Let $m,n$ be natural numbers. Then prove the commutativity of the product $n\times m = m\times n$.

Let $$m,n$$ be natural numbers. Then $$n\times m = m\times n$$.

MY ATTEMPT (EDIT)

Lemma 1

We shall need first the following result: $$m\times 0 = 0$$.

Let us prove it by induction on $$m$$. Indeed, one has that $$0\times 0 = 0$$, by the definition of multiplication by $$0$$ on the left. Let us assume the proposition holds for $$m$$, that is to say, $$m\times 0 = 0$$, and we shall prove it to $$m\texttt{+}\texttt{+}$$. Indeed, one has \begin{align*} (m\texttt{+}\texttt{+})\times 0 = (m\times 0) + 0 = 0 + 0 = 0 \end{align*}

And we are done.

Lemma 2

We shall prove that $$m\times(n\texttt{+}\texttt{+}) = m\times n + m$$ by induction on $$m$$.

To start with, notice that $$0\times(n\texttt{+}\texttt{+}) = 0 = 0\times n + 0$$, and the base case is done. Let us assume that $$m\times(n\texttt{+}\texttt{+}) = m\times n + m$$, and prove it holds that $$(m\texttt{+}\texttt{+})\times(n\texttt{+}\texttt{+}) = (m\texttt{+}\texttt{+})\times n + m\texttt{+}\texttt{+}$$: \begin{align*} (m\texttt{+}\texttt{+})\times(n\texttt{+}\texttt{+}) & = m\times(n\texttt{+}\texttt{+}) + n\texttt{+}\texttt{+} = m\times n + m + n \texttt{+}\texttt{+}\\\\ & = (m\times n + n\texttt{+}\texttt{+}) + m = (m\times n + n)\texttt{+}\texttt{+} + m\\\\ & = ((m\texttt{+}\texttt{+})\times n)\texttt{+}\texttt{+} + m = ((m\texttt{+}\texttt{+})\times n + m)\texttt{+}\texttt{+}\\\\ & = (m\texttt{+}\texttt{+})\times n + m\texttt{+}\texttt{+} \end{align*} And we are done.

Proposition

Based on the previous result, we shall prove the proposed statement by induction on $$n$$. According to lemma 1, one has that $$0\times m = m\times 0 = 0$$. Let us assume that $$n\times m = m\times n$$ and let us prove it for $$n\texttt{+}\texttt{+}$$. Second lemma 2, we have that \begin{align*} (n\texttt{+}\texttt{+})\times m = n\times m + m = m\times n + m = m\times(n\texttt{+}\texttt{+}) \end{align*}

And we are done.

Any comments or contributions on the solution?

• Everything you've written is correct. Remember you want to prove $(n++)\times m=m\times(n++)$. So we compute both sides to verify the equality: you already computed $(n++)\times m$, on the other hand $m\times(n++)=m\times n+m$. – EBO Mar 5 at 6:46
• The problem consists in the fact that such property hasn't been proven yet. – BrickByBrick Mar 5 at 18:55
• In the argument above I didn't use that property. What I tried to say is: Let's say we want to prove $a=b$, where $a$ and $b$ are expressions or whatsoever. One way to go is to prove $a=c$ and $b=c$ so that we can now conclude $a=b$. I hope I'm making myself clear. – EBO Mar 6 at 0:15

$$(n++)\times m=(n\times m)+ m = (m\times n)+m=m+(m\times n)=m\times (n++)$$
• The thing is that it hasn't been proven that $(n\texttt{+}\texttt{+})\times m = m\times(n\texttt{+}\texttt{+})$ yet. – BrickByBrick Mar 5 at 18:35