# Rational Numbers with Denominator Power of Two form Integral Domain

## Theorem

Let $\Q$ denote the set of rational numbers.

Let $S \subseteq \Q$ denote the set of set of rational numbers of the form $\dfrac p q$ where $q$ is a power of $2$:

$S = \set {\dfrac p q: p \in \Z, q \in \set {2^m: m \in \Z_{\ge 0} } }$

Then $\struct {S, +, \times}$ is an integral domain.

## Proof

From Rational Numbers form Integral Domain we have that $\struct {\Q, +, \times}$ is an integral domain.

Hence to demonstrate that $\struct {S, +, \times}$ is an integral domain, we can use the Subdomain Test.

We have that the unity of $\struct {\Q, +, \times}$ is $1$.

Then we note:

$1 = \dfrac 1 1$

and:

$1 = 2^0$

and so $1 \in S$.

Thus property $(2)$ of the Subdomain Test is fulfilled.

It remains to demonstrate that $\struct {S, +, \times}$ is a subring of $\struct {\Q, +, \times}$, so fulfilling property $(2)$ of the Subdomain Test.

Hence we use the Subring Test.

We note that $S \ne \O$ as $1 \in S$.

This fulfils property $(1)$ of the Subring Test.

Let $x, y \in S$.

Then:

 $\displaystyle x + \paren {-y}$ $=$ $\displaystyle \dfrac a {2^p} + \dfrac b {2^q}$ for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$ $\displaystyle$ $=$ $\displaystyle \dfrac {a 2^q - b 2^p} {2^p 2^q}$ Definition of Rational Addition $\displaystyle$ $=$ $\displaystyle \dfrac {a 2^q - b 2^p} {2^{p + q} }$ $\displaystyle$ $\in$ $\displaystyle S$

This fulfils property $(2)$ of the Subring Test.

Then:

 $\displaystyle x \times y$ $=$ $\displaystyle \dfrac a {2^p} \times \dfrac b {2^q}$ for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$ $\displaystyle$ $=$ $\displaystyle \dfrac {a b} {2^p 2^q}$ Definition of Rational Multiplication $\displaystyle$ $=$ $\displaystyle \dfrac {a b} {2^{p + q} }$ $\displaystyle$ $\in$ $\displaystyle S$

This fulfils property $(3)$ of the Subring Test.

Hence the result.

$\blacksquare$