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Show that $\mathbb{F}(a,b) = \mathbb{F}(b,a)$

This question is presented to me like this. I suppose $\mathbb{F}$ is a field. As I understand, $\mathbb{F}(p)$, if we take for example the field $\mathbb{Q}$ and $p = \sqrt{2}$, should be the intersection of all fields that contain $\mathbb{Q}$ and $\sqrt{2}$ and are contained in $\mathbb{R}$. Am I right? But where did $\mathbb{R}$ came from? I don't have idea. Maybe because $\sqrt{2}\in\mathbb{R}$, but I don't know for sure. Also, there's no specification for what $a,b$ are, so I'm even more confused. The only thing I can say is that I'm studying Field Theory.

If we interpret $\mathbb{F}(a,b)$ as being the intersection of all fields that contain $\mathbb{F}$, contain $a,b$, and are contained in something, then it should have no difference from $\mathbb{F}(b, a)$

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    $\begingroup$ If this is literally all you are given, it is an incredibly poorly written question. $\endgroup$ – Eric Wofsey Sep 17 '17 at 0:34
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Lacking any more context, here is what I would guess is the intended question:

Let $\mathbb{F}$ be a field and let $K$ be an extension field of $\mathbb{F}$. Prove that if $a,b\in K$, then $\mathbb{F}(a)(b)=\mathbb{F}(b)(a)$. Here $\mathbb{F}(a)$ denotes the smallest subfield of $K$ containing $\mathbb{F}$ and $a$, and then $\mathbb{F}(a)(b)$ denotes the smallest subfield of $K$ containing $\mathbb{F}(a)$ and $b$, and similarly for $\mathbb{F}(b)(a)$.

The other interpretation would be, as you suggest, that $\mathbb{F}(a,b)$ is the smallest subfield of $K$ containing $\mathbb{F}$, $a$, and $b$, but as you note, that interpretation makes the problem excessively trivial. (The definition of $\mathbb{F}(a)(b)$ I gave above is equivalent, but with my definition it at least takes a little effort to verify that $\mathbb{F}(a)(b)$ and $\mathbb{F}(b)(a)$ are the same thing.)

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