Question

prove that √3 is an irrational number​

Answer 1

$\underline{\underline{\red{\sf{GIVEN:}}}}$

• A Irrational number√3 is given to us.

$\underline{\underline{\red{\sf{TO\: PROVE:}}}}$

• √3 is a Irrational number.

$\underline{\underline{\red{\sf{ANSWER:}}}}$

Given number to us is √3 and we have to prove that it is a Irrational number.

So , on the contrary let us assume that √3 is a Rational number , and as per definition of Rational number it can be expressed in the form of p/q where p and q are integers and q ≠ 0.

Also HCF of p and q is 1 , that is p and q are co - primes .

As per our assumption we can write √3 as ,

$\tt{\implies \sqrt{3}=\dfrac{p}{q}}$

$\tt{\implies (\sqrt{3})^{2}=(\dfrac{p}{q})^2 }$

$\tt{\implies 3 = \dfrac{p^2}{q^2}}$

$\purple{\tt{\leadsto p^2 = 3q^2}}$ .........(i)

• This implies that 3 is a factor of p² .So by the Fundamental Theorem of Arithmetic we can say that 3 is factor of p also.

$\tt{\mapsto p = 3k (let) }$ ..........(ii)

$\underline{\pink{\sf{Putting\: value\: of \:(ii) \:in \:(i) }}}$

$\tt{\implies (3k)^2 = 3q^2}$

$\tt{\implies9k^2=3q^2 }$

$\tt{\implies q^2 = \dfrac{\cancel{9}}{\cancel{3}}}$

$\tt{\implies q^2 = 3k}$

• This implies that 3 is a factor of q² .So by the Fundamental Theorem of Arithmetic we can say that 3 is factor of q also.

This contradicts our assumption that p and q are co- primes . Hence our assumption was wrong , √3 is not a Rational number.

$\underline{\green{\bf{Hence\:\sqrt{3}\:is\:a\: Irrational\: number.}}}$

$\orange{\sf{\underset{\green{Hence\:Proved}}{\underbrace{\sqrt{3}\:is\:is\: Irrational \:number}}}}$