Proving $3n^2 - 4n \in \Omega(n^2)$

Question

Proving $3n^2 - 4n \in \Omega(n^2)$

Attempt:

$3n^2 - 4n \geq cn^2$

$n(3n-4) \geq cn^2$

$(3n-4) \geq cn$

$3n - 4 - cn \geq 0$

$n(3-c) \geq 4$

$n \geq \frac{4}{3-c}$

would do I choose for $n_0$ and $c$ to satisfy this proof?

Answer 1

Note that the final inequality is equivalent to the first one if $c<3$ (you divided by $(3-c)$). Then, according to your work, by taking for example $c=1$ and $n_0=4/(3-c)=2$, it follows that $$3n^2 - 4n \geq cn^2 \quad\forall n\geq n_0$$ Hence we may conclude that $3n^2 - 4n \in \Omega(n^2)$.

Answer 2

$3n^2 - 4n \in \Omega(n^2)$ means:

there are $c>0$ and $n_0 \in \mathbb N$ such that $3n^2-4n \ge cn^2$ for $n \ge n_0$.

Since $\frac{3n^2-4n}{n^2}= 3-\frac{4}{n} \to 3$ as $n \to \infty$, there is $n_0$ such that $\frac{3n^2-4n}{n^2} \ge 2$ for $n \ge n_0$.

Hence $3n^2-4n \ge 2n^2$ for $n \ge n_0$.

This shows that $3n^2 - 4n \in \Omega(n^2)$.