Prove the cuberoot of 2 is irrational I need to prove the cube root is irrational. I followed the proof for the square root of $2$ but I ran into a problem I wasn't sure of. Here are my steps:


*

*By contradiction, say $ \sqrt[3]{2}$ is rational

*then $ \sqrt[3]{2} = \frac ab$ in the lowest form, where $a,b \in \mathbb{Z}, b \neq 0$

*$2b^3 = a^3 $

*$b^3 = \frac{a^3}{2}$

*therefore, $a^3$ is even

*therefore, $2\mid a^3$,

*therefore, $2\mid a$

*$\exists k \in \mathbb{Z}, a = 2k$ 

*sub in: $2b^3 = (2k)^3$

*$b^3 = 4k^3$, therefore $2|b$ 

*Contradiction, $a$ and $b$ have common factor of two


My problem is with step 6 and 7. Can I say that if $2\mid a^3$ , then $2\mid a$. If so, I'm gonna have to prove it. How??
 A: Your proof is fine, once you understand that step 6 implies step 7:  

This is simply the fact odd $\times$ odd $=$ odd. (If $a$ were odd, then $a^3$ would be odd.)

Anyway, you don't need to assume that $a$ and $b$ are coprime:

Consider $2b^3 = a^3$. Now count the number of factors of $2$ on each side: on the left, you get an number of the form $3n+1$, while on the right you get an a number of the form $3m$. These numbers cannot be equal because $3$ does not divide $1$.

A: The Fundamental Theorem of Arithmetic tells us that every positive integer $a$ has a unique factorization into primes $p_1^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n}$.
You have $ 2 \mid a^3$, so $2 \mid (p_1^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n})^3 = p_1^{3\alpha_1}p_2^{3\alpha_2} \ldots p_n^{3\alpha_n}$.
Since primes are numbers that are only divisible by 1 and themselves, and 2 divides one of them, one of those primes (say, $p_1$) must be $2$.
So we have $2 \mid a^3 = 2^{3\alpha_1}p_2^{3\alpha_2} \ldots p_n^{3\alpha_n}$, and if you take the cube root of $a^3$ to get $a$, it's $2^{\alpha_1}p_2^{\alpha_2} \ldots p_n^{\alpha_n}$. This has a factor of 2 in it, and therefore it's divisible by 2.
A: This is not, probably, the most convincing or explanatory proof, and this certainly does not answer the question, but I love this proof.
Suppose that $ \sqrt[3]{2} = \frac p q $. Then $ 2 q^3 = p^3 $. This means $ q^3 + q^3 = p^3 $. The last equation has no nontrivial integer solutions due to Fermat's Last Theorem.
A: If $p$ is prime, and $p\mid a_1a_2\cdots a_n$ then $p\mid a_i$ for some $i$.
Now, let $p=2$, $n=3$ and $a_i=a$ for all $i$.
A: For the sake of contradiction, assume $ \sqrt[3]{2}$ is rational.
We can therefore say $ \sqrt[3]{2} = a/b$ where $a,b$ are integers, and $a$ and $b$ are coprime (i.e. $a/b$ is fully reduced).
2=$a^{3}/b^{3}$
$2b^{3} = a^{3}$
Hence $a$ is an even integer.
Like all even integers, we can say $a=2m$ where $m$ is an integer.
2$b^{3} = (2m)^{3}$
$2b^{3} = 8m^{3}$
$b^{3} = 4m^{3}$
So $b$ is also even. This completes the contradiction where we assumed $a$ and $b$ were coprime. 
Hence, $ \sqrt[3]{2}$ is irrational.
A: A different approach is using polynomials and the rational root theorem. Since $\sqrt[3]2$ is a root of $f(x)=x^3-2$, it is enough to show that if $f(x)$ has no rational roots, then $\sqrt[3]2$ is irrational.
By the rational root theorem, possible roots are $x=\pm 1$ or $x=\pm2$
Next check that $f(-2)$, $f(-1)$, $f(1)$, $f(2)$ ,$\not= 0$
$$f(-2)=-10\not= 0$$
$$f(-1)=-3\not= 0$$
$$f(1)=-1\not= 0$$
$$f(2)=6\not= 0$$
So since none of these possible rational roots are equal to zero, $\sqrt[3]2$ is irrational.
