If $a$ is a complex number s.t. $a\notin \mathbb R$, then $\mathbb R(a)=\mathbb C$? If $a$ is a complex number s.t. $a\notin \mathbb R$, then $\mathbb R(a)=\mathbb C$?
I'm asked to give a proof or a counterexample. I'm a bit confused on the notation of $\mathbb R(a)$, what does this mean exactly? I'm also unclear on what exactly the direction my proof neeeds to go in (assuming it is true). Though limited, my thoughts so far are as follows:
Since $a\notin \mathbb R$ then $a=x+iy$ where $y\ne 0$.
Elements in $\mathbb R$ are of the form $\{a+b\sqrt{2}|a,b\in \mathbb Q\}$.
Similarly, for $\mathbb C$, $\{a+b\sqrt{-2}|a,b\in \mathbb Q\}$.
Am I trying to show that plugging in some $a=x+yi$ in $\mathbb R$ that I can write it in the form of $\{a+b\sqrt{-2}|a,b\in \mathbb Q\}$? 

For $a \in \Bbb C \setminus \Bbb R$, we indeed have $\Bbb R(a) = \Bbb C$; indeed, we have $i \in \Bbb R(a)$; this may be seen as follows:
With such $a$, we have 
$0 < \bar a a \in \Bbb R \subset \Bbb R(a); \tag{1}$
since $\Bbb R(a)$ is a field and $a \ne 0$, we further have
$\bar a = a^{-1} (\bar a a) \in \Bbb R(a); \tag{2}$
thus
$a - \bar a \in \Bbb R(a); \tag{3}$
writing
$a = a_r + i a_i \tag{4}$
with $a_r, a_i \in \Bbb R$ and $a _i \ne 0$ (since $a \ne \Bbb R$),
we see that
$a - \bar a = a_r + i a_i - (a_r - i a_i)= 2 i a_i, \tag{5}$
whence via (3)
$2 i a_i \in \Bbb R(a), \tag{6}$
which yields
$i \in \Bbb R(a); \tag{7}$
thus for $x, y \in \Bbb R \subset \Bbb R(a)$,
$x + i y \in \Bbb R(a). \tag{8}$
(8) clearly shows that $\Bbb C \subset \Bbb R(a)$; $\Bbb R(a) \subset \Bbb C$ is obvious; thus $\Bbb R(a) = \Bbb C$.  QED.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
