Question

Prove that √(5 ) is irrational.​

Answer 1

Answer:

Let √5 be a rational number.

then it must be in form of $\frac{p}{q}$ where, q≠0 ( p and q are co-prime)

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

$\sqrt{5} \times q = p$

Suaring on both sides,

$5 {q}^{2} = {p}^{2} \: \: \: \: \: \: \: \: \: \: ...(1)$

${p}^{2}$ is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

${p}^{2} = 25 {c}^{2} \: \: \: \: \: \: \: \: \: \: \: ...(2)$

Put p² in equation (1)

$5 {q}^{2} = 25 {c}^{2}$

${q}^{2} = 5 {c}^{2}$

So, q is divisible by 5.

.

Thus p and q have a common factor of 5.

So, there is a contradiction as per our assumption.

We have assumed p and q are co-prime but here they a common factor of 5.

The above statement contradicts our assumption.

Therefore, √5 is an irrational number.