Proving that $ \lfloor \lceil x \rceil \rfloor = \lceil x \rceil $ for all real numbers $ x $.

Question

Let $ \mathsf{LHS} = \lfloor \lceil x \rceil \rfloor $ and $ \mathsf{RHS} = \lceil x \rceil $.

Let us call $ n = \lceil x \rceil $.

Case 1: $ n $ is even, i.e., there exists a $ k \in \mathbb{Z} $ such that $ n = 2 k $.

I’m not sure how to go on or if I’m setting this up right.

You have to show that, if $n=\lceil x\rceil$, then $\lfloor n\rfloor=n$? If $n$ is the floor of some number then $n$ is its own ceiling? If $n$ is in the range of the floor function, then $n$ is a fixed point of the ceiling function? It would help a lot if I knew (a) what is the range of the floor function, and (b) what are the fixed points of the ceiling function. Beats me. You have any ideas? — bof, Mar 1, 2015 at 8:30

Matt Samuel · Accepted Answer · 2014-11-13 23:40:39Z

13

$\lceil x\rceil$ is an integer, and if $n$ is an integer then $\lfloor n\rfloor=n$.

answered Nov 13, 2014 at 23:40

Matt Samuel

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Jytug · Accepted Answer · 2014-11-13 23:43:32Z

9

First notice that $ \left \lfloor{n}\right \rfloor = n$ for all integers $ n$. And $ \lceil x \rceil $ is an integer

answered Nov 13, 2014 at 23:43

Jytug

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2 Answers 2