Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
This is not the nature of transcendental numbers so it is not done. –  quanta May 16 '11 at 22:58
1  
Only for very large values of $\pi$. –  muntoo May 17 '11 at 5:11
2  
I'm reminded of the "temperature" of Pickover curlicue fractals: 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) –  deoxygerbe May 17 '11 at 5:20
1  
@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
2  
@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 4 '11 at 2:29

5 Answers 5

up vote 20 down vote accepted

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.

share|improve this answer
    
When I click on the check mark, nothing happens. -OP Mike Jones –  Mike Jones May 18 '11 at 10:55
    
I'm not sure what the trouble could be. I flagged for moderator attention. But please don't worry about it... –  JDH May 18 '11 at 11:19
    
@Mike: I could be wrong, but it sounds like the problems you're having are client-side, not problems with the software. Have you tried clearing your cookies and/or cache? –  Qiaochu Yuan May 18 '11 at 14:15

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.

share|improve this answer
    
Is there any lower bound on the irrationality measure of $\pi$ better than $2$? –  Srivatsan Sep 3 '11 at 23:57

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

share|improve this answer

$\pi$ 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.

share|improve this answer

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)

share|improve this answer
5  
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 5:32
    
@Gerry: Thanks. I was not aware that. It is interesting to know that $\sqrt[3]{2}$ has no nice pattern. –  user17762 May 30 '11 at 17:06
1  
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 0:01
    
@Gerry: Well written. +1 –  user02138 Sep 3 '11 at 23:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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