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