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Transcendent numbers

  1. Apr 14, 2005 #1

    jp

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    Transcendental numbers

    I was trying to find out more about transcendental numbers , but I couldn't find more about it other than that they can't be drawn...<--which is still a little confusing because pi is a transcendent number but if you draw a circle w/ radius of 0.5, the circumference is pi. I'm assuming transcendental numbers can't be drawn in a straight segment.

    Can someone help me out?
     
    Last edited: Apr 14, 2005
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  3. Apr 14, 2005 #2

    HallsofIvy

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    Depends on what you mean by "drawn". Line segments of length equal to a transcendental number (given a "unit" length) cannot be constructed with compass and straight edge. That isn't really a characteristic of transcendental numbers- any number which not algebraic of order a power of two cannot be constructed.
     
  4. Apr 14, 2005 #3
    Hmm, I'm not sure what you mean by order two, but you can construct some algebraic numbers of degree greater than 2. Any number which involves a finite amount of square root extractions, for example, can be constructed.
    In answer to the OP, transcendental numbers are defined as not being the root of any finite polynomial with rational coefficients. In other words, a transcendental number is a real number that is Not an algebraic number (Akin to defining irrationals to be real numbers that are not rational numbers).
     
    Last edited: Apr 14, 2005
  5. Apr 15, 2005 #4

    Curious3141

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    I think what HallsofIvy was referring to was the degree of the algebraic number. This is defined as the minimum degree of the polynomial equation with integral coefficients (and with RHS zero) that has that algebraic number as one of its root(s). So the degree of sqrt(2) is 2, cube root(2) is 3, etc. The degree of a rational number is one, since a linear equation can be formed to give a rational root.

    Halls was saying that algebraic numbers with degrees that are not powers of 2 are not constructible. This means degrees other than 2, 4, 8, 16, etc.

    The numbers with finite square root extractions all have degrees that are powers of 2, so actually you and Halls are in agreement. :smile:
     
    Last edited: Apr 15, 2005
  6. Apr 15, 2005 #5

    Hurkyl

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    Also, note that not every number of order a power of 2 is constructible -- for instance, take a solution to any 8-th degree polynomial that isn't solvable with radicals.
     
  7. Apr 15, 2005 #6

    HallsofIvy

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    I didn't say "order of 2", I said "order a power of 2"- 4, 8, 16, etc. A number is "algebraic of order n" if it can be found as a root of a polynomial equation with integer coefficients of degree n but no such polynomial of lower degree. Saying that a number is "algebraic of order a power of 2" is exactly the same as saying it can be calculated by "a finite amount of square root extractions".
     
  8. Apr 15, 2005 #7
    Ah, gotcha. 'Twas too late last night for me to be reading anything. :smile:
     
  9. Apr 15, 2005 #8
    Transcendental numbers have historically been a hard thing to get a handle on. Historically they were known to exist for well over a hundred years before anyone could find one (Pi of course was suspected of being transcendental but a proof is pretty evasive). In the 1800's Liouville was the first to find a proof that certain numbers were transcendental. I would recomend trying to find a proof of Liouville's approximation theory. It is very approachable, you probably only need some calculus to understand it, and it gives a recipe for builiding a whole family of family of transcendental numbers. If I remember correctly the easiest example is [tex]\sum_{n=1} 1\times 10^{-n!}[/tex]
    0.110001000000000000000001000...
    ie a 1 in every n!'th position.

    Liouville numbers are only a small subset of the transcendentals but they are a set you can get a handle on and actually see why they are transcendental.

    G'luck finding out more
    Steven
     
  10. Apr 15, 2005 #9
    Not according to what I've heard. According to Wikipedia, in 1844 Liouville showed that they exist by finding his example. http://en.wikipedia.org/wiki/Transcendental_number

    In 1874 Cantor's diagonal proof demonstrated the existence of transcendental numbers, and is often said to be a nonconstructive, pure existence proof. But I don't accept that, in fact I've written a Java applet to print out a number constructed via Cantor's proof: http://www.chronon.org/Applets/transcendental.html
     
  11. Apr 15, 2005 #10

    Hurkyl

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    Cantor's proof is nonconstructive because he didn't do the details necessary to turn it into a constructive proof. :tongue2:

    I do have a complaint about your program -- it looks like you're only printing out one digit per (irreducible?) polynomial... but what you really want to do is print out one digit per root of each (irreducible?) polynomial, otherwise you're missing some of the algebraic numbers.
     
