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From Barry Mazur's Imagining Numbers: (particularly the square root of minus fifteen), p. 97:

Suppose that you have some unknown operation that allows, as its "input," any two positive whole numbers $M$ and $N$ and produces, as a result, another whole number, which we will denote $M*N$. [...] Suppose further that we know our mystery operation satisfies the two simple laws $(a)$ and $(b)$:

(a) For any positive whole number $N$, $$ 1*N=N $$

(b) For any three positive whole numbers $A$, $B$, and $C$, $$ A*C+B*C=(A+B)*C; $$ that is, the distributive law holds.

Then we can show that our mystery operation $*$ is none other than multiplication.

[...] A formal treatment of our subject would offer, at this point, a proof that these laws - (a) and (b) - do indeed characterize the operation of multiplication, as defined, say, by the creeping strategy.

Question: Formally and precisely, what's the result we would prove ?

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What Mazur is saying is that any function $\mu: \mathbb N \to \mathbb N$ satisfying $\mu(1,n)=n$ and $\mu(a,c)+\mu(b,c)=\mu(a+b,c)$ must be ordinary multiplication given by $\mu(a,b)=a\cdot b$.

This follows by induction: $$ \mu(1+b,c)=c+\mu(b,c)=\mu(1,c)+b\cdot c=c+b\cdot c=(c+1)\cdot b $$ with the base case $$ \mu(1,n)=n=1\cdot n $$

This is usually the definition of multiplication from the Peano axioms.

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  • $\begingroup$ Are you sure about this? Mazur is talking about multiplication in $\mathbb{Z}$, not in $\mathbb{N}$. I am aware of the recursive definition of multiplication in $\mathbb{N}$ and of it's existence and unicity given by a Recursion Theorem for Peano structures. $\endgroup$
    – Guest
    Commented Oct 23, 2015 at 13:27
  • $\begingroup$ @Guest, well, the text says "positive whole numbers" and so it's about $\mathbb N$. At least, it follows that $\star$ is ordinary multiplication on $\mathbb N$. Extending the argument to $\mathbb Z$ should be easy. $\endgroup$
    – lhf
    Commented Oct 23, 2015 at 13:31
  • $\begingroup$ My bad, I misread and got confused. I'll leave the question as it is anyway. $\endgroup$
    – Guest
    Commented Oct 23, 2015 at 13:36

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