Show that $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2}\}$ is a vector space over $\mathbb{Q}.$ 
Show that $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2}\} $ is a vector space over $\mathbb{Q}.$

I understand that posting questions without showing your work is looked down upon, however, speaking freely, I don’t know how to even begin the demonstration. I know how to demonstrate something is a vector space, I don’t know how to operate with $\mathbb{Q}[\sqrt{2}]$.
 A: An $F$-vectorspace is an abelian group equipped with an $F$-action i.e. it is a set $V$ with a binary operation $+$ such that $(V,+)$ is an abelian group and another operation $\cdot : F \times V \to V$ satisfying field action axioms. I am sure you've seen the definition of a vectorspace before, but I believe it is helpful to go back to basics. Does this set
$$\mathbb{Q} = \{a+b\sqrt{2} : a,b \in \mathbb{Q}\}$$
satisfy the axioms of a $\mathbb{Q}$-vectorspace? What is addition? What is scalar multiplication?
A: This might get you started:
We have a field $\mathbb{Q}$ with the usual interpretation of $\oplus$ and $\odot$.
I write $\oplus$ and $\odot$, which might look confusing, but I think it helps, because we actually want to seperate the usual $+$ and $\cdot$ you know from $(\mathbb{Q},+,\cdot)$ here. But, as it turns out, they are just the same.
Why is it still helpful? 
Because we want to know, where the structure comes from (laws of a associativity, commutativity, distributivity) and you already know, that these hold for the usual $+$ and $\cdot$.
First of all we have to show, that $\oplus:\mathbb{Q}[\sqrt{2}]\times\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{2}]$, $(x,y)\mapsto x+y$ is 'closed' under addition. So when you add two elements from $\mathbb{Q}[\sqrt{2}]$ then the sum is an element of $\mathbb{Q}[\sqrt{2}]$.

Note again, that $+$ is the sign you already know by heart.

So let $x,y\in\mathbb{Q}[\sqrt{2}]$. By definition of this set, it is $x=a+\sqrt{2}b$ and $y=c+\sqrt{2}d$, with $a,b,c,d\in\mathbb{Q}$.
Now we have to show, that $x+y\in\mathbb{Q}[\sqrt{2}]$.
That means it has to be $x+y=a'+\sqrt{2}b'$ for some $a',b'\in\mathbb{Q}$.
And finding them is a simple task, since
$\underbrace{\oplus(x,y)}_{\text{Again, this ist just notation}}=\underbrace{(a+\sqrt{2}b)+(c+\sqrt{2}d)}_{\text{by definition of the binary operation $\oplus$}}=\underbrace{(a+c)}_{=a'}+\sqrt{2}\underbrace{(b+d)}_{=b'}$
So indeed we have $x+y\in\mathbb{Q}[\sqrt{2}]$.

Note that we use for the last equality, that $+$ and $\cdot$ (where I used the common notation of $x\cdot y=xy$) already has these properties of associativity, commutativity and distributivity.

There are a bunch of axioms to show, all are pretty much really simple, if you just try.
Let me show one more:
We have to give a neutral element, with regards to $+$, for $(\mathbb{Q}[\sqrt{2}],\oplus,\odot)$ to be a vectorspace.
This is given by $0+0\sqrt{2}=0$
Because it is $\oplus(x,0)=(a+\sqrt{2}b)+(0+\sqrt{2}0)=(a+0)+\sqrt{2}(b+0)=a+\sqrt{2}b=x$
And $\oplus(0,x)=(0+\sqrt{2}0)+(a+\sqrt{2}b)=(0+a)+\sqrt{2}(0+b)=a+\sqrt{2}b=x$

Note, that $0$ already has this property in the field $(\mathbb{Q}, +,\cdot)$.

The other axioms can be shown this way.
I hope this little guide helps.
