I'm trying to prove without the well ordering principle that no integer $m$ exists such that $n < m < n + 1$ for positive integers $m$ and $n$.
I know there's a proof here that uses the well ordering principle, but I'm trying to prove it without using the well ordering principle because this rundown of the proof that the natural numbers follow the well ordering principle uses this property. All help would be appreciated. Thanks.
Let $n$ be some integer. Then $n$ exists. Then $n+1$ is also an integer and also exists.
$$n,n+1\in\mathbb{Z}$$
Assume that there exists some other integer $m$ such that $n < m $. This is fine so far. Further assume that $m$ has the property $ m < n + 1$. Then it is the case that $m$ and $n$ satisfy the following:
$$ n < m < n + 1 $$
Since we are assuming that the above is True, then we can do algebra on it. So subtract $n$ from all sides:
$$ n - n < m - n < n - n + 1\\ $$
Of course $n-n=0$ and $n-n+1=1$. Thus
$$ 0 < m - n < 1 $$
This is equivalent to saying "$m-n$ is less than 1 and is also greater than zero". Here's another way: $m-n>0$ and $m-n<1$. Further, we can conclude that since $m$ and $n$ are integers, then $m-n$ is an integer. I mean just look: $5-4=1$, $18-1=17$, $1000-865=135$, etc. Let's just rename $m-n$ for simplicity. Let $p=m-n$. Therefore since $m$ and $n$ are integers $p$ is an integer and $0<p<1$.
Such a simple proof!
However there is one problem! There is no integer $p$ between the integers 0 and 1. We have a contradiction! Either The integer $m$ doesn't exist (absurd, or course there are integers such that $m>n$) or the additional assumption that $m<n+1$ is incompatible with the definition of $m$. [INSERT REFERENCE TO INVISIBLE, PINK UNICORNS OR SQUARE CIRCLES].
Suppose you doubt the above and think "keep an open mind". I would appreciate your skepticism and ask you to to write down the integer that it could be. You would not be able to give that number because the integers are the set $\{0,1,2,3,4,...\}$ and there is no integer $p$ way off the list that I wrote that has the property $0<p<1$. Suppose you still doubt or even ask how to you really know that? I would say because all $p$ that I did not list above are all significantly greater than 1 and we are looking for a $p$ less than 1. We dont want some integer that is greater that 1.
Suppose you wanted me to check this fact and further suppose that one of the $p$ that I didn't list might satisfy $0<p<1$. By the definition of $m$ and $n$ we know that $p>0$. We assumed that there had to be at least one number that satisfied $p<1$. We can also note that since 1 is less than all other integers, then our assumption directly implies that $p$ is too. $p<1$. But all other numbers that I did not list are at least greater than 4! So therefore we must reject the idea that there exists integers between any two consecutive integers. Skepticism warranted or not, we can safely conclude that the integers don't have invisible integers.
This fact is also guaranteed by the well ordering principle. It is equivalent for me to write out this whole mini-essay or just say to you "The well ordering principles states that all non-empty sets contain a least element. Let M be a set of all integers $m$ that satisfy $n<m<n+1$ for all $n$. Either M is empty and there exists no such $m$ or you can definitely write or otherwise tell me what the smallest integer in M is."