For all naturals $n: nI've been having trouble proving that there are no natural numbers between two consecutive natural numbers.
 The axioms and definitions I'm using are the same of this question: https:\math.stackexchange.com/questions/2896530/prove-that-if-m-n-are-natural-and-mn-then-m-n-is-natural
My attempt. I tried by induction: ($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 $m-n \in N$
 A: 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 $0\notin N$.

$$(\forall m\in N)\left (m\neq 1 \implies m-1\in N\right )\tag{1} $$

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 
$$m-(k+1) = (m-k) -1\in N. $$
A: These types of problems can be surprisingly hard since it's not clear what can be taken for granted and what can't.
I will give it a go taking the following as given:


*

*$N = \{0,1,2,\ldots\}$

*if $n \in N$ and $n \not= 0$, then there exists $m \in N$ with $m+1=n$

*the well-ordering principle: any nonempty set of natural numbers has a least element

*the usual ordering and arithmetic properties of $N$

*your claim that $0 < x < 1$ implies $x \notin N$.


Define $E = \{x \in N \mid \exists n \in N\ n < x < n+1\}$.The goal is to show that $E$ is empty. Assume to the contrary that $E \not= \emptyset$. Then $E$ has a least element $x$. By definition of the set $E$ there exists $n \in N$ with $n < x < n+1$.
How large can $n$ be? If $n \not = 0$ then there exists $m \in N$ with $n=m+1$. This leads to
$$ m+1 < x < m+2$$
which in turn gives $$m < x-1 < m+1.$$
This is precisely what is meant by membership in $E$, giving you that $x-1 \in E$. 
This contradicts the minimal nature of $x \in E$. We made two hypotheses: first that $E$ is nonempty, and second that $n \not = 0$. It follows that the most recent hypothesis fails, so we conclude that $n = 0$ and $0 < x < 1$. This, however contradicts the fifth axiom listed above, so the original hypothesis $E \not= \emptyset$ fails.
Thus $E = \emptyset$ and no such $x$ exists.
