I've been having trouble proving that there are no natural numbers between two consecutive natural numbers.
The axioms and definitions I'm using are the same of this question: https:\math.stackexchange.com/questions/2896530/prove-that-if-m-n-are-natural-and-mn-then-m-n-is-natural
My attempt. I tried by induction: ($x$ is a real)
Basis: $0<x<1 \implies x \notin N$
This can be easily proved by noting that the union of $\{0\}$ and $[1, + \infty [ $ is inductive and contains 0 thereore is a superset of naturals and easy to show that $0<x<1 $ implies that $x$ is not in that set therefore $x$ is not natural.
I can't proceed, no idea how to prove that the hypothesis for some $k$ implies for $k+1$ since I can't enumerate all numbers before $k$ like in the basis case.
Note that I haven't yet proved that if $m,n \in N$ and $m>n$ then $m-n \in N$
Best Answer
Assume the claim holds for $k-1 < x < k$, where $k\geq 1$. Now suppose there is a natural number $k< y <k+1$. Then $0<y-k<1$ must also be a natural number since $y>k$. This contradicts the case $k=1$.
Assuming here $0\notin N$.
Proof. Suppose for a contradiction $m-1\notin N$. The set $N\setminus \{m\}$ is inductive. Indeed, $1\in N\setminus \{m\}$ and if $n\in N\setminus \{m\}\subset N$, then $n+1\in N$. In particular $n+1\neq m$, thus $n+1\in N\setminus \{m\}$. But now $N \subseteq N\setminus\{m\}$, a contradiction.
Proof of claim. Let $m\in N\setminus \{1\}$. We will show by induction that for all $n<m$ it holds that $m-n\in N$. The base case is establised by $(1)$. Suppose $m-k\in N$. Then for the case $k+1<m$ we may again note by $(1)$ that $$m-(k+1) = (m-k) -1\in N. $$