Question

Prove that √3 is an irrational number.

Answer 1

Let √3 be rational and its simplest form be \(\frac{a}b\).

Then, a, b are integers with no common factors other than 1 and b ≠ 0. 

Now, 

√3 = \(\frac{a}b\) ⇒ 3 = \(\frac{a^2}{b^2}\) [on squaring both sides] 

⇒ 3b² = a² ….(1) 

⇒ 3 divides a² [since 3 divides 3b² ] 

⇒ 3 divides a [since 3 is prime, 3 divides a² ⇒ 3 divides a] 

Let a = 3c for some integer c. 

Putting a = 3c in equation (1), we get 

3b² = 9c² ⇒ b = 3c² 

⇒ 3 divides b² [since 3 divides 3c² ] 

⇒ 3 divides b [since 3 is prime, 3 divides b² ⇒ 3 divides b] 

Thus, 3 is a common factor of both a, b. 

But this contradicts the fact that a, b have no common factor other than 1. 

The contradiction arises by assuming √3 is rational. 

Hence, √3 is rational.

Answer 2

Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.

√3 = p/q

⇒ p = √3 q

By squaring both sides, we get,

p² = 3q² 

p² / 3 = q² ------- (1)

(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p², then a divides p, where a is a positive integer)

Here 3 is the prime number that divides p², then 3 divides p and thus 3 is a factor of p.

Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,

(3c)² / 3 = q²

9c²/3 =  q² 

3c²  =  q² 

c²  =  q² /3 ------- (2)

Hence 3 is a factor of q (from 2)

Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is a contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.