Question

Prove that √3 is sn irrational nubet

Answer 1

• Let us assume that $\sqrt{3}$ is a rational number.

⇒ $\sqrt{3}$ = $\dfrac{a}{b}$

Here a and b are co-prime numbers.

Squaring on both sides

⇒ $(\sqrt{3})^{2}$ = ${ \bigg(\dfrac{a}{b} \bigg)}^{2}$

⇒ 3 = $\dfrac{ {a}^{2} }{ {b}^{2} }$

⇒ 3b² = a² _______ (eq 1)

Here 3 divides a² and 3 divide a also.

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⇒ a = 3c

Here c is integer.

Squaring on both sides

⇒ a² = (3c)²

⇒ a² = 9c²

⇒ 3b² = 9c²

⇒ b² = 3c²

Here 3 divide b² and 3 divide b also.

Therefore a and b have common factor i.e. 3.

This means that a and b have no common factor other than 1.

So, our assumption is wrong.

$\bold{\sqrt{3} \: is \: irrational \: number}$

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