I need help on how to prove this:
If x is an arbitrary real number, prove that there are integers m and n such that m < x < n.
Thank you
WLOG for any $\displaystyle x>0$ consider $\displaystyle k=1$. By the Archimedean Ordering on $\displaystyle \mathbb{R}$ there shall exist $\displaystyle n$ such that $\displaystyle n\cdot 1 > x$. Hence we can find $\displaystyle n>x$. That completes the second inequality.
For any $\displaystyle x>0$ consider $\displaystyle x-1$. I claim we can choose an integer $\displaystyle m$ such that $\displaystyle x-1\leq m <x$.
Above we can find an integer $\displaystyle n>x$. Hence consider the interval $\displaystyle [-n,n]$. Define the set of integers $\displaystyle S=\{ k \in \mathbb{Z} | x-1 \leq k \leq n\}$. This is a finite non-empty set. So choose $\displaystyle m=\min S$. I leave it to you to prove that $\displaystyle m$ is the desired integer.