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$$ \left\lfloor \frac{n+3}{2} \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor +1$$ Is it correct? Thanks in advance.

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closed as off-topic by user147263, Travis, dustin, graydad, voldemort Jan 21 '15 at 4:05

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  • $\begingroup$ Have you plugged in an odd number and even number to check it? $\endgroup$ – turkeyhundt Jan 21 '15 at 1:34
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    $\begingroup$ Consider $n = 1$. $\endgroup$ – A.S Jan 21 '15 at 1:35
  • $\begingroup$ Consider $n=3$, $n=5$ $\endgroup$ – Eleven-Eleven Jan 21 '15 at 1:36
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    $\begingroup$ A useful fact is that whenever $n$ is an integer, $\lfloor x+n\rfloor=\lfloor x\rfloor+n$. (Try to prove it; it’s not hard.) Here that means that $$\left\lfloor\frac{n+3}2\right\rfloor = \left\lfloor\frac{n+1}2+1\right\rfloor = \left\lfloor\frac{n+1}2\right\rfloor+1\;,$$ and you can easily check that this is not the same as $\left\lfloor\frac{n}2\right\rfloor+1$. (Of course as Andrew Salmon pointed out, you can also check very directly that your proposed equality doesn’t hold in general. $\endgroup$ – Brian M. Scott Jan 21 '15 at 1:38
  • $\begingroup$ It would involve $\Bigl\lfloor\dfrac{n+1}2\Bigr\rfloor=\Bigl\lfloor\dfrac n2\Bigr\rfloor$. Is that always true? $\endgroup$ – Bernard Jan 21 '15 at 1:39
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$$ \left\lfloor \frac{n+3}{2} \right\rfloor = \left\lfloor \frac{n+1}{2} +1 \right\rfloor = \left\lfloor \frac{n+1}{2} \right\rfloor +1 $$

so your proposition is true iff $$\left\lfloor \frac{n+1}{2} \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor $$

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The correct formula is:

$$\left\lfloor \frac{n+3}{2} \right\rfloor=\left\lfloor \frac{n+1}{2} +1 \right\rfloor=\left\lfloor \frac{n+1}{2} \right\rfloor +1 $$

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