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Answer :

Let us suppose that \(3 + 2\sqrt{5}\) is rational

So there should be two co- prime integers (p and q, where q \(\ne\) 0 ) such that:

\(3 + 2\sqrt{5} = {{p}\over{q}} \)

\(\Rightarrow 2\sqrt{5} ={{p}\over{q}} - 3 \)

\(\Rightarrow 2\sqrt{5} ={{p}\over{q}} - 3 \)

\(\Rightarrow \sqrt{5} =\frac{1}{2}(\frac{p}{q} - 3)\) .................(i)

p and q are integers which means \(\frac{1}{2}(\frac{p}{q} - 3)\) is a rational number.

But we know \(\sqrt{5}\) is irrational.

This is not possible since LHS \(\ne\) RHS.

That proves that our assumption that \(3 + 2\sqrt{5}\) is rational is wrong.

Hence it is proved that \(3 + 2\sqrt{5}\) is irrational.

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