Difficulty proving the commutative law of multiplication on $\mathbb{N}$

Question

I am trying to prove the commutative law of multiplication for the natural numbers using the Peano axioms. I have already defined addition and order with the successor function. My definition of multiplication is as follows:

• $0 \times n = 0$
• $s(n) \times m = (n \times m) + m$

I want to now show that $n \times m = m \times n$ for all $m,n \in \mathbb{N}$. I attempt to do this as follows:

1. Show that $n \times 0 = 0$ --- (completed)
2. Show that $m \times s(n) = (n \times m) + m$ --- (this is where I have trouble)
3. Given 1. and 2. show that $n \times m = m \times n$ --- (completed)

So, step 2 is the only place I have a problem. This is my attempt at proving it.

Prove $m \times s(n) = (n \times m) + m$ by induction on $m$. For the base case, let $m = 0$. Then on LHS $0 \times s(n) = 0$ from def'n of multiplication. On RHS, ($n \times 0) + 0 = 0 + 0 = 0$ from 1. LHS = RHS so base case is proven

For the inductive step, assume $m \times s(n) = (n \times m) + m$, and attempt to prove $s(m) \times s(n) = (n \times s(m)) + s(m)$.

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

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

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

????

It seems no matter how I try and manipulate this, I can't prove the inductive step. Am I missing something here?

NOTE: This should not require the associative or distributive law to already be proven, as these aren't asked to be proven until later on in the book I am reading. I am using Tao's Analysis I, and am essentially trying to prove Lemma 2.3.2.

Answer 1

It seems that what I wanted to prove in step 2. was not what I needed. What I need to prove in step 2. is $n \times s(m) = (n \times m) + n$. This I could prove without a problem, and therefore can now prove commutativity of multiplication.

Answer 2

First let us prove, that $m \times s(0)=m$ by induction.

$0 \times s(0)=0$ by definition.

Now $s(m) \times s(0)=m \times s(0)+s(0)=m+s(0)$, by induction hypothesis. Therefore: $s(m) \times s(0)=s(m)$.

Now we prove by induction on $n$, that $n \times (a+b)=n\times a+n\times b$ for all the choices of $a,b$. It is trivially true for $n=0$.

Furthermore $s(n) \times (a+b)=n\times (a+b)+(a+b)=na+nb+a+b$, by induction hypothesis and therefore $s(n)(a+b)=s(n)a+s(n)b$.

Now that we have these two results, we can finally prove your 2.

We have $m \times s(n)=m\times (n+s(0))= m\times n+m \times s(0)=m\times n +m$