**Question** : Prove that the following are irrationals :

(i) \(1\over \sqrt{2}\)

(ii) \(7\sqrt{5}\)

(iii) \(6 + \sqrt{2}\)

**Solution** :

**(i) ** Let us assume, to the contrary, that \(1\over \sqrt{2}\) is rational. That is, we can find co-prime integers p and q (q \(\ne\) 0) such that

\(1\over \sqrt{2}\) = \(p\over q\) or \(1\times \sqrt{2}\over \sqrt{2}\times \sqrt{2}\) = \(p\over q\) or \(\sqrt{2}\over 2\) = \(p\over q\)

or \(\sqrt{2}\) = \(2p\over q\)

Since p and q are integers \(2p\over q\) is rational, and so \(\sqrt{2}\) is rational.

But this contradicts the fact that \(\sqrt{2}\) is irrational.

so, we conclude that \(1\over \sqrt{2}\) **is an irrational.**

**(ii)** Let us assume, to the contrary, that \(7\sqrt{5}\) is rational. That is, we can find co-prime integers p and q (q \(\ne\) 0) such that \(7\sqrt{5}\) = \(p\over q\).

So, \(\sqrt{5}\) = \(p\over 7q\)

Since p and q are integers, \(p\over 7q\) is rational and so is \(\sqrt{5}\).

But this contradicts the fact that \(\sqrt{5}\) is irrational. So, we conclude that **\(7\sqrt{5}\) is an irrational.**

**(iii)** Let us assume, to the contrary, that \(6 + \sqrt{2}\) is rational. That is, we can find co-prime integers p and q (q \(\ne\) 0) such that

\(6 + \sqrt{2}\) = \(p\over q\) or 6 – \(p\over q\) = \(\sqrt{2}\)

or \(\sqrt{2}\) = 6 – \(p\over q\)

Since p and q are integers, 6 – \(p\over q\) is rational, and so \(\sqrt{2}\) is rational.

But this contradicts the fact that \(\sqrt{2}\) is irrational.

so, we conclude that \(6 + \sqrt{2}\) **is an irrational.**