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I’ve often wondered about this, and I conjecture the affirmative, based mainly on that it is so much easier to prove the transcendence of $e$ than that of $\pi$.

I would be surprised if, just as numbers form a linearly ordered classes ranging from algebraic of degree $n$ and culminating in the class of transcendental numbers, the transcendental numbers themselves are not further divided into a linearly ordered classes, with $\pi$ belonging to the top class, and $e$ belonging to some class below it, so this is really a reference request. Could someone please cite chapter and verse where this is done?

Mike Jones
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    This is not the nature of transcendental numbers so it is not done. – quanta May 16 '11 at 22:58
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    Only for very large values of $\pi$. – Mateen Ulhaq May 17 '11 at 05:11
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    I'm reminded of the "temperature" of Pickover curlicue fractals: http://mathworld.wolfram.com/CurlicueFractal.html for various numbers. (I do not think the pictures for $\pi$ and $\gamma$ are correct, and do not correspond to my memories of the Pickover book I first saw them in) – graveolensa May 17 '11 at 05:20
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    @muntoo: If you subtact the smallest value of π from the largest value of π, you get the root of the cause of the topological disconnectedness of the set of pure imaginary numbers:) – Mike Jones May 19 '11 at 20:38
  • "I would be surprised if … the transcendental numbers themselves are further divided into a linearly ordered classes" — I think you have a "not" missing somewhere. – ShreevatsaR May 30 '11 at 05:19
  • I would reason that if $\pi$ is more transcendental than $e$, then it should be lot easier to prove that $\pi$ is transcendental compared to $e$, and not the other way round. Roughly, if I am looking for a red object, it would help me if the it is very red as compared to less red. – Srivatsan Sep 02 '11 at 22:10
  • @ShreevatsaR: (belatedly) corrected. Thanks for pointing that out. – Mike Jones Sep 03 '11 at 22:57
  • @Srivatsan Narayanan: At first it might seem so, but it seems to me that what you offer is an instance of what Samuel Taylor Coleridge called a surface contradiction. After all, I think we would all agree that π is more irrational that the square root of 2, but it is much easier to prove that the square root of 2 is irrational than it is to prove that π is irrational. – Mike Jones Sep 03 '11 at 22:59
  • @Mike Jones, I actually think that $\pi$ is less irrational. For instance, the irrationality measure of $\sqrt{2}$ is exactly $2$, whereas that of $\pi$ is strictly bigger than $2$ (I believe, I do not know for sure). See thei's answer in this regard. – Srivatsan Sep 03 '11 at 23:57
  • There are several "irrationality measures" (as mentioned in the answers), but I doubt that any such measure is related to the difficulty of $proving$ irrationality. – leonbloy Sep 04 '11 at 00:43
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    @muntoo: unfortunately, $\pi$ is decreasing. Underwood Dudley published a study in the Journal of Recreational Mathematics where he looked at the values people computed for it over the years, and found that it has been decreasing on average by 0.000773 per year. :-) – tzs Sep 04 '11 at 02:29
  • @tzs It is hypothesized that this is caused by the drifting of the earth's magnetic north pole. As this claim has been disproven a total of $\pi$ times, it is further theorized that it will eventually become true once $\pi=0$. – Mateen Ulhaq Sep 04 '11 at 03:43

5 Answers5

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This question at MathOverflow has several answers discussing several senses in which one can say that one real number is more irrational than another, including irrationality measures and other hierarchies of complexity for real numbers.

Community
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JDH
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There are different kinds of irrationality measures:

http://mathworld.wolfram.com/IrrationalityMeasure.html

As you can see in the table above $e$ is not more irrational than algebraic numbers, but it is not clear if $\pi$ is more irrational, but both are not very irrational compared to Liouville numbers.

Phira
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You might be interested in Mahler's classification of transcendental numbers, which you can start to read about at http://en.wikipedia.org/wiki/Transcendental_number#Mahler.27s_classification

Gerry Myerson
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$\pi$ belongs to the probably "least transcendental" (and most algebraically structured) category of non-algebraic numbers, that of period integrals. Periods can be associated with algebraic varieties defined over Q, and consequently carry some algebraic structure.

zyx
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This might be a subset of the previous answers but I think this is still worth writing it as a separate answer.

One reason to believe $e$ is less transcendental than $\pi$ is that there is a nice pattern in the continued fraction of $e$ as opposed to that of $\pi$.

$e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,14,1,1,16,\ldots]$ whereas there is no such pattern for $\pi$. (We can find patterns in $\pi$ as well if we look at generalized continued fractions where the "numerators" in the continued fraction need not be $1$ always)

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    This would also make $e$ less transcendental than the cube root of 2 (which, so far as anyone knows, has no nice pattern in its continued fraction). This is a bit of a worry, since $e$ is, after all, transcendental, while cube root of 2 isn't. – Gerry Myerson May 30 '11 at 05:32
  • @Gerry: Thanks. I was not aware that. It is interesting to know that $\sqrt[3]{2}$ has no nice pattern. –  May 30 '11 at 17:06
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    careful - I only said that $\it so\ far\ as\ anyone\ knows\ $ the continued fraction has no nice pattern. No one is expecting a nice pattern to turn up, but no one has proved that it can't happen. – Gerry Myerson May 31 '11 at 00:01
  • @Gerry: Well written. +1 – user02138 Sep 03 '11 at 23:51