The Euclidean Algorithm in Circle/Sphere Packings

Total Page:16

File Type:pdf, Size:1020Kb

The Euclidean Algorithm in Circle/Sphere Packings The Euclidean Algorithm in Circle/Sphere Packings Arseniy (Senia) Sheydvasser October 25, 2019 I Choose a circle C with center (x0; y0) and radius R. I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I Choose a circle C with center (x0; y0) and radius R. Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I Choose a circle C with center (x0; y0) and radius R. I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions Circle/Sphere Inversions Question n Let Γ be a subgroup of M¨ob(R ), and S an n-sphere. What does the orbit Γ:S look like? Can we compute it effectively? Circle/Sphere Inversions Definition n M¨ob(R ) is the group generated by n-sphere reflections in n R [ f1g. Circle/Sphere Inversions Definition n M¨ob(R ) is the group generated by n-sphere reflections in n R [ f1g. Question n Let Γ be a subgroup of M¨ob(R ), and S an n-sphere. What does the orbit Γ:S look like? Can we compute it effectively? Motivation Question What analogs of the Apollonian circle packing are there? Motivation Question n What do hyperbolic quotient manifolds H =Γ look like? 3.0 2.5 2.0 1.5 1.0 0.5 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 SL(2; Z) SL(2; Z[i]) Motivation 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) I an orientation-reversing . M¨obiustransformation. Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) SL(2; H)={±1g I an orientation-reversing . M¨obiustransformation. Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) ??? an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) SL(2; H)={±1g I an orientation-reversing . M¨obiustransformation. ??? 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 Inverses are given as follows: a b−1 dz −bz = c d −cz az What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Inverses are given as follows: a b−1 dz −bz = c d −cz az What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 Inverses are given as follows: a b−1 dz −bz = c d −cz az 3 I Every orientation-preserving element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. 3 I Every orientation-reversing element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. M¨ob(R3) as SLz(2; H) 3 + I There is an action on R [ f1g = H [ f1g defined by a b :z = (az + b)(cz + d)−1 c d M¨ob(R3) as SLz(2; H) 3 + I There is an action on R [ f1g = H [ f1g defined by a b :z = (az + b)(cz + d)−1 c d 3 I Every orientation-preserving element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. 3 I Every orientation-reversing element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. I We will ask that Γ is arithmetic. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations. I Roughly, an arithmetic group is the set of integer solutions to that set of polynomial equations. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations. I Roughly, an arithmetic group is the set of integer solutions to that set of polynomial equations. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I We will ask that Γ is arithmetic. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I We will ask that Γ is arithmetic. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations.
Recommended publications
  • Survey of Modern Mathematical Topics Inspired by History of Mathematics
    Survey of Modern Mathematical Topics inspired by History of Mathematics Paul L. Bailey Department of Mathematics, Southern Arkansas University E-mail address: [email protected] Date: January 21, 2009 i Contents Preface vii Chapter I. Bases 1 1. Introduction 1 2. Integer Expansion Algorithm 2 3. Radix Expansion Algorithm 3 4. Rational Expansion Property 4 5. Regular Numbers 5 6. Problems 6 Chapter II. Constructibility 7 1. Construction with Straight-Edge and Compass 7 2. Construction of Points in a Plane 7 3. Standard Constructions 8 4. Transference of Distance 9 5. The Three Greek Problems 9 6. Construction of Squares 9 7. Construction of Angles 10 8. Construction of Points in Space 10 9. Construction of Real Numbers 11 10. Hippocrates Quadrature of the Lune 14 11. Construction of Regular Polygons 16 12. Problems 18 Chapter III. The Golden Section 19 1. The Golden Section 19 2. Recreational Appearances of the Golden Ratio 20 3. Construction of the Golden Section 21 4. The Golden Rectangle 21 5. The Golden Triangle 22 6. Construction of a Regular Pentagon 23 7. The Golden Pentagram 24 8. Incommensurability 25 9. Regular Solids 26 10. Construction of the Regular Solids 27 11. Problems 29 Chapter IV. The Euclidean Algorithm 31 1. Induction and the Well-Ordering Principle 31 2. Division Algorithm 32 iii iv CONTENTS 3. Euclidean Algorithm 33 4. Fundamental Theorem of Arithmetic 35 5. Infinitude of Primes 36 6. Problems 36 Chapter V. Archimedes on Circles and Spheres 37 1. Precursors of Archimedes 37 2. Results from Euclid 38 3. Measurement of a Circle 39 4.
