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olya-2409 [2.1K]
3 years ago
12

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

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0

\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.

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