Proving m mod d = n mod d with Quotient Remainder Theorem

[cmajor47]

Homework Statement

Prove that is m, n, and d are integers and d divides (m-n) then m mod d = n mod d.

Homework Equations

Quotient Remainder Theorem: Given any integer n and positive integer d, there exists unique integers q and r such that n=dq + r and 0$$\leq$$r<d and n mod d = r.

The Attempt at a Solution

Proof: Let m, n, d $$\in$$ Z st d divides (m-n)
$$\exists$$ k $$\in$$ Z st m=dk + r
$$\exists$$ j $$\in$$ Z st n=dj + s
m-n=(dk + r)-(dj + s)
=dk+r-dj+s
=d(k-j)+(r-s)

Am I going along with the proof correctly? I don't know where to go from this point and would really appreciate some help.

 

[CompuChip]

I'm not sure this is the easiest proof, but it looks correct.

You have now shown that
if m = r mod d, n = s mod d, then (m - n) mod d = (r - s)
But you also know that d | (m - n) which you haven't used yet. So what does that tell you about (m - n) mod d?