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.
