Quadratic Integers over 2 form Subdomain of Reals/Proof 1

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

Closure
Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.

Then:

So both the operations $+$ and $\times$ are closed on $\Z \sqbrk {\sqrt 2}$.

Associativity
We have that addition and multiplication are associative on $\R$.

Therefore it follows from Restriction of Associative Operation is Associative that they are also associative on $\Z \sqbrk {\sqrt 2}$.

Commutativity
We have that addition and multiplication are commutative on $\R$.

Therefore it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Z \sqbrk {\sqrt 2}$.

Identities
We have:

and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$.

So $\paren {0 + 0 \sqrt 2}$ is the identity for $+$ on $\Z \sqbrk {\sqrt 2}$.

Then:

and similarly for $\paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}$.

So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Z \sqbrk {\sqrt 2}$.

Inverses
We have:

and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$.

So $\paren {-a + \paren {-b} \sqrt 2}$ is the inverse of $\paren {a + b \sqrt 2}$ for $+$ on $\Z \sqbrk {\sqrt 2}$.

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

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

Divisors of Zero
By Real Numbers form Field, $\R$ 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.