Jtdylktuy

Let $a$ and $b$ be positive integers.

If $b$ is even, then we have $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil frac{a+b}{2} rightrceil = a$$

I think the equality also hold when $b$ is odd. What could be a proof for it?

If $a$ and $b$ are both odd (or both even), then $a-b$ and $a+b$ are both even, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2}.$$

Otherwise, if exactly one of $a$ and $b$ is odd, then $a-b$ and $a+b$ are both odd, and thus $$leftlfloor frac{a-b}{2} rightrfloor = frac{a-b}{2} - frac12 quadtext{and}quad leftlceil frac{a+b}{2} rightrceil = frac{a+b}{2} + frac12.$$

In either case, it's easy to check that your equation holds.

BTW, as noted by Gareth McCaughan, your equation in fact holds for all real numbers $b$, as long as $a$ is an integer. One fairly simple way to show this is to note that $frac{a+b}{2} = a - frac{a-b}{2}.$ Thus, we can rewrite your equation as $$leftlfloor frac{a-b}{2} rightrfloor + leftlceil a - frac{a-b}{2} rightrceil = a.$$

Since $a$ is an integer (by assumption), and since $lceil k + x rceil = k + lceil x rceil$ for any integer $k$, we can extract $a$ from the ceiling term to get $$leftlfloor frac{a-b}{2} rightrfloor + a + leftlceil - frac{a-b}{2} rightrceil = a,$$ and finally, by applying the identity $lceil -x rceil = - lfloor x rfloor$, rewrite this as $$leftlfloor frac{a-b}{2} rightrfloor + a - leftlfloor frac{a-b}{2} rightrfloor = a.$$ Cancelling the floor terms then just leaves the identity $a = a$.

On the other hand, as also noted by Gareth, your equation cannot hold for any non-integer $a$, since its left-hand side is always an integer.

The equality only depends on the parity of $a+b$, as it is the same as that of $a-b$. Then

$$leftlfloorfrac02rightrfloor+leftlceilfrac02rightrceil=0$$
and
$$leftlfloorfrac12rightrfloor+leftlceilfrac12rightrceil=1$$
are enough as a proof.

This actually works whatever the value of $b$ -- it doesn't need to be an integer. Clearly it's true when $b=0$. Now imagine changing $b$ smoothly from $0$ to its final value. When does the value of our expression change? Precisely when $(a+b)/2$ or $(a-b)/2$ passes an integer; that is, when $apm b$ is an even integer; that is, when $b$ differs from $a$ by an even integer. When this happens, both terms change in opposite ways, so the expression as a whole doesn't change its value. So by the time $b$ reaches its final value, our expression still hasn't changed.

(On the other hand, if $a$ isn't an integer then the equation never holds because one side is an integer and the other isn't.)

If $a$ and $b$ are both odd then we have $a=2m+1$ and $b=2n+1$ where $m$ and $n$ are positive integers.

Then, we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m+1)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m+1)+(2n+1)}2rightrceil\
&=leftlfloorfrac{2m-2n}2rightrfloor+ leftlceilfrac{2m+2n+2}2rightrceil\
&=lfloor m-nrfloor + lceil m+n+1rceil\
&= m-n + m+n+1tag{$*$}\
&= 2m+1\
&=aend{align}

We can get to $(*)$ because $m$ and $n$ are integers and so their floor (or ceiling) is just the number inside (two integers added or subtracted will always give an integer answer).

If $a$ is even and $b$ is odd, then we have $a=2m$ and $b=2n+1$, again for $m$ and $n$ integers.

So we have begin{align}leftlfloorfrac{a-b}2rightrfloor+leftlceilfrac{a+b}2rightrceil &= leftlfloorfrac{(2m)-(2n+1)}2rightrfloor+ leftlceilfrac{(2m)+(2n+1)}2rightrceil\
&=leftlfloorfrac{2m-2n-1}2rightrfloor+ leftlceilfrac{2m+2n+1}2rightrceil\
&=leftlfloor m-n-frac 12rightrfloor + leftlceil m+n+frac12rightrceil\
&= m-n -1+ m+n+1tag{$dagger$}\
&= 2m\
&=aend{align}

Here we get to $(dagger)$ because we round the first half down and the second half up to their respective nearest integers

As $s:=a+b$ and $a-b$ have the same parity, we can write

$$leftlfloorfrac{s-2b}2rightrfloor+leftlceilfrac s2rightrceil=leftlfloorfrac s2rightrfloor+leftlceilfrac s2rightrceil-b=s-b=a.$$