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This question already has an answer here:

Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$ for all real $x$ and $y$.

If anybody could post a simple solution (no complicated abstract theories or calc :D) to the above question, I would greatly appreciate it. Thanks!

Also, if possible, the solution shouldn't wander too much away from floor, ceiling, fraction functions, and other related functions.

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marked as duplicate by Semiclassical, B. Goddard, Misha Lavrov, Milo Brandt, Zain Patel Apr 22 '17 at 3:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I would say this: break $x = [x] + \{ x\}, y = [y] + \{ y\}$, and substitute in the above expression. Once you get a known expression, work backwards. $\endgroup$ – астон вілла олоф мэллбэрг Apr 22 '17 at 2:27
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Let $X=m+s$ and $y=n+t$ where $m,n\in\mathbb Z$ and $0\le s,t\lt1.$ Then

$$\lfloor2x\rfloor+\lfloor 2y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor x+y\rfloor$$ $$\iff\lfloor2m+2s\rfloor+\lfloor2n+2t\rfloor\ge\lfloor m+s\rfloor+\lfloor n+t\rfloor+\lfloor m+n+s+t\rfloor$$ $$\iff2m+\lfloor2s\rfloor+2n+\lfloor2t\rfloor\ge m+\lfloor s\rfloor+n+\lfloor t\rfloor+m+n+\lfloor s+t\rfloor$$ $$\iff\lfloor2s\rfloor+\lfloor2t\rfloor\ge\lfloor s+t\rfloor.$$ Without loss of generality, suppose $s\ge t.$ Then $2s\ge s+t,$ so $$\lfloor2s\rfloor+\lfloor2t\rfloor\ge\lfloor2s\rfloor\ge\lfloor s+t\rfloor.$$

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HINT

Let $x = n + r$ with $n$ an integer and $0 \le r <1$

Similarly, $y = m + s$ with $m$ an integer and $0 \le s < 1$

Now there are 4 cases to consider:

  1. $0 \le r < \frac{1}{2}$ and $0 \le s < \frac{1}{2}$

  2. $0 \le r < \frac{1}{2}$ and $\frac{1}{2} \le s <1$

  3. $\frac{1}{2} \le r < 1 $ and $0 \le s < \frac{1}{2}$

  4. $\frac{1}{2} \le r < 1$ and $\frac{1}{2} \le s <1$

Just to show case 2:

$\lfloor 2x \rfloor + \lfloor 2y \rfloor = \lfloor 2n + 2r \rfloor + \lfloor 2m + 2s \rfloor = 2n + 0 + 2m + 1$

$\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y\rfloor = n + m + n + m + \lfloor s +r \rfloor$ where $\lfloor s + r \rfloor \le 1$

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  • $\begingroup$ Can you continue with your case work? I always get stuck with case work, so I need some help here. $\endgroup$ – Quantum Pizza Apr 22 '17 at 2:47
  • $\begingroup$ @QuantumPizza Just work out the expressions and verify that the equation is satified.... i did case 2 as an example $\endgroup$ – Bram28 Apr 22 '17 at 2:59

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