The Division Algorithm

[kjm789]

Excuse me for my brain fart. I have a question regarding the proof of the division algorithm, specifically a question when proving uniqueness.

Recall:

(Division Algorithm)

Let a and b be integers, with b > 0. Then there exist unique integers q and r such that a = b*q + r where 0 <= r < b.

When proving uniqueness:

Uniqueness of q and r.

Suppose there exists integers r, r', q, and q' such that a = b*q + r, 0 <= r < b and a = b*q' + r', 0 <= r' < b. Then b*q + r = b*q' + r'. Assume that r' >= r. From the last equation we have b(q - q') = r' - r ; Therefore b must divide r'-r and 0 <= r' - r <= r' < b. This is only possible if r'- r = 0.

This is where may questioned exists. I am able to see that r' - r is a b multiple, but r' - r = 0. What axioms or properties am I missing?? It deductively makes sense that r' - r must be equal to 0 but if I was required to elaborate more on why this is, this is where my concern lies.

Could someone help me out with the understanding?? Thanks in advance

 

[Ishuda]

Excuse me for my brain fart. I have a question regarding the proof of the division algorithm, specifically a question when proving uniqueness.

Recall:

(Division Algorithm)

Let a and b be integers, with b > 0. Then there exist unique integers q and r such that a = b*q + r where 0 <= r < b.

When proving uniqueness:

Uniqueness of q and r.

Suppose there exists integers r, r', q, and q' such that a = b*q + r, 0 <= r < b and a = b*q' + r', 0 <= r' < b. Then b*q + r = b*q' + r'. Assume that r' >= r. From the last equation we have b(q - q') = r' - r ; Therefore b must divide r'-r and 0 <= r' - r <= r' < b. This is only possible if r'- r = 0.

This is where may questioned exists. I am able to see that r' - r is a b multiple, but r' - r = 0. What axioms or properties am I missing?? It deductively makes sense that r' - r must be equal to 0 but if I was required to elaborate more on why this is, this is where my concern lies.

Could someone help me out with the understanding?? Thanks in advance

Not quite sure of your question. You say that you see that r'-r is a multiple of b. However, you don't see why r'-r is zero because of that. Is that correct?

If so, then we know that all positive multiples of b are greater than or equal to b. So, if r'-r is a multiple of b, r'-r is either zero or greater than or equal to b. But r'-r can't be greater than or equal to b since both r' and r are non-negative and thus their difference is less than or equal to r' which is less than b. Thus r'-r is zero.