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Of course, if you plug the cube root of 9 into a calculator, you get an endless stream of digits. However, how does one prove this on paper?

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Suppose that $9^{1/3}=m/n$ with $m$, $n$ integers with ${\rm GCD}(m,n)=1$. This may be assumed because if $d$ is an integer divisor of both $m$ and $n$, then $m/n=(m/d)/(n/d)$. Then $$ 9n^3=m^3 $$ so that $3$ divides $m^3$, hence $m$ since $3$ is prime. Thus $3^3=27$ divides both sides of the equality so that $3$ divides $n^3$, hence $n$ for the same reason as above. This contradicts the assumption that $m$ and $n$ are coprime.

This arguent generalizes immediately to showing that the $n$-th root of an integer which is not an $n$-th power of an integer is not rational.

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    $\begingroup$ Note that $m,n\in\mathbb{Z}$ just means $m$ and $n$ are integers, and ${\rm GCD}(m,n)=1$ means the fraction $m/n$ is in lowest terms. In the interest of making the posting comprehensible to a broad audience, I'd have written it that way. $\endgroup$ – Michael Hardy Jan 25 '12 at 17:32
  • $\begingroup$ @MichaelHardy: Agreed. Especially since it's not otherwise obvious that ${\rm GCD}(m,n)=1$ can be assumed without loss of generality. $\endgroup$ – ruakh Jan 25 '12 at 18:42
  • $\begingroup$ I assumed that was elementary enough to left it implicit. Anyway I edited my answer if that can improve readability for a beginner. $\endgroup$ – Andrea Mori Jan 26 '12 at 8:46
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By the rational root test, if $$P(x) = a_nx^n+\cdots +a_0$$ is a polynomial with integer coefficients, and $\frac{u}{v}$ is a rational number with $\gcd(u,v)=1$ such that $P(\frac{u}{v})=0$, then $u$ divides $a_0$ and $v$ divides $a_1$.

(This holds in any GCD-domain, even if there is no unique factorization into primes)

(If you don't understand the last parenthetical comment, don't worry; it's a nod towards those who know some more abstract algebra)

Look at $x^3-9$. If $\frac{u}{v}$ is a root, then $$\begin{align*} \frac{u^3}{v^3}-3&=0\\ \frac{u^3}{v^3} &=9\\ u^3&=9v^3\\ v^3 &\text{divides }u^3\\ v &\text{divides }u^3\\ v &\text{divides }1 &&\text{(since }\gcd(u,v)=1\text{)} \end{align*}$$ hence any rational root must be an integer.

But if $a$ is a positive integer, $a\leq 2$ implies $a^3\leq 8$, and $a\geq 3$ implies $a^3\geq 27$. So there are no integers with cube $9$.

More generally, if you look at $x^n-b$ with $b$ an integer, it has rational roots if and only if it has integer roots, if and only if $b$ is a perfect $n$th power.

Even more generally, rational roots of monic polynomials with integer coefficients ("monic" means "leading coefficient equal to $1$) are necessarily integers (and must divide the constant term).

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Assume the cube root of $9$ is rational, then it can be written as $p/q$ for integers $p$ and $q$ such that $p$ and $q$ share no common divisor.

$$\begin{align*}(p/q)^3 &= 9\\ &\implies (p^3)/(q^3) = 9\\ &\implies p^3 = 9 * (q^3) \end{align*}$$

Now, $p$ must be a multiple of $3$, else $p^3$ wouldn't be a multiple of 3 (or 9). Let $r$ be the integer such that $(3r)^3 = p^3$

So, $$\begin{align*} (3r)^3 &= 9(q^3)\\ &\implies 3 * 3 * 3 * r^3 = 9(q^3)\\ &\implies q^3 = 3 r^3 \end{align*}$$ hence $q$ is also a multiple of $3$.

If $p$ and $q$ are both multiples of $3$, then $p/q$ is not the simplest way to express the cubed root of $9$ and so we have a contradiction.

Therefore, the cubed root of $9$ must be irrational

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  • $\begingroup$ Try to use latex to make your answer more readable:) $\endgroup$ – sxd Jan 25 '12 at 16:53
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    $\begingroup$ To anyone desiring to flesh this out to a complete proof: It is important to show why if $3$ divides $p^3$ (resp. $q^3$), then $3$ divides $p$ (resp. $q$). $\endgroup$ – Jonas Meyer Jan 25 '12 at 17:02
  • $\begingroup$ +1 for explaining where the number 3 as a factor of p came from. $\endgroup$ – Dan Neely Jan 25 '12 at 18:25
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This is essentially the same proof I gave in my answer here.

Suppose $9^{\frac{1}{3}}$ is rational. Then $3^2n^3 = m^3$ for some natural numbers $n$ and $m$. On left side of the equation, the power of $3$ is of the form $3k + 2$ and on the right side it is of the form $3l$. This is a contradiction, because each integer greater than one has a unique prime factorization by the fundamental theorem of arithmetic. Thus $9^{\frac{1}{3}}$ is not rational.

This same proof also works for a more general case. Let $p$ be prime and $n \geq 2$ an integer. Then $\sqrt[n]{p^k}$ is irrational when $n$ does not divide $k$. Just like before, assuming that $\sqrt[n]{p^k}$ is rational leads to a situation where we have a number with two different prime factorizations. One factorization has $p$ of power divisible by $n$, while the other has $p$ of power not divisible by $n$.

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I typically use Bézout’s theorem http://mathworld.wolfram.com/BezoutsTheorem.html. And here is an example of how to use it http://rtybase.blogspot.com/2011/06/regarding-square-root-of-n.html which can be easily extrapolated to a more general case.

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