Liouville's Theorem (Transcendental Nos)

[QIsReluctant]

I am reading a book on transcendental number theory and have come across (and proven) the following:

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

\frac{c}{q^{d}} \leq \left|\alpha - \frac{p}{q}\right|

is satisfied.

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

Let \alpha be real. For all c>0 and positive integers d, let there be some rational number p/q satisfying

\frac{c}{q^{d}} > \left|\alpha - \frac{p}{q}\right|

Then \alpha is transcendental.

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 \alpha 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 \alpha is either rational (i.e., the negative of irrational) or not algebraic degree d.

So \alpha 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 \alpha = \frac{p_{0}}{q_{0}}, and if we always pick that same number for the rational number, the result holds ... but \alpha is NOT transcendental!

I wrote in the margin of my book that we may have to make \alpha 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 \alpha 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?

 

[elvishatcher]

I wrote in the margin of my book that we may have to make \alpha irrational for the book's statement to work.

You said initially "Let \alpha 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 \alpha is irrational it seems like there isn't a problem

 

[QIsReluctant]

You said initially "Let α 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 α is irrational it seems like there isn't a problem

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.

 

[elvishatcher]

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

 

[voko]

You need to prove irrationality, then apply your result.

 

[micromass]

The statement

Let \alpha be real. For all c>0 and positive integers d, let there be some rational number p/q satisfying

\frac{c}{q^{d}} > \left|\alpha - \frac{p}{q}\right|

Then \alpha is transcendental.

is obviously false for rational numbers. Indeed, we can take \alpha=p/q.

 

[QIsReluctant]

You need to prove irrationality, then apply your result.

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)?

The statement

Let α be real. For all c>0 and positive integers d, let there be some rational number p/q satisfying ...
Then α is transcendental.

is obviously false for rational numbers. Indeed, we can take α=p/q.

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.