Q:

Prove that $\mathbb{Q}[\sqrt{p}]$ is a field between $\mathbb{Q}$ and $\mathbb{R}$

I really have trouble understanding a task. 
We've got  $p\in$ P, while P are all prime numbers. Now we construct a field $$\mathbb{Q}[\sqrt{p}]:=\{x+y\sqrt{p}:x,y \in \mathbb{Q}\}$$ The Task is to show that this is a field "in between" $\mathbb{Q}$ and $\mathbb{R}$. I was explained that this means we have to show:
$$\mathbb{Q} \subseteq \mathbb{Q}[\sqrt{p}] \subseteq \mathbb{R}$$
So that means, the field that we just constructed has to contain every rational number, plus  some (not all) of the real numbers. My question is: 
I found out that $\sqrt{p}$ is always an irrational number $\notin \mathbb{Q}$
If i multiply an irrational number by a a rational number, i will get an irrational number , right?
And if i add a rational number to an irrational number i still get a an irrational number as the result? So doesn't this field we constructed there just contain irrational numbers (This is probably not the case, but why)? 
And if i got that right, how can i show that this field contains all rational numbers? (After that i will also have to show that $\mathbb{Q}[\sqrt{p}]$ is a field, but i think i can do that by myself then.)
I am confused.

A:

By the way this is written, I assume that you have already shown that $\mathbb{Q}(\sqrt{p})$ is a field. Then consider $\alpha=x+y\sqrt{p}\in\mathbb{Q}(\sqrt{p})$ such that $y=0$, then this number is $\alpha=x\in\mathbb{Q}$ by definition of your proposed intermediary field, and thus it contains the rationals.
Showing that this is a subset of $\mathbb{R}$ consists of showing an arbitrary element is in the reals. $\mathbb{Q}\subset\mathbb{R}$, and all numbers of the form $x\sqrt{p}\in\mathbb{R}$, and since the reals are closed under addition, we know that any number in your intermediary field is in $\mathbb{R}$.

