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

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$n = ⌊ x ⌋ \Leftrightarrow n \leq x < n + 1 \Leftrightarrow {x - 1 < n n \leq x \Leftrightarrow x - 1 < ⌊ x ⌋ \leq x$

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

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

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

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Question ID: mtid-5-stid-8-sqid-736-qpid-621