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

User Asthomas
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\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.

User Shailesh Sonare
by
7.6k points

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