For $x\in \mathbb{R}$ let's define $[x]$ as: $$ [x] = max \{ k\in \mathbb{Z}: k\leq x \} $$ and $[x]^{*}$ as: $$ [x]^{*} = min \{ k\in \mathbb{Z}: k\geq x \}. $$ Show that: $$ [x]^{*} = -[-x]. $$
So far I have managed to notice that: $$ [x]^{*} -1 \leq x < [x]^{*} $$ so: $$ -[x]^{*} \leq -x < 1- [x]^{*} $$ I believe that it might be the right direction here, but I have a problem to manipulate it further to get the $-[-x]$ form.
On
Well, when in trouble set variables.
If $x \in \mathbb Z$ then $x = \lfloor x \rfloor = \lceil x \rceil$[*] and the result is trivial.
Otherwise
Let $\lfloor x \rfloor = k$. So $k < x < k + 1$.[**]
So $\lceil x \rceil = k + 1$.
$-k - 1 < - x < -k$
So $\lfloor -x \rfloor = -k -1$
An $\lceil -x \rceil = -k$.
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[* and **] We probably shouldn't take these for granted but they're easy though tedious to prove via the archimedian property and the transitivity of order.
Note that for a subset of integers $A$ the following holds $$\min A=-\max\{-x:x\in A\}$$ Therefore $$[x]^*=-\max\{-k:k\ge x\}\cap\mathbb Z=-\max\{-k:-k\le -x\}\cap\mathbb Z\\ =-\max\{k:k\le -x\}\cap\mathbb Z=-[-x]$$