Hurwitz's Theorem (Number Theory)/Lemma 1

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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

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