Question

prove that $\sqrt{5}$ is not a rational number. Hence, prove that 2 - $\sqrt{5}$ is also irrational.

Answer 1

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

Hence, $\sqrt{5}$ can be written as $\frac{a}{b}$

where both $aandb$ are co-primes.

$\sqrt{5}=\frac{a}{b}$

$\Rightarrow \sqrt{5}b=a$

$\Rightarrow 5b^2=a^2$

$\therefore \frac{a^2}{5}=b^2$ ......... $1$

We know that, if a number $p$ divides $q^2$, it will divide $q$ as well.

Here, $5$ divides $a^2$. Hence, it must divide $a$ as well.

$\therefore \frac{a}{5}=c$ ($c=$ any integer)

$a=5c$ ..... $2$

From $1$ and $2$, we get:

$25c^2=5b^2$

$b^2=5c^2$

$\frac{b^2}{5}=c^2$

Again, $5$ divides $b^2$. Hence, it will also divide $b$.

Hence, $5$ is a factor of both $a$ and $b$.

Therefore $a$ and $b$ are not co-primes.

So, what we assumed is not true.

Hence, $\sqrt{5}$ is irrational.

Now, according to the properties of irrational numbers, the sum or difference of a rational number and an irrational number is always an irrational number.

Hence, $(2-\sqrt{5})$ is also irrational.