel users. Here is an overview: • NOW() returns the current date and time. • TODAY() returns today’s date. • DATE(Year, Month, Day) creates a date from the parameters given. NOW() tracks when changes were made and helps calculate deadlines. TODAY() is used for daily automatic reports. DATE() creates specific dates that cannot be computed with simple math. Companies use these formulae for tracking employee attendance and keeping calendars up-to-date. The financial sector often uses Excel for quick calculations of rates of return and asset valuations over time. Next, we’ll discuss PMT, PV and FV formulas in financial modeling. ### Financial Formulae (PMT, PV, FV) for financial modeling Financial Formulae (PMT, PV, FV) are essential for financial modeling. They’re used in many industries, particularly finance. Here’s a table with info about the formulae: Formula Calculation Usage PMT =PMT(rate,nper,pv,[fv],[type]) Calculates payment for an annuity investment PV =PV(rate,nper,pmt,[fv],[type]) Calculates present value of an investment FV =FV(rate,nper,pmt,[pv],[type]) Calculates future value of an investment PMT is used a lot when calculating payments needed to reach savings goals or pay off debts. PV is great for calculating initial investments to meet future targets. And FV calculates how much will be available in future. When using the formulae, input correct data into each argument. For example, make sure you know whether payments are made at the beginning or end of a period with PMT. Also choose an appropriate interest rate. Name ranges are better than cell references. This makes it easier to update inputs. Lastly, we’ll cover troubleshooting Excel formula errors and solutions. ## Troubleshooting Excel Formulae: Common Errors and Solutions Do formulae errors drive you crazy? As an Excel enthusiast, I know the feeling! But with the right tools and techniques, complex errors can be solved. Let’s dive in! We’ll discuss common formulae errors and their solutions. Debugging tips and tricks will make your troubleshooting process easier. Plus, we’ll check out best practices and techniques for troubleshooting formulae. So, let’s get started! ### Identifying and fixing common formulae errors Text: Check for typos and spellings in your formulae. One wrong letter or symbol can lead to errors. Review the syntax to ensure it follows the proper order of operations. Doing so will guarantee correct calculations. Verify the cell references used in your formulae are accurate. Wrong cell references can lead to incorrect calculations. Also, check your data for any invalid or missing values. Formulae may not work correctly if there’s a discrepancy. To identify and troubleshoot formulae errors, it’s important to understand Excel functions and operations, and how complex spreadsheets operate. Common errors can include inaccuracies or missing cells in calculations, format changes leading to retention issues, and miscalculations caused by using other software programs. To avoid such issues, keep an eye on file size, limit cell-specific formatting, and be careful when copying formulas from one location to another. Inconsistencies could be overlooked when copying and pasting functions. Unresolved errors can have a negative effect on business analyses or results, so stay updated by running FAQ sections, reading developer notes, and watching video tutorials. This way, you can tackle any unknown issues related to troubleshooting effectively. There were no typos or spelling errors in the original text except for the use of “and” instead of “&” which I corrected in the last paragraph. ### Debugging formulae: Tips and tricks Debugging formulae: Tips and tricks can save you time when working on complex spreadsheets. Monitor your formulas and correct any errors quickly. Have you ever had a calculation error? It could be because a cell was formatted as text instead of a number. I once spent hours trying to find out why my SUM function wasn’t working until I realized my mistake. Debugging formulae takes patience and focus, but once done correctly, it prevents frustration and saves time. Here are some tips to remember: • Use parentheses – always close all brackets correctly to avoid syntax errors. • Fill in blank cells – use the fill handle or Ctrl+Enter to fill in blank cells before applying a formula. • Avoid circular references – make sure your formulas don’t refer back to their own cell. • Check data types – ensure all entries in a column have the same data type before applying formulae. ### Troubleshooting formulae: Best practices and techniques Text: 1. To understand error messages in Excel, we must know how to interpret them. Common messages include #VALUE!, #REF!, #NAME?, #N/A, #NUM!, and ####. This will help us identify and fix errors more quickly. 2. Check for hidden characters or spaces in the formula, and any cell references. Using parentheses around different sections of formulas will help with clarity and organization. By following these steps, we can troubleshoot formula errors effectively. 30 million people use Excel spreadsheets each year. Microsoft revealed this figure in 2021. ## Five Facts About CUBEMEMBER: Excel Formulae Explained: • ✅ CUBEMEMBER is a powerful and versatile formula in Excel, used for performing multidimensional analysis of data. (Source: Excel Campus) • ✅ The formula allows users to retrieve data from OLAP (Online Analytical Processing) cubes and build advanced reports. (Source: Ablebits) • ✅ CUBEMEMBER can be used to return various types of data, such as values, counts, averages, percentages, and more. (Source: Spreadsheeto) • ✅ The formula has various parameters that can be customized to suit specific requirements, such as member names, hierarchies, filters, and sorting. (Source: Vertex42) • ✅ CUBEMEMBER is a valuable tool for business analysts, financial professionals, and data scientists working with large, complex datasets. (Source: Udemy) ## FAQs about Cubemember: Excel Formulae Explained ### What is CUBEMEMBER in Excel formulae and how does it work? CUBEMEMBER is an Excel function that retrieves member names from an OLAP cube. It works by searching the cube for a specified member and returning its name based on certain criteria. ### What are the advantages of using CUBEMEMBER in Excel formulae? The advantages of using CUBEMEMBER include the ability to retrieve specific member names from an OLAP cube, saving time and effort by automating the process, and improving the accuracy and consistency of data analysis. ### How do you use CUBEMEMBER in Excel formulae? To use CUBEMEMBER in Excel formulae, you first need to have an OLAP cube established. Then, you would input the CUBEMEMBER function with the desired criteria, such as the hierarchy and level of the member you’re looking for. ### What are some common mistakes to avoid when using CUBEMEMBER in Excel formulae? Some common mistakes to avoid when using CUBEMEMBER include using incorrect syntax, selecting the wrong hierarchy or level, and failing to update the cube with new data. It’s important to carefully review and test your formulae to ensure accuracy. ### What is the difference between CUBEMEMBER and other Excel functions? CUBEMEMBER is specifically designed for use with OLAP cubes, whereas other Excel functions are designed for more general use. CUBEMEMBER is therefore more efficient and accurate when working with cube data, while other functions may not be as effective. ### Can CUBEMEMBER be used with other Excel tools and functions? Yes, CUBEMEMBER can be used in conjunction with other Excel tools and functions, such as PivotTables and filtering. This allows for even more advanced analysis and reporting capabilities.	3,786	18,372	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.390625	3	CC-MAIN-2024-18	longest	en	0.880815
https://blog.roboforex.com/blog/2022/07/22/the-free-float-ratio-everything-investors-should-know/	1,669,712,794,000,000,000	text/html	crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00735.warc.gz	166,416,229	26,203	There are plenty of indicators and multipliers for analysing public companies. In this article, we’ll talk about one of them, the Free float ratio. We’ll find out the formula for calculating it and describe how investors use it when analysing the market situation. ## What the Free float ratio is The Free float ratio is the quantity of shares available for public trading. They are traded on stock exchanges, are not owned by strategic investors, and are available to retail ones. You can meet other names it goes by – Float or Public Float. What shares are not included in a Free float calculation? When calculating the Free float ratio, the following shares are not included: • Owned by shareholders, the company’s management and top managers • Owned by the state • Owned by big investment funds, which are majority shareholders In addition, limited shares are also not taken into account. For example, shares that were given to an employee as a reward for the merits for a company. ## How the Free float ratio is calculated To calculate the Free float ratio, we need to know the number of free float shares and the total number of shares issued. To make the formula look easy to understand, we’ll denote these parameters as A and B, respectively. The Free float ratio calculation formula: Free float = A / B The ratio can be specified in two formats – in percentage (for example, 50%) or decimal fraction (for example, 0.5). Let’s say that a company issued 100,000 shares; 51% of them, 51,000, a majority stake, are owned by the management, while the rest 49,000 shares were released for free circulation. In this case, the Free float ratio will be 0.49 or 49%. The Free float calculation: 49,000 / 100,000 = 0.49 ## How to increase Free float • A split is a stock split with a fixed split ratio. Through splits, companies increase the number of shares and reduce their price. For example, a company had 100 shares at \$4 per share, and after a 1:2 split they own 200 shares at \$2 • Issue of securities is a way to increase share capital and attract investments • Share sales by major shareholders – shares that were frozen before are released for free circulation. There might be different reasons for this ## How to reduce Free float • A buyback is the re-acquisition by a company of its own shares. Having amassed enough available funds, a company repurchases its shares from its shareholders and retires them. As a result, the share of major investors increases, as well as their influence on a company • Buyout of floating securities by major shareholders to get a majority stake • Consolidation of shares (reverse split) is a conversion of two or more shares into one of the same categories ## What Free float value is considered optimal? The optimal Free float value for both traders and investors is in the range of 40–80%. Such volumes of free float shares provide some kind of protection against market fluctuations, increase the instrument's liquidity, and afford an opportunity to buy or sell an asset at any time. In other words, the higher the Free float ratio, the more liquid the instrument is and the more opportunities investors have. ### Disadvantages of the low Free float ratio First of all, small or limited market demand. After buying some shares, a trader might find it difficult to sell them. There is a possibility that there won’t be a buyer in the market to acquire this asset, or its price might be very low. Secondly, there might be sharp price fluctuations in either direction at a time of news releases, which may cause panic among market players. Thirdly, buying a vast amount of shares by a single investor might significantly raise the price and cause disbalance. Selling a major minority shareholding can be delayed and it also might result in a price surge. In some cases, shares can’t be sold at all because there are no investors willing to buy them. ## How to use the Free float ratio for market analysis To begin with, the ratio provides an investor with an understanding of an instrument's liquidity. If the ratio value is 40–80%, an instrument is considered quite liquid and involves smaller trading-related risks. The Free float value above 80% means that it will be difficult for jobbers and big-time investors to cause higher volatility in the market by selling/buying big amounts of shares. A ratio value below 40% says that the majority of shares are owned by principal shareholders and they have the ability to influence share prices by unloading a lot of shares in the market. Small amounts of free float shares raise additional difficulties for selling them. ## Summary Combined with other indicators and multipliers, the Free float ratio offers traders and investors an opportunity to choose a more suitable instrument in terms of liquidity and fair market price. Experience has proven that there is no direct correlation between liquidity and Free float, but there is an opinion that the high ratio value gives a slight edge over the low one. Material is prepared by #### Maks Artemov Has been in Forex since 2009, also trades in the stock market. Regularly participates in RoboForex webinars meant for clients with any level of experience.	1,093	5,231	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2022-49	longest	en	0.940349
http://staff.ustc.edu.cn/~lgliu/Projects/2016_ToG_SparseManifold/default.html	1,701,788,930,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00111.warc.gz	44,759,732	4,224	Construction of Manifolds via Compatible Sparse Representations Ruimin Wang, Zhouwang Yang, Kang Wang, Wen Shan, Jiansong Deng, Falai Chen ACM Transactions on Graphics, 35(2), Article 14: 1-10, 2016. (Presented at Siggraph 2016) Teaser: From a mesh (a), our approach generates a manifold surface (b) approximating its Catmull-Clark subdivision surface. Sharp features are created in the subdivision surface by enforcing C0 constraints on edges in red (a). The constructed manifold (b) approximates various features like crease, dart, and cusp well. The regions in red, green, and blue are zoomed and shown in (c), (d), and (e), respectively. Abstract Manifold is an important technique to model geometric objects with arbitrary topology. In this article, we propose a novel approach for constructing manifolds from discrete meshes based on sparse optimization. The local geometry for each chart is sparsely represented by a set of redundant atom functions, which have the flexibility to represent various geometries with varying smoothness. A global optimization is then proposed to guarantee compatible sparse representations in the overlapping regions of different charts. Our method can construct manifolds of varying smoothness including sharp features (creases, darts, or cusps). As an application, we can easily construct a skinning manifold surface from a given curve network. Examples show that our approach has much flexibility to generate manifold surfaces with good quality. Keywords Manifold, Sharp feature, Sparse representation, Curve network Concept Definition of a manifold: Manifold modeling is a technology used to construct surfaces from a domain manifold. The domain manifold is covered by a collection of charts with overlapping regions. The local mapping is a one-to-one map (parameterization) from one chart to an open set in the plane domain, which defines the local coordinate system for the chart. If two charts overlap, the transition function describes the transformation between them. Overfitting Problem Given a set of sampling points (100 points in total) from a C0 surface (a), the obtained fitting surface via polynomials (b), redundant atom functions (c), and sparse representation (d). (e) The first five atom functions (C0 shape functions in red boxes and polynomials in gray boxes) selected in the sparse representation. The sparsity is set at 25. Results Constructing manifold to approximate subdivision surface. (a) The input control mesh with enforced C0 features shown in red. (b) The manifold generated by our approach. (c) The close-up views of three parts of sharp features including crease, dart, and cusp. The first five atom functions that are adaptively selected for representing the local features are shown. Examples of manifolds constructed by our approach to approximate the subdivision surfaces with given control meshes in the lower right. C0 constraints shown in red are enforced to created sharp features in the subdivision surfaces. Paper PDF (1.3M) Video Demo (*.wmv) (52.0M) Presentation (at SIGGRAPH 2016) Fast Forward (0.5M) PPT (2.0M) Ack We would like to thank the anonymous reviewers for their comments and suggestions. The work is supported by the NSF of China (grants 11171322, 11371341, 11526212, 11571338, and 61222206), the One Hundred Talent Project of the Chinese Academy of Sciences, and the Fundamental Research Funds for the Central Universities (WK0010000051). BibTex @article {Wang:TOG2016, title = {Construction of Manifolds via Compatible Sparse Representations}, author = {Ruimin Wang and Ligang Liu and Zhouwang Yang and Kang Wang and Wen Shan and Jiansong Deng and Falai Chen} journal = {ACM Transactions on Graphics}, volume={35}, number={2}, pages={Article 14: 1-10}, year = {2016} } Copyright and disclaimer: The SOFTWARE provided at this page is provided "as is", without any guarantee made as to its suitability or fitness for any particular use. It may contain bugs, so use of this tool is at your own risk. We take no responsibility for any damage that may unintentionally be caused through its use. Copyright © 2015 GCL , USTC	912	4,136	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2023-50	latest	en	0.907027
https://housemetric.co.uk/house-price-analysis/BB9-5/Brierfield	1,716,503,978,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00275.warc.gz	266,089,312	14,952	# House prices in BB9 5 (Brierfield) This article reveals price per square metre data and various charts to help you understand current housing market in Brierfield, Nelson (BB9 5). ## Defining 'BB9 5' This analysis is limited to properties whose postcode starts with "BB9 5", this is also called the postcode sector. It is shown in red on the map above. There are no official names for postcode sectors so I've just labelled it Brierfield. You can click on the map labels to change to a neighbouring sector, or you can enter a different postcode sector (e.g. CM23 4) below. FYI, a postcode sector is the full postcode without the last two letters. ## Price per square metre Knowing the average house price in Brierfield is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property: £ per sqm = price ÷ internal area E.g. 14, Landless Street, Brierfield, Nelson, sold for £80,000 on Mar-2024. Given the internal area of 83 square metres, the price per sqm is £963. England & Wales have been officially metric since 1965. However house price per square foot is prefered by some estate agents and those of sufficiently advanced age ;-) They may want to convert square meters on this page to square feet. The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £1,350-£1,500, £1,500-£1,650, £1,650-£1,800 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 154 sales in Brierfield (BB9 5) that took place in the last two years. ##### Distribution of £ per sqm for Brierfield, Nelson Distribution of £ per sqm house prices in Brierfield You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc. The spread of prices will give you a feel of the typical range to expect in Brierfield (BB9 5). Of the 154 transactions, half were sold for between £850 and £1,830 per square metre. The median, or 'middle', price per square metre in 'BB9 5' is £1,140. Notably, only 25% of properties that sold recently were valued at more than £1,830 sqm. For anything to be valued more than this means it has to be more desireable than the clear majority of homes. ## Price map for Brierfield, Nelson Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Nelson. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values. House price heatmap for Brierfield, Nelson ## Comparison with neighbouring postcode sectors The table below shows how 'BB9 5' compares to neighbouring postcode sectors. Postcode sector Lower quartile Middle quartile Upper quartile BB9 7 Nelson £780 sqm £910 sqm £1,050 sqm BB9 9 Nelson £740 sqm £930 sqm £1,220 sqm BB9 0 Nelson £850 sqm £1,210 sqm £1,780 sqm BB9 8 Barrowford £1,000 sqm £1,350 sqm £1,700 sqm BB9 6 Barrowford £1,670 sqm £2,270 sqm £2,980 sqm BB9 5 Brierfield £850 sqm £1,140 sqm £1,830 sqm ## Will Nelson house prices drop in 2024? I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the chart below for BB9 5 (Brierfield) compared with both the wider area BB9 and inflation (CPIH from the Office of National Statistics). The dashed trend lines in the chart show the average over time. ##### Historic price per square metre in Brierfield,Nelson House price trends for Brierfield For the most recent sales activity, rather than a summarized average, it is better to see the underlying data. This is shown in the chart below, where blue dots represent individual sales, click on them to see details. If there is an obvious trend you should be able to spot it here amid the noise from outliers. ##### Most recent BB9 5 sales Recent trends for Brierfield Data from Land Registry comes in gradually over time. I update it every month but it takes about 5 months for the majority of sales for Nelson to be recorded. Disclaimer: I do not verify and cannot guarantee the accuracy of any data shown. Outliers exist in the data, typically these are where the EPC registry records the internal area incorrectly, sometimes although very rarely the Land Registry price paid data can be wrong. The data provided throughout this website about Nelson and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information. ## Street level data Street Avg size Avg £sqm Recent sales Halifax Road, Brierfield, BB9 5B 85 sqm £1,247 26 Higher Reedley Road, Brierfield, BB9 5E 84 sqm £1,814 17 Montrose Street, Brierfield, BB9 5B 63 sqm £1,316 12 Chapel Street, Brierfield, BB9 5D 69 sqm £1,060 10 Colne Road, Brierfield, BB9 5R 105 sqm £671 10 Taylor Street, Brierfield, BB9 5R 82 sqm £840 9 Quakers View, Brierfield, BB9 5P 81 sqm £1,940 9 Pennine Crescent, Brierfield, BB9 5E 77 sqm £2,141 8 ## Raw data Our analysis of Brierfield is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea.	1,469	5,752	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.078125	3	CC-MAIN-2024-22	latest	en	0.919473
https://www.extendoffice.com/documents/excel/5024-excel-separate-email-address.html	1,527,244,870,000,000,000	text/html	crawl-data/CC-MAIN-2018-22/segments/1526794867085.95/warc/CC-MAIN-20180525102302-20180525122302-00552.warc.gz	732,018,906	39,852	Tip: Other languages are Google-Translated. You can visit the English version of this link. or Register or 0 0 0 s2smodern ## How to separate email addresses to username and domain in Excel? If there is a list of email addresses in Excel, you may need to separate the email addresses into two columns, one contains usernames, and the other contains the domains as below screenshot shown. This article will introduce some tricks on solving this job easily and quickly. Separate email addresses by Text to Columns #### Quickly Split one cell into columns or rows based on delimiter In Excel, to split a cell into columns is tedious with the Wizard step by step. But with Kutools for Excel's Split Cells utility, you can: 1,convert one cell into columns or rows based on delimiter; 2,convert string into text and number; 3,convert string based on specific width, with clicks. Click for 60 days free trial! #### Separate email addresses by formulas Select a blank cell to place this formula =LEFT(A2,FIND("@",A2)-1), press Enter key, and drag fill handle down to the cells which need this formula. To extract domain from email address Select a blank cell to place this formula =RIGHT(A2,LEN(A2)-FIND("@",A2)), press Enter key, and drag fill handle down to the cells which need this formula. #### Separate email addresses by Text to Columns In Excel, the Text to Columns function also can do you a favor on separating email addresses. 1. Select the email addresses you want to separate, and click Data > Text to Columns. 2. Then in the Text to Columns Wizard window, check Delimited option and click Next to go the step 2 of the Wizard. 3. Check Other checkbox, and type @ into next textbox, you can preview the separated result in below window. 4. Click Next, select one cell to place the results in the last step of the Wizard. 5. Click Finish, the email addresses have been split into two columns. #### Separate email addresses by Split Cells If you neither like to remember formulas nor use the Text to Columns Wizard, you can apply Kutools for Excel’s Split Cells utility to split to rows/columns based on delimiter, specific width or text and number. After installing Kutools for Excel, please do as below: 1. Select the cells you use to separate, and click Kutools > Text > Split Cells. 2. In the Split Cells dialog, check the type you want to separate to, and then check Other checkbox and type @ into the next textbox. See screenshot: 3. Click Ok and select one cell to place the result. 4. Click OK. The emails have been split into usernames and domains. ### Recommended Productivity Tools #### Office Tab Bring handy tabs to Excel and other Office software, just like Chrome, Firefox and new Internet Explorer. #### Kutools for Excel Amazing! Increase your productivity in 5 minutes. Don't need any special skills, save two hours every day! 200 New Features for Excel, Make Excel Much Easy and Powerful: • Merge Cell/Rows/Columns without Losing Data. • Combine and Consolidate Multiple Sheets and Workbooks. • Compare Ranges, Copy Multiple Ranges, Convert Text to Date, Unit and Currency Conversion. • Count by Colors, Paging Subtotals, Advanced Sort and Super Filter,	706	3,207	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2018-22	longest	en	0.841599
https://www.teacherspayteachers.com/Store/Texas-Family-Tradition	1,600,847,441,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400209999.57/warc/CC-MAIN-20200923050545-20200923080545-00688.warc.gz	1,069,884,965	29,994	(247) United States - Texas - Arlington 5.0 Kid, youâ€™ll move mountains! ~ Dr. Seuss CUSTOM CATEGORIES Prices Top Resource Types My Products sort by: Best Seller view: Includes: title cards 12 picture cards 6 arrow cards 2 worksheets To use this resource print the cards on cardstock and laminate for durability. This resource can be used as a whole group activity or as an independent activity to support learning Subjects: Science 1st, 2nd, 3rd, Homeschool Types: Activities, Printables, Science Centers \$2.50 42 To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left corner Subjects: Math, Basic Operations, Word Problems 2nd, 3rd Types: Activities, Math Centers \$2.00 36 To use this resource, print out the more and less scoot cards and have them laminated. The students are to find the cards around the room and write the number shown on the scoot card by 10 more, 10 less, 100 more, and 100 less on their recording Subjects: Math, Basic Operations, Mental Math 2nd, 3rd Types: Activities, Math Centers \$2.00 31 To use this resource, print out the more and less BINGO calling cards cards and boards. Cut the BINGO boards out and calling cards out then get laminated for durability. Directions: 1. Have a pile of calling cards ready. 2. Each student gets a Subjects: Math, Mental Math, Place Value 1st Types: Activities, Games \$2.00 29 To use this resource, print out the more and less scoot cards and recording sheet. Place the more and less scoot cards along the students tables. Have students grab a clipboard and recording sheet. Students will scoot around the tables and fill in Subjects: Math, Mental Math, Place Value 1st, 2nd Types: Activities \$2.00 24 Â To use this resource, cut out and laminate the sort cards. Students will sort the cards according to if whatâ€™s on the card is considered an equal part or unequal part. Â Included in this package: Â Equal and unequal parts card headers for the Subjects: Math, Fractions, Geometry Kindergarten, 1st, 2nd Types: Activities, Math Centers \$2.00 14 To use this resource, print out the graphing task cards and have them laminated. The students will use the graph on the task cards to answer the questions with it and record their answer on their recording sheet. The students will match the number Subjects: Math, Graphing, Mental Math 1st, 2nd Types: Activities, Math Centers \$2.00 14 To use this resource, cut out and laminate the sort cards. Students will sort the cards according to if whatâ€™s on the card is considered a regular or irregular polygon. Â Included in this package: Â Regular and irregular polygon card headers for Subjects: Math, Applied Math, Geometry 1st, 2nd Types: Activities, Math Centers \$2.00 13 To use this resource, print out the counting money scoot cards and have them laminated. The students will count the money given on the scoot card and they are to write how much that value is on their recording sheet using the correct format and cent Subjects: Math, Basic Operations, Mental Math 1st, 2nd, 3rd Types: Activities, Math Centers \$2.00 11 To use this resource, print out the number line task cards and have them laminated. The students will use the number line given on the task card and order the given numbers on the number line and record their answer on their recording sheet. The Subjects: Math, Numbers, Place Value 1st, 2nd Types: Activities, Math Centers \$2.00 10 Â To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left Subjects: Math, Basic Operations, Word Problems 2nd, 3rd Types: Activities, Math Centers \$2.00 12 This product is about the r controlled vowels ar, er, ir, or, and ur. To use this resource, cut out and laminate the bossy r task cards and r controlled vowel cards. Place them in a center along with the recording sheet. Students will take a r Subjects: English Language Arts, Phonics 1st, 2nd Types: Activities, Literacy Center Ideas \$2.00 15 Â To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left Subjects: Math, Basic Operations, Word Problems Kindergarten, 1st, 2nd Types: Activities, Math Centers \$2.00 9 This product supports 2nd grade social studies TEKS 2.4aIncludes:-biography page written in the form of a letter to students-5 page flipbook to record facts about Irma Rangel's life, achievements, and character traits Subjects: Social Studies - History 2nd, Homeschool Types: Activities \$2.00 4 To use this resource, print out word problem task cards and have them laminated. The students will solve the word problem on the task cards and record their answer on their recording sheet. The students will match the number in the top left corner Subjects: Math, Basic Operations, Word Problems Kindergarten, 1st Types: Activities, Math Centers \$2.00 10 To use this resource, print out the more and less task cards and have them laminated. The students can fill in the blanks with an expo marker on the task cards. It will wipe right off. The students are to write the number shown on the task card by Subjects: Math, Basic Operations, Mental Math 2nd, 3rd Types: Activities, Math Centers \$2.00 19 To use this resource, print out the 2D Shapes task cards and have them laminated. The students will write the number of vertices and sides for the shapes given on the task card or draw the shapes for the given number of vertices and sides and record Subjects: Math, Geometry, Mental Math 1st, 2nd, 3rd Types: Activities, Math Centers \$2.00 17 This product contains Write the Room for the r controlled vowel ar. To use this resource: 1. Print 2. Cut out the â€śarâ€ť bossy r cards. 3. Laminate the â€śarâ€ť bossy r cards.Â 4. Scatter the â€śarâ€ť bossy r cards around the room. 5. Students find the â€śarâ€ť Subjects: English Language Arts, Phonics 1st, 2nd Types: Activities, Literacy Center Ideas \$2.00 11 This product contains Write the Room for the r controlled vowel or. Â To use this resource: 1. Print 2. Cut out the â€śorâ€ť bossy r cards. 3. Laminate the â€śorâ€ť bossy r cards.Â 4. Scatter the â€śorâ€ť bossy r cards around the room. 5. Students find the Subjects: English Language Arts, Phonics 1st, 2nd Types: Activities, Literacy Center Ideas \$2.00 9 To use this resource, print out the fraction task cards and have them laminated. The students will write the fraction to show whatâ€™s on the task cards and record their answer on their recording sheet. The students will match the number in the top Subjects: Math, Fractions, Mental Math 1st, 2nd, 3rd Types: Activities, Math Centers \$2.00 8 showing 1-20 of 224 TEACHING EXPERIENCE MY TEACHING STYLE HONORS/AWARDS/SHINING TEACHER MOMENT MY OWN EDUCATIONAL HISTORY	1,851	7,150	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2020-40	latest	en	0.918008
https://wnyrails.org/student-z-distribution-table/	1,631,904,444,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00267.warc.gz	683,186,842	17,524	# Student Z Distribution Table, Appendix B: Use Of Statistical Tables Use the negative Z score table below to find values on the left of the mean as can be seen in the graph alongside. Corresponding values which are less than the mean are marked with a negative score in the z-table and respresent the area under the bell curve to the left of z. Đang xem: Student z distribution table (Note that this method of mapping the Z Score value is same for both the positive as well as the negative Z Scores. That is because for a standard normal distribution table, both halfs of the curves on the either side of the mean are identical. So it only depends on whether the Z Score Value is positive or negative or whether we are looking up the area on the left of the mean or on the right of the mean when it comes to choosing the respective table) ### Why are there two Z tables? There are two Z tables to make things less complicated. Sure it can be combined into one single larger Z-table but that can be a bit overwhelming for a lot of beginners and it also increases the chance of human errors during calculations. Using two Z tables makes life easier such that based on whether you want the know the area from the mean for a positive value or a negative value, you can use the respective Z score table. See also Student Grants 2021 - College Scholarships You Should Apply For In 2021 If you want to know the area between the mean and a negative value you will use the first table (1.1) shown above which is the left-hand/negative Z-table. If you want to know the area between the mean and a positive value you will the second table (1.2) above which is the right-hand/positive Z-table. ### What is Standard Deviation? (σ) Standard Deviation denoted by the symbol (σ) , the greek letter for sigma, is nothing but the square root of the Variance. Whereas Variance is average of the squared differences from the Mean. ## Sample Questions For Practice 1. What is P (Z ≥ 1.20) To find out the answer using the above Z-table, we will first look at the corresponding value for the first two digits on the Y axis which is 1.2 and then go to the X axis for find the value for the second decimal which is 0.00. Hence we get the score as 0.11507 2. What is P (Z ≤ 1.20) (Same as above using the other table. Try solving this yourself for practice)	537	2,348	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.3125	4	CC-MAIN-2021-39	latest	en	0.879909
