Divisibility in Z. Definition Let a, b Z be integers. We say that b divides a, if there exists c Z such that a = b c; we write b a.

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1 Divisibility in Z Tomáš Madaras 2016 Definition Let a, b Z be integers. We say that b divides a, if there exists c Z such that a = b c; we write b a. The divisibility of integers is thus a kind of relation between two numbers, that is, it is a binary relation on Z (for every two integers, either the first divides the second, or not).

2 Basic properties of divisibility of numbers with respect to 0 and 1: for all a Z, it holds a 0 0 a a = 0 a 1 a { 1, 1} 1 a Other properties: reflexivity: ( a Z) a a transitivity: ( a, b, c Z) (b a a c) b c ( a, b, c, u, v Z) (b a b c) b au ± cv ( a, b, c Z) b a b ac ( a, b Z, a 0) b a b a

3 Warning The relation of divisibility is not symmetric (the fact that b a does not imply a b); if b a and a b, then a = b. Warning If b a ± c, then neither b a nor b c holds in general: , but neither 4 17 nor 4 3. Similarly, if b ac, then neither b a nor b c holds in general: , but neither 4 10 nor 4 2.

4 Theorem (on Euclidean division) For each a Z, b Z + there exists the unique pair q, r of integers such that a = bq + r and 0 r < b. The idea of the proof: we find the smallest multiple of b which is less than a (the corresponding multiplies is equal to q); the difference of a and this multiple is then equal to r. Proof: Put q = a (where u is the floor of u, that is, the integer b c satisfying c u < c + 1). Then from which we get a b 1 < q a b a b < bq a bq a < bq + b Take r = a bq. Then the previous inequality yields 0 r < b.

5 Suppose now that, for some a, b Z, the pair q, r is not determined uniquely, that is, there exist two distinct pairs q, r and q, r such that a = bq + r = bq + r with 0 r < b, 0 r < b. If r = r, then also q = q, so assume (without loss of generality) that r > r. By subtracting these equalities, we obtain 0 = b(q q) + (r r) This gives b r r, but then b r r r, a contradiction.

6 Definition The number d Z is called common divisor of the numbers a, b Z, if d a and d b. The number d Z is called greatest common divisor of a, b Z, if 1 d a d b 2 ( d Z) (d a d b) d d The set of all common divisors of a, b Z is denoted by D(a, b); D(a) = D(a, a) denotes the set of all divisors of a. Warning The number 2 satisfies both above conditios for the greatest common divisor of the numbers 10 and 4, but these conditions are satisfied also by -2; in general, if d is greatest common divisor of a, b, then also d is. The greatest positive common divisor of a, b is denoted by (a, b).

7 Lemma Let a, b Z. If there exist integers u, v Z such that a = bu + v, then D(a, b) = D(b, v). The idea of the proof: we show that each number dividing both a and b, divides also v, and each number dividing both b and v, divides also a. Proof: let d D(a, b). Then d a, d b, from which d a bu = v. Hence d D(b), d D(v), so d D(b, v). Thus D(a, b) D(b, v) holds. Conversely, if t D(b, v), then t b, t v, hence t bu + v = a. Thus t D(b), t D(a) givingt D(a, b). We obtain then D(b, v) D(a, b). These set inclusions imply that D(a, b) = D(b, v).

8 Euclid algorithm INPUT: the integers a, b Z +, b 0 OUTPUT: the greatest positive common divisor of a, b 1 Find integers q 1, r 2 such that a = bq 1 + r 2 and 0 r 2 < b. 2 If r 2 > 0, then find integers q 2, r 3 such that b = r 2 q 2 + r 3 and 0 r 3 < r 2 else END 3 Repeat: If r k > 0, then find integers q k, r k+1 such that r k 1 = r k q k + r k+1 and 0 r k+1 < r k else END 4 END: r n+1 = 0, r n > 0 r n 2 = r n 1 q n 1 + r n r n 1 = r n q n Return r n = (a, b).

9 Example Find the greatest common divisor of 2076 and Using the Euclid algorithm (for a = 2076, b = 1776), we obtain 2076 = (q 1 = 1, r 2 = 300) 1776 = (q 2 = 5, r 3 = 276) 300 = (q 3 = 1, r 4 = 24) 276 = (q 4 = 11, r 5 = 12) 24 = (q 5 = 2, r 6 = 0) Hence (2076, 1776) = 12.

10 The proof of algorithm finiteness: for integers b, r 2, r 3,..., r n, according to algorithm, we get b > r 2 > r 3 > > r n > 0 hence, there are finitely many of them (we have n b). The proof of algorithm correctness: From the equality r n 1 = r n q n we get r n D(r n 1 ), from r n 2 = r n 1 q n 1 + r n and from r n D(r n 1 ) we have r n D(r n 2 ); in general, from r k 1 = r k q k + r k+1 and from r n D(r k ) we obtain r n D(r k 1 ). Thus, we have r n D(r n 1 ), r n D(r n 2 ),..., r n D(r 2 ), r n D(b), r n D(a). Hence r n D(b) D(a) = D(a, b). Let d D(a, b), d > 0 is a common divisor of a, b. Then

11 from the equality a = bq 1 + r 2 we have D(a, b) = D(b, r 2 ) from the equality b = r 2 q 2 + r 3 we have D(b, r 2 ) = D(r 2, r 3 ) from the equality r k 1 = r k q k + r k+1 we have D(r k 1, r k ) = D(r k, r k+1 ) from r n 1 = r n q n we have D(r n 1, r n ) = D(r n, 0) = D(r n ). Thus d D(r n ), so d r n. This gives, using the definition of the greatest common divisor, that r n = (a, b).

