Prove that $\lceil 2x\rceil =\lceil x\rceil +\lceil x+1/2\rceil$ -1 for all $x$ in $\mathbb R$

Question

I don't know how to approach the problem. I have searched for different ceiling and floor properties but none of them seemed to help. Or simply I just don't know how to tackle the question.

Answer 1

This is false when $x=\tfrac12$.

Answer 2

Just do it in cases:

There is a unique integer $n$ and a unique real $y$ where $0\le y < 1$ and $x = n+y$.

Case 1: $y=0$.

Then $\lfloor x \rfloor = \lceil x \rceil = n$.

$\lfloor 2x \rfloor = \lceil 2x \rceil = 2n$

$\lfloor x+\frac 12 \rfloor = n; \lceil x +\frac 12 \rceil = n+1$

So $\lfloor 2x \rfloor = \lfloor x \rfloor+\lfloor x+\frac 12 \rfloor$

but $\lceil 2x \rceil \ne \lceil x \rceil + \lceil x +\frac 12 \rceil$.

Case 2: $0 < y < \frac 12$

Then $\lfloor x \rfloor = n; \lceil x \rceil = n+1$.

$\lfloor 2x \rfloor = 2n; \lceil 2x \rceil = 2n+1$

$\lfloor x+\frac 12 \rfloor = n; \lceil x +\frac 12 \rceil = n+1$

So $\lfloor 2x \rfloor = \lfloor x \rfloor+\lfloor x+\frac 12 \rfloor$

but $\lceil 2x \rceil \ne \lceil x \rceil + \lceil x +\frac 12 \rceil$.

Case 3: $y=\frac 12$

Then $\lfloor x \rfloor= n; \lceil x \rceil = n+1$.

$\lfloor 2x \rfloor = \lceil 2x \rceil = 2n+1$

$\lfloor x+\frac 12 \rfloor = \lceil x +\frac 12 \rceil = n+1$

So $\lfloor 2x \rfloor = \lfloor x \rfloor+\lfloor x+\frac 12 \rfloor$

but $\lceil 2x \rceil \ne \lceil x \rceil + \lceil x +\frac 12 \rceil$.

Case 4: $\frac 12 < y < 1$.

Then $\lfloor x \rfloor= n; \lceil x \rceil = n+1$.

$\lfloor 2x \rfloor = 2n+1; \lceil 2x \rceil = 2n+2$

$\lfloor x+\frac 12 \rfloor = n+1 \lceil x +\frac 12 \rceil = n+2$

So $\lfloor 2x \rfloor = \lfloor x \rfloor+\lfloor x+\frac 12 \rfloor$

but $\lceil 2x \rceil \ne \lceil x \rceil + \lceil x +\frac 12 \rceil$.

So your equation is never true.

But it is always true for floor functions.