Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that. a = bq + r and 0 r < b.


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1 Theorem (The division theorem) Suppose that a and b are integers with b > 0. There exist unique integers q and r so that a = bq + r and 0 r < b. We re dividing a by b: q is the quotient and r is the remainder, which is why 0 r < b. Proof. This is an existence and uniqueness proof, so there are two parts: show that q and r exist, and show that they are unique. We deal with existence first. First suppose that a 0. I ll give two proofs in this case, one by (strong) induction, and another using the method from the book.
2 Existence proof, #1. We fix b and proceed by strong induction on a. If 0 a < b, then let q = 0 and r = a. This deals with the base case. Now for the inductive step, assume that a b and that the statement is true for all nonnegative integers k < a: whenever 0 k < a, assume there are integers q and r such that k = bq + r and 0 r < b. We want to prove the same thing for a. Since a b, the number k = a b is nonnegative, so there are integers q and r such that a b = bq + r and 0 r < b. Therefore we let q = q + 1 and r = r : a = b(q + 1) + r = bq + r and 0 r < b. This completes the inductive step, and hence the proof.
3 Existence proof, #2. We would like to find an integer q so that bq a < b(q + 1); then we will let r = a bq, so a = bq + r and 0 r < b. As above, suppose that a 0. Let A = {k Z : k 0, bk a}. This is nonempty (since 0 A) and it is finite (since its elements are integers between 0 and a/b). Therefore it has a maximum element. Let q = max A. Then we have bq a < b(q + 1), as desired.
4 The rest of the existence proof. If a < 0, we apply the previous argument to a, which is positive: there are numbers q 1 and r 1 so that a = bq 1 + r 1 and 0 r 1 < b. Then a = bq 1 r 1. If r 1 = 0, this is good enough: let q = q 1 and r = 0. If 0 < r 1 < b, then a = b( q 1 1) + (b r 1 ). So let q = q 1 1 and r = b r 1. We have now established that for all a Z, there exist integers q and r such that a = bq + r and 0 r < b.
5 The uniqueness part of the proof. To finish the proof, we need to prove uniqueness: q and r are the only integers that satisfy these conditions. So suppose that a = bq 1 + r 1 and a = bq 2 + r 2, with 0 r i < b for i = 1, 2. These last inequalities mean that b < r 2 r 1 < b. The first two equalities imply that bq 1 + r 1 = bq 2 + r 2, so b(q 1 q 2 ) = r 2 r 1, so b r 2 r 1. Since b < r 2 r 1 < b, the only possibility is that r 2 r 1 = 0 and therefore q 1 q 2 = 0. That is, r 1 = r 2 and q 1 = q 2. This finishes the proof of uniqueness, and hence the proof of the theorem.
6 Corollary If an integer n is a square, then n = 4q or n = 4q + 1 for some q Z. (Saying that n is a square means k Z : n = k 2. ) For example, is not a square. Proof. Suppose that n = k 2. By the division theorem, there is an integer q so that k = 4q or k = 4q + 1 or k = 4q + 2 or k = 4q + 3. Therefore k 2 is either 16q 2 = 4(4q 2 ) or 16q 2 + 8q + 1 = 4(4q 2 + 2q) + 1 or 16q q + 4 = 4(4q 2 + 4q + 1) or 16q q + 9 = 4(4q 2 + 6q + 2) + 1. No matter what k is, the remainder of k 2 is either 0 or 1 after division by 4.
7 Greatest common divisors Suppose that a and b are integers, at least one of which is nonzero. Recall that their greatest common divisor is the unique positive integer d such that Lemma d a and d b if c a and c b, then c d. If b is positive and b a, then gcd(a, b) = b. Proof. Well, b a by assumption, and b b automatically, so b satisfies the first condition in the definition. If c a and c b, then c b, so b satisfies the second condition.
8 Lemma Suppose that a and b are nonzero, and suppose that there are integers q and r with a = bq + r. Then gcd(a, b) = gcd(b, r). Recall the notation D(a), meaning the set of divisors of a. Proof. This is easiest to prove somewhat indirectly. We will prove that D(a) D(b) = D(b) D(r). (*) That is, the set of common divisors of a and b is the same as the set of common divisors of b and r. Given (*), we would have gcd(a, b) = max(d(a) D(b)) = max(d(b) D(r)) = gcd(b, r). Therefore proving (*) will complete the proof.
9 Proof, continued. To prove D(a) D(b) = D(b) D(r), we have to prove that each side is a subset of the other. Suppose that c is in D(a) D(b). This means that c a and c b. We need to prove that c b and c r, so we really just need to prove that c r. Since r = a bq, if c a and c b, then c r. Now suppose that c b and c r. We need to prove that c a. Since a = bq + r, this is immediate.
10 is a recursive method for computing the gcd of any two integers. It works like this: To compute gcd(a, b), set a 0 = a and a 1 = b. Divide a 0 by a 1 with remainder (i.e., use the division theorem) to get integers q 1 and r 1 such that a 0 = a 1 q 1 + r 1 and 0 r 1 < a 1. If r 1 = 0, stop: the algorithm is done and gcd(a, b) = gcd(a 0, a 1 ) = a 1 = b (by the first lemma). Otherwise, set a 2 = r 1 and repeat with a 1 and a 2, etc. So if a 0, a 1,..., a k have been defined, then there are integers q k and r k such that a k 1 = a k q k + r k and 0 r k < q k. If r k = 0, stop: gcd(a, b) = a k. Otherwise, by the second lemma, gcd(a k 1, a k ) = gcd(a k, r k ), so set a k+1 = r k and continue.
11 Notice that when using the Euclidean algorithm, a 1 > a 2 > a 3 >..., and all of these are nonnegative, so the algorithm can t run forever. (This is an important feature of any welldesigned algorithm: you should be able to prove that it eventually stops and returns the correct answer.)
12 Example Suppose a = 8 and b = 5. Then divide 8 by 5: 8 = quotient 1, remainder 3. So (8, 5) = (5, 3). Now divide 5 by 3: 5 = quotient 1, remainder 2, so (5, 3) = (3, 2), so divide 3 by 2: 3 = quotient 1, remainder 1, so (3, 2) = (2, 1). Dividing 2 by 1 will leave remainder zero, so the last nonzero remainder, namely 1, is the gcd: (5, 8) = 1.
13 Example Let a = 144, b = 200. Then 144 = so (144, 200) = (200, 144), 200 = so (200, 144) = (144, 56), 144 = so (144, 56) = (56, 32), 56 = so (56, 32) = (32, 24), 32 = so (32, 24) = (24, 8), 24 = so (24, 8) = 8. Thus (144, 200) = 8.
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