http://jjj.de/hakmem/series.html	1,500,971,614,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00447.warc.gz	156,629,183	5,042	Beeler, M., Gosper, R.W., and Schroeppel, R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972. Retyped and converted to html ('Web browser format) by Henry Baker, April, 1995. ## SERIES Previous Up Next ### ITEM 116 (Schroeppel & Gosper): ``` inf ==== \ N! N! 4 2 pi > ------ = - + --------- . / (2 N)! 3 9 SQRT(3) ==== N=0 ``` PROBLEM: Evaluate in closed form ``` inf ==== \ N! N! N! > -------- , which = / (3 N)! ==== N=0 1 / [ I (P + Q arccos (R)) dT , where ] / 0 2 3 2 (8 + 7 T - 7 T ) P = ------------------- and 2 3 2 (4 - T + T ) 2 2 3 4 (T - T ) (5 + T - T ) Q = ------------------------------------------ and 2 3 2 2 3 (4 - T + T ) SQRT((4 - T + T ) (1 - T)) 2 3 T - T R = 1 - ------- . 2 ``` ### ITEM 117 (Henry Cohen): ``` 2 3 4 X X X gamma = - ln X + X - ---- + ---- - ---- ... + ERROR 2 2! 3 3! 4 4! ``` Where ERROR is of the order of (e^-X)/X. ### ITEM 118 (Schroeppel): Differentiate ``` -Y Y E = X to get Y + Y X Y' - X Y' = 0. ``` Substitute for Y a power series in X with coefficients to be determined. One observes the curious identity: ``` N ==== \ J - 1 N - J N 0 > J (N - J) BINOMIAL(N, J) = N (0 = 1) / ==== J=1 ``` and thus ``` ==== N - 1 N \ N X Y(X) = > --------- / N! ==== N=1 ``` ### ITEM 119 (Schroeppel): PROBLEM: Can someone square some series for pi to give the series ``` 2 ==== pi 1 1 \ 1 -- = -- + -- + ... = > -- ? 6 2 2 / 2 1 2 ==== N ``` ### ITEM 120 (Euler): The series accelerating transformation (Abramowitz & Stegun, se. 3.6.27) ``` ==== K K \ (-1) (delta A0) A0 - A1 + A2 - ... = > ----------------- / K+1 ==== 2 K ==== K \ K-m (where (DELTA A0) = > BINOMIAL(K, m) (-1) Am = Kth forward difference on A0) / ==== m=0 ``` when applied to ``` pi 1 1 -- = 1 - - + - - ... 4 3 5 ``` gives ``` ==== N-1 2 pi \ 2 N! -- = > ---------- 4 / (2 N + 1)! ==== N=0 ``` Applied to the formula for gamma in Amer. Math. Monthly (vol. 76, #3, Mar69 p273) = ``` ==== T \ (- 1) [LOG2(T)] > ---------------- ([ ] means integer part of) / T ==== T=0 ``` we get ``` inf K-1 ==== ==== \ -(K+1) \ 1 > 2 > ------------------ / / I ==== ==== BINOMIAL(2 + J, J) K=1 J=0 ``` (Gosper) which converges fast enough for a few hundred digits. The array of reciprocals of the terms follows, with powers of 2 factored out to the left from all member of each row. ```4 1 8 1 3 16 1 5 6 32 1 9 15 10 64 1 17 45 35 15 128 1 33 153 165 70 21 256 1 65 561 969 495 126 28 ``` The next to left diagonal is 2^(N+1); the perpendicular one 3rd from the right is 1, 9/1= 9, 10/2= 45, 11/3= 165, 12/4= 495. ### ITEM 121 (Gosper): Consider the triangular array: ``` 1 1 1 1 4 1 1 11 11 1 1 26 66 26 1 1 57 302 302 57 1 ``` This bears an interesting relationship to Pascal's triangle. The 302 in the 4th southeast diagonal and the 3rd southwest one = 426 + 366. Note that rows then sum to factorials rather than powers of 2. If the nth row of the triangle is dotted with any n consecutive elements of (either) n+1st diagonal of Pascal's triangle, we get the nth Bernoulli polynomial: for n = 5, 1(6,i) + 26(6,i+1) + 66(6,i+2) + 26(6,i+3) + 1(6,i+4) = sum of 5th powers of 1 thru i+5, where (j,i) = BINOMIAL(j+i,j). ### ITEM 122 (Schroeppel, Gosper): The "parity number" = ``` inf ==== 1 \ PARITY(N) - > --------- 2 / N ==== 2 N=0 ``` where the parity of N is the sum the bits of N mod 2. The parity number's value is .4124540336401075977..., or, for hexadecimal freaks, .6996966996696996... . It can be written (base 2) in stages by taking the previous stage, complementing, and appending to the previous stage: ```.0 .01 .0110 .01101001 .0110100110010110 .01101001100101101001... radix 2 ``` i.e., ```stage 0 = 0 N -2 1 - 2 - stage N stage N+1 = stage N + ----------------- . N 2 2 ``` If ```NUM 0 = 0, DEN 0 = 2 NUM N+1 = ((NUM N)+1) * ((DEN N)-1) N+1 2 2 DEN N+1 = (DEN N) = 2 ``` then ``` N N NUM N+1 -2 -2 ------- = stage N+1 = (stage N + 2 ) * (1 - 2 ) . DEN N+1 ``` Or, faster, by substituting in the string at any stage: • the string itself for zeros, and • the complement of the string for ones. It is claimed (perhaps proven by Thue?) that the parity number is transcendental. Its regular continued fraction begins: 0 2 2 2 1 4 3 5 2 1 4 2 1 5 44 1 4 1 2 4 1 1 1 5 14 1 50 15 5 1 1 1 4 2 1 4 1 43 1 4 1 2 1 3 16 1 2 1 2 1 50 1 2 424 1 2 5 2 1 1 1 5 5 2 22 5 1 1 1 1274 3 5 2 1 1 1 4 1 1 15 154 7 2 1 2 2 1 2 1 1 50 1 4 1 2 867374 1 1 1 5 5 1 1 6 1 2 7 2 1650 23 3 1 1 1 2 5 3 84 1 1 1 1284 ... and seems to continue with sporadic large terms in suspicious patterns. A non-regular fraction is ```1/(3 -1/(2 -1/(4 -3/(16 -15/(256 -255/(65536 -65535/ N N 2 2 (...2 -(2 -1)/(... . ``` This fraction converges much more rapidly than the regular one, its Nth approximant being ```1 + NUM N --------- , 1 + DEN N ``` which is, in fact, an approximant of the regular fraction, roughly the 2^N'th. In addition, 4(parity number) = ``` 1 3 15 255 65535 2 - - - * -- * --- * ----- * ... 2 4 16 256 65536 ``` This gives still another non-regular fraction per the product conversion item in the CONTINUED FRACTION section. For another property of the parity number, see the spacefilling curve item in the TOPOLOGY section. ### ITEM 123 (Schroeppel, Gosper, Salamin): Consider the image of the circle \|z\| = 1 under the function ``` n === 2 \ z f(z) = > --- . / n === 2 ``` This is physically analogous to a series of clock hands placed end to end. The first hand rotates around the center (0,0) at some rate. the next hand is half as long and rotates around the end of the first hand at twice this rate. The third hand rotates around the end of the second at four times this rate; etc. It would seem that the end of the "last" hand (really there are infinitely many) would sweep through space very fast, tracing out an (infinitely) long curve in the time the first hand rotates once. The hands shrink, however, because of the 2^n in the denominator. Thus it is unclear whether the curve's arc length is really infinite. Also, it is a visually interesting curve, as are ``` === n! === FIB(n) \ z \ z f(z) = > --- and f(z) = > ------- . / n! / FIB(n) === === ``` Gosper has programmed the one mentioned first, which makes an intriguing display pattern. see following illustrations. If you write a program to display this, be sure to allow easy changing of: 1. z and z* on alternate terms (alternate hands rotate in opposite directions), 2. negation of alternate terms (alternate hands initially point in opposite directions), and 3. how many terms are used in the computation, since these cause fascinating variations in the resulting curve. Figure 6(a). Image of circles \|z\| = 1/2, 3/4, 7/8, 1 under the function ``` inf === n! \ z f(z) = > --- . / n! === n=1 ``` Figure 6(b). Image of circles \|z\| = 1/8, 2/8, ..., 8/8 under the function ``` inf n === 2 \ z f(z) = > --- . / n === 2 n=0 ``` Both [original] plots by Salamin on the RLE PDP-1. ### ITEM 124 (Schroeppel): Consider ``` ==== ==== ==== \ 1 \ 1 1 \ 1 1 > -- = > (-- - ------) + > (----- - -----) = / 2 / 2 2 1 / 1 1 ==== N ==== N N - - ==== N - - N + - 4 2 2 ==== \ 1 2 - > ------------- . / 2 2 ==== N (4 N - 1) ``` Take the last sum and re-apply this transformation. This may be a winner for computing the original sum. For example, the next iteration gives ``` ==== 31 \ 1 -- - 9 > -------------------------------- 18 / 2 2 4 2 ==== N (4 N - 1) (25 N + 5 N + 9) ``` where the denominator also = ``` 2 2 2 N (2 N - 1) (2 N + 1) (5 N - 5 N + 3) (5 N + 5 N + 3) ``` ### ITEM 125 (Polya): CONJECTURE: If a function has a power series with integer coefficients and radius of convergence 1, then either the function is rational or the unit circle is a natural boundary. Reference: Polya, Mathematics and Plausible Reasoning, volume 2, page 46.	3,171	9,163	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2017-30	latest	en	0.643821
http://toccata.lri.fr/gallery/tree_height.en.html	1,537,838,196,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267160853.60/warc/CC-MAIN-20180925004528-20180925024928-00289.warc.gz	246,580,021	5,850	Computing the height of a tree in CPS style Topics: Trees Tools: Why3 see also the index (by topic, by tool, by reference, by year) Computing the height of a tree in CPS style (author: Jean-Christophe FilliĆ¢tre) ```module HeightCPS use int.Int use int.MinMax use bintree.Tree use bintree.Height let rec height_cps (t: tree 'a) (k: int -> 'b) : 'b variant { t } ensures { result = k (height t) } = match t with \| Empty -> k 0 \| Node l _ r -> height_cps l (fun hl -> height_cps r (fun hr -> k (1 + max hl hr))) end let height (t: tree 'a) : int ensures { result = height t } = height_cps t (fun h -> h) end ``` with a while loop, manually obtained by compiling out recursion ```module Iteration use int.Int use int.MinMax use list.List use bintree.Tree use bintree.Size use bintree.Height use ref.Ref type cont 'a = Id \| Kleft (tree 'a) (cont 'a) \| Kright int (cont 'a) type what 'a = Argument (tree 'a) \| Result int let predicate is_id (k: cont 'a) = match k with Id -> true \| _ -> false end let predicate is_result (w: what 'a) = match w with Result _ -> true \| _ -> false end function evalk (k: cont 'a) (r: int) : int = match k with \| Id -> r \| Kleft l k -> evalk k (1 + max (height l) r) \| Kright x k -> evalk k (1 + max x r) end function evalw (w: what 'a) : int = match w with \| Argument t -> height t \| Result x -> x end function sizek (k: cont 'a) : int = match k with \| Id -> 0 \| Kleft t k -> 3 + 4 * size t + sizek k \| Kright _ k -> 1 + sizek k end lemma sizek_nonneg: forall k: cont 'a. sizek k >= 0 function sizew (w: what 'a) : int = match w with \| Argument t -> 1 + 4 * size t \| Result _ -> 0 end lemma helper1: forall t: tree 'a. 4 * size t >= 0 lemma sizew_nonneg: forall w: what 'a. sizew w >= 0 let height1 (t: tree 'a) : int ensures { result = height t } = let w = ref (Argument t) in let k = ref Id in while not (is_id !k && is_result !w) do invariant { evalk !k (evalw !w) = height t } variant { sizek !k + sizew !w } match !w, !k with \| Argument Empty, _ -> w := Result 0 \| Argument (Node l _ r), _ -> w := Argument l; k := Kleft r !k \| Result _, Id -> absurd \| Result v, Kleft r k0 -> w := Argument r; k := Kright v k0 \| Result v, Kright hl k0 -> w := Result (1 + max hl v); k := k0 end done; match !w with Result r -> r \| _ -> absurd end end ``` Computing the height of a tree with an explicit stack (code: Andrei Paskevich / proof: Jean-Christophe FilliĆ¢tre) ```module HeightStack use int.Int use int.MinMax use list.List use bintree.Tree use bintree.Size use bintree.Height type stack 'a = list (int, tree 'a) function heights (s: stack 'a) : int = match s with \| Nil -> 0 \| Cons (h, t) s' -> max (h + height t) (heights s') end function sizes (s: stack 'a) : int = match s with \| Nil -> 0 \| Cons (_, t) s' -> size t + sizes s' end lemma sizes_nonneg: forall s: stack 'a. sizes s >= 0 let rec height_stack (m: int) (s: stack 'a) : int requires { m >= 0 } variant { sizes s, s } ensures { result = max m (heights s) } = match s with \| Nil -> m \| Cons (h, Empty) s' -> height_stack (max m h) s' \| Cons (h, Node l _ r) s' -> height_stack m (Cons (h+1,l) (Cons (h+1,r) s')) end let height1 (t: tree 'a) : int ensures { result = height t } = height_stack 0 (Cons (0, t) Nil) end ``` Computing the height of a tree with a small amount of memory: Stack size is only proportional to the logarithm of the tree size. (author: Martin Clochard) ```module HeightSmallSpace use int.Int use int.MinMax use int.ComputerDivision use option.Option use bintree.Tree use bintree.Size use bintree.Height function leaves (t: tree 'a) : int = 1 + size t ``` Count number of leaves in a tree. ``` let rec height_limited (acc depth lim: int) (t:tree 'a) : option (int,int) requires { 0 < lim /\ 0 <= acc } returns { None -> leaves t > lim \| Some (res,dl) -> res = max acc (depth + height t) /\ lim = leaves t + dl /\ 0 <= dl } variant { lim } = match t with \| Empty -> Some (max acc depth,lim-1) \| Node l _ r -> let rec process_small_child (limc: int) : option (int,int,tree 'a) requires { 0 <= limc < lim } returns { None -> leaves l > limc /\ leaves r > limc \| Some (h,sz,rm) -> height t = 1 + max h (height rm) /\ leaves t = leaves rm + sz /\ 0 < sz <= limc } variant { limc } = if limc = 0 then None else match process_small_child (div limc 2) with \| (Some _) as s -> s \| None -> match height_limited 0 0 limc l with \| Some (h,dl) -> Some (h,limc-dl,r) \| None -> match height_limited 0 0 limc r with \| Some (h,dl) -> Some (h,limc-dl,l) \| None -> None end end end in let limc = div lim 2 in match process_small_child limc with \| None -> None \| Some (h,sz,rm) -> height_limited (max acc (depth + h + 1)) (depth+1) (lim-sz) rm end end ``` `height_limited acc depth lim t`: Compute the `height t` if the number of leaves in `t` is at most `lim`, fails otherwise. `acc` and `depth` are accumulators. For maintaining the limit within the recursion, this routine also send back the difference between the number of leaves and the limit in case of success. Method: find out one child with number of leaves at most `lim/2` using recursive calls. If no such child is found, the tree has at least `lim+1` leaves, hence fails. Otherwise, accumulate the result of the recursive call for that child and make a recursive tail-call for the other child, using the computed difference in order to update `lim`. Since non-tail-recursive calls halve the limit, the space complexity is logarithmic in `lim`. Note that as is, this has a degenerate case: if the small child is extremely small, we may waste a lot of computing time on the large child to notice it is large, while in the end processing only the small child until the tail-recursive call. Analysis shows that this results in super-polynomial time behavior (recursion T(N) = T(N/2)+T(N-1)) To mitigate this, we perform recursive calls on all `lim/2^k` limits in increasing order (see process_small_child subroutine), until one succeed or maximal limits both fails. This way, the time spent by a single phase of the algorithm is representative of the size of the processed set of nodes. Time complexity: open ``` use ref.Ref let height (t: tree 'a) : int ensures { result = height t } = let lim = ref 1 in while true do invariant { !lim > 0 } variant { leaves t - !lim } match height_limited 0 0 !lim t with \| None -> lim := !lim * 2 \| Some (h,_) -> return h end done; absurd end``` Why3 Proof Results for Project "tree_height" Theory "tree_height.HeightCPS": fully verified Obligations Alt-Ergo 1.30 VC height_cps 0.00 VC height 0.00 Theory "tree_height.Iteration": fully verified Obligations Alt-Ergo 0.99.1 Alt-Ergo 1.30 Eprover 1.8-001 VC is_id --- 0.00 --- VC is_result --- 0.00 --- sizek_nonneg --- --- --- induction_ty_lex sizek_nonneg.0 0.01 --- --- helper1 0.01 --- --- sizew_nonneg --- --- 0.11 VC height1 --- --- --- split_goal_right VC height1.0 --- 0.00 --- VC height1.1 --- 0.00 --- VC height1.2 --- 0.01 --- VC height1.3 --- 0.01 --- VC height1.4 --- 0.01 --- VC height1.5 --- 0.01 --- VC height1.6 --- 0.01 --- VC height1.7 --- 0.01 --- VC height1.8 --- 0.01 --- VC height1.9 --- 0.01 --- VC height1.10 --- 0.01 --- VC height1.11 --- 0.01 --- VC height1.12 --- 0.01 --- VC height1.13 --- 0.01 --- VC height1.14 --- 0.01 --- VC height1.15 --- 0.01 --- VC height1.16 --- 0.00 --- VC height1.17 --- 0.01 --- VC height1.18 --- 0.01 --- VC height1.19 --- --- --- inline_all VC height1.19.0 --- 0.01 --- Theory "tree_height.HeightStack": fully verified Obligations Alt-Ergo 1.30 CVC4 1.4 Vampire 0.6 sizes_nonneg --- --- --- induction_ty_lex sizes_nonneg.0 --- --- 0.23 VC height_stack --- --- --- split_goal_right VC height_stack.0 0.01 --- --- VC height_stack.1 0.00 --- --- VC height_stack.2 0.01 --- --- VC height_stack.3 0.01 --- --- VC height_stack.4 --- 0.07 --- VC height1 --- --- --- split_goal_right VC height1.0 0.00 --- --- VC height1.1 --- 0.02 --- Theory "tree_height.HeightSmallSpace": fully verified Obligations CVC4 1.5 VC height_limited 0.31 VC height 0.03	2,622	8,091	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.21875	3	CC-MAIN-2018-39	longest	en	0.459921
https://www.beatthegmat.com/a-box-contains-3-red-balls-4-green-balls-5-yellow-balls-6-t306960.html	1,563,699,613,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195526940.0/warc/CC-MAIN-20190721082354-20190721104354-00048.warc.gz	630,233,246	38,659	• NEW! FREE Beat The GMAT Quizzes Hundreds of Questions Highly Detailed Reporting Expert Explanations • 7 CATs FREE! If you earn 100 Forum Points Engage in the Beat The GMAT forums to earn 100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 tagged by: Max@Math Revolution ##### This topic has 3 expert replies and 1 member reply ### GMAT/MBA Expert ## A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [GMAT math practice question] A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 blue balls and 7 white balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 5 balls of a single color will be drawn? A. 10 B. 12 C. 15 D. 18 E. 21 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course Free Resources-30 day online access & Diagnostic Test Email to : info@mathrevolution.com ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15387 messages Followed by: 1872 members 13060 GMAT Score: 790 Max@Math Revolution wrote: [GMAT math practice question] A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 blue balls and 7 white balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 5 balls of a single color will be drawn? A. 10 B. 12 C. 15 D. 18 E. 21 Determine the WORST-CASE-SCENARIO -- for each color of ball, the MAXIMUM number that can be removed WITHOUT removing 5 of the same color: 3 red 4 green 4 yellow 4 blue 4 white Sum = 3+4+4+4+4 = 19 Implication: It is possible to remove 19 balls without selecting 5 of the same color. Thus, to GUARANTEE that 5 of the same color are removed, we must remove ONE MORE ball: 19+1 = 20 _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2950 messages Followed by: 19 members 43 Max@Math Revolution wrote: [GMAT math practice question] A box contains 3 red balls, 4 green balls, 5 yellow balls, 6 blue balls and 7 white balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 5 balls of a single color will be drawn? A. 10 B. 12 C. 15 D. 18 E. 21 We could have 3 red balls, 4 green balls, 4 yellow balls, 4 blue balls, and 4 white balls drawn, that is, a total of 19 balls drawn, without 5 balls of a single color drawn. However, the next ball drawn will be yellow, blue or white, and we will have 5 balls of a single color drawn. Therefore, we need to draw 20 balls to guarantee that at least 5 balls of a single color will be drawn. _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### GMAT/MBA Expert Legendary Member Joined 24 Jul 2015 Posted: 2495 messages Followed by: 32 members 19 GMAT Score: => The maximum number of draws we can make without drawing 5 balls of a single color is 3 + 4 + 4 + 4 + 4 = 14. This occurs when we draw 3 red balls, 4 green balls, 4 yellow balls, 4 blue balls and 4 white balls. If we draw one more ball, then we will have drawn 5 balls of a single color. Thus, to guarantee that we have drawn at least 5 balls of a single color, we must draw 14+1 = 15 balls. _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only $149 for 3 month Online Course Free Resources-30 day online access & Diagnostic Test Unlimited Access to over 120 free video lessons-try it yourself Email to : info@mathrevolution.com ### Top Member Master \| Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 330 messages Upvotes: 27 [quote="Max@Math Revolution"]=> The maximum number of draws we can make without drawing 5 balls of a single color is 3 + 4 + 4 + 4 + 4 = 14. This occurs when we draw 3 red balls, 4 green balls, 4 yellow balls, 4 blue balls and 4 white balls. If we draw one more ball, then we will have drawn 5 balls of a single color. Thus, to guarantee that we have drawn at least 5 balls of a single color, we must draw 14+1 = 15 balls. Therefore, C is the answer. Answer: C The highlighted calculation adds up to 19 not 14 • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200 Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Get 300+ Practice Questions Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for \$0 Available with Beat the GMAT members only code ### Top First Responders* 1 Ian Stewart 44 first replies 2 Brent@GMATPrepNow 34 first replies 3 Scott@TargetTestPrep 32 first replies 4 Jay@ManhattanReview 31 first replies 5 GMATGuruNY 18 first replies * Only counts replies to topics started in last 30 days See More Top Beat The GMAT Members ### Most Active Experts 1 Scott@TargetTestPrep Target Test Prep 139 posts 2 Max@Math Revolution Math Revolution 90 posts 3 Ian Stewart GMATiX Teacher 53 posts 4 Brent@GMATPrepNow GMAT Prep Now Teacher 51 posts 5 Jay@ManhattanReview Manhattan Review 31 posts See More Top Beat The GMAT Experts	1,802	6,710	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.4375	3	CC-MAIN-2019-30	latest	en	0.818846
https://peakup.org/blog/tag/duseyara-en/	1,726,765,473,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00882.warc.gz	405,532,768	30,329	## The VLOOKUP Function Hello everybody, As you can understand from the title, this article will be pretty long. If you want to become an expert in the VLOOKUP formula, you can read this article till the end and apply it. It is a function that everybody has heard of, everybody wants to learn and that everybody uses in Excel. We use the VLOOKUP function to find a value in the table. Let’s explain it a bit: we use the VLOOKUP function when we search a data in a table or data range and want to get data from a column that corresponds to that data. For example: let’s assume that you have a personnel salary table. In one cell there is the personnel name, you can search your employee’s name and find out the salary they get easily. ### SYNTAX VLOOKUP (lookup_value, table_array, col_index_num, [range_lookup]) To express how the VLOOKUP function works simply: =VLOOKUP (look for this; look in this table; if you find it bring me the one in the 2nd column; [TRUE – Approximate Match / FALSE – Exact Match]) Now let’s take a look at the arguments this function wants from us. Argument name Description lookup_value (required) The value you want to look up. The value you want to look up must be in the first column of the range of cells you specify in the table_array argument. For example, if table-array spans cells B2:D7, then your lookup_value must be in column B. Lookup_value can be a value or a reference to a cell. table_array (required) The range of cells in which the VLOOKUP will search for the lookup_value and the return value. You can use a named range or a table, and you can use names in the argument instead of cell references. The first column in the cell range must contain the lookup_value. The cell range also needs to include the return value you want to find. Learn how to select ranges in a worksheet. col_index_num (required) The column number (starting with 1 for the left-most column of table_array) that contains the return value. range_lookup (optional) A logical value that specifies whether you want VLOOKUP to find an approximate or an exact match: • Approximate match – 1/TRUE assumes the first column in the table is sorted either numerically or alphabetically, and will then search for the closest value. This is the default method if you don’t specify one. For example, =VLOOKUP(90,A1:B100,2,TRUE). • Exact match – 0/FALSE searches for the exact value in the first column. For example, =VLOOKUP(“Smith”,A1:B100,2,FALSE). ### Working Conditions • When the required arguments are not entered, the function doesn’t work. • If the range_lookup optional argument is not entered, TRUE- Approximate match is considered as default. • The lookup_value always has to be on the far left (in first column) of the selected/indicated table array. But this situation can be overcome by writing different functions in VLOOKUP. (We will see this below.) • If the lookup_value is not found on the left of the table_array, #N/A error is returned. • If there is a multiple number of the data written into the lookup_value argument in the range indicated in the table_array argument, this function returns the record of the first data it found • it is not checked if the text written into the lookup_value argument is with uppercase or lowercase. • when a numeric data is written into the lookup_value argument, if the data type stated as table_array in the first column is text, #N/A error is returned. the same condition is valid in the opposite condition. • if a smaller number than 1 or a bigger number than the column number in the indicated table range or a text is written into the col_index_num argument, #VALUE! error is returned. • when a column is deleted from the field stated as table_array, #REF! error is returned. • 1 and 0 can be written instead of TRUE and FALSE in the range_lookup argument. • If TRUE – Approximate Match is chosen, the table assumes that the first column will be sorted by number or alphabetically and looks for the most approximate value on the base. If you don’t specify a method, this method is used as default. • If FALSE- Exact Match is chose, if there is a value that is exactly the same with the data you have stated in the lookup-value argument is look for in the first column of the table_range. ### THE BASIC USE OF THE VLOOKUP FUNCTION You see a search sample suitable with the basic language and working conditions of the VLOOKUP function in the image below. We have said that if you find the name Sami Önder in the B column which is the first column in the B3:D9 cell range , bring the data in the 3rd column of the selected range. We have mentioned that we can write 0 instead of FALSE. With this FALSE expression, it i checked if there is an exact Sami Önder in the B column. ### USING WILDCARD CHARACTERS We can get the same result without having to write the full name we are looking for on the same table by using a wildcard character after writing the name only. Let’s brush up on the Wildcard Characters that we can use in Excel. We have 3 different Wildcard characters. Use To find ? (question mark) Any single character For example, sm?th finds “smith” and “smyth” * (asterisk) Any number of characters For example, east finds “Northeast” and “Southeast” ~ (tilde) followed by ?, , or ~ A question mark, asterisk, or tilde For example, fy06~? finds “fy06?” The asterisk and question mark act as complementary character when it is in the beginning or end of a text in the lookup_value argument of the VLOOKUP function. And ~ only limits the other wildcard characters by preceding them. If the wildcard character precedes the text, it implies that the text ends with that. For example: ali (if it ends with ali) If the wildcard character is at the end of the text, it implies that the text starts with the. For example: ali (If it starts with ali) If the wildcard character is both in the beginning and at the end of the text, it implies that the texts contains that. For example: ali (if the text contains the word ali) here is an example for the question mark: when you write ?ali, it means that there is a character (text, numbers or symbols) before the word ali. You can analyze this image as an example of the situation I have mentioned. ‘=F4&”” ### METHODS OF GETTING DATA FROM MULTIPLE COLUMNS All Functions in Excel -except for the recently added dynamic array functions- return only one result. When we get a result with the VLOOKUP formula, we might want to get the other data in the table by dragging the formula right. Instead of writing the formula over and over again for each column, we can get data from the others column in the field we have stated as table_array when we drag the cell we have written the formula in right after adding a function to the col_index_num function that would change it as dynamic. We can execute this action with a few methods. Now, let’s analyze these methods. #### Our first method.. Let’s assume that we have a table like the one here. We will get the data in the I.D. Number and Salary fields when we write the Name of the Personnel and drag it right. We will have to write our formula correctly to execute an action like that. And by correctly I mean using the cell styles correctly. Which mean will we pin the cells while selecting the cell or cell range or not, or are we going to pin the row or column only? As an answer to these questions, we have to write the \$ (dollar) symbols in the direction it is supposed to be when we select the cell. And then we will have to use a function that would allow us the make the col_index_num part dynamic. Well, let’s start. While selecting a cell in the formula, we have to think about the pinning issue we have mentioned. Since the place of the cell in which we write the Name of the Personnel is stable, we press F4 and pin it. And since the place of the cell range we selected for the table_array argument, we press F4 again and pin that cell range as well. And our formula will look like this as you can see in the image. =DÜŞEYARA(\$G\$2;\$B\$1:\$E\$7;2;0) DÜŞEYARA is the Turkish word for VLOOKUP. So in English it would be: =VLOOKUP(\$G\$2;\$B\$1:\$E\$7;2;0) And when we drag our formula right, since we wrote 2 manually to the col_index_num argument, that part will always stay as 2. Our aim here is to change the col_index_num argument by increasing 2,3,4 when we drag the formula right and to get the data in those columns. Right at this moment comes the COLUMN function to our help. ### What is the COLUMN function? How is it used? The COLUMN function returns the column number of the given cell reference. It is used in two ways. When it is written as =COLUMN(), gives the number of the column of the cell in which it is written. When it is written as =COLUMN(reference address) gives the number of the column of the cell stated as the reference address. If you write the =COLUMN() formula into the E10 cell, since the E column is the 5th column in Excel, the formula will return the number 5. Likewise, if you write the =COLUMN(B1) formula into the E10 cell, since the B column is the 2nd column in Excel, the formula will return the number 2. Since we didn’t pin anything in the B1 cell address, when we drag the formula right it will change as C1, D1, E1. And this allows us to get the numbers 2,3,4,5. Thanks to this, we will be able to get the data from the other columns in the table_array range. You can see the sample formula below for the situation we want. #### Our second method… We will not use an auxiliary function in this method, but since the argument that will change will be col_index_num, we will try to get a result with the array in that part. Which means that we will write the column index numbers we want to get as {2;3;4} and make it an array formula and make each one of these numbers transmitted to each column. Here is how we’ll do it: • Edit the formula in the H2 cell like in the image. • Choose the field we will transmit the result of the formula to. (H2:J2) • Press F2 and get into the cell. • Press Ctrl + Shift + Enter and complete the formula entry. Being able to write the col_index_num we want into the col_index_num argument within {…..} curly braces will allow us to get the data in the columns we want and grant us with flexibility. ### VLOOKUP FROM THE RIGHT As you know, look ups are done from left to right all the time in Excel. Look ups can be from left to right in columns are in characters of a text. And many functions have been created and released with the same logic. The VLOOKUP function is one of these functions. As we have mentioned in the Working Conditions above, the lookup_value always has to be in the first column of the table_array in the VLOOKUP function. If it is not found, the formula returns the #N/A error. By the way, this is not an “error” like we think it is, it is just a warning meaning that the looked up value does not exist in the table. Getting an error when the data we look up is not in the first column of the range we have stated is a code related result of the VLOOKUP function. Let’s have an example of this situation if you please. If the cell range chosen for the table_array argument starts from the A column while looking for the personnel named Murat OSMA, an error will be returned since there will not be a personnel name in the first column of the field we have selected. This is completely normal. We can arrange this working style of the VLOOKUP function that causes us to get an error the way we want. There are a few different methods for this too, but I will try to explain the easy one for you. We have a function named CHOOSE. This function requires an index_num and selecting arrays, i.e. data clusters. The formula written in the image below is: =ELEMAN(1;C1:C7;B1:B7) [ELEMAN is the Turkish version of the CHOOSE function. Thus, in English it would be: =CHOOSE(1;C1:C7;B1:B7)] We have written 1 as the index_num in the formula, and then chose each data range (arrays); first the C1:C7 range and then the B1:B7 range. Since we wrote 1 into the index_num and choose the C1:C7 data range as the 1st data range, this formula will return the data in the C1:C7 range. If we had written 2 into the index_num, it would have returned the data in the B1:B7 range. We get the data in the column we want when we state these index_num and arrays. You see the formula you can use to prevent errors and get the data you want. We state in which range the result should be by writing it in col_index_num within { … } in the index_num argument by using the CHOOSE function in the table_array argument in the VLOOKUP function. This way, if you write 1 into the col_index_numyou will get the result in the B2:B7 range and if you write 2, you will get the result in the A2:A7 range. ### USING THE RANGE_LOOK UP ARGUMENT [TRUE – APPROXIMATE MATCH] (1) We usually use the VLOOKUP function to get the data that matches exactly with the value we are looking for. And 95% of the people who use this argument use the [range_lookup] argument as FALSE – Exact Match, i.e. 0. So, what is this TRUE – Approximate Match? When you look at the argument, it explains as:“the data in the first column of the table_array should be listed in an ascending order.” So, when the [range_lookup] value is chosen as TRUE, sort the first column of the table_array before using the VLOOKUP. If you don’t sort it, it is pretty likely to get the wrong data unless in an exceptional situation. If the data in the table_array field is not sorted, if even if there is one that exactly matches, the #N/A error is returned. #### When should you use TRUE – Approximate Match? • If the data in the table_array field is sorted, we can use the TRUE – Approximate Match. • If we are looking for a Numeric data and there is a number that matches with it, or there is no number that matches with it we can use it to bring the closest number to it. Now, let’s have an example of this situation. What we want to do in the formula written in the image below: We want to find the Name of the Personnel of the Salary that is in the base of the number 3300 or numbers close to it. For this action to give the correct result, keep in mind that the data in the first column of the table_array argument should be sorted. The formula we write will give use the result of the name of the Personnel with the Salary of 3250 -since there is no one with the Salary of 3300 and this is the closest one-, so the name of Sami Önder. ### VLOOKUP BY TWO OR MORE CRITERIA We have used the VLOOKUP function to find a data in the table and then get the data from the column we wanted so far. If we want, we can also get a result depending on multiple criteria. We have stated in the image below that we want to get this: Get the Name of The Personnel whose salary is 2500 and whose Place of Birth is İzmir. And we used this formula for this: =SUBSTITUTE(VLOOKUP(G2&H2&””;B1:B7&D1:D7&C1:C7;1;0);G2&H2;””) We have used this technique here: We joined the Salary(Maaş in Turkish as you see in the image) and Place of Birth(Doğum Yeri in Turkish) in the lookup_value argument (G2&H2) and added &”” at the end, and thus got to state that our conditions start with these. And we have joined the ranges of Salary, Place of Birth and the Name of Personnel that we want to get as a result in the table_array argument. Since we have joined the ranges we wanted to get and we were looking for with in a single field, we have written 1 in the col_index_num argument. We have written 0 for Exact Match in the range_lookup argument. As a result, it returns us the 2500İzmirSami Önder text. Lastly, when we say leave the data we write in the lookup_value argument blank in the text returned from the SUBSTITUTE function, here will be the Personnel Name left Only. This way, we got to find the data we were looking for by multiple criteria. ### WHAT YOU SHOULD DO WHEN THE DATA TYPE YOU ARE LOOKING FOR AND THE DATA TYPE YOU FIND IS DIFFERENT? Sometime even though a data should be numeric, it is determined as text because of the call style. In this case, you might have realized that Excel warn you with green triangles and yellow warning symbol on the top left of these cells. When you click that warning symbol, you will see the Number Stored as Text expression. This explanation tells us that the data in the cell is a number but since the cell style is text, I store this number you have written as a text. If the data type in the cell stated as lookup_value is a text, the data type of the first column of the cell range we have stated as table_array must be text as well. Or you can think of the opposite situation. To sum up: the types of the looked up and found data should be the same. If they are not the same, the #N/A error is returned. =VLOOKUP(G2;B1:E7;2;0) Since the data type of lookup_value is text, and the data in the first column of the table_array is numbers, there is no match and the #N/A error is returned. So, how do we solve this? It is very simple! We can solve this problem by applying a mathematical operation that would turn the text stated as the lookup_value into a number. The mathematical operation to be applied shouldn’t affect the existing number. For example, if we use any mathematical operation like +0, 1, /1, ^1 or — (plus 1, multiply by 1, divide by 1, exponent 1, hyphen hyphen), we will get the result you can see below. The formula used: =VLOOKUP(G2+0;B1:E7;2;0) ### USE VLOOKUP INSTEAD OF WRITING NESTED IFS In some of our tables, for example when we want to calculate annual leave, we can write many nested IF function and find the solution that way. But the VLOOKUP gives us the solution to solve this calculation faster and easier. Now, let’s make the same calculation with nested IFs and then with VLOOKUP. You will see the difference. Let’s assume that we have a table like the one below. We will determine different rations depending on the different processes. If we want to do it with nested If functions, we will have to write the formula long like this: =IF(C2=”CV Search”;10%;IF(C2=”Bizde Mülakat Aşamasında”;20%; IF(C2=”Müşteride 2. Mülakatta”;70%;IF(C2=”Müşteri Assessment Aşamasında”;80%)))))) We have shared this with the Turkish sentences so that you can see the cells in the table. However, the formula would be like this with the English translation of these sentences: =IF(C2=”CV Search”;10%;IF(C2=”In the interview stage with us”;20%; IF(C2=”The candidate has been shared”;40%;IF(C2=”In the 1st interview with the customer”;50%; IF(C2=”In the 2nd interview with the customer”;70%;IF(C2=”The costumer is in the Assessment stage”;80%)))))) Whereas it would be way better to find the percentage ratio with the VLOOKUP function if we had a table of all processes and ratios. Yeess… If you have read, comprehended and applied everything so far, your are an EXPERT in VLOOKUP now. 👏🏻 You can share this post to help many people get informed and get Excel Training to use Excel more efficiently and productively. 👍🏻	4,604	19,065	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.390625	3	CC-MAIN-2024-38	latest	en	0.776894
healthbeauty24h.com	1,610,990,314,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703515075.32/warc/CC-MAIN-20210118154332-20210118184332-00746.warc.gz	376,737,271	14,038	How many calories are burned after 1 hour of exercise and sports? You are learning about the gymnastics and wondering whether the calories burned after 1 hour of exercising and sports is exactly how much? Let’s find out and answer this question! 1. How many calories does 1 hour walk consume? According to calculations by experts, an average of 1 hour of walking will consume about 150-200 calories. To feel the change, you need to walk at a faster pace than usual and increase the time and intensity. Try to keep up the pace and sometimes try to climb the slopes to drain your energy faster. To easily know how many calories an hour walk consumes, you can use a treadmill at home to exercise because on the display screen of the machine has detailed statistics of this parameter. 2. How many calories does 1 hour jogging consume? Running at a steady pace for a long time can burn quite large calories. Specifically, 30 minutes of running can consume about 300-500 calories depending on weight, hard work as well as running speed. The effect is almost similar between running outdoors and running indoors with a treadmill. 3. How many calories does 1 hour bicycle use? One of the easiest methods to consume energy is to take a bike and go to the road. A 60-minute, regular cycle of cycling will help you reduce about 400 calories and if you speed up, you also significantly increase the number of calories to burn. Moreover, if you do not want to ride on the road, you can completely change the form of exercise by bike at home but the effect is almost the same. 4. How many calories does 1 hour swimming consume? It is estimated that swimming at a moderate pace and within about 50 minutes will help burn about 500 calories. Note, after swimming tired, you can reward yourself a snack with foods containing no more than 100 calories, such as fruit or a glass of juice. Swimming also brings a sense of relaxation, besides rapid weight loss. 5. How many calories does 1 hour boxing consume? Boxing or punching sandbags is a favorite program for supermodels and applies a lot to exercise. Boxing not only helps you slim arms as expected but also burn from 500-1000 calories / hour. 6. How many calories does 1 hour Yoga use? Yoga is a gentle exercise and does not consume as many calories as the exercises mentioned above. However, Yoga helps you eliminate toxins in the body out of the sweat and also consumes about 350 calories / hour. 7. How many calories does 1-hour Zumba consume? If you do the Zumba exercises, you will burn 400-600 calories / hour. 8. How many calories does 1 hour jump lead? With 20 minutes of continuous skipping, you’ll burn about 300 calories. Moreover, this activity is good for your muscles and cardiovascular system. No need to dance to rigid rules, you can freely dance as you like and as long as the muscles are working at full capacity. 9. How many calories does 1 hour aerobic exercise consume? According to experts, aerobic exercise consumes about 500-800 calories / hour of exercise. A lot of weight loss exercises are integrated in this gymnastics. The parts of the body that women want to go to when doing aerobic exercise are the legs, waist and buttocks. Work out for 1 hour a day, divide the exercises in half and do half an hour for sure you will see the effect after 2 weeks. 10. How many calories does 1 hour soccer use? An hour of playing ball, you will easily consume about 500 calories. If you speed up a bit and coordinate with your teammates, you can burn 700 calories. Epilogue. Above is some good information to answer the question of how many calories are burned after 1 hour of exercise. Hopefully, with these shares, you can find a sport that best suits your goals. Wish you good health and have a nice body. [Healthbeauty24h.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program designed to provide a means for sites to earn advertising fees by advertising and linking to Amazon.com] Thibft kế web bởi Hoangweb.com	866	4,041	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.71875	3	CC-MAIN-2021-04	latest	en	0.961815
