Proof that $\lfloor x + x\sqrt3 \rfloor = x + \lfloor x\sqrt3 \rfloor$ Where $x \in N$ I'm not sure if this is true or not. I was going to prove it via induction. Here is what I have so far.
Base Case (x=0): $\lfloor 0 + 0\rfloor = 0 = 0 + \lfloor 0 \rfloor$
Induction Hypothesis (IH): $\lfloor x + x\sqrt3\rfloor = x + \lfloor x\sqrt3 \rfloor$
Induction Step: $\lfloor x + 1 + (x + 1)\sqrt3\rfloor = \lfloor x + (x + 1)\sqrt3\rfloor + 1$.
This is where I get stuck. Any help? I'm not convinced that this is even true, but I can't think of a counter example.

Write $x\sqrt 3=m+\epsilon $ where $m\in \mathbb N$ and $0<\epsilon <1$. 
Then, on one hand, we have  
$\lfloor x+x\sqrt 3\rfloor =\lfloor (x+m)+\epsilon \rfloor=x+m$
and on the other
$x+\lfloor x\sqrt 3\rfloor =x+\lfloor m+\epsilon \rfloor =x+m$
so the two sides are equal.
