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I need help proving that if $m|a \lor m|b \Rightarrow m|ab$.

I was thinking that $m|a$ and $m|b$ is equivalent to $a = km$ and $b = lm$. Thus $ab = (kl)m^2$. Since this is a multiple of $m$ then $m|(kl)m^2$ and $m|ab$. But I'm not sure if that proves it - also with the $\lor$ in the premise.

Any advice or hints would be greatly appreciated.

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  • $\begingroup$ What you proved is that $m|a$ and $m|b$ implies $m|ab$. This does not prove what you want, you want to show $m|a$ OR $m|b$ implies $m|ab$. Try writing as two cases. i.e, "If $m|a$, ... Then $m|ab$", "If $m|b$, ..., then $m|ab$, thus if $m|a$ or $m|b$ then $m|ab$". $\endgroup$ – Robin Carlier Oct 30 '19 at 16:11
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Welcome to Maths SX!

You almost have it. You just forgot that, as it is or in the assertion, you can only use one translation as an equality. Explicitly, suppose you're in the case $m\mid a$, i.e. $a=km$ for some $k$. Then $$ab=(km)b=(kb)m\quad\text{by associativity and commutativity}.$$ If we're in the case $m\mid b$, just exchange the roles of $a$ and $b$.

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In your argument, you seem to be assuming that $m$ divides both $a$ and $b$. Let $a,b, m \in \mathbb{Z}$. Supose, with no loss of generality, that $m|a$. Then there exists an integer $k$ such that $a = km$. Thus, $ab=kbm$ and, once $k$ and $b$ are both integers, $kb$ is an integer. This proves $m|ab$ (you only need to take $bk$ instead of $k$.

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Why on earth do they insist on making symbolic logic so isolated and abstract? $\lor$ means "OR".

So if either $m|a$ or $m|b$. If $m|a$ then there is a $k$ so that $a = mk$ and $ab = m(kb)$. And if $m|b$ then there is an $l$ so that $b = ml$ and $ab = (al)m$. Either way, $m|ab$.

There shouldn't be any worry about what magical and arcane incantations to satisfy the symbols. The symbols are cases, so handle the cases, one after another.

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