TITLE: How to define addition through multiplication?
QUESTION [31 upvotes]: One might define multiplication $\bullet$ on $\mathbb Z$ as follows:
$\bullet: \mathbb Z\times \mathbb N\ni (a,b) \mapsto a+\cdots+a\in \mathbb Z$ where we add $b$ times.
But suppose we are in a universe where we can only multiply. How would one define addtion, or could one even define it? 
Silly approach 1: $\log(e^ae^b)=\log(e^{a+b})=a+b$, but this assumes existence of $\log$ and $e$ and is rather unsatisfying.
Approach 2: If we could find a formula for $a+1$ (where $a\in \mathbb N$), then we could successively extend this notion to get an addition function $+:\mathbb N\times\mathbb N\to \mathbb N$. But we can only define $a+1$ through multiplication and the only given parameter is $a$, so $a+1$ is be a product of $a$, hence $a\mid (a+1)$. But this is impossible for $a>1$. So it seems we cannot define an addition function using only multiplication, which is again unsatisfying.
Does anybody have an idea whether this is at all possible? What if we could use a (possibly infinite) product of real numbers to define $a+1$?

REPLY [14 votes]: Another, simpler way to show that addition can't be defined from multiplication in "any nice way" (to be defined below): note that the structure $M=(\mathbb{N}, \times)$ has lots of automorphisms. In fact, its automorphism group is $S_\omega$, the group of all permutations of a countably infinite set: automorphisms of $M$ correspond exactly to rearranging the primes.
By contrast, the structure $(\mathbb{N}, +)$ has no automorphisms at all (easy exercise). So addition certainly can't be defined in $(\mathbb{N}, \times)$ by any kind of sentence $\phi$ which doesn't distinguish between isomorphic structures (and this is a basic requirement of every abstract logic I've seen). In particular, first-order sentences certainly have this property, but so do second-order sentences and  infinitary sentences, so this isn't a problem with first-order logic: the issue is really that addition contains "more" information than multiplication.

An extension of this argument: consider the case of $\mathbb{R}$ instead of $\mathbb{N}$. Now $(\mathbb{R}, +)$ does have automorphisms (scale by a constant factor), so the above logic doesn't quite work; but there's an easy fix. Consider the map $x\mapsto x^3$; this is an automorphism of $(\mathbb{R}, \times)$, but does not respect addition since in general $$ (x+y)^3\not=x^3+y^3. $$ So again, addition is not definable from multiplication over $\mathbb{R}$ by any kind of isomorphism-respecting sentence. This argument is more robust, and will work in every context I can think of where addition and multiplication make sense.