1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Liouville's Theorem (Transcendental Nos)

  1. Aug 3, 2012 #1
    I am reading a book on transcendental number theory and have come across (and proven) the following:

    Let [itex]\alpha[/itex] be an irrational, algebraic number of degree d. Then there exists a positive constant c, depending only on [itex]\alpha[/itex], such that for every rational number [itex]\frac{p}{q}[/itex], the inequality

    [itex]\frac{c}{q^{d}} \leq \left|\alpha - \frac{p}{q}\right|[/itex]

    is satisfied.

    Where I come across problems is in the contrapositive. The book gives the following as the contrapositive:

    The problem is that, whenever I tried to create the counterpositive to the first theorem, I got a weaker result than the book: namely, that [itex]\alpha[/itex] could be transcendental OR rational.

    I reversed and negated to arrive at the following contrapositive:

    If for ALL c>0 and a PARTICULAR positive integer d the opposite inequality holds, then [itex]\alpha[/itex] is either rational (i.e., the negative of irrational) or not algebraic degree d.

    So [itex]\alpha[/itex] could well be rational and not contradict this statement for any d. Furthermore, even if we were to look at the book's version, all we have to do is pick a certain rational value [itex]\alpha = \frac{p_{0}}{q_{0}}[/itex], and if we always pick that same number for the rational number, the result holds ... but [itex]\alpha[/itex] is NOT transcendental!

    I wrote in the margin of my book that we may have to make [itex]\alpha[/itex] irrational for the book's statement to work. If that wording is changed, then the theorem seems to work just fine. Incidentally, I see no reason why [itex]\alpha[/itex] has to be real. Is this wording a mistake on the book's part, or is there some essential piece of information that I'm failing to take into account?
     
  2. jcsd
  3. Aug 3, 2012 #2
    You said initially "Let [itex]\alpha[/itex] be an irrational, algebraic number of degree d." So it seems like this condition would still apply to the book's statement. So, if the book's statement works when [itex]\alpha[/itex] is irrational it seems like there isn't a problem
     
  4. Aug 3, 2012 #3
    No, the stipulation doesn't apply to both statements, just the first one. The second α is only required to be real.

    Let me elaborate on my reasoning. I see the initial stipulation as the "if" in an if-then statement.

    Original statement:
    IF (α is irrational AND α is algebraic degree d) THEN the statement applies with that d.

    Contrapositive:
    IF NOT (the statement with a certain d) THEN NOT (α is irrational AND α is algebraic degree d)

    or, equivalently,

    If the opposite inequality is true for d, then either α is rational OR α is not algebraic degree d.


    So unless I am not considering things correctly (Elvis, perhaps I misunderstood you?), I am not only arriving at a weaker result but also not understanding the need for α to be real.
     
  5. Aug 3, 2012 #4
    Yeah, I see what you're saying. It seems like the contrapositive would just be if the inequality does not hold, then α is not irrational and of degree d (which is, I believe, what you are saying). So just based on looking at it and the definition of contrapositive I seem to have the same problem, and the actual details of the math here are over my head, I think. Sorry I can't help. Good luck, though
     
  6. Aug 4, 2012 #5
    You need to prove irrationality, then apply your result.
     
  7. Aug 4, 2012 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    The statement

    is obviously false for rational numbers. Indeed, we can take [itex]\alpha=p/q[/itex].
     
  8. Aug 4, 2012 #7
    So, in effect, you're saying that α has to be proven irrational before we can conclude that it's transcendental (i.e., the same thing I proposed)?

    That's what I thought. Now, on Wikipedia (link: http://en.wikipedia.org/wiki/Liouville_number, I read that zero is a lower, non-inclusive bound for the error -- which would disqualify the obvious example you gave. However, I don't understand how that zero gets there from taking the contrapositive, and the proof given on Wikipedia is a little too dense before I've had my coffee. :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Liouville's Theorem (Transcendental Nos)
  1. Liouville's Theorem (Replies: 8)

Loading...