How to prove $\sqrt[3]{5}$ is irrational?
I know this question was asked many time before but I can't understand the way the people explained the question so is it possible for someone to re-explain this question but with easy wording so I can understand better.
It's exactly the same as proving $\sqrt2 $ is irrational.
Suppose $5 = ({\frac a b})^3$ where $a, b$ are integers and $gcd(a,b) = 1)$ [i.e. the fraction is in lowest terms].
The $5b^3 = a^3$ so 5 divides $a^3$ but as 5 is prime (indivisible) it follows 5 divides $a$. So $a = 5a'$ for some integer $a'$.
So $5b^3 = (5a')^3 = 125a'^3$ so $b^3 = 25a'^3$.
So $25$ divides $b^3$. So $5$ divides $b^3$. 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 $\sqrt[3]{5}$ is irrational.