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:

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

Then:

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

Hence the result.