# Numbers of Type Integer a plus b root 3 form Integral Domain

## Theorem

Let $\R$ denote the set of real numbers.

Let $S \subseteq \R$ denote the set of set of real numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are integers:

- $S = \set {a + b \sqrt 3: a, b \in \Z}$

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

## Proof

From Real Numbers form Integral Domain we have that $\struct {\R, +, \times}$ 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 $\struct {\R, +, \times}$ is $1$.

Then we note:

- $1 = 1 + 0 \times \sqrt 3$

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 $\struct {\R, +, \times}$, 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 $x, y \in S$ such that:

- $x = a + b \sqrt 3$

- $y = c + d \sqrt 3$

Then:

\(\displaystyle x + \paren {-y}\) | \(=\) | \(\displaystyle \paren {a + b \sqrt 3} - \paren {c + d \sqrt 3}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a - c} + \paren {b \sqrt 3 - d \sqrt 3}\) | Definition of Real Addition | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a - c} + \paren {b - d} \sqrt 3\) | |||||||||||

\(\displaystyle \) | \(\in\) | \(\displaystyle S\) |

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

Then:

\(\displaystyle x \times y\) | \(=\) | \(\displaystyle \paren {a + b \sqrt 3} \paren {c + d \sqrt 3}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a c + a d \sqrt 3 + b c \sqrt 3 + 3 b d\) | Definition of Real Multiplication | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a c + 3 b d} + \paren {a d + b c} \sqrt 3\) | |||||||||||

\(\displaystyle \) | \(\in\) | \(\displaystyle S\) |

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

Hence the result.

$\blacksquare$

## Sources

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