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 $q \ne 0$.
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
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental