Answer to Question #271839 in Discrete Mathematics for Jaishree

Question

Show that if x is a real number, then x − 1 < ⌊x⌋ ≤ x ≤

⌈x⌉ < x + 1.

Answer 1

$n=\lfloor x\rfloor\quad \Leftrightarrow \quad n\leq x<n+1 \quad \Leftrightarrow \quad \begin{cases} x-1<n \\ n\leq x \end{cases} \quad \Leftrightarrow \quad x-1< \lfloor x\rfloor \leq x$

$m=\lceil x\rceil\quad ⇔\quad m-1<x\leq m \quad \Leftrightarrow \quad \begin{cases} x\leq m \\ m<x+1 \end{cases} \quad \Leftrightarrow \quad x\leq \lceil x\rceil <x+1$

So, $x-1<\lfloor x \rfloor \leq x\leq \lceil x \rceil <x+1$