All About Division

1. All About Division

2. Definition • A nonzero integer t is a divisor of an integer s if there is an integer u such that s = tu. • If t is a divisor of s, we write t | s, read 't divides s'

3. Vocabulary • We also say that s is a multiple of t. • A prime is an integer greater than 1 whose only positive divisors are 1 and itself. • A positive integer with divisors other than itself and 1 is composite.

4. Example • 8|24 because 24 = 8*3 8 is a divisor of 24. 24 is a multiple of 8. 24 is not prime. 24 is composite.

5. Theorem 0.1 Division Algorithm • Let a and b be integers with b > 0. • There exist unique integers q and r with the property that a = bq + r, where 0 ≤ r < b

6. a qb (q+1)b My Proof (Existence) • Consider every multiple of b. • Since a is an integer, it must lie in some interval [qb,(q+1)b). • Set r = a – qb. • Note that r is an integer with 0 ≤ r < b and a = qb + r as required.

7. Proof: (Uniqueness) • Suppose a = q1b + r1and a = q2b + r2where 0 ≤ r1,r2 < b. We may suppose that r1 ≥ r2. • Then 0 ≤ r1 – r2 < b, But r1 – r2 = (a – q1b ) – (a – q2b) = (q2 – q1)b • Hence r1 – r2 is a non-negative multiple of b that is strictly less than b. • It follows that r1 – r2 = 0. • So r1 = r2and then q1 = q2 as required.

8. Note • In our existence proof, we said that "a must lie" in an interval … • How do we know that? It is not a consequence of algebra(+ - * \) or of order ( < = > ) • It is a consequence of the well ordering principle.

9. a–bk r=a–bq a Gallian's proof • S = {a–bk | k is an integer, a–bk > 0} • By the WOP, there is a smallest element of S, call it r.

10. It remains to show that r < b. To get a contradiction, suppose r ≥ b. Then r – b ≥ 0. But r = a – bk for some integer k, So r – b = a – (k+1)b ≥ 0 Then r – b is in S and smaller than r. This contradiction shows r < b.

11. Given a, b find q, r • Divide 38 by 7: • Write: 38 = 5*7 + 3, so q = 5, r = 3 • Divide -38 by 7: • Write: -38 = -6*7 + 4, so q = -6, r = 4