# Math Help - supremum and infimum

1. ## supremum and infimum

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

2. Originally Posted by PersaGell
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 $x>0$ consider $k=1$. By the Archimedean Ordering on $\mathbb{R}$ there shall exist $n$ such that $n\cdot 1 > x$. Hence we can find $n>x$. That completes the second inequality.

For any $x>0$ consider $x-1$. I claim we can choose an integer $m$ such that $x-1\leq m .

Above we can find an integer $n>x$. Hence consider the interval $[-n,n]$. Define the set of integers $S=\{ k \in \mathbb{Z} | x-1 \leq k \leq n\}$. This is a finite non-empty set. So choose $m=\min S$. I leave it to you to prove that $m$ is the desired integer.