Question

Show that 5 + $\sqrt{5}$ is an irrational number.

Answer 1

Let us assume that 5+$\sqrt{5}$ is a rational number.

Then, $5+\sqrt{5}=\frac{a}{b}$, where a and b are co-prime integers (b$\ne$0).
So,
$$\begin{gathered} 5+\sqrt{5}=\frac{a}{b} \\ \text{Or},\sqrt{5}=\frac{a}{b}-5 \\ \text{Or},\sqrt{5}=\frac{a-5b}{b} \\ \mathrm{Since}a\mathrm{and}b\mathrm{are}\mathrm{integers},\frac{a-5b}{b}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \\ \text{So},\sqrt{5}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{nu}\mathrm{mber}. \\ \mathrm{But}\mathrm{this}\mathrm{contradict}s\mathrm{the}\mathrm{fact}\mathrm{that}\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \mathrm{This}\mathrm{contradiction}\mathrm{arsies}\mathrm{because}\mathrm{of}\mathrm{our}\mathrm{wrong}\mathrm{assumption}\mathrm{that}5+\sqrt{5}\mathrm{is}\mathrm{a}\mathrm{rational}\mathrm{number}. \\ \mathrm{So},5+\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$