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
`(a)/(b)` where
`a,b\in\mathbb{Z}` and
`\text{gcd}(a,b)=1`.

`\sqrt[3]{12}=(a)/(b)\\\\12=(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.