# Sprague's Property of Root 2

## 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.