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 $\lfloor x\rfloor = n$

$n \leq x &lt; n+1$
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.
$n\leq x&lt; n+1$ , then $f(x)=n$ for all $x$ 

I) If $n&lt;x&lt; \frac{2n+1}{2}$

$f(x)+f(x+0.5)= n+n=2n=f(2x)$

since $x+0.5&lt;\frac{2n+1}{2} + \frac{1}{2}=n+1$

II) If  $n+1&gt;x\geq \frac{2n+1}{2}$ 

$f(x)+f(x+0.5)= n+n+1=2n+1=f(2x)$

Since $x+0.5\geq\frac{2n+1}{2} + \frac{1}{2}=n+1$