Prove that \(3 + 2\sqrt{5}\) is irrational.

Solution :

Let us assume, to the contrary, that \(3 + 2\sqrt{5}\) is an irrational number.

Now, let \(3 + 2\sqrt{5}\) = \(a\over b\), where a and b are coprime and b \(ne\) 0.

So, \(2\sqrt{5}\) = \(a\over b\) – 3  or  \(\sqrt{5}\) = \(a\over 2b\) – \(3\over 2\)

Since a and b are integers, therefore

\(a\over 2b\) – \(3\over 2\)  is a rational number.

\(\therefore\)  \(\sqrt{5}\)  is an irrational number.

But \(\sqrt{5}\)  is an irrational number.

This shows that our assumption is incorrect.

So, \(3 + 2\sqrt{5}\) is an irrational number.

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