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 $\alpha$.

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

(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


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