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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.

  1. Assume $2^{1/2}$ is irrational.
  2. $2^{1/3} * 2^{x} = 2^{1/2} \Rightarrow x = 1/6$.
  3. $2^{1/3} * {2^{1/2}}^{1/3} = 2^{1/2}$.
  4. if $2^{1/2}$ is irrational, then ${2^{1/2}}^{1/3}$ is irrational.
  5. $2^{1/3} = 2^{1/2} / {2^{1/2}}^{1/3}$.
  6. $2^{1/3}$ equals an irrational number divided by an irrational number.
  7. $2^{1/3}$ is an irrational number.
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16  
Re steps 6-7, note that an irrational number divided by an irrational number is not always an irrational number. –  Did Dec 14 '11 at 20:14
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Do you know a proof that $\sqrt2$ is irrational? If you do, try to adapt it to the case of $\sqrt[3]{2}$. –  Did Dec 14 '11 at 20:15
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Why not mimic the proof that $2^{1/2}$ is irrational? Assume that $2^{1/3} = p/q$ with $\gcd(p,q)=1$. Then $2=p^3/q^3$; therefore, $2q^3 = p^3$. Go from there. –  Arturo Magidin Dec 14 '11 at 20:16
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For any irrational number $r$, $\frac{r}{r}$ is rational... –  user5137 Dec 14 '11 at 20:22
    
@DidierPiau, I suppose any given irrational number divided by itself is 1, which is rational. hmmph. I'll take another crack at it with gcd algorithm tonight. –  bwkaplan Dec 14 '11 at 20:22

9 Answers 9

up vote 49 down vote accepted

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.

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I've never seen this proof. I think this is nicer than the well-known one. –  the symplectic camel Dec 14 '11 at 20:28
    
@JackManey, I like this proof because it reduces the possibilities to a small number (four) of cases that can be checked to see if they are zeros of the polynomial. I'll give this question some time, but this'll probably be the answer I accept. –  bwkaplan Dec 14 '11 at 21:04
    
@bwkaplan - Thank you. :) Another advantage of using the rational root test--along with Euclid's Lemma--is that it's just as easy to show that $\sqrt[m]{p}$ is irrational for any prime $p$ and any $m\geq 2$. –  user5137 Dec 14 '11 at 21:13
1  
It is simple with the usual proof too: If $pb^m = a^m$, then $p$ divides $a^m$ and so $p$ divides $a$. Thus $p$ divides $b^m$ too, as $m \geq 2$. So $p$ divides $b$, which is a contradiction when we choose $\gcd(a,b) = 1$. Also works with the version I posted. –  Mikko Korhonen Dec 14 '11 at 22:24
    
@Karatug The rational root test implies that rational roots of integer coefficient polynomials must be integral if the polynomial is monic, i.e. has leading coefficient = 1. Thus if the polynomial has no integral roots then every root is irrational. See here for much further discussion. –  Bill Dubuque Dec 14 '11 at 23:29

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''

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you're right, as I noted in the comments to the OP. I would argue on philosophical grounds that a proof by contradiction isn't as strong as another that is constructed, but that is not a mathematical objection, so I'll let it lie... –  bwkaplan Dec 14 '11 at 20:28
    
Well, I'll respond to your argument "on philosophical grounds": it's a complete mystery as to what the complement $\mathbb{R} \setminus \mathbb{Q}$ is * really * like - e.g. number-theoretic conjectures about algebraic independence of various constants. In other words, all we know about "irrational" numbers is that they're "not rational" - it's not clear what we * really * gain after completing with respect to the usual metric. –  StackQs Dec 14 '11 at 21:51
    
bwkaplan: math.andrej.com/2010/03/29/… –  sdcvvc Dec 14 '11 at 23:25

The polynomial $X^3-2$ is irreducible over $\mathbb Q$ by Eisenstein's criterion, hence has no rational root.

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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.

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4  
Of all the many proofs, this is the one I like best. –  Lubin Dec 15 '11 at 0:51

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.

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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.

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

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 - precisely as we did above. Namely, using the same notation, we have $$\begin{eqnarray} \rm r^{n+1} &=&\rm\: e\ r^n +\: f(r),\quad deg\ f < n,\quad e\in\mathbb Z,\quad f(x)\in \mathbb Z[x] \\ 0 &=&\rm\: (a - b\ r)\ (d\ r^n +\: c\ r^{n-1}) \\ \Rightarrow\ \ 0 &=&\rm\: (ad\!-\!b\,c)\ r^n + ac\ r^{n-1}\! - de\color{#0A0}{\bf b}\ r^n\ \ -\ \ \ bd\,f(r),\quad\!\! so\ \ \ ad\!-\!bc = \color{#C00}{\bf 1}\ \ yields \\ \Rightarrow\ \ 0 &=&\rm\ \ \ \ \color{#C00}{\bf 1}\cdot r^n\quad +\ \ \ (ac\ \ \ \,-\,\ \ \ de\color{blue}{\bf a})\ r^{n-1}\! - bd\ f(r),\ \ \ by\ \ \ b\:r=a\:\Rightarrow\: \color{#0A0}{\bf b}\:r^n = \color{blue}{\bf a}\:r^{n-1} \\ \end{eqnarray}$$

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 (a monic case of the Rational Root Test).

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1  
Nice use of Bezout :-) (+1) –  robjohn Dec 15 '11 at 8:46
    
I would add that the theorm is correct for any k∈Z not only k=3. This can be seen by decomposing a and b to primes. if r is not in Z than there is a prime p such that p appear in the factorization of b but not the factorization of a. raising by k (a/b) you get p^k in b but not the a thus a^k/b^k is not in Z (contradiction) –  Belgi Dec 17 '11 at 0:08
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@Belgi As I mentioned in the remark, the proof immediately generalizes to higher degree monic polynomials, which is far more general than the k'th root case that you mention. –  Bill Dubuque Dec 17 '11 at 5:42

Since $\mathbb Z$ is a UFD it is integrally closed and a rational solution of $x^3-2=0$ would be an integer.

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4  
Note to readers unfamiliar with the terms. This is simply a highbrow way of saying: apply the special case of the Rational Root Test where the polynomial is monic, i.e. leading coefficient $= 1$. For a little-known simple inductive proof of this result see my answer here. –  Bill Dubuque Dec 15 '11 at 2:17
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Dear @Bill, highbrow is in the eye of the beholder: I consider "integrally closed" to be a basic and easy concept (as demonstrated by your comment!) . Anyway, there are now many correct proofs, some very explicit, some less so. The user may now choose the one most appropriate to their comfort level: isn't that wonderful? –  Georges Elencwajg Dec 15 '11 at 9:00
    
Yes, of course I agree. Many of my proofs promote highbrow views. My point here is to emphasize that in this case the equivalent lowbrow view is very well-known, viz. the monic case of the rational root test. –  Bill Dubuque Dec 15 '11 at 14:27
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By the way, I blush to confess that I could not resist the challenge of posting two one-line answers... –  Georges Elencwajg Dec 15 '11 at 14:43
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Who can blame you? The remarkable confluence of many divisibility notions in $\rm\:\mathbb Z\:$ leads to many interesting views of elementary results. My favorite "highbrow" view is the elegant one-line proof that results from employing Dedekind's notion of conductor ideal. I wrote much about that in my posts to Gerry Myerson in this 2009/20/5 sci.math thread. Alas, this nice view is not as well known as it should be, e.g. it escaped Estermann and Niven. Click on thread title on top to see the whole sci.math thread. –  Bill Dubuque Dec 15 '11 at 16:42

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.

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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.

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11  
+1 for Fermat's Last Theorem. Nice! –  Matthew Crumley Dec 15 '11 at 19:14
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Dear Eric: in order to make your wonderful answer self-contained, I would suggest you add the proof of that Theorem in a short edit: I guess it would just take you about a minute. (Unless the box for the answer is too narrow, of course.) –  Georges Elencwajg Dec 16 '11 at 0:10
    
(+1) This is beautiful. –  Parth Kohli Feb 3 '13 at 13:25

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