# Proof using modulo

Suppose $$n|m$$, then $$n\cdot d=m,~d\in\Bbb Z$$.

If $$a\equiv b\mod m$$, then $$a\equiv b\mod{n\cdot d}$$.
Additionally, $$d\equiv d\mod{n\cdot d}$$.

So we know $$ad\equiv bd\mod{n\cdot d}$$.
$$\therefore a\equiv b\mod n$$

Would this be a valid proof?

• What is the statement that you're proving? – Yanko Oct 4 at 21:37
• I'm trying to prove that if n|m where n, m are integers greater than 1, and if a === b (mod m) then a === b (mod n) – Paul Silvestri Oct 4 at 21:45
• Ok then your argument is valid. But perhaps you need to prove the last line (like, why can you divide everything by $d$, did you prove this in class?) – Yanko Oct 4 at 21:47
• I don't think so. So I would need to prove that? – Paul Silvestri Oct 4 at 21:48
• I think you do, if there's a $k$ such that $ad= (nd)\cdot k + bd$ what can you do next? – Yanko Oct 4 at 21:49