<s>[INST] 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 $A$.
 [/INST] Hints:
After realizing that $\;0=0+0\cdot\sqrt2\;,\;\;1=1+0\cdot\sqrt2\;$, we see all the axioms of a field that are inherited to subsets are fulfilled in $\;A\;$ since the operations used are exactly the same as the ones in $\;\Bbb R\;$ .
The only thing thus that is left to show is closedness of operations and existence of multiplicative inverse. For the first ones you already were commented enough, as as for the last one think of
$$a,b\in\Bbb Q\;,\;\;a\neq 0\;\;or\;\;b\neq 0\implies\frac1{a+b\sqrt2}=\frac1{a+b\sqrt2}\cdot\frac{a-b\sqrt2}{a-b\sqrt2}$$
and now do the maths in the rightmost expression and use the fact the rationals is a field.
</s>