TITLE: Use an induction argument to prove that for any natural number $n$, the interval $(n,n+1)$ does not contain any natural number.
QUESTION [7 upvotes]: Use an induction argument to prove that for any natural number $n$, the interval $(n,n+1)$ does not contain any natural number. 

I don't know where I could go with an induction argument. I was thinking of proving that if $s\in (n,n+1)$, where $s$ is a natural number, then $s-n$ is a natural number which lies in $(0,1)$, which is impossible as all natural numbers are bounded below by $1$. 
However, this assumes that natural numbers are closed under addition. Also, this does not use induction. 
Any pointers for this question?

REPLY [0 votes]: Firstly, recall that
$$
m \leq n
\stackrel{\text{df}}{\iff}
(m,n \in \textbf{N}) ~ \text{and} ~ (\exists k \in \textbf{N})(n = m + k)
$$
and
$$
m < n \stackrel{\text{df}}{\iff} (m \leq n) ~ \text{and} ~ (m \neq n).
$$
It is easy to prove that $ \leq $ is a reflexive and transitive relation on $ \textbf{N} $. To establish its anti-symmetry, we need a lemma.

Lemma. If $ n \in \textbf{N} \setminus \{ 0 \} $, then there exists an $ m \in \textbf{N} $ such that $ n = m + 1 $.

Proof. Let
$$
A = \{ n : \textbf{N} \mid (n = 0) ~ \text{or} ~ (\exists m \in \textbf{N})(n = m + 1) \}.
$$
Clearly, $ 0 \in A $, and if $ n \in A $, then $ n + 1 \in A $ also, because there exists an $ m \in \textbf{N} $ such that $ n + 1 = m + 1 $. Therefore, by the Principle of Induction, $ A = \textbf{N} $, which yields the lemma. $ \qquad \square $

Theorem. $ \leq $ is an anti-symmetric relation on $ \textbf{N} $.

Proof. Let $ m,n \in \textbf{N} $, and suppose that $ m \leq n $ and $ n \leq m $. Then there exist $ k,l \in \textbf{N} $ such that
$$
n = m + k \qquad \text{and} \qquad
m = n + l.
$$
Consequently,
$$
  n
= m + k
= (n + l) + k
= n + (l + k).
$$
By the Cancellation Law for $ + $, we find that $ l + k = 0 $. If $ k = 0 $, then $ l = l + k = 0 $. If $ k \neq 0 $, then the lemma says that there exists a $ j \in \textbf{N} $ such that $ k = j + 1 $, so
$$
  0
= l + k
= l + (j + 1)
= (l + j) + 1.
$$
This is a contradiction because $ 0 $ is not a successor. Therefore, $ k = 0 = l $, which yields $ m = n $. $ \qquad \square $

Proposition. Let $ m \in \textbf{N} $. Then there does not exist an $ n \in \textbf{N} $ such that $ m < n < m + 1 $.

Proof. Assume the contrary, i.e., there exists an $ n \in \textbf{N} $ such that $ m < n < m + 1 $. Then there exists a $ k \in \textbf{N} $ such that $ n = m + k $. As $ m \neq n $, we have $ k \neq 0 $. By the lemma, there exists a $ j \in \textbf{N} $ such that $ k = j + 1 $, so
$$
     m + 1
\leq (m + 1) + j
=    m + (1 + j)
=    m + (j + 1)
=    m + k
=    n.
$$
We thus have $ n < m + 1 \leq n $. As $ \leq $ is an anti-symmetric relation on $ \textbf{N} $, we obtain $ n = m + 1 $, contradicting $ n < m + 1 $. $ \qquad \square $
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