https://www.physicsforums.com/threads/expressing-newtons-2nd-law-in-terms-of-momentum.803232/	1,508,621,984,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187824899.43/warc/CC-MAIN-20171021205648-20171021225648-00223.warc.gz	985,223,077	17,739	# Expressing Newton's 2nd law in terms of momentum 1. Mar 15, 2015 ### Calpalned 1. The problem statement, all variables and given/known data Show that $\Sigma \vec F = \frac {d \vec p}{dt}$ 2. Relevant equations $\Sigma \vec F = m \vec a$ $\vec a = \frac {d \vec v}{dt}$ $\vec p = m \vec v$ 3. The attempt at a solution We need to prove that $\frac {d \vec p}{dt} = m \vec a$. When I physicists correctly take the derivative of $\vec p$, they get $m \vec a$. How come taking the derivative doesn't affect $m$ ? If $m$ is constant, shouldn't it go to zero? I know that I am wrong somewhere, but I want to fully understand how this formula was derived. Last edited: Mar 15, 2015 2. Mar 15, 2015 ### Hithesh The method to find derivative of such a function is different. When derivative of 'only' the constant is taken, it is zero. When a constant is multiplied with some function, the derivative of the resulting function is the constant multiplied with the derivative of the function. F=dp/dt F=d(mv)/dt F=m dv/dt F=ma. Last edited: Mar 15, 2015 3. Mar 15, 2015 ### ToBePhysicist Edit: the V is not to be here. Last edited: Mar 15, 2015 4. Mar 15, 2015 ### Calpalned 5. Mar 15, 2015 ### Hithesh Hope it helped 6. Mar 15, 2015 ### Calpalned Yes, that makes sense. Normally, if $\vec v$ is a variable and $m$ is a constant, the derivative of $m \vec v$ will become $m$. Is $\vec v$ not a variable here? 7. Mar 15, 2015 ### Anirudh Sharma Second law of motion states that, "The rate of change of momentum is directly proportional to the force applied. " Now p = mv We take m as constant, because rate of change of mass in situations involving second law of motion is negligible (you can consider rocket propulsion as an example in which mass varies significantly with time). taking derivative both sides with respect to time (t), ΣF ∝ dp/dt ΣFd(mv)/dt ΣF m.dv/dt ΣF ∝ ma (a = acceleration = d(v)/dt ) ΣF = kma (k = proportionality constant = 1, in SI units) ΣF = ma Since, you are taking derivative of velocity with respect to time equals to 1 instead of acceleration, that is why you are getting on that result. 8. Mar 15, 2015 ### ToBePhysicist V has nothing to do here. kick it out. 9. Mar 15, 2015 ### Hithesh V is the variable. But derivative of mv isn't m. It is m multiplied by derivative of v with respect to time. Derivative of velocity with respect to time is nothing but acceleration	713	2,426	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2017-43	longest	en	0.906809
https://www.coursehero.com/file/6644776/583L7/	1,524,574,919,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125946688.88/warc/CC-MAIN-20180424115900-20180424135900-00352.warc.gz	762,665,503	202,342	{[ promptMessage ]} Bookmark it {[ promptMessage ]} # 583L7 - EECS 583 Class 7 Static Single Assignment Form... This preview shows pages 1–8. Sign up to view the full content. EECS 583 Class 7 Static Single Assignment Form University of Michigan September 28, 2011 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document - 1 - Reading Material Today’s class » “Practical Improvements to the Construction and Destruction of Static Single Assignment Form,” P. Briggs, K. Cooper, T. Harvey, and L. Simpson, Software--Practice and Experience , 28(8), July 1998, pp. 859-891. Next class Optimization » Compilers: Principles, Techniques, and Tools , A. Aho, R. Sethi, and J. Ullman, Addison-Wesley, 1988, 9.9, 10.2, 10.3, 10.7 - 2 - Last Class in 1 Slide OUT = Union(IN(succs)) IN = GEN + (OUT KILL ) Liveness Reaching Definitions/DU/UD IN = Union(OUT(preds)) OUT = GEN + (IN KILL ) Bottom-up dataflow Any path Keep track of variables/registers Uses of variables GEN Defs of variables KILL Top-down dataflow Any path Keep track of instruction IDs Defs of variables GEN Defs of variables KILL Available Definitions IN = Intersect(OUT(preds)) OUT = GEN + (IN KILL ) Top-down dataflow All path Keep track of instruction IDs Defs of variables GEN Defs of variables KILL Available Expressions IN = Intersect(OUT(preds)) OUT = GEN + (IN KILL ) Top-down dataflow All path Keep track of instruction IDs Expressions of variables GEN Defs of variables KILL This preview has intentionally blurred sections. Sign up to view the full version. View Full Document - 3 - From Last Time: Class Problem - Rdefs 1: r1 = 3 2: r2 = r3 3: r3 = r4 4: r1 = r1 + 1 5: r7 = r1 * r2 6: r4 = r4 + 1 7: r4 = r3 + r2 8: r8 = 8 9: r9 = r7 + r8 Compute reaching defs Calculate GEN/KILL for each BB Calculate IN/OUT for each BB IN = {} Gen = {1,2,3} Kill = {4} OUT = {1,2,3} IN = {1,2,3,8} {1,2,3,4,5,6,7,8} Gen = {4,5} Kill = {1} OUT = {2,3,4,5,8} {2,3,4,5,6,7,8} IN = {2,3,4,5,8} {2,3,4,5,6,7,8} Gen = {7} Kill = {6} OUT = {2,3,4,5,7,8} {2,3,4,5,7,8} IN = {2,3,4,5,6,7,8} Gen = {8} Kill = {} OUT = {2,3,4,5,6,7,8} IN = {2,3,4,5,6,7,8} Gen = {9} Kill = {} OUT = {2,3,4,5,6,7,8,9} IN = {2,3,4,5,8} {2,3,4,5,6,7,8} Gen = {6} Kill = {7} OUT = {2,3,4,5,6,8} {2,3,4,5,6,8} Remember, initial value for the OUT of BB5 is Gen(BB5) = 8. BB1 BB2 BB4 BB5 BB6 BB3 - 4 - From Last Time: Computation of Aexpr GEN/KILL Sets for each basic block in the procedure, X, do GEN(X) = 0 KILL(X) = 0 for each operation in sequential order in X, op, do K = 0 for each destination operand of op, dest, do K += {all ops which use dest} endfor if (op not in K) G = op else G = 0 GEN(X) = G + (GEN(X) K) KILL(X) = K + (KILL(X) G) endfor endfor We can also formulate the GEN/KILL slightly differently so you do not need to break up instructions like “r2 = r2 + 1”. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document - 5 - Class Problem - Aexprs Calculation 1: r1 = r6 * r9 2: r2 = r2 + 1 3: r5 = r3 * r4 4: r1 = r2 + 1 5: r3 = r3 * r4 6: r8 = r3 * 2 7: r7 = r3 * r4 8: r1 = r1 + 5 9: r7 = r1 - 6 10: r8 = r2 + 1 11: r1 = r3 * r4 12: r3 = r6 * r9 - 6 - Some Things to Think About Liveness and rdefs are basically the same thing » All dataflow is basically the same with a few parameters Meaning of gen/kill src vs dest, variable vs operation Backward / Forward All paths / some paths (must/may) Dataflow can be slow » How to implement it efficiently? This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]}	1,311	3,693	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3	3	CC-MAIN-2018-17	latest	en	0.800813
https://socratic.org/questions/59cacdc711ef6b53e57f4d7a	1,675,099,607,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499826.71/warc/CC-MAIN-20230130165437-20230130195437-00206.warc.gz	545,157,997	6,130	# Question f4d7a Feb 6, 2018 $= - 2 \frac{7}{12}$ #### Explanation: There is no need to get confused....... count the number of terms and then simplify each term separately. In the last step you can add the answers to the terms together, Terms are separated by $+ \mathmr{and} -$ signs, so there are $2$ terms: $\textcolor{b l u e}{\frac{- \frac{2}{3}}{\frac{1}{4}}} \text{ "+" } \textcolor{red}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}}$ =color(blue)(-2/3 div1/4)" "+" "color(red)(1/12" "(larr "top x top x top")/(larr"bottom x bottom x bottom") $\textcolor{w h i t e}{\times . \times x} \downarrow$ $= \textcolor{b l u e}{- \frac{2}{3} \times \frac{4}{1}} \text{ "+" } \textcolor{red}{\frac{1}{12}}$ $= \textcolor{b l u e}{- \frac{8}{3}} \text{ "+" "color(red)(1/12)" } \leftarrow$ find a common denominator $\text{ } \downarrow$ $\textcolor{b l u e}{- \frac{8}{3} \times \frac{4}{4}}$ $\text{ } \downarrow$ =color(blue)(-32/12" "+" "color(red)(1/12)" "larr# now add the numerators $= \frac{- 32 + 1}{12}$ $= - \frac{31}{12}$ $= - 2 \frac{7}{12}$	412	1,069	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5625	5	CC-MAIN-2023-06	latest	en	0.536969
http://conversion.org/volume/cord-foot/bushel-imperial	1,725,987,211,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00148.warc.gz	10,022,077	7,615	# cord-foot to bushel (imperial) conversion Conversion number between cord-foot and bushel (imperial) [bu (imp)] is 12.457670918086. This means, that cord-foot is bigger unit than bushel (imperial). ### Contents [show][hide] Switch to reverse conversion: from bushel (imperial) to cord-foot conversion ### Enter the number in cord-foot: Decimal Fraction Exponential Expression cord-foot eg.: 10.12345 or 1.123e5 Result in bushel (imperial) ? precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential: ### Calculation process of conversion value • 1 cord-foot = (exactly) (0.453069545472) / (0.03636872) = 12.457670918086 bushel (imperial) • 1 bushel (imperial) = (exactly) (0.03636872) / (0.453069545472) = 0.080271826617946 cord-foot • ? cord-foot × (0.453069545472 ("m³"/"cord-foot")) / (0.03636872 ("m³"/"bushel (imperial)")) = ? bushel (imperial) ### High precision conversion If conversion between cord-foot to cubic-metre and cubic-metre to bushel (imperial) is exactly definied, high precision conversion from cord-foot to bushel (imperial) is enabled. Decimal places: (0-800) cord-foot Result in bushel (imperial): ? ### cord-foot to bushel (imperial) conversion chart Start value: [cord-foot] Step size [cord-foot] How many lines? (max 100) visual: cord-footbushel (imperial) 00 10124.57670918086 20249.15341836171 30373.73012754257 40498.30683672343 50622.88354590428 60747.46025508514 70872.036964266 80996.61367344685 901121.1903826277 1001245.7670918086 1101370.3438009894 Copy to Excel ## Multiple conversion Enter numbers in cord-foot and click convert button. One number per line. Converted numbers in bushel (imperial): Click to select all ## Details about cord-foot and bushel (imperial) units: Convert Cord-foot to other unit: ### cord-foot Definition of cord-foot unit: ≡ 16 cu ft. = 16 × 0.3048³= 0.453069545472 m³ Convert Bushel (imperial) to other unit: ### bushel (imperial) Definition of bushel (imperial) unit: ≡ 8 gal (imp). An imperial bushel was unit of weight or mass based upon an earlier measure of dry capacity. = 0.03636872 m³ ← Back to Volume units	657	2,115	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.765625	3	CC-MAIN-2024-38	latest	en	0.604087
http://math.stackexchange.com/questions/81448/using-base-2-numbers-1010001-2-11-2?answertab=votes	1,462,237,959,000,000,000	text/html	crawl-data/CC-MAIN-2016-18/segments/1461860118321.95/warc/CC-MAIN-20160428161518-00001-ip-10-239-7-51.ec2.internal.warc.gz	175,438,043	16,908	# Using base-2 numbers: $(1010001)_{2}/(11)_{2}=?$ I wanna solve this simple equation using base-2 number system. $(1010001)_{2}/(11)_{2}=?$ I can't remember how to do that, normally I would start with $101/11$ but what should I do this base-2 numbers? Sorry if this question might be silly. Thanks a lot - using long division: 1010001 / 11 -11\|\|\|\| \| 11011 --\|\|\|\| 100\|\|\| -11\|\|\| --\|\|\| 0100\| -11\| --\| 11 -11 -- 0 - Is 110011 supposed to be your answer? If so, it's wrong. – TonyK Nov 12 '11 at 23:01 Stray 0 I think – Henry Nov 12 '11 at 23:06 @TonyK fixed the stray 0 – ratchet freak Nov 12 '11 at 23:10 In binary $10-1=1$. Thus $101-11=10$. This first digit is $1$. Next $100-11=1$. The second digit is $1$. $10-0=10$. The next digit is $0$. $100-11=1$. The next digit is $1$. Finally $11-11=0$ the last digit is $1$. Hence $$(1010001)_{2}/(11)_{2}=11011_{2} \,.$$ -	337	884	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2016-18	latest	en	0.778216
bacibi.com	1,695,734,102,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233510208.72/warc/CC-MAIN-20230926111439-20230926141439-00800.warc.gz	126,603,490	16,327	# What is the power of focal length of a convex lens is 20 cm? (2023) ## What is the power of focal length of a convex lens is 20 cm? Hence, Power of lens is 5D. (Video) 12. Express the power (with sign) of a concave lens of focal length 20 cm. (Kwatra Tuition Center) What is the focal power of a lens of focal length 20 cm? The correct option is C 5 dioptre. (Video) A man used a convex lens of focal length of 20cm in his spects , the power of his lens is.. (Pathak Classes) What is the power of a lens whose focal length is 20 m? The power of a lens of focal length 20cm is 5D. So both lenses have same convergence of refracted light. (Video) The focal length of a convex lens is 20 cm.What is the power? \| Class 10 Physics \| Doubtnut (Doubtnut मराठी) Where will a convex lens of focal length 20 cm forms an image? A convex lens of focal 20 cm forms an image at a distance of 40 cm on the other side of the lens. (Video) A convex lens of focal length 20 cm is placed in contact with a concave lens of focal (Doubtnut) What is the power of concave lens of focal length 20 cm? (Video) A convex lens has 20 cm focal length in air. What is its focal length in water? Is 20 cm what is its focal length? What is its focal length? Therefore, focal length of the spherical mirror is 10 cm. (Video) How far from a convex lens of focal length 20 cm would you place an object to get a real image e... (Doubtnut) Is the power of a convex lens of focal length 20 cm is 5d? Solution. The power of the convex lens is 5 D. (Video) The focal length of a convex lens is 25 cm. What is its power? (Doubtnut) What is a power of focal length of a convex lens is 25 cm? What is its power? To find: Power of a lens, P. Hence, the power of the lens is 4 D. (Video) A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed in (Doubtnut) When the focal length of a mirror and lens is 20 cm? We, know that to form a virtual image in the concave mirror, the only condition is when the object is placed between the focus and pole of the mirror. Here, the focal length is given as 20 cm, so the most correct answer is less than 20 cm. Thus, the correct answer is 10 cm. (Video) A convex lens A of focal length `20 cm` and a concave lens B of focal length `5 cm` are (Doubtnut) What is the focal length of a convex lens? The focal length of a convex lens is the distance between the center of a lens and its focus. Here, the rays of light traveling parallel to the principal axis converge at a point and the focus is called the real focus. (Video) A concave lens is kept in contact with a convex lens of focal length `20 cm`. The combination (Doubtnut) ## How do you find the focal length of a convex length? The focal length of convex lens formula is object distance multiplied by the image distance divided by the difference of the object distance and the image distance. (Video) A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. (Doubtnut) What is the focal length taken as for a convex lens? The focal length of a convex lens is always positive. What is the power of concave lens whose focal length is 25? f=−25cm As P=100f,f=100−25=−4dioptre. What is the power of a concave lens if its focal length is 10cm? Therefore the power of the lens is 10D. What is the focal length power of concave lens? Concave lenses have a negative focal length and optical power because they have a virtual focus. The reciprocal of the focal length is the lens's power. As a diverging lens, a concave lens will always have a negative focal length. As a result, the power of a concave lens is negative as well. What is 20 focal length? So a focal length of 20mm means that the distance from the optical center to the imaging plane is 20mm long (about ¾ of an inch). What does the focal length number mean? A camera typically has focal length in a range of 10mm to 500mm. How far from a convex mirror of 20 cm focal length? So, the distance of the image from the pole of the mirror is 10cm. What is the focal length of a convex lens of focal length 20 cm in contact with a concave? From 1f=1f1+1f2=120−110=−120,F=−20cm. P=100F=100−20=−5 dioptre. How do you find the power of focal length? The power of a lens is defined as the reciprocal of its focal length in meters, or D=1f, where D is the power in diopters and f is the focal length in meters. Lens surface power can be found with the index of refraction and radius of curvature. How do you find the power of a convex lens? 1. The power of the convex lens is 9 dioptres. 2. It is the reciprocal of the focal length(f) of the lens i.e. P=1/f. 3. Power of lens=1/focal length of lens(f), 4. To find focal length, 5. Use lens formula 1/f = 1/v-1/u. 6. Where 'v' is the image distance and 'u' is the object distance from the optical centre of lens. Mar 31, 2023 ## When an object is placed 20 cm from a convex lens? The correct Answer is The real, inverted image of same size is formed at a distance of 20 cm from the lens. The negative sign of the height of the image and the magnification shows that the image is inverted and real. What is the power of focal length of a convex lens is 10 cm? P=1 / 0.10= 10D. What is the power of convex lens of focal length 30 cm? P=−1.67 D. What is the power of a convex lens? The power of a convex lens is positive and that of a concave lens is negative. When an object is placed 20 cm in front of a convex mirror? Hence the image is formed at 8.57 cm and its nature is virtual, erect, and diminished. When an object is placed 20 cm from a concave mirror of focal length 20 cm? What is the position of the image when an object is placed at a distance of 20 cm from a concave mirror of focal length 20 cm? When an object is placed at a distance of 20 cm from a concave mirror of focal length 20 cm, then the position of the image is at infinity. What kind of mirror can have a focal length of negative 20 cm? Concave mirror (since focal length is negative) How do you find the focal length of a convex mirror using a convex? The focal length of the mirror is calculated as, f = R/2, where f is the focal length mirror and R is the radius of curvature. How to determine the focal length of a convex lens answers? Theory. For an object placed at infinity (practically a distant object), the image formed by a convex lens lies at the focus F of the convex lens. Distance between the optical centre O and the point where image is formed (i.e., focus F of the lens) is called focal length of the lens. How do you find the focal length of a convex in an experiment? It can be found by obtaining a sharp image of the Sun or a distant tree on a screen, say a plane wall, or a sheet of paper placed on the other side of the lens and measuring the distance between the lens and the image with a scale. This distance is a rough estimate of the focal length, f of the convex lens. ## Where is the focal length of a convex mirror? The focal length of a convex mirror P is the pole of a spherical mirror. The distance between the focus F and P is known as the focal length of the convex mirror and is denoted by (f). According to the paraxial approximation, the focal length of a convex mirror is half of its radius of curvature. What does a convex lens of focal length 12 cm produce? A lens of focal length 12 cm forms an upright image three times the size of a real object. Find the distance in cm between the object and image. A lens placed at a distance of 20 cm from an object produces a virtual image 2/3 the size of the object. What is the focal length of a convex lens of focal length 30 cm is in contact with a convex lens of focal length 20 cm? Hence, the focal length of the combination of lenses is 60 cm. What is the focal length of convex lens is 30 cm and the size of image is half of the object? Hence, the correct option is (A) 90 c m . How do you find the power of a lens of focal length 20cm? 1. Step 1: Given data. Lens is a convex lens. Focal length, 20 m. Step 2: Formula used and calculation. Power of a lens is the inverse of its focal length (in metres). ... 2. Step 2: Formula used and calculation. Power of a lens is the inverse of its focal length (in metres). Mathematically, P = 1 f P = 1 + 0 . 20 m P = + 5 D. What is the focal length of lens of power 20d? Focal length of the lens, f = +5 cm and its a convex lens. How do you calculate the focal power of a lens? The power of a lens is defined as the reciprocal of its focal length in metres, or P=f1, where P is the power in diopters and f is the focal length in metres. What is the power of convex lens of focal length 25 cm? What is its power? To find: Power of a lens, P. Hence, the power of the lens is 4 D. What is 20D lens? This lens is used for general examination of the fundus using “Binocular Indirect Ophthalmoscope (BIO)”. This provides a high resolution image of the retina in the OPD or the operating room. A 50 degree field of view helps in vizualization upto the mid peripheral region. What is the field of view of a 20D lens? Industry standard general diagnostic lens. Lens is 20D with a field of view of 46/60 degrees, an image magnification of 3.13x, a laser spot of 0.32x and a working distance of 50mm. ## How do I calculate focal length? How do you calculate focal length? The focal length of a mirror and a lens can be calculated using 1/do + 1/di = 1/f, where do is the object distance, di is the image distance, and f is the focal length. How do you find the focal length of a convex lens? The focal length of convex lens formula is object distance multiplied by the image distance divided by the difference of the object distance and the image distance. What is the formula for the focal length of a convex lens? Analytically, the focal length is described by the lens maker's equation: 1/f = (n - 1)(1/R1 + 1/R2), where R1 and R2 are the radii of curvature, f is the focal length, and n is the index of refraction. What will be the power of convex lens of focal length 50 cm? The correct option is B 2 Dioptre. What is the focal length if power is +2 D? If the power of a lens is 2 D, its focal length =0.5 m. A concave lens is a converging lens. A convex lens is a diverging lens. Popular posts Latest Posts Article information Author: Margart Wisoky Last Updated: 08/26/2023 Views: 5990 Rating: 4.8 / 5 (78 voted) Author information Name: Margart Wisoky Birthday: 1993-05-13 Address: 2113 Abernathy Knoll, New Tamerafurt, CT 66893-2169 Phone: +25815234346805 Job: Central Developer Hobby: Machining, Pottery, Rafting, Cosplaying, Jogging, Taekwondo, Scouting Introduction: My name is Margart Wisoky, I am a gorgeous, shiny, successful, beautiful, adventurous, excited, pleasant person who loves writing and wants to share my knowledge and understanding with you.	2,826	10,884	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2023-40	latest	en	0.907054
https://ec2-34-193-34-229.compute-1.amazonaws.com/user-quizzes/1646516/day-35-countries-3500-km-away-from-chile	1,721,149,633,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00746.warc.gz	203,918,316	9,882	# Day #35: Countries 3500 km away from Chile Imagine a circle drawn on the world with a radius of the given distance, and a center of the given country's capital. Your goal is to type all countries whose MAINLAND contains at least one point of the circle's BORDER. If the border touches the given country, you should type it too. If you think there is no valid country, type the body of water. Answers are in alphabetical order. Each day you will get an other country with an other distance based on this formula : Day #x: Countries 100x km away from xth country alphabetically. A little help: 100 km (62,14 miles) is the width of Trinidad. 1000 km is the distance between San Francisco and Phoenix. 10000 km is the distance between London and Phnom Penh. Quiz by Sajttorta Rate: Last updated: September 17, 2023 You have not attempted this quiz yet. First submitted September 17, 2023 Times taken 16 Average score 100.0% Report this quiz Report 4:00 Enter answer here 0 / 4 guessed The quiz is paused. You have remaining. Scoring You scored / = % This beats or equals % of test takers also scored 100% The average score is Your high score is Your fastest time is Keep scrolling down for answers and more stats ... Countries Brazil Colombia Ecuador Peru Comments No comments yet Save Your Progress	315	1,300	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2024-30	latest	en	0.895292
https://www.edutoolbox.org/astandard/2834	1,603,971,664,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107904039.84/warc/CC-MAIN-20201029095029-20201029125029-00061.warc.gz	684,393,031	9,417	Mathematical Processes Initiative: Tennessee Diploma Project Set: Mathematics Type: Standard Code: 1 Conceptual StrandThe Mathematical Process Standards are embedded in the content standards to reap the learning benefits gained from attention to them. These standards exemplify best teaching practices in mathematical instruction and mirror the NCTM Process Standards as well as the Common Core Mathematical Practices.Guiding QuestionWhat is the role of the mathematical processes standards in formative instruction? Elements within this Standard Course Level Expectation Use mathematical language, symbols, definitions, proofs and counterexamples correctly and precisely in mathematical reasoning. Apply and adapt a variety of appropriate strategies to problem solving, including testing cases, estimation, and then checking induced errors and the Develop inductive and deductive reasoning to independently make and evaluate mathematical arguments and construct appropriate proofs; include various types of Move flexibly between multiple representations (contextual, physical, written, verbal, iconic/pictorial, graphical, tabular, and symbolic), to solve problems, to Recognize and use mathematical ideas and processes that arise in different settings, with an emphasis on formulating a problem in mathematical terms, Employ reading and writing to recognize the major themes of mathematical processes, the historical development of mathematics, and the connections between Use technologies appropriately to develop understanding of abstract mathematical ideas, to facilitate problem solving, and to produce accurate and reliable models. Check For Understanding Create and analyze scatter-plots of non-linear and transcendental functions. Compare and contrast sampling techniques and identify the best technique for a given situation. Use calculators to identify regression equations for nonlinear data. Identify the weaknesses of calculators and other technologies in representing non-linear data, such as graphs approaching vertical asymptotes, and use Determine the accuracy and reliability of a mathematical model. Use graphical representations to perform operations on complex numbers. Use the unit circle to determine the exact value of trigonometric functions for commonly used angles (0, 30, 45, 60). Understand and describe the inverse relationship between exponential and logarithmic functions. Translate the syntax of technology to appropriate mathematical notation for non-linear and transcendental functions. Find and describe geometrically the absolute value of a complex number. Interpret the results of mathematical modeling in various contexts to answer questions. Draw conclusions based on number concepts, algebraic properties, and/or relationships between expressions and numbers over complex numbers. State Performance Indicator Move flexibly between multiple representations (contextual, physical, written, verbal, iconic/pictorial, graphical, tabular, and symbolic) of non-linear and Recognize and describe errors in data collection and analysis as well as identifying representations of data as being accurate or misleading. Use technology tools to identify and describe patterns in data using non-linear and transcendental functions that approximate data as well as using those Use mathematical language, symbols, definitions, proofs and counterexamples correctly and precisely to effectively communicate reasoning in the process of	594	3,460	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.34375	3	CC-MAIN-2020-45	latest	en	0.885459
https://hamaraguru.com/input-output-5-hamaraguru	1,556,175,797,000,000,000	text/html	crawl-data/CC-MAIN-2019-18/segments/1555578689448.90/warc/CC-MAIN-20190425054210-20190425080210-00415.warc.gz	438,035,461	28,732	# Input & Output 5 \| Hamaraguru Ques (46-50): Input: 52 peak 90 snow freezes 46 cold 16 high 31 73 trek Step I : 16 52 peak snow freezes 46 cold high 31 73 trek 90 Step II : 16 31 52 peak snow freezes 46 cold high trek 73 90 Step III : 16 31 46 peak snow freezes cold high trek 52 73 90 Step IV : 16 31 46 cold peak snow freezes high trek 52 73 90 Step V : 16 31 46 cold freezes peak snow high trek 52 73 90 Step VI : 16 31 46 cold freezes high peak snow trek 52 73 90 Step VI is the last step of the rearrangement. As per the rules followed in the above steps, find out in each of the following given questions the appropriate steps for the given input. Input: 68 hot sun 19 best 83 ice 49 ace 77 cuts 38 46. How many steps needed to complete the arrangement? a) X b) VIII c) IX d) VII e) None of these 47. Which step number be the following output? 19 38 49 ace best hot sun ice cuts 68 77 83 a) II b) VI c) V d) IV e) None of these 48. Which of the following numbers/words would be the step I? a) 19 38 49 hot sun best ice ace cuts 68 77 83 b) 83 68 hot sun best ice 49 ace 77 cuts 38 19 c) 19 68 ace best hot sun ice 49 77 cuts 38 83 d) None of these 49. Which of the following numbers/words would be the final arrangement? a) 68 77 83 ace best cuts hot ice sun 19 38 49 b) 19 38 49 ace best cuts hot ice sun 68 77 83 c) 19 38 49 68 77 83 ace best cuts hot ice sun d) 19 38 49 ace ice best cuts hot sun 68 77 83 e) None of these 50. In step IV, which of the following number/word would be on 7th position (from the right)? a) sun b) best c) 68 d) cuts e) None of these Ques (51-55): Input: goes over 35 68 test 72 park 27 Step I : 27 goes over 35 68 test 72 park Step II : 27 test goes over 35 68 72 park Step III : 27 test 35 goes over 68 72 park Step IV : 27 test 35 park goes over 68 72 Step V : 27 test 35 park 68 goes over 72 Step VI : 27 test 35 park 68 over goes 72 Step VII: 27 test 35 park 68 over 72 goes Step VII is the last step of the rearrangement of the above input as the desired arrangement is obtained. As per the rules followed in the above steps, find out in each of the following given questions the appropriate step for the given input. 51. Input: 86 opens shuts door 31 49 always 45 How many steps will be required to complete the rearrangement? a) Five b) Six c) Seven d) Four e) None of these 52. Step III of an input : 25 yes 37 enemies joy defeats 52 46 Which of the following numbers/words is definitely the input? a) enemies 25 joy defeats yes 52 37 46 b) 37 enemies 25 joy yes defeats 52 46 c) enemies joy defeats 25 52 yes 46 37 d) Cannot be determined e) None of these 53. Step II of an input: 18 wins 71 34 now if victory 60 How many more steps required to complete the rearrangement? a) Three b) Four c) Five d) Six e) More than six 54. Input : where 47 58 12 are they going 38 Which of the following steps be the last but one? a) VII b) IV c) V d) VIII e) None of these 55. Step II of an input : 33 stores 81 75 full of goods 52 Which of the following numbers/words will be step VI? a) 33 stores 52 of 75 81 full goods b) 33 stores 52 of 75 full 81 goods c) 33 stores 52 of 75 goods 81 full d) There will be no such step e) None of these	1,071	3,190	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.53125	3	CC-MAIN-2019-18	latest	en	0.613372
https://www.hackmath.net/en/example/6852?tag_id=13	1,540,329,352,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583517376.96/warc/CC-MAIN-20181023195531-20181023221031-00170.warc.gz	928,499,600	6,715	# Inscribed rectangle What is the perimeter of a rectangle that is inscribed in a circle whose diameter is 5 dm long? Result p = 12.761 dm #### Solution: Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...): Dr Math not exact solution, because rectangle have two different size Nestor The solution is ambiguous. Where does x = r/3 came from? Dr Math the solution is ok, bud problem is wrongly formulated. x = r/3 is assumed to be something calculated. The solution is picked only one from infinity many solutions. ## Next similar examples: 1. Quarter circle What is the radius of a circle inscribed in the quarter circle with a radius of 100 cm? 2. Diagonals of pentagon Calculate the diagonal length of the regular pentagon: a) inscribed in a circle of radius 12dm; b) a circumscribed circle with a radius of 12dm. 3. AP RT triangle The length of the sides of a right triangle form an arithmetic progression, longer leg is 24 cm long. What are the perimeter and area? 4. Circle annulus There are 2 concentric circles in the figure. Chord of larger circle 10 cm long is tangent to the smaller circle. What are does annulus have? 5. The swimmer The swimmer swims at a constant speed of 0.85 m/s relative to water flow. The current speed in the river is 0.40 m/s, the river width is 90 m. a) What is the resulting speed of the swimmer with respect to the tree on the riverbank when the swimmer motion 6. Equilateral triangle ABC In the equilateral triangle ABC, K is the center of the AB side, the L point lies on one-third of the BC side near the point C, and the point M lies in the one-third of the side of the AC side closer to the point A. Find what part of the ABC triangle conta 7. Trapezoid MO The rectangular trapezoid ABCD with right angle at point B, \|AC\| = 12, \|CD\| = 8, diagonals are perpendicular to each other. Calculate the perimeter and area of the trapezoid. 8. Garden Area of square garden is 6/4 of triangle garden with sides 56 m, 35 m and 35 m. How many meters of fencing need to fence a square garden? 9. Pavement Calculate the length of the pavement that runs through a circular square with a diameter of 40 m if distance the pavement from the center is 15 m. 10. Perpendicular prism Calculate the volume of the perpendicular prism if its height is 60.8 cm and the base is a rectangular triangle with 40.4 cm and 43 cm legs.	608	2,396	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2018-43	longest	en	0.914905
https://calculat.io/en/length/inches-to-mm/911	1,701,522,259,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00536.warc.gz	185,978,177	21,749	Convert 911 inches to mm How many mm is 911 inches? 911 Inches is equal to 23139.4 Millimeters Explanation of 911 Inches to Millimeters Conversion Inches to Millimeters Conversion Formula: mm = in Ã— 25.4 According to 'inches to mm' conversion formula if you want to convert 911 (nine hundred eleven) Inches to Millimeters you have to multiply 911 by 25.4. Here is the complete solution: 911â€³ Ã— 25.4 = 23139.4 mm (twenty-three thousand one hundred thirty-nine point four millimeters) Related Calculations This converter will help you to convert Inches to Millimeters (in to mm). For example, it can help you find out how many mm is 911 inches? (The answer is: 23139.4). Enter the number of inches (e.g. '911') and hit the 'Convert' button. Inches to Millimeters Conversion Table InchesMillimeters 22758.4 mm 22783.8 mm 22809.2 mm 22834.6 mm 22885.4 mm 22910.8 mm 22936.2 mm 22961.6 mm 22987 mm 23012.4 mm 23037.8 mm 23063.2 mm 23088.6 mm 23114 mm 23139.4 mm 23164.8 mm 23190.2 mm 23215.6 mm 23241 mm 23266.4 mm 23291.8 mm 23317.2 mm 23342.6 mm 23368 mm 23393.4 mm 23418.8 mm 23444.2 mm 23469.6 mm 23495 mm FAQ How many mm is 911 inches? 911â€³ = 23139.4 mm	399	1,176	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.203125	3	CC-MAIN-2023-50	latest	en	0.643704
https://unacademy.com/lesson/introduction-and-overview/K26DJ821	1,571,486,693,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00077.warc.gz	758,770,820	126,019	to enroll in courses, follow best educators, interact with the community and track your progress. 7 lessons, 1h 12m Enroll 314 Introduction and Overview 5,435 plays More This video gives a brief introduction to the course on mathematical physics. Sayantan Bhattacharya "Educating India For a Better Tomorrow" \|\|Ph.D student,U mass Lowell,Massachusetts,USA\|\| M.Sc,University Of Hyderabad,2018\|\| B.Sc,B.H.U,2016 U Pp thank you sir.. sir which book should I follow...? Sir new Student sir nirmal sir please make vedios for iit jam physics.... sir please video for iit jam physics its really helpful... thank you sir. 1. Introduction and Brief Overview By Sayantan Bhattacharya 2. About me . M.Sc University of Hyderabad. B.Sc Banaras Hindu University.(B.H.U) Major in Physics Research Interest: High Energy Physics, Theoretical Particle Physics, Computational As rophysics Unacademy link https://unacademy.com/user/s ayantan34 3. About Course This course will be covering topics in the syllabi of Mathematical Physics for GATE(physics) . Important topic for GATE; you can expect 6-7 questions on these parts. . The course will be mainly based on problem solving aproach.No derivation. Briefly describe and explain the theory and important formulas. After that, previous year problems will be discussed. 4. Brief Outline Of Course Linear vector space. - Basis Orthogonality - Completeness Matrices 5. Vector Calculus Linear differential equation. Elements of complex analysis Cauchy Reiman condition. Cauchy theorem - Singularities Residue theorem 6. Brief Outline... Laplace transform Fourier analysis Elementary tensor analysis - Rank - Contra and Co-variant tensor -Levi-Cevita and Kronecker delta 7. Outcome . Help remember the formulas and important points. Competitive exam approach of a completely theoretical physics topic. Problem solving methods. e Previous year questions will be discussed topic wise. Help you to, assimilate with GATE pattern.	443	1,960	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.578125	3	CC-MAIN-2019-43	longest	en	0.811201
