• September 27th 2009, 04:52 PM
koukou8617
Assume $a \equiv b (mod n)$, then by definition $a = s \cdot n$ and $b = r \cdot n$ for $s,r \in \mathbb{N}$.
Now, you can see that $b-a = rn - sn = (r-s)n = t \cdot n$ for some $t \in \mathbb{N}$, thus $n|(b-a)$.
Now, for the second part: $n|(b-a) \Rightarrow b-a = k \cdot n \Rightarrow b = kn + a \Rightarrow b \equiv a (mod n)$.