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.

  • $\begingroup$ the definition of $f(x) = \lfloor x \rfloor$ is $f(n+\epsilon) = n$ for any $n \in \mathbb{Z}, \epsilon \in [0;1[$ so $f(n+x) = n+f(x)$ for any $n \in \mathbb{Z}$ (this is what you need to prove, for example by induction) $\endgroup$
    – reuns
    Feb 7 '16 at 21:03
  • $\begingroup$ Since $x\in\Bbb N$, this question makes a good proof: Number Theory Question on the floor function $\endgroup$
    – user228113
    Mar 10 '16 at 9:30

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 $


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.

  • $\begingroup$ That's exactly what I did. $\endgroup$
    – AlohaSine
    Feb 7 '16 at 22:19
  • $\begingroup$ Kind of nasty to downvote my answer isn't it? JUust because didn't read your answer. But now that I did, I found an error, so I wll downvote you also. $\endgroup$ Feb 7 '16 at 22:25
  • $\begingroup$ sigh. At least I gave you the reason I downvoted you $\endgroup$
    – AlohaSine
    Feb 7 '16 at 22:26
  • $\begingroup$ you wrote this: $\lfloor u \rfloor = u - \epsilon$ and it's not right. You really should watch the downvoting. Like I said II did not read your answer. Downvotes should be for bad proofs not for bruised egos. $\endgroup$ Feb 7 '16 at 22:27
  • $\begingroup$ Fine, I apologize. But remember the feature is not a personal insult. Can you kindly explain how that it is an error? Suppose we take $u=6.4$. Then the floor is $6$ and $\epsilon$ is 0.4. $\endgroup$
    – AlohaSine
    Feb 7 '16 at 22:32

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