Liouville's Theorem (Number Theory)

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 $q \ne 0$.

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 $q \ne 0$.

Source of Name

This entry was named for Joseph Liouville.