Prove that 2√3 is an irrational number.

Question

Prove that \(2 \sqrt{3}\) is an irrational number.

Answer 1

Let us assume that √3 is a rational number which can be expressed in the form of p/q, where p and q are integers, q ≠ 0 and p and q are co prime that is HCF(p, q) = 1.

We have,

√3 = p/q ⇒ √3q = p......(1)

⇒ 3q² = p² (squaring both sides)

⇒ p² is divisible by 3

⇒ p is divisible by 3......(2)

Therefore, for an integer r,

p = 3r

⇒ √3q = 3r (from eq. (1))

⇒ 3q² = 9/3 r² (squaring both sides)

⇒ q² = 9/3 r²

⇒ q² = 3r²

⇒ q² is divisible by 3

⇒ q is divisible by 3......(3)

From equations 2 and 3, we get that 3 is the common factor of p and q which contradicts that p and q are co prime. This means that our assumption was wrong.

Thus √3 is an irrational number.

Now, since multiplication of a rational number with an irrational number is an irrational number.

Hence 2√3 is an irrational number.