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Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer.
I can't resist: Suppose $2^{\frac{1}{3}}=\frac{n}{m}$. Then $$2m^3=n^3,$$ or in other words $$m^3+m^3=n^3.$$ But this contradicts Fermats Last Theorem.
Just use the rational root test on the polynomial equation $x^3-2=0$ (note that $\sqrt[3]{2}$ is a solution to this equation). If this equation were to have a rational root $\frac{a}{b}$ (with $a,b\in \mathbb{Z}$ and $b\not=0$), then $b\vert 1$ and $a\vert 2$. Thus, $\frac{a}{b}\in\{\pm 1,\pm 2\}$. However, none of $\pm 1,\pm 2$ are solutions of $x^3-2=0$. Therefore the equation $x^3-2=0$ has no rational solutions and $\sqrt[3]{2}$ is irrational.
Alternatively, suppose we have $\sqrt[3]{2}=\frac{a}{b}$ for some $a,b\in \mathbb{Z}$, $b\not=0$, and $\gcd(a,b)=1$. Then, rearranging and cubing, we have $2b^3=a^3$. Therefore $a^3$ is even....what does that say about $a$? What, in turn, does that say about $b$? It's really not that different from the classic proof that $\sqrt{2}$ is irrational.
Suppose $2^{1/3}$ is rational. Then $2 \cdot m^3 = n^3$ for some $m, n \in \mathbb{N}$ . Looking at the left side, the power of two in the prime factorization of $2 \cdot m^3$ is of the form $3k + 1$. On the right side, it must be of the form $3l$. This is a contradiction, because the factorizations on both sides must be the same by the fundamental theorem of arithmetic. Thus $2^{1/3}$ cannot be rational.
The polynomial $X^3-2$ is irreducible over $\mathbb Q$ by Eisenstein's criterion, hence has no rational root.
Surprise: irrationality proofs of cube roots follow from irrationality proofs of square roots!
Theorem $\ $ If $\rm\ r^3\: =\: \color{#0A0}m\in \mathbb Z\ $ then $\rm\ r\in \mathbb Q\ \Rightarrow\ r\in\mathbb Z$
Proof $\quad\ \rm r = a/b \in \mathbb Q,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad-bc \;=\; \color{#C00}{\bf 1}\;$ for some $\:\rm c,d \in \mathbb{Z}\;\;$ by Bezout.
So $\rm\ 0\: =\: (a\!-\!br)\: (dr^2\!+cr) \: =\: \color{#C00}{\bf 1}\cdot r^2 + ac\ r\, - bd\color{#0A0}m \ $ so $\rm\ r\in\mathbb Z\ $ by the quadratic case. $ $ QED
Remark $\ $ This degree reduction generalizes to higher degree. If $\rm\ r = a/b \in \mathbb Q\ $ is the root of a monic polynomial $\in \mathbb Z[x]\:$ of degree $> 1$ then we can construct a lower degree monic polynomial having $\rm\:r\:$ as root - exactly as above. Namely, using the same notation, we have $$\begin{eqnarray} \rm r^{n+1} &=&\rm\: e\ r^n +\: f(r),\quad deg\ f < n,\quad e\in\mathbb Z,\quad f(x)\in \mathbb Z[x] \\[.2em] 0\, &=&\rm\: (a - b\ r)\ (d\ r^n +\: c\ r^{n-1})\ \ \text{so expanding, using above value of } r^{n+1}\ yields\\[.2em] \Rightarrow\ \ 0\, &=&\rm\: (ad\!-\!b\,c)\ r^n +\, ac\ r^{n-1}\! - de\color{#0A0}{\bf b}\ r^n\ \ -\ \, bd\,f(r),\quad\!\! so\ \ \ ad\!-\!bc = \color{#C00}{\bf 1}\ \ yields \\[.2em] \Rightarrow\ \ 0\, &=&\rm\qquad\quad \color{#C00}{\bf 1}\cdot r^n + (ac\ \ \ \,-\ \ \ de\color{#c0f}{\bf a})\, r^{n-1}\! - bd\ f(r),\ \ by\,\ \ \color{#0A0}{\bf b}\,r^n = \color{#c0f}{\bf a}\,r^{n-1}\ {\rm by}\ \ b\,r=a \\ \end{eqnarray}$$
Thus by induction on $\rm\,n\,$ we may assume $\rm\,n = 0,\,$ so $\rm\, r\ =\ e\in\mathbb Z.\:$ Hence a rational root of a monic integer coefficient polynomial is integral if rational (monic case of Rational Root Test).
Your $\#6 \Rightarrow \#7$ makes no sense: for example, $1= \frac{\sqrt{2}}{\sqrt{2}}$ but that doesn't mean $1$ is irrational.
It's better to argue by contradiction: suppose $2^{1/3}$ was a rational number. Then it's equal to $\frac{a}{b}$ for some integers $a,b \in \mathbb{Z}$, $b \neq 0$, $\text{gcd}(a,b)=1$. Ok, so then $\frac{a^3}{b^3}=2$ which means $a^3 = 2 b^3$. This shows that $a^3$ is an even integer, so $2$ divides into it. But if $2$ divides into $a \times a \times a$ for an integer $a$ then $2$ must divide into each $a$, so $a^3$ is really divisible by $2^3=8$. But that means $2b^3$ is divisible by $8$ as well, so $b^3$ must be divisible by $4$. In particular, $b$ must be divisible by $2$. But now we have $a$ and $b$ both divisible by $2$, which contradicts $\text{gcd}(a,b,)=1$!
This seems fishy, to be sure, but it works. The proof shows that the quantity $2^{1/3}$ is able to evade ``being a fraction''
Since $\mathbb Z$ is a UFD it is integrally closed and a rational solution of $x^3-2=0$ would be an integer.
Suppose that $2^{1/3}$ is some rational number a/b in lowest terms. Then $2a^3 = b^3$. Consider this equation mod 7. Cubes are 0, 1 or 6 mod 7. So both a and b must be 0 mod 7, contradicting that they are in their lowest terms.
There's nothing special about this proof, apart from the fact that it comes directly from the algorithm "Consider the equation mod the least prime which is congruent to 1 mod the h.c.f. of the powers appearing in the equation". This works surprisingly often because Fermat's Little Theorem ensures the powers take on few values mod that prime.
Let $x=2^{1/3}$ be a rational $\frac{p}{q}$ where $p$ and $q$ are natural numbers having no common factors.
Then $x^3 = 2$, and $x = \frac{x^3}{x^2} = \frac{2}{(2^{1/3})^2} = \frac{2}{\big(\frac{p}{q}\big)^2}$
Hence $x = \frac{p}{q} = \frac{2q^2}{p^2}$.
Since $\frac{p}{q}$ is in its lowest terms, then the second denominator $p^2$ is a multiple of $q$, which is a contradiction unless $q=1$.
But if $q=1$, then $x=2^{1/3}=p$, a natural number, so $x$ is a natural number as well. But $x^3=2$, and no natural number is equal to 2 when cubed. Hence, we have a contradiction, and so $x$ must be irrational, as required.
