ceiling functions inequality

Question

Please, help me in solving this ceiling function inequality.

$ \lceil n/4 \rceil \ge 3$

I know the formal definiton of the ceiling functions:

$\lceil x \rceil = n$ iff $n-1< x \le n $

Should I apply this rule and how?

Thank you.

Answer 1

Yes, you should use the definition. However, the definition should be $\lceil x \rceil = n$ iff $n-1 < x \leq n$. (Otherwise the ceiling of an integer could take on two different values - e.g. $2-1\leq 2\leq 2$ and $3-1 \leq 2 \leq 3$ so $\lceil 2\rceil = 2$ and $\lceil 2 \rceil =3$, which is obviously not what we want).

Now you want to find which $n$ satisfy $\lceil n/4 \rceil \geq 3$. This happens if and only if $n/4>3-1$. To see why, suppose $\lceil n/4\rceil =k \geq 3$. Then $k-1 < n/4 \leq k$, so $n/4 >k-1 \geq 3-1$. Conversely, suppose $n/4>3-1=2$. Let $k=\lceil n/4 \rceil$. Then $k-1<n/4 \leq k$. Since we also have $n/4>2$, it follows that $k \geq n/4>2$, and thus $\lceil n/4 \rceil = k \geq 3$.

So the inequality becomes $2<n/4$, and thus $n>8$.

Answer 2

$$ \lceil x \rceil = \min\{k \in \mathbb{Z} \mid k \geq x\}$$

So, basically:

$$\left\lceil {n\over 4} \right\rceil \ge 3 \iff {n\over 4} > 2 \iff n > 8$$