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Let $\lfloor a \rfloor$ be the least element of {$k \in \Bbb Z : k\ge a$} and $\lceil a \rceil$ the greatest element of {$k \in \Bbb Z : k\le a$}. How can I prove that for all $a \in \Bbb R$,

$\lfloor -a \rfloor=-\lceil a \rceil$ ?

I have tried using the definitions of the roof and ceiling functions: $\lfloor a \rfloor: =n≤x<n+1$ and $\lceil a \rceil:=m−1<x≤m$.

Thanks.

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  • $\begingroup$ I'd suggest looking at graphical translations as an argument. HINT: what happens to f(x) when you replace x with -x? $\endgroup$ – The Great Duck Mar 4 '16 at 5:09
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If $\,n\in\Bbb Z\,$ then $\,n\le \lfloor -a\rfloor\!\iff\! n\le -a\!\iff\!-n\ge a\!\iff\!-n \ge \lceil a\rceil\!\iff\! n\le -\lceil a\rceil$

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Let $a=b+r$ where $0\le r<1$ and $b$ is the integral part

Case $\#1$ If $r>0,$

$\displaystyle\lceil a \rceil=b+1$

and $\displaystyle-a=-b-r=-b-1+(1-r)\implies\lfloor-a\rfloor=-b-1 $

Case $\#2$ If $r=0,$

$\displaystyle\lceil a \rceil=b$

and $\displaystyle-a=-b\implies\lfloor-a\rfloor=-b $

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