1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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 $

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.