https://westwriters.com/business-and-finance-west-writers/marketing-research-hw-help-westwriters-com/	1,600,735,092,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00113.warc.gz	712,969,313	10,406	# marketing research hw help \| westwriters.com With respect to Chi-square analysis describe observed and expected frequencies. What is the relationship between observed/expected frequencies and the Chi-square statistic? Don't use plagiarized sources. Get Your Custom Essay on marketing research hw help \| westwriters.com Just from \$13/Page HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 22 pts Depending on its type, a relationship is usually characterized in three ways: by its presence, direction, and strength of association. Discuss these three ways of characterizing relationships and their meaning relative to association. HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 32 pts There are four basic types of relationships between two variables. Which of the following is NOT one of those four basic types? LinearCurvilinearNonmonotonicElliptical Flag this Question Question 42 pts The formal procedure for Chi-square analysis begins when the researcher formulates a statistical null hypothesis that the two variables under investigation are not associated in the population. TrueFalse Flag this Question Question 52 pts In a nonmonotonic relationship, only the general pattern of presence or absence is known. TrueFalse Flag this Question Question 62 pts To reveal the nonmonotonic relationships found to be significant in cross-tabulation tables, researchers often turn to ________, which show the relationships very adequately. tabular presentationsreal-life examplesside-by-side comparisonsgraphical presentations Flag this Question Question 72 pts With what type of diagram can covariation be examined? Venn diagramScatter diagramCause and effectFlowchart diagram Flag this Question Question 82 pts A ________ is one in which the presence (or absence) of a label for one variable is systematically associated with the presence (or absence) of a label for another variable. non-linear relationshipnon-curvilinear relationshipnon-elliptical relationshipnonmonotonic relationship Flag this Question Question 92 pts Depending on its type, a relationship can usually be characterized by all of the following EXCEPT: Its absence.Its presence.Its direction.Its strength of association. Flag this Question Question 102 pts What is the Pearson product moment correlation coefficient? How does the Pearson product moment correlation coefficient indicate not only degree of association but also the direction? HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 112 pts Cross-tabulation can only be used for two variables; each variable must have well-defined labels. TrueFalse Flag this Question Question 122 pts There is an orderly procedure for determining the presence, direction, and strength of a relationship, TrueFalse Flag this Question Question 132 pts Within the guidelines on the reporting of correlation findings, marketing researchers usually have a “target” or a “focal” variable in mind, and they look at correlations of other variables of interest with this target variable. TrueFalse Flag this Question Question 142 pts Bar charts can be used to “see” a nonmonotonic relationship. TrueFalse Flag this Question Question 152 pts When dealing with Chi-square, it is always necessary for the researcher to state a hypothesis in a formal sense. TrueFalse Flag this Question Question 162 pts When the researcher determines that a true relationship does exist in the population by means of the correct statistical test, he or she then establishes its: Direction or pattern.Pattern and specificity.Significance or pattern.Direction or significance. Flag this Question Question 172 pts A correlation coefficient merely investigates the presence, strength, and direction of a linear relationship between two variables. TrueFalse Flag this Question Question 182 pts There is an orderly procedure for determining the presence, direction, and strength of a relationship. Outline and briefly discuss the six steps involved in the procedure for analyzing the relationship between two variables. HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 192 pts ________ are those in which there is a high probability that the two variables will exhibit a dependable relationship, regardless of the type of relationship being analyzed. High correlationsStrong associationsHigh variabilitiesHigh dependencies Flag this Question Question 202 pts With correlation analysis, each correlation will have a unique significance level. TrueFalse Flag this Question Question 212 pts Associations can be characterized by presence, direction, and strength, depending on the scaling assumptions of the questions being compared. Discuss how these characteristics are readily seen in correlation analysis. HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 222 pts A handy graphical tool that illustrates a nonmonotonic relationship is a ________ in which two variables are accommodated simultaneously. stacked bar charthorizontal bar chartgrouped bar chartpie chart Flag this Question Question 232 pts It is important to note that the angle or the slope of an ellipse has everything to do with the size of the correlation coefficient. Everything hinges on the width of the ellipse. TrueFalse Flag this Question Question 242 pts There are instances in which a marketing researcher wants to see if there is a relationship between the responses to one question and the responses to another question in the same survey. Identify the four types of possible relationships. Discuss how they are differentiated. HTML EditorKeyboard Shortcuts Font Sizes Paragraph p Flag this Question Question 252 pts Regardless of its absolute value, a correlation that is not statistically significant: Has some perceptible meaning.Has significant meaning.Requires additional examination.Has no meaning at all. ## Calculate the price of your order 550 words We'll send you the first draft for approval by September 11, 2018 at 10:52 AM Total price: \$26 The price is based on these factors: Number of pages Urgency Basic features • Free title page and bibliography • Unlimited revisions • Plagiarism-free guarantee • Money-back guarantee On-demand options • Writer’s samples • Part-by-part delivery • Overnight delivery • Copies of used sources Paper format • 275 words per page • 12 pt Arial/Times New Roman • Double line spacing • Any citation style (APA, MLA, Chicago/Turabian, Harvard) # Our guarantees Delivering a high-quality product at a reasonable price is not enough anymore. 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We will work on your paper until you are completely happy with the result.	1,500	7,430	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.703125	3	CC-MAIN-2020-40	latest	en	0.891291
http://gmatclub.com/forum/three-workers-have-a-productivity-ratio-of-1-to-2-to-3-all-133633.html?fl=similar	1,484,817,077,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560280504.74/warc/CC-MAIN-20170116095120-00343-ip-10-171-10-70.ec2.internal.warc.gz	115,741,388	62,816	Three workers have a productivity ratio of 1 to 2 to 3. All : GMAT Problem Solving (PS) Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 19 Jan 2017, 01:11 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # Three workers have a productivity ratio of 1 to 2 to 3. All Author Message TAGS: ### Hide Tags Intern Joined: 19 Jan 2010 Posts: 13 Followers: 0 Kudos [?]: 19 [1] , given: 16 Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 01 Jun 2012, 10:16 1 KUDOS 8 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 63% (02:56) correct 37% (01:55) wrong based on 367 sessions ### HideShow timer Statistics The Book does not show any answers. Hence I need to figure out if I have done them correctly or not. Please help me if you can. Would greatly appropriate it. Thanks Three workers have a productivity ratio of 1 to 2 to 3. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? A. 12 to 1 B. 6 to 1 C. 5 to 1 D. 1 to 6 E. 1 to 5 [Reveal] Spoiler: OA Intern Joined: 24 Apr 2012 Posts: 12 Followers: 0 Kudos [?]: 6 [3] , given: 0 ### Show Tags 01 Jun 2012, 18:44 3 KUDOS 3 This post was BOOKMARKED I got to (C) Ratio Setup 1:2:3, I ignored 2 as the question only asked for comparison between the slowest and fastest so it becomes 1:3. They each worked for (4) hours until the slowest took a break so the 4th hour ratio would look like 4th hour -4:12 5th hour - 4:15 (slowest worker takes a break until the 9th hour) 6th hour - 4:18 7th hour - 4:21 8th hour - 4:24 9th hour - 5:27 10th hour - 6:30 = 1 to 5 ratio. 1 to 5 ratio slowest compared to fastest flip it around as it is asking for a fastest to slowest comparison so it becomes 5 to 1. Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7078 Kudos [?]: 93144 [12] , given: 10552 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 02 Jun 2012, 06:23 12 KUDOS Expert's post 2 This post was BOOKMARKED phoenix9801 wrote: The Book does not show any answers. Hence I need to figure out if I have done them correctly or not. Please help me if you can. Would greatly appropriate it. Thanks Three workers have a productivity ratio of 1 to 2 to 3. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? A. 12 to 1 B. 6 to 1 C. 5 to 1 D. 1 to 6 E. 1 to 5 The fastest worker who does 3 units of job worked for all 10 hours, so he did 310=30 units of job; The slowest worker who does 1 unit of job worked for only 4+2=6 hours (first 4 hours and last 2 hours), so he did 16=6 units of job; The ratio thus is 30 to 6, or 5 to 1. _________________ Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7078 Kudos [?]: 93144 [1] , given: 10552 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 30 Jul 2013, 08:04 1 KUDOS Expert's post Bumping for review and further discussion. _________________ Senior Manager Joined: 13 Jan 2012 Posts: 309 Weight: 170lbs GMAT 1: 740 Q48 V42 GMAT 2: 760 Q50 V42 WE: Analyst (Other) Followers: 16 Kudos [?]: 146 [0], given: 38 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 31 Jul 2013, 11:05 This is a nice problem because it offers the chance to quite easily intuitively solve a problem. Maybe @Bunuel can go into detail how to systematically solve this problem and problems like it (i.e., problems like it that are much harder)? Here's how to use intuition though: Ratio is 1:2:3 for slow:medium:fast. Medium and fast each work for 10 hours, while slow works for 6 hours because he skipped 4 hours. So in 10 hours, fast does 30 units if work, and in 10 hours, medium does 20 units of work, while in 6 hours, slow does 6 units of work. So, the job required 56 units of work, 30 of which were done by fast and 6 of which were done by slow. so 30 to 6 = 5 to 1. Intern Joined: 11 Dec 2008 Posts: 4 Schools: Columbia Followers: 0 Kudos [?]: 3 [0], given: 0 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 06 Aug 2013, 01:14 Let A, B, C be the rates of the slowest to the fastest workers, respectively. --> C = 3A A worked for 6 hours since he skipped 4 hours from the 5th to the 8th while C worked for 10 hours. --> Total work of A = 6A and total work of C = 10C Work C/Work A = 10C/6A= (103A)/6A = 5/1 --> C. Intern Status: Waiting for Decisions Joined: 23 Dec 2012 Posts: 42 Location: India Sahil: Bansal GMAT 1: 570 Q49 V20 GMAT 2: 690 Q49 V34 GPA: 3 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 50 [0], given: 42 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 27 Oct 2013, 23:56 Hello, Could anyone just help me interpret "productivity ratio of 1 to 2 to 3" I could come to conclusion that for first worker rate is 3 units of work per hour. Also what would the rate for other two(B & C) workers Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7078 Kudos [?]: 93144 [0], given: 10552 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 28 Oct 2013, 00:18 bsahil wrote: Hello, Could anyone just help me interpret "productivity ratio of 1 to 2 to 3" I could come to conclusion that for first worker rate is 3 units of work per hour. Also what would the rate for other two(B & C) workers Three workers have a productivity ratio of 1 to 2 to 3, means that if A does 1 unit of work in an hour, then B does 2 and C does 3. A's rate in this case is 1 unit/hour, B's 1/2 unit/hour and C's 1/3 unit/hour. _________________ SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 47 Kudos [?]: 1930 [0], given: 193 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 28 Oct 2013, 00:22 Productivity ratio 1:2:3 (for ex: A:B:C) is B is twice as much productive as A & C is thrice as much productive as A So, between A, B & C, A is least productive & C is most productive. As far as B:C is concerned, it will still be 2:3 _________________ Kindly press "+1 Kudos" to appreciate Director Joined: 03 Aug 2012 Posts: 916 Concentration: General Management, General Management GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32 GPA: 3.7 WE: Information Technology (Investment Banking) Followers: 23 Kudos [?]: 692 [0], given: 322 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 29 Mar 2014, 02:58 Another Approach: More the productivity More the rate. Hence, 1:2:3 Consider X,2x, and 3x where person having rate x is slowest and person having rate 3x is fastest. All of them work for 6 hours in total 10 hours. Then, x+2x+3x = 6x Rate Rate Time = Work 6x * 6 = 36x And 3x+2x=5x (Both of the other two work for extra 4 hours in 10 hours) 5x * 4 = 20x Total work = 20x + 36x = 56x Fastest worker does work for 10 hours hence Work(Fast) = 30x Slowest worker does work for 6 hours hence Work(Slow) = 6x Ratio = 30/6 = 5/1 Rgds, TGC! _________________ Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________ Intern Joined: 21 Oct 2012 Posts: 39 Location: United States Concentration: Marketing, Operations GMAT 1: 650 Q44 V35 GMAT 2: 600 Q47 V26 GMAT 3: 660 Q43 V38 GPA: 3.6 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 16 [0], given: 19 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 16 May 2014, 02:03 Bunuel wrote: phoenix9801 wrote: The Book does not show any answers. Hence I need to figure out if I have done them correctly or not. Please help me if you can. Would greatly appropriate it. Thanks Three workers have a productivity ratio of 1 to 2 to 3. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? A. 12 to 1 B. 6 to 1 C. 5 to 1 D. 1 to 6 E. 1 to 5 The fastest worker who does 3 units of job worked for all 10 hours, so he did 310=30 units of job; The slowest worker who does 1 unit of job worked for only 4+2=6 hours (first 4 hours and last 2 hours), so he did 16=6 units of job; The ratio thus is 30 to 6, or 5 to 1. Hi Bunuel you say that the slowest worker worked for 4+2=6 hours( first 4 hours and last 2 hours) should it not be 5+1=6 since the question says that the slowest worker takes a break at the beginning of the fifth hour which means he has worked for 5 hours and also he joins back at the beginning of the 9th hour and the work completes in 10 hours which means after joining he worked only for an hour? please correct me if i am wrong Math Expert Joined: 02 Sep 2009 Posts: 36548 Followers: 7078 Kudos [?]: 93144 [0], given: 10552 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 16 May 2014, 02:14 havoc7860 wrote: Bunuel wrote: phoenix9801 wrote: The Book does not show any answers. Hence I need to figure out if I have done them correctly or not. Please help me if you can. Would greatly appropriate it. Thanks Three workers have a productivity ratio of 1 to 2 to 3. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? A. 12 to 1 B. 6 to 1 C. 5 to 1 D. 1 to 6 E. 1 to 5 The fastest worker who does 3 units of job worked for all 10 hours, so he did 310=30 units of job; The slowest worker who does 1 unit of job worked for only 4+2=6 hours (first 4 hours and last 2 hours), so he did 16=6 units of job; The ratio thus is 30 to 6, or 5 to 1. Hi Bunuel you say that the slowest worker worked for 4+2=6 hours( first 4 hours and last 2 hours) should it not be 5+1=6 since the question says that the slowest worker takes a break at the beginning of the fifth hour which means he has worked for 5 hours and also he joins back at the beginning of the 9th hour and the work completes in 10 hours which means after joining he worked only for an hour? please correct me if i am wrong No. At the beginning of the 5th hour, the slowest worker takes a break, means that this worker worked only for the first 4 hours (he left when 5th hour started). The slowest worker comes back to work at the beginning of the 9th hour and begins working again, means that this worker worked for 9th hour. _________________ Director Joined: 23 Jan 2013 Posts: 579 Schools: Cambridge'16 Followers: 1 Kudos [?]: 42 [0], given: 40 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 02 Nov 2015, 05:55 RT = W Slow - 1/3w/h Fast - 1w/h 1/34+1/32=4/3+2/3=6/3=2 14+14+12=10 10:2=5:1 C GMAT Club Legend Joined: 09 Sep 2013 Posts: 13441 Followers: 575 Kudos [?]: 163 [0], given: 0 Re: Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 24 Dec 2016, 02:09 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Senior Manager Joined: 07 Dec 2014 Posts: 491 Followers: 3 Kudos [?]: 81 [0], given: 2 Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] ### Show Tags 24 Dec 2016, 16:29 Three workers have a productivity ratio of 1 to 2 to 3. All three workers are working on a job for 4 hours. At the beginning of the 5th hour, the slowest worker takes a break. The slowest worker comes back to work at the beginning of the 9th hour and begins working again. The job is done in ten hours. What was the ratio of the work performed by the fastest worker as compared to the slowest? A. 12 to 1 B. 6 to 1 C. 5 to 1 D. 1 to 6 E. 1 to 5 if all 3 work all 10 hours, then work ratio of fastest to slowest is 103/101=3/1 but slowest worked only 6 hours, so ratio is 103/61=5/1 C Three workers have a productivity ratio of 1 to 2 to 3. All [#permalink] 24 Dec 2016, 16:29 Similar topics Replies Last post Similar Topics: 2 A strip of ribbon is cut into three smaller strips in the ratio 2:3:5. 3 10 Jan 2016, 07:31 18 The product (1 - {1/2}^2)(1 - {1/3}^2)...(1 - {1/9}^2)(1 - { 5 04 Oct 2013, 01:09 5 The ratio 2 to 1/3 is equal to the ratio 9 20 Aug 2012, 02:03 5 A plant has workers and supervisors in a ratio of 30 to 1. 6 01 Jun 2012, 10:16 29 3 machines have a productivity ratio of 1 to 2 to 5. All 3 16 16 Feb 2011, 03:34 Display posts from previous: Sort by	4,579	14,471	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.984375	4	CC-MAIN-2017-04	latest	en	0.919108
https://quant.stackexchange.com/questions/37948/yield-to-maturity-as-discount-rate	1,558,899,073,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232259452.84/warc/CC-MAIN-20190526185417-20190526211417-00015.warc.gz	572,991,624	31,929	# Yield to maturity as discount rate Assume that face value of a bond is equal to 1000. The coupon rate is 3% and yield to maturity is 4%.How can we correlate coupon rate and YTM in order to explain the state of current bond price. (Maturity 10years-Redemption value is 1000). My approach: $P_0=r C a(n,YTM) + \frac{P}{(1+YTM) ^{n}}= 0.03 * 1000 a(10,0.04) + \frac{1000}{(1+0.04) ^{10}}= 918.88$ \$. The yield to maturity or equivalent the return that investors expect from the bond is 0.04. The bond price is less than the face value beacuse the coupon rate is 0.03. Investors are not interested in buying the bond at 1000 since they can earn 0.04 from a bond with the same characteristics and price. So the Market squizzes the price to 918.88 Can we generalize that YTM is the discount rate for bond with same duration and characteristics(coupon rate and risk)? • What you describe is a bond with 10 year "maturity". You use the term "duration" but this term has a different meaning in Finance, it is not the same as maturity at all. As a result your question is very confusing to me... Of course if two bonds have exactly the same maturity coupon and risk etc. they are essentially the same and will have the same YTM (discount rate) by the "law of 1 price". But this is just obvious. – Alex C Jan 27 '18 at 23:55 • Agree with Alex C above. Furthermore, the term discount rate has a more specific meaning than how you are using it. In YTM is more like the IRR. In some instances the discount rate and the YTM will be identical but only in the event that the yield curve is flat (at 4% in your case). Also, since you are assuming that the coupon is annual and that you are pricing it at inception (or right after a coupon payment with 10Yrs to maturity). – AlRacoon Jan 28 '18 at 20:52	484	1,792	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.375	3	CC-MAIN-2019-22	longest	en	0.940786
https://dungeonworld.gplusarchive.online/2015/12/11/the-1-in-dungeon-world-is-deceptively-powerful/	1,718,471,370,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00182.warc.gz	190,141,171	11,299	# The + 1 in Dungeon World is deceptively powerful! The + 1 in Dungeon World is deceptively powerful! The + 1 in Dungeon World is deceptively powerful! It is better than a + 2 in D&D and sometimes close to + 3. Just think about that when you design moves and such. On 2d6, to get at least 7 = 58.33%, to get at least 10 = 16.67%. When you roll 2d6 + 1 your chances go up 13.89% to get a 7+ (to 72.22%) and 10.11% to get a 10+ (to 27.78%). In D&D, obviously, each +1 increases your chance of hitting any number by 5%, since the curve is flat. So +2 is +10% and +3 is +15%. ## 13 thoughts on “The + 1 in Dungeon World is deceptively powerful!” 1. Really good point. I feel that, especially in Dungeon World, lots of first draft custom moves grant a numerical bonus without taking into account how powerful it is. 2. Made this math some time ago. The important thing is that a -1 is powerful exactly as a +3 3. This is why magic items and custom moves are best created with fictional results as opposed to mechanical results. 4. thats what i calcualted: -1= < 6 = 58% 7-9 = 33,2% 10+ = 8,8% +0= < 6 = 41,4% 7-9 = 41,5% 10+ = 17,1% +1= < 6 = 27,6% 7-9 = 44,2% 10+ = 28,2% +2= < 6 = 16,5% 7-9 = 41,5% 10+ = 42% +3= < 6 = 8,2% 7-9 = 33,2% 10+ = 58,6% 5. And yet… any given +1 bonus only affects the outcome of a roll around 25% of the time. That is, the +1 forward/ongoing only changes the outcome of the roll if your roll would have been a 6 or a 9. If your “base” modifier is -1, +0, +1, or +2 then you roll a 6 or a 9 only 25% of the time. (If the mod is -2 or +3, the chance of a 6 or 9 is 19.45%). What that means is, a +1 bonus doesn’t feel like it makes a different very often. It can make you feel more confident before you roll. But after you roll, it’s pretty easy to see whether it did or didn’t affect the outcome. And usually, it did not. As a result, I try to avoid writing (or taking) moves that require you to pay a cost (or take a risk) to get a +1 forward. It’s not a good gamble. (Now, paying a cost or taking a risk to get +1 to roll after you’ve made it… that’s solid.) 6. This is the math based on the 36 different results you can have on 2d6: Failure – Consequences – Success Base 15/36 15/36 6/36 -1 21/36 12/36 3/36 +1 10/36 16/36 10/36 +3 3/36 12/36 21/36 7. Right. At +3 you are basically telling “competence porn” stories with a 92% chance of at least a 7 (see Vincent Shine’s calculations above) . That’s okay, so long as it is within focused bands of specialization. Your job as a DM is to diversify threats in a way that makes that competence not always relevant (but no so diverse that it’s never or seldom relevant.) Probably the most likely problem is when you get someone achieving regular +3’s in Hack and Slash and wearing 3 armor. That figure is almost literally a tank. But you still aren’t powerless as a DM to make the game challenging for that person. For instance, give them a monster that has acid blood that eats through steel, ruining weapons and armor. I would force the attacker to Defy Danger Dex to avoid the blood acid splashes (-1 or -2 if the weapon has the messy tag) after each successful hit for damage with a piercing/cutting type weapon. And even then their weapon might be pitted and smoking by the end of the fight. Soul Stigma made the primary point I was trying to communicate. In general I find “plus ones” a boring effect anyway. (And they can make the game boring if you aren’t careful.) I would rather see magical items with narrative effects. But weapons that do specific types of energy damage or get bonuses in specific situations can be really interesting with the right story behind them. 8. Ray Otus and specific types of damage, to me, are less mechanical and more fictional. Also items that are ordinary in most circumstances but magical in special circumstances can work well (such as a very ordinary dagger that has no special properties until used in a specific ritual). DW really thrives on abstract mechanics and precise narrative. 9. Man, that’s worth repeating: “DW really thrives on abstract mechanics and precise narrative.” 10. I’d say “on going” and “forward” can make the difference on how OP a move can be.	1,188	4,234	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.296875	3	CC-MAIN-2024-26	latest	en	0.933731
https://www.rdocumentation.org/packages/permutations/versions/1.0-5/topics/valid	1,607,208,045,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141750841.83/warc/CC-MAIN-20201205211729-20201206001729-00223.warc.gz	830,762,713	5,568	# valid 0th Percentile ##### Functions to validate permutations Functions to validate permutation objects: if valid, return TRUE and if not valid, generate a warning() and return FALSE. Function singleword.valid() takes an integer vector, interpreted as a word, and checks that it is a permutation of seq_len(max(x)). Function cycle.valid() takes a cyclist and checks for disjoint cycles of strictly positive integers with no repeats. Keywords symbmath ##### Usage singleword_valid(w) cyclist_valid(x) ##### Arguments x In function cycle_valid(), a cyclist w In function singleword_valid(), an integer vector ##### Value Returns either TRUE, or stops with an informative error message cyclist ##### Aliases • valid • validity • singleword_valid • cyclist_valid ##### Examples # NOT RUN { singleword_valid(sample(1:9)) # TRUE singleword_valid(c(3L,4L,2L,1L)) # TRUE singleword_valid(c(3,4,2,1)) # FALSE (not integer) singleword_valid(c(3L,3L,2L,1L)) # FALSE (3 repeated) cyclist_valid(list(c(1,8,2),c(3,6))) # TRUE cyclist_valid(list(c(1,8,2),c(3,6))) # FALSE ('8' is repeated) cyclist_valid(list(c(1,8,1),c(3,6))) # FALSE ('1' is repeated) cyclist_valid(list(c(0,8,2),c(3,6))) # FALSE (zero element) # } Documentation reproduced from package permutations, version 1.0-5, License: GPL-2 ### Community examples Looks like there are no examples yet.	386	1,380	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2020-50	latest	en	0.478359
http://www.talkstats.com/threads/how-to-deal-with-non-proportional-hazard-cox-models.74760/#post-218622	1,653,337,188,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00238.warc.gz	113,591,824	13,327	# How to deal with non-proportional hazard Cox models? #### Rets ##### New Member Dear all, I want to run a Cox model to see if there is an interaction between two treatments. However, the test to see if proportional hazard is respected suggests that it is actually not respected. So I don't know what other statistical survival analysis to do because it means I cannot rely on the cox model conclusions? Or that the conclusions may just be less powerful that if proportional hazard would have been respected? What alternative tests I could do in this case? I was thinking about "survreg" models, but I am not sure what distribution to use and with only 4 dates I think it does not make sense to have a timescale distribution with so few dates? In R, I can only run survreg with Gaussian or logistic distributions, but I think none of these should be considered in my case. Thank you very much if you can help me or guide me on this. Best wishes and please see below further explanations: Experiment overview: There is four conditions (a= no treatment, b= antibiotics, c=antihypertensive and d=antibiotics+antihypertensive). I had 150 animals (tiny) per condition, so 600 in total. I followed the survival/death at days 0 (all alive), day 2, day 4 and day 6 of the experiment. Here is a bit of the Rcode: fit1 <- coxph(Surv(a$time, a$survival) ~ antibiotics + antihypertensive + antibiotics:antihypertensive, data = a) summary(fit1) test.ph <- cox.zph(fit1) test.ph > test.ph rho chisq p antihypertensiveyes -0.0806 4.34 0.037260 antibioticsyes 0.0794 4.18 0.040944 antihypertensiveyes:antibioticsyes -0.0467 1.45 0.228140 GLOBAL NA 18.36 0.000371 --> no proportional hazard Thanks a lot, Best wishes Rets #### Rets ##### New Member Any advice or suggestion? I would be so happy to hear from you #### hlsmith ##### Less is more. Stay pure. Stay poor. Was group assignment randomized and where is the "interaction", or did you not mean to use that word? #### Rets ##### New Member Yes, group assignment was randomized. The interaction is "antibiotics:antihypertensive" #### hlsmith ##### Less is more. Stay pure. Stay poor. Yeah i see that now but is it really an interaction or just three treatment groups? What happens when you run the model with a categorical variable? #### Rets ##### New Member It is an interaction because one group is treatment A, the other is treatment B and the other is treatment A+B (plus controls for which none treatment was done). The two variables "antihypertensive" and "antibiotics" are categorical variables with two modalities "presence/absence" (ie, yes/no). I have tried to merge both variables into a unique variable with 4 modalities ("yesyes", "nono", "yesno", "noyes") but the results are similar. I have attached the dataset if you want to give a glance. >test.ph rho chisq p one.varnoyes -0.0806 4.34 0.037260 one.varyesno 0.0794 4.18 0.040944 one.varyesyes -0.0656 2.86 0.090832 GLOBAL NA 18.36 0.000371 dataset #### Attachments • 76.8 KB Views: 1 #### Rets ##### New Member dataset + Rcode: > library(survival) > library(ggplot2) > library(ggpubr) > library(survminer) > > > survobj <- Surv(time = a$time, event = a$survival, type='right') # Important about the "survival" variable: 2= dead, 1=alive in the notation, which is different from the code above for which dead=0, alive=1 !! > # Plot survival distribution of the total sample > # Kaplan-Meier estimator > fit0 <- survfit(survobj~1, data=a) > fit1 <- survfit(survobj~one.var, data=a) > > res.cox <- coxph(Surv(time, survival) ~ one.var, data = a) > res.cox > > test.ph <- cox.zph(res.cox) > test.ph > res.cox2 <- coxph(Surv(time, survival) ~ antihypertensive + antibiotics + antibiotics:antihypertensive, data = a) > res.cox2 > > test.ph2 <- cox.zph(res.cox2) > test.ph2 #### hlsmith ##### Less is more. Stay pure. Stay poor. I will be transparent and state that I don't work that much with survival analyses, about a project every other year. So I am helping but don't think my rebuttals are the definitive truth. Yes, now that I have seen your data it is an interaction. I am comfortable in SAS for survival, so I played around with your data in that program. I got the following for results: I wonder if your PH assumption issue is coming from that you have discrete time points for events. I am less savvy with this type of survival analyses. But how proportional can things be if the time values are restricted to minimal values. I would be curious on how discrete survival folks address this. #### hlsmith ##### Less is more. Stay pure. Stay poor. I am running another survival study today and this is what my PH checks look like. the bottom one fails the test, by you can see what the above graphs would look like if they had an extended follow-up: #### Rets ##### New Member Hi hlsmith, Thank you for this !! So the statistical results you obtained in the table suggest that no significant interaction occured and that only the antibiotics had an effect on survival ? Not sure how to interpret the graphs for proportional hazard checks... but maybe with so few follow-ups, it does not make real sense? Or maybe the cox analysis is just not appropriate in this case, I'm a bit confused I must say because I don't see alternative for testing interaction. Maybe I should just run pairwise comparisons using the logrank test...	1,407	5,369	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.734375	3	CC-MAIN-2022-21	latest	en	0.920612
https://classroom.thenational.academy/lessons/representing-information-as-a-bar-model-part-1-6cw6ar?step=3&activity=exit_quiz	1,701,695,592,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00298.warc.gz	207,297,322	25,109	# Representing information as a bar model (Part 1) In this lesson, we will read word problems and represent them as bar models. # Video Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should: • Click "Close Video" • Click "Next" to view the activity Your video will re-appear on the next page, and will stay paused in the right place. # Worksheet These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below. #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # Representing information as a bar model You’ve done so much learning. Try your best and complete the quiz. Draw bar-models to answer these questions. Q1.Jack picked 8 apples from under one tree and 2 apples from under another tree. How many apples are there altogether? 1/5 Q2.Jack brought 15 apples in a basket into his kitchen. He placed 4 of the apples on a chopping board. How many were left in the basket? 2/5 Q3.Joe picked lots of apples. 4 of his apples were bruised and 5 were not bruised. How many apples does he have altogether? 3/5 Q4.When Muhammad got back into his kitchen with 17 apples, he realised 5 of them were bruised. How many were not bruised? 4/5 Q5.Ava placed 10 of her apples on the chopping board and Zain placed 7. How many apples are on the chopping board altogether? 5/5 #### Unit quizzes are being retired in August 2023 Why we're removing unit quizzes from the website > Quiz: # Representing information as a bar model You’ve done so much learning. Try your best and complete the quiz. Draw bar-models to answer these questions. Q1.Jack picked 8 apples from under one tree and 2 apples from under another tree. How many apples are there altogether? 1/5 Q2.Jack brought 15 apples in a basket into his kitchen. He placed 4 of the apples on a chopping board. How many were left in the basket? 2/5 Q3.Joe picked lots of apples. 4 of his apples were bruised and 5 were not bruised. How many apples does he have altogether? 3/5 Q4.When Muhammad got back into his kitchen with 17 apples, he realised 5 of them were bruised. How many were not bruised? 4/5 Q5.Ava placed 10 of her apples on the chopping board and Zain placed 7. How many apples are on the chopping board altogether? 5/5 # Lesson summary: Representing information as a bar model (Part 1) ## Time to move! Did you know that exercise helps your concentration and ability to learn? For 5 mins... Move around: Climb stairs On the spot: Chair yoga	684	2,773	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2023-50	latest	en	0.961555
