TITLE: The real numbers are a field extension of the rationals?
QUESTION [13 upvotes]: In preparing for an upcoming course in field theory I am reading a Wikipedia article on field extensions.  It states that the complex numbers are a field extension of the reals.  I understand this since $\mathbb R(i) = \{ a + bi : a,b \in \mathbb R\}$.
Then the article states that the reals are a field extension of the rationals.  I do not understand how this could be.  What would you adjoin to $\mathbb Q$ to get all the reals?  The article doesn't seem to say anything more about this.  Is there a way to explain this to someone who has yet to take a course in field theory?

REPLY [2 votes]: Several people here have already noted that, while $\mathbb{Q}$ is a subfield of $\mathbb{R}$, this fact isn't "caused" by the same explanation as $\mathbb{R}$ being a subfield of $\mathbb{C}$. Pithily, $\mathbb{C}$ is an algebraically complete algebraic extension of $\mathbb{R}$ while $\mathbb{R}$ is a metric-complete metric extension of $\mathbb{Q}$.
If $N$ is a norm on a ring $R$, we can define Cauchy sequences in $R$ with respect to $N$ as those $(x_n)$ with $\forall \delta \in \mathbb{R}^+ \exists N\in\mathbb{N} \forall m,\,n \in \mathbb{N} (m,\,n > N \to N(x_m-x_n)<\delta)$. We can also define null sequences in $R$ with respect to $N$, viz. $\forall \delta \in \mathbb{R}^+ \exists N\in\mathbb{N} \forall n \in \mathbb{N} (n > N \to N(x_n)<\delta)$. We call Cauchy sequences $(x_n),\,(y_n)$ equivalent if $(x_n-y_n)$ is a null sequence. We can think of equivalent Cauchy sequences as "having the same limit", even if that limit does not exist in $R$. We say $R$ is metric-complete if it contains its Cauchy sequences' limits; for the choice $N(x)=|x|$, $\mathbb{R}$ is metric-complete but $\mathbb{Q}$ does not. Indeed, just as we may identify $\mathbb{C}=\mathbb{R}[i]$ with $i^2=-1$, we can identify $\mathbb{R}$ with the set of equivalence classes on $\mathbb{Q}$ with $N(x)=|x|$. Each real number is one such equivalence class. For example, $\sqrt{2}$ is the set of Cauchy sequences $(x_n)$ in $\mathbb{Q}$ for which $x_n^2 \to 2$.
If you want something to which to compare the $\mathbb{Q}$-to-$\mathbb{R}$ extension, you can consider the $p$-adic numbers. These are obtained the same way, but with a different choice of $N$. The trivial norm $$N\left(x\right)=\left\{ \begin{array}{cc}
0 & x=0\\
1 & x\neq0
\end{array}\right.$$ obtains only eventually constant Cauchy sequences, which are equivalent iff their eventually constant values match, so $\mathbb{Q}$ is metric-complete with respect to this choice. It can be shown the only other norms on $\mathbb{Q}$ are the $p$-adic norms; for fixed $p\in\mathbb{P}$ define $$\text{ord}_p x=\inf {k\in\mathbb{Z}|p^k x\in\mathbb{Z}},\,N(x)=p^{-\text{ord}_p x}.$$Now we get different Cauchy sequences and different equivalence classes of them (as we have different null sequences), and the $p$-adic numbers $\mathbb{Q}_p$ differ from the reals (as well as the $q$-adic numbers for $q\in\mathbb{P}$ with $p\neq q$). 
All metric completions of $\mathbb{Q}$ are also field extensions. They can all be algebraically extended by introducing an imaginary unit; there are also complex $p$*-adic numbers* $\mathbb{C}_p$.