Proving $\lfloor 2x \rfloor = \lfloor x \rfloor + \lfloor x+0.5\rfloor$

I can intuitively see that this is true, but I'm having a very hard time proving it. I'm actually not even quite sure where to begin. I tried using the inequalities that define the floor function, and I'm pretty sure this is the way to go, but I'm failing to make any kind of progress.

I appreciate any help.

Hint: split it in to cases $x\in[n,n+\frac{1}{2})$ and $x\in[n+\frac{1}{2},n+1)$ for some integer $n$.

Case 1: The fractional part of $x$ is less than $1/2$, i.e. $x=n+a$ where $n$ is an integer and $0\le a<1/2$. Then $\lfloor 2x\rfloor = \lfloor 2n+2a \rfloor = 2n$. But also $\lfloor x\rfloor=n$ and $\lfloor x +0.5 \rfloor = n$. Since $2n=n+n$, the equality holds.

Case 2: The fractional part of $x$ is more than $1/2$, i.e. $x=n+a$ where $n$ is an integer and $1/2\le a<1$. I leave the remainder as an exercise; let me know if you need more help.

You can "draw" an integer in and out of the floor function ($\lfloor x+n\rfloor=\lfloor x\rfloor+n$). Then it suffices to establish the property for $x\in[0,\frac12)$ and $x\in[\frac12,1)$, because the three floors remain constant in these intervals.

• $x\in[0,\frac12)\implies0=0+0$,

• $x\in[\frac12,1)\implies1=0+1$.

Simple as that.