Floor, Ceil, Fractional Part

We know that

$\hspace{13pt}0 \leq \{x\} < 1\newline\Rightarrow 0 \leq 2\{x\} < 2\newline\Rightarrow 0 \leq \lfloor x \rfloor + \lceil x \rceil < 2$

Consider two cases.

Case 1 : $x$ is an integer.

Then $\{x\} = 0$

$\Rightarrow \lfloor x \rfloor + \lceil x \rceil = 0\newline\Rightarrow 2x = 0\hspace{50pt}[\because \lfloor x \rfloor = \lceil x \rceil = x]\newline\Rightarrow x = 0$

Case 2 : $x$ is not an integer.

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

Now, $2\{x\} = \lfloor x \rfloor + \lceil x \rceil = 2\lfloor x \rfloor + 1\newline \Rightarrow \{x\} = \lfloor x \rfloor + \large\frac{1}{2}$ $\newline\Rightarrow \{x\} = \large\frac{1}{2}$ $\newline \Rightarrow \lfloor x \rfloor = 0\newline \Rightarrow x = \lfloor x \rfloor + \{x\} = \large\frac{1}{2}$

So all the solutions are $0$ and $\large\frac{1}{2}$

Sum is $\large\frac{1}{2}$

There exists an integer n such that n <= x <= n+1. There also exists a number A in [0,1) such that x = n + A. The equation can then be rewritten:

n + (n+1) = 2*A

2n + 1 = 2*A

2n = 2*A - 1

But, A being in [0,1) means that 2 A - 1 lies in [-1,2), so that 2 n = -1, 0, or 1 (since n is an integer). However, the only solution that permits n to be an integer is 2n = 0. This means 2 A-1 = 0 or A = .5. Hence, x = n + A = 0 + .5 = .5