Liouville's Theorem (Number Theory)

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x$ be an irrational number that is algebraic of degree $n$.

Then there exists a constant $c > 0$ (which can depend on $x$) such that:

$\size {x - \dfrac p q} \ge \dfrac c {q^n}$

for every pair $p, q \in \Z$ with $q \ne 0$.


Corollary

Liouville numbers are transcendental.


Proof

Let $r_1, r_2, \ldots, r_k$ be the rational roots of a polynomial $P$ of degree $n$ that has $x$ as a root.

Since $x$ is irrational, it does not equal any $r_i$.

Let $c_1 > 0$ be the minimum of $\size {x - r_i}$.

If there are no $r_i$, let $c_1 = 1$.

Now let $\alpha = \dfrac p q$ where $\alpha \notin \set {r_1, \ldots, r_k}$.

Then:

\(\ds \map P \alpha\) \(\ne\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \size {\map P \alpha}\) \(\ge\) \(\ds \frac 1 {q^n}\) as $\map P \alpha$ is a multiple of $\dfrac 1 {q^n}$
\(\ds \leadsto \ \ \) \(\ds \size {\map P x - \map P \alpha}\) \(\ge\) \(\ds \frac 1 {q^n}\) because $\map P x = 0$

Suppose:

$\ds \map P x = \sum_{k \mathop = 0}^n a_k x^k$

Then:

\(\ds \map P x - \map P \alpha\) \(=\) \(\ds \sum_{k \mathop = 0}^n a_k x^k - \sum_{k \mathop = 0}^n a_k \alpha^k\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n a_k \paren {x^k - \alpha^k}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k \paren {x^k - \alpha^k}\) $x^0 - \alpha^0 = 0$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n a_k \paren {x - \alpha} \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i\) Difference of Two Powers
\(\ds \) \(=\) \(\ds \paren {x - \alpha} \sum_{k \mathop = 1}^n a_k \sum_{i \mathop = 0}^{k - 1} x^{k - 1 - i} \alpha^i\)


Case 1: Let $\size {x - \alpha} \le 1$.

Then:

\(\ds \size \alpha - \size x\) \(\le\) \(\ds \size {x - \alpha}\) Reverse Triangle Inequality
\(\ds \leadsto \ \ \) \(\ds \size \alpha - \size x\) \(\le\) \(\ds 1\) Transitive
\(\ds \leadsto \ \ \) \(\ds \size \alpha\) \(\le\) \(\ds \size x + 1\) rearranging


Therefore:

\(\ds \size {\map P x - \map P \alpha}\) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \alpha^i}\) Triangle Inequality
\(\ds \) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1 - i} \paren {\size x + 1}^i}\) substituting $\size x + 1$ for $\alpha$
\(\ds \) \(\le\) \(\ds \size {x - \alpha}\sum_{k \mathop = 1}^n \size {a_k} \sum_{i \mathop = 0}^{k - 1} \size {x^{k - 1} \paren {\frac {\size x + 1} x}^i}\) moving $ x^{-i}$ over
\(\ds \) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^{k - 1} } \sum_{i \mathop = 0}^{k - 1} \size {\paren {1 + \frac 1 x}^i}\) moving $x^{k - 1}$ out of the sum
\(\ds \) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^{k - 1} } \frac {\paren {1 + \frac 1 x}^k - 1} {\paren {1 + \frac 1 x} - 1}\) Sum of Geometric Sequence
\(\ds \) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k x^k} \paren {\paren {1 + \frac 1 x}^k - 1}\)
\(\ds \) \(\le\) \(\ds \size {x - \alpha} \sum_{k \mathop = 1}^n \size {a_k} \paren {\paren {\size x + 1}^k - \size x^k}\) Distribute

To summarize:

$\size {\map P x - \map P \alpha} \le \size {x - \alpha} c_x$

where:

$\ds c_x = \sum_{k \mathop = 1}^n \size {a_k} \paren {\paren {\size x + 1}^k - \size x^k}$

So for such $\alpha$:

$\size {x - \alpha} \ge \dfrac {\size {\map P x - \map P \alpha} } {c_x} \ge \dfrac 1 {c_x q^n}$

$\Box$}


Case 2: Let $\size {x - \alpha} > 1$.

Then:

$\size {x - \alpha} > 1 \ge \dfrac 1 {q^n}$

$\blacksquare$


Examples

Example: $\sqrt 2$

Applying Liouville's Theorem (Number Theory) to $\sqrt 2$:

$\size {\sqrt 2 - \dfrac p q} \ge \dfrac c {q^2} \implies 0 \lt c \le 6 - 4 \sqrt 2$

for every pair $p, q \in \Z$ with $q \ne 0$.


Also see


Source of Name

This entry was named for Joseph Liouville.


Sources