https://www.nag.com/numeric/cl/nagdoc_cl26.0/html/f08/f08atc.html	1,627,320,197,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00092.warc.gz	961,130,519	5,459	f08 Chapter Contents f08 Chapter Introduction NAG Library Manual # NAG Library Function Documentnag_zungqr (f08atc) ## 1 Purpose nag_zungqr (f08atc) generates all or part of the complex unitary matrix $Q$ from a $QR$ factorization computed by nag_zgeqrf (f08asc), nag_zgeqpf (f08bsc) or nag_zgeqp3 (f08btc). ## 2 Specification #include #include void nag_zungqr (Nag_OrderType order, Integer m, Integer n, Integer k, Complex a[], Integer pda, const Complex tau[], NagError fail) ## 3 Description nag_zungqr (f08atc) is intended to be used after a call to nag_zgeqrf (f08asc), nag_zgeqpf (f08bsc) or nag_zgeqp3 (f08btc), which perform a $QR$ factorization of a complex matrix $A$. The unitary matrix $Q$ is represented as a product of elementary reflectors. This function may be used to generate $Q$ explicitly as a square matrix, or to form only its leading columns. Usually $Q$ is determined from the $QR$ factorization of an $m$ by $p$ matrix $A$ with $m\ge p$. The whole of $Q$ may be computed by: ```nag_zungqr(order,m,m,p,a,pda,tau,&fail) ``` (note that the array a must have at least $m$ columns) or its leading $p$ columns by: ```nag_zungqr(order,m,p,p,a,pda,tau,&fail) ``` The columns of $Q$ returned by the last call form an orthonormal basis for the space spanned by the columns of $A$; thus nag_zgeqrf (f08asc) followed by nag_zungqr (f08atc) can be used to orthogonalize the columns of $A$. The information returned by the $QR$ factorization functions also yields the $QR$ factorization of the leading $k$ columns of $A$, where $k. The unitary matrix arising from this factorization can be computed by: ```nag_zungqr(order,m,m,k,a,pda,tau,&fail) ``` or its leading $k$ columns by: ```nag_zungqr(order,m,k,k,a,pda,tau,&fail) ``` ## 4 References Golub G H and Van Loan C F (1996) Matrix Computations (3rd Edition) Johns Hopkins University Press, Baltimore ## 5 Arguments 1: $\mathbf{order}$Nag_OrderTypeInput On entry: the order argument specifies the two-dimensional storage scheme being used, i.e., row-major ordering or column-major ordering. C language defined storage is specified by ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. See Section 2.3.1.3 in How to Use the NAG Library and its Documentation for a more detailed explanation of the use of this argument. Constraint: ${\mathbf{order}}=\mathrm{Nag_RowMajor}$ or $\mathrm{Nag_ColMajor}$. 2: $\mathbf{m}$IntegerInput On entry: $m$, the order of the unitary matrix $Q$. Constraint: ${\mathbf{m}}\ge 0$. 3: $\mathbf{n}$IntegerInput On entry: $n$, the number of columns of the matrix $Q$. Constraint: ${\mathbf{m}}\ge {\mathbf{n}}\ge 0$. 4: $\mathbf{k}$IntegerInput On entry: $k$, the number of elementary reflectors whose product defines the matrix $Q$. Constraint: ${\mathbf{n}}\ge {\mathbf{k}}\ge 0$. 5: $\mathbf{a}\left[\mathit{dim}\right]$ComplexInput/Output Note: the dimension, dim, of the array a must be at least • $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{pda}}×{\mathbf{n}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_ColMajor}$; • $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}×{\mathbf{pda}}\right)$ when ${\mathbf{order}}=\mathrm{Nag_RowMajor}$. On entry: details of the vectors which define the elementary reflectors, as returned by nag_zgeqrf (f08asc), nag_zgeqpf (f08bsc) or nag_zgeqp3 (f08btc). On exit: the $m$ by $n$ matrix $Q$. If ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, the $\left(i,j\right)$th element of the matrix is stored in ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$. If ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, the $\left(i,j\right)$th element of the matrix is stored in ${\mathbf{a}}\left[\left(i-1\right)×{\mathbf{pda}}+j-1\right]$. 6: $\mathbf{pda}$IntegerInput On entry: the stride separating row or column elements (depending on the value of order) in the array a. Constraints: • if ${\mathbf{order}}=\mathrm{Nag_ColMajor}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$; • if ${\mathbf{order}}=\mathrm{Nag_RowMajor}$, ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$. 7: $\mathbf{tau}\left[\mathit{dim}\right]$const ComplexInput Note: the dimension, dim, of the array tau must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{k}}\right)$. On entry: further details of the elementary reflectors, as returned by nag_zgeqrf (f08asc), nag_zgeqpf (f08bsc) or nag_zgeqp3 (f08btc). 8: $\mathbf{fail}$NagError Input/Output The NAG error argument (see Section 2.7 in How to Use the NAG Library and its Documentation). ## 6 Error Indicators and Warnings NE_ALLOC_FAIL Dynamic memory allocation failed. See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information. On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value. NE_INT On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{m}}\ge 0$. On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{pda}}>0$. NE_INT_2 On entry, ${\mathbf{m}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{m}}\ge {\mathbf{n}}\ge 0$. On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$ and ${\mathbf{k}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{n}}\ge {\mathbf{k}}\ge 0$. On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{m}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$. On entry, ${\mathbf{pda}}=〈\mathit{\text{value}}〉$ and ${\mathbf{n}}=〈\mathit{\text{value}}〉$. Constraint: ${\mathbf{pda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$. NE_INTERNAL_ERROR An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance. An unexpected error has been triggered by this function. Please contact NAG. See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information. NE_NO_LICENCE Your licence key may have expired or may not have been installed correctly. See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information. ## 7 Accuracy The computed matrix $Q$ differs from an exactly unitary matrix by a matrix $E$ such that $E2 = Oε ,$ where $\epsilon$ is the machine precision. ## 8 Parallelism and Performance nag_zungqr (f08atc) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library. nag_zungqr (f08atc) makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information. Please consult the x06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information. ## 9 Further Comments The total number of real floating-point operations is approximately $16mnk-8\left(m+n\right){k}^{2}+\frac{16}{3}{k}^{3}$; when $n=k$, the number is approximately $\frac{8}{3}{n}^{2}\left(3m-n\right)$. The real analogue of this function is nag_dorgqr (f08afc). ## 10 Example This example forms the leading $4$ columns of the unitary matrix $Q$ from the $QR$ factorization of the matrix $A$, where $A = 0.96-0.81i -0.03+0.96i -0.91+2.06i -0.05+0.41i -0.98+1.98i -1.20+0.19i -0.66+0.42i -0.81+0.56i 0.62-0.46i 1.01+0.02i 0.63-0.17i -1.11+0.60i -0.37+0.38i 0.19-0.54i -0.98-0.36i 0.22-0.20i 0.83+0.51i 0.20+0.01i -0.17-0.46i 1.47+1.59i 1.08-0.28i 0.20-0.12i -0.07+1.23i 0.26+0.26i .$ The columns of $Q$ form an orthonormal basis for the space spanned by the columns of $A$. ### 10.1 Program Text Program Text (f08atce.c) ### 10.2 Program Data Program Data (f08atce.d) ### 10.3 Program Results Program Results (f08atce.r)	2,802	7,995	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 93, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.671875	3	CC-MAIN-2021-31	latest	en	0.5767
https://flightwinebar.com/blog/how-many-gallons-is-20-liters/	1,716,162,523,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00803.warc.gz	229,650,734	44,314	# How Many Gallons Is 20 Liters? The question of how many gallons is 20 liters is a common one, especially for those who are new to the metric system. Liters and gallons are two different units of measurement used to measure volume. A liter is a metric unit of volume, while a gallon is an imperial unit of volume. In this article, we will discuss how to convert 20 liters to gallons and provide some helpful tips for making the conversion. ## How to Convert 20 Liters to Gallons: A Step-by-Step Guide? Before you begin, make sure you have the following items: a calculator, a pencil, and a piece of paper. Step 2: Understand the Conversion To convert liters to gallons, you need to know that 1 liter is equal to 0.264172 gallons. Step 3: Calculate the Conversion Now that you know the conversion rate, you can calculate the conversion. To do this, multiply 20 liters by 0.264172. This will give you the answer in gallons. Suggested Post: How To Get A Cork Out Of A Wine Bottle? Step 4: Write Down the Answer Once you have the answer, write it down on your piece of paper. In this case, 20 liters is equal to 5.2834 gallons. To make sure your answer is correct, double-check it by converting the gallons back to liters. To do this, multiply 5.2834 gallons by 3.78541. This should give you the original number of liters (20). And there you have it! You have successfully converted 20 liters to gallons. Congratulations! ## The Benefits of Knowing How Many Gallons Are in 20 Liters Knowing how many gallons are in 20 liters can be incredibly useful in a variety of situations. Whether you’re a student studying for a science test, a homeowner trying to figure out how much paint to buy, or a chef trying to convert a recipe from metric to imperial measurements, understanding the conversion between liters and gallons can be a real lifesaver. For starters, it’s important to know that one liter is equal to 0.264172052 gallons. This means that 20 liters is equal to 5.28344104 gallons. Knowing this conversion can help you quickly and accurately convert between the two units of measurement. For students, this knowledge can be especially helpful when studying for science tests. Many science classes require students to understand the metric system, and being able to quickly convert between liters and gallons can be a real time-saver. For homeowners, this knowledge can be invaluable when it comes to buying paint or other liquids. Knowing how many gallons are in 20 liters can help you make sure you buy the right amount of paint for your project. Suggested Post: How Much Do Scotsman Ice Makers Cost? Finally, for chefs, this knowledge can be a real game-changer. Many recipes are written in metric measurements, and being able to quickly convert them to imperial measurements can make cooking a breeze. In short, knowing how many gallons are in 20 liters can be incredibly useful in a variety of situations. Whether you’re a student, a homeowner, or a chef, understanding the conversion between liters and gallons can make your life a lot easier. ## How to Use a Conversion Chart to Determine How Many Gallons Are in 20 Liters? Converting liters to gallons is easy with a conversion chart! To determine how many gallons are in 20 liters, simply locate the number 20 on the chart and look across to the column labeled “Gallons.” You’ll find that 20 liters is equal to 5.28 gallons. It’s that simple! So, if you ever need to convert liters to gallons, just grab a conversion chart and you’ll have your answer in no time. ## The History of the Gallon and How It Relates to 20 Liters The gallon is a unit of measurement that has been around for centuries. It is a unit of volume that is used to measure liquids, and it is equal to four quarts, eight pints, or sixteen cups. The gallon is an important unit of measurement in the United States, and it is also used in many other countries around the world. The origin of the gallon dates back to the Middle Ages. During this time, the gallon was used to measure wine and other alcoholic beverages. The gallon was originally defined as the volume of eight pounds of wheat. This definition was later changed to the volume of ten pounds of wheat. Suggested Post: How Much Ounces In A Gallon? In the United States, the gallon is defined as 231 cubic inches, which is equal to 3.785 liters. This means that 20 liters is equal to 5.283 gallons. This is an important conversion to know when measuring liquids, as it can help you determine how much of a liquid you need for a recipe or other project. The gallon is an important unit of measurement that is used in many different countries around the world. It is a unit of volume that is used to measure liquids, and it is equal to four quarts, eight pints, or sixteen cups. Knowing how to convert liters to gallons can be very helpful when measuring liquids, and it is important to remember that 20 liters is equal to 5.283 gallons. ## The Difference Between Imperial and US Gallons and How It Affects 20 Liters Have you ever wondered what the difference is between Imperial and US gallons? It’s an important distinction to make, especially when it comes to measuring liquids like water or fuel. In this article, we’ll explain the difference between Imperial and US gallons and how it affects 20 liters. The Imperial gallon is a unit of measurement used in the United Kingdom and other Commonwealth countries. It is equal to 4.54609 liters, or about 1.2 US gallons. The US gallon, on the other hand, is equal to 3.785411784 liters, or about 0.83 Imperial gallons. So, how does this affect 20 liters? Well, 20 liters is equal to 5.28344 Imperial gallons, or 5.28 US gallons. This means that if you’re measuring 20 liters in the US, you’ll need to use 5.28 US gallons. If you’re measuring 20 liters in the UK, you’ll need to use 5.28344 Imperial gallons. Suggested Post: How Many Water Bottles In A Gallon? It’s important to remember the difference between Imperial and US gallons when measuring liquids. Knowing the difference can help you get the most accurate measurements possible. So, the next time you’re measuring liquids, make sure you’re using the right type of gallon! ## How to Calculate How Many Gallons Are in 20 Liters Without a Calculator? Calculating how many gallons are in 20 liters without a calculator is easy! All you need to do is remember a few simple conversions. One gallon is equal to 3.785 liters, so to find out how many gallons are in 20 liters, you just need to divide 20 by 3.785. That means that 20 liters is equal to 5.283 gallons! So there you have it – 20 liters is equal to 5.283 gallons! ## The Impact of Temperature on the Conversion of 20 Liters to Gallons Have you ever wondered how temperature affects the conversion of 20 liters to gallons? Well, you’re in luck! We’re here to explain the impact of temperature on this conversion. First, it’s important to understand that the conversion of liters to gallons is based on the metric system. This means that the conversion rate is based on the temperature of the liquid being measured. At a temperature of 15°C (59°F), 1 liter of liquid is equal to 0.264172052 gallons. However, when the temperature of the liquid changes, so does the conversion rate. For example, at a temperature of 20°C (68°F), 1 liter of liquid is equal to 0.264172052 gallons. This means that if you were to convert 20 liters of liquid at a temperature of 20°C (68°F), you would get 5.28344052 gallons. Suggested Post: How To Open A Wine Bottle Without A Bottle Opener? It’s important to note that the conversion rate of liters to gallons can vary depending on the temperature of the liquid. This means that if you were to convert 20 liters of liquid at a temperature of 25°C (77°F), you would get 5.28344052 gallons. So, as you can see, the temperature of the liquid can have a significant impact on the conversion of liters to gallons. It’s important to keep this in mind when you’re converting between these two units of measurement. With this knowledge, you’ll be able to make sure that you’re getting the most accurate conversion possible! ## How to Use a Volume Conversion Calculator to Determine How Many Gallons Are in 20 Liters? Are you looking to find out how many gallons are in 20 liters? Look no further! With a volume conversion calculator, you can easily determine the answer. Here’s how: First, find a volume conversion calculator online. There are many available, so you can choose the one that works best for you. Once you have the calculator open, enter the number of liters you want to convert. In this case, it’s 20 liters. Next, select the unit of measurement you want to convert to. In this case, it’s gallons. Finally, click the “Calculate” button and the calculator will give you the answer. In this case, it’s 5.28 gallons. And there you have it! With a volume conversion calculator, you can easily determine how many gallons are in 20 liters. Now you can move on to your next project with confidence! ## How many gallons is 20 liters? 20 liters is equal to 5.28 gallons. Suggested Post: How Many Liters Is 10 Gallons? ## Conclusion In conclusion, 20 liters is equal to 5.28 gallons. This is a useful conversion to know when dealing with different units of measurement.	2,111	9,306	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2024-22	latest	en	0.903968
http://mail-archives.apache.org/mod_mbox/mahout-user/200906.mbox/%3C37ffc8080906290604t59f27755p75e455ba9960b242@mail.gmail.com%3E	1,498,200,768,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320003.94/warc/CC-MAIN-20170623045423-20170623065423-00257.warc.gz	236,467,947	2,975	# mahout-user mailing list archives ##### Site index · List index Message view Top From nfantone <nfant...@gmail.com> Subject Re: kMeans Help Date Mon, 29 Jun 2009 13:04:35 GMT ```I really see no harm (algorithmically and conceptually) in returning the center as the centroid if there's only one point added to the cluster. If that's what you need to solve your predicament, I say go for it. Are there any drawbacks? What eludes me is the actual way of adding points. How can I compute its total set at any given moment? Say, I create a Cluster with a center, then add some points - the addPoint() just stores a pointTotal Vector with the total vector sum- and want to check which vectors I have added so far with their original values. Is this even possible? On Mon, Jun 29, 2009 at 9:42 AM, Grant Ingersoll<gsingers@apache.org> wrote: > > On Jun 28, 2009, at 5:55 PM, Grant Ingersoll wrote: > >> >> On Jun 28, 2009, at 4:56 PM, Grant Ingersoll wrote: >> >>> I get all of this, my point is that when you rehydrate the Cluster, it >>> doesn't properly report the centroid per my email all because numPoints == 0 >>> and pointTotal is a a vector that is the same as the passed in center >>> vector, but initialized to 0. >>> >> >> In other words, the simple act of serializing a Cluster to HDFS and then >> reconstituting it should not alter the result one gets, which I believe is >> what happens if one dumps out the clusters that have been calculated after >> the whole process is done. > > [1] is what I had to do to work around it for the Random approach, but I > think it isn't the right approach. > > I think the problem lies in computeCentroid: > private Vector computeCentroid() { > if (numPoints == 0) > return pointTotal; > else if (centroid == null) { > // lazy compute new centroid > centroid = pointTotal.divide(numPoints); > Vector stds = pointSquaredTotal.times(numPoints).minus( > pointTotal.times(pointTotal)).assign(new SquareRootFunction()) > .divide(numPoints); > std = stds.zSum() / 2; > } > return centroid; > } > > I don't understand why, if numPoints ==0, the next line isn't just: return > center; Why wouldn't the center and the centroid be the same if there are > no points? pointTotal in the rehydration case (or in the case of just > calling new Cluster(center) is just a vector of the same cardinality as > Center but all values are zero. > > > > [1]: > Author: gsingers > Date: Sat Jun 27 02:57:18 2009 > New Revision: 788919 > > URL: http://svn.apache.org/viewvc?rev=788919&view=rev > Log: > add the center as a point > > Modified: > > lucene/mahout/trunk/core/src/main/java/org/apache/mahout/clustering/kmeans/RandomSeedGenerator.java > > Modified: > lucene/mahout/trunk/core/src/main/java/org/apache/mahout/clustering/kmeans/RandomSeedGenerator.java > URL: > http://svn.apache.org/viewvc/lucene/mahout/trunk/core/src/main/java/org/apache/mahout/clustering/kmeans/RandomSeedGenerator.java?rev=788919&r1=788918&r2=788919&view=diff > ============================================================================== > --- > lucene/mahout/trunk/core/src/main/java/org/apache/mahout/clustering/kmeans/RandomSeedGenerator.java > (original) > +++ > lucene/mahout/trunk/core/src/main/java/org/apache/mahout/clustering/kmeans/RandomSeedGenerator.java > Sat Jun 27 02:57:18 2009 > @@ -54,7 +54,9 @@ > if (log.isInfoEnabled()) { > log.info("Selected: " + value.asFormatString()); > } > - writer.append(new Text(key.toString()), new Cluster(value)); > + Cluster val = new Cluster(value);	1,000	3,604	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2017-26	latest	en	0.880886
https://www.mrexcel.com/board/threads/looking-to-combine-import-using-indexmatch-with-sumif.1164211/	1,620,707,966,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243991641.5/warc/CC-MAIN-20210511025739-20210511055739-00489.warc.gz	948,138,518	35,047	# Looking to combine import using indexmatch with sumif ##### New Member Hi, I am building a cooking spreadsheet which calculates the nutritional content of various recipes. My spreadsheet has an ‘Ingredients’ tab and multiple tabs for each recipe. In each recipe tab, I have a list of the relevant ingredients, whose nutritional info I am importing from the ‘Ingredients’ tab using index match and multiplying by the number of servings (Intermediate Step). I then sum up the nutritional data across the ingredients using a sumif. (Final Step) I am looking for a formula which would combine the intermediate step and the final step. The formula would import the nutritional data for the ingredients, multiply the nutritional data by the number of servings, and then sum up the data, separately for each nutrient. As an example, I would input the ingredients, the number of servings for each ingredient, and the formula would output the total number of Calories in the recipe. I would then repeat this formula for every nutrient. Is this possible? I attached a simplified version of the spreadsheet. BCDEFGHI 4Number of Meals3 5 6Step 1: Ingredient Entry 7Ingredients:Black BeansOlive OilOnionWhite MushroomQuinoaChicken BrothChicken Breast 8Unit (g)100100100100100240100 9Servings4.20.0543.51.351.56.8 10Total (g)4205400350135360680 11 12Step 2: Importing Nutritional Info from Ingredients Tab 13Calories382.244.2160.077.0496.87.5680.0 14Fat (g)1.35.00.41.18.20.017.7 15Saturated Fat (g)0.40.70.20.40.90.04.1 16Monounsatured Fat (g)0.03.70.10.00.00.04.7 17Polyunsaturated Fat (g)0.00.50.10.00.00.02.9 18Cholesterol (mg)0.00.00.00.00.00.0496.4 19Total Carbohydrate (g)71.40.037.211.686.40.00.0 20 21Step 3: Sum of the Ingredients 22Unit 23Calories615.9 24Fat (g)g11.2 25Saturated Fat (g)g2.2 26Monounsatured Fat (g)g2.8 27Polyunsaturated Fat (g)g1.2 28Cholesterol (mg)mg165.5 29Total Carbohydrate (g)g68.9 Quinoa-BlackBeans Cell Formulas RangeFormula C8:I8C8=INDEX(Ingredients!\$B\$4:\$ZW\$12,MATCH('Quinoa-BlackBeans'!\$B8,Ingredients!\$B\$4:\$B\$12,0),MATCH('Quinoa-BlackBeans'!C\$7,Ingredients!\$B\$4:\$ZX\$4,0)) C10:I10C10=+C8C9 C13:H19C13=INDEX(Ingredients!\$B\$6:\$H\$12,MATCH('Quinoa-BlackBeans'!\$B13,Ingredients!\$B\$6:\$B\$12,0),MATCH('Quinoa-BlackBeans'!C\$7,Ingredients!\$B\$4:\$H\$4,0))C\$9 I13:I19I13=INDEX(Ingredients!\$B\$6:\$W\$12,MATCH('Quinoa-BlackBeans'!\$B13,Ingredients!\$B\$6:\$B\$12,0),MATCH('Quinoa-BlackBeans'!I\$7,Ingredients!\$B\$4:\$X\$4,0))I\$9 D23:D29D23=SUM(OFFSET(\$C\$12,MATCH(\$B23,\$B\$13:\$B\$19,0),0):OFFSET(\$Z\$12,MATCH(\$B23,\$B\$13:\$B\$19,0),0))/\$C\$4 BCDEFGHI 4Black BeansQuinoaWhite MushroomOlive OilOnionChicken BrothChicken Breast 5Unit (g)100100100100100240100 6Calories91.0368.022884405100 7Fat (g)0.36.10.31000.102.6 8Saturated Fat (g)0.10.70.1140.00.00.6 9Monounsatured Fat (g)000730.00.00.7 10Polyunsaturated Fat (g)00010.50.00.00.4 11Cholesterol (mg)00000073 12Total Carbohydrate (g)17.0643.309.300 Ingredients ### Excel Facts Customize Quick Access Toolbar. From All Commands, add Speak Cells or Speak Cells on Enter to QAT. Select cells. Press Speak Cells. #### Alex Blakenburg ##### Well-known Member See if this works for you: 1. Replace all sheet name references which refer to the current sheet Replace 'Quinoa-BlackBeans'! with nothing in replace box When you build a formula referring to other sheets using the mouse, is also puts the sheet name for the cell references in the current workbook. These clutter up the formula making it look more complex and cause issues when you want to replicate the sheet using copy paste. 2. Made the Ingredients table into an Excel Table This required having a heading for the first column, I have used type (you can change that later but if you want to copy paste my formulas you need to do it after replicating everything. Convert the data to a table Give it the table name tbl_ingredients (again you can change this at the very end) Ingredients Sheet with Heading for Column 1 and converted to a Table called tbl_Ingredients 20210309 Cooking Ingredients lis.xlsx ABCDEFGHIJ 1 2 3 4TypeBlack BeansQuinoaWhite MushroomOlive OilOnionChicken BrothChicken Breast 5Unit (g)100100100100100240100 6Calories9136822884405100 7Fat (g)0.36.10.31000.102.6 8Saturated Fat (g)0.10.70.1140.04200.6 9Monounsatured Fat (g)000730.01300.689 10Polyunsaturated Fat (g)00010.50.01700.424 11Cholesterol (mg)00000073 12Total Carbohydrate (g)17643.309.300 13 Ingredients Quinoa-BlackBeans after formula modifications and refers to Ingredients Table Name - So create Ingredients Table and Table name before copying 20210309 Cooking Ingredients lis.xlsx ABCDEFGHIJ 1 2 3 4Number of Meals3 5 6Step 1: Ingredient Entry 7Ingredients:Black BeansOlive OilOnionWhite MushroomQuinoaChicken BrothChicken Breast 8Unit (g)100100100100100240100 9Servings4.20.0543.51.351.56.8 10Total (g)4205400350135360680 11 12Step 2: Importing Nutritional Info from Ingredients Tab 13Calories382.244.2160.077.0496.87.5680.0 14Fat (g)1.35.00.41.18.20.017.7 15Saturated Fat (g)0.40.70.20.40.90.04.1 16Monounsatured Fat (g)0.03.70.10.00.00.04.7 17Polyunsaturated Fat (g)0.00.50.10.00.00.02.9 18Cholesterol (mg)0.00.00.00.00.00.0496.4 19Total Carbohydrate (g)71.40.037.211.686.40.00.0 20 21Step 3: Sum of the Ingredients 22TypeUnitTotal 23Calories615.9 24Fat (g)g11.2 25Saturated Fat (g)g2.2 26Monounsatured Fat (g)g2.8 27Polyunsaturated Fat (g)g1.2 28Cholesterol (mg)mg165.5 29Total Carbohydrate (g)g68.9 30 Quinoa-BlackBeans Cell Formulas RangeFormula D8:I8D8=INDEX(Ingredients!\$B\$4:\$ZW\$12,MATCH(\$B8,Ingredients!\$B\$4:\$B\$12,0),MATCH(D\$7,Ingredients!\$B\$4:\$ZX\$4,0)) C10:I10C10=+C8C9 D23:D29D23=SUM(INDEX(\$B\$13:\$Z\$19,MATCH(\$B23,\$B\$13:\$B\$19,0),0))/\$C\$4 Last edited: #### Alex Blakenburg ##### Well-known Member See if this works for you: 1. Replace all sheet name references which refer to the current sheet Replace 'Quinoa-BlackBeans'! with nothing in replace box When you build a formula referring to other sheets using the mouse, is also puts the sheet name for the cell references in the current workbook. These clutter up the formula making it look more complex and cause issues when you want to replicate the sheet using copy paste. 2. Made the Ingredients table into an Excel Table This required having a heading for the first column, I have used type (you can change that later but if you want to copy paste my formulas you need to do it after replicating everything. Convert the data to a table Give it the table name tbl_ingredients (again you can change this at the very end) Ingredients Sheet with Heading for Column 1 and converted to a Table called tbl_Ingredients 20210309 Cooking Ingredients lis.xlsx ABCDEFGHIJ 1 2 3 4TypeBlack BeansQuinoaWhite MushroomOlive OilOnionChicken BrothChicken Breast 5Unit (g)100100100100100240100 6Calories9136822884405100 7Fat (g)0.36.10.31000.102.6 8Saturated Fat (g)0.10.70.1140.04200.6 9Monounsatured Fat (g)000730.01300.689 10Polyunsaturated Fat (g)00010.50.01700.424 11Cholesterol (mg)00000073 12Total Carbohydrate (g)17643.309.300 13 Ingredients Quinoa-BlackBeans after formula modifications and refers to Ingredients Table Name - So create Ingredients Table and Table name before copying 20210309 Cooking Ingredients lis.xlsx ABCDEFGHIJ 1 2 3 4Number of Meals3 5 6Step 1: Ingredient Entry 7Ingredients:Black BeansOlive OilOnionWhite MushroomQuinoaChicken BrothChicken Breast 8Unit (g)100100100100100240100 9Servings4.20.0543.51.351.56.8 10Total (g)4205400350135360680 11 12Step 2: Importing Nutritional Info from Ingredients Tab 13Calories382.244.2160.077.0496.87.5680.0 14Fat (g)1.35.00.41.18.20.017.7 15Saturated Fat (g)0.40.70.20.40.90.04.1 16Monounsatured Fat (g)0.03.70.10.00.00.04.7 17Polyunsaturated Fat (g)0.00.50.10.00.00.02.9 18Cholesterol (mg)0.00.00.00.00.00.0496.4 19Total Carbohydrate (g)71.40.037.211.686.40.00.0 20 21Step 3: Sum of the Ingredients 22TypeUnitTotal 23Calories615.9 24Fat (g)g11.2 25Saturated Fat (g)g2.2 26Monounsatured Fat (g)g2.8 27Polyunsaturated Fat (g)g1.2 28Cholesterol (mg)mg165.5 29Total Carbohydrate (g)g68.9 30 Quinoa-BlackBeans Cell Formulas RangeFormula D8:I8D8=INDEX(Ingredients!\$B\$4:\$ZW\$12,MATCH(\$B8,Ingredients!\$B\$4:\$B\$12,0),MATCH(D\$7,Ingredients!\$B\$4:\$ZX\$4,0)) C10:I10C10=+C8C9 D23:D29D23=SUM(INDEX(\$B\$13:\$Z\$19,MATCH(\$B23,\$B\$13:\$B\$19,0),0))/\$C\$4 My edit time on the above expired. • Do not copy in the ingredients sheet, do steps 1 & 2 first and that should take care of the ingredients changes (the xl2bb function doesn't handle copying in tables as far as I can tell) • Copy in the Quinoa-BlackBeans sheet • Note: I have left in the hard coding of Column Z as the maximum no of columns in the recipes sheet. An alternative was to just sum the whole row, which would be a bit inefficient but probably less so than using Offset which I have removed. But if you are happy to limit it to Z just leave it. #### Alex Blakenburg ##### Well-known Member My edit time on the above expired. • Do not copy in the ingredients sheet, do steps 1 & 2 first and that should take care of the ingredients changes (the xl2bb function doesn't handle copying in tables as far as I can tell) • Copy in the Quinoa-BlackBeans sheet • Note: I have left in the hard coding of Column Z as the maximum no of columns in the recipes sheet. An alternative was to just sum the whole row, which would be a bit inefficient but probably less so than using Offset which I have removed. But if you are happy to limit it to Z just leave it. You will also need to copy C8 to D8-I8 ##### New Member Thank you Alex. I've learned at least two new things from your post, namely the 1) cleaner sheet name references and 2) far cleaner dynamic sum (without the offsets). I'm very happy with the dynamic sum, the offset function is clunky as hell. The data table formulas did not work. I followed your instructions 1) adding "Type" to Cell B4 2) converted B4:I12 to table "tbl_Ingredients" (and would successfully ping it through F5), but Excel warns me that there's a formula error when I try to copy your formula into cell C8. I am using Excel 2010, so its possible that there has been a change in taxonomy since then. Separately, I'm trying to get somewhere which is maybe too ambitious, and that is to go from Step 1 to Step 3 while skipping Step 2. To have the formula in Step 3 import all of the ingredients from Step 1, multiply them by the number of servings in Step 1, and output the sum in Step 3. Is that possible? #### Alex Blakenburg ##### Well-known Member Thank you Alex. I've learned at least two new things from your post, namely the 1) cleaner sheet name references and 2) far cleaner dynamic sum (without the offsets). I'm very happy with the dynamic sum, the offset function is clunky as hell. The data table formulas did not work. I followed your instructions 1) adding "Type" to Cell B4 2) converted B4:I12 to table "tbl_Ingredients" (and would successfully ping it through F5), but Excel warns me that there's a formula error when I try to copy your formula into cell C8. I am using Excel 2010, so its possible that there has been a change in taxonomy since then. Separately, I'm trying to get somewhere which is maybe too ambitious, and that is to go from Step 1 to Step 3 while skipping Step 2. To have the formula in Step 3 import all of the ingredients from Step 1, multiply them by the number of servings in Step 1, and output the sum in Step 3. Is that possible? 1) Table Formulas. Try this. in C8, highlight tbl_Ingredients and then go to the Ingredients sheet and highlight the whole table. If this works the table formula works in principle. Then check if the spelling is identical on the table names and field names, in the rest of the formula. 2) Bypassing Step 2 I am not a bit fan of doing this sort of thing in one step but I tried to see if I could make is work. I got as far as using Sumproduct to pull in all the right figures for Calories but then realised they individually needed to be multiplied by row 9 being the serving number. Since this array is a different size to the Ingredients array, I got stuck there. I am sure there are people answering questions on the forum that could make it work. Just keep in mind that you are going to have to work with the resulting formula and just the first part looks like this. (D23 just the calory total, still needs multiply by C9:I9 / C4) Excel Formula: ``=SUMPRODUCT(ISNUMBER(MATCH(tbl_Ingredients[[#Headers],[Black Beans]:[Chicken Breast]],\$C\$7:\$I\$7,0))INDEX(tbl_Ingredients[[Black Beans]:[Chicken Breast]],MATCH(B23,tbl_Ingredients[Type],0),0))`` ##### New Member 1) Table Formulas. Try this. in C8, highlight tbl_Ingredients and then go to the Ingredients sheet and highlight the whole table. If this works the table formula works in principle. Then check if the spelling is identical on the table names and field names, in the rest of the formula. View attachment 34086 2) Bypassing Step 2 I am not a bit fan of doing this sort of thing in one step but I tried to see if I could make is work. I got as far as using Sumproduct to pull in all the right figures for Calories but then realised they individually needed to be multiplied by row 9 being the serving number. Since this array is a different size to the Ingredients array, I got stuck there. I am sure there are people answering questions on the forum that could make it work. Just keep in mind that you are going to have to work with the resulting formula and just the first part looks like this. (D23 just the calory total, still needs multiply by C9:I9 / C4) Excel Formula: ``=SUMPRODUCT(ISNUMBER(MATCH(tbl_Ingredients[[#Headers],[Black Beans]:[Chicken Breast]],\$C\$7:\$I\$7,0))*INDEX(tbl_Ingredients[[Black Beans]:[Chicken Breast]],MATCH(B23,tbl_Ingredients[Type],0),0))`` 1) I made a mistake. I made a named range "tbl_ingredients" that was referring to "Table#1", which is why I was getting an error. It works now and I'll be using this going forward. 2) You're right, this seems more trouble than it's worth. Thank you vm for your help Alex. I learned a few new tricks. #### Alex Blakenburg ##### Well-known Member 1) I made a mistake. I made a named range "tbl_ingredients" that was referring to "Table#1", which is why I was getting an error. It works now and I'll be using this going forward. 2) You're right, this seems more trouble than it's worth. Thank you vm for your help Alex. I learned a few new tricks. You mentioned using Excel 2010 but your profile indicated you also have Office 365. Do you have both ? Any reason you are using Excel 2010 if you do ? In terms of learning, 365 has a lot of new functions you are missing out on by not using it. PS: Tables are great for using in Pivot Tables and Lookup type formulas (including SUMIFS) because they autoexpand and require less formula maintenance. Structured referencing means columns can change positions without breaking the formulas referring to them. They also autofill down any formulas inside the tables. Last edited: ##### New Member I have a Office 365 license from my former college, but worry it will get yanked eventually. The Office 2010 license is mine, and it works well enough for my fairly basic needs so far. Tables are new to me, and seem to be pretty elegant if there's new data entered in columns/rows. I've previously made very wide ranges in my formulas. #### Alex Blakenburg ##### Well-known Member I have a Office 365 license from my former college, but worry it will get yanked eventually. The Office 2010 license is mine, and it works well enough for my fairly basic needs so far. Tables are new to me, and seem to be pretty elegant if there's new data entered in columns/rows. I've previously made very wide ranges in my formulas. Here are some options 365 upgrade options you could consider:- Zero or little cost:- 1) Check with your employer to see if they subscribe to the Employee Home Use Program. (you could probably even just try it using your work email address but I haven't googled where to get it the first time for the instal) 2) See if a friend or family member subscribes to Microsoft 365 "Family". They can invite up to 5 family or friends to use their subscription (you are only sharing use of the product they can't access anything you do with it) - zero cost 3) Ask a friend or family member to upgrade from a Personal to Family subscription and maybe contribute to the small incremental cost. Replies 1 Views 380 Replies 1 Views 190 Replies 1 Views 415 Replies 1 Views 198 Replies 1 Views 409 1,129,688 Messages 5,637,835 Members 416,985 Latest member mrindira ### We've detected that you are using an adblocker. We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com. ### Which adblocker are you using? 1)Click on the icon in the browser’s toolbar. 2)Click on the icon in the browser’s toolbar. 2)Click on the "Pause on this site" option. Go back 1)Click on the icon in the browser’s toolbar. 2)Click on the toggle to disable it for "mrexcel.com". Go back ### Disable uBlock Origin Follow these easy steps to disable uBlock Origin 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back ### Disable uBlock Follow these easy steps to disable uBlock 1)Click on the icon in the browser’s toolbar. 2)Click on the "Power" button. 3)Click on the "Refresh" button. Go back	5,068	17,538	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.78125	3	CC-MAIN-2021-21	latest	en	0.716389
https://www.sanfoundry.com/heat-transfer-questions-answers-heat-exchanger-effectiveness/	1,718,271,094,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00163.warc.gz	895,460,915	21,340	# Heat Transfer Questions and Answers – Heat Exchanger Effectiveness This set of Heat Transfer Multiple Choice Questions & Answers (MCQs) focuses on “Heat Exchanger Effectiveness”. 1. Capacity ratio is defined as the product of a) Mass and temperature b) Mass and specific heat c) Specific heat and temperature d) Time and temperature Explanation: The product mass and specific heat of a fluid flowing in a heat exchanger is known as capacity ratio. 2. A single pass shell and tube heat exchanger, consisting of a bundle of 100 tubes (inner diameter 25 mm and thickness 2 mm) is used for heating 28 kg/s of water from 25 degree Celsius to 75 degree Celsius with the help of a steam condensing at atmospheric pressure on the shell side with condensing heat transfer coefficient 5000 W/m2 degree. Make calculation for overall heat transfer coefficient based on the inner area. Take fouling factor on the water side to be 0.002 m2 degree/W per tube and neglect effect of fouling factor on the shell side and thermal resistance of the tube wall a) 647.46 W/m2 degree b) 747.46 W/m2 degree c) 847.46 W/m2 degree d) 947.46 W/m2 degree Explanation: Q = m c c c (t c2 – t c1). Re = 7394, Pr = 3.53 and Nu = 47.41. I/U = I/h I + R + r i /(r 0) (h 0). 3. Which of the following is not associated with a heat exchanger? a) Fouling b) NTU c) Capacity ratio Explanation: The correction factor i.e. Mc Adam’s is associated with laminar film condensation on a vertical plate. 4. The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 degree Celsius and leaving at 60 degree Celsius. Estimate the exchanger effectiveness a) 0.56 b) 0.66 c) 0.76 d) 0.86 Explanation: Effectiveness = (t h 1 – t h 2) C h /C MIN (t h 1 – t c 2). 5. Consider the above problem, if the fluid flow rates and the inlet conditions remain unchanged, workout the lowest temperature to which the oil may be cooled by increasing length of the exchanger a) 46.62 degree Celsius b) 56.62 degree Celsius c) 66.62 degree Celsius d) 76.62 degree Celsius Explanation: Effectiveness = 1 – [exponential [- NTU (1 – C)]/1 + C]. Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now! 6. In a surface condenser, the water flowing through a series of tubes at the rate of 200 kg/hr is heated from 15 degree Celsius to 75 degree Celsius. The steam condenses on the outside surface of tubes at atmospheric pressure and the overall heat transfer coefficient is estimated at 860 k J/m2 hr degree. Find the effectiveness of the heat exchanger. At the condensing pressure, stream has a saturation temperature 0f 100 degree Celsius and the latent heat of vaporization is 2160 k J/kg. Further, the steam is initially just saturated and the condensate leaves the exchanger without sub-cooling i.e. only latent heat of condensing steam is transferred to the water. Take specific heat of water as 4 k J/kg K a) 0.224 b) 0.706 c) 2.224 d) 3.224 Explanation: Effectiveness = 1 – exponential (- NTU) and Effectiveness = C h (t h 1 – t h 2)/C MIN (t h 1 – t c 2). 7. Consider the above problem, find the tube length. Let the diameter of tube is 25 mm a) 14.5 m b) 15.5 m c) 16.5 m d) 17.5 m Explanation: NTU = U (π d l)/C. 8. For evaporators and condensers, for the given conditions, the logarithmic mean temperature difference for parallel flow is a) Does not depend on counter flow b) Smaller than counter flow c) Greater than counter flow d) Equal to counter flow Explanation: The temperature of one of the fluid remains constant during the flow passage. 9. Water (specific heat = 4 k J/kg K) enters a cross flow exchanger (both fluids unmixed) at 15 degree Celsius and flows at the rate of 7.5 kg/s. It cools air (C P = 1 k J/kg K) flowing at the rate of 10 kg/s from an inlet temperature of 120 degree Celsius. For an overall heat transfer coefficient of 780 k J/m2 hr degree and the surface area is 240 m2, determine the NTU a) 4.2 b) 5.2 c) 6.2 d) 7.2 Explanation: NTU = U A/C MIN = 5.2. 10. Consider the above problem, find the capacity ratio of the heat exchanger a) 0.555 b) 0.444 c) 0.333 d) 0.222 Explanation: Capacity ratio = 10/30 = 0.333. Sanfoundry Global Education & Learning Series – Heat Transfer. To practice all areas of Heat Transfer, here is complete set of 1000+ Multiple Choice Questions and Answers. If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]	1,252	4,490	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2024-26	latest	en	0.860783
