Solutions Math 308 Homework 9 11/20/2018. Throughout, let a, b, and c be non-zero integers.

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1 Throughout, let a, b, and c be non-zero integers. Solutions Math 308 Homework 9 11/20/ Consider the following statements: i. a is divisible by 3; ii. a is divisible by 9; iii. a is divisible by 12; iv. a = 24; v. a 2 is divisible by 3; vi. a is even and divisible by 3. Which conditions are necessary for a to be divisible by 6? Which are sufficient? Which are necessary and sufficient? Answer. Recall, a statement A is necessary for 6 a if 6 a A. Namely, whenever a = 6k for some k Z implies A. And A is sufficient if A 6 a. Namely, if A, then a = 6k for some k Z. i. 3 a: Necessary, but not sufficient. Why: If a = 6k = 3(2k), then 3 a. But if a = 3, then 3 a but 3 a. ii. 9 a: Not necessary, and not sufficient. Why: If a = 6, then 6 a but 9 a. And if a = 9, then 9 a but 6 9. iii. 12 a: Sufficient, but not necessary. Why: If a = 6, then 6 a but 12 a. But if a = 12k = 6(2k), then 6 a. iv. a = 24: Sufficient, but not necessary. Why: If a = 6, then 6 a but a 24. But if a = 24 = 6(4), then 6 a. v. 3 a 2 : Necessary, but not sufficient. Why: We showed that 3 a if and only if 3 a 3. Thus this is equivalent to i. vi. a is even and divisible by 3: Necessary and sufficient. Why: If a is even and 3 a, then a = 2k for some k Z and 3 a, so that 3 2k. But since 3 is prime and 3 2, we have 3 k. So k = 3l, meaning that a = 2(3l) = 6l. Therefore 6 a. And if 6 a, then a = 6m = 2 3 m for some m Z. Thus a is even and 3 a. 2. Run the Euclidean algorithm with a = 30, b = 12. Answer. 30 = =

2 2 3. Recall from lecture that executing the Euclidean algorithm for a = 100 and b = 36 gives the following equations: 100 = , (E1) 36 = , (E2) 28 = , (E3) 8 = (E4) (a) Follow these steps to express 4 as an integer combination of 100 and 36, i.e., find (possibly negative) integers x and y such that 100x + 36y = 4: (i) Use equation (E3) to express 4 as an integer combination of 8 and 28 (find integers x and y such that 8x + 28y = 4). (ii) Use equation (E2) to express 8 as an integer combination of 28 and 36. (iii) Use equation (E1) to express 28 as an integer combination of 36 and 100. (iv) Plug your equation from part (ii) into your equation in part (i), expanding and simplifying, to express 4 as an integer combination of 28 and 36. (v) Plug your equation from part (iii) into your equation in part (iv), expanding and simplifying, to express 4 as an integer combination of 36 and 100. (b) Make an argument (write an informal proof) justifying the following claim: For any positive integers a and b, there exist integers x and y satisfying gcd(a, b) = ax + by. Answer. We have 4 = , 8 = , and 28 = So 4 = 28 ( ) 3 = = 4 ( ) 3 36 = In general, we can use this strategy to solve for any r i in terms of r i 1 and r i 2. Then by iteratively substituting, we can solve for any r i in terms of a and b. And since gcd(a, b) is the last non-zero r i, this means that we can solve for gcd(a, b) in terms of a and b. At reach step, we are taking integral combinations of the r j s and so the result is an integral combination of a and b.

3 3 4. Consider Euclid s Lemma and its proof from Chapter 28 of How to think... : Euclid s Lemma. Suppose that n, a, and b are natural numbers. If n ab and gcd(n, a) = 1, then n b. Proof. Since gcd(n, a) = 1, there exist integers k and l such that kn + la = 1. Thus knb+lab = b. We obviously have n knb. We also have n ab, so n lab. Thus n knb+lab, i.e. n b. (a) Analyze the theorem statement: give examples, non-examples, assumptions and conclusions, compare to other results, etc. Compare to the statement given in class. Answer. You answers will vary based on the examples you give. But the assumptions are that that n, a, and b are natural numbers, that n ab, and that gcd(n, a) = 1; and the conclusion is that n b. This is similar to one of our alternate definitions of primes (where n is prime). (b) Identify in the proof (here) where the hypotheses were used. Answer. Proof. Since gcd(n, a) = 1, there exist integers k and l such that kn + la = 1. Thus knb + lab = b. We obviously have n knb. We also have n ab, so n lab. Thus n knb + lab, i.e. n b. (c) What theorems/lemmas/etc. were used in the proof? Answer. We used Theorem 28.7, which says that gcd(n, a) = kn + la for some k, l Z. We also used Theorem 27.5, which says that if n b and n c, then n (xb + yc) for all integers x and y. (d) Compare/contrast this proof to the proof from class. (e) Analyze what happens when we drop the hypothesis that gcd(n, a) = 1. Answer. If gcd(n, a) = d > 1, then we still have integers k and l such that kn + la = d, so that knb + lab = bd. This still implies that n bd. But this does not imply that n b. For example, if n = 12, a = 9, and b = 4, then n ab, but n b. We do have, though, that d = gcd(12, 9) = 3, so that n bd. Using hypotheses Using hypotheses

