Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that 
for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm \leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j \leq [nx] < mn + j + 1 = [nx]+1$.