Prove that the following are irrationals :

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.

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