Why is $K:=\{x+\sqrt{2}y : x,y\in \mathbb{Q}\}$ a field? 
Why is $K:=\{x+\sqrt{2}y : x,y\in \mathbb{Q}\}$ a field?

I know that if $K$ is a sub-field of a field (in this case it should be $\mathbb{R}$?), it is a field itself.
Firstly, have to prove that if $a,b\in K \Longrightarrow a+b\in K$ $\land$ $a\cdot b \in K$
I am having trouble figuring out what $a$ and $b$ are since $x+\sqrt{2}y$ is an addition... any suggestions?
 A: If
$K = \{x + y\sqrt 2, \; x, y \in \Bbb Q \} \tag 1$
and
$a \in K, \tag 2$
then there exist
$a_1, a_2 \in \Bbb Q \tag 3$
such that
$a = a_1 + a_2\sqrt 2; \tag 4$
likewise,
$b \in K \tag 5$
takes the form
$b = b_1 + b_2\sqrt 2, \; b_1, b_2 \in \Bbb Q; \tag 6$
then
$a + b = (a_1 + a_2\sqrt 2) + (b_1 + b_2\sqrt 2) = (a_1 + b_1) + (a_2 + b_2)\sqrt 2 \in K \tag 7$
since
$a_1 + b_1, a_2 + b_2 \in \Bbb Q; \tag 8$
also,
$ab = (a_1 + a_2\sqrt 2)(b_1 + b_2\sqrt 2) = a_1b_1 + a_1b_2\sqrt 2 + a_2b_1\sqrt 2 + a_2b_2(\sqrt 2)^2$
$= a_1b_1 + a_1b_2\sqrt 2 + a_2b_1\sqrt 2 + 2a_2b_2 = (a_1b_1 + 2a_2b_2) + (a_1b_2+ a_2b_1)\sqrt 2 \in K, \tag 9$
since again
$a_1b_1 + 2a_2b_2, a_1b_2+ a_2b_1 \in \Bbb Q, \tag{10}$
valid since $\Bbb Q$ is itself a field, hence closed under addition and multiplication.  Clearly
$0 = 0 + 0(-\sqrt 2) \in K, \tag{11}$
and
$a = a_1 + a_2\sqrt K \Longleftrightarrow -a = -a_1 - a_2\sqrt 2 \in K; \tag{12}$
(11) and (12) affirm that $K$ contains an additive identity element and that every element of $K$ is possessed of an additive inverse, also in $K$.  Furthermore,
$1 = 1 + 0\sqrt 2 \in K, \tag{13}$
i,e,, $K$ contains a multipliciative identity as well.
So $(K, +)$ is an abelian group, also closed under multiplication, with multiplicative identity $1$.  Commutativity, associativity, and distributivity of these operations hold in $K$ by virtue of the fact they are inherited from $\Bbb R$.
$K \subsetneq \Bbb R \tag{14}$
is a sub-ring.  $K$ is thus itself a commutative unital ring, and we only need show every $a \in K$ has a multiplicative inverse to prove $K$ is in fact a field.  To this end we observe that for any
$0 \ne a = a_1 + a_2\sqrt 2 \in K, \tag{15}$
$a_1^2 - 2a_2^2 \ne 0; \tag{16}$
otherwise
$a_1^2 = 2a_2^2, \tag{17}$
whence
$2 = \dfrac{a_1^2}{a_2^2}; \tag{18}$
but this is impossible since it yields
$\sqrt 2 = \dfrac{\vert a_1 \vert}{\vert a_2 \vert}, \tag{19}$
which asserts $\sqrt 2 \in \Bbb Q$; therefore (16) binds and we may write
$(a_1 + a_2\sqrt 2)\dfrac{a_1 - a_2\sqrt 2}{a_1^2 - 2a_2^2} = \dfrac{a_1^2 - 2a_2^2}{a_1^2 - 2a_2^2} = 1, \tag{20}$
so that
$(a_1 + a_2\sqrt 2)^{-1} = \dfrac{a_1 - a_2\sqrt 2}{a_1^2 - 2a_2^2}; \tag{21}$
we now may conclude that $K$ is indeed a field, since $(K, \cdot)$ forms an abelian group.
A: Here is  a fast  way to prove  all non-zero elements of $K$ have multiplicative inverses in $K$, without computations:
Observe $K$ is a two-dimensional $\mathbf Q$-vector  space and for any non-zero $x\in K$, consider the map
\begin{align}
m_x:K& \longrightarrow K,\\
y&\longmapsto xy.
\end{align}
This is a $\mathbf Q$-linear map, and as we are in a subring of the field $\mathbf C$, this linear map is injective. However, in a finite dimensional vector space,  injectivity of an endomorphism is equivalent to its surjectivity. Thus $1$ is attained, i.e. there exists $y\in K$ such that $xy=1$ thereby proving the existence of multiplicative inverses.
A: Think about $\mathbb{C}$ here.
You note a complex number $z\in\mathbb{C}$ as $z=a+ib$.
An element $k\in K$ can be written as $k=x+\sqrt{2}y$
So to verify, that it is closed under addition, would go like this:
Take $k,k'\in K$. Then $k=x+\sqrt{2}y$ and $k'=x'+\sqrt{2}y'$.
Now $k+k'=x+\sqrt{2}y+x'+\sqrt{2}y'=(x+x')+\sqrt{2}(y+y')$. 
So it is of the form which is required to be an element of $K$. 
Note that $x+x'$ and $y+y'$ are elements of $\mathbb{Q}$.
Similar you can show, that $k\cdot k'$ is an element of $K$.
The only harder part is to show, that $k\neq 0$ (so $x,y\neq 0$) has an inverse $k^{-1}$.
But if you make yourself clear what is to show, I think you can do it.
This might be worth looking at:
Show that $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2}\}$ is a vector space over $\mathbb{Q}.$
A: Well, first you can view the elements $x+\sqrt 2y$  of ${\Bbb Q}(\sqrt 2)$ as formal sums and prove that this set forms a field.
Second you can show that this field is isomorphic to the quotient ring ${\Bbb Q}[x]/\langle x^2-2\rangle$.