    [Show full text]
  • Math 3010 § 1. Treibergs First Midterm Exam Name
    Math 3010 x 1. First Midterm Exam Name: Solutions Treibergs February 7, 2018 1. For each location, fill in the corresponding map letter. For each mathematician, fill in their principal location by number, and dates and mathematical contribution by letter. Mathematician Location Dates Contribution Archimedes 5 e β Euclid 1 d δ Plato 2 c ζ Pythagoras 3 b γ Thales 4 a α Locations Dates Contributions 1. Alexandria E a. 624{547 bc α. Advocated the deductive method. First man to have a theorem attributed to him. 2. Athens D b. 580{497 bc β. Discovered theorems using mechanical intuition for which he later provided rigorous proofs. 3. Croton A c. 427{346 bc γ. Explained musical harmony in terms of whole number ratios. Found that some lengths are irrational. 4. Miletus D d. 330{270 bc δ. His books set the standard for mathematical rigor until the 19th century. 5. Syracuse B e. 287{212 bc ζ. Theorems require sound definitions and proofs. The line and the circle are the purest elements of geometry. 1 2. Use the Euclidean algorithm to find the greatest common divisor of 168 and 198. Find two integers x and y so that gcd(198; 168) = 198x + 168y: Give another example of a Diophantine equation. What property does it have to be called Diophantine? (Saying that it was invented by Diophantus gets zero points!) 198 = 1 · 168 + 30 168 = 5 · 30 + 18 30 = 1 · 18 + 12 18 = 1 · 12 + 6 12 = 3 · 6 + 0 So gcd(198; 168) = 6. 6 = 18 − 12 = 18 − (30 − 18) = 2 · 18 − 30 = 2 · (168 − 5 · 30) − 30 = 2 · 168 − 11 · 30 = 2 · 168 − 11 · (198 − 168) = 13 · 168 − 11 · 198 Thus x = −11 and y = 13 .
    [Show full text]
  • CSU200 Discrete Structures Professor Fell Integers and Division
    CSU200 Discrete Structures Professor Fell Integers and Division Though you probably learned about integers and division back in fourth grade, we need formal definitions and theorems to describe the algorithms we use and to very that they are correct, in general. If a and b are integers, a ¹ 0, we say a divides b if there is an integer c such that b = ac. a is a factor of b. a | b means a divides b. a | b mean a does not divide b. Theorem 1: Let a, b, and c be integers, then 1. if a | b and a | c then a | (b + c) 2. if a | b then a | bc for all integers, c 3. if a | b and b | c then a | c. Proof: Here is a proof of 1. Try to prove the others yourself. Assume a, b, and c be integers and that a | b and a | c. From the definition of divides, there must be integers m and n such that: b = ma and c = na. Then, adding equals to equals, we have b + c = ma + na. By the distributive law and commutativity, b + c = (m + n)a. By closure of addition, m + n is an integer so, by the definition of divides, a | (b + c). Q.E.D. Corollary: If a, b, and c are integers such that a | b and a | c then a | (mb + nc) for all integers m and n. Primes A positive integer p > 1 is called prime if the only positive factor of p are 1 and p.
    [Show full text]
  • History of Mathematics Log of a Course
    History of mathematics Log of a course David Pierce / This work is licensed under the Creative Commons Attribution–Noncommercial–Share-Alike License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ CC BY: David Pierce $\ C Mathematics Department Middle East Technical University Ankara Turkey http://metu.edu.tr/~dpierce/ [email protected] Contents Prolegomena Whatishere .................................. Apology..................................... Possibilitiesforthefuture . I. Fall semester . Euclid .. Sunday,October ............................ .. Thursday,October ........................... .. Friday,October ............................. .. Saturday,October . .. Tuesday,October ........................... .. Friday,October ............................ .. Thursday,October. .. Saturday,October . .. Wednesday,October. ..Friday,November . ..Friday,November . ..Wednesday,November. ..Friday,November . ..Friday,November . ..Saturday,November. ..Friday,December . ..Tuesday,December . . Apollonius and Archimedes .. Tuesday,December . .. Saturday,December . .. Friday,January ............................. .. Friday,January ............................. Contents II. Spring semester Aboutthecourse ................................ . Al-Khw¯arizm¯ı, Th¯abitibnQurra,OmarKhayyám .. Thursday,February . .. Tuesday,February. .. Thursday,February . .. Tuesday,March ............................. . Cardano .. Thursday,March ............................ .. Excursus.................................
