Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$. 
Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$.


completely stuck on it. can I get some help please.
 A: Hint: Show that the (topological!) closure of $K$ is a subfield $L$ of $\mathbb{C}$.  Since $K$ contains $\mathbb{Q}$, $L$ contains $\mathbb{R}$, the topological closure of $\mathbb{Q}$.  Since $[\mathbb{C}:\mathbb{R}] = 2$ and $L$ is assumed to contain an element of $\mathbb{C} \setminus \mathbb{R}$, we must have $L = \mathbb{C}$.  
A: Since $K$ is not contained in $\mathbb{R}$, there's some $z \in K - \mathbb{R}$. Thus $1$ and $z$ form a basis of $\mathbb{C}$ as a $\mathbb{R}$-vector space. Now, show that if you take any basis of $\mathbb{C}$ as a $\mathbb{R}$-vector space, and consider the subspace of $\mathbb{C}$ as a $\mathbb{Q}$-vector space, then this subspace will be dense in $\mathbb{C}$.
A: $K$ must contain $1$, thus also $\mathbb{N}$ by addition, and every $\frac{1}{n}$ for $n$ in $\mathbb{N}$ by taking the inverse.   
So $\mathbb{Q}$ is contained in $K$.
$K$ contains a complex(non-real) number $x$, so it contains $\mathbb{Q}+x\mathbb{Q}$, which is dense in $\mathbb{C}$.
Proof of density: let $z\in\mathbb{C}$. ($1$,$x$) is a base of $\mathbb{C}$ seen as a real vector space, so $z=a+bx$ with $a$ and $b$ real numbers. $\mathbb{Q}$ being dense in $\mathbb{R}$, you have $a_n$ and $b_n$ in $\mathbb{Q}$ which converge toward $a$ and $b$ respectively.
So $a_n+b_n x$ lives in $\mathbb{Q}+x\mathbb{Q}$ and converges toward $z$.
A: Hint: Recall that $\Bbb Q$ is dense in $\Bbb R$. If $K$ is not a subfield of $\Bbb R$ then it contains some element $x\notin\Bbb R$ such that $x^2\in\Bbb R$. Show that $K$ contains something which looks a bit like $\Bbb Q[i]$.
