Proof that $\lfloor x + x\sqrt3 \rfloor = x + \lfloor x\sqrt3 \rfloor$ Where $x \in N$

Question

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.

Answer 1

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 $

Answer 2

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.