Proving the following set of real numbers is a field Show that the following set $A$ of real numbers under addition and multipication is a field:
$A = {a + b\sqrt{2} : a,b \ \text{rational}}$
I am not sure if I am right but here is what I have thus far:
Closure: Let $a_{1} + b_{1}\sqrt{2}$, $a_{2} + b_{2}\sqrt{2}\in \mathbb{R}$. Then $(a_{1} + b_{1}\sqrt{2})+ (a_{2} + b_{2}\sqrt{2}) = (a_{1} + a_{2}) + (b_{1} + b_{2})\sqrt{2}\in A$.
Now, $(a_{1} + b_{1}\sqrt{2})\times (a_{2} + b_{2}\sqrt{2}) = (a_{1}\times a_{2} + 2\times b_{1}\times b_{2}) + (a_{1}\times b_{2} + b_{1}\times a_{2})\sqrt{2}\in A$
So both addition and multipication are closed on $A$
Associativity: As addition and multipication are associative on $\mathbb{R}$ it follows from Restriction of Operation Associativity that they are also associative on the set $A$.
Commutativity: As addition and multipication are commutative on $\mathbb{R}$ it follows from Restriction of Operation Commutativity that they are also commutative on the set $A$.
Identity: We have $(a + b\sqrt{2}) + (0 + 0\sqrt{2}) = (a + 0) + (b + 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(0 + 0\sqrt{2}) + (a + b\sqrt{2})$. So, $(0 + 0\sqrt{2})$ is the identity for addition on $A$. Now, for multipication we have $(a + b\sqrt{2})(1 + 0\sqrt{2}) = (a\times 1 + 2\times b\times 0) + (b\times 1 + a\times 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(1 + 0\sqrt{2})(a + b\sqrt{2})$. So, $(1 + 0\sqrt{2})$ is the identity for multipication on $A$.
Inverses: $(a + b\sqrt{2}) + (-a + (-b)\sqrt{2}) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2}$ and similarly for $(-a + (-b)\sqrt{2}) + (a + b\sqrt{2})$. So, $(-a + (-b)\sqrt{2})$ is the inverse of $(a + b\sqrt{2})$ for addition on $A$.\
For product inverse consider the difference of squares: $(a + b\sqrt{2})(a - b\sqrt{2}) = a^{2} - 2b^{2}$ which leads to $(a + \sqrt{2})(\frac{a - b\sqrt{2}}{a^{2} - 2b^{2}}) = 1 = 1 + 0\sqrt{2}$. So, demonstrating that the product inverse of $(a + b\sqrt{2})$ is $\frac{a}{a^{2} - 2b^{2}} - \frac{b\sqrt{2}}{a^{2} - 2b^{2}}$ as $a,b$ are rational, it follows that so are $\frac{a}{a^{2} - 2b^{2}}$ and $\frac{b}{a^{2} - 2b^{2}}$. So, the product inverse $(a + b\sqrt{2})\in A$
Distributivity: We have the Real Multipication Distributes over addition, so by Restriction of Operation Distrbutibity, multipication is distributive over addition on $A$.

Please can we use the fact that it is closed under addition, subtraction, multiplication and division to prove that it is a field. 
