Question

Prove that cube root of 4 is irrational

Answer 1

Since a 3 = 4 b 3 , we have 4 ∣ a 3 , which implies that is even as is prime. If you deduce any contradiction from the hypothesis that 4 1 3 is rational, it means that it is indeed irrational.

Answer 2

Answer:

Suppose ∛ 4 is a rational number,

Thus, by the property of rational number,

$\sqrt[3]{4} = \frac{p}{q}$

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

By cubing on both sides,

$4 = \frac{p^3}{q^3}$

$4q^3 = p^3$ -------(1)

Thus, 4 is the multiple of $p^3$

⇒ p is an even number,

Let for any k,

p = 2k

By substituting this value in equation (1),

⇒ $4 q^3 = 8 k^3$

⇒ $q^3 = 2 k^3$

⇒ 2 is the multiple of $q^3$

⇒ q is also an even number,

⇒ p and q are not distinct,

Therefor there is a contradiction in our assumption,

∛ 4 is not a rational number,

⇒ ∛ 4 is an irrational number.

Hence, proved.