By definition, $\Q \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers. Thus $+$ and $\times$ on $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ are well-defined. === Closure === Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Q \sqbrk {\sqrt 2}$. Then: {{begin-eqn}} {{eqn | l = \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}       | r = \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2       | c =  }} {{eqn | o = \in       | r = \Q \sqbrk {\sqrt 2}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}       | r = \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2       | c =  }} {{eqn | o = \in       | r = \Q \sqbrk {\sqrt 2}       | c =  }} {{end-eqn}} So both $+$ and $\times$ are closed on $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} === Associativity === As addition and multiplication are associative on $\R$ it follows from Restriction of Associative Operation is Associative that they are also associative on $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} === Commutativity === As addition and multiplication are commutative on $\R$ it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} === Identity === We have: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}       | r = \paren {a + 0} + \paren {b + 0} \sqrt 2       | c =  }} {{eqn | r = a + b \sqrt 2       | c =  }} {{end-eqn}} and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$. So $\paren {0 + 0 \sqrt 2}$ is the identity for $+$ on $\Q \sqbrk {\sqrt 2}$. Then: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}       | r = \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2       | c =  }} {{eqn | r = a + b \sqrt 2       | c =  }} {{end-eqn}} and similarly for $\paren {1 + 0 \sqrt 2} \times \paren {a + b \sqrt 2}$. So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} === Inverses === We have: {{begin-eqn}} {{eqn | l = \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}       | r = \paren {a - a} + \paren {b - b} \sqrt 2       | c =  }} {{eqn | r = 0 + 0 \sqrt 2       | c =  }} {{end-eqn}} 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 $\Q \sqbrk {\sqrt 2}$. Calculating the product inverse of $\paren {a + b \sqrt 2}$ is less trivial.  From Difference of Two Squares we have: :$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$ which leads to: :$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1 = 1 + 0 \sqrt 2$ so demonstrating that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} - \dfrac {b \sqrt 2} {a^2 - 2 b^2}$. As $a$ and $b$ are rational, it follows that so are $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$. So the product inverse of $\paren {a + b \sqrt 2}$ is an element of $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} === Distributivity === We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Q \sqbrk {\sqrt 2}$. {{qed|lemma}} The result follows by putting all the pieces together. {{qed}}