Prove that $2^{1/3}$ is irrational. Proof. Assume by way of contradiction that $\sqrt[3] 2$ is rational. Then $\sqrt[3] 2$ can be written as $a/b$ where $ a,b\in\mathbb{N},b\neq0$. Then $(a^3/b^3 = 2) = (a^3 = 2b^3)$. Thus $2|a^3$ and by the lemma, $2|a$. Then $a=2x $ where $x\in\mathbb{Z} $. Substituting $a=2x$ into $a^3 = 2b^3$ we get $8x^3 = 2b^3$, simplifying we get $4x^3 = b^3$. Which implies $4|b^3$ and then by the lemma, $4|b$. This is a contradiction because if a is a multiple of $2$ and $b$ is multiple of $4$, then they are not in lowest terms. Hence $\sqrt[3] 2$ is irrational.
I think I am good up til the $2|a^3$ part. I don't know if I can say that, "by the lemma $2|a$". Any tips or help is much appreciated!
 A: Since $2|a^3$, then $2|a$ because $2$ is prime (Euclid's Lemma).
Later on when you say $4|b^3$, you can argue that this implies $2|b^3$.
By the same argument as before, $2|b$ (as mentioned in the comments). This contradicts that $\text{gcd}(a,b)=1$ (you will need to state this as well in your proof because the fraction $\dfrac{a}{b}$ must be reduced to lowest terms).
A: The present problem may be resolved if we can show that for primes $p$ with
$p \mid a^n$, $n \ge 1$, then $p \mid a$.
We note that if
$p \not \mid a \tag 0$
then
$\gcd(p, a) = 1; \tag 1$
for the only divisors of $p$ are $p$ and $1$; and if $p \not \mid a$, we are left with $1$ as the sole common divisor of $p$ and $a$; now in light of (1) we have, via Bezout's identity,
$xp + ya = 1, \tag 2$
for some
$x, y \in \Bbb Z; \tag 3$
multiplying (2) by $a^{n - 1}$ yields
$xpa^{n - 1} + ya^n = a^{n - 1}, \tag 4$
which since $p \mid a^n$ yields
$p \mid a^{n - 1}; \tag 5$
we now repeat the argument, replacing $a^{n - 1}$ with $a^{n - 2}$; then from (2),
$xpa^{n - 2} + ya^{n - 1} = a^{n - 2}; \tag 6$
now, again by virtue of (4) we find
$p \mid a^{n - 2}; \tag 7$
continuing in this manner we reach
$p \mid a \tag 8$
after $n - 1$ repititions.  But this stands in contradiction to (0), which must thus be false; hence (8) binds.
