# 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.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Exercises: Chapter $1$: Exercise $1 \ \text{(iii)}$