# Algorithm Theta Notation : How constant $c_2 \geq 1/2$ is derived from the inequality $c_1 \leq 1/2 - 3/n \leq c_2$

I was learning ϴ(n) notation in my course "Asymptotic Analysis for Algorithms" when I encountered the following example:

For any non-negative constants $c_1\geq 0,c_2\geq 0,n\geq n_0$ we have the following inequality:

$$c_1\leq\frac{1}{2}-\frac{3}{n}\leq c_2$$

For these constants to satisfy this inequality, will be $c_2\geq\frac{1}{2}$ when $n\geq 1$. This is the part I'm having issues with.

I tried to derive it assuming $n \geq 1$ and I found:

\begin{align*} c_2 &\geq \frac{1}{2} - \frac{3}{n} \\ \implies c_2 &\geq \frac{1}{2} - \frac{3}{1} \\ \implies c_2 &\geq -\frac{5}{2}` \end{align*}

I managed to show the left inequality holds when $n\geq 7$ and $c_1\leq \frac{1}{14}$.

• $3/n$ is positive, so $1/2-3/n$ will be smaller than $1/2$. In other words, $1/2-3/n \leq 1/2$. Thus you can choose $c_2$ to be anything bigger than this upper bound of $1/2$ if you want it to be an upper bound too. So $1/2-3/n \leq c_2$ if $c_2 \geq 1/2$. – Antonio Vargas Nov 2 '16 at 7:49

The point is $-3/n$ is increasing to zero, so for large $n$ $1/2-3/n$ is approximately (but less than) $1/2$
In other words $1/2-3/n \ge 1/2-3/1$ is wrong, so you cannot derive the second inequality from the first one.