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??
Best Answer
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.