Question

Explain why the cube root of 64 is rational and the cube root of 1 2 is not.

Answer 1

$\sqrt[3]{64}=4$ so it's obvious, I guess :)

Now, let's prove $\sqrt[3]{12}$ is not rational number.

Proof by contradiction.

Let assume $\sqrt[3]{12}$ is a rational number. Therefore it can be expressed as a fraction  $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$ and $\text{gcd}(a,b)=1$.

$\sqrt[3]{12}=\dfrac{a}{b}\\\\12=\dfrac{a^3}{b^3}\\\\a^3=12b^3$

$12b^3$ is an even number, so $a^3$ must also be an even number, and therefore also $a$ must be an even number. So, we can say that $a=2k$, where $k\in\mathbb{Z}$.

$(2k)^3=12b^3\\\\8k^3=12b^3\\\\2k^3=3b^3$

Since $2k^3$ is an even number, then also $3b^3$ must be an even number. 3 is odd, so for $3b^3$ to be an even number, $b^3$ must be an even number, and therefore $b$ is an even number.

But if both a and b are even numbers, then it contradicts our earlier assumption that $\text{gcd}(a,b)=1$. Therefore $\sqrt[3]{12}$ is not a rational number.