Solving the inequality $0< \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ algebraically

Question

I'm not particularly familiar working with the ceiling function, particularly not algebra with it. I'm trying to solve the above inequality algebraically, specifically when $x\in \mathbb{Z}^{+}$. I've plotted the graph $y= \lceil \frac{x}{2} \rceil - \lceil \frac{x+1}{3} \rceil$ and examined where the curve is greater than $0$. The weird part about this curve is that when I plug in $x=5$, algebraically I obtain a difference of 1, but the plot shows me that the difference is 0, which is why I'm hesitant to trust the graph, I also know that in a proof I can't just put the graph. In terms of what I've tried algebraically, I've tried to consider 6 cases, depending on the equivalence class of $x \pmod{6}$, but again I get stuck because I feel as though I'm missing some properties of how the ceiling function behaves. If anyone could help me solve this I'd really appreciate it!

Answer 1

Working modulo $6$ there are $6$ residue classes $[0]_6,[1]_6,[2]_6,[3]_6,[4]_6,[5]_6$, thus we can write every element in $\mathbb{Z}$ as $6n+r$ for some $n\in\mathbb{Z}$ and $r=0,1,2,3,4,5$.

The ceiling function has the property that it rounds up to the nearest integer thus $\lceil \text{integer} + d \rceil = \text{integer} +1$ where $0\leq d <1$.

$\lceil\frac{6n}{2}\rceil = 3n$

$\lceil\frac{6n+1}{2}\rceil =\lceil3n+\frac{1}{2}\rceil = 3n+1$

$\lceil\frac{6n+2}{2}\rceil =\lceil3n+1\rceil = 3n+1$

$\lceil\frac{6n+3}{2}\rceil =\lceil3n+1+\frac{1}{2}\rceil = 3n+2$

$\lceil\frac{6n+4}{2}\rceil = 3n+2$

$\lceil\frac{6n+5}{2}\rceil =\lceil3n+2+\frac{1}{2}\rceil = 3n+3$

similarly

$\lceil\frac{6n+1}{3}\rceil = 2n+1$

$\lceil\frac{6n+2}{3}\rceil = 2n+1$

$\lceil\frac{6n+3}{3}\rceil = 2n+1$

$\lceil\frac{6n+4}{3}\rceil = 2n+2$

$\lceil\frac{6n+5}{3}\rceil = 2n+2$

$\lceil\frac{6n+6}{3}\rceil = 2n+2$

for $n \in \mathbb{Z}$

and

$3n-2n-1=n-1$

$3n+1-2n-1=n$

$3n+2-2n-2=n$

$3n+3-2n-2=n+1$

From which it is equal to zero on multiple occasions

$\lceil \frac{6}{2} \rceil - \lceil \frac{7}{3} \rceil =0$ ; $x=6$

$\lceil \frac{1}{2} \rceil - \lceil \frac{2}{3} \rceil =0$ ; $x=1$

$\lceil \frac{2}{2} \rceil - \lceil \frac{3}{3} \rceil =0$ ; $x=2$

$\lceil \frac{3}{2} \rceil - \lceil \frac{4}{3} \rceil =0$ ; $x=3$

$\lceil \frac{4}{2} \rceil - \lceil \frac{5}{3} \rceil =0$ ; $x=4$

Maybe you were looking at $x=6$?

Answer 2

If $x$ is integer then you write $x=6k+r$ where $r\in\{0,1,2,3,4,5\}$. Then you get $$0<3k+\lceil{r\over 2} \rceil -2k- \lceil{r+1\over 3} \rceil$$

Now check what happens when you put $r=0,1,...$

• If $r=0$ then we get $0<k+0 - 1 $...
• If $r=1$ then we get $0<k+1 - 1 $...
• If $r=2$ then we get $0<k+1 - 1 $...
• $\vdots $

Answer 3

From the inequalities $u\le\lceil u\rceil\lt u+1$, we have

$$\left\lceil x\over2\right\rceil-\left\lceil x+1\over3\right\rceil\gt{x\over2}-\left({x+1\over3}+1 \right)={1\over6}(x-8)$$

The OP's desired (strict) inequality thus holds for certain as soon as $x\ge8$. It remains to check the remaining cases, $x=1,2,\ldots,7$. Checking these one at a time, one finds that strict inequality occurs only for $x=7$ and $x=5$:

$$\left\lceil 7\over2\right\rceil-\left\lceil 8\over3\right\rceil=4-3=1\gt0$$

and

$$\left\lceil 5\over2\right\rceil-\left\lceil 6\over3\right\rceil=3-2=1\gt0$$

The other values all turn out to be equal to $0$.

Added later: I recall being dissatisfied when I first wrote this answer on its reliance on explicit evaluations of $\lceil x/2\rceil$ and $\lceil(x+1)/3\rceil$ for $1\le x\le7$. It finally occurs to me there's a way to simplify things, at least a bit:

The ceiling function $\lceil x/2\rceil$ increases by $1$ to the right of $x=2$, $4$, and $6$, while $-\lceil(x+1)/3\rceil$ dereases by $1$ to the right of $x=2$ and $5$. Thus the combined function $\lceil x/2\rceil-\lceil(x+1)/3\rceil$, which equals $1-1=0$ for small positive (real) values of $x$, doesn't change at all until it gets past $x=4$, at which point it goes up and down by $1$ over the next three intervals, so it's $1$ at $x=5$, back to $0$ at $x=6$, and back to $1$ at $x=7$.