Question

Prove that root3 is irrational

Answer 1

Answer:

Let us assume on the contrary that

√3 is a rational number.

Then, there exist positive integers a and b such that

√3 = a/b where, a and b, are co-prime i.e.

there HCF is 1

Now,

√3 = a

b

⇒3= a²

b²

⇒3b² =a²

⇒3 divides a² [∵3 divides 3b²]

⇒3 divides a...(i)

⇒a=3c for some integer c

⇒a²=9c²

⇒3b²=9c² [∵a² =3b²]

⇒b²=3c²

⇒3 divides b² [∵3 divides 3c²]

⇒3 divides b...(ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence,b√3 is an irrational number.

[❤✔]✌

Answer 2

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2