Let assume contrary that √7 is a rational number, so it can be written in \(\frac pq \) form.

Let √7 = \(\frac pq \), gcd(p, q) = 1, p, q \(\in \mathbb{Z}\), q \(\ne \) 0

⇒ p = √7q

⇒ p² = 7q²     ......(1)

⇒ 7 divides p²

⇒ 7 divides p

⇒ p = 7m, m \(\in \mathbb{Z}\)

⇒ p² = 49m²

⇒ 7q² = 49m²     (From(1))

⇒ q² = 7m²

⇒ 7 divides q²

⇒ 7 divides q

\(\because \) 7 divides both p & q.

\(\therefore\) 7 is a factor of both p & q.

\(\therefore\) gcd(p, q) = 7 or multiple 7.

⇒ gcd(p, q) = 1

which is contradiction the fact that 

gcd(p, q) = 1

Hence, our assumption is wrong.

Hence, √7 is an irrational number.

Let 2-√7 is rational number.

⇒ -2 - √7 = \(\frac pq \),  q \(\ne \) 0, p, q \(\in \mathbb{Z}\)

⇒ \(\sqrt 7 = 2 - \frac pq = \frac{2q - p}{q} \in \mathbb Q\)     (\(\because \) 2q - p \(\in \mathbb{Z}\) & q \(\in \mathbb{Z}\),  q \(\ne \) 0)

which is contradiction 

(\(\because \) √7 is irrational and \(\frac {2q-p}q \) is rational And rational \(\ne \) irrational)

Hence, our assumption is wrong.

Hence, 2 -√7 is an irrational number.