Quadratic Integers over 2 form Ordered Integral Domain

Theorem

Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:

$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$

that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.

Then the algebraic structure:

$\struct {\Z \sqbrk {\sqrt 2}, +, \times}$

where $+$ and $\times$ are conventional addition and multiplication on real numbers, is an ordered integral domain.

Proof

We have that Quadratic Integers over 2 form Subdomain of Reals.

We also have that such numbers are real.

The result follows from Real Numbers form Ordered Integral Domain.

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$\blacksquare$

Sources

• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $10$