All About Division Definition A nonzero integer t

All About Division

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'

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. /

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.

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

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

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.

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.

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

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.

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