Am I a wisecrack or a fool?

"Show that if $a$, $b$, $m$, and $n$ are integers such that $m0$, $n0$, $n|m$, and $a\equiv b (\mod m)$, then $a\equiv b (\mod n)$."

Proof: By definition, we have that $a\equiv b (\mod m)\implies m|(a-b)$. Since $n|m$, it follows that $n|(a-b)$. Therefore, $a\equiv b (\mod n)$. $\blacksquare$

This is too short to be true. What do you think?

Question

Am I a wisecrack or a fool?

"Show that if $a$, $b$, $m$, and $n$ are integers such that $m>0$, $n>0$, $n|m$, and $a\equiv b (\mod m)$, then $a\equiv b (\mod n)$."

Proof: By definition, we have that $a\equiv b (\mod m)\implies m|(a-b)$. Since $n|m$, it follows that $n|(a-b)$. Therefore, $a\equiv b (\mod n)$. $\blacksquare$

This is too short to be true. What do you think?

anonymous · Answer

Seems true to me.

anonymous · Answer

If you want to make it longer, you can add an intermediate step to prove that if $n|m$ and $m|k$ then $n|k$.

anonymous · Answer

Although I don't see any good reason for doing that! :D

anonymous · Answer

I mean, I can always write a more rigorous proof, but I don't know if this one would be any better than the one above:

Proof: By definition, we have that $a\equiv b\mod m\implies a=b+km$, $\exists k\in\mathbb{Z}$. Since $n|m$, we have $m=cn$, $\exists c\in\mathbb{Z}$. Substituting yields $a=b+km\implies a=b+kcn$. But because $kc\in\mathbb{Z}$, it follows that $a\equiv b\mod n$. $\blacksquare$

anonymous · Answer

Both of them look good to me. The second one is more rigorous as you said, which is better I think.

anonymous · Answer

Thank you. :)

anonymous · Answer

You're welcome! :)