1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.