MATH 25 CLASS 5 NOTES, SEP

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1 MATH 25 CLASS 5 NOTES, SEP Contents 1. A brief diversion: reltively prime numbers 1 2. Lest common multiples 3 3. Finding ll solutions to x + by = c 4 Quick links to definitions/theorems Euclid s Lemm (importnt!) 1. A brief diversion: reltively prime numbers Before continuing with the study of liner equtions, we mke brief detour to tlk bout some useful properties of reltively prime numbers nd number relted to gcds. Recll tht two integers, b re reltively prime if gcd(, b) = 1. At this point, we know enough to prove some very importnt fcts bout reltively prime numbers: Proposition 1. Let, b be two reltively prime numbers, nd let c be some integer. If bc, then c. Proof. We know tht bc. Becuse, b re reltively prime, we know tht x+by = 1 hs (infinitely) mny integer solutions. Select one of them. Multiply this eqution by c: cx + bcy = c. Notice tht cx, nd since bc by ssumption, bcy. Therefore, c. This simple result is of fundmentl importnce. Notice tht we used our knowledge bout when x + by = d hs solutions in n essentil wy to prove this proposition. Another importnt observtion is tht the bove proposition requires tht, b be reltively prime in order to be true. Cn you think of n exmple where, b re not reltively prime, nd where bc but c? Finlly, one specil cse of the bove proposition deserves mention. Suppose = p is prime number ( number divisible only by 1 nd itself). Then the bove proposition cn be rewritten in the following wy: Lemm 1 (Euclid s Lemm). Let p be prime, nd let, b be two integers. If p b, then p or p b. Proof. If p, there is nothing to prove, so suppose p. Then gcd(, p) = 1, since the only divisors of p re 1 nd p, while p does not divide. An ppliction of the previous proposition shows tht p b. 1

2 2 MATH 25 CLASS 5 NOTES, SEP Exmple. This exmple shows tht the originl proposition (nd Euclid s Lemm) cn be flse when their ssumptions re not true. For instnce, if = 4, b = 6, so tht gcd(, b) = 2, then we cn choose c = 2. Then bc = 12, so bc, but c. This exmple lso works to show why p must be prime in Euclid s Lemm; notice tht = 4 is not prime, yet b, c. The previous proposition nd lemm re one of the most importnt pplictions of our knowledge of when x + by = d hs integer solutions. It is well worth lerning their sttements nd proofs thoroughly. Here re severl other useful propositions: Proposition 2 (Corollry 1.11 of the text). If, b re reltively prime integers, nd c, b c, then b c. Proof. Since gcd(, b) = 1, there exist integers x, y such tht x + by = 1. Multiply this eqution by c : cx + bcy = c. Since b c, (b) cx, nd since c, (b) bcy. Therefore (b) c. Proposition 3. [Exercise 1.8 of the text] Let, b be two integers. If c is divisor of, b, then c gcd(, b). Proof. We know tht there is pir of integers x, y such tht x + by = gcd(, b). Since c, b, this implies tht c gcd(, b). Proposition 4 (Corollry 1.10 of the text). Let, b be two integers, nd let m be positive integer. Then gcd(m, mb) = m gcd(, b). Proof. Clerly m gcd(, b) gcd(m, mb), becuse m gcd(, b) divides both m nd mb. For the reverse inequlity, gin there re two integers x, y such tht x + by = gcd(, b). Multiplying this eqution by m, we get mx+mby = m gcd(, b). However, this is only possible if gcd(m, mb) m gcd(, b), which in prticulr implies tht gcd(m, mb) m gcd(, b), s desired. Proposition 5 (Corollry 1.10 of the text). Let, b be two integers, nd let d, b. Then gcd( d, b gcd(, b) ) =. In prticulr, d d gcd(, b) nd b re reltively gcd(, b) prime. Proof. Agin, there exist integers x, y which stisfy x + by = gcd(, b). Divide this eqution by d: d x + b gcd(, b) y =. d d Since /d, b/d re integers, this sys tht gcd(/d, b/d) gcd(, b)/d. On the other hnd since gcd(, b), b, gcd(, b)/d /d, b/d. So gcd(, b)/d gcd(/d, b/d), nd therefore we hve equlity. As you cn see, we re getting lot of milege out of the fct tht x + by = d hs integer solutions x, y if nd only if gcd(, b) d. Let s conclude this section with n exmple illustrting these propositions. Exmples.

