Question

Prove that $\sqrt{5}$ is irrational.

Answer 1

Let root $\sqrt{5}$ be rational.

Then it must in the form of $\frac{p}{q}$ [$q$ is not equal to $0$ and $p,q$ are co-prime].

Hence, $\sqrt{5}=\frac{p}{q}$

$⟹\sqrt{5}q=p$.

Squaring on both sides,

$⟹5q^2=p^2$ ------ (1)

$⟹q^2=\frac{p^2}{5}$.

Therefore, $p^2$ is divisible by $5$

and $p$ is divisible by $5$. ------{If $p$ is a prime no. and $p$ divides $a^2$, then $p$ divides $a$ also, where $a$ is a positive integer}

Then, $p=5c$ [$c$ is a positive integer].

Squaring on both sides,

$⟹p^2=25c^2$ --------- (2)

Now, substitute for $p^2$ in (1),

we get, $5q^2=25c^2$

$⟹q^2=5c^2$.

Therefore, $q^2$ is divisible by $5$

and $q$ is divisible by $5$.

Thus $q$ and $p$ have a common factor $5$.

There is a contradiction.

Therefore, $p$ and $q$ are not co-prime.

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