4 5. Prove the following. (a) We have a b if and only if a b. Proof. If a b, then b = ka for some k Z. Thus b = ( k)( a), giving that a b. And if a b, then b = la for some l Z. Thus b = ( l)a, giving that a b. 4 (b) If δ is a common divisor of a and b, then δ gcd(a, b). Proof. Let d = gcd(a, b) so that there are some x, y Z such that xa + yb = d. Now, if δ a and δ b, then there are also some k, l Z such that a = kδ and b = lδ. Thus d = x(kδ) + y(lδ) = (xk + yl)δ. So since xk + yl Z, we have δ d, as desired. (c) If a > 4 is not prime, then a (a 1)!. Proof. We have shown that every integer has a prime factorization, so let p be a prime such that p a. Since a is not itself prime, we have p < a. Now, since p a, we have some b Z such that a = pb. And since p > 1, we have 2 b a 1. Now, if b p, then (a 1)! = ( ) i = pb j = (pb)m = am. 1 i a 1 1 j a 1 j b,p } {{ } m So a (a 1)!, as desired. Otherwise, b = p 2, i.e. a = p 2. But if a > 4, then p 3. Then 2p < a and 2p p. So (a 1)! = ( ) i = p(2p) j = p 2 (2m) = a(2m). 1 i a 1 Thus a (a 1)!, as desired. 1 j a 1 j p,2p } {{ } m Note: originally there was a typo saying that a 4 instead of a > 4. But you can check that this must be a typo since (4 1)! = 3! = 6, which is not a multiple of Use strong induction to prove the division algorithm: For any a, b Z with b 0, there are unique integers q and r satisfying a = bq + r and 0 r < b. Your answer should be a clean-up of Lemma 28.1 in the book.

5 7. An integer l is called a common multiple of non-zero integers a and b if a l and b l. The smallest positive such l is called the least common multiple of a and b, denoted lcm(a, b). For example, lcm(3, 7) = 21 and lcm(12, 66) = 132. (a) Compute lcm(12, 8), lcm(30, 20), lcm( 10, 22), and lcm(9, 10). Answer. lcm(12, 8) = 24, lcm(30, 20) = 60, lcm( 10, 22) = 110, lcm(9, 10) = (b) Prove that if a m and b m, then lcm(a, b) m. Proof. We will see in our proof of part (d) that if L = gcd(a,b), then L = lcm(a, b). We will do this by proving that L divides any common multiple of a and b. Thus this statement follows from the proof of part (d). ab Note: we call this a porism. It s like a corollary, but it doesn t follow from the statement; it follows from the proof. (c) Prove that for any r Z, we have lcm(ra, rb) = r lcm(a, b). Proof. Let l = lcm(a, b). Since a lcm(a, b) and b lcm(a, b), there are x, y Z such that ax = l and by = l. So rax = rl and rby = rl. So rl is a common multiple of ra and rb. And if m > 0 is a common multiple of ra and of rb, then there are x, y Z such that Then letting n = m r, we have rax = m and rby = m. n = ±ax and n = ±by, so that n is a positive integer and is common multiple of a and b. Therefore n l. And therefore m = r n > r l. So l is the least (positive) common multiple. Note: Again, there was a typo in the original statement, saying that lcm(ra, rb) = r lcm(a, b). But if r < 0, then r lcm(a, b) < 0, which violates the definition of least common multiple.

6 (d) Show that if a, b > 0, then ab = gcd(a, b)lcm(a, b). Proof. Let g = gcd(a, b), and let x, y Z such that gx = a and gy = b. Consider L = ab g. We have that L = ab g = a b g = ay and L = ab g = ba g = bx, so that L Z and L is a common (integer) multiple of a and b. To see that it is the least such multiple, let m be any positive common multiple of a and b. Let u, v Z satisfy g = ua + vb. Since a m and b m, we have Therefore, m = as and m = bt for some s, t Z. m = (m/g)g = (m/g)ua + (m/g)vb ( ab = (bt/g)ua + (as/g)vb = tu g ( ) ab = (tu + sv) = (tu + sv)l. g ) + sv ( ) ab g So since lu + kv Z, we have L m. And since L, m > 0, we must have lu + kv 1, so that m L = lu + kv m. Thus L is the least common multiple of a and b, as desired. 6