For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m<x<n$ For $x\in\mathbb R$. Prove that there exists $m,n \in \mathbb Z$ such that $m&lt;x&lt;n$.


Assume that $m &lt; n$. I apply first Archimedian property to get $x &lt; n$. But I need to show that $m &lt; x$ to complete proof?

Could you provide some hints?
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 $m\lt x$.
Look at $[x]-1$ and $[x]+1$ where $[x]$ denotes the integer part of $x$.