How to define addition through multiplication?

Solution 1:

The following is a partial answer for $\mathbb{N}$, and the usual notion of (first-order) definability, which is more restrictive than the notion you are implicitly using.

It is an old result of Mostowski that if $S$ is the set of true sentences of number theory that do not use addition, then $S$ is recursive. This is an analogue of the more famous result about the decidability of Presburger arithmetic. There we look at sentences that do not use multiplication.

If addition were definable from multiplication, then the set of true number theoretic sentences would be recursive. But we know that this is not the case.

Thus addition cannot be defined from multiplication in the restricted setting we have described.

Remark: Your definition of multiplication in terms of addition is not first-order. The issue is with the $\dots$, combined with "$b$ times." Replacing this with a more formal recurrence does not make it first-order.

Solution 2:

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.