<s>[INST] Proving $\sqrt[3]{2}$ is irrational without using prime factorization Prove that $\sqrt[3]{2}$ is irrational without using prime factorization.
The standard proof that $\sqrt[3]{2}$ is irrational uses prime factorization in an essential way. So I wondered if there is a proof that does not use it.
This was inspired by the fact that I know two proofs that $\sqrt{2}$ is irrational that do not use  prime factorization.
The first uses
$$\sqrt{2}=\sqrt{2}\frac{\sqrt{2}-1}{\sqrt{2}-1}=\frac{2-\sqrt{2}}{\sqrt{2}-1}
$$
to show that if $\sqrt{2} = \dfrac{a}{b}$ then
$$\sqrt{2}=\frac{2-\dfrac{a}{b}}{\dfrac{a}{b}-1}=\frac{2b-a}{a-b}
$$
is a rational $\sqrt{2}$ with a smaller denominator.
The second uses
$$(x^2-2y^2)^2=(x^2+2y^2)^2-2(2xy)^2
$$
and $3^2-2\cdot 2^2 = 1$ to show that $x^2-2y^2=1$ has arbitrarily large solutions and this contradicts $\sqrt{2}$ being rational.
I have not been able to extend either of these proofs to $\sqrt[3]{2}$. Results that  I do not consider "legal" in solving this problem include Fermat's Last Theorem (which definitely uses unique factorization) and the  rational root theorem (which uses unique factorization
in its proof).
 [/INST] Assume that $\sqrt[3]{2}=\dfrac{a}{b}$ with $a,b\in\mathbb{Z^+}$
We will have $\dfrac{a^3}{b^3}=2\Leftrightarrow a^3=2b^3$ $(*)$
The above implies that $a$ must be divisible by $2$ (if $a$ is odd then $a^3$ is odd), let $a=2a_1$ with $a_1\in\mathbb{N}$, then $a^3=2b^3\Leftrightarrow 8a_1^3=2b^3\Leftrightarrow4a_1^3=b^3$
The above implies that $b$ must be divisible by $2$ (if $b$ is odd then $b^3$ is odd), let $b=2b_1$ with $b_1\in\mathbb{N}$, then $4a_1^3=b^3\Leftrightarrow 4a_1^3=(2b_1)^3\Leftrightarrow 4a_1^3=8b_1^3\Leftrightarrow a_1^3=2b_1^3$.
This will continue for an infinite loop, let $n\in\mathbb{Z^+}$ and $a_n=2a_{n+1}$ then 
$$a^3=2b^3\Leftrightarrow a_1^3=2b_1^3\Leftrightarrow a_2^3=2b_2^3\Leftrightarrow a_3^3=2b_3^3\Leftrightarrow\cdots\Leftrightarrow a_n^3=2b_n^3\Leftrightarrow a_{n+1}^3=2b_{n+1}^3$$
Because $a_{n+1}\in\mathbb{Z^+}$, 
$$2|a_{n},b_{n}\Rightarrow2^2|a_{n-1},b_{n-1}\Rightarrow2^3|a_{n-2},b_{n-2}\Rightarrow\cdots\Rightarrow 2^{n+1}|a,b$$
$a,b$ must be divisible by $2^{n+1}$ for all $n\in\mathbb{N}$, this is only possible if $a=b=0$, which makes $\sqrt[3]{2}=\dfrac{a}{b}$ undefined.
By contradicition, $\sqrt[3]{2}$ is irrational.
Alternatively... 
If $\sqrt[3]{2}$ is rational then it can be expressed in the form $\sqrt[3]{2}=\dfrac{a}{b}$ with $a,b\in\mathbb{Z^+}$ and $\dfrac{a}{b}$ is an irreducible fraction or $GCD(a,b)=1$.
Because we have proven that both $a$ and $b$ are even numbers, $\dfrac{a}{b}$ now becomes reducible, which also contradicts the condition above. This will be a much quicker conclusion compared to the "infinite loop" method.
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