    [Show full text]
  • Discrete Mathematics Homework 3
    Discrete Mathematics Homework 3 Ethan Bolker November 15, 2014 Due: ??? We're studying number theory (leading up to cryptography) for the next few weeks. You can read Unit NT in Bender and Williamson. This homework contains some programming exercises, some number theory computations and some proofs. The questions aren't arranged in any particular order. Don't just start at the beginning and do as many as you can { read them all and do the easy ones first. I may add to this homework from time to time. Exercises 1. Use the Euclidean algorithm to compute the greatest common divisor d of 59400 and 16200 and find integers x and y such that 59400x + 16200y = d I recommend that you do this by hand for the logical flow (a calculator is fine for the arith- metic) before writing the computer programs that follow. There are many websites that will do the computation for you, and even show you the steps. If you use one, tell me which one. We wish to find gcd(16200; 59400). Solution LATEX note. These are separate equations in the source file. They'd be better formatted in an align* environment. 59400 = 3(16200) + 10800 next find gcd(10800; 16200) 16200 = 1(10800) + 5400 next find gcd(5400; 10800) 10800 = 2(5400) + 0 gcd(16200; 59400) = 5400 59400x + 16200y = 5400 5400(11x + 3y) = 5400 11x + 3y = 1 By inspection, we can easily determine that (x; y) = (−1; 4) (among other solutions), but let's use the algorithm. 11 = 3 · 3 + 2 3 = 2 · 1 + 1 1 = 3 − (2) · 1 1 = 3 − (11 − 3 · 3) 1 = 11(−1) + 3(4) (x; y) = (−1; 4) 1 2.
    [Show full text]
  • Math 105 History of Mathematics First Test February 2015
    Math 105 History of Mathematics First Test February 2015 You may refer to one sheet of notes on this test. Points for each problem are in square brackets. You may do the problems in any order that you like, but please start your answers to each problem on a separate page of the bluebook. Please write or print clearly. Problem 1. Essay. [30] Select one of the three topics A, B, and C. Please think about these topics and make an outline before you begin writing. You will be graded on how well you present your ideas as well as your ideas themselves. Each essay should be relatively short|one to three written pages. There should be no fluff in your essays. Make your essays well-structured and your points as clearly as you can. Topic A. Explain the logical structure of the Elements (axioms, propositions, proofs). How does this differ from earlier mathematics of Egypt and Babylonia? How can such a logical structure affect the mathematical advances of a civilization? Topic B. Compare and contrast the arithmetic of the Babylonians with that of the Egyptians. Be sure to mention their numerals, algorithms for the arithmetic operations, and fractions. Illustrate with examples. Don't go into their algebra or geometry for this essay. Topic C. Aristotle presented four of Zeno's paradoxes: the Dichotomy, the Achilles, the Arrow, and the Stadium. Select one, but only one, of them and write about it. State the paradox as clearly and completely as you can. Explain why it was considered important. Refute the paradox, either as Aristotle did, or as you would from a modern point of view.
    [Show full text]
  • Ruler and Compass Constructions and Abstract Algebra
    Ruler and Compass Constructions and Abstract Algebra Introduction Around 300 BC, Euclid wrote a series of 13 books on geometry and number theory. These books are collectively called the Elements and are some of the most famous books ever written about any subject. In the Elements, Euclid described several “ruler and compass” constructions. By ruler, we mean a straightedge with no marks at all (so it does not look like the rulers with centimeters or inches that you get at the store). The ruler allows you to draw the (unique) line between two (distinct) given points. The compass allows you to draw a circle with a given point as its center and with radius equal to the distance between two given points. But there are three famous constructions that the Greeks could not perform using ruler and compass: • Doubling the cube: constructing a cube having twice the volume of a given cube. • Trisecting the angle: constructing an angle 1/3 the measure of a given angle. • Squaring the circle: constructing a square with area equal to that of a given circle. The Greeks were able to construct several regular polygons, but another famous problem was also beyond their reach: • Determine which regular polygons are constructible with ruler and compass. These famous problems were open (unsolved) for 2000 years! Thanks to the modern tools of abstract algebra, we now know the solutions: • It is impossible to double the cube, trisect the angle, or square the circle using only ruler (straightedge) and compass. • We also know precisely which regular polygons can be constructed and which ones cannot.
