Question

Prove that $\frac{2+\sqrt{3}}{5}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Answer 1

Let us assume that $\frac{2+\sqrt{3}}{5}$ is a rational number.

Thus, $\frac{2+\sqrt{3}}{5}$ can be represented in the form of $\frac{p}{q}$, where p and q both are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} \frac{2+\sqrt{3}}{5}=\frac{p}{q} \\ \Rightarrow 2+\sqrt{3}=5\left(\frac{p}{q}\right) \\ \Rightarrow \sqrt{3}=\frac{5p}{q}-2 \\ \Rightarrow \sqrt{3}=\frac{5p-2q}{q} \end{gathered}$$

since, $\frac{5p-2q}{q}$ is rational ⇒ $\sqrt{3}$ is rational.

But, it is given that $\sqrt{3}$ is an irrational number.

Therefore, our assumption is wrong.

Hence, $\frac{2+\sqrt{3}}{5}$ is an irrational number.