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.

We don't need induction. Note $\lfloor n + u \rfloor = \lfloor u \rfloor + n$ for natural $n$. This follows since $\lfloor u \rfloor = u - \epsilon$ for $\epsilon \in [0,1)$.
So we get  
$\lfloor n + u \rfloor = \lfloor n + \epsilon + \lfloor u \rfloor \rfloor$ = $n + \lfloor u \rfloor$ since $\lfloor u \rfloor + n  ≤ n + \epsilon + \lfloor u \rfloor < \lfloor u \rfloor + n + 1 $
