How can I prove $n - 2\sqrt{n} = \Theta(n)$

Question

I want to prove the following $$n - 2\sqrt{n} = \Theta(n)$$

It's $n - 2\sqrt{n} \leq n = O(n)$

How can I prove the same for $\Omega(n)$

Answer 1

Note that $$ n - 2\sqrt{n} \geq \frac 12 n \iff\\ n \geq 4 \sqrt{n} \iff\\ \sqrt{n} \geq 4 \iff\\ n \geq 16 $$ That is, for $n \geq 16$, we have $n - 2\sqrt{n} \geq k n$ for all $n \geq 16$ when $k = 1/2$. Thus, we have $n- 2\sqrt{n} = \Omega(n)$.

---

Alternatively, we can concisely prove that $n - 2\sqrt{n} = \Theta(n)$ (or, in particular, that $n -2\sqrt n \sim n$) by simply noting that $$ \lim_{n \to \infty} \frac{n - 2\sqrt{n}}{n} = 1 $$