Question

prove that under root3 is irrational​

Answer 1

Let us suppose that √3 is rational.

$\sqrt{3} = \frac{x}{y}$

where x and y are integers,HCF(x,y)=1

Squaring both sides

$3 = \frac{ {x}^{2} }{ {y}^{2} } \\ {x}^{2} = 3 {y}^{2} .......(1)$

$\implies {x}^{2}$ is divisible by 3

$\implies$ x is divisible by 3.....(3)

$x = 3m$

Squaring both sides

${x}^{2} = 9 {m}^{2} .......(2)$

Put (2) in (1)

$9 {m}^{2} = 3 {y}^{2} \\ {y}^{2} = 3 {m}^{2}$

$\implies {y}^{2}$ is divisible by 3

$\implies$ y is divisible by 3.....(4)

From (3) and (4)

HCF(x,y)=3

Our supposition is wrong.

So, √3 is irrational

Hence proved.

Answer 2

Explanation:-

Prove that $\sqrt{3}$ is an irrational number.

Proof :-

Let us assume that$\sqrt{3}$is a rational number then, it can be expressed in the form of p/q.

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

Then,

→$\sqrt{3}= \dfrac{p}{q}$

→$q\sqrt{3} = p$

• Squaring on both side.

→$(q\sqrt{3})^2 = p^2$

→$3q^2 = p^2$

• here 3 divides p².

Let p = 3r for some integer r.

• Put p = 3r.

→$3q^2 = (3r)^2$

→$3q^2 = 9r^2$

→$q^2 = 3r^2$

• here 3 divides q².

• 3 divides both p and q so , 3 is a common factor of both p and q.

But p and q has no other factor than 1.

This start contradiction due to our wrong assumption.

hence,

$\sqrt{3}$ is an irrational number.