proof of commutativity of multiplication for natural numbers using Peano's axiom How do you prove commutativity of multiplication using peano's axioms.I know we have to use induction and I have already proved n*1=1*n.But I cant think of how to prove the inductive step.
 A: Show that if an operator $\star$ satisfies the defining equations for multiplication, i.e. $0 \star n = 0$ and $(m + 1)\star n = n + m \star n$, then $\star$ is multiplication (this is a straightforward induction).
Then show that the operator defined by $m \star n = n \times m$ satisfies the defining equations for multiplication, and therefore $m \star n = m \times n$, so $n \times m = m \times n$.
That is, show that $n \times 0 = 0$ and $n \times (m + 1) = n + n \times m$, and you're done.
Unfortunately, that's just not easy. $n \times 0 = 0$ is not too bad, you can use induction on $n$, but the only proof I have of the latter statement uses the associativity and commutativity of addition, which both have to be proven by induction as well.
A: Honestly, proving this using Peano's axioms is not very educational, however logically valid it may be. After pondering this question for too long a while, I finally just sat down and wrote out a proof which I haven't been able to find anywhere. Here it is:
First we need the definition of multiplication for natural numbers (which nearly nobody actually knows). For any two natural numbers $n$ and $m$, $n*m$ $\space$:= $\space$  $\underbrace{n+n+...+n}_m$ , and this definition is crucial to a structured view of mathematics. Now, by removing 1 from each summand and summing all those removals into a new summand, since there are $m$ summands, we obtain $\underbrace{(n-1)+(n-1)+...+(n-1)}_m + m$ , then, repeating the process with the old summands, we obtain $\underbrace{(n-2)+(n-2)+...+(n-2)}_m + m + m$, and we keep repeating the process until we obtain $\underbrace{(n-n)+(n-n)+...+(n-n)}_m + \underbrace{m + m+...+m}_n$ , which will happen because n is finite, and this result is identically $\underbrace{0+0+...+0}_m + \underbrace{m + m+...+m}_n$ $\space$ = $\space$ $0 +  \underbrace{m + m+...+m}_n$ $\space$ = $\space$ $\underbrace{m + m+...+m}_n$ $\space$ = $\space$ $m*n$ $\space \implies \space n*m$ = $m*n$ .
That proves multiplicative commutativity for any particular choice of $n$ and $m$. To prove multiplicative commutativity for ALL choices of $n$
 and $m$, we must show that $(n+1)*m$ = $m*(n+1)$, $\space$ $n*(m+1)$ = $(m+1)*n$,$\space$ and $(n+1)*(m+1)$ = $(m+1)*(n+1)$ $\space$ (the "for ALL" generality follows by recursion).
I created this account just to answer your question, so I hope you like that response. 
A: Using the following Peano-like axioms to define + (addition), · (multiplication) and 1, where x and y are variables of type natural number and S denotes the successor function: (Note: Ax. is an abbreviation for "Axiom".)
Ax. x + 0     =   x
Ax. x + S(y)  =   S(x + y)
Ax. x · 0     =   0
Ax. x · S(y)  =   x  +  x · y
Ax. 1 = S(0)
One may then prove that · is commutative by first establishing the following theorems:
Th. (0) S(x) = x + 1 (does not require induction)
Th. (1) + is commutative (i. e. x + y = y + x)
Th. (2) 0 · x = 0
Th. (3) 1 is a left identity of · (i. e. 1 · x = x)
Th. (4) · right-distributes over + (i. e. (x + y) · z = x · z + y · z)
Theorem (1) can be proved easily by induction over y if you first establish the theorems 0 + x = x and S(x) + y = S(x + y). (Which may both be shown by induction) Theorem (4) may be proved by induction over z. (Just saying in case it is not evident.)
And then the proof of multiplication's commutativity by induction over y can go like this: (Hints for the justifications of each successive manipulation are indicated between curly brackets.)
The base case: x · 0 = 0 · x

    x · 0
= { First multiplication axiom }
    0
= { Theorem (2) }
    0 · x

Now, by postulating that the induction hypothesis x · y = y · x is true for some natural number y, we may prove that x · S(y) = S(y) · x is implied:

    x · S(y)
= { Second multiplication axiom }
    x  +  x · y
= { Induction hypothesis }
    x  +  y · x
= { Theorem (3) }
    1 · x  +  y · x
= { Theorem (4) }
    (1 + y) · x
= { Theorems (1) and (0) }
    S(y) · x

And that completes the proof.
