For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m
For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m<x<n$.
Assume that $m < n$. I apply first Archimedian property to get $x < n$. But I need to show that $m < x$ to complete proof?
Could you provide some hints?
 A: From the wording, it looks as if we can assume that the reals have the Archimedean property. So there is a natural number $n$ such that $x\lt n$. It remains to deal with the "$m$" part of the assertion.
If $x\ge 0$, let $m=-1$. Then $m\lt x\lt n$.
Now suppose that $x\lt 0$. There  is a natural number $w$ such that $|x|=-x\lt w$. Let $m=-w$. Show that $m\lt x$.
A: So what is the statement of the archemedian property.
By my book it is:a) for $x, y \in \mathbb R$, $x > 0$ then there is positive integer such that $nx > y$ b) for $x, y \in \mathbb R$, $x < y$ there exists $q\in \mathbb Q$ so that $x < q <y$.
We won't use b.
So for any $y$, $1 > 0$ there exist $n > |y|$. So $-n < -|y|$.  
So either $y \ge 0$ and $|y| = y$ and $y > -y$ and $-n < -y < y < n$ and $-n < y < n$, ore $y \le 0$ and $|y| = -y$ and $-y > y$ and $-n < -|y| = y < |y| = -y < n$ and $-n < y < n$.
Which is actually a very weak and unsatisfying result.  We know there exists an $n$ so that $n \le x < n+1$.
It might be a good idea to understand why the archemdiean principal holds and that it could just as easily been written as: for all $x \in \mathbb R$ there is an $n \in \mathbb Z$ such that $n \le x < n + 1$.
Let $S = \{n \in \mathbb Z| n < x$.  If $S$ is empty then $x$ is a lower bound for $\mathbb Z$.  Let $v \inf \mathbb Z$.  Then for any $0 < \epsilon < 1$ there is a $n \in \mathbb Z$ such that $v \le n < v + \epsilon$.  But then $v - 1 \le n - 1  < v - 1 + \epsilon < v$.  But $n-1 \in \mathbb Z$ and $n-1 < \inf \mathbb Z$ which is a contradiction. 
This proves there are integers below $x$ and also that $\mathbb Z$ is not bounded below.  An exact equivalent argument  can show $\mathbb Z$ is not bounded above and there are integers above $x$.
So that proves your weak statement.  But what about my strong statement that there exist $n$ so that $n \le x < n+1$.
If $x \in \mathbb Z$ this is trivial: $n = x < n+1$.
So let's assume $x \not \in \mathbb Z$.
Let's go back to our  $S = \{m \in \mathbb Z| m > x\}$ and $v = \inf S$.  $x$ is lower bound so $x \le v$.  If $v$ is greatest lower bound, s for any $0 < \epsilon < 1$ there is an $n \in \mathbb Z$ so that $v \le n < v+ \epsilon$.  If $v < n$ then that exist another integer $n_2$ so that $v \le n_2 < n$. But $0 < n - n_2 < \epsilon < 1$ which is impossible.  So $v \in \mathbb Z$.  And $x \le v$ and $v - 1 \not \in S$ so $v-1 \le x \le v$.
Which if $x \not \in \mathbb Z$ proves $n = v-1 < x < v= n+1$.
And if $x \in \mathbb Z$ we trivially have $n = x < n+1$.
A: tl;dr
If you can assume that the integers are not bounded above or below then that means for any $x \in \mathbb R$ that $x$ is neither an upper bound nor a lower bound of the integers.
Thus there are integers larger and smaller than $x$.  
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Now to actually prove the integers are not bounded is another issue altogether....
A: Look at $[x]-1$ and $[x]+1$ where $[x]$ denotes the integer part of $x$.
