The real numbers are a field extension of the rationals?

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?

Solution 1:

Saying "the reals are an extension of the rationals" just means that the reals form a field, which contains the rationals as a subfield. This does not mean that the reals have the form $\mathbb{Q}(\alpha)$ for some $\alpha$; indeed, they do not. You have to adjoin uncountably many elements to the rationals to get the reals.

Solution 2:

For $\mathbb{R}$ to be field extension of $\mathbb{Q}$, all we need is that $\mathbb{R}$ is a field containing $\mathbb{Q}$ as a subfield. That's definitely true.

The construction is a bit more delicate and analytic in nature: $\mathbb