Question

show that root 3 is to an inational number​

Answer 1

Step-by-step explanation:

Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1. Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p. ... Therefore, the root of 3 is irrational.

Answer 2

Let's us assume that $\sf{\sqrt{3}}$ is rational then , there exists positive integers a & b such that $\sf{\sqrt{3}} = {\dfrac{a}{b}}$ where , a & b are co primes i.e HCF is 1

Now ,

$\sf{\sqrt{3}} = {\dfrac{a}{b}}$

⇒ 3 = $\sf{\dfrac{a²}{b²}}$

⇒ 3b² = a²

⇒ 3 divides a²

⇒ 3 divides a ───── ( 1 )

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a = 3c [ for some integer c ]

⇒ a² = 9c²

⇒ 3b² = 9c² [ ∴ a² = 3b² ]

⇒ b² = 3c²

⇒ 3 divides b² [ ∴ 3 divides 3c² ]

⇒ 3 divides b ───── ( 2 )

From ( 1 ) and ( 2 ) , we observe that a & b have 3 as a common factor . But , this contradicts the fact that a & b are co primes . This means that our assumption is incorrect .

Hence , $\sf{\sqrt{3}}$ is irrational .