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


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.


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?

  • $\begingroup$ 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 $m\times(n++)=m\times n+m$. $\endgroup$ – EBO Mar 5 at 6:46
  • $\begingroup$ The problem consists in the fact that such property hasn't been proven yet. $\endgroup$ – BrickByBrick Mar 5 at 18:55
  • $\begingroup$ 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. $\endgroup$ – EBO Mar 6 at 0:15

$(n++)\times m=(n\times m)+ m = (m\times n)+m=m+(m\times n)=m\times (n++)$

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

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