12 Lemma (Bézout identity) If a, b N, then there exist integers u, v Z such that (a, b) = au + bv. The idea of the proof: we take all integer linear combinations of a, b; then the smallest positive such combination is equal to (a, b). Proof: Consider the set H = {ax + by x, y Z}. Let d be the smallest positive integer from this set, that is, d = au + bv > 0 for some u, v Z. By theorem of Euclidean division, for x, y Z, there exist integers q, r, 0 r < d such that ax + by = dq + r. Then r = ax + by dq = ax + by (au + bv)q = a(x uq) + b(y vq) = ax + by H. Hence r H and r < d, which implies r = 0. From this we obtain that d divides any number from H, and thus d divides both a and b (because a = a 1 + b 0 H, b = a 0 + b 1 H.) Further, if d > 0, d D(a, b), then d d (since d = au + bv), thus d d. This yields d = (a, b).

13 Example Express the greatest common divisor of 2076 and 1776 in the form 2076u v for u, v Z. From the Euclid algorithm for a = 2076, b = 1776, we obtain equalities 2076 = = = = from which we obtain the following expressions: 300 = = = =

14 Example (cont.) From the penultimate equality, we substitute the expression for 24 to the last equality, thus obtaining the expression of 12 using the integers 276 and 300: 12 = = 276 ( ) 11 = 300 ( 11) Into this equality, we substitute the expression of 276 using 1776 and 300: 12 = 300 ( 11) = 300 ( 11) + ( ) 12 = ( 71) Finally, we substitute 300 expressed by 2076 and 1776: 12 = ( 71) = ( ) ( 71) = 2076 ( 71) Hence 12 = (2076, 1776) = 2076 ( 71)

15 Definition The integers a, b Z are called coprime, if their greatest common divisor equals 1. Lemma The integers a, b Z are coprime if and only if there exist u, v Z such that 1 = au + bv. Proof: If a, b are coprime, then, by Bézout identity, there exist integers u, v such that 1 = (a, b) = au + bv. Conversely, let there exist integers u, v Z such that 1 = au + bv, and let (a, b) = d > 1. Since d a, d b, we have a = d a, b = d b and 1 = au + bv = (d a )u + (d b )v = d(a u + b v) from which we obtain that d 1, a contradiction.

16 Theorem Let a, b, c Z be integers. Then: a 1 if b 0, then (a, b), b are coprime (a, b) 2 if a, b are coprime and a, c are coprime, then a, bc are coprime 3 if a bc a a, b are coprime, then a c 4 if a c, b c and a, b are coprime, then ab c Proof: 1) By Bézout identity, there exist u, v Z such that (a, b) = au + bv; this implies 1 = au + bv (a, b) = a (a, b) u + b (a, b) v Since the numbers implies that ( a (a, b), a (a, b), b (a, b) b (a, b) ) = 1. are integers, the last equality

17 2) If a is coprime both with b and with c, then there exist integers u, v, x, y such that 1 = au + bv 1 = ax + cy From the second equality we get b = axb + cyb and, after substituting the b into the first equality, we obtain 1 = au + (axb + cyb)v = a(u + xbv) + (bc)yv = ax + (bc)y which means that (a, bc) = 1. 3) We have (a, b) = 1, hence there exist u, v Z such that 1 = au + bv. From this we get c = acu + bcv. Since a acu and a bcv (as, by theorem assumptions, a bc), we have a acu + bcv = c. 4) Since a c, b c, we have c = ma = nb for some m, n Z. Further, (a, b) = 1, so there exist u, v Z such that 1 = au + bv. From this we obtain that c = auc + bvc = au(nb) + bv(ma) = ab(nu + mb), hence ab c.

18 Definition The integer n Z is called least common multiple of a, b Z, if 1 a n b n 2 ( n Z) (a n b n ) n n The least common multiple of a, b is denoted by [a, b]. Applications: the summation of fractions (common denominator)

19 Theorem For all a, b Z, a, b 0, it holds [a, b] = ab (a, b). Proof: Let m = ab (since (a, b) a, (a, b) b, m is a common (a, b) multiply of a and b) and let M be a common multiple of a, b. From the previous theorems, we have that there exist integers u, v such that (a, b) = au + bv. Then M m = M ab (a,b) = M (a, b) ab = M(au + bv) ab = M b u + M a v. Since M is a multiple of both a and b, the fractions M a, M b are integers in fact, thus the right side of the considered equality is an integer. This implies that m M, so m M. From the definition of the least common multiple we therefore obtain that m = [a, b].