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$\begingroup$

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.
  • 18
    Re steps 6-7, note that an irrational number divided by an irrational number is not always an irrational number.2011-12-14
  • 9
    Do you know a proof that $\sqrt2$ is irrational? If you do, try to adapt it to the case of $\sqrt[3]{2}$.2011-12-14
  • 12
    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.2011-12-14
  • 7
    For any irrational number $r$, $\frac{r}{r}$ is rational...2011-12-14
  • 0
    @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.2011-12-14
  • 0
    The gcd algorithm?? What for? I do not see the link. As I said: do you know a proof that $\sqrt2$ is irrational? If yes, you could append it to the end of your post and we could start from there.2011-12-14
  • 0
    Can't resist this one - see the first answer in http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts. I don't mean to confuse the OP or anyone else with the link - this is purely for humor.2011-12-14
  • 0
    You can't prove something starting by assuming what you want to prove.2011-12-15
  • 0
    @missingno, I didn't. I assumed $2^{1/2}$ was irrational; my intention was to prove $2^{1/3}$ was irrational. Surely you can see the difference now.2011-12-15

9 Answers 9

6

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.

85

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.

  • 55
    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.)2011-12-16
  • 3
    There's always one :P2014-09-28
51

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.

  • 1
    I've never seen this proof. I think this is nicer than the well-known one.2011-12-14
  • 1
    @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.2011-12-14
  • 0
    @bwkaplan - Thank you. :) Another advantage of using the rational root test--along with [Euclid's Lemma](http://en.wikipedia.org/wiki/Euclid's_lemma)--is that it's just as easy to show that $\sqrt[m]{p}$ is irrational for any prime $p$ and any $m\geq 2$.2011-12-14
  • 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.2011-12-14
  • 0
    @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](http://groups.google.com/group/sci.math/msg/b547bca171fc24be) for much further discussion.2011-12-14
22

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

22

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.

  • 5
    Of all the many proofs, this is the one I like best.2011-12-15
21

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).

  • 2
    Nice use of Bezout :-) (+1)2011-12-15
  • 0
    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)2011-12-17
  • 2
    @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.2011-12-17
  • 0
    [See also here](http://math.stackexchange.com/a/1183042/242) for another proof of the Rational Root Test by induction on degree.2015-03-09
10

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''

  • 0
    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...2011-12-14
  • 0
    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.2011-12-14
  • 0
    bwkaplan: http://math.andrej.com/2010/03/29/proof-of-negation-and-proof-by-contradiction/2011-12-14
10

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

  • 5
    Note to readers unfamiliar with the terms. This is simply a highbrow way of saying: apply the special case of the [Rational Root Test](http://en.wikipedia.org/wiki/Rational_root_theorem) where the polynomial is *monic*, i.e. leading coefficient $= 1$. For a little-known simple inductive proof of this result see my answer here.2011-12-15
  • 4
    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?2011-12-15
  • 0
    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.2011-12-15
  • 0
    Dear @Bill, I am very happy that we see eye to eye on these matters and I'm looking forward to reading more of your always interesting posts.2011-12-15
  • 2
    By the way, I blush to confess that I could not resist the challenge of posting two one-line answers...2011-12-15
  • 1
    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.](http://groups.google.com/group/sci.math/msg/b547bca171fc24be) 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.2011-12-15
  • 0
    You're right, Bill: although I know what a conductor is, I hadn't seen it used in this context before. Thanks for the link, in which I was pleasantly surprised to see that other frequent contributor Arturo Magidin. This is a small world....2011-12-15
6

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.

  • 0
    Wonderful, not even using unique factorisation!2015-04-11