# Liouville's Theorem (Transcendental Nos)

1. Aug 3, 2012

### 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:

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?

2. Aug 3, 2012

### elvishatcher

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

3. Aug 3, 2012

### QIsReluctant

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.

4. Aug 3, 2012

### 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

5. Aug 4, 2012

### voko

You need to prove irrationality, then apply your result.

6. Aug 4, 2012

### micromass

Staff Emeritus
The statement

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

7. Aug 4, 2012

### QIsReluctant

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.