• anonymous
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?
Mathematics

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