Rational Numbers with Denominator Power of Two form Integral Domain

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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:

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

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


Then:

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

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


Hence the result.

$\blacksquare$


Sources

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