I have already read the answer to the following question. I have a different proof and I don't know if it is correct. The textbook also doesn't provide solutions and I am trying to self-study (I am asking questions here so I guess its not really self-studying) so I have no way of verifying my solutions. Thank you very much for your time, I appreciate any feedback!
The following can be assumed:
The Peano Axioms. The most relevant one for this answer is the Principle of Mathematical Induction.
The properties of addition on the natural numbers (commutativity, associativity, cancellation).
Definition 2.3.1 (Multiplication of natural numbers): Let $m$ be a natural number. To multiply zero to $m,$ we define $0×m:= 0.$ Now suppose inductively that we have defined how to multiply $n$ to $m.$ Then we can multiply $n++$ to $m$ by defining $(n++)×m:=(n×m)+m.$
The product of two natural numbers is a natural number.
I am trying to prove:
Lemma 2.3.2 (Multiplication is commutative): Let $n,m$ be natural numbers. Then $n×m=m×n.$
Proof: First, we prove the following:
Lemma 1: If $k$ is a natural number, then $k\times 0=0.$
Proof of Lemma 1: For the base case $(k=0),$ we have $$0\times 0=0$$ by definition. Next, take any natural number $k$ and assume that $k\times 0=0.$ For $k++,$ we have $$(k++)\times0=(k\times0)+0.$$ By the inductive hypothesis, the above is equal to $0+0=0.$
$\blacksquare$
Next, we induct on $n.$ For the base case $(n=0)$, we have $0\times m=m\times 0=0$ via the definition of multiplication and lemma 1. Now, take any natural number $n$ and assume that for all $m,$ $n\times m=m\times n.$ For $n++$, we need to show that $$(n++)\times m=m\times(n++).$$ However, we can instead show that $$((n++)\times m)+(n++)=(m\times(n++))+(n++).$$ If the above statement is true then so will the desired one because of the Cancellation Law. So we have \begin{align}L.H.S.&=((n++)\times m)+(n++)\\&=(n\times m)+m+(n++) \tag{by definition}\\&=(n\times m)+n+(m++) \tag{rearranging}\\&=(m\times n)+n+(m++) \tag{Inductive Hypothesis}\\&=((m++)\times n)+(m++)\tag{by definition}\\&=(n\times (m++))+(m++)\tag{Inductive Hypothesis}\\&=(n++)\times (m++) \tag{by definition}.\end{align} For the other side, we have \begin{align}R.H.S.&=(m\times (n++))+(n++)\\&=(m++)\times(n++) \tag{by definition}.\end{align} Since $R.H.S.=L.H.S.,$ this closes the induction and completes the proof.
$\square$
