[Math] How to define addition through multiplication

arithmeticnumber theory

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 $a+1$?

Best Answer

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.

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