prove that if x is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x + \frac{1}{2}\rfloor$ prove that if $x$ is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor + \lfloor x + \frac{1}{2}\rfloor$
I've seen the proof in my discrete math book, but it was very confusing. I don't understand why they considered those cases and why they chose $\frac{1}{2}$ specifically. It would be great if someone can explain better or provide a much clearer proof.
Proof from the book:
Let x = n + $\epsilon$ where n is an integer and 0 $\le\epsilon \lt 1$
case 1: Consider 0 $\le\epsilon \lt \frac{1}{2}$
2x = 2n + 2$\epsilon$ and $\lfloor$2x$\rfloor$ = 2n because 0 $\le 2\epsilon \lt 1$
Similarly, x + $\frac{1}{2}$= n + ( $\frac{1}{2}$+ $\epsilon$), so $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n, 
because 0 < $\frac{1}{2}$ + $\epsilon$ < 1.
Consequently,$\lfloor$2x$\rfloor$ = 2n and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$ = n + n = 2n.
case 2: $\frac{1}{2}$ ≤ $\epsilon$< 1.
2x = 2n + 2$\epsilon$ =(2n + 1) + (2$\epsilon$ − 1).
Because 0 ≤ 2$\epsilon$ − 1 < 1, it follows that $\lfloor$2x$\rfloor$ = 2n + 1. Because$\lfloor$x + $\frac{1}{2}$$\rfloor$ = $\lfloor$n + ( $\frac{1}{2}$+ $\epsilon$)$\rfloor$ = $\lfloor$n + 1 + ($\epsilon$ − $\frac{1}{2}$ )$\rfloor$ and 0 ≤ $\epsilon$ − $\frac{1}{2}$ < 1,
it follows that $\lfloor$x + $\frac{1}{2}$$\rfloor$ =n + 1. 
Consequently, $\lfloor$2x$\rfloor$ = 2n + 1 and $\lfloor$x$\rfloor$ + $\lfloor$x + $\frac{1}{2}$$\rfloor$= n + (n + 1) = 2n + 1. 
This concludes the proof.

One has to distinguish these cases since in this example "something special" happens with the floor function when the variable $x$ passes an integer multiple of ${1\over2}$. Nevertheless it can be said that the proof in your book is much too complicated.
Instead begin with the remark that both sides of the stated equation increase by $2$ when $x$ increases by $1$. It is therefore sufficient to prove it for $0\leq x<1$.
If $0\leq x<{1\over2}$ then $x+{1\over2}<1$ and therefore $\lfloor2x\rfloor=0=\lfloor x\rfloor+\bigl\lfloor x+{1\over2}\bigr\rfloor$.
Similarly: If ${1\over2}\leq x<1$ then $\lfloor2x\rfloor=1=0+1=\lfloor x\rfloor+\bigl\lfloor x+{1\over2}\bigr\rfloor$.
