For all naturals $n: n<x<n+1 \implies x \notin N$

Question

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 $m-n \in N$

Answer 1

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 $k=1$.

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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 $$m-(k+1) = (m-k) -1\in N. $$

Answer 2

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.