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.
$\begingroup$ I believe the real trick here is to choose your induction variable wisely. You have chosen to fix $n$ and prove that $n\cdot i = i \cdot n$ by induction on $i$. This might not be the easy approach. Two other approaches that stand out are to use induction on $n$ where the induction hypothesis is that $i\cdot j = j \cdot i$ for all $i, j$ with $i + j \leq i$, or alternatively for all $i, j$ with $\max\
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.
answered May 23, 2014 at 16:10 Ben Millwood Ben Millwood 14.4k 3 3 gold badges 34 34 silver badges 62 62 bronze badges $\begingroup$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 $m$, $n*m$ $\space$:= $\space$ $\underbrace_m$ , 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 $\underbrace_m + m$ , then, repeating the process with the old summands, we obtain $\underbrace_m + m + m$, and we keep repeating the process until we obtain $\underbrace_m + \underbrace_n$ , which will happen because n is finite, and this result is identically $\underbrace_m + \underbrace_n$ $\space$ = $\space$ $0 + \underbrace_n$ $\space$ = $\space$ $\underbrace_n$ $\space$ = $\space$ $m*n$ $\space \implies \space n*m$ = $m*n$ .
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).
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