# Prove for natural numbers that $n\times m = 0$ if and only if at least one of $n,m$ is equal zero.

Let $$n,m$$ be natural numbers. Then $$n\times m = 0$$ if and only if at least one of $$n,m$$ is equal zero. In particular, if $$n$$ and $$m$$ are both positive, then $$nm$$ is also positive.

MY ATTEMPT

Suppose otherwise, that is to say, $$n\times m = 0$$ and $$n\neq0$$ and $$m\neq 0$$. Thus $$n = a\texttt{+}\texttt{+}$$ and $$m = b\texttt{+}\texttt{+}$$. Consequently, \begin{align*} n\times m = (a\texttt{+}\texttt{+})\times m = a\times m + m \geq m = b\texttt{+}\texttt{+} > 0 \end{align*} a contradiction. Thus we conclude that $$n = 0$$ or $$m = 0$$.

As to the second statement, one has that \begin{align*} (n > 0)\wedge(m > 0) \Longrightarrow n\times m = (a\texttt{+}\texttt{+})\times m = a\times m + m \geq m = b\texttt{+}\texttt{+} > 0 \end{align*} and we are done.

Any comments on the proposed solution?