Question

Prove
$\sqrt[3]{5}$
is irrational by contradiction method​

Answer 1

Answer:

It's exactly the same as proving √2 is irrational.

Suppose 5 = ()³ where a, b are integers and gcd(a, b) = 1) [i.e. the fraction is in lowest terms].

The 56³ a³ so 5 divides a³ but as 5 is = prime (indivisible) it follows 5 divides a. So a = 5a' for some integer a'.

So 56³ = (5a')³= 125a/³ so 6³ 25a'³ =

So 25 divides 6³. So 5 divides b³. But as 5 is prime (indivisible) it follows that 5 divides b.

So you have 5 divides a AND we have 5 divides b. But that contradicts gcd(a, b) = 1 and that the fraction is in lowest terms.

We must conclude such a fraction is impossible and 5 is irrational.