Originally Posted by

**gate13** Good day to all. I have been asked to prove the following proposition:

**Let m and n be integers. Use the contradiction method to prove that the quotient and the remainder of the division of m by n are unique.**

The following is a rough proof I have come up with. I was wondering if someone could offer their opinion on my proof, if I have made glaring mistakes and such. Any input would be greatly appreciated. I have omitted the theorems used, to avoid making the post heavier than it already is.

**Proof:**

We assume that the quotient and remainder are not unique. Then

m = q1n + r1 and m = q2n + r2

m=m

q1n + r1=q2n + r2

n(q1-q2) = -(r1-r2)

This implies the equality 0 = 0 since zero is the only number equal to its negative.

And what has this fact to do with what you got above?? You didn't get something like $\displaystyle x=-x$ , you got $\displaystyle n(q_1-q_2)=-(r_1-r_2)$ and thus you cannot deduce from this that $\displaystyle n(q_1-q_2)=0$ . I'd try using $\displaystyle r_i=0\,\,\,or\,\,\,|r_i|<q_i\,,\,\,i=1,2$ .

Tonio

Therefore n(q1-q2) = 0 => n = 0 or (q1-q2) = 0

if n = 0 => r1 = r2 and we have q1 = ((n-r1)/n) and q2 = ((n-r1)/n) i.e. q1 = q2

if (q1-q2) = 0 => q1 = q2 and -(r1-r2) = 0 => r1 = r2

Q.E.D.