Question

Prove that $\sqrt{5}$ is irrational and hence prove that $2-\sqrt{5}$ is also irrational.

Answer 1

Let us assume to the contrary that $\sqrt{5}$ is a rational number.

It can be expressed in the form of $\frac{p}{q}$

where $p$ and $q$ are co-primes and $q\ne 0.$

$$\begin{gathered} \Rightarrow \sqrt{5}=\frac{p}{q} \\ (Squaringonboththesides) \\ \Rightarrow 5=\frac{p^2}{q^2} \\ \Rightarrow 5q^2=p^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(1\right) \end{gathered}$$

This means that $5$ divides $p^2$ This means that $5$divides $p$ because each factor should appear two times for the square to exist.

So we have $p=5r$

where r is some integer.

$$\begin{gathered} \Rightarrow p^2=25r^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right) \\ fromequation\left(1\right)and\left(2\right) \\ \Rightarrow 5q^2=25r^2 \\ \Rightarrow q^2=5r^2 \end{gathered}$$

Where $q^2$ is multiply of $5$ and also $q$ is multiple of $5$.

Then$p,q$ have a common factor of $5$. This runs contrary to their being co-primes. Consequently,$\frac{p}{q}$ is not a rational number. This demonstrates that $\sqrt{5}$ is an irrational number.

Since$\sqrt{5}$is an irrational number, the difference between the rational number and irrational number is always an irrational number.

∴$2-\sqrt{5}$ is also an irrational number.