For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$ 
For all $x$ which are real numbers, prove that
  $$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$

I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
 A: I am assuming $\lfloor x\rfloor = n$
Let's take two cases - one - the fractional part of $x$ is less than $0.5$  LHS will be $2n$ RHS will be $n+n$
case two - The fractional part of $x$ is greater than $0.5$LHS will be $2n+1$ RHS will be $n+n+1$
Hence, proved.
A: Let $ x = i + d $ where $ 0 \le d < 1 $
$ \lfloor 2x \rfloor = \lfloor 2i + 2d \rfloor = 2i + \lfloor 2d \rfloor $
$ \lfloor x \rfloor + \lfloor x + 0.5 \rfloor = 2i + \lfloor d \rfloor + \lfloor d + 0.5 \rfloor $
So we reduced the problem to just show for $ d $
Suppose $ 0 \le d < 0.5 $, then both sides are just $ 2d $
Suppose $ 0.5 \le d < 1 $, then both sides are just $ d + 1 $.
QED $ \blacksquare $
A: I will try to show
$\lfloor nx \rfloor = \sum_{k=0}^{n-1} \lfloor  x+\frac{k}{n} \rfloor$.
Let
$m = \lfloor x \rfloor$
and
$d = x - m$,
so
$0 \le d < 1$.
Let
$j
=\lfloor nd \rfloor
$,
so
$0 \le j \le n-1
$
and
$\frac{j}{n}
\le d
< \frac{j+1}{n}
$.
If
$0 \le k \le n-j-1$,
$\begin{array}\\
m
&\le m+d+\frac{k}{n}\\
&< m+\frac{n-j}{n}+\frac{j}{n}\\
&= m+\frac{n}{n}\\
&= m+1\\
\end{array}
$
so
$\lfloor m+d+\frac{k}{n} \rfloor
=m
$.
If
$n-j \le k \le n-1$,
$\begin{array}\\
m+d+\frac{k}{n}
&\ge m+\frac{n-j}{n}+\frac{j}{n}\\
&= m+\frac{n}{n}\\
&= m+1\\
\end{array}
$
and
$\begin{array}\\
m+d+\frac{k}{n}
&\lt m+\frac{n-j+1}{n}+\frac{n-1}{n}\\
&= m+\frac{2n-j}{n}\\
&= m+2-\frac{j}{n}\\
\end{array}
$
so
$\lfloor m+d+\frac{k}{n} \rfloor
=m+1
$.
Therefore
$\begin{array}\\
\sum_{k=0}^{n-1} \lfloor x+\frac{k}{n}  \rfloor
&=\sum_{k=0}^{n-1} \lfloor m+d+\frac{k}{n}  \rfloor\\
&= \sum_{k=0}^{n-j-1} \lfloor m+d+\frac{k}{n} \rfloor
+\sum_{k=n-j}^{n-1} \lfloor m+d+\frac{k}{n} \rfloor\\
&= (n-j)m+j(m+1)\\
&= nm+j\\
\end{array}
$
and
$\lfloor nx \rfloor
=\lfloor n(m+d) \rfloor
=nm+\lfloor nd \rfloor
=nm+j
$.
We are done.
A: $n\leq x< n+1$ , then $f(x)=n$ for all $x$ 
I) If $n<x< \frac{2n+1}{2}$
$f(x)+f(x+0.5)= n+n=2n=f(2x)$
since $x+0.5<\frac{2n+1}{2} + \frac{1}{2}=n+1$
II) If  $n+1>x\geq \frac{2n+1}{2}$ 
$f(x)+f(x+0.5)= n+n+1=2n+1=f(2x)$
Since $x+0.5\geq\frac{2n+1}{2} + \frac{1}{2}=n+1$
