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' / 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. / 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. / 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 / 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. / 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. / 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. a My Proof (Existence) qb (q+1)b

7. Proof: (Uniqueness) / Suppose a = q 1 b + r 1 and a = q 2 b + r 2 where 0 ≤ r 1,r 2 < b. We may suppose that r 1 ≥ r 2. / Then 0 ≤ r 1 – r 2 < b, But r 1 – r 2 = (a – q 1 b ) – (a – q 2 b) = (q 2 – q 1 )b / Hence r 1 – r 2 is a non-negative multiple of b that is strictly less than b. / It follows that r 1 – r 2 = 0. / So r 1 = r 2 and then q 1 = q 2 as required. / Suppose a = q 1 b + r 1 and a = q 2 b + r 2 where 0 ≤ r 1,r 2 < b. We may suppose that r 1 ≥ r 2. / Then 0 ≤ r 1 – r 2 < b, But r 1 – r 2 = (a – q 1 b ) – (a – q 2 b) = (q 2 – q 1 )b / Hence r 1 – r 2 is a non-negative multiple of b that is strictly less than b. / It follows that r 1 – r 2 = 0. / So r 1 = r 2 and then q 1 = q 2 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. / 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. 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. / S = {a–bk | k is an integer, a–bk > 0} / By the WOP, there is a smallest element of S, call it r. a a–bkr=a–bq

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. 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 / 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