0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.