## anonymous 4 years ago 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?

1. anonymous

Seems true to me.

2. anonymous

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

3. anonymous

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

4. anonymous

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 $$a\equiv b\mod n$$. $$\blacksquare$$

5. anonymous

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

6. anonymous

Thank you. :)

7. anonymous

You're welcome! :)