# Proof that each natural number has a unique successor

I've proven that every positive natural number has a unique predecessor using Peano's axioms. But now, I was wondering how I could prove that every natural number has a unique successor using the same axiomatic system. Of course, by Peano's axioms, if $$n\in\mathbb{N}$$ then $$\sigma(n)\in\mathbb{N}$$. But how do I show that this natural number is unique?

• $\sigma$ is supposed to be a function.
– ameg
Commented May 26 at 14:43
• @ameg so is the unique successor fact implicitly assumed since $\sigma$ is a function? Commented May 26 at 14:45
• Right. It's an unstated axiom.
– MJD
Commented May 26 at 15:16
• Another unstated axiom is the existence if $0$. It's quite consistent with the other axioms for there to exist no natural numbers at all.
– MJD
Commented May 26 at 15:38
• MJD, in what sense are these axioms "unstated"? Normally, the existance of 0 (or 1) is the first axiom. Then there is an set of axioms about a successor function that should exist and have such-and-such properties. Commented May 26 at 16:08

The successor function, $$\sigma(n)$$, defines the successor of a number $$n$$. You don't need to prove that it's unique - it's a function - i.e., for any $$n$$ this function gives just one value (usually called $$n+1$$) which will be defined as the successor. The only successor.
After understanding that, you might wander now whether is is possible that two different numbers $$m$$ and $$n$$ have the same successor $$k$$. This is not possible because of the predecessor uniqueness that you said you proved: both $$m$$ and $$n$$ are predecessors of $$k$$ so if the predecessor is unique, they must be equal.