Question

Prove that `sqrt(6)` is an irrational number.

Answer 1

Given: Let `sqrt(6)` be rational, so `sqrt(6) = a/b` in lowest terms with integers a, b, b ≠ 0, gcd(a, b) = 1.

To Prove: `sqrt(6)` is irrational (i.e., no such integers a, b in lowest terms exist).

Proof [Step-wise]:

1. Assume `sqrt(6) = a/b` with integers a, b, gcd(a, b) = 1.

2. Square both sides: `6 = a^2/b^2`, so a² = 6b².

3. From a² = 6b² we see a² is divisible by 2.

By the standard lemma “if a prime p divides a² then p divides a”, 2 divides a.

4. Hence, a is even: write a = 2k for some integer k.

Substitute into a² = 6b²: 

(2k)² = 6b²

⇒ 4k² = 6b²

⇒ 2k² = 3b²

5. The right side 3b² is divisible by 3, so the left side 2k² is divisible by 3.

Since 3 does not divide 2, it must divide k² and by the same lemma 3 divides k.

Thus, 3 divides k, so 3 divides a = 2k.

6. Therefore, both 2 and 3 divide a, so 6 divides a: a = 6t for some integer t.

Substitute into a² = 6b²: 

(6t)² = 6b²

⇒ 36t² = 6b²

⇒ 6t² = b²

7. From b² = 6t² we conclude, by the same reasoning, that 2 divides b and 3 divides b, so 6 divides b.

Thus, a and b have a common factor 6, contradicting gcd(a, b) = 1.

Since assuming `sqrt(6)` is rational leads to a contradiction, the assumption is false.