Let a and b be integers and n positive integer. Proof that a and b has the same remainder when divided by n if and only if there is integer k such that a-b=kn
I tried like this by contradiction: a and b are integers and n is positive interger.
Assuming there is no integer k such that a - b = kn (If a and b has the same remainder, a - b = kn which means a is congruent to b modulo n, where k is integer.), then a - b = kn + r , where r is integer more than 0, and less than n. (if r is 0 or n, a - b = kn) If a - b = kn + r, then a is not congruent to b mod n and therefore a and b doesnt have the same remainder.
¬∃k∈Z(a-b=kn -> a-b=kn+r) , 1 <= r < n
->a-b/n = k+r
∴ a ≡ b (mod n) ⇔ ∃k∈Z: a-b=kn
Is it correct? I know there is different ways to solve this but I wanted to try by contradiction for a practice and also for translating it for symbols. If it doesn't make sense, I would like to know how to do it as contradiction proof.
I'm afraid your collection of symbols makes little sense. Mathematical proofs are not sequences of symbols where it's very easy to get lost.
Write the proof in plain language and only after you're sure it works, translate it in stenography, if you are required to.
There's no need to use contradiction.
Suppose $a$ and $b$ have the same remainder when divided by $n$, that is, $$ a=nx+r,\qquad b=ny+r $$ Then $a-b=(nx+r)-(ny+r)=n(x-y)$. So we can take $k=x-y$.
Conversely, suppose $a-b=nk$. Write $b=nq+r$, with $0\le r<n$. Then $$ a=nk+b=n(k+q)+r $$ and uniqueness of remainder ends the proof.