Quadratic Integers over 2 form Subdomain of Reals

Theorem
Let $\Z \left[{\sqrt 2}\right]$ denote the set:
 * $\Z \left[{\sqrt 2}\right] := \left\{{a + b \sqrt 2: a, b \in \Z}\right\}$

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

Then the algebraic structure:
 * $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$

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

Proof
Clearly $\Z \left[{\sqrt 2}\right] \subseteq \R$ where $\R$ is the set of real numbers, so we immediately have that addition and multiplication are well-defined.

Closure
Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \left[{\sqrt 2}\right]$.

Then:

So both addition and multiplication are closed on $\Z \left[{\sqrt 2}\right]$.

Associativity
As addition and multiplication are associative on $\R$ it follows from Restriction of Operation Associativity that they are also associative on $\Z \left[{\sqrt 2}\right]$.

Commutativity
As addition and multiplication are commutative on $\R$ it follows from Restriction of Operation Commutativity that they are also commutative on $\Z \left[{\sqrt 2}\right]$.

Identities
We have:

and similarly for $\left({0 + 0 \sqrt 2}\right) + \left({a + b \sqrt 2}\right)$.

So $\left({0 + 0 \sqrt 2}\right)$ is the identity for $+$ on $\Z \left[{\sqrt 2}\right]$.

Then:

and similarly for $\left({a + b \sqrt 2}\right) \times \left({1 + 0 \sqrt 2}\right)$.

So $\left({1 + 0 \sqrt 2}\right)$ is the identity for $\times$ on $\Z \left[{\sqrt 2}\right]$.

Inverses
We have:

and similarly for $\left({-a + \left({-b}\right) \sqrt 2}\right) + \left({a + b \sqrt 2}\right)$.

So $\left({-a + \left({-b}\right) \sqrt 2}\right)$ is the inverse of $\left({a + b \sqrt 2}\right)$ for $+$ on $\Z \left[{\sqrt 2}\right]$.

We have no need to investigate inverses for $\times$ (which is convenient as $\times$ happens not to be closed for inverses on $\Z \left[{\sqrt 2}\right]$).

Distributivity
We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Z \left[{\sqrt 2}\right]$.

Divisors of Zero
By Field of Real Numbers, $\R$ is a field and so by Field is Integral Domain is itself an integral domain.

Hence it has no proper zero divisors.

The result follows by putting all the pieces together.