# 2 The Euclidean algorithm

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1 2 The Euclidean algorithm Do you understand the number 5? 6? 7? At some point our level of comfort with individual numbers goes down as the numbers get large For some it may be at 43, for others, 4 In any case, the most basic way of understanding a natural number is nowing its prime factorization Ironically, there is no easy way to determine if a number is prime or not, or what its prime factors are For instance, say you had a random 5 digit number: There is no simple way to determine if it is prime or not, or what its divisors (other than 1 obviously) are, short of trying to divide it by all possible smaller numbers: 2, 3, 4, 5, 6, etc (Of course it suffices to try to divide it by all primes smaller than it, but to put this in to practice, you already need to now which smaller numbers are primes and which are not) On the other hand, it turns out to be easy to determine the gcd (greatest common divisor) of two numbers This may seem somewhat paradoxical at first, but having two numbers instead of just one lets you compare them In fact, if we thin about what it should mean to understand something (a number), one good interpretation is to understand how it is similar and how it is different from other things (other numbers) And, from a multiplicative point of view, the gcd tells us precisely what two numbers have in common Furthermore, the gcd will help us with the problems of understanding primes and prime factorization 21 The gcd by subtraction Let a, b, d N First note that if d is a common divisor of a and b, ie, for some a, b N, then a = a d, b = b d, a b = a d b d = (a b )d so d is a divisor of a b Similarly, if d is a common divisor of a b and b, then it is also a divisor of a = (a b) b Hence the common divisors of a and b are the same as the common divisors of a b and b In particular, gcd(a, b) = gcd(b, a b) Euclid used this idea to mae an efficient algorithm to determine gcd(a, b) The Euclidean algorithm goes as follows Set Then we inductively compute stopping only when we have a 1 = max {a, b}, b 1 = min {a, b} a i+1 = max {b i, a i b i }, b i+1 = min {b i, a i b i }, a = b (This procedure produces smaller and smaller pairs of natural numbers so must eventually terminate by descent The max/min business is to ensure we always have a i b i so that the a i b i appearing in the next step is positive) 9

2 Example Do a = 15, b = 6 The reason this wors is as follows Since gcd(a, b) = gcd(b, a b), we have gcd(a, b) = gcd(a 1, b 1 ) = gcd(a 2, b 2 ) = = gcd(a, b ) = gcd(a, a ) = a Example Do a = 18, b = 5 If gcd(a, b) = 1, we say a and b are relatively prime Exercise 21 Exercises 211, 213, 215 Exercise 22 Compute gcd(84, 63) using the above method Write out each step 22 The gcd by division with remainder A more efficient version of the Euclidean algorithm is as follows Set halting when we have a pair Then a 1 = max {a, b}, b 1 = min {a, b}, a i+1 = b i, b i+1 = remainder in a i /b i, a, b with b a gcd(a, b) = b This algorithm is essentially the same as the subtraction version, but the division can do several steps of subtraction at once Example Do a = 18, b = 5 Write a and b in binary Suppose a > b and a is n bits (binary digits) long Then the remainder in a/b has at most n 1 bits, so this algorithm will terminate at most n steps In other words, if max a, b < 2 n+1, then we can determine gcd(a, b) in at most n steps (called logarithmic time) This is as efficient as one could hope for A computer could handle numbers thousands of digits long in a fraction of a second On the other hand, even with very advanced algorithms, a modern computer might tae up to a year to factor a 200 digit number Another advantage of the division version is it can deal with other number systems Eg, if you want to compute gcd(17, 4 + i) in Z[i], you can divide 17 by 4 + i (and get 4 i exactly), but subtraction gives you nothing Exercise 23 Compute gcd(42, 8) using the division method Write out each step 10

3 23 Linear representation of the gcd If we go bac to the subtraction version of the Euclidean algorithm, it is clear that at each step a i and b i are (integral) linear combinations of a and b Hence gcd(a, b) = a = ma + nb (1) for some m, n Z Often it will be helpful to determine not just gcd(a, b) but also the above m and n This can be done through a variety of equivalent methods, sometimes called the extended Euclidean algorithm We will present the tableau method, which is more efficient than the one in the text Consider the example from the text: a = 34, b = 19 The idea is to use a little linear algebra, and is similar to matrix row reduction, but we build a table, starting with the following two rows For clarification I will write the underlying equation on the right, though in practice you will omit this The entries running down the x column will just be the successive numbers a 1, b 1, b 2,, b from the division algorithm The m and n entries for the b i row will just be the coefficients needed for ma + nb = b i For example, here b 2 = a 1 b 1, so the next row will just be obtained by subtracting the second from the first (do this to each column) to get We do this again to get a 1 b = a 1 b = a + 2 b = 4 Now 4 goes into 15 3 times, so we should subtract 3 times the last row from the previous row to get a 1 b = a + 2 b = a 7 b = 3 11

