Mathematics

The key is proof by contradiction using the Rational Root Theorem. If $\sqrt[3]{5}$ were rational, you could write it as a fraction $\frac{p}{q}$ in lowest terms, then cube both sides to get $p^3 = 5q^3$. From here, divisibility arguments on prime factors will force both $p$ and $q$ to be divisible by 5, contradicting the assumption that the fraction is fully reduced—so $\sqrt[3]{5}$ must be irrational.

The key is proof by contradiction using the Rational Root Theorem. If $\sqrt[3]{5}$ were rational, you could write it as a fraction $\frac{p}{q}$ in lowest terms, then cube both sides to get $p^3 = 5q^3$. From here, divisibility arguments on prime factors will force both $p$ and $q$ to be divisible by 5, contradicting the assumption that the fraction is fully reduced—so $\sqrt[3]{5}$ must be irrational.