Prove the cube root of $36$ is irrational.

Question

Prove the cube root of $36$ is an irrational number.

Proof by contradiction: let it equal $\frac{a}{b}$, where $\frac{a}{b}$ is in its simplest form and $b$ does not equal $0$.

$36 = \frac{a^3}{b^3}$

Therefore, $36$ is a factor of a cubed. Since $36 = 2 \times2 \times 3 \times 3$ then by prime factorization, we can deduce that using Euclid’s Lemma if $gcd(a, b) = 1$ and $a | bc$ then $a | c$, such that $2| a$ and $3| a$.

Ok, I understand all of this. Now, what is next? Do I say then let $a= 2c$? I tried that and couldn't prove it. The same happens if I, say, let $a= 3c$. Can I just multiply $2$ by $3$ and state that $6| a$? If I use $a = 6c$, then I know how to finish this problem. I don't want the solution. I just need to know the reasoning behind using $a= 6c$.

Answer 1

Since $36b^3=a^3$, you see that $2\mid a$, so $a=2x$. After simplifying, $$ 9b^3=2x^3 $$ which implies that $2\mid b$. Contradiction, because by assumption $\gcd(a,b)=1$.

Answer 2

$$36 = \frac{a^3}{b^3}$$ $$6^2b^3 = a^3$$ $$2^23^2b^3 = a^3$$ This implies that a is even, or
Let a = 2c $$ 2^2 3^2 b^3 = 8c^3$$ $$3^2b^3 = 2c^3$$ b is also also even which means gcd (a,b) is not 1 which is a contradiction.

Answer 3

Note that: $$\sqrt[3]{36}=\frac{6}{\sqrt[3]{6}}=\frac ab \iff \sqrt[3]{6}=\frac{6b}{a}.$$ Cube root of $6$ is irrational (see here), so is the cube root of $36$.

Answer 4

Thanks all. I figured out what to do.This was more a clarity issue in explanation. No need to comment anymore. I've used the method I mentioned above and have worked out how to explain it concisely.