# Proof that $\sqrt[3]{3}$ is irrational

Proof that $\sqrt[3]{3}$ (Just need a check)

Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function is a 1-1 function. Does this work?

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Why is that $3b = a$? –  Makoto Kato Sep 27 '12 at 5:10

No, $3b^3=a^3$ does not imply that $a=3b$. In fact if $a=3b$, then $a^3=(3b)^3=3^3b^3=27b^3$. What $3b^3=a^3$ does imply is that $3\mid a^3$, which then further implies that $3\mid a$. Then you have $a=3m$ for some integer $m$, and $3b^3=a^3=(3m)^3=27m^3$. Divide through by $3$ to get $b^3=9m^3$. Now $3\mid b^3$; what does this tell you about $b$? If you had begun by assuming (as you certainly may) that $\frac{a}b$ was in lowest terms, how would this be a contradiction?

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Let $n \ge 2$ be an integer. Let $m \ge 2$ be a square free integer. I claim that $\sqrt[n]{m}$ is irrational. The claim that $\sqrt[3]{3}$ is irrational follows immediately.

Suppose $\sqrt[n]{m} = \frac{a}{b}$, where gcd$(a, b) = 1$. Then $b^n m = a^n$. Let $p$ be a prime number such that $p|m$. Since $p|a^n, p|a$. Hence $p^n|a^n$. Since $m$ is square free, $p^{n-1}|b^n$. Since $n \ge 2, p|b^n$. Hence $p|b$. Since gcd$(a, b) = 1$, this is a contradiction.

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Below are a few ways to prove $\,3^{1/3}\,$ is irrational.

By the Rational Root Test, any rational root of $\rm\:x^3 - 3\:$ is an integer, contradiction.

$\rm\,3^{1/3} = a/b\:\Rightarrow\: a^3 = 3b^3\,$ contra unique factorization: the number of $3$'s in $\rm\,a^3\,$ is a multiple of $3$, versus one more than a multiple of $3$ in $\rm\,3b^3.$

Irrationality proofs for cube roots follow from irrationality proofs for 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.

Thus $\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

This inductive proof generalizes to higher-degree.

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Supposing $\sqrt[3]{3}$ is a rational solution of an equation, we can assume $x=\sqrt[3]{3}$ and $x^3=3$. But $x^3=3 \Rightarrow x^3-3 =0$. The rational-root theorem tells us that for a polynomial $f(x) = a_nx^n + a_{n−1}x^{n−1} + \ldots + a_1x + a_0$ all of whose coefficients ($a_n$ through $a_0$) are integers, the real rational roots are always of the form $\frac{p}{q}$, where $p$ is an integer divisor of $a_0$ and $q$ is an integer divisor of $a_n$. In this equation the possible rational roots could be $3,-3,1,-1$ and there wasn't any $\sqrt[3]{3}$. So $\sqrt[3]{3}$ is an irrational number.

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