Question

Proof that 3√6 is not a rational number

Answer 1

Answer:

Suppose 3√6 is a rational number,

Thus, by the property of rational number,

We can write,

$3 \sqrt{6} = \frac{p}{q}$

Where, p and q are distinct  integers and q ≠ 0,

$3 \sqrt{6} q = p$

By squaring both sides,

$54 {q}^{2} = {p}^{2}$ ------(1)

⇒ p² is a multiple of 54,

⇒ p is a multiple of 54

Thus, we can write,

p = 54a, where a is any number,

From equation (1),

$54 {q}^{2} = 2916 {a}^{2}$

${q}^{2} = 54 {a}^{2}$

⇒ q² is a multiple of 54,

⇒ q is a multiple of 54

Therefore, p and q are not distinct,

Which is a contradiction,

⇒ 3√6 is not a rational number,

Thus, 3√6 is an irrational number

Hence, proved......