Thread: Division Algorithm question

The division algorithm states
where and
Prove

I shall attempt a proof by contradiction
Assuming
Let where

So




Placing it into the original equation




And as and , then
So , but this is a contradiction to the division algorithm
So

Is this alright?

N.B.
The case where works similarly. You end up with , which is also a contradiction

The problem statement does not say that b >= a. If a > b, then q = 0 and r = b.

Let r=\frac{mb}{n} where m<n<2m

Are m and n integer? Why do they exist?

I have not read your proof to the end, but I think there is an easier direct proof. Suppose b >= a. If a <= b/2, then r < a <= b/2. If a > b/2, then q = 1 and r = b - a < b/2.

Sorry about that omission, but it is true that b>a>0 for this question.

Yes, m and n are integers, they exist because any integer can be expressed as the fraction of another integer, and I chose to represent r as a fraction of b.

But I understand your proof, it is easier.
Thanks