All About Division
Definition
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
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
a and b be integers with b > 0.
There exist unique integers q and r
Let
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)
a = q1b + r1 and a = q2b + r2 where
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 = r2 and then q1 = q2 as required.
Suppose
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