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For any natural numbers $a,b,c$, we have that $(a\times b)\times c = a\times(b\times c)$.

MY ATTEMPT

We shall prove it by induction on $c$. For $c = 0$, one has that \begin{align*} (a\times b)\times 0 = 0 = a\times 0 = a\times(b\times 0) \end{align*}

Then we are going to assume that $(a\times b)\times c = a\times(b\times c)$ and prove the proposed relation holds for $c\texttt{+}\texttt{+}$. Indeed, one has that \begin{align*} (a\times b)\times(c\texttt{+}\texttt{+}) & = (a\times b)\times c + a\times b = a\times(b\times c) + a\times b\\\\ & = a\times((b\times c) + b) = a\times(b\times(c\texttt{+}\texttt{+})) \end{align*} and we are done.

Any comments or theoretical clarification?

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    $\begingroup$ For natural numbers I would write $a\cdot b$ and not $a\times b$ because this usually is the cross product which need not be associative, see here. $\endgroup$ – Dietrich Burde Mar 5 at 19:54

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