There's a short way to prove this. k\in \mathbb{Z})" alt="x\equiv y \pmod{a} \Leftrightarrow x=y+ka\:\k\in \mathbb{Z})" />
Thus, using the fact that for all , it follows that
There's a short way to prove this. k\in \mathbb{Z})" alt="x\equiv y \pmod{a} \Leftrightarrow x=y+ka\:\k\in \mathbb{Z})" />
Thus, using the fact that for all , it follows that
how would one prove the last thing? I've seen it before but I've never seen the proof...