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:

1. By contradiction, say $ \sqrt[3]{2}$ is rational
2. then $ \sqrt[3]{2} = \frac ab$ in the lowest form, where $a,b \in \mathbb{Z}, b \neq 0$
3. $2b^3 = a^3 $
4. $b^3 = \frac{a^3}{2}$
5. therefore, $a^3$ is even
6. therefore, $2\mid a^3$,
7. therefore, $2\mid a$
8. $\exists k \in \mathbb{Z}, a = 2k$
9. sub in: $2b^3 = (2k)^3$
10. $b^3 = 4k^3$, therefore $2|b$
11. 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??

Solution 1:

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.

Solution 2:

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 $i$.

Solution 3:

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 $1$.

Solution 4:

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.