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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.

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    $\begingroup$ "If i multiply an irrational number by a a rational number, i will get an irrational number , right?" yes, unless the rational number happens to be 0. $\endgroup$ – Gregory J. Puleo Nov 11 '14 at 22:29
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    $\begingroup$ Well your field contains all rational numbers because you can take $y=0$. $\endgroup$ – xavierm02 Nov 11 '14 at 22:29
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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 $\mathbb{R}$.

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  • $\begingroup$ Yes i got it now. Weird that i didn't consider an y=0, thank you. $\endgroup$ – Falco Winkler Nov 11 '14 at 22:34

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