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Line of real numbers-transcendental numbers

  1. Mar 2, 2005 #1
    Hi everybody,
    I would like to ask two things:

    1)What is the line of real numbers? Is it just a graphical way of representing the set of real numbers?

    2)How is the existence of transcendental numbers explained? I mean, if we look at the number line, and thinking of the Dedekind cuts, why only very few of the irrationals are transcendental numbers? Do we know how many transcendental numbers exist? And is there some explanation for their position in the number line?

    Thanks
     
    Last edited: Mar 2, 2005
  2. jcsd
  3. Mar 2, 2005 #2

    HallsofIvy

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    1) Yes, the "number line" is just a graphical way of representing the set or real numbers.

    2) "why only very few of the irrationals are transcendental numbers?"
    Excuse me? In a very specific sense, almost all real numbers ARE transcendental! That is, the set of all transcendental numbers is far larger than the set of algebraic numbers (non-transcendental numbers which includes all rational numbers). As far as "how many irrational numbers exist": an "uncountable" number.
    Basically, "countable" means that there is a one-to-one correspondence between the set and the set of all counting numbers (1, 2, 3, etc.). "Uncountable" means there are far too many to "count"- to set into a correspondence with even all counting numbers.
    Their position? In any interval of real numbers, no matter how small, there exist and infinite (indeed "uncountable") number of irrationals and and infinite (but countable) number of rationals.
     
  4. Mar 2, 2005 #3

    jcsd

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    1) It's a geometrical way of represnting the real numbers so we have a line where each poitn correponds to a real number. Of course when preople talk about the rela line quite often they're not specifically refering to the geomtrical represemtation, instead they are talking about the real numbers as an ordered set/metric space/topological space/etc.

    2) As Hall ofIvy says, the transcendental numbers have the cardinality of the continuum (i.e. the same cardianlity as the rela numbers) wheras the algebraic (non-transcendental) numbers have a cardinailty of aleph-0, the same cardianltiy as the natural numbers.
     
  5. Mar 2, 2005 #4
    I made a mistake in my initial post. I have edited it, so the question is how many transcendental numbers (and not just irrational) exist? However, you mention that almost all real numbers are transcendental! I don't think that is correct. Transcendental numbers examples: pi, e ...
    And I think that irrationals, like rationals are a countable set, but i am not sure of that.
    Actually i have just found this from MathWorld:
    From Gelfond's theorem, a number of the form a^b is transcendental (and therefore irrational) if a is algebraic<>0,1 and b is irrational and algebraic.
     
    Last edited: Mar 2, 2005
  6. Mar 2, 2005 #5
    Show that the set of all algebraic real numbers is countable (easy). A transcendental number is a non-algebraic real number. Thus, imply that the set of transcendental numbers is uncountable.
    Similarly, show that the set of all rational numbers is countable. Thus imply that the set of all irrational numbers, as the complement of the rationals in the reals, is uncountable.
     
    Last edited: Mar 2, 2005
  7. Mar 2, 2005 #6
    You are right, rationals are countable and reals are not, so irrationals can't be countable. As for the set of all algebraic real numbers, if it is countable, then transcendentals are not countable.
    Thanks for your help
     
  8. Mar 2, 2005 #7
    If you have shown that rationals are countable, then along with the fundamental theorem of algebra you have all the machinery you need to show that algebraic numbers are countable. :)
     
  9. Mar 2, 2005 #8

    CTS

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    Then to show that the almost all real numbers are irrational, show the rationals have measure zero. Start with rationals [tex]h/k[/tex] with [tex]1<h<k[/tex]. Enclose them in the intervals [tex](h/k - \epsilon/k^3, h/k + \epsilon/k^3)[/tex]. Because h and k are not necessarily relatively prime, we are duplicating many rationals and the total length of all these intervals is less than [tex]\sum_{k=1}^\infty \sum_{h=1}^k \frac{2\epsilon}{k^3} = \epsilon*\frac{\pi^2}{3}[/tex] which can be made as small as you want. For the rest of the positive rationals, notice that each [tex]h/k<1[/tex] is [tex]k/h>1[/tex] which can be enclosed in an interval in the same way as above. Extension to zero, 1, and the negative rationals is obvious.
     
  10. Mar 3, 2005 #9
    Thanks for your answers. To make something clear: I haven't proved myself that rationals are countable, I jsut remembered a proof I had read some time ago. Generally i am not very familiar with these concepts so i can't think of way to prove that algebraic numbers are countable. Also, i didn't understand much of what CTS is writing. It's probably things i haven't studied yet. Anyway, if there is quite a clear way to count algebraic numbers, i would really like to read it.
     
  11. Mar 3, 2005 #10

    CTS

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  12. Mar 10, 2005 #11
    Well, a somewhat more intuitive thing is that the computable numbers (all those for which there exists an algorithm by which the decimal representation of the number can be computed to arbitrary precision - which can be stated more rigorously) form a countable set. Since there's at least one algorithm to calculate each of them, you simply need to notice that it must be possible to number the algorithms (in other words, to form a bijection with the naturals), although this is obviously not rigorous enough to be called a proof!

    The computable numbers clearly include the rationals (long division), the algebraics (simply approximate the solution to the corresponding polynomial equation), and some transcendentals (pi, e, etc.). Note that most real numbers are not computable, in the same sense as most numbers are transcendental, since the computable numbers are countable and the real numbers are not.

    In some sense, the set of numbers that we will ever need is thus countable.
     
    Last edited: Mar 10, 2005
  13. Mar 11, 2005 #12

    HallsofIvy

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    CTS is using a little overkill- one can prove that any countable set has "measure 0" under any measure.
     
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