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Suppose we have that prove that $mk|n \implies m|n \text{ and } k|n$. This is my idea so far. Since by definition of divisibility $m$ divides $n$ if there exists $q$ such that $n = mq$. Let $q = k$ (from assumption), then $m|n$. I can use similar argument for $k|n$ case. Is my reasoning correct? Also would the reverse implication be true? I think it would be false since i can use same arguement and show that even if $m|n$ and $k|n$ there is a chance that $mk > n$ and thus it wont be true. An example would be 2|4 and 4|4 but $(2\cdot 4) |4$ is not true

Bill Dubuque
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UnKnoWnZ
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    The counterexample for the reverse is correct (simpler let $\,m = k = n > 1\,$ so $n\mid n$ but $n^2\nmid n$), but the argument for the forward direction is incorrect. See the link for correct proofs. – Bill Dubuque Oct 14 '20 at 21:43
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    A correct proof starts from $\,mk\mid n,\,$ i.e. $ (mk)q = n\,$ for $\,q\in\Bbb Z,\,$ then concludes by deducing that $\,m,k\mid n.\ $ – Bill Dubuque Oct 14 '20 at 21:51

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