4 With one more step we are done: a 1 b = a + 2 b = a 7 b = a + 9 b = 1 We now we are done now because the last b j (x = 1) divides the previous b j (x = 3) Hence the tableau method has shown two things: gcd(a, b) = 1 and gcd(a, b) = 5a + 9b Exercise 24 Exercise 232 (You may use either the tableau method, the method in the text, or any other equivalent method you lie Just write down each step and explain your method if it is not one of two mentioned above) 24 Primes and factorization Proposition 21 Let n N, n > 1 Then has a prime factorization n = p 1 p 2 p where each p i is prime (Here the p i s are not necessarily distinct) Proof By definition, n is either prime or n = ab for some a, b > 1 If a and b are prime, we are done If not, then we repeat this with a and b, until we have reduced all factors to products of primes This process terminates by descent, and we are done Note the existence of a prime factorization relies only on the definition of prime and Fermat s descent, and not any any arithmetic of N Hence this will hold for any number system in which descent holds (such as N or Z, but not Q or R) Contrast that to the following Proposition 22 (Prime divisor property) Let p be prime If p ab then p a or p b Proof Suppose p a Then gcd(a, p) = 1 since p is prime From Section 23, we have for some m, n Z Hence 1 = ma + np b = mab + npb Since p mab (by assumption) and p npb (trivially), we have p b, which proves the proposition 12

5 This seemingly evident fact relies on the arithmetic of N (or Z if you will) in particular the fact that gcd(a, b) is a linear combination of a and b (which we proved with via the Euclidean algorithm) We will see that in other number systems, such as Z[ 5] = { a + b 5 : a, b Z }, this prime divisor property is not true The prime divisor property allows us to prove another basic result which we normally tae for granted (This result and the previous are in fact equivalent) Proposition 23 (Fundamental theorem of arithmetic; aa Unique prime factorization; aa Unique factorization) Let n N, n > 1 Then the prime factorization of n is unique, up to reordering the primes Proof (By contradiction) Suppose n has two distinct factorizations p 1 p = q 1 q l By cancelling out any common factors on each side, we may assume none of the p i s equal any of the q j s Now p 1 divides the product on the left, hence p 1 q 1 q l So by the prime divisor property, p 1 q 1 or p 1 q 2 q 3 q l But p 1 q 1 because p 1 q 1 and q 1 is prime Hence p 1 q 2 q l = p 1 q 2 or p 1 q 3 q l Similarly p 1 q 2, and continuing this argument we eventually get p 1 q l 1 q l = p 1 q l 1 or p 1 q l But both of these are impossible, contradicting our supposition We will often be woring with integers and not just natural numbers, so it may be helpful to restate it in terms of integers It is not explicitly stated this way in the text at this point Fundamental theorem of arithmetic (for integers) Let n Z, n 0 Then n can be expressed in a unique way (up to reordering) as n = up e 1 1 pe 2 2 pe where u is a unit (±1), p i N is prime, e i N and 0 (Note we allow the possibility that no primes appear ( = 0) in the factorization to include the units ±1 in the statement) This result is called the fundamental theorem of arithmetic (though not by the text), because it plays such an important role in number theory (which is sometimes also called arithmetic) Its importance was first pointed out by Gauss (because he noticed that it doesn t hold for Z[ 5], which maes Z[ 5] much harder to wor with) However, it will hold in some other number systems, such as the Gaussian integers Z[i] (the statement will be essentially identical to our second version) 13

