Let $ m \in \mathbb{N}$ . Show that $ \nexists n \in \mathbb{N}$ such that $ m < n < s(m)$ , where $ s$ is the successor function.

Here’s my proof using only the Peano Axioms I was introduced. I’d appreciate someone to check my work.

Let’s prove it by contradiction. Suppose that $ \exists n \in \mathbb{N}$ such that (1) $ m < n$ and (2) $ n < s(m)$ .

From the definition of order we have that:

(1) implies that $ \exists c \in \mathbb{N} \setminus \{0\}$ such that $ m + c = n$ .

(2) imples that $ \exists c’ \in \mathbb{N} \setminus \{0\}$ such that $ n + c’ = s(m)$ .

Therefore $ $ \begin{align*} n + c’ + c &= s(m) + c\ &= s(m + c)\ &= s(n)\ \end{align*} $ $

Since $ c’ \neq 0$ and $ c \neq 0$ , we know that $ c’ + c \neq 0$ and therefore $ \exists k \in \mathbb{N}$ such that $ s(k) = c’ + c$ . Hence:

$ $ \begin{align*} n + s(k) &= s(n)\ s(n + k) &= s(n) \rightarrow n + k = n \rightarrow k = 0 \end{align*} $ $

Since we’ve concluded that $ k=0$ , we have that $ s(k) = s(0) = 1 = c + c’$ , where follows that $ c$ or $ c’$ needs to be equals to $ 0$ , and that is a contradiction.

I also have one more question… After defining what a function is, can’t we conclude that $ s : \mathbb{N} \setminus \{0\} \rightarrow \mathbb{N}$ is actually $ s(n) = n + 1$ ?

If so, using that the proof would’ve been more immediate…

Any kind of comments and critics are highly appreciated! Thank you!