How to prove that doesnt exist a natural number such that is equal to it successor from Peano axioms?

Question

Im getting a hard time trying to prove the general for any natural number $n$ such that

$$\nexists n\in\Bbb N: S(n)= n$$

From the second Peano axiom we know that

$$\nexists n\in\Bbb N: S(1)= n$$

and from the third axiom that

$$S(a)=S(b)\to a=b$$

I really dont have a clear clue about how to prove this. Someone tell me that I must use the "axiom" number 5, the induction. But the induction need the existence of a second natural number more than $1$ (or more than $0$ using the axioms that use it) that hold the implicit difference on the axiom 2. Example:

$$S(1)\ne 1 \to S(1)=2$$

But now, how to prove that $S(2)\ne 2$? Thank you in advance.

Answer 1

The successor map $\sigma : \mathbb{N} \rightarrow \mathbb{N}$ is injective. This is an axiom. In addition, as far as the base case goes, $\sigma(1) \neq 1$; this is also an axiom since $1$ is not the successor of another number.

Now, we employ induction. Suppose for some $n$, $\sigma(n) \neq {n}$. This implies $$\sigma(\sigma(n)) \neq \sigma(n)$$ since otherwise would contradict the function being injective.

Answer 2

Ok, I solved it using the third axiom. If $S(1)=2$ and $S(2)=2$ then we have that $S(1)=S(2)$ and by the third axiom we have that $1=2$, a contradiction, so $S(2)\ne 2$.

After all it was so simple but I didnt see in first place. What a dumb.

To complete the answer for any number $n$ is enough to see that the natural numbers are defined through the successor function over $1$ (or $0$, depending of the version of the Peano axioms).

So, necesarily, if $S(2)\ne 2$ and $S(1)\ne 1$ then $S(n)\ne n$ by induction.