2
$\begingroup$

I'm having trouble following one step in the proof of Liouville's theorem on approximation of real algebraic numbers, from Murty and Rath's book "Transcendental Numbers".

The step is:

$$|\alpha-\frac{p}{q}|\geq \frac{1}{|a_n|q^n\prod^n_{j=2}|\alpha_j-p/q|}$$

I'm happy to provide more context, but I suspect it won't be neccesary. Feel free to ask for a sketch of the rest of the proof.

The notation is:

  • $\alpha$ is the real algebraic number of degree $>1$ we are trying to approximate.

  • $p/q$ is any rational number with $(p,q)=1$, $q>0$.

  • $\alpha_2,...,\alpha_n$ are the other roots of the minimal polynomial of $\alpha$.

$\endgroup$
  • $\begingroup$ What is $a_n\ $? $\endgroup$ – Jack M Jun 8 '16 at 14:29
2
$\begingroup$

(I rename $a_n$ as $c_n$)

the minimal polynomial of an algebraic number $\alpha = \alpha_1$ is $$P(x) = \sum_{k=0}^n c_k x^k = c_n \prod_{j=1}^n (x-\alpha_j)$$ where by definition of an algebraic number, $P \in \mathbb{Z}[x]$ : the coefficients $c_k \in \mathbb{Z}$, and is irreducible over $\mathbb{Q}[x]$ : the roots $\alpha_j \not \in \mathbb{Q}$.

for some rational number $p/q$ : $$P(p/q) = c_n \prod_{j=1}^n (p/q-\alpha_j)$$ and your inequality reduces to $$\frac{|P(p/q)|}{|c_n|} = \prod_{j=1}^n |p/q-\alpha_j| \ge \frac{q^{-n}}{|c_n|}$$

since $$q^n P(p/q) = \sum_{k=0}^n c_k p^k q^{n-k}$$ it reduces to $$|\sum_{k=0}^n c_k p^k q^{n-k}| \ge 1$$ which is obvious since the LHS is a non-zero integer

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.