Real Numbers of Type Rational a plus b root 2 form Field

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

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

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

where $+$ and $\times$ are conventional addition and multiplication on real numbers, is a field.

Corollary
The field $\left({\Q \left[{\sqrt 2}\right], +, \times}\right)$ is a subfield of $\left({\R, +, \times}\right)$.

Proof of Main Result
Clearly $\Q \left[{\sqrt 2}\right] \subseteq \R$ where $\R$ is the set of real numbers, so we immediately have that $+$ and $\times$ on $\left({\Q \left[{\sqrt 2}\right], +, \times}\right)$ are well-defined.

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

Then:

So both $+$ and $\times$ are closed on $\Q \left[{\sqrt 2}\right]$.

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

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

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 $\Q \left[{\sqrt 2}\right]$.

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 $\Q \left[{\sqrt 2}\right]$.

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 $\Q \left[{\sqrt 2}\right]$.

Calculating the [Definition:Product Inverse|product inverse]] of $\left({a + b \sqrt 2}\right)$ is less trivial.

From Difference of Two Squares we have:
 * $\left({a + b \sqrt 2}\right) \left({a - b \sqrt 2}\right) = a^2 - 2b^2$

which leads to:
 * $\left({a + b \sqrt 2}\right) \left({\dfrac {a - b \sqrt 2} {a^2 - 2b^2}}\right) = 1 = 1 + 0 \sqrt 2$

so demonstrating that the product inverse of $\left({a + b \sqrt 2}\right)$ is $\dfrac a {a^2 - 2b^2} - \dfrac {b \sqrt 2} {a^2 - 2b^2}$.

As $a$ and $b$ are rational, it follows that so are $\dfrac a {a^2 - 2b^2}$ and $\dfrac b {a^2 - 2b^2}$.

So the product inverse of $\left({a + b \sqrt 2}\right)$ is an element of $\Q \left[{\sqrt 2}\right]$.

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

The result follows by putting all the pieces together.

Proof of Corollary
As stated in the proof of the main result, numbers of the form $a + b \sqrt 2$ are real.

So $\Q \left[{\sqrt 2}\right] \subseteq \R$.

Then we have from the main result that $\left({\Q \left[{\sqrt 2}\right], +, \times}\right)$ is a field.

Hence the result.