# Proof of an equation containing the ceiling and floor functions

Given $$n = r- p + 1,$$ where $$n,r,$$ and $$p$$ are all positive integers and $$1\le p \lt r \le n$$. I'm trying to prove that $$\lfloor \frac{r+p}2 \rfloor - p + 1 = \lceil\frac n2\rceil$$. I tried first to examine the four cases that arise depending on whether each of p and r is odd or even, but I couldn't reach anything.

I also tried to prove it directly using some properties of the floor and ceiling functions:

$$\lfloor \frac{r+p}2 \rfloor - p + 1 = \lfloor \frac{r+p-2p+2}2\rfloor = \lfloor \frac{r-p+2}2 \rfloor$$.

And we know that $$\lceil \frac n2 \rceil = \lceil \frac{r-p+1}2 \rceil.$$ Therefore, the only thing left to prove is that $$\lfloor r-p+2 \rfloor = \lceil r-p+1 \rceil$$.

I'm not seeing how to continue, perhaps I did something wrong on the way?

• If $r-p+1=n\geq r$ then $p\leq1$. Combining this with $1\leq p$ we find that $p=1$. Commented Feb 4, 2020 at 9:23
• I see, so now that we have $p = 1$ , how do we prove that $\lfloor \frac{r+1}2 \rfloor = \lceil \frac r2 \rceil$ ? I'm still a beginner with these functions sorry if my question is silly :p Commented Feb 4, 2020 at 9:38

As commented we have $$p=1$$ so that proving: $$\lfloor\frac{r+1}2\rfloor=\lceil\frac{r}2\rceil\tag1$$ for integer $$r$$ is enough.

We discern two cases:

• $$r$$ is odd, or equivalently $$r=2m+1$$ for some integer $$m$$.

Then $$\frac{r+1}2=m+1$$ so that the LHS of $$(1)$$ equals $$m+1$$.

So it remains to prove that the RHS of $$(1)$$, which is $$\lceil m+\frac12\rceil$$ also equals $$m+1$$ (which is obviously true by definition).

• $$r$$ is even, or equivalently $$r=2m$$ for some integer $$m$$.

Try this yourself.