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Homework Help: 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.

    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
    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:
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