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Prove that $$\left\lfloor \frac {n+3}{2} \right\rfloor=\left\lceil \frac {n+2}{2}\right\rceil$$

I have tried to solve this on my own, and I want to check my solution.

My steps: Set $x=\left\lfloor \frac {n+3}{2}\right\rfloor $ then, for an integer x $$ \frac {n+3}{2}=x+\epsilon , 0 \le \epsilon \lt 1$$ $$ n=2x+2 \epsilon-3 $$ We substitute in the right hand side and get $$ \left\lceil \frac {2x+2\epsilon -1}{2} \right\rceil = \left\lceil x + \epsilon - \frac {1}{2}\right\rceil $$ Now, sine x is an integer we can write the last statement as $$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil $$

Now , I played around with the inequality. I have subtracted a half from all the sides so I got $$ -\frac {1}{2} \le \epsilon -\frac {1}{2} \lt \frac {1}{2} $$

I have drawn the number line and found that the inequality turns to $$ -1 \lt \epsilon -\frac {1}{2} \le 0 $$ (Since we are dealing with integers)

Now, we see that $\left\lceil \epsilon - \frac {1}{2}\right\rceil = 0$

So, we proved that$$ x +\left\lceil \epsilon - \frac {1}{2}\right\rceil = x $$ Which the same as the left hand side. Is my procedure correct??

This is the solution that I have found

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Cheese Cake
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Noussa
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2 Answers2

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You can pull integers out of floor/ceiling, so that setting $n=2k$ or $n=2k+1$ and pulling $k+1$

$$\left\lfloor \frac {2k+3}{2}\right \rfloor=\left\lceil \frac {2k+2}{2}\right\rceil$$ is equivalent to

$$\left\lfloor \frac {1}{2}\right \rfloor=\left\lceil \frac {0}{2}\right\rceil,$$

and

$$\left\lfloor \frac {2k+1+3}{2}\right \rfloor=\left\lceil \frac {2k+1+2}{2}\right\rceil$$ is equivalent to

$$\left\lfloor \frac {2}{2}\right \rfloor=\left\lceil \frac {1}{2}\right\rceil.$$

  • I think you are explaining the alternative answer. Thanx, but I want to know is my procedure correct? Because I understood the other answer. – Noussa Mar 19 '19 at 16:21
  • @Noussa: your approach is hard to follow and not really "checkable". As $n$ and $x$ are integers, the condition $0\le\epsilon<1$ is weird. Look for something simpler, and remember: you may pull an integer ouf of a floor/ceiling. –  Mar 19 '19 at 16:24
  • Ok, Thank you – Noussa Mar 19 '19 at 16:43
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If $n$ is an integer, then $\epsilon$ is either $0$ or $\frac12$ and hence $$ \bigg\lceil\epsilon-\frac12\bigg\rceil=0. $$ Otherwise, the identity doesn't hold. For example, $n=4.3$ and then $$ \bigg\lfloor\frac{4.3+3}{2}\bigg\rfloor=\lfloor3.65\rfloor=3, \bigg\lceil\frac{4.3+2}{2}\bigg\rceil=\lceil3.25\rceil=4. $$

xpaul
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