6 25 Consequences of unique prime factorization Here we will explore some simple consequences of unique factorization First we will mae the following observation: n is a square n = p 2e 1 1 p 2e Proof ( ) If n = m 2, write m = p e 1 1 pe ( ) Clearly n = (p e 1 1 pe ) 2 This observation does not require unique prime factorization, but the next proposition does (Pay attention how Unfortunately the text is not careful in pointing out where unique factorization is used and how it is required here) Proposition 24 If ab is a square and gcd(a, b) = 1, then a and b are squares Proof If ab is square, write ab = p 2e 1 1 p 2e Now by unique factorization, the prime factorizations of a and b must be of the form a = p f 1 1 pf, b = pg 1 1 pg where f i, g i 0 and f i + g i = 2e i If a and b share no common factors, then by reordering our primes, we can write a = p 2e 1 1 p 2e j j, b = p 2e j+1 j+1 p 2e Hence a and b are squares If you don t have unique factorization, then it could happen that there are distinct primes p, q and r such that p 2 = qr, so that the proposition would be false Proposition 25 If N is nonsquare, then N is irrational Proof We prove the contrapositive Suppose N = a b is rational Write and a = p m 1 1 p m b = p n 1 1 pn where m i, n i 0 (Note this is not the unique prime factorization, but by allowing 0 s in the exponents, we can write a and b as products of the same primes) Then Hence N is a square N = a2 b 2 = p2(m 1 n 1 ) 1 p 2(m n ) Exercise 25 Does Proposition 25 (the irrational square roots result in Section 25) require that the prime factorization is unique? Explain 14

7 Exercise 26 Prove the following (slighly generalized) assertion from Section 25: Unique prime factorization implies that each prime power divisor of a natural number n (ie, a prime power that divides n) actually appears in the prime factorization of n (Hint: it s easy, but you need to write down what it means to be a divisor) The previous exercise implies that a common prime power divisor of a and b appears in the prime factorization of each Hence, gcd(a, b) is the product of the largest common prime powers in the prime factorizations of a and b By largest common prime power we mean for each prime occuring in both a and b, the largest power of it which divides both (The text does not tal about prime powers here, maing its statements somewhat vague) Example Do a = , b = Thus it is easy to compute the gcd if you now the prime factorizations We can rewrite this mathematically as follows Write a = p m 1 1 p m, b = pn 1 1 pn where m i, n i 0 (again not the unique factorization as exponents of 0 are allowed) in terms of common primes Then gcd(a, b) = p min(m 1,n 1 ) 1 p min(m,n ) Similarly one has lcm(a, b) = p max(m 1,n 1 ) 1 p max(m,n ) Exercise 27 Exercises 251 and Linear Diophantine equations Note: This section could have been done right after Section 23, and might be more natural there The simplest Diophantine equations (remember Diophantine equations? number theory?) are the binary (2-variable) linear ones: the alleged subject of ax + by = c, a, b, c Z (2) For fixed a, b, c the first thing to as is, is there a solution (in integers)? If so, determine all solutions Graphically, this is a line in the plane with rational slope a b and rational y-intercept c b : y = a b x + c b (assuming b 0) But we will not solve it graphically it is easier to use the divisor theory we developed Proposition 26 Equation (2) has a solution (in integers) if and only if gcd(a, b) c Proof ( ) If there is a solution, then gcd(a, b) ax and gcd(a, b) by = gcd(a, b) c ( ) If gcd(a, b) c, we can write c = gcd(a, b)d and by Section 23, for some m, n Z gcd(a, b) = am + bn = c = gcd(a, b)d = amd + bnd exhibiting a solution of x = md and y = nd where m and n are found via the extended Euclidean algorithm (Section 23) 15

8 Note the homogeneous equation ax + by = 0 (3) has infinitely many solutions Specifically, let d = gcd(a, b) and write a = a d, b = b d Then all integer solutions are given by { (x, y) = (b, a ) : Z } Proposition 27 Suppose (2) has a solution (x 0, y 0 ) (which we obtained in the proof of the proposition above) All solutions to (2) are given by (x 0, y 0 ) + (x, y) where (x, y) is a solution to (3) Proof You should have seen this proof in linear algebra already ( ) Suppose (x 1, y 1 ) is another solution to (2) We want to show it is of the desired form Then (ax 1 + by 1 ) (ax 0 + by 0 ) = c c = 0 Hence (x, y) = (x 1 x 0, y 1 y 0 ) is a solution to (3) ( ) Suppose (x, y) is a solution to (3) Then so (x 0, y 0 ) + (x, y) is a solution to (2) a(x 0 + x) + b(y 0 + y) = (ax 0 + by 0 ) + (ax + by) = c + 0 = c Exercise 28 Exercises 261, 262, 263, *The vector Euclidean algorithm 28 *The map of relatively prime pairs We will sip these sections I do not find them particularly interesting, however they are used in Conway s graphical theory of quadratic forms (see Chapter 5), which I do find interesting, but I do not plan to cover 29 Discussion Read it if you want it I made all the comments I wanted to already 16

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