  12. Apr 17, 2005 #11
    No it does go through all of the roots of each polynomial - or at least all of the positive real roots.
     
  13. Apr 17, 2005 #12
    I think you can construct transcendental numbers

    Probabilistically :

    Straight edge : just draw a segment on a flat paper at random, this will be with probability 1 a transcendental number

    Surely :

    Compass : take the unit length on the compass, roll the paper into a clyinder of radius unity, take this distance with the compass on the most curved direction, and unfold it after

    (There is always a need to specify in which geometry we are working and which operations with the surface are allowed)
     
  14. Apr 17, 2005 #13
    Except that when someone refers to what can be constructed with a compass and a straight edge, you should assume (unless they've stated otherwise) that they're refering to the restrictions used in ancient greece.
     
  15. Apr 17, 2005 #14

    mathwonk

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    kelinwolf, drawing a line segment on a piece of paper with say a pencil, does not even come close to specifying a number. it is just a smudge of graphite particles, with no well defined beginning or end.
     
  16. Apr 17, 2005 #15
    What do they didn't have to restrict the proposed construction ?
     
  17. Apr 18, 2005 #16
    Solutions must be exact - finding a solution that has probability one of being correct isn't good enough. And you can't deform the drawing surface (and even if you could, your deformation would be invalid since it uses a cylinder, which you don't have).
     
  18. Apr 18, 2005 #17

    HallsofIvy

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    Historically, the Greeks, while very good at geometry, didn't have a very good numeration system (basically, they just used letters to represent numbers with complicated combinations for large numbers or fractions). They thought of line segments as representing numbers: if a given line segment represents "1" then a line segment twice that long represents "2", etc. Using a marked ruler or other aid was the same as assuming you already "knew" numbers such 1/3 or [itex]\sqrt{2}[/itex]. The compass and straight edge required no such knowledge. It is easy to see that you can construct all integers, all rational numbers, and "Pythagoras" gives square roots.
     
    Last edited: Aug 5, 2008
  19. Aug 5, 2008 #18
    Hi there!

    I hope I'm not drifting far away from the point :0 I have the following question:

    We know pi and e are indefinite non-periodic decimal fractions, as many others. But pi and e can easily be defined as limits :) My question is, if any indefinite non-perodic decimal fraction could be defined as a limit and how :) ?

    here's one: 5,134625376845828297... (chosen totally random)

    best regards, marin
     
  20. Aug 6, 2008 #19

    HallsofIvy

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    Yes, all real numbers can be defined in terms of limits. One standard method of 'constructing' the real numbers from the rationals is this:

    Consider the set of all increasing sequences of rational numbers having an upper bound. Define two such sequences, {an} and {bn}, to be "equivalent" if the sequence {an- bn} converges to 0. It is easy to see that is an equivalence relation and so defines equivalence classes on the set of all such sequences. We define the real numbers to be these equivalence classes. If r is a rational number, we identify r with the equivalence class containing the constant sequence an= r for all n. That construction has the virtue of making it particulary easy to prove the "monotone convergence property": that all increasing, bounded sequences of real numbers converge. We can do the same thing with the set of all Cauchy sequences. That construct gives the "Cauchy criterion", that all Cauchy sequences of real numbers converge, easily.

    Of course, in the simplest sense, every decimal expansion of a real number is such a limit. If x= 0.a1[/sup]a2a3... then x is the limit of the sequence 0.a1, 0.a1a2, 0.a1a2a3, ... Of course, every number in that sequence is a rational number because it is a terminating decimal. Note also, that this sequence is both "increasing, bounded" and a Cauchy sequence.
     
  21. Aug 6, 2008 #20

    Integral

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    Note that this thread had not been posted to for over 3 yrs until Rodolfos post yesterday. Don't expect replies to ancient posts.
     
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