    [Show full text]
  • The Distance Geometry of Music
    The Distance Geometry of Music Erik D. Demaine∗ Francisco Gomez-Martin† Henk Meijer‡ David Rappaport§ Perouz Taslakian¶ Godfried T. Toussaint§k Terry Winograd∗∗ David R. Wood†† Abstract We demonstrate relationships between the classical Euclidean algorithm and many other fields of study, particularly in the context of music and distance geometry. Specifically, we show how the structure of the Euclidean algorithm defines a family of rhythms that encompass over forty timelines (ostinatos) from traditional world music. We prove that these Euclidean rhythms have the mathematical property that their onset patterns are distributed as evenly as possible: they maximize the sum of the Euclidean distances between all pairs of onsets, viewing onsets as points on a circle. Indeed, Euclidean rhythms are the unique rhythms that maximize this notion of evenness. We also show that essentially all Euclidean rhythms are deep: each distinct distance between onsets occurs with a unique multiplicity, and these multiplicities form an interval 1, 2, . , k − 1. Finally, we characterize all deep rhythms, showing that they form a subclass of generated rhythms, which in turn proves a useful property called shelling. All of our results for musical rhythms apply equally well to musical scales. In addition, many of the problems we explore are interesting in their own right as distance geometry problems on the circle; some of the same problems were explored by Erd˝osin the plane. ∗Computer Science and Artificial Intelligence Laboratory, Massachusetts Institute of Technology,
    [Show full text]
  • Math 3110: Summary of Linear Diophantine Equations January 30, 2019
    Math 3110: Summary of Linear Diophantine Equations January 30, 2019 Goal: Given a; b; c 2 Z, to find all integer solutions (x; y) to ax + by = c. The Homogeneous Case The term homogeneous means the number after the = is zero. Theorem 1 (Homogeneous Case). Let a; b 2 Z. The set of integer solutions (x; y) to the equation ax + by = 0 is bk −ak ; : k 2 : gcd(a; b) gcd(a; b) Z Proof. See Course Notes of Mon, Jan 28 (used the gcd(a; b) · lcm(a; b) = ab theorem). The principle of adding solutions If (x; y) is a solution to ax + by = c, and (x0; y0) is a solution to ax + by = c0, then (x + x0; y + y0) is a solution to ax + by = c + c0. 1 Extended GCD Replacement Theorem 2 (Extended GCD Replacement). Let A; B 2 Z. Let k 2 Z, and write C = A + kB. Then 1. gcd(A; B) = gcd(B; C) 2 2. Let a; b 2 Z. Suppose gcd(a; b) = 1. Suppose that (xA; yA) 2 Z is a solution to ax + by = A and (xB; yB) is a solution to ax + by = B. Then (xA + kxB; yA + kyB) is a solution to ax + by = C. Proof. Item 1 is the GCD Replacement Theorem (proven in class) for A, B and C = A + kB. The additional statement, Item 2, is a consequence of the principle of adding solutions. Extended Euclidean Algorithm The regular Euclidean Algorithm is written in regular typeface (to match your course notes), and the Extended Euclidean Algorithm includes the extra italicized parts.
    [Show full text]
  • And the Golden Ratio
    Chapter 15 p Q( 5) and the golden ratio p 15.1 The field Q( 5 Recall that the quadratic field p p Q( 5) = fx + y 5 : x; y 2 Qg: Recall too that the conjugate and norm of p z = x + y 5 are p z¯ = x − y 5; N (z) = zz¯ = x2 − 5y2: We will be particularly interested in one element of this field. Definition 15.1. The golden ratio is the number p 1 + 5 φ = : 2 The Greek letter φ (phi) is used for this number after the ancient Greek sculptor Phidias, who is said to have used the ratio in his work. Leonardo da Vinci explicitly used φ in analysing the human figure. Evidently p Q( 5) = Q(φ); ie each element of the field can be written z = x + yφ (x; y 2 Q): The following results are immediate: 88 p ¯ 1− 5 Proposition 15.1. 1. φ = 2 ; 2. φ + φ¯ = 1; φφ¯ = −1; 3. N (x + yφ) = x2 + xy − y2; 4. φ, φ¯ are the roots of the equation x2 − x − 1 = 0: 15.2 The number ring Z[φ] As we saw in the last Chapter, since 5 ≡ 1 mod 4 the associated number ring p p Z(Q( 5)) = Q( 5) \ Z¯ consists of the numbers p m + n 5 ; 2 where m ≡ n mod 2, ie m; n are both even or both odd. And we saw that this is equivalent to p Proposition 15.2. The number ring associated to the quadratic field Q( 5) is Z[φ] = fm + nφ : m; n 2 Zg: 15.3 Unique Factorisation Theorem 15.1.