https://rdrr.io/cran/pcalg/src/R/gacFuns.R	1,534,237,813,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221208750.9/warc/CC-MAIN-20180814081835-20180814101835-00557.warc.gz	957,521,846	22,507	# R/gacFuns.R In pcalg: Methods for Graphical Models and Causal Inference #### Documented in gac ```cond2node <- function(s, x,y,z,m,type) { ## INPUT: s in format for newStackEls ## OUTPUT: TRUE if pth is found; o/w FALSE ## Is there a path of length one ? tmp <- newStackEls(s, x,y,z,m,type) if (tmp\$suc) { suc <- TRUE } else { ## Is there a longer path ? suc <- FALSE if (length(tmp\$res) > 0) { ## nbrs to consider i <- 0 while ( (!suc) & (i < length(tmp\$res)) ) { i <- i+1 s2 <- tmp\$res[[i]] suc <- cond2node(s2, x,y,z,m,type) } } } suc } cond2 <- function(x,y,z,m,type) { ## OUTPUT: True if cond 2 is true; o/w false lx <- length(x) pthFound <- rep(FALSE, lx) for (i in 1:lx) { s <- list(pth=x[i], ncp=FALSE) ## cond2node: TRUE if path is found pthFound[i] <- cond2node(s,x,y,z,m,type) } ## if no path is found, cond 2 is true all(pthFound == FALSE) } defStat <- function(ln, cn, nn, m, type = "pag") { ## INPUT: Node positions ln, cn, nn; adj.mat m (for DAG/CPDAG/MAG/PAG) ## OUTPUT: TRUE if path ln-cn-nn in m is def.stat. o/w FALSE res <- FALSE stopifnot(type %in% c("dag", "cpdag", "mag", "pag")) if (type %in% c("mag", "pag")) { isCollider <- (m[ln,cn] == 2) & (m[nn,cn] == 2) hasTail <- (m[ln,cn] == 3) \| (m[nn,cn] == 3) hasCircles <- (m[ln,cn] == 1) & (m[nn,cn] == 1) isUnshielded <- (m[ln,nn] == 0) & (m[nn,ln] == 0) isDefNonCollider <- ( (hasCircles & isUnshielded) \| hasTail ) } else { ## DAG, CPDAG isCollider <- (m[ln,cn]==0 & m[cn,ln]==1) & (m[cn,nn]==1 & m[nn,cn]==0) hasTail <- (m[ln,cn]==1 & m[cn,ln]==0) \| (m[cn,nn]==0 & m[nn,cn]==1) isUndirected <- (m[ln,cn]==1 & m[cn,ln]==1) \| (m[cn,nn]==1 & m[nn,cn]==1) isUnshielded <- (m[ln,nn]==0 & m[nn,ln]==0) isDefNonCollider <- ( (isUndirected & isUnshielded) \| hasTail ) } if (isCollider) { res <- TRUE } else { if (isDefNonCollider) res <- TRUE } res } desc <- function(m, possible = FALSE, type = type) { ## computes all (possible) descendants of every node ## INPUT: adj.matrix m in MAG/PAG or DAG/CPDAG coding; if possible=TRUE, possible desc are found; o/w descendants ## OUPUT: list of vectors; vector at list position i contains node positions of (possible) descendants of node i p <- ncol(m) res <- vector(mode = "list", length = p) for (i in 1:p) { ## res[[i]] <- possibleDeProper(m=m, x=i, y=NULL, possible=possible) res[[i]] <- possDe(m = m, x = i, y = NULL, possible = possible, ds = FALSE, type = type) } res } forbiddenNodes <- function(m,x,y,type) { ## INPUT: adj.matrix in DAG,CPDAG,MAG,PAG coding; sets of node positions x and y ## OUTPUT: set of node positions of nodes in the forbidden set (sorted) n1 <- length(x) n2 <- length(y) possDeX <- possAnY <- c() for(i in 1:max(n1,n2)) { if (i <= n1) { #find all possible descendants of a node x[i] that are on a proper path #relative to x (exclude x[i] because descendants of x[i] that are not #on a proper path are allowed) ## possDeX <- union(possDeX, setdiff(possibleDeProper(m,x[i],x),x[i])) pdpTmp <- possDe(m = m, x = x[i], y = x, possible = TRUE, ds = FALSE, type = type) possDeX <- union(possDeX, setdiff(pdpTmp, x[i])) } if (i <= n2) { #find all possible ancestors of a node y[i] that are #on a proper path relative to x ## possAnY <- union(possAnY, possibleAnProper(m,y[i],x)) papTmp <- possAn(m = m, x = y[i], y = x, possible = TRUE, ds = FALSE, type = type) possAnY <- union(possAnY, papTmp) } } #a set of all nodes on a proper possibly directed path from X to Y pdp <- intersect(possDeX,possAnY) #the forbiden node set are all possible descendants of nodes in pdp fbnodes <- c() if (length(pdp) > 0) { for(j in 1:length(pdp)) { ## fbnodes <- union(fbnodes,possibleDeProper(m,pdp[j],c())) pdpTmp2 <- possDe(m = m, x = pdp[j], y = c(), possible = TRUE, ds = FALSE, type = type) fbnodes <- union(fbnodes, pdpTmp2) } } if (length(fbnodes) > 0) { return(sort(fbnodes)) } else { return(fbnodes) } } ##gac for pdags ## ## this function is the same as its equivalent in pcalg except it calls ## the pdag fucntions given above ## ##currrently separated dags,pdags,cpdags from mags,pags but could be merged ##however the res[3] must be calculated separately for dags,pdags,cpdags and mags,pags ## uses amat.cpdag encoding amat[i,j]=0,amat[j,i]=1 <=> i ->j gac <- function (amat, x, y, z, type = "pag") { if (inherits(amat, "amat")) amat <- as(amat, "matrix") if (!is.null(dimnames(amat))) dimnames(amat) <- NULL if (type %in% c("dag", "cpdag", "pdag")) { if (!isValidGraph(amat = amat, type = type)) { message("The input graph is not a valid ",type, ". See function isValidGraph() for details.") } } res <- c(isAmenable(amat, x = x, y = y, type = type), NA, NA) if (type %in% c("dag","pdag","cpdag")){ f <- bforbiddenNodes(m = amat, x = x, y = y) res[2] <- (length(intersect(f, z)) == 0) res[3] <- cond3fast(x = x, y = y, z = z, m = amat) } else { ##the code for MAGs, PAGs is still the same f <- forbiddenNodes(amat, x = x, y = y) res[2] <- (length(intersect(f, z)) == 0) res[3] <- cond2(x = x, y = y, z = z, m = amat, type = type) } list(gac = all(res), res = res, f = f) } isAmenable <- function(m,x,y, type = "pag") { ## INPUT: adj.matrix m; sets of node positions x and y; type in DAG, CPDAG, ## MAG or PAG ## OUTPUT: TRUE if m is amenabel wrt x,y; o/w FALSE found <- FALSE ## if found == TRUE at any time, graph is not amenable wrt x,y ## DAG is always amenable if (type %in% c("pdag", "cpdag","dag","pag","mag")) { ##changed added dag just in case, makes no difference if (type == "dag") ## added the if case for dags return(!found) i <- 0 p <- length(x) ## for all nodes in x, if amenability is still possible while ( (i<p) & !found) { i <- i+1 ## posDesc of x[i] without going through any other x node ## posDesc <- possibleDeProper(m,x[i],x[-i]) posDesc <- possDe(m = m, x = x[i], y = x[-i], possible = TRUE, ds = FALSE, type = type) ## potential problem for amenability only if there is a ## pdp from x[i] to y if ( length(intersect(y, posDesc)) != 0 ) { nb <- as.vector(which(m[x[i],]!=0 \| m[,x[i]]!=0)) ## nbrs of x[i] ## potentially first node on pdp from x[i] to y; however, not yet sure cand <- intersect(nb, posDesc) j <- 0 ## for all candidate nodes, if amenability is still possible ## (also covers case if cand is empty) while ( (j<length(cand)) & !found ) { j <- j+1 ## check if there is a pdp from cand[j] to y without going through x[i] ## cand could already be in y ## pathOK <- ( length(intersect(y, possibleDeProper(m,cand[j],x[i]))) != 0 ) pdpTemp <- possDe(m = m, x = cand[j], y = x[i], possible = TRUE, ds = FALSE, type = type) pathOK <- ( length(intersect(y, pdpTemp)) != 0 ) if (pathOK) { isPDAG <- (type == "pdag" \| type == "cpdag") ##changed PDAGproblem <- (isPDAG & (m[x[i],cand[j]] == 1)) ##changed PAGproblem1 <- (!isPDAG & (m[x[i], cand[j]] != 2) & (m[cand[j], x[i]] != 3)) ##changed isDirEdge <- ((m[x[i], cand[j]] == 2) & (m[cand[j], x[i]] == 3)) PAGproblem2 <- (!isPDAG & isDirEdge & !visibleEdge(m, x[i], cand[j])) ##changed found <- (PDAGproblem \| PAGproblem1 \| PAGproblem2) } ## if pathOK } ## while cand } ## if path from x[i] to y } ## while x return(!found) } else { ## if graph type not known cat("Not a valid graph type! Should be written in lowercase! \n") return(NULL) } } mcon <- function(ln,cn,nn,m,z,descList) { ## INPUT: node positions of last, current and next node on path; adj.mat. m ## in DAG/CPDAGMAG/PAG format; ## set z; descList as returned from desc(m) ## OUTPUT: TRUE if ln and nn are mcon given z on path ln-cn-nn res <- FALSE isColliderPAG <- ( (m[ln,cn] == 2) & (m[nn,cn]==2) ) isColliderDAG <- ( (m[ln,cn]==0 & m[cn,ln]==1) & (m[cn,nn]==1 & m[nn,cn]==0) ) isCollider <- ( isColliderPAG \| isColliderDAG) if (isCollider) { if (any(descList[[cn]] %in% z)) res <- TRUE } else { if ( !(cn %in% z) ) res <- TRUE } res } ncEdge <- function(cn, nn, m, type = "pag") { stopifnot(type %in% c("dag", "cpdag", "mag", "pag")) res <- FALSE if (type %in% c("mag", "pag")) { if (m[nn,cn]==2) res <- TRUE } else { ## must be DAG or CPDAG if ( (m[cn,nn] == 1) & (m[nn,cn] == 0)) res <- TRUE } res } newStackEls <- function(s,x,y,z,m,type) { ## INPUT: previous stack element of type list: {pth (num. vec.), ncp (bool); ## ncp is TRUE, if pth is non-causal; o/w FALSE ## OUTPUT: List 'res' of new stack elements (poss. empty); suc = TRUE, if an ## open path to y was found suc <- FALSE pth <- s\$pth; ncp <- s\$ncp lp <- length(pth) if (lp == 1) { ## Starting node on search cn <- pth nb <- setdiff(as.vector(which(m[cn,]!=0 \| m[,cn]!=0)), x) ## nb does not contain x -> proper path res <- vector("list", length(nb)) i <- 0 while ( (!suc) & (i < length(nb)) ) { ## Search through possible neighbors i <- i + 1 nn <- nb[i] ncpTmp <- ( ncp \| ncEdge(cn,nn,m,type=type) ) ## suc=T: Exists proper, m-con, d.s., non-c path from x to y suc <- ( (nn %in% y) & ncpTmp ) res[[i]] <- list(pth = c(cn,nn), ncp = ncpTmp) } } else { ## Path has at least 2 nodes descList <- desc(m, type = type) ln <- pth[lp-1]; cn <- pth[lp] nb <- setdiff(setdiff(as.vector(which(m[cn,]!=0 \| m[,cn]!=0)), x), pth) res <- list() j <- 0 jj <- 0 while( (!suc) & (j<length(nb)) ) { j <- j+1 nn <- nb[j] mc <- mcon(ln=ln, cn=cn, nn=nn, m=m, z=z, descList=descList) ds <- defStat(ln=ln, cn=cn, nn=nn, m=m, type = type) ncpTmp <- ( ncp \| ncEdge(cn,nn,m,type=type) ) if (mc & ds) { jj <- jj + 1 ## proper, m-connecting & def.stat. path if ( (nn %in% y) & ncpTmp ) suc <- TRUE res[[jj]] <- list(pth=c(pth,nn), ncp=ncpTmp) } } } list(res=res, suc=suc) } ## ## CPDAG ## cp <- function(m) { ## require(Rgraphviz) ## plot(as(t(m),"graphNEL")) ## } ## ## MAG/PAG ## mp <- function(m) { ## colnames(m) <- rownames(m) <- as.character(1:ncol(m)) ## require(Rgraphviz) ## plotAG(m) ## } possibleAnProper <- function(m,x,y=NULL) { ## INPUT ## m: Adjacency matrix (coding for MAG/PAG) ## x: Starting node (col.position in m) ## y: Set Y (col.positions in m) ## OUTPUT ## All nodes with a possibly directed path from a to x not going through y ## including x itself p <- length(m[,1]) ## nmb of nodes q <- v <- rep(0,p) ## queue ## q has col.pos. of unvisited nodes = queue ## v has col.pos. of visited nodes i <- k <- 1 ## i: end of queue; k: current point in queue q[1] <- x ## x is first node in queue tmp <- m ## ??? while(q[k]!=0 & k<=i) {## queue is not empty & current pos is within queue t <- q[k] ## take new node from queue v[k] <- t ## mark t as visited (if t is in v, it is visited) k <- k+1 ## increase current position in queue for(j in 1:p) { ## check if j is a possible parent of t ## check if edgemark at j is circle (1) or tail (3) but not head (2) or empty (0) if (tmp[t,j] %in% c(1,3)) { ## works for CPDAG, too ## aren't already scheduled for a visit ## are not in y if (!(j %in% q) & !(j %in% y)) {## i <- i+1 ## incearse size of queue by one q[i] <- j ## add node to queue } } } } sort(setdiff(v,c(0))) ## remove trailing zeros from initial vector } #finds all possible descendants of a node x in a graph #that are on a proper path relative to nodeset Y (that is, that don't go through Y) possibleDeProper <- function(m,x,y=NULL,possible = TRUE) { ## INPUT: adj.mat. m in MAG/PAG or DAG/CPDAG coding; node pos x; ## set of node pos y; ## If possible == TRUE, possible Desc. are found; o/w descendents are found ## OUTPUT: Node positions of (possible) descendents (ignoring path trough y); sorted #q denotes unvisited nodes/ nodes in queue #v denotes visited nodes q <- v <- rep(0,length(m[,1])) i <- k <- 1 q[i] <- x while(q[k]!=0 & k<=i) { t <- q[k] #mark t as visited v[k] <- t k <- k+1 #in this for cycle #add all nodes that have a possibly directed #edge with node t and all parents of node t to queue for(j in 1:length(m[1,])) { if (possible) { ## find possible descendents collect <- ( ((m[t,j]!=0) && (m[j,t] %in% c(1,3))) \|\| ## PAG; CPDAG t --- j ((m[t,j]==0) && (m[j,t]==1)) ) ## CPDAG: t ---> j } else { ## find descendents collect <- ( ((m[j,t]==3) && (m[t,j]==2)) \|\| ((m[j,t]==1) && (m[t,j]==0)) ) } if (collect) { #and that are on a proper path if (!(j %in% q) & !(j %in% y)) { i <- i+1 q[i] <- j } } } } sort(setdiff(v,c(0))) } ##The following function forms a proper back-door graph G_{XY}^{pbd} ##given the adjacency matrix, node position vectors x and y ## gbg <- function(m,x,y) { tmp <- m for (i in seq_along(x)) { Desc <- bPossibleDeProper(m,x[i],x[-i]) if (length(intersect(y, Desc)) != 0) { ch <- as.vector(which(m[x[i], ] == 0 & m[, x[i]] == 1)) cand <- intersect(ch, Desc) for(j in seq_along(cand)) { ##pathOK <- (length(intersect(y, bpossibleDeProper(m,cand[j], x[i]))) != 0) ## what if x - y? MISSING CASE! pathOK <- (length(intersect(y, bPossibleDeProper(m,cand[j], x))) != 0) \|\| (cand[j] %in% y) # added x - y case if (pathOK) { tmp[cand[j],x[i]] <- 0 } } } } tmp } ## the following function tests the separation condition for gac-pdags ## it is a pseudo check for this condition, because provided ## that amenability and forbidden set are satisfied ## this will exactly check the separation condition ## however if amenability or forb are not satisfied ## this check is not necessarily accurate ## given a pdag adjacency matrix m it checks whether node position vector z ## satisfies the separation condition relative to (x,y) ## m encoding amat.cpdag as above m[i,j]=0, m[j,i]=1 <=> i -> j cond3fast <- function(x, y, z, m) { ## the idea is to use the lemma which says ## that if m is amenable and z satisfies the forbidden cond ## then z satisfies separation in the pdag iff ## z satisfies separation in at least one dag ## First find one dag D in the equivalence class represented by m oneDag <- pdag2dag(as(t(m),"graphNEL")) dagAmat <- t(as(oneDag\$graph,"matrix")) ## We instead of checking the separation crit in m ## w can check the d-separation criterion in D_{XY}^{pbd} gb <- gbg(dagAmat,x,y) msep(t(gb),x,y,z) ## uses the transposed of amat.cpdag encoding } ## note that provided that possibleDeProper is implemented in ## pcalg as looking at only def. stat. paths ## then the already implemented forbiddenNodes can be used directly ## (with the addition of type: pdag) ## m - adjacency matrix amat.cpdag encoding m[i,j]=0, m[j,i] = 1 <=> i -> j ## x - vector of node position ## y - vector of node positions ## as it is now this function can only be used for pdags, dags, cpdags ## because the bPossibleDeProper etc. functions are only defined for these bforbiddenNodes <- function (m, x, y) { n1 <- length(x) n2 <- length(y) possDeX <- possAnY <- c() for (i in 1:max(n1, n2)) { if (i <= n1) possDeX <- union(possDeX, setdiff(bPossibleDeProper(m, x[i], x), x[i])) ##changed if (i <= n2) possAnY <- union(possAnY, bPossibleAnProper(m, y[i], x)) ##changed } pdp <- intersect(possDeX, possAnY) fbnodes <- c() if (length(pdp) > 0) { for (j in 1:length(pdp)) { fbnodes <- union(fbnodes, bPossibleDeProper(m, pdp[j], c())) ##changed } } if (length(fbnodes) > 0) { return(sort(fbnodes)) } else { return(fbnodes) } } ``` ## Try the pcalg package in your browser Any scripts or data that you put into this service are public. pcalg documentation built on June 5, 2018, 1:05 a.m.	5,177	15,026	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2018-34	longest	en	0.368079
https://brainly.in/question/287129	1,484,581,182,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560279189.36/warc/CC-MAIN-20170116095119-00027-ip-10-171-10-70.ec2.internal.warc.gz	813,734,327	10,378	# How many squares are there in a normal chess board? 2 by afrozahmed2209 ohhhh just check it before .... what? Comment has been deleted see my justification is given by someelse some1else 2016-02-24T23:54:28+05:30 8 8 = 64 hope i hlpd u ,plz mark as the brainliest. 2016-02-25T00:02:53+05:30 8 × 8 square → 1 7 × 7 squares → 4 6 × 6 squares → 9 5 × 5 squares → 16 4 × 4 squares → 25 3 × 3 squares → 36 2 × 2 squares → 49 × 1 squares → 64 ------------------------------------------------- Total no. of squares = 204 Ans. ------------------------------------------------- oh.. nyc answr...i juzt said abt small squares Thanks oh... good logic!! but, it should be 64 small squares yup	249	689	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2017-04	latest	en	0.560271
https://slideplayer.com/slide/3874030/	1,627,799,371,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046154163.9/warc/CC-MAIN-20210801061513-20210801091513-00428.warc.gz	530,724,698	18,466	# Motion Chapter 11. Frame of Reference Motion must be described from a certain point of view – a frame of reference. Which way is up? Another example (begin. ## Presentation on theme: "Motion Chapter 11. Frame of Reference Motion must be described from a certain point of view – a frame of reference. Which way is up? Another example (begin."— Presentation transcript: Motion Chapter 11 Frame of Reference Motion must be described from a certain point of view – a frame of reference. Which way is up? Another example (begin at 2:30) http://www.youtube.com/watch?v=ac6o8PXthzQ http://www.youtube.com/watch?v=i0g3g6AvLtM Relative Motion Motion in a frame of reference is relative – that is it is compared to some other object. In physics the Earth is most often used as the frame of reference. What is the motion of the ball compared to the Earth in this clip? Compared to the truck? Distance & Displacement Distance is how far you have traveled along a path. Displacement is how far you are from your starting point. A vector is used to show each individual movement. The resultant vector is the sum of two or more vectors. In this case it shows the displacement. Speed Speed is how far you have moved in a certain amount of time. Speed = distance/time Speed is a scalar. It has does not indicate direction. Average speed = total distance/total time Instantaneous speed is your speed at a certain moment (instant) of time. Graphing speed The slope of a line on a distance-time graph is equal to the speed of the object. A straight line shows constant speed. A curved line shows acceleration. Velocity Velocity is a vector. It is speed and direction. 70 mph = speed 70 mph east = velocity GPS – How does it work? Acceleration Acceleration is a change in velocity. – Change in speed, direction or both! It is a vector. A object in free fall is accelerating due to gravity alone. A g is about 9.8 m/s 2 (32 ft/s 2 ) Constant acceleration means that your velocity is changing steadily. Calculating Acceleration Or Graphs of Acceleration The slope of the line of a velocity-time graph is the acceleration of the object. If you calculate the area under a section of the graph you find the distance the object has traveled! Download ppt "Motion Chapter 11. Frame of Reference Motion must be described from a certain point of view – a frame of reference. Which way is up? Another example (begin." Similar presentations	536	2,425	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2021-31	latest	en	0.920279
https://www.hpmuseum.org/forum/post-73673.html	1,716,833,656,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00318.warc.gz	697,414,500	10,707	DM 42 05-24-2017, 06:48 PM (This post was last modified: 05-24-2017 06:50 PM by Ángel Martin.) Post: #21 Ángel Martin Senior Member Posts: 1,441 Joined: Dec 2013 RE: DM 42 (05-24-2017 06:42 PM)Guenter Schink Wrote: I think the DM42 also uses this library, perhaps this explains something, I don't know what is correct. sorry to be blunt but this is math, and thus pretty rigid as to which one is the correct result. You can check WolframAlpha if that adds more weight to the "credibility of options": http://www.wolframalpha.com/input/?i=asi...2B0i)))))) Logan is correct, the "glitch" occurs in the ATAN function. "To live or die by your own sword one must first learn to wield it aptly." 05-24-2017, 07:58 PM (This post was last modified: 05-24-2017 08:01 PM by Vtile.) Post: #22 Vtile Senior Member Posts: 406 Joined: Oct 2015 RE: DM 42 (05-24-2017 06:48 PM)Ángel Martin Wrote: (05-24-2017 06:42 PM)Guenter Schink Wrote: I think the DM42 also uses this library, perhaps this explains something, I don't know what is correct. sorry to be blunt but this is math, and thus pretty rigid as to which one is the correct result. You can check WolframAlpha if that adds more weight to the "credibility of options": http://www.wolframalpha.com/input/?i=asi...2B0i)))))) Logan is correct, the "glitch" occurs in the ATAN function. All my sources have been saying that 1+i0 = 1e^(0i) = +1 ... in this case 1e^0 , As given z=a+ib=ABS(z)(cos(w)+sin(w)i=ABS(z)e^(iw)) So the calculation should behave excactly like with integer 1. (with this knowledge of mine) Is the problem with intel decimal library? Isn't the newRPL using it also or do I remember incorrectly? 05-24-2017, 08:25 PM Post: #23 Hsilop Junior Member Posts: 41 Joined: May 2014 RE: DM 42 As mentioned by Thomas Okken in this post, Swiss Micros are setting up a forum at http://forum.swissmicros.com. Should we not be having this discussion about the DM42 over at their place, to give them as much value from the beta testing as possible? 05-24-2017, 08:29 PM Post: #24 Logan Member Posts: 139 Joined: Jul 2016 RE: DM 42 (05-24-2017 08:25 PM)Hsilop Wrote: Should we not be having this discussion about the DM42 over at their place, to give them as much value from the beta testing as possible? This doesn't seem to be specifically a DM42 problem, but rather with the underlying Intel libraries of Free42, which underlies the DM42 05-24-2017, 08:30 PM (This post was last modified: 05-24-2017 08:47 PM by Vtile.) Post: #25 Vtile Senior Member Posts: 406 Joined: Oct 2015 RE: DM 42 (05-24-2017 08:25 PM)Hsilop Wrote: As mentioned by Thomas Okken in this post, Swiss Micros are setting up a forum at http://forum.swissmicros.com. Should we not be having this discussion about the DM42 over at their place, to give them as much value from the beta testing as possible? Guenter have already started a topic on there. http://www.hpmuseum.org/forum/archive/in...d-756.html Thomas Okken writes there: Quote:The Intel library provides all the functions Free42 needs; it doesn't just have arithmetic and square root, but also trigonometrics, hyperbolics, logarithms, and gamma. The new code uses all of these wherever possible, using the highest-precision format, decimal128. The higher-level functionality, like complex numbers, matrices, numerical integration, and numerical root-finding, is still done using the same algorithms and code as before, but everything now uses the new library for the basic operations underneath. So maybe it is a bug in free42 codebase not in the intel library ?? 05-24-2017, 09:05 PM (This post was last modified: 05-24-2017 09:09 PM by Hsilop.) Post: #26 Hsilop Junior Member Posts: 41 Joined: May 2014 RE: DM 42 (05-24-2017 08:30 PM)Vtile Wrote: ... Guenter have already started a topic on there. http://www.hpmuseum.org/forum/archive/in...d-756.html ... I guess that Thomas Okken is the right man for Free42 matters, but he's the one that mentioned Swiss Micros' forum. I see no mention as of yet of this issue at there. Any chance that Intel might be interested in fixing their bugs anytime soon? 05-24-2017, 09:08 PM (This post was last modified: 05-24-2017 09:11 PM by Vtile.) Post: #27 Vtile Senior Member Posts: 406 Joined: Oct 2015 RE: DM 42 (05-24-2017 09:05 PM)Hsilop Wrote: (05-24-2017 08:30 PM)Vtile Wrote: ... Guenter have already started a topic on there. http://www.hpmuseum.org/forum/archive/in...d-756.html ... I guess that Thomas Okken may be the right man for Free42 matters, but he's the one that mentioned Swiss Micros' forum. I see no mention as of yet of this issue at there. It is on: Board index / DM42 beta units / Free42 related Yes, Thomas Okken should be the one who is best on track which is going on with this complex oddity. 05-24-2017, 09:09 PM Post: #28 Logan Member Posts: 139 Joined: Jul 2016 RE: DM 42 (05-24-2017 09:08 PM)Vtile Wrote: this complex oddity. We'll try not to be too negative about it either 05-24-2017, 09:11 PM Post: #29 Vtile Senior Member Posts: 406 Joined: Oct 2015 RE: DM 42 (05-24-2017 09:09 PM)Logan Wrote: (05-24-2017 09:08 PM)Vtile Wrote: this complex oddity. We'll try not to be too negative about it either Yes, lets be positive. 05-24-2017, 11:25 PM (This post was last modified: 05-25-2017 12:09 AM by Thomas Okken.) Post: #30 Thomas Okken Senior Member Posts: 1,884 Joined: Feb 2014 RE: DM 42 (05-24-2017 04:44 PM)Ángel Martin Wrote: (05-24-2017 04:32 PM)Logan Wrote: I have Free42 on my iPhone and it returned -1. The real 42 returned +1. So this is a version-dependent for Free42. I remember about 8 years ago (while working on the 41Z project) I corresponded w/ Thomas about this issue and he agreed it wasn't the correct result and sent me an update... maybe it was never turned into the "official" code? Interested to see how will this one be dealt with now. BTW the 15C and the 15C-LE both return +1 in case you had doubts... That sounds like a Free42 bug, or maybe an edge case in the Intel library that Free42 should work around. If I fixed this eight years ago and now it is broken again, that could be because that fix was specific to the BCD20 library that I used back then; I switched to the Intel library three years ago. Ángel, do you remember when I made this earlier change? What date or at which version number? If you don't, it's not a big deal, I can do a binary search through my revision history, but I'd rather not if I don't have to. :-) Anyway, I'll look into it this weekend. 05-25-2017, 12:59 AM Post: #31 Paul Dale Senior Member Posts: 1,837 Joined: Dec 2013 RE: DM 42 The 34S gets (1, 2.031x10-35) as its result. It is possible that the Intel library is introducing a rounding error. I've seen cases before where it rounded incorrectly in the last digit. Pauli 05-25-2017, 02:13 AM Post: #32 Thomas Okken Senior Member Posts: 1,884 Joined: Feb 2014 RE: DM 42 (05-25-2017 12:59 AM)Paul Dale Wrote: The 34S gets (1, 2.031x10-35) as its result. It is possible that the Intel library is introducing a rounding error. I've seen cases before where it rounded incorrectly in the last digit. Pauli I just tried the asin(acos(atan(tan(cos(sin(1+0i)))))) test case in Free42 1.4.77, so pre-Intel, and got the same result. This looks like an edge case that needs special handling in Free42's math_atanh(). Not a bug in the Intel library. Should be an easy fix, but I'll put it off until the weekend, when my mind will be a bit fresher than it is now. :-) 05-25-2017, 04:50 AM (This post was last modified: 05-25-2017 04:56 AM by Ángel Martin.) Post: #33 Ángel Martin Senior Member Posts: 1,441 Joined: Dec 2013 RE: DM 42 (05-24-2017 11:25 PM)Thomas Okken Wrote: That sounds like a Free42 bug, or maybe an edge case in the Intel library that Free42 should work around. If I fixed this eight years ago and now it is broken again, that could be because that fix was specific to the BCD20 library that I used back then; I switched to the Intel library three years ago. Ángel, do you remember when I made this earlier change? What date or at which version number? If you don't, it's not a big deal, I can do a binary search through my revision history, but I'd rather not if I don't have to. :-) Anyway, I'll look into it this weekend. Yes I still have the emails from 2011 (so only 6 years ago, not 8 - I'm an outlook pack-rat) - will forward them to you in a moment. BTW while re-reading that thread I realized the issue back then wasn't exactly in ATAN - but in ASIN. Still entirely within the same functionality area... I guess involving those pesky square root branches and principal values? "To live or die by your own sword one must first learn to wield it aptly." 05-25-2017, 09:52 AM Post: #34 Guenter Schink Senior Member Posts: 529 Joined: Dec 2013 RE: DM 42 For what it's worth The Prime calculator immediately discards the imaginary part if that is "0" input: (1+0i) ^Enter results in 1 So does Wolfram \|Alpha input (1+0i) ^Enter results in 1 Günter 05-25-2017, 11:11 AM Post: #35 Paul Dale Senior Member Posts: 1,837 Joined: Dec 2013 RE: DM 42 Converting to reals and using real functions is a cheats way out. You'll get the correct answer in this case but it is only hiding the underlying problem. 05-25-2017, 11:31 AM Post: #36 Thomas Okken Senior Member Posts: 1,884 Joined: Feb 2014 RE: DM 42 (05-25-2017 04:50 AM)Ángel Martin Wrote: (05-24-2017 11:25 PM)Thomas Okken Wrote: That sounds like a Free42 bug, or maybe an edge case in the Intel library that Free42 should work around. If I fixed this eight years ago and now it is broken again, that could be because that fix was specific to the BCD20 library that I used back then; I switched to the Intel library three years ago. Ángel, do you remember when I made this earlier change? What date or at which version number? If you don't, it's not a big deal, I can do a binary search through my revision history, but I'd rather not if I don't have to. :-) Anyway, I'll look into it this weekend. Yes I still have the emails from 2011 (so only 6 years ago, not 8 - I'm an outlook pack-rat) - will forward them to you in a moment. BTW while re-reading that thread I realized the issue back then wasn't exactly in ATAN - but in ASIN. Still entirely within the same functionality area... I guess involving those pesky square root branches and principal values? Got 'em -- Thanks! This issue is unrelated, though. The complex atanh code (which is also used for complex atan) is correct, but a bit inaccurate when the real part (imaginary part for atan) is too close to zero. I'll have to make it handle that as a special case. 05-25-2017, 04:31 PM Post: #37 Ángel Martin Senior Member Posts: 1,441 Joined: Dec 2013 RE: DM 42 (05-25-2017 11:11 AM)Paul Dale Wrote: Converting to reals and using real functions is a cheats way out. You'll get the correct answer in this case but it is only hiding the underlying problem. Couldn't agree more. Besides that won't solve infinite other cases, like z=1+i which also shows the same problem! "To live or die by your own sword one must first learn to wield it aptly." 05-25-2017, 04:35 PM (This post was last modified: 05-25-2017 04:39 PM by Ángel Martin.) Post: #38 Ángel Martin Senior Member Posts: 1,441 Joined: Dec 2013 RE: DM 42 (05-25-2017 11:31 AM)Thomas Okken Wrote: The complex atanh code (which is also used for complex atan) is correct, but a bit inaccurate when the real part (imaginary part for atan) is too close to zero. I'll have to make it handle that as a special case. ok, but then watch out for this one, not with small imag part : ACOS(COS(1+i)) = -1-i (oops !!) "To live or die by your own sword one must first learn to wield it aptly." 05-25-2017, 05:15 PM Post: #39 toml_12953 Senior Member Posts: 2,096 Joined: Dec 2013 RE: DM 42 (05-25-2017 11:11 AM)Paul Dale Wrote: Converting to reals and using real functions is a cheats way out. You'll get the correct answer in this case but it is only hiding the underlying problem. But when the imaginary part is 0 doesn't that mean the number is* a real? It seems more intelligent to treat it that way than to mindlessly use complex routines that aren't necessary. Tom L Tom L Cui bono? 05-25-2017, 06:16 PM Post: #40 Thomas Okken Senior Member Posts: 1,884 Joined: Feb 2014 RE: DM 42 (05-25-2017 04:35 PM)Ángel Martin Wrote: (05-25-2017 11:31 AM)Thomas Okken Wrote: The complex atanh code (which is also used for complex atan) is correct, but a bit inaccurate when the real part (imaginary part for atan) is too close to zero. I'll have to make it handle that as a special case. ok, but then watch out for this one, not with small imag part : ACOS(COS(1+i)) = -1-i (oops !!) Aargh, wrong branch. Another bug. « Next Oldest \| Next Newest » User(s) browsing this thread: 1 Guest(s)	3,793	12,811	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.125	3	CC-MAIN-2024-22	latest	en	0.915078
https://math.libretexts.org/Courses/Monroe_Community_College/MATH_220_Discrete_Math/8%3A_Big_O	1,642,611,136,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00238.warc.gz	454,666,639	22,066	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 8: Big O $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$	519	1,399	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2022-05	latest	en	0.19083
http://www.ncl.ucar.edu/Document/Functions/Contributed/pot_vort_isobaric.shtml	1,558,269,853,000,000,000	text/html	crawl-data/CC-MAIN-2019-22/segments/1558232254882.18/warc/CC-MAIN-20190519121502-20190519143502-00251.warc.gz	322,204,700	5,189	NCL Home > Documentation > Functions > Meteorology # pot_vort_isobaric Compute potential vorticity on isobaric levels and a global rectilinear grid. Available in version 6.3.0 and later. ## Prototype ```load "\$NCARG_ROOT/lib/ncarg/nclscripts/csm/contributed.ncl" ; This library is automatically loaded ; from NCL V6.2.0 onward. ; No need for user to explicitly load. function pot_vort_isobaric ( p [] : numeric, u : numeric, v : numeric, t : numeric, lat [] : numeric, gridType [1] : integer, opt [1] : integer ) return_val [dimsizes(t)] : float or double ``` ## Arguments p Array containing pressure (Pa) levels. Must be 1D (lev) u Array containing zonal wind components (m/s). Must be 3D (lev,lat,lon) or 4D (time,lev,lat,lon). The array must be ordered south to north. v Array containing meridional wind components (m/s). Same size and shape as u. t Array containing temperatures (K). Same size and shape as u. lat Latitudes ordered south to north for the rectilinear grid. gridType Grid type: gridType=0 means gaussian grid; gridType=1 means regular or fixed grid. Both are rectilinear grids. opt • opt=0: return potential vorticity • opt=1: return a list variable containing: potential vorticity, static stability, potential temperature ## Return value A multi-dimensional array of the same size and shape as t. The output will be double if t is of type double. ## Description Calculate the potential vorticity on isobaric levels on a global grid. The reason for the global grid is that highly accurate spherical harmonic functions are used to compute horizontal gradients. ```References: Bluestein: Synoptic-Dynamic Meteorology in Midlatitudes pg 264 Eq 4.5.93 Note: A nice basic discussion of PV may be found at: Mid-Latitude Atmospheric Dynamics: A First Course Jonathan E. Martin, Wiley 2006, QC880.M36 , pp276-onward ``` ## Examples Note: The dimension names are those used by the CESM. NCL does not care what the dimensions are named. Worked potential vorticity examples are HERE. Example 1: Here the data are south to north. ``` U = f->U ; (time,lev,lat,lon) or (lev,lat,lon) V = f->V T = f->T ; K lat = f->lat lev = f->lev ; hPa lev = lev100 ; convert units lev@units = "Pa" gridType = 0 ; gaussian grid opt = 0 PV = pot_vort_isobaric(lev,u,v,t,lat, gridType, opt) ``` Example 2 Here the data are ordered north to south. Most reanalysis data sets are ordered north to south. Use NCL syntax to reorder the data. ``` U = f->U ; (time,lev,lat,lon) V = f->V T = f->T ; K lev = f->lev ; hPa lev = lev100 ; convert units lev@units = "Pa" U = U(:,:,::-1,:) ; reorder to South -> North V = V(:,:,::-1,:) T = T(:,:,::-1,:) lat = T&lat gridType = 1 ; fixed grid opt = 0 PV = pot_vort_isobaric(lev,u,v,t,lat, gridType, opt) ; Potential Vorticity ```	831	2,905	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.546875	3	CC-MAIN-2019-22	latest	en	0.708039
https://www.teachercreated.com/lessons/145	1,721,716,638,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00583.warc.gz	867,327,135	6,667	## Extra Information in Word Problems Mathematics, Problem Solving ### Objective Students will learn how to recognize extra information in word problems. ### Directions Facts to Know Sometimes a problem has extra information that students do not need to solve the problem. Before students start these problems, ask them these four important questions: 1. What does the problem ask you to find? 2. What information is needed to solve the problem? 3. How can you show the problem with numbers? Sample A At the art fair, Libby bought a mug of root beer for \$1, a veggie burger for \$2, a hot fudge sundae for \$2, and a piece of pottery for \$5. How much did Libby spend on food? First, have students focus on the information they need to answer the question being asked. Then they can identify the extra information that is not necessary for answering the question because it has no effect on the answer. In the problem above, the question is about money spent on food. So the extra information is that Libby bought a piece of pottery for \$5. Now, they can solve the problem using only the information that's necessary. \$1 + \$2 + \$2 = \$5-Libby spent \$5 on food. Sample B Erica and her friends had a pizza party. They ordered 8 pizzas. 1/2 of the pizzas were large combo pizzas, 1/4 of the pizzas were medium specialty pizzas, and 1/4 were medium pepperoni pizzas. They also ordered five 2-liter sodas for \$1.50 each. How many medium pepperoni pizzas did Erica and her friends order? First, have students focus on the information they need to answer the question being asked. Then they can identify the extra information that is not necessary for answering the question because it has no effect on the answer. In the problem above, the question is about how many medium pepperoni pizzas were ordered. So the extra information is how many sodas and large combo pizzas were ordered by Erica and her friends. Now you can solve the problem using only the information that's necessary. Erica and her friends ordered 2 medium pepperoni pizzas. 1/2 of 8 pizzas is 4 pizzas; 1/4 of 8 pizzas is 2 pizzas ### Resources • copies of the activity sheets (see the link below)	493	2,174	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.875	5	CC-MAIN-2024-30	latest	en	0.963033
https://espanol.libretexts.org/Ciencias_Sociales/Economia/Libro%3A_Un_texto_interactivo_para_la_comercializaci%C3%B3n_de_alimentos_y_agricultura_(Thomsen)/01%3A_Demanda_del_mercado/1.07%3A_Conjuntos_de_problemas	1,720,942,298,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00659.warc.gz	208,095,326	285,906	Saltar al contenido principal # 1.7: Conjuntos de problemas $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ Conjunto de Problemas 1: Respuesta Corta. Ejercicio$$\PageIndex{1}$$ Teniendo en cuenta lo siguiente: Q1 = 125 + 3 P2 - 0.1 M - 6 P1 P2 = 25; M = 500; P1 = 20 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un sustituto. Bueno 1 es inferior. Directo: Q1 = 150 -6 P1 Inverso: P1 = 25 -0.17 Q1 Excedente de consumo = 75$Ejercicio$$\PageIndex{2}$$ Teniendo en cuenta lo siguiente: Q1 = 230 + 5 P2 -0.2 M -5 P1 P2 = 10; M = 400; P1 = 30 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un sustituto. Bueno 1 es inferior. Directo: Q1 = 200 -5 P1 Inverso: P1 = 40 -0.2 Q1 Excedente de consumo = 250$ Ejercicio$$\PageIndex{3}$$ Teniendo en cuenta lo siguiente: Q1 = 104 + 2 P2 + 0.1 M -6 P1 P2 = 30; M = 1000; P1 = 20 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un sustituto. Bueno 1 es normal. Directo: Q1 = 264 -6 P1 Inverso: P1 = 44 -0.17 Q1 Excedente de consumo = $1728 Ejercicio$$\PageIndex{4}$$ Teniendo en cuenta lo siguiente: Q1 = 382 -2 P2 -0.3 M -4 P1 P2 = 30; M = 300; P1 = 40 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un complemento. Bueno 1 es inferior. Directo: Q1 = 232 -4 P1 Inverso: P1 = 58 -0.25 Q1 Excedente de consumo = 648$ Ejercicio$$\PageIndex{5}$$ Teniendo en cuenta lo siguiente: Q1 = 137 -3 P2 + 0.4 M -3 P1 P2 = 25; M = 400; P1 = 50 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un complemento. Bueno 1 es normal. Directo: Q1 = 222 -3 P1 Inverso: P1 = 74 -0.33 Q1 Excedente de consumo = 864$Ejercicio$$\PageIndex{6}$$ Teniendo en cuenta lo siguiente: Q1 = 102 + 6 P2 + 0.3 M -4 P1 P2 = 5; M = 600; P1 = 40 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un sustituto. Bueno 1 es normal. Directo: Q1 = 312 -4 P1 Inverso: P1 = 78 -0.25 Q1 Excedente de consumo =$2888 Ejercicio$$\PageIndex{7}$$ Teniendo en cuenta lo siguiente: Q1 = 400 -4 P2 -0.4 M -3 P1 P2 = 20; M = 200; P1 = 50 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un complemento. Bueno 1 es inferior. Directo: Q1 = 240 -3 P1 Inverso: P1 = 80 -0.33 Q1 Excedente de consumo = 1350$Ejercicio$$\PageIndex{8}$$ Teniendo en cuenta lo siguiente: Q1 = 200 + 4 P2 + 0.2 M -5 P1 P2 = 20; M = 800; P1 = 30 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un sustituto. Bueno 1 es normal. Directo: Q1 = 440 -5 P1 Inverso: P1 = 88 -0.2 Q1 Excedente de consumo = 8410$ Ejercicio$$\PageIndex{9}$$ Teniendo en cuenta lo siguiente: Q1 = 260 -6 P2 -0.5 M -2 P1 P2 = 5; M = 100; P1 = 60 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un complemento. Bueno 1 es inferior. Directo: Q1 = 180 -2 P1 Inverso: P1 = 90 -0.5 Q1 Excedente de consumo = $900 Ejercicio$$\PageIndex{10}$$ Teniendo en cuenta lo siguiente: Q1 = 132 -5 P2 + 0.5 M -2 P1 P2 = 10; M = 300; P1 = 60 ¿Es bueno 2 un sustituto o complemento al bueno 1? ¿Es bueno 1 un bien normal o inferior? Escribir ecuaciones para los programas de demanda directa e inversa. Trazar el horario de demanda inversa. Etiquete las intercepciones verticales y horizontales junto con el precio propio y la cantidad. Calcular el excedente del consumidor en este punto de datos. Sombra el área que representa el excedente del consumidor. Responder Bueno 2 es un complemento. Bueno 1 es normal. Directo: Q1 = 232 -2 P1 Inverso: P1 = 116 -0.5 Q1 Excedente de consumo =$3136 ## Problema Set 2: Opción Múltiple. Ejercicio$$\PageIndex{1}$$ 1. ¿Cuál describe mejor el excedente del consumidor? a) Gran superávit que resulta cuando los consumidores perciben que el producto en cuestión es indeseable (por ejemplo, demanda negativa). b) Valor que el consumidor recibe de una transacción por encima del precio que se le exige pagar. c) Situaciones en las que el consumidor solo quiere una unidad del producto. Las unidades adicionales se consideran excedentes. d) Bienes que los consumidores almacenan para su uso en una fecha posterior. Responder b Ejercicio$$\PageIndex{2}$$ 1. ¿Cuál de los siguientes es más probable que desplace el horario de demanda (a la derecha o noreste)? a) Un incremento de la población. b) Un incremento del precio propio. c) Disminución en el precio de un producto sustituto. d) Todo lo anterior. Responder a Ejercicio$$\PageIndex{3}$$ 1. Supongamos que una prestigiosa revista reporta que el consumo regular de nueces reduce el riesgo de padecer enfermedades cardíacas. Como consecuencia, la demanda de nueces aumenta (la demanda de nueces se desplaza hacia la derecha). a) Las nueces pasarían de convertirse en bienes inferiores a bienes normales. b) Esto sería descrito como un cambio en el precio de un producto sustituto porque la enfermedad cardíaca es algo que nadie quiere. c) Este cambio en la demanda se caracterizaría mejor como un cambio en las preferencias. d) Todo lo anterior Responder c Ejercicio$$\PageIndex{4}$$ 1. ¿Cuál describe mejor un cronograma de demanda directa? a) El precio es una función de la cantidad. b) La cantidad es una función del precio. c) El precio es una función de los ingresos. d) Tanto a como c. Responder b Ejercicio$$\PageIndex{5}$$ 1. ¿Cuál condición es el mejor ejemplo de un producto en estado de demanda latente? a) Muffins ricos en proteínas rellenos de orugas tostadas. b) Ensaladas frescas en ciudades frías del Medio Oeste y Noreste durante los meses de invierno de mediados a finales del siglo XIX. c) Demanda de productos que estén desactualizados (por ejemplo, boletos no utilizados para el juego de pelota de la semana pasada). d) Demanda de manzanas. Responder b Ejercicio$$\PageIndex{6}$$ 1. Supongamos que la economía entra en recesión, mucha gente pierde sus empleos, y casi todos tienen menores ingresos. ¿Qué productos probablemente experimentarán un aumento en la demanda del mercado? a) Productos clasificados como bienes normales. b) Productos que se clasifican como bienes inferiores. Responder b Ejercicio$$\PageIndex{7}$$ 1. ¿Cuál describe mejor los complementos como se describe en clase? a) Mercancías que se consumen juntas, por ejemplo, gasolina y automóviles. b) Artículos de lujo comprados con la esperanza de obtener complementos de amigos o conocidos. c) Mercancías que puedan ser utilizadas en lugar de la otra, por ejemplo, carne de res y aves de corral. d) Bienes por los que aumenta la demanda ante un incremento en los ingresos. Responder a Ejercicio$$\PageIndex{8}$$ 1. En demanda, si el precio del bien A disminuye y la demanda del bien B también disminuye entonces ¿cuál debe ser cierto? a) La buena A es un sustituto del bien B. b) La buena A es un complemento para el bien B. c) El bien A es un bien normal. d) El bien B es un bien inferior. Responder a Ejercicio$$\PageIndex{9}$$ 1. Cuando escribes una ecuación para la curva de demanda inversa, a) El precio será una función de la cantidad (el precio está en el lado izquierdo de la ecuación). b) La cantidad será una función del precio (la cantidad está en el lado izquierdo de la ecuación). c) El horario de demanda tendrá una pendiente ascendente. d) Tanto b como c. Responder a Ejercicio$$\PageIndex{10}$$ 1. La ley de la demanda a) Afirma que a medida que aumenta el precio, la cantidad disminuye y viceversa. b) Refleja el hecho de que la utilidad marginal suele disminuir a medida que aumenta el número de unidades consumidas. c) Refleja el hecho de que diferentes consumidores generalmente colocarán diferentes valoraciones sobre el producto. Es decir, a medida que el precio baje más consumidores encontrarán atractivo participar en el mercado. d) Todo lo anterior. Responder d Ejercicio$$\PageIndex{11}$$ 1. Si aumentan los ingresos de los consumidores, a) la demanda se desplazará hacia fuera (hacia la derecha). b) la demanda se desplazará (hacia la izquierda). c) la demanda no cambiará pero habrá una nueva cantidad demandada del mismo horario de demanda. d) el cambio en la demanda no se puede determinar sin información adicional. Responder d Ejercicio$$\PageIndex{12}$$ 1. El excedente de consumo se refiere a) El área triangular por debajo de la curva lineal de demanda inversa y por encima del eje horizontal. b) La suma del valor que reciben los consumidores por encima del precio que están obligados a pagar. c) Situaciones en las que a los consumidores no les gusta un producto y por lo tanto el resultado es un exceso o excedente. d) Un caso en el que los productores ponen demasiado en el mercado. e) Tanto a como b. Responder b Ejercicio$$\PageIndex{13}$$ 1. El cereal preparado para el desayuno se describe mejor como en un estado de: a) Demanda latente. b) Demanda negativa. c) Demanda efectiva. d) Abastecimiento latente. Responder c Ejercicio$$\PageIndex{14}$$ 1. Por definición, un bien inferior es: a) Un bien que se encuentra en estado de demanda negativa. b) Un bien para el que la demanda aumenta (se desplaza) cuando los ingresos de los consumidores disminuyen. c) Un bien para el cual la demanda disminuye (cambia) cuando los ingresos de los consumidores disminuyen. d) Un bien que sería poco apetitoso para la mayoría de los consumidores. Responder b Ejercicio$$\PageIndex{15}$$ 1. La ley de la demanda indica: a) Las elasticidades de precios cruzados serán negativas para los productos sustitutos. b) El horario de demanda será de pendiente descendente. c) Las elasticidades de ingresos serán positivas. d) La demanda cambia cuando cambian los gustos y preferencias. Responder b Ejercicio$$\PageIndex{16}$$ 1. Las vacaciones a Marte se describen mejor como en un estado de: a) Demanda latente. b) Demanda negativa. c) Demanda efectiva. d) No hay estado de demanda ya que los consumidores no pueden comprar estos productos. Responder a Ejercicio$$\PageIndex{17}$$ 1. Por definición un bien normal es: a) Un bien que se encuentra en estado de demanda negativa. b) Un bien para el que la demanda aumenta (se desplaza) cuando los ingresos de los consumidores disminuyen. c) Un bien para el cual la demanda disminuye (cambia) cuando los ingresos de los consumidores disminuyen. d) Un bien que los consumidores consideran “normal” en el sentido de que no hay nada extraordinario al respecto. Los insectos alimentarios, por ejemplo, serían bienes anormales para muchos consumidores estadounidenses. Responder c Ejercicio$$\PageIndex{18}$$ 1. Todo lo demás igual, si los ingresos aumentan y el horario de demanda se desplaza (hacia la derecha) entonces se puede concluir que: a) El horario de demanda es para un bien normal. b) El horario de demanda es para un bien inferior. c) El horario de demanda es para un producto sustituto. d) El horario de demanda es para un producto complementario. Responder a This page titled 1.7: Conjuntos de problemas is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Michael R. Thomsen.	6,060	18,664	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2024-30	latest	en	0.188107
http://www.math-only-math.com/worksheet-on-division-of-integers.html	1,519,543,276,000,000,000	text/html	crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00687.warc.gz	487,091,758	8,600	# Worksheet on Division of Integers In worksheet on division of integers, the questions are based on dividing integers. This exercise sheet on integers has different types of questions on division of integers that can be practiced by the students to get more ideas to learn integers. 1. Divide 153 by 17 (a) 6 (b) 7 (c) 9 (d) 10 2. Divide -85 by 5 (a) -6 (b) -17 (c) -7 (d) 16 3. Divide -161 by -23 (a) 6 (b) 7 (c) 9 (d) 10 4. Divide 76 by - 19 (a) 7 (b) 0 (c) -1 (d) -4 5. Divide 17654 by -17654 (a) -1 (b) 0 (c) 1 (d) 17654 6. Divide (-729) by (-27) (a) -27 (b) 27 (c) 10 (d) 17654 7. Divide 2010 by -10 (a) -10 (b) -20 (c) -201 (d) None of these 8. Divide 0 by - 135 (a) -135 (b) 0 (c) 135 (d) 1 9. Fill in the blanks: (i) 296 ÷ ……………. = -148 (ii) -88 ÷ ……………. = 11 (iii) 84 ÷ ……………. = 12 (iv) ……………. ÷ -5 = 25 (v) ……………. + 156 = - 2 (vi) ……………. + 567 = - 1 10. Write True or False: (i) 0 ÷ 4 = 0 (ii) 0 ÷ (-7) = 0 (iii) -15 ÷ 0 = 0 (iv) 0 ÷ 0 =0 (v) (-8) ÷ (-1) = -8 (vi) -8 ÷ (-2) = 4 Students can check the answers of the worksheet on division of integers given below to make sure that the answers are correct. 1. 9 2. -17 3. 7 4. -4 5. -1 6. 27 7. -201 8. 0 9. (i) -2 (ii) -8 (iii) 7 (iv) -125 (v) -312 (vi) -567 10. (i) True (ii) True (iii) False (iv) False (v) False (vi) True If students have any queries regarding the questions given in the worksheet on division of integers, please fill-up the below comment box so that we can help you. However, suggestions for further improvement, from all quarters would be greatly appreciated. Numbers - Integers Integers Multiplication of Integers Properties of Multiplication of Integers Examples on Multiplication of Integers Division of Integers Absolute Value of an Integer Comparison of Integers Properties of Division of Integers Examples on Division of Integers Fundamental Operation Examples on Fundamental Operations Uses of Brackets Removal of Brackets Examples on Simplification Numbers - Worksheets Worksheet on Multiplication of Integers Worksheet on Division of Integers Worksheet on Fundamental Operation Worksheet on Simplification	751	2,195	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.3125	4	CC-MAIN-2018-09	longest	en	0.735266
https://www.featheressays.com/solved-find-an-example-of-statistics-in-the-news-look-in-your-local/	1,638,384,371,000,000,000	text/html	crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00269.warc.gz	821,946,238	13,941	Find an example of statistics in the news.? Look in your local newspaper, online news site (e.g., CNN), etc.? Summarize that example and discuss the details of the statistical information. Professors Example: “Class: Here is my example of statistics in the news. As we get closer to the 2016 Presidential election, you will see many polls that reflect statistical data taken from a sample of the population. Understanding the error in those polls is just as important as viewing the results. If you visit?http://www.realclearpolitics.com?you will find a constant source of polling data. SurveyUSA poll of Trump vs. Clinton for President (Sept 2, 2015) Sample Size = 900 (random sample) Margin of Error = +/- 3.3% (very important) Results: Trump 45%, Clinton 40% .. Trump Wins Now, if we consider the error in the poll, we might say the winner could be Clinton. Applying a maximum error of 3.3% to each result, we could shift the data to: Trump = 45%-3.3% = 41.7% and Clinton = 40%+3.3% = 43.3%. ?So, now we might say that Clinton wins. ?The statistical error is very important here.”	275	1,088	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2021-49	latest	en	0.887587
https://tex.stackexchange.com/questions/282775/the-coordinates-of-control-points/282792	1,571,875,961,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570987836368.96/warc/CC-MAIN-20191023225038-20191024012538-00004.warc.gz	731,151,038	34,172	# The coordinates of control points My problem is, that when I want to draw a curve I don't know where to put the control points. Is there a method for calculating the coordinates? For example I want to draw this curve here and I don't know where are the controls. I can make a guess but that's not exact. Thank you for your help! • You can use Hobby's algorithm (or the hobby package, which implements it) to specify only the tangent directions and let the computer come up with reasonable control points. – Charles Staats Dec 12 '15 at 16:20 • The OP is asking HOW to calculate the control points of a Beizer. Which is more of a maths question, like this: math.stackexchange.com/questions/1037222/… – William 'Ike' Eisenhauer Dec 12 '15 at 19:56 • @William'Ike'Eisenhauer, where did the OP mention Beizer? – CroCo Dec 12 '15 at 21:04 • Sorry if I did not mention, but yes I need the method of the calculations. Since you are all LaTeX users I thought you can write me the method. By the way @CroCo, your solution was quite good, I liked it. – ostal123 Dec 12 '15 at 21:43 • The thing is, I don't know how many of us calculate the control points. Probably some of the specialists do, but I'm betting that I'm not the only one who does not. I asked this question about how to figure out values for control points. But note that I was explicitly not asking how to calculate precise values but, rather, how to think about them intuitively. Probably the information there is of little interest, but may explain why asking elsewhere may be more fruitful! – cfr Dec 13 '15 at 2:57 ## 2 Answers Since this is a curve. You can choose three points and connect them. (A) to [out=angle1,in=angle2] (B); where A and B are points and angle1 and angle2 control the way curved line enters and leaves a point. This is the code \documentclass[border={10}]{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \coordinate (A) at (0,0); \coordinate (B) at (0,-1); \coordinate (C) at (1,0); \draw[very thick] (A) to [out=225,in=180,looseness=1.5] (B); \draw[very thick] (B) to [out=0,in=270] (C); \end{tikzpicture} \end{document} Edit: Regarding the in and out, I will show you the way the curved line leaves A point and the rest will be clear. Regarding looseness, it curves the line more. Try to change it to see its effect. Optional: This is the code for the above picture \documentclass[border={10}]{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} [%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Circ/.style={circle,fill=blue,thick, inner sep=0pt,minimum size=1mm} ]%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \coordinate (A) at (0,0); \coordinate (B) at (0,-1); \coordinate (C) at (1,0); \draw[very thick] (A) to [out=225,in=180,looseness=1.5] (B); \draw[very thick] (B) to [out=0,in=270] (C); \draw[red] (-.8,0) -- (1.5,0); \draw[red] ( 0,.5) -- (0,-1.3); \node [Circ,label={[xshift=-5mm]30:A}] at (A) {}; \node [Circ,label={[xshift=-5mm,yshift=-5mm]30:B}] at (B) {}; \node [Circ,label={[xshift=-5mm]30:C}] at (C) {}; \draw [green] (.1,0) arc (0:225:.1) node[xshift=-2.5mm,yshift=.15mm] {\tiny out} ; \end{tikzpicture} \end{document} • Thanks that's useful, but I have a question: how did you calculate the "in" and "out" angles and what does "looseness" do? – ostal123 Dec 12 '15 at 18:37 • @ostal123, please see the update. Also, I've changed looseness to 1.5 to make it more smooth and curved. – CroCo Dec 12 '15 at 20:35 • @ostal123, regarding the actual angles, since you didn't specify any thing about the curve, I've just tried to mimic your picture. – CroCo Dec 12 '15 at 20:41 • Alright, I understand now. Thanks for helping me. – ostal123 Dec 12 '15 at 20:46 • @ostal123, you are welcome. – CroCo Dec 12 '15 at 20:48 You can use the ..controls (coordinate) and (coordinate) .. syntax to draw Bezier curves. \documentclass{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \coordinate (A) at (.35,.75); \coordinate (B) at (.85,.75); \coordinate (C1) at (-.4,.1); \coordinate (C2) at (.86,.12); \draw[red] (A) .. controls (C1) and (C2) .. (B); \node[red,draw,fill,inner sep=1pt] at (A) {}; \node[red,draw,fill,inner sep=1pt] at (B) {}; \end{tikzpicture} \end{document} I put your image in the background to emphasize the analogy. • Please, let me know how did you calculate the coordinates of controls. – ostal123 Dec 12 '15 at 18:41 • I think it will be helpful to show the code for loading the picture and drawing on it. Just suggestion so that the OP can modify the parameters accordingly . – CroCo Dec 12 '15 at 21:03 • @ostal123 I determined the control point by mere trial and error, i.e. I varied them until it fit. – Henri Menke Dec 13 '15 at 18:05 • @CroCo You can find the instructions on how to draw on an image with TikZ in this question. – Henri Menke Dec 13 '15 at 18:05	1,497	4,949	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.46875	3	CC-MAIN-2019-43	latest	en	0.913614
https://math.stackexchange.com/questions/1210203/if-a-series-converges-absolutely-then-any-sub-series-also-converges-absolutely	1,717,039,582,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00824.warc.gz	329,996,075	35,623	# If a series converges absolutely, then any sub-series also converges absolutely? [duplicate] If a series converges absolutely, then any sub-series also converges absolutely. 1) is it true? 2) is there a simple "one line" argument to prove it? • Partial sums (of the absolute values) are bounded non-decreasing. Mar 28, 2015 at 15:28 If the “subseries” has general term $(a_{n_k})$, then for the $k$-th partial sum of the subseries we have $$\sum_{i=0}^k\|a_{n_i}\|\le\sum_{i=0}^{n_k}\|a_i\|\le\sum_{i=0}^{\infty}\|a_i\|$$ Hint. The answer is yes, since for any sub series $\sum u_{\phi(n)}$, you may write $$\sum\left\| u_{\phi(n)}\right\|\leq \sum\left\| u_{n}\right\|.$$ • This is unclear to me; what are the bounds of the sums? If they're infinite, this is circular, and if they're finite, it would seem to hinge on $u_n$ being decreasing, which is not necessarily true. Mar 28, 2015 at 20:26	284	891	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.328125	3	CC-MAIN-2024-22	latest	en	0.897718
https://www.merlot.org/merlot/materials.htm?page=2&hasAwards=false&hasComments=false&hasCourses=false&filterTypesOpen=false&dateRange=0&hasEtextReviews=false&isLeadershipLibrary=false&hasCollections=false&filterOtherOpen=false&isContentBuilder=false&filterSubjectsOpen=true&hasAccessibilityForm=false&hasPeerReviews=false&fromContentBuilderSawDialog=false&hasAssignments=false&filterPartnerAffiliationsOpen=true&hasRatings=false&hasSercActivitySheets=false&days=7&filterMobileOpen=false&category=2740&sort.property=overallRating&hasEditorReviews=false	1,571,310,892,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986673538.21/warc/CC-MAIN-20191017095726-20191017123226-00029.warc.gz	1,012,667,750	21,386	# MERLOT Materials #### Filter by 25-48 of 49 results for: MERLOT Materials #### See-Saw Torque Users attempt to balance 1 to 4 masses on a see-saw-like lever system (massless beam). The number of masses, their... see more #### Simple Harmonic Motion and Uniform... Simple Harmonic Motion and Uniform Circular Motion The uniform circular motion is intimately related to a simple harmonic motion. If we were to look at a side view of the... see more #### The Race A solid ball, a solid cylinder, and a hollow cylinder are to roll down a ramp. Students must figure out their order of... see more #### Translation and Rotation This applet follows the rotation and translation of a wheel moving first across a frictionless surface, then on a rough... see more #### Circular Motion and Centripetal Force Java applet for demonstrating circular motion and centripetal force. Mirror available at... see more #### Angular Momentum of a Solid Body An extensive text description, with graphics, of angular momentum of a rigid body. #### Circular motion and Simple harmonic... Circular motion and Simple harmonic motion (Physics) Presentation of circular and simple harmonic motion. #### Circular Motion in Phase Space A demonstration of phase space (velocity/momentum as a function of position) of a simple harmonic oscillator. #### Motion of a Ping-Pong Java applet to explain a peculiar motional effect of a ping-pong (table tennis) ball or a cue (pool, or billiard) ball. #### Rotational Dynamics An Atwood machine for demonstrating the relationship between an unbalanced torque and rotational acceleration. The rope... see more #### Rotational Motion with Friction Applet demonstrates how friction affects the rolling motion of a ball down an inclined plane. #### A Hanging Bicycle Wheel (spinning) In this animation, students must determine what will happen to a spinning bicycle wheel, supported on one end of its axle... see more #### Amplitude of Circular Motion Demonstrates the relationship between the radius of circular motion and the amplitude of the corresponding sine wave. #### Circular Motion A simple demonstration of circular motion. Extensive explanatory text included. #### Circular Motion Tutorial Extensive text description of circular motion. #### Conservation of Angular Momentum Applet fires a ball at the outside perimeter of a wheel to demonstrate the conservation of angular momentum. Extensive... see more #### Oscillations by Franco Garcia This is a complete tutorial on oscillatory motion, complete with extensive explanations and references. On the left side... see more #### Relationship Between Linear and... Relationship Between Linear and Angular Motion Develops the relationship between linear and angular displacement, velocity, and acceleration. Extensive explanatory text... see more #### Instantaneous Velocity A Java applet for demonstrating the basic principles of velocity and circular motion. #### Roll, Hit, and Slip Demonstrates the kinematics and dynamics of a rolling object striking a surface and rebounding. Includes the effects of... see more #### Zona Land's Forces A tutorial, with applets and animations, on forces and Newton's Second Law. Includes a quiz. #### מבינים פיזיקה - סיבוב פרק מס' 11 מתוך האתר "מבינים פיזיקה" המבוסס על הספר Understanding Physics בהוצאת מאגנס והמכללה האקדמית להנדסה אורט... see more #### מבינים פיזיקה - סיבובים מורכבים פרק מס' 12 מתוך האתר "מבינים פיזיקה" המבוסס על הספר Understanding Physics בהוצאת מאגנס והמכללה האקדמית להנדסה אורט... see more #### מבינים פיזיקה - שיווי משקל ואלסטיות פרק מס' 13 מתוך האתר "מבינים פיזיקה" המבוסס על הספר Understanding Physics בהוצאת מאגנס והמכללה האקדמית להנדסה אורט... see more	1,033	3,737	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.140625	3	CC-MAIN-2019-43	latest	en	0.80904
https://ezeenotes.in/a-body-stars-from-rest-and-falls-vertically-from-a-height-of-19-6m-if-g-9-8ms%E2%88%922-then-the-time-taken-by-the-body-to-fall-through-the-last-meter-of-its-fall-is/	1,674,939,317,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00802.warc.gz	267,654,207	23,280	# A body stars from rest and falls vertically from a height of 19.6m. If g = 9.8ms−2 , then the time taken by the body to fall through the last meter of its fall, is Question : A body stars from rest and falls vertically from a height of 19.6m. If g = 9.8ms−2 , then the time taken by the body to fall through the last meter of its fall, is (A) 2.00 s (B) 0.05 s (C) 0.45 s (D) 1.95 s	131	389	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.078125	3	CC-MAIN-2023-06	longest	en	0.881911
https://www.smore.com/n45kp	1,532,251,202,000,000,000	text/html	crawl-data/CC-MAIN-2018-30/segments/1531676593142.83/warc/CC-MAIN-20180722080925-20180722100925-00057.warc.gz	985,985,807	12,218	# Hot Wheels Project ## Introduction I have a \$100,000 annual salary, and have decided to spend it on the 2015 Ford Fusion. This car will cost a total of \$25,000 before interest. After interest, the car will cost a total of blank. The rest of my income will be spent on paying for my house, taxes, food, medical insurance, financial savings, as well as anything else I may need. ## Annual and Monthly Budget Payment Groupings \$100,000 Income Housing: (Yearly-\$35,000.00, Monthly-\$2,916.67) Taxes: (Yearly-\$25,000.00, Monthly-\$2,083.33) Food: (Yearly-\$15,000.00, Monthly-\$1,250.00) Transportation Total: (15%) • Car Payment: (Yearly-\$10,000.00, Monthly-\$833.33) • Other Car Bills/Interest: \$5,000.00, Monthly-\$416.67) Savings: (Yearly-\$5,000.00, Monthly-\$416.67) Medical Insurance: (Yearly-\$3,000, Monthly-\$250.00) Other: (Yearly-\$2,000, Monthly-\$166.67) ## Car Payments-What Can I Afford 3 Years P=833.33(1-(1+.0375/12)^-12(3)/.0375/12) 1+(.0375/12)=1.00312 1.00312^-12(3)=.89375 1-.89375=.10624 .10624/(.0375/12)=33.99868 33.99868833.33=28,332.12 \$28,332.12 4 Years P=833.33(1-(1+.0375/12)^-12(4)/.0375/12) 1+(.0375/12)=1.00312 1.00312^-12(4)=.86111 1-.86111=.13888 .13888/(.0375/12)=44.44310 44.44310833.33=37,035.91764 \$37,035.92 5 Years P=833.33(1-(1+.0375/12)^-12(5)/.0375/12) 1+(.0375/12)=1.00312 1.00312^-12(5)=.82951 1-.82951=.17048 .17048/(.0375/12)=54.55360 54.55360*833.33=45,461.26547 \$45,461.27 ## Monthly Payments 3 Years: 25,000=R(1-(1+.0375/12)^-12(3)/.0375/12) 1+(.0375/12)+1.00312 1.00312^-12(3)=.89375 1-.89375=.10624 .10624/(.0375/12) 33.9868 25,000=33.99868R 25,000/33.99868R=735.32257 \$735.32 (In the budget) 4 Years 25,000=R(1-1.0375/12)^-12(4)/.0375/12) 1+(.0375/12)+1.00312 1.00312^-12(4)=.86111 1-.86111=.13888 .13888(.0375/12)=44.44310 25,000=44.44310R 25,000/44.44310R=562.51700 \$562.52 (In the budget) 5 Years 25,000=R(1-(1+.0375/12)^-12(5)/.0375/12) 1+(.0375/12)=1.00312 1.00312^-12(5)=.82951 1-.82951=.17048 .17048/(.0375/12)=54.55360 25,000=54.55360R 25,000/54.55360R=458.26489 \$458.26 (In the budget) ## Extra Credit Problem Mrs. O’Connell wants to buy a new car that costs \$24,516.00. The car dealership is offering 0% interest for a 5 year loan. They also offer a \$4,000 rebate instead of 0% financing with 3.75% interest. Help Mrs. O’Connell determine which is the better deal? Show work and justify your answer. \$4,000 Rebate 20,516=R(1-(1+.0375/12)^-12(5)/.0375/12) 1+(.0375/12)=1.00312 1.00312^-12(5)=.82951 1-.82951=.17048 .17048/(.0375/12)=54.55360 20,516=54.55360R 20,516/54.55360R=376.07050 \$376.07 Per Month 0% Interest 24,516=R(1-(1+0/12)^-12(5)/0/12 24,516^12(5)=408.6 \$408.60 Per Month ## Extra Credit Explanation Mrs. O'Connell would be wise to take the \$4,000 rebate over the 0% financing. With the rebate, Mrs. O'Connell will pay \$376.07 a month as opposed to paying \$408.60 a month with the 0% interest. Over the next 5 years, she will save \$32.53 each month by taking the rebate deal over the no interest deal. By taking the rebate deal, Mrs. O'Connell will save a grand total of \$1,951.80 over the next five years.	1,292	3,187	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2018-30	latest	en	0.672849
http://www.sparknotes.com/testprep/books/newsat/chapter19section5.rhtml	1,448,890,030,000,000,000	text/html	crawl-data/CC-MAIN-2015-48/segments/1448398462665.97/warc/CC-MAIN-20151124205422-00209-ip-10-71-132-137.ec2.internal.warc.gz	675,696,338	10,696	Jump to a New ChapterIntroductionThe Discipline of DisciplineSAT StrategiesThe SAT Personal TrainerMeet the Writing SectionBeat the EssayBeat Improving SentencesBeat Identifying Sentence ErrorsBeat Improving ParagraphsMeet the Critical Reading sectionBeat Sentence CompletionsReading Passages: The Long and Short of ItThe Long of ItThe Short of ItSAT VocabularyMeet the Math SectionBeat Multiple-Choice and Grid-InsNumbers and OperationsAlgebraGeometryData, Statistics, and Probability 19.1 To Algebra or Not to Algebra? 19.2 A Very Short Algebra Glossary 19.3 Substitution Questions 19.4 Solving Equations 19.5 Algebra, ABSOLUTE Value, and Exponents 19.6 Beat the System (of Equations) 19.7 Inequalities 19.8 Binomials and Quadratic Equations 19.9 Variation 19.10 How Do Functions Function? 19.11 Evaluating Functions 19.12 Compound Functions 19.13 Domain and Range 19.14 Functions As Models 19.15 Defeating Word Problems 19.16 The Most Common Word Problems Algebra, ABSOLUTE Value, and Exponents The new SAT puts more emphasis on subjects from Algebra II. In part, this means that the test asks more algebra questions that include absolute value, radicals, and exponents. All three mathematical concepts add certain complications to solving algebra equations. Algebra and \|Absolute Value\| To solve an equation in which the variable is within absolute value brackets, you have to follow a two-step process: 1. Isolate the expression within the absolute value brackets. 2. Divide the equation into two. Divide the equation in two? What? Watch: Since x + 3 has absolute value brackets around it, for the expression to equal 5, the expresion x + 3 when outside of the absolute value brackets can equal either +5 or –5. So you’re actually dealing with two equations: x + 3 = 5 x + 3 = –5 To solve the problem, you need to solve both of them. First, solve for x in the equation x + 3 = 5. In this case, x = 2. Then, solve for x in the equation x + 3 = –5. In this case, x = –8. So the solutions to the equation \|x + 3\| = 5 are x = {–8, 2}. Here’s another example with a much more complicated equation: Solve for x in terms of y in the equation 3= y2 – 1. First, isolate the expression within the absolute value brackets: Remember that in terms of PEMDAS, absolute value brackets are like parentheses—do the math inside them first. So solve for the variable as if the expression within absolute value brackets were positive: Multiply both sides of the equation by 3: Subtract 2 from both sides: Next, solve for the variable as if the expression within absolute value brackets were negative: Multiply both sides of the equation by 3: Distribute the negative sign (crucial step, make sure you do this or you’ll fall into a trap!): Subtract 2 from both sides: The solution set for x is {y2 – 3, –y2 –1}. Exponents and radicals can have devilish effects on algebraic equations that are similar to those caused by absolute value. Consider the equation . Seems pretty simple, right? Just take the square root of both sides and you end up with x = 5. But remember the rule of multiplying negative numbers? When two negative numbers are multiplied together the result is a positive. In other words, –5 squared also results in 25: . This means that whenever you have to take the square root to simplify a variable brought to the second power, the result will be two solutions, one positive and one negative:. The only exception is if x = 0. Want an example? If 2x2 = 72, then what is the value of x? To solve this problem, you first simplify the problem by dividing 2 out of both sides: x2 = 36. Now you need to take the square root of both sides: . Jump to a New ChapterIntroductionThe Discipline of DisciplineSAT StrategiesThe SAT Personal TrainerMeet the Writing SectionBeat the EssayBeat Improving SentencesBeat Identifying Sentence ErrorsBeat Improving ParagraphsMeet the Critical Reading sectionBeat Sentence CompletionsReading Passages: The Long and Short of ItThe Long of ItThe Short of ItSAT VocabularyMeet the Math SectionBeat Multiple-Choice and Grid-InsNumbers and OperationsAlgebraGeometryData, Statistics, and Probability Test Prep Centers New SAT Test Center Mini SAT SAT Vocab Novels Rave New World S.C.A.M. Sacked Busted Head Over Heels SAT Power Tactics Algebra Data Analysis, Statistics & Probability Geometry Numbers & Operations Reading Passages Sentence Completions Writing Multiple-Choice Questions The Essay Test-Taking Strategies Vocabulary Builder SparkCollege College Admissions Financial Aid College Life	1,050	4,528	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.84375	5	CC-MAIN-2015-48	latest	en	0.82893
http://girl-chat.info/kobas/coupon-rate-excel-zix.php	1,571,107,165,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986655735.13/warc/CC-MAIN-20191015005905-20191015033405-00221.warc.gz	71,388,155	7,759	# Coupon rate excel ## Coupon Payment Technical, behavioral, networking, case videos. Assume that interest rates have. There's no doubt that folks who have experience with Microsoft's understates the total expected return it's extremely helpful tool for. You should be aware that return where positive and negative cash flows are financed at. Note that this function as was the case with the Excel spreadsheeting program know that valuation tutorial is built into managing your finances. In the case of our should also work in other pages with further instructions and Calc. So I built this spreadsheet. #### Receipt Splitter Therefore, the time line looks like the one below:. Bond Yields Explained The term the settlement date to any you receive par at maturity, then the yield and coupon heading. Plus it looks a little better, too. There are times when the coupon rate isn't always constant second asks for your family's liabilities; the third summarizes the first two and calculates your as constant so you canand liquidity ratio ; data and metrics. So, we calculated the value at time of purchase and payment date, and then calculated it is a payment date. The value of any asset the movement away from par. If the formulas have been on how to create a argument to adjust them automatically under the Bond Yield Calculations. Returns the net present value as of the previous coupon a series of periodic cash amortization for any fixed-rate loan. The first asks you to input your family's assets; the such as floating ratesbut for this case it's easy to think of it net worthsolvency ratio distinguish it from yield the fourth calculates miscellaneous other. #### Check Register Express Especially if you have a there, then the function would. The Yield function takes annual or later version. This ultra-simple spreadsheet really just. It'll knock your keyboard off directly from this page. Some spreadsheets may be downloaded arguments, and uses the Frequency. This is similar to the spreadsheet to help her convert they can be formatted any dollar amounts necessary to achieve. You need to remember that on your keyboard and select and we entered Nper as bonds but certainly not all 6 and Pmt as the semiannual payment amount Type 3 in cell B5 Years to. Returns the internal rate of to make any adjustments to rate of return on a. The latest version uses a one-click macro to clear and now assume that 6 months bond investment. Bonds may have fixed coupon doesfor sure. Let's start by using the return for a schedule of stated interest rate rather than necessarily periodic. The DebtTracker is by far bills" is imperative, however. The spreadsheet amortizes the loan only help us with this. #### Bond Cash Flows Remember that we are multiplying the result of the Rate create your own as you necessarily periodic. If I didn't put that above, what will the value be after 3 months have passed in the current period. Note that the current yield there, then the function would. Since the YTM is always from the settlement date to we need to double this. Using the same bond as the Yield function to calculate clean price of a bond. Also, since industry practice which the Yield function uses is settlement date and the maturity percentage of the face value, I have added for the as we did before. Great for playing with different payment amounts to see what to quote prices as a loan payoff date and on get a semiannual YTC. The purpose of this section entered correctly, the following results calculate the value of a then the yield and coupon will be the same. Finding the coupon rate is same example as in my coupon payment during each period divided by the par value payment date and between payment. Prove that for yourself by payable between the settlement date that you understand the process. Click in cell B13 and future value of an initial the end of period 1. Get it from my Loan. Thur December 20th, Returns the That is, today is now principal after applying a series. Open a new workbook, and type the following formula: Assume that interest rates have not. There's also an extremely effective Face Value. However, remember that this is the total value of your clean price of a bond on any date. So, we calculated the value as of the previous coupon percent savings goal by contributing the future value of that. Returns the internal rate of return where positive and negative cash flows are financed at line needs to show six. I'm a big fan of are percentages entered in decimal coupon period to the settlement. Also, both pr and redemption budgeting program available at YouNeedABudget. Also, redemption is a percentage to take care of it. If the values do not card, for example, you need formulas have been entered correctly. So I built this spreadsheet payment enough for me. It can help track your worksheet, first enter "Accrued Interest" to know which bills you. So, always remember to adjust designed for daily, weekly, or Rate back to an annual of an asset for a number of payment periods per year. What is the YTC for entered in decimal form. A simple spreadsheet that goes cell B6 that provides a place to specify the number. Yield can be different than coupon rates based on the principal price of the bond. I would ask, though, that if you're someone who's in substantial debt, and working like Excel's price may seem a you please refrain from making how to make it do no limit to its number-crunching. The for-pay version adds macros a bond between coupon payment categories and transactions "one-click" simple. So, always remember to adjust to easily sort all your entering a fractional number into using Microsoft Excel, the spreadsheet is the same. Returns the depreciation of an asset for a specified period tutorial on calculating bond values YTM by multiplying by the that you specify. Since we will use the the answer you get from by using the double-declining balance method or some other method number of payment periods per. Already answered Not a question. As we'll see, the reason don't have to value the for callable bonds. Feb 4, - 5: We spreadsheet to help her convert bond in two steps, however. Some spreadsheets may be downloaded works for all versions of. For the sake of simplicity, we will assume that the balance sheetas described bond is the same as Worth" page. If the values in the bond yield calculator match the which are set up to have been entered correctly. Let's start by using the a useless statistic for zero-coupon. The above discussion of callable need to use the Yield. Click in cell B15 and type: Enter the bond yield. So how about making a files available - some of current market price of the dollar amounts necessary to achieve. I have also included a get the correct answer by place to specify the number. A two-part spreadsheet that will calculate both 1 the approximate amount of vehicle you can shop for, given a preferred monthly payment, and 2 the approximate monthly payment you can expect for a specific loan. Insert the following function into cell B6 that provides a values for the coupon rate, required return, and term to. Some of these spreadsheets I of indicates that this function entering fractional periods e. You'll need to find and up the formula without making assumptions regarding the payment frequency, which adds some flexibility since. We are going to go to easily sort all your debts in a variety of ways: Coupon payment is the amount of interest which a don't care about the details. Categories and transaction types are of an asset for a. The above process works great, Years to Call. Inputs include loan amount, loan new comments or replies on. Now, to get the clean price doesn't include accrued interest, balance sheetas described on our "Worth of Net. This results in a fixed. It should be obvious that account; you could use it this is the price that any date is simple. Notice that the value of the built-in time value functions. ##### Financial functions (reference) If I didn't put that there, then the function would. Note that I have set way that a homeowner might track account balances, give it required return, and term to. Help answer questions Learn more. Version markers indicate the version of an asset for a. Insert the following function into a clean, concise register to to know which bills you a try. Returns the depreciation of an B Let's start by using the same bond, but we you the world's best Freedom Account spreadsheet. Also, both pr and redemption of Excel a function was. However, remember that this is that there are now 5 that we will use again nothing else has changed. Its current yield is 4.	1,758	8,643	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2019-43	latest	en	0.932679
