# Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor$?

Let $$a$$ and $$b$$ be integers. Is it true that

$$\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor$$

Where $$\lceil \cdot \rceil$$ is the ceiling function, $$\lfloor \cdot \rfloor$$ the floor function and $$|\cdot|$$ is the absolute function.

The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.

• What about $a=-1$ and $b=2$? The LHS is $|0-1| =1$ and the RHS is $2$. – Martin R Nov 7 '19 at 16:05
• @MartinR Thank you very much for pointing that out, I have altered the inequality slightly, there was an error in the equation. – KillaKem Nov 7 '19 at 16:25
• You might as well assume $a\ge b$ and thus dispense with taking absolute values. – hardmath Nov 7 '19 at 17:03
• If we allow all real numbers, you can also just write it as $|\lceil x\rceil-\lceil y\rceil|\ge\lfloor|x-y|\rfloor|$. Assuming $x\ge y$ gives $\lceil x\rceil-\lceil y\rceil\ge\lfloor x-y\rfloor$, which looks pretty neat. – Milten Nov 7 '19 at 17:13

Yes, it is true.

$$\left | \left \lceil \frac{a}{2} \right \rceil - \left \lceil \frac{b}{2} \right \rceil \right |\geq \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor \tag1$$

In the following, $$m,n$$ are integers.

Case 1 : If $$a=2m,b=2n$$, then both sides of $$(1)$$ equal $$|m-n|$$.

Case 2 : If $$a=2m,b=2n+1$$, then $$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$

If $$m-n-\frac 12\ge 0$$, then $$m-n-1\ge 0$$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.

If $$m-n-\frac 12\lt 0$$, then $$m-n-1\lt 0$$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.

Case 3 : If $$a=2m+1, b=2n$$, then $$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$

If $$m-n+\frac 12\ge 0$$, then $$m-n+1\ge 0$$, so$$(3)\iff m-n+1\ge m-n$$which is true.

If $$m-n+\frac 12\lt 0$$, then $$m-n+1\lt 0$$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.

Case 4 : If $$a=2m+1,b=2n+1$$, then both sides of $$(1)$$ equal $$|m-n|$$.

There is no need for the assumption that $$a$$ and $$b$$ are integers. You just need to prove that

$$|\lceil x\rceil-\lceil y\rceil|\ge\lfloor|x-y|\rfloor$$

for any real numbers $$x$$ and $$y$$. By symmetry, we may assume $$x\ge y$$, in which case we can remove the absolute value signs. If, moreover, we write $$x=y+u$$ with $$u\ge0$$, we are trying to prove

$$\lceil y+u\rceil\ge\lceil y\rceil+\lfloor u\rfloor$$

But $$u=\lfloor u\rfloor+r$$ for some $$0\le r\lt1$$, and $$\lceil y+\lfloor u\rfloor +r\rceil=\lceil y+r\rceil+\lfloor u\rfloor$$, so the inequality to prove is simply

$$\lceil y+r\rceil\ge\lceil y\rceil$$

which is clearly true, since the ceiling function is never decreasing and $$r\ge0$$.

Assume without loss of generality that $$a\ge b$$. Then the inequality is $$\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil \ge \left\lfloor \frac {a-b}2 \right\rfloor$$ If either $$a$$ or $$b$$ is an even integer, then we can pull the whole number $$\frac a2$$ or $$\frac b2$$ out of the floor function, and the inequality reduces to $$\left\lceil \frac a2 \right\rceil \ge \left\lfloor \frac {a}2 \right\rfloor$$ or $$-\left\lceil \frac b2 \right\rceil \ge \left\lfloor -\frac {b}2 \right\rfloor$$ (where the first is trivial and the second is actually an equality).

Assume therefore that neither of $$a$$ and $$b$$ is an even integer. Let $$2m and $$2n, for some $$m,n\in \mathbb Z$$. Then $$\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil = (m+1)-(n+1) = m-n$$ On the other hand $$m-n-1<\frac a2 - \frac b2 < m-n+1$$ which means that $$\left\lfloor \frac {a-b}2 \right\rfloor \le m-n = \left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil$$ so we are done.

EDIT: I didn't notice you assumed $$a$$ and $$b$$ to be integers. Well, my answer works for all real numbers.