Hurwitz's Theorem (Number Theory)/Lemma 1
Lemma
Let $\xi$ be an irrational number.
Let $A \in \R$ be a real number strictly greater than $\sqrt 5$.
Then there may exist at most a finite number of relatively prime integers $p, q \in \Z$ such that:
- $\size {\xi - \dfrac p q} < \dfrac 1 {A \, q^2}$
Proof
We will take as our example of such an irrational number:
- $\xi = \dfrac {\sqrt 5 - 1} 2$
This is equal to $\phi - 1$, where $\phi$ is the Golden mean.
Aiming for a contradiction, suppose that there exist an infinite number of $p, q$ with $p \perp q$ such that:
- $\size {\xi - \dfrac p q} < \dfrac 1 {A \, q^2}$
Then there exist an infinite number of $p, q$ with $p \perp q$ such that:
- $\xi = \dfrac p q + \dfrac \delta {q^2}$
where:
- $A > \sqrt 5$
Therefore:
- $\dfrac 1 A < \dfrac 1 {\sqrt 5}$
and:
- $0 < \size \delta < \dfrac 1 A < \dfrac 1 {\sqrt 5}$
Hence:
\(\ds \dfrac \delta {q^2}\) | \(=\) | \(\ds \xi - \dfrac p q\) | ||||||||||||
\(\ds \dfrac \delta q\) | \(=\) | \(\ds q \xi - p\) | multiplying through by $q$ | |||||||||||
\(\ds \dfrac \delta q\) | \(=\) | \(\ds q \paren{\dfrac {\sqrt 5 - 1} 2 } - p\) | substituting $\xi = \dfrac {\sqrt 5 - 1} 2$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac \delta q - \dfrac {q \sqrt 5} 2\) | \(=\) | \(\ds -\dfrac q 2 - p\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren{\dfrac \delta q - \dfrac {q \sqrt 5} 2 }^2\) | \(=\) | \(\ds \paren {-\dfrac q 2 - p}^2\) | squaring both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\delta^2} {q^2} - \delta \sqrt 5 + \dfrac {5 q^2} 4\) | \(=\) | \(\ds \dfrac {q^2} 4 + pq + p^2\) | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac {\delta^2} {q^2} - \delta \sqrt 5\) | \(=\) | \(\ds p^2 + p q - q^2\) |
Over the interval ${-\dfrac 1 {\sqrt 5} } < \delta < \dfrac 1 {\sqrt 5}$, the left hand side of $(1)$ takes on values:
- $\dfrac 1 {5 q^2} - 1 < \dfrac {\delta^2} {q^2} - \delta \sqrt 5 < \dfrac 1 {5 q^2} + 1$
At the same time, the right hand side of $(1)$ is always an integer.
Thus, for the equality to hold infinitely many times, it must hold at zero:
\(\ds p^2 + p q - q^2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds 4p^2 + 4p q - 4q^2\) | \(=\) | \(\ds 0\) | multiplying through by $4$ | |||||||||||
\(\ds 4p^2 + 4p q\) | \(=\) | \(\ds 4q^2\) | ||||||||||||
\(\ds 4p^2 + 4p q + q^2\) | \(=\) | \(\ds 4q^2 + q^2\) | adding $q^2$ to both sides - Completing the Square | |||||||||||
\(\ds \paren {2 p + q}^2\) | \(=\) | \(\ds 5 q^2\) | ||||||||||||
\(\ds \paren {2 p + q}\) | \(=\) | \(\ds \pm \sqrt 5 q\) | taking square roots of both sides | |||||||||||
\(\ds \paren {2 \dfrac p q + 1}\) | \(=\) | \(\ds \pm \sqrt 5\) | dividing through by $q$ | |||||||||||
\(\ds \dfrac p q\) | \(=\) | \(\ds \dfrac {-1 \pm \sqrt 5} 2\) | isolating the rational number $\dfrac p q$ |
Hence $\dfrac {-1 \pm \sqrt 5} 2$, is a rational number.
Hence $\sqrt 5$ is also rational.
But by Square Root of Prime is Irrational, $\sqrt 5$ is irrational.
This is a contradiction.
Hence by Proof by Contradiction there cannot be an infinite number of such $p, q$.
Hence the result.
$\blacksquare$
Source of Name
This entry was named for Adolf Hurwitz.
Sources
- 1979: G.H. Hardy and E.M. Wright: An Introduction to the Theory of Numbers (5th ed.): $11.8$: The measure of the closest approximation to an arbitrary irrational: Theorem $194$