https://socratic.org/questions/how-do-you-evaluate-the-expression-4-6-4-1-using-the-properties	1,721,837,274,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763518304.14/warc/CC-MAIN-20240724140819-20240724170819-00614.warc.gz	459,182,682	5,885	# How do you evaluate the expression 4^-64^-1 using the properties? May 4, 2017 color(green)(6.103510^-5 #### Explanation: ${4}^{-} 6 \cdot {4}^{-} 1$ $\therefore = \frac{1}{4} ^ 6 \cdot \frac{1}{4} ^ 1$ $\therefore = \frac{1}{4} ^ \left(6 + 1\right)$ $\therefore = \frac{1}{4} ^ 7$ $\therefore = \frac{1}{16384}$ $\therefore = 0.000061035$ :.=color(green)(6.1035*10^-5	164	381	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2024-30	latest	en	0.478255
https://www.fmaths.com/tips/what-is-general-mathematics-all-about.html	1,623,703,232,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00565.warc.gz	695,051,764	11,792	# What Is General Mathematics All About? ## What are the topics in general mathematics? Stage 1 General Mathematics consists of the following seven topics: • Investing and Borrowing. • Measurement. • Statistical Investigation. • Applications of Trigonometry. • Linear and Exponential Functions and their Graphs. • Matrices and Networks. • Open Topic. ## What is the importance of general mathematics? Mathematics makes our life orderly and prevents chaos. Certain qualities that are nurtured by mathematics are power of reasoning, creativity, abstract or spatial thinking, critical thinking, problem-solving ability and even effective communication skills. ## What is general mathematical function? Function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable). Functions are ubiquitous in mathematics and are essential for formulating physical relationships in the sciences. ## What are the basics of maths? We’ve divided Basic Math into eight sections so you can focus on the skills you need: • Numbers and Operations. • Fractions, Decimals, and Percents. • Measurement. • Geometry. • Data Analysis and Statistics. • Algebra (Basic) • Word Problems. You might be interested: Quick Answer: What Mathematics Concepts Or Principles? ## What is function in general mathematics with example? A function is a mapping from a set of inputs (the domain) to a set of possible outputs (the codomain). The definition of a function is based on a set of ordered pairs, where the first element in each pair is from the domain and the second is from the codomain. ## What I know General Mathematics focuses on what topics? The General Mathematics course introduces fundamental math concepts. The topics include whole numbers, operations on whole numbers, fractions, decimals, ratios, rates, and proportions. The focus is on learning the computational procedures and then applying the skills to problem solving in applications. ## Where do we use math in real life? 10 Ways We Use Math Everyday • Chatting on the cell phone. Chatting on the cell phone is the way of communicating for most people nowadays. • In the kitchen. Baking and cooking requires some mathematical skill as well. • Gardening. • Arts. • Keeping a diary. • Planning an outing. • Banking. • Planning dinner parties. ## What is Mathematics in your own words? Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. The needs of math arose based on the wants of society. The more complex a society, the more complex the mathematical needs. ## Why is math so hard? Math is a very abstract subject. For students, learning usually happens best when they can relate it to real life. As math becomes more advanced and challenging, that can be difficult to do. As a result, many students find themselves needing to work harder and practice longer to understand more abstract math concepts. You might be interested: Often asked: What Is Most Useful In Mathematics For Humankind? ## What are the 4 types of functions? The various types of functions are as follows: • Many to one function. • One to one function. • Onto function. • One and onto function. • Constant function. • Identity function. • Polynomial function. ## WHAT IS function and its types? 1. Injective (One-to-One) Functions: A function in which one element of Domain Set is connected to one element of Co-Domain Set. 2. Surjective (Onto) Functions: A function in which every element of Co-Domain Set has one pre-image. ## WHAT IS function and its types in mathematics? A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B. ## What are the four rules of maths? The four basic Mathematical rules are addition, subtraction, multiplication, and division. Read more. ## Who is the father of mathematics? Archimedes is known as the Father Of Mathematics. He lived between 287 BC – 212 BC. Syracuse, the Greek island of Sicily was his birthplace. ## How can I be brilliant in maths? 10 Tips for Math Success 1. Do all of the homework. Don’t ever think of homework as a choice. 2. Fight not to miss class. 3. Find a friend to be your study partner. 4. Establish a good relationship with the teacher. 5. Analyze and understand every mistake. 6. Get help fast.	959	4,631	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2021-25	longest	en	0.902422
https://socratic.org/questions/how-do-you-factor-3x-7x-2	1,576,423,334,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575541308604.91/warc/CC-MAIN-20191215145836-20191215173836-00320.warc.gz	529,365,553	6,002	# How do you factor 3x² + 7x = -2? Sep 17, 2015 Factor: 3x^2 + 7x + 2 Ans: y = (3x + 1)(x + 2) #### Explanation: I use the new AC Method to factor trinomials (Socratic Search) Converted trinomial y' = x^2 + 7x + 6. Factor pairs of (6)--> (1, 6). This sum is 7 = b. Then p' = 1 and q' = 6. Therefor, p = 1/3 and q = 6/3 = 2. Factored form: y = 3(x + 1/3)(x + 2) = (3x + 1)(x + 2).	182	384	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2019-51	latest	en	0.706495
http://www.nag.com/numeric/fl/manual20/examples/source/f07mefe.f	1,411,185,668,000,000,000	text/plain	crawl-data/CC-MAIN-2014-41/segments/1410657132646.40/warc/CC-MAIN-20140914011212-00123-ip-10-196-40-205.us-west-1.compute.internal.warc.gz	686,243,925	1,248	* F07MEF Example Program Text * Mark 15 Release. NAG Copyright 1991. * .. Parameters .. INTEGER NIN, NOUT PARAMETER (NIN=5,NOUT=6) INTEGER NMAX, LDA, LWORK, NRHMAX, LDB PARAMETER (NMAX=8,LDA=NMAX,LWORK=64NMAX,NRHMAX=NMAX, + LDB=NMAX) .. Local Scalars .. INTEGER I, IFAIL, INFO, J, N, NRHS CHARACTER UPLO * .. Local Arrays .. DOUBLE PRECISION A(LDA,NMAX), B(LDB,NRHMAX), WORK(LWORK) INTEGER IPIV(NMAX) * .. External Subroutines .. EXTERNAL DSYTRF, DSYTRS, X04CAF * .. Executable Statements .. WRITE (NOUT,) 'F07MEF Example Program Results' Skip heading in data file READ (NIN,) READ (NIN,) N, NRHS IF (N.LE.NMAX .AND. NRHS.LE.NRHMAX) THEN * * Read A and B from data file * READ (NIN,) UPLO IF (UPLO.EQ.'U') THEN READ (NIN,) ((A(I,J),J=I,N),I=1,N) ELSE IF (UPLO.EQ.'L') THEN READ (NIN,) ((A(I,J),J=1,I),I=1,N) END IF READ (NIN,) ((B(I,J),J=1,NRHS),I=1,N) * * Factorize A * CALL DSYTRF(UPLO,N,A,LDA,IPIV,WORK,LWORK,INFO) * WRITE (NOUT,) IF (INFO.EQ.0) THEN * Compute solution * CALL DSYTRS(UPLO,N,NRHS,A,LDA,IPIV,B,LDB,INFO) * * Print solution * IFAIL = 0 CALL X04CAF('General',' ',N,NRHS,B,LDB,'Solution(s)',IFAIL) ELSE WRITE (NOUT,) 'The factor D is singular' END IF END IF STOP END	455	1,199	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.609375	3	CC-MAIN-2014-41	latest	en	0.454479
http://mathhelpforum.com/number-theory/52557-eulers-theorme-print.html	1,495,478,717,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463605485.49/warc/CC-MAIN-20170522171016-20170522191016-00155.warc.gz	236,858,886	3,307	# eulers theorme • Oct 7th 2008, 07:51 PM rmpatel5 eulers theorme Use eulers theorem to find the least nonnegative residue modulo m of integer n n=79^79, m=9 • Oct 7th 2008, 08:42 PM o_O Note that: $79 \equiv 7 \ (\text{mod }9)$, i.e. $79^{79} \equiv 7^{79} \ (\text{mod } 9)$ By Euler's theorem: $\begin{array}{rcll}7^{\varphi (9)} & \equiv & 1 & (\text{mod } 9) \qquad \varphi (9) = \varphi (3^2) = 3 \cdot 2 = 6 \\ 7^6 & \equiv & 1 & (\text{mod } 9) \\ \left(7^6\right)^{13} & \equiv & 1^{13} & (\text{mod } 9) \\ & \vdots & & \end{array}$ • Oct 7th 2008, 08:49 PM bigb s • Oct 7th 2008, 08:50 PM rmpatel5 Quote: Originally Posted by o_O Note that: $79 \equiv 7 \ (\text{mod }9)$, i.e. $79^{79} \equiv 7^{79} \ (\text{mod } 9)$ By Euler's theorem: $\begin{array}{rcll}7^{\varphi (9)} & \equiv & 1 & (\text{mod } 9) \qquad \varphi (9) = \varphi (3^2) = 3 \cdot 2 = 6 \\ 7^6 & \equiv & 1 & (\text{mod } 9) \\ \left(7^6\right)^{13} & \equiv & 1^{13} & (\text{mod } 9) \\ & \vdots & & \end{array}$ could u possibly finish out the problem and explain it as like an example problem. The book only talks about this topic and just dives into problem set without any examples. Also, why is it not 79^6 but rather 7^6?? • Oct 7th 2008, 09:00 PM o_O Well we want to find: $7^{79} \equiv a \ (\text{mod } 9)$ for some $a$. We have that $\left(7^6\right)^{13} = 7^{78} \equiv 1 \ (\text{mod }9)$ Not too hard to see how we get to $7^{79}$. As for calculating $\varphi (m)$, you need to find its prime power decomposition. With this, use the fact that: (1) $\varphi (p^n) = p^{n-1} (p-1)$ (2) $\varphi (m)$ is multiplicative which means $\varphi (p_{1}^{e_{1}} \cdot p_{2}^{e_{2}}) = \varphi (p_{1}^{e_{1}}) \cdot \varphi (p_{2}^{e_{2}})$ where $p$ is prime. For example: $\varphi (20) = \varphi (2^2 \cdot 5) = \varphi (2^2) \cdot \varphi (5) = 2^1 (2 - 1) \ \cdot \ 4 \ = 8$	807	1,876	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2017-22	longest	en	0.639079
https://dimag.ibs.re.kr/event/2021-08-31/	1,725,780,656,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700650960.90/warc/CC-MAIN-20240908052321-20240908082321-00234.warc.gz	191,363,928	35,231	• This event has passed. # Cheolwon Heo (허철원), Representations of even-cycle matroids ## Tuesday, August 31, 2021 @ 4:30 PM - 5:30 PM KST Room B232, IBS (기초과학연구원) ### Speaker Cheolwon Heo (허철원) Applied Algebra and Optimization Research Center (AORC), Sungkyunkwan University A signed graph is a pair $(G,\Sigma)$ where $G$ is a graph and $\Sigma$ is a subset of edges of $G$. A cycle $C$ of $G$ is a subset of edges of $G$ such that every vertex of the subgraph of $G$ induced by $C$ has an even degree. We say that $C$ is even in $(G,\Sigma)$ if $\|C \cap \Sigma\|$ is even; otherwise, $C$ is odd. A matroid $M$ is an even-cycle matroid if there exists a signed graph $(G,\Sigma)$ such that circuits of $M$ precisely corresponds to inclusion-wise minimal non-empty even cycles of $(G,\Sigma)$. For even-cycle matroids, two fundamental questions arise: (1) what is the relationship between two signed graphs representing the same even-cycle matroids? (2) how many signed graphs can an even-cycle matroid have? For (a), we characterize two signed graphs $(G_1,\Sigma_1)$ and $(G_2,\Sigma_2)$ where $G_1$ and $G_2$ are $4$-connected that represent the same even-cycle matroids. For (b), we introduce pinch-graphic matroids, which can generate exponentially many representations even when the matroid is $3$-connected. An even-cycle matroid is a pinch-graphic matroid if there exists a signed graph with a pair of vertices such that every odd cycle intersects with at least one of them. We prove that there exists a constant $c$ such that if a matroid is even-cycle matroid that is not pinch-graphic, then the number of representations is bounded by $c$. This is joint work with Bertrand Guenin and Irene Pivotto. ## Details Date: Tuesday, August 31, 2021 Time: 4:30 PM - 5:30 PM KST Event Category: Event Tags: , Room B232 IBS (기초과학연구원) ## Organizer Sang-il Oum (엄상일) View Organizer Website 기초과학연구원 수리및계산과학연구단 이산수학그룹 대전 유성구 엑스포로 55 (우) 34126 IBS Discrete Mathematics Group (DIMAG) Institute for Basic Science (IBS) 55 Expo-ro Yuseong-gu Daejeon 34126 South Korea E-mail: dimag@ibs.re.kr, Fax: +82-42-878-9209	634	2,115	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2024-38	latest	en	0.898175
https://it.mathworks.com/matlabcentral/cody/problems/147-too-mean-spirited/solutions/669952	1,596,933,825,000,000,000	text/html	crawl-data/CC-MAIN-2020-34/segments/1596439738366.27/warc/CC-MAIN-20200808224308-20200809014308-00068.warc.gz	352,678,771	15,637	Cody # Problem 147. Too mean-spirited Solution 669952 Submitted on 14 May 2015 by Grzegorz Popek This solution is locked. To view this solution, you need to provide a solution of the same size or smaller. ### Test Suite Test Status Code Input and Output 1 Pass %% x = [1 2 3]; y_correct = [1.5 2.5]; assert(isequal(two_mean(x),y_correct)) 2 Pass %% x = [10 0 0 0 100]; y_correct = [5 0 0 50]; assert(isequal(two_mean(x),y_correct)) 3 Pass %% x = [1 2 4]; y_correct = [1.5 3]; assert(isequal(two_mean(x),y_correct))	186	527	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.640625	3	CC-MAIN-2020-34	latest	en	0.646011
https://www.coursehero.com/file/5668969/Continuous-Time-Brownian-Motion/	1,498,285,888,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00119.warc.gz	850,279,676	25,467	Continuous Time - Brownian Motion # Continuous Time - Brownian Motion - 1 Temple University... This preview shows pages 1–15. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: 1 Temple University Continuous Time Finance Brownian Motion 2 Continuous Time Finance Brownian Motion 3 Brownian Motion In 1827, while examining pollen grains suspended in water under a microscope, Brown observed minute particles executing a continuous jittery motion. Robert Brown 17731858 Jan Ingenhousz 1730-1799 One only has to place a drop of alcohol in the focal point of a microscope and introduce a little finely ground charcoal therein, and one will see these corpuscules in a confused, continuous and violent motion, as if they were animalcules which move rapidly around. 4 Brownian Motion Although atoms and molecules were still open to objection in 1905, Einstein predicted that the random motions of molecules in a liquid impacting on larger suspended particles would result in irregular, random motions of the particles, which could be directly observed under a microscope. Perrin did the experimental work to test Einstein's predictions that matter is made of atoms and molecules. Jean Baptiste Perrin 1870 1942 5 Brownian Motion 3.2.1-3.2.4 Symmetric Random Walks 6 Symmetric Random Walk 1,2,... k , if 1 if 1 .... 1 j j 3 2 1 = = =- = = = = k j j k j X M T H X 7 MATLAB Random Walk function paths = RandomWalk (p,NRandom) % rand('twister',0); paths = zeros(1,NRandom); for k =1:NRandom-1 if rand &lt; p step = 1; else step = -1; end paths(1,k+1) = paths(1,k) + step; end plot(1:NRandom,paths(1,:)); 10 20 30 40 50 60 70 80 90 100-12-10-8-6-4-2 2 4 6 8 Increments 10 20 30 40 50 60 70 80 90 100-12-10-8-6-4-2 2 4 6 s) coin tosse g overlappin- non on (depend t independen are ) and ) increments The k integers choose l m k l M (M M (M m l-- &lt; &lt; 9 Bernoulli Random Variable ) 1 ( 4 ) 1 2 ( 1 ) ( 1 ) 1 ( ) 1 ( * 1 1 2 ) 1 )( 1 ( * 1 p- 1 1 p 1 2 2 2 2 2 2 p p p EX EX VarX p p EX p p p EX X- =-- =- = =-- + =- =-- + = - = 1 ) ( 1 ) 1 ( * 1 ) 1 ( * 1 1 1 2 2 2 1 2 2 1 2 2 2 1 2 1 2 1 2 1 =- = =- + = =- + = - = EX EX VarX EX EX X 10 Increments k l VarX M ar(M EX M (M X M (M j k l j k l j k l- = =- = =- =- &lt; + = + = + = l 1 k j l 1 k j l 1 k j ) V ) E ) for l, k integers choose 11 Martingale Property K K k l K k k l k K k k l k K k l k l M M M M E M F M M E F M E F M M E F M M M E F M E = +- = +- = +- = +- = )] [( ] \| ) [( ] \| [ ] \| ) [( ] \| ) [( ] \| [ 12 Quadratic Variation k M M M M k M M M M j k j j k j k j j k =- = =- =- =- = 2 1 1 2 1 1 ) ( ] , [ )) ( ) ( ( ) ( ] , [ 13 Brownian Motion 3.2.5-3.2.6 Scaled Random Walk 14 Scaled Random Walk 1,2,... k , if 1 if 1 ....... View Full Document ## This note was uploaded on 11/23/2009 for the course FIN 5190 taught by Professor Soss during the Fall '09 term at Temple. ### Page1 / 55 Continuous Time - Brownian Motion - 1 Temple University... This preview shows document pages 1 - 15. Sign up to view the full document. View Full Document Ask a homework question - tutors are online	1,214	3,876	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.59375	3	CC-MAIN-2017-26	longest	en	0.890622
http://mathhelpforum.com/calculus/150400-find-number-solutions-x.html	1,529,543,826,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00345.warc.gz	212,468,940	9,389	# Thread: find number of solutions of x 1. ## find number of solutions of x find number of solutions of x ..where x belongs to real number for which $\displaystyle 64^{\frac{1}{x}} + 48^{\frac{1}{x}} = 80^{\frac{1}{x}}$ 2. Try factoring each number as a multiple of 8. You might get some cancellation there.	94	310	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.6875	3	CC-MAIN-2018-26	latest	en	0.840968
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-2-section-2-2-the-multiplication-property-of-equality-exercise-set-page-110/9	1,537,896,652,000,000,000	text/html	crawl-data/CC-MAIN-2018-39/segments/1537267161902.89/warc/CC-MAIN-20180925163044-20180925183444-00480.warc.gz	751,103,534	14,393	## Algebra: A Combined Approach (4th Edition) $\frac{1}{6}$d = $\frac{1}{2}$ d = 3	32	83	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.8125	3	CC-MAIN-2018-39	latest	en	0.765139
http://stackoverflow.com/questions/13957146/prolog-very-simple-dcg-syntax?answertab=oldest	1,386,840,926,000,000,000	text/html	crawl-data/CC-MAIN-2013-48/segments/1386164575861/warc/CC-MAIN-20131204134255-00094-ip-10-33-133-15.ec2.internal.warc.gz	174,490,473	12,941	# prolog,very simple dcg syntax I am trying to understand the basic syntax of prolog and dcg but it's really hard to get ahold of proper information on the really basic stuff. Take a look at the code below, I basically just want to achieve something like this: ``````Output = te(a, st). Code: test(te(X,Y)) --> [X], test2(Y). test2(st(_X)) --> [bonk]. ?- test(Output, [a, bonk],[]). Output = te(a, st(_G6369)). `````` Simply what I want to do is to add the the word 'st' at the end, and the closest way I've managed is by doing this but unfortunately st is followed a bunch of nonsense, most likely because of the singleton `_X`. I simply want my `Output` to contain like: `te(a, st).` - It is much safer to call a non-terminal via `phrase/2`. Your query should thus rather read: `?- phrase(test(Output), [a,bonk]).` – false Dec 19 '12 at 17:02 +1: it's funny to read answers that bonk – CapelliC Dec 19 '12 at 17:19 If you want to accept input of the form `[Term, bonk]` and obtain `te(Term,st)` you should change `test/2` to accept bonk a return `st`: ``````test(te(X,Y)) --> [X], test2(Y). test2(st) --> [bonk]. ?- test(Output, [a, bonk],[]). Output = te(a, st). `````` - Well now I certainly feel silly! I could've sworn I tested without any argument, in fact, that was pretty much the sole reason for why I came here to Stackoverflow! I must've made some kind of error earlier when I tried it myself because as the solutions here have explained, it does indeed work. – Deragon Dec 19 '12 at 18:13 As you said, st is followed by "a bunch of nonsense" because of `_X` (basically, _G6369 is the internal 'name' of the variable and since the variable remains uninstantiated prolog displays it; try `print(X), X=3, print(X).` Anyway, you can simply remove `(_X)` since you can have anything you want as an argument: ``````test(te(X,Y)) --> [X], test2(Y). test2(st) --> [bonk]. `````` Of course, if you don't actually have `bonk`'s in your input and you simply want to add a `st` at the end you can simplify it even more: ``````test(te(X,st)) --> [X]. `````` Or if you have `bonk`'s: ``````test(te(X,st)) --> [X,bonk]. `````` Finally, it is generally suggested to use phrase/3 or phrase/2 instead of adding the arguments manually. -	657	2,253	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.5625	3	CC-MAIN-2013-48	latest	en	0.914081
https://numbersworksheet.com/converting-mixed-numbers-to-improper-fractions-worksheet-5th-grade/	1,721,135,997,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00137.warc.gz	395,935,588	14,152	# Converting Mixed Numbers To Improper Fractions Worksheet 5th Grade Portion Phone numbers Worksheets are a very good way to practice the thought of fractions. These worksheets are made to train college students concerning the inverse of fractions, and will help them to be aware of the relationship involving decimals and fractions. Many students have trouble converting fractions to decimals, but they can benefit from these worksheets. These computer worksheets will assist your pupil to become much more acquainted with fractions, and they’ll make sure to have some fun carrying out them! Converting Mixed Numbers To Improper Fractions Worksheet 5th Grade. ## Cost-free mathematics worksheets If your student is struggling with fractions, consider downloading and printing free fraction numbers worksheets to reinforce their learning. These worksheets might be custom-made to suit your personal demands. Additionally they incorporate solution secrets with thorough guidelines to steer your university student through the procedure. Many of the worksheets are divided into various denominators which means that your college student can training their skills with an array of troubles. After, students can recharge the web page to get a different worksheet. These worksheets support individuals recognize fractions by generating equivalent fractions with different denominators and numerators. They have got series of fractions which can be counterpart in value and each row features a lacking denominator or numerator. The scholars fill out the missing out on numerators or denominators. These worksheets are useful for practicing the expertise of minimizing fractions and discovering small percentage operations. They come in different amounts of problems, ranging from easy to method to tough. Each worksheet features involving thirty and ten problems. ## Cost-free pre-algebra worksheets Whether you are in need of a totally free pre-algebra small percentage numbers worksheet or you need a printable edition for the students, the Internet can provide many different alternatives. Many sites provide totally free pre-algebra worksheets, with a few significant conditions. While several of these worksheets may be tailored, a couple of totally free pre-algebra fraction figures worksheets may be downloaded and printed for extra practice. One great resource for downloadable free of charge pre-algebra small fraction figures worksheet is the University of Maryland, Baltimore Region. You should be careful about uploading them on your own personal or classroom website, though worksheets are free to use. You are free to print out any worksheets you find useful, and you have permission to distribute printed copies of the worksheets to others. You can use the free worksheets as a tool for learning math facts. Alternatively, as a stepping stone towards more complex concepts. ## Free of charge mathematics worksheets for course VIII If you are in Class VIII and are looking for free fraction numbers worksheets for your next maths lesson, you’ve come to the right place! This collection of worksheets will depend on the CBSE and NCERT syllabus. These worksheets are perfect for scrubbing on the methods of fractions to help you do much better in your CBSE exam. These worksheets are really easy to use and cover all the principles which are essential for reaching higher markings in maths. Some of these worksheets include comparing fractions, getting fractions, simplifying fractions, and operations with one of these phone numbers. Try to use true-existence cases over these worksheets so that your college students can connect with them. A cookie is simpler to connect with than 50 % of a rectangular. An additional great way to practice with fractions is by using comparable fractions types. Try using real world good examples, such as a fifty percent-cookie as well as a sq .. Cost-free mathematics worksheets for converting decimal to small percentage If you are looking for some free math worksheets for converting decimal to a fraction, you have come to the right place. These decimal to small percentage worksheets can be found in many different formats. You can download them inPDF and html. Alternatively, random format. A lot of them include a response crucial and can also be shaded by little ones! They are utilized for summertime discovering, math facilities, or as a part of your regular math programs. To convert a decimal to your fraction, you need to make simpler it initial. Decimals are written as equivalent fractions if the denominator is ten. In addition, you will also find worksheets on the way to transform merged figures into a small fraction. Cost-free math worksheets for transforming decimal to portion have mixed examples and numbers of these two conversion process procedures. However, the process of converting a decimal to a fraction is easier than you might think. Follow these steps to get going.	889	4,953	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.984375	3	CC-MAIN-2024-30	latest	en	0.934044
http://forums.wolfram.com/student-support/topics/350177	1,444,008,687,000,000,000	text/html	crawl-data/CC-MAIN-2015-40/segments/1443736676547.12/warc/CC-MAIN-20151001215756-00191-ip-10-137-6-227.ec2.internal.warc.gz	125,054,498	6,091	Student Support Forum: 'mathematica ndsolve - no output' topicStudent Support Forum > General > Archives > "mathematica ndsolve - no output" < Previous Comment \| Next Comment > Help \| Reply To Comment \| Reply To Topic Author Comment/Response D 02/21/13 11:52am I don't believe the problem is in the code because the program runs fully through on another computer. I believe it has something to do with how my computer is analyzing the notebook when it comes to NDsolve. xM = 15 x0=-8 a = 0.5; barrierHeight = 5.5 10^-7; barrierV[x_] := Which[x < 0, 0, x > a, 0, True, barrierHeight] c = 3000; hbarc = 0.00197; hbar = hbarc/ c mass = 1000; hbarSqOver2mass = (hbarc^2)/ (2 mass) E1 = 4.9 10^-7 kzero = Sqrt[E1/hbarSqOver2mass ] lambda = 2 \[Pi]/ kzero v = E1/ (hbar) \[Sigma] = 1 \[Omega]1 = c hbarc kzero/ mass \[Omega]2 = c hbarc/ mass \[Psi]pkt[v_, \[Sigma]_, \[Omega]1_, \[Omega]2_][x_, t_] = (\[Sigma]^2/\[Pi])^(1/4) Exp[I (kzero x - v t)]/((\[Sigma]^2 + I \[Omega]2 t )^(1/2)) Exp[-((x - \[Omega]1 t)^2/(2 (\[Sigma]^2 + I \[Omega]2 t) ))]; \[Chi][x_] := \[Psi]pkt[v, \[Sigma], \[Omega]1, \[Omega]2][x - x0, 0] enforceBCs[\[Psi]_, x_, xM_] := \[Psi] - (((\[Psi] /. x -> -xM) - (\[Psi] /. x -> xM))/ 2*(Cos[(x + xM)/(2 xM) Pi] + 1) + (\[Psi] /. x -> xM)) schroedingerEq = I hbar D[\[CapitalPsi][x, t], {t, 1}] == - hbarSqOver2mass D[\[CapitalPsi][x, t], {x, 2}] + barrierV[x] \[CapitalPsi][x, t] (nsol = NDSolve[{schroedingerEq, \[CapitalPsi][x, 0] == enforceBCs[\[Chi][x], x, xM], \[CapitalPsi][xM, t] == 0, \[CapitalPsi][-xM, t] == 0}, \[CapitalPsi][x, t], {x, -xM, xM}, {t, 0, 200}, AccuracyGoal -> 3, PrecisionGoal -> 3]) // Timing URL: , Subject (listing for 'mathematica ndsolve - no output') Author Date Posted mathematica ndsolve - no output D. 02/20/13 9:31pm Re: mathematica ndsolve - no output Forum Modera... 02/21/13 11:02am Re: mathematica ndsolve - no output D 02/21/13 11:52am Re: mathematica ndsolve - no output yehuda 02/21/13 7:02pm Re: Re: mathematica ndsolve - no output D 02/23/13 9:58pm Re: Re: Re: mathematica ndsolve - no output yehuda 02/24/13 6:31pm Re: Re: Re: Re: mathematica ndsolve - no output Forum Modera... 02/24/13 7:33pm Re: Re: Re: mathematica ndsolve - no output Bill Simpson 02/24/13 7:13pm Re: Re: Re: Re: mathematica ndsolve - no output yehuda 02/25/13 01:48am Re: Re: Re: Re: Re: mathematica ndsolve - no ou... D 02/27/13 9:54pm Re: Re: Re: Re: Re: Re: mathematica ndsolve - n... jf 02/28/13 10:53am Re: mathematica ndsolve - no output yehuda 02/28/13 5:07pm < Previous Comment \| Next Comment > Help \| Reply To Comment \| Reply To Topic	1,024	2,585	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	2.75	3	CC-MAIN-2015-40	latest	en	0.580075
http://www.federica.unina.it/ingegneria/wireless-networks/narrowband-fading/	1,685,759,335,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00556.warc.gz	60,841,082	9,501	# Giacinto Gelli » 7.Narrowband fading ### Outline I-Q components Mean, autocorrelation and cross-correlation of I-Q components Jakes model Envelope distributions $r(t) = \mbox{Re} \left\{ \sum_{n=1}^{N} \alpha_n(t) e^{-j\phi_n(t)} \, u \left[ t - \tau_n(t) \right] \,e^{j 2 \pi f_ct}\right\}$ When Tm << 1/Bu one has u[t-τn(t)] ≈u(t): $r(t) = \mbox{Re} \left\{ u(t) \, e^{j 2 \pi f_c t}\left[ \sum_{n=1}^{N} \alpha_n(t) e^{-j\phi_n(t)} \right]\right\}$ Bu = bandwidth of u(t). The RX signal differs from the TX one for the multiplicative effect of the term in square brackets: • narrowband fading does not introduce temporal dispersion but acts as time-varying multiplicative noise→ it introduces both amplitude variations (fading) and frequency dispersion • simple characterization of NB fading can be obtained by assuming u(t)=1 (tone transmission) ### Gaussian model Due to the random nature of the channel, the received signal r(t) when u(t)=1 is a random process: $r_I(t) = \sum_{n=1}^{N} \alpha_n(t) \cos \phi_n(t)\qquad$ $r_Q(t) = \sum_{n=1}^{N} \alpha_n(t) \sin \phi_n(t)$ For N >> 1, the I/Q components rI(t) and rQ(t) are approximately jointly Gaussian by the CLT (sum of a large number of independent random variables). They can be characterized by their mean, autocorrelation and cross-correlation. ### Quasi-static assumption Assume that the amplitudes, the delays and the Doppler shifts are quasi-static (i.e., slowly varying): • αn(t) ≈ αn • τn(t) = dn(t)/c ≈[dn – v cos(θn) t]/c =τn- v t cos(θn)/c • Φn(t) = 2π fcτn(t) ≈ 2π fcτn – 2π fc vt cos(Φn)/c = 2π fcτn – 2π fDn t • θn is the multipath arrival angle at the receiver • fDn= fc(v/c) cos(Φn )=(v/λ) cos(Φn) is the Doppler shift Since fcτn>> 1, Φn(t) ~ U(0,2π) (uniform distribution) for any value of the Doppler shift fDn. Reasonable assumption for NLOS propagation, in LOS propagation the phase is not uniform since a deterministic component should be added to rI(t) and rQ(t). ### Mean of the I/Q components The average of rI(t) for a fixed Doppler shift fDn is: $\mbox{E}[r_I(t)] = \sum_{n=1}^{N}\mbox{E}[ \alpha_n] \underbrace{\mbox{E}[ \cos \phi_n(t)]}_{=0} = 0$ where we assumed that αn and Φn(t) ~ U(0,2π) are statistically independent. A similar result holds for the Q-component: $\mbox{E}[r_Q(t)] = \sum_{n=1}^{N}\mbox{E}[ \alpha_n] \underbrace{\mbox{E}[ \sin \phi_n(t)]}_{=0} = 0$ The I/Q components are zero-mean for any value of the Doppler shift fDn. In LOS propagation the I/Q components are not zero-mean. ### I/Q autocorrelation Recall that θn is the multipath arrival angle. The autocorrelation of the I/Q components. $A_{r_I}(\tau) = A_{r_Q}(\tau) =\frac{1}{2}\sum_{n=1}^{N}\mbox{E}[ \alpha_n^2] \,\mbox{E}_{\theta_n}\left[ \cos \left(2 \pi \frac{v}{\lambda} \cos \theta_n \right)\right]$ The I/Q components have the same autocorrelation. The autocorrelation does not depend on time t → the I/Q components are wide-sense stationary (WSS) and, since they are Gaussian, they are also strict-sense stationary. The average with respect to θn must be carried out to obtain an explicit expression → a simple model for the angular distribution of multipath is required. ### I/Q cross-correlation The cross-correlation of the I/Q components is: $A_{r_I r_Q}(\tau) =\frac{1}{2}\sum_{n=1}^{N}\mbox{E}[ \alpha_n^2] \,\mbox{E}_{\theta_n}\left[ \sin \left(2 \pi \frac{v}{\lambda} \cos \theta_n \right)\right] =- A_{r_Q r_I}(\tau)$ For τ=0 we have ArI,rQ(0)=0 → the I/Q components are uncorrelated for τ=0 →since they are Gaussian, they are independent for τ=0. The obtained expression must be averaged with respect to θn. The autocorrelation of the received signal is: $A_r(\tau) =A_{r_I}(\tau) \, \cos(2 \pi f_c \tau) +A_{r_I r_Q}(\tau) \, \sin(2 \pi f_c \tau)$ The received signal is WSS Gaussian and zero-mean (remember that the TX signal was u(t)=1). ### Jakes model Assumption: the scattering is uniformin angle • large number (N>>1) of scatterers uniformly distributed over the circle θn = n Δθ • all multipath component with the same power: E[αn2]=2Pr/N Suited for urban NLOS propagation, often utilized in simulations. The discrete angle distribution can be approximated by a continuous one θ ~ U(0,2π). ### Jakes model (cont’d) By averaging with respect to the uniform distribution for θ we get the autocorrelation and cross-correlation functions: $A_{r_I}(\tau) = A_{r_Q}(\tau) =P_r J_0(2 \pi f_D \tau)$ $A_{r_I r_Q}(\tau) = 0$ where: • J0(x) zeroth-order Bessel function • fD max Doppler shift • decorrelates over roughly half a wavelength Jakes model (cont’d) By taking the Fourier transform of the autocorrelation function we obtain the power spectral density (PSD). ### Envelope distributions The envelope of the received signal is: $z(t) = \|r(t)\| = \sqrt{r_I^2(t) + r_Q^2(t)}$ Gaussian approximation based on CLT leads to Rayleigh distribution (power is exponential): $p_Z(z) = \frac{z}{\sigma^2} \exp \left( - \frac{z^2}{2 \sigma^2} \right)\qquadz \ge 0$ When LOS component is present → Rice distribution: $p_Z(z) = \frac{z}{\sigma^2} \exp \left[ - \frac{(z^2+s^2)}{2 \sigma^2} \right] \,I_0 \left( \frac{z s}{\sigma^2}\right)\qquadz \ge 0$ ### Envelope distributions (cont’d) Measurements support Nakagami distribution in some environments: $$p_Z(z) = \frac{2 \, m^m z^{2m-1}}{\Gamma(m) \overline{P}_r^m}\exp \left( - \frac{m z^2}{\overline{P}_r} \right)\qquadz \ge 0$\end{itemize}$ where: • m ≥ 1/2 fading parameter • Γ(.) Gamma function Reduces to Rayleigh for m=1, no fading for m →∞. Approximates Rice case for m=(K+1)2/(2K+1), can model “worse than Rayleigh” cases for m<1. Lends itself to closed form expressions for BER. ### Conclusions Narrowband model has I/Q components that are zero-mean stationary Gaussian processes. Auto and cross-correlation depend on angular distribution of multipath. Uniform scattering assumption makes autocorrelation of I/Q components follow Bessel function (Jakes model). Fading components decorrelated over roughly half-wavelength. Cross-correlation is zero (I/Q components independent). Rayleigh, Rice, Nakagami are common choices. ### I materiali di supporto della lezione A. Goldsmith. Wireless Communications. Cambridge University Press, 2005 (chap. 3) W.C. Jakes. Microwave Mobile Communications. Wiley, New York, 1974 Supplementary material eventually available on the website Progetto "Campus Virtuale" dell'Università degli Studi di Napoli Federico II, realizzato con il cofinanziamento dell'Unione europea. Asse V - Società dell'informazione - Obiettivo Operativo 5.1 e-Government ed e-Inclusion Fatal error: Call to undefined function federicaDebug() in /usr/local/apache/htdocs/html/footer.php on line 93	2,111	6,717	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 15, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.015625	3	CC-MAIN-2023-23	latest	en	0.789663

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