    [Show full text]
  • Euclid's Algorithm
    Euclid’s Algorithm Sudeshna Chakraborty, Mark Heller, Alex Phipps. Faculty Advisor: Dr. Ivan Soprunov, Department of Mathematics, Cleveland State University Introduction Proof continued Worst Case The Euclidean algorithm, also known as Euclid’s algorithm, is an Earlier we determined that the distance between the numbers algorithm for finding the greatest common divisor (GCD) between 1 GCD(a, b) b ≥ a > 0 was an important factor in determining the runtime, so this leads two numbers. The GCD is the largest number that divides two to the question what is the worst possible distance? Consecutive numbers without a remainder. The GCD of two numbers can be 2 b = q1 × a + r1 a > r1 ≥ 0 Fibonacci numbers give the worst possible run time. Fibonacci found by making a list of factors for the two numbers, and finding × numbers are recursive in nature adding the two previous the largest factor that is in both sets. This works well for small 3 a = q2 r1 + r2 r1 > r2 ≥ 0 numbers to get the next number. This leads to each q value (the numbers, but it can become quite tedious and time consuming 4 r = q × r + r r > r ≥ 0 same q value shown in figure 2) being 1 and the remainder when for larger numbers. To address this problem, Euclid’s algorithm 1 3 2 3 2 3 they are divided being the previous Fibonacci number. The can be used, which allows for the GCD of large numbers to be … example below demonstrates this on two small Fibonacci found much faster. Euclid’s algorithm uses the principle that the numbers 8 and 5: GCD of a set of two numbers does not change if you replace the 1: 8 = 1 × 5 + 3; n rn - 1 = qn + 1 × rn + 0 larger of the two with the remainder when you divide the larger of 2: 5 = 1 × 3 + 2; the two by the smaller.
    [Show full text]
  • Square Roots of Matrices J.B
    Volume 16 Part 1 February 1992 A SCHOOL MATHEMATICS MAGAZINE FUNCTION is a mathematics magazine addressed principallyto~~d~ntsmthe:upper forms of.secondary schools. .. ...... It is a's.peci~l interest' j()umal for those who are .inter.es.t~inth~th:~m~tics. Windsurfers, chess-players and gardeners all have magazines that cater to theiikter~sts. FUNCfION is a counterpart of these. Coverage is wide - pure mathematics, statIstiCS, computer science and applications of mathematics are all included. Recent issues have carried articles on advances in mathematics,- news items on mathematics and its ~pplications, special interest matters, such as computer chess, problems and solutions, discussions, cover diagrams, even cartoons. * **** Articles, correspondence, .problems {with or without solutions) 'and other material for publication are invited. Addressthetn to: 1'beJ3dit()fS; ~SFI<?N, .... ' D~p,8ft:JA~P:f'of .. ¥athematics, M911~$h.···yniyersity, Gl~yton; 'V'ietona, 3168. Alternatively·corresPondence'·Jllay.be-:acidressednindividually to any of the editors at the mathematics departments of the institution's listed on the· inside back cover. FUNCfION is pUbli~hed five times~y~~,app~aripg in 'Fe~ruary, April, June, August, October. Price for five. issues .(:i.n.cl\l,c!!n~ .. pQ~ta.ge): $17.00*; .si:n.g1e issues ·$4.00. Payments should be .. sent to theBusiness':1\I1aJl~gerat the above .address: .cheques and money orders should be made payable to Monash::Wnjversity. Enquiries about advertising should be directed to the business· manager. *$8.00 for bona fide secondary or tertiary students. ** * ** .FUNCTION Volume 16 Part 1 (Founder editor: G.B_. Preston) CONTENTS About Function 2 The Front Cover Michael A.B.
    [Show full text]