3 MATH 25 CLASS 5 NOTES, SEP We sw tht gcd(994, 399) = 7. Therefore, the only common divisors of 994, 399 re 1, 7 (Proposition 3). As 994 = 7 142, 399 = 7 57, we lso see tht gcd(142, 57) = 1. (Proposition 5) Proposition 2 cn be flse if gcd(, b) 1. For instnce, if = 6, b = 9, nd c = 18, then c, b c, but b = 54 c. 2. Lest common multiples Recll tht multiple of n integer is ny number of the form n, where n Z. Given two numbers, b, we cll the smllest positive integer which is both multiple of, b the lest common multiple of, b. This number is often written lcm(, b), or sometimes [, b], lthough gin the ltter nottion cn be mbiguous, since it lso mens the closed intervl from to b. There is the obvious generliztion of this definition to list of more thn two numbers. Exmple. Let = 8, b = 12. Then the lest common multiple of, b is 24, since 24 is the smllest number tht is multiple of both, b. How re the lcm nd gcd of two nonzero numbers, b relted? Notice tht gcd(8, 12) = 4, for exmple. A bit of experimenttion will probbly led you to the clim tht gcd(, b)lcm(, b) = b. Let s prove this: Proof. We cn ssume tht, b re positive, since gcd, lcm re unchnged if we chnge the signs of, b. First notice tht becuse gcd(, b), b, we know tht / gcd(, b) is n integer, nd similrly, b/ gcd(, b) is n integer. Therefore, gcd(, b) b = b gcd(, b) = b gcd(, b) shows tht b/(gcd(, b)) is common multiple of, b. Therefore, lcm(, b) b/(gcd(, b)). Now we wnt to show tht the opposite inequlity is true. Suppose tht c is the lest common multiple of, b. Then we cn write c = n = bm for some integers n, m. In prticulr, this mens tht n bm, m n. First notice tht gcd(n, m) = 1. This must be true becuse if gcd(n, m) > 1, then we cn divide both n, m by their gcds to obtin new integers n, m, with (n, m ) = 1, nd n = bm is still common multiple of, b which is smller thn c, contrdicting the fct tht c is the lest common multiple of, b. Since gcd(n, m) = 1, we cn pply the first proposition we lerned to see tht n b, m. Let 1 = /m, b 1 = b/n. However, we know tht n = bm, so this tells us tht 1 = b 1. Cll this number d. Notice tht d is common divisor of, b. Therefore, d gcd(, b). But this implies tht b d b gcd(, b). Since b/d = c, this shows tht c b/(gcd(, b)) s desired. Putting the two inequlities we ve proved together, we hve c = b/(gcd(, b)), s desired. Exmple. Going bck to = 994, b = 399, since gcd(994, 399) = 7, lcm(994, 399) = /7 =

4 4 MATH 25 CLASS 5 NOTES, SEP We conclude with proposition which is the mirror imge of Proposition 3. Proposition 6 (Exercise 1.14 of the text). Let c be common multiple of, b. Then c is multiple of lcm(, b). Proof. Write l = lcm(, b). Since c l, Eucliden division of c by l gives n eqution c = lq + r, where 0 r < l. But since, b c, l, this mens, b r, which shows tht r is common multiple of, b. Since l is the lest common multiple, we must hve r = 0, which mens tht c is multiple of l = lcm(, b), s desired. The converse to the bove proposition is obviously true tht is, ny multiple of lcm(, b) is itself common multiple of nd b. Let s conclude by going bck to fmilir exmple. Exmple. We clculted tht lcm(994, 399) = Therefore ny common multiple of 994 nd 399 is multiple of Finding ll solutions to x + by = c The Eucliden lgorithm gives us wy to find pir of integer solutions x, y to x + by = c, s long s gcd(, b) c. However, it would be idel to know how to find ll the solutions to this eqution, insted of just one. The following proposition tells us just how to do this: Proposition 7 (Theorem 1.13 of the text). Let, b be nonzero integers, nd c n integer which is multiple of gcd(, b) = d. Let x 0, y 0 be one pir of integer solutions to x + by = c. Then the set of ll integer solutions x, y to the eqution x + by = c hs the form (1) x = x 0 + b d n, y = y 0 d n, where n is ny integer. (In prticulr when n = 0 we get the initil pir x 0, y 0.) Proof. We will begin by checking tht every pir of integers x, y stisfying Eqution 1 stisfies x + by = c. Plug in the two equtions from Eqution 1 into x + by = c: (x 0 + bd ) n + b (y 0 ) d n = x 0 + b d n + by 0 b d n = x 0 + by 0 = c. In the lst equlity, we used the fct tht x 0, y 0 ws solution to x + by = c. We now wnt to prove the converse sttement, tht ny solution x, y is of the form given by Eqution 1. So suppose x, y re integers such tht x + by = c. Since x 0 + by 0 = c s well, we hve x 0 + by 0 = x + by, or (x 0 x) = b(y y 0 ). Both sides re divisible by d = gcd(, b), so divide both sides of this eqution by d: d (x 0 x) = b d (y y 0).

5 MATH 25 CLASS 5 NOTES, SEP Recll tht /d, b/d re reltively prime. Since /d, b/d re reltively prime nd (b/d) (/d)(x x 0 ), we must hve (b/d) (x x 0 ). In other words, there is n integer n such tht b d n = x x 0, or x = x 0 + b d n. Plugging in this expression for x into the previous eqution, we obtin Solving for y, we get b d d n = b d (y y 0). y = y 0 d n. Exmples. Going to our fvorite exmple of = 994, b = 399, we found the solution x = 2, y = 5 to 994x + 399y = 7. Since gcd(, b) = d = 7, nd /d = 142, b/d = 57, the previous proposition tells us tht every solution to 994x + 399y = 7 is given by x = n, y = 5 142n, where n Z. Notice tht this proposition works on the eqution x + by = c even when c is lrger thn gcd(, b). For exmple, consider the eqution 4x + 6y = 4. It is obvious tht x = 1, y = 0 gives n integer solution. We hve = 4, b = 6, gcd(, b) = d = 2, so /d = 2, b/d = 3. Then the previous proposition tells us tht every pir of integer solutions hs the form x = 1 + 3n, y = 2n. In generl, it is esy to check your nswer by plugging in your expressions for x, y into the eqution x + by = c nd checking tht you get true eqution. In prticulr, ny ns which pper should end up cnceling out.