So i got this guided question that basically proves why liouville's constant is transcendental. Here is the Question: let $\alpha=\sum_{k=1}^\infty 10^{-k!}$ be liouville's constant, in this drill we will see that $\alpha$ is transcendental in $\mathbb{Q}$.
Prove that $\alpha$ is irrational.
Let $f=a_nx^n+...+a_0 \in \mathbb{Z}[x]$, and let $\frac{p}{q}$ be a rational number which is not a root of $f$. prove that $|f(\frac{p}{q})|\geq \frac{1}{q^n}$
- Now we assume that $\alpha$ is a root of $f$ and $\frac{p}{q}\in(\alpha-1,\alpha+1)$ and closer to $\alpha$ more that to any other root of $f$. Because $f$ is a polynomal that implies that exits a constant $M$ that for every $y\in(\alpha-1,\alpha+1)$, $|f^{'}(y)|\leq M$. prove that $|\frac{p}{q}-\alpha|\geq\frac{1}{Mq^n}$.
- Conclude that for every $r>n$ there exits only a finite number of rationals $\frac{p}{q}$ so: $|\frac{p}{q}-\alpha|<\frac{1}{q^r}$.
- Explain why this is a contradiction to the definition of $\alpha$ and conclude that $\alpha$ is transcendental.
I managed to solve the first three parts of the drill (in part 3 i used the mean-value theorem). My problem is the other two steps (4 and 5) , I would appreciate it if someone will help me with those last steps. Thanks for the helpers. In general, how would you prove that there are only a finite number of rationals in a case similar to this