Sprague's Property of Root 2

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Theorem

Let $S = \sequence {s_n}$ be the sequence of fractions defined as follows:

Let the numerator of $s_n$ be:

$\floor {n \sqrt 2}$

where $\floor x$ denotes the floor of $x$.

Let the denominators of the terms of $S$ be the (strictly) positive integers missing from the numerators of $S$:

$S := \dfrac 1 3, \dfrac 2 6, \dfrac 4 {10}, \dfrac 5 {13}, \dfrac 7 {17}, \dfrac 8 {20}, \ldots$


Then the difference between the numerator and denominator of $s_n$ is equal to $2 n$.


Proof

Denote the numerators of the terms of $S$ as $\sequence {N_n}$.

Denote the denominators of the terms of $S$ as $\sequence {D_n}$.

From the definition:

$\sequence {N_n}$ is a Beatty sequence, where $\sequence {N_n} = \BB_{\sqrt 2} = \sequence{\floor{n \sqrt 2} }_{n \mathop \in \Z_{> 0} }$
$\sequence {N_n}$ and $\sequence {D_n}$ are complementary Beatty sequences.

Then by Beatty's Theorem, $\sequence {D_n}$ is a Beatty sequence.

Define $\sequence {D_n} = \BB_y = \sequence{\floor{n y} }_{n \mathop \in \Z_{> 0} }$.

Then we have:

\(\ds \dfrac 1 {\sqrt 2} + \dfrac 1 y\) \(=\) \(\ds 1\) Beatty's Theorem
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 y\) \(=\) \(\ds 1 - \dfrac 1 {\sqrt 2}\)
\(\ds \) \(=\) \(\ds \dfrac {\sqrt 2 - 1} {\sqrt 2}\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \dfrac {\sqrt 2} {\sqrt 2 - 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\sqrt 2 \paren {\sqrt 2 + 1} } {\paren {\sqrt 2 - 1} \paren {\sqrt 2 + 1} }\)
\(\ds \) \(=\) \(\ds 2 + \sqrt 2\)

So we have $s_n = \dfrac {N_n} {D_n} = \dfrac {\floor {n \sqrt 2} } {\floor {n \paren {2 + \sqrt 2} } }$.

The difference between the numerator and denominator of $s_n$ is $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$.

We have:

\(\ds \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}\) \(=\) \(\ds \floor {2 n + n \sqrt 2} - \floor {n \sqrt 2}\)
\(\ds \) \(=\) \(\ds 2 n + \floor {n \sqrt 2} - \floor {n \sqrt 2}\) Floor of Number plus Integer
\(\ds \) \(=\) \(\ds 2 n\)

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Roland Percival Sprague.


Sources

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