Is there a direct proof that $\pi$ is not constructible, that is, that squaring the circle cannot be done by rule and compass?

Of course, $\pi$ is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that $\pi$ is not a root of polynomial of degree $2^n$ with rational coefficients?

The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that $\pi$ is not rational. Does any one know of a proof along these lines or any other direct proof?

I just want a direct proof that $\pi$ is not constructible without appealing to transcendence.

  • 10
    $\begingroup$ So far as I know, there is no proof that $\pi$ doesn't have degree $2^n$ that doesn't also prove $\pi$ is transcendental. I leave this as a comment, rather than an answer, because I don't know how to substantiate it. $\endgroup$ – Gerry Myerson Jan 30 '12 at 0:20
  • 6
    $\begingroup$ @lhf: I don't see how knowing that $\pi$ is irrational helps. There are lots of irrational algebraic numbers. $\endgroup$ – Zhen Lin Jan 30 '12 at 0:30
  • 4
    $\begingroup$ @Zhen Lin: What lhf means is, I think, consider a tower of quadratic extensions, starting from Q. Note that $\pi\not\in \mathbb Q$ as an induction basis and then, by some magic induction argument, if $\pi\not\in F\implies \pi\not\in F(u)$, where $u^2\in F$. $\endgroup$ – Myself Jan 30 '12 at 0:49
  • 6
    $\begingroup$ @Marc: that's still not the question being asked in the first sentence. Most irreducible polynomials of degree $2^n$ do not have constructible numbers as their roots; in fact, generically they have Galois group $S_{2^n}$. $\endgroup$ – Qiaochu Yuan Jan 30 '12 at 22:52
  • 7
    $\begingroup$ Ha. Dave Renfro's sci.math post was a response to one of mine. I guess when the conversation turns to $\pi$, things naturally keep coming 'round.... $\endgroup$ – Gerry Myerson Jul 4 '12 at 4:00

I've just read in the book The Number $\pi$ by Eymard and Lafon that no such proof is known.

“The proof that it is impossible to square the circle does not involve direct demonstration of the non-constructibility of the number $\pi$. As far as we are aware, it is not known how to do this! One proves that it is not algebraic, which is much more restrictive, then one uses the fact that a constructible number is algebraic.” [§4.2, p. 134]

  • 1
    $\begingroup$ BTW, it's a wonderful book! $\endgroup$ – lhf Oct 8 '12 at 18:59
  • $\begingroup$ I see that the book was written $8$ years ago, maybe the information is not up to date, though I doubt that it's not [+1] $\endgroup$ – Belgi Oct 9 '12 at 11:09
  • 2
    $\begingroup$ @Belgi, I think that if things had changed since, we'd have heard! $\endgroup$ – lhf Oct 9 '12 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.