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The way I tried to proved it seemed to on a weaker version of what I was trying to prove in the first place. It is clearly true but it isn't obvious to me how to prove it.

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Let $r, s\in R$, $m, n\in\mathbb{N}$ be given. Then $$\begin{align*} (mr)(ns) &= (r + \ldots + r)(s + \ldots + s)\text{ ($m$ and $n$ times, respectively)}\\ &= (r(s + \ldots + s) + \ldots + r(s + \ldots + s))\text{ by distributivity}\\ &= rs + \ldots + rs\text{ ($mn$ times) by distributivity}\\ &= mn(rs) \end{align*}$$

Generalizing to $m, n\in\mathbb{Z}$ is straightforward.

It should be noted that we are freely using the associativity of $+$ without comment above.

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    $\begingroup$ Good answer. A remark: we're freely using associativity of addition here (depending on how rigorous you want to be in your proof). $\endgroup$
    – Dave
    Commented Nov 28, 2019 at 0:14
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    $\begingroup$ Good point, edited to reflect that. $\endgroup$
    – nbritten
    Commented Nov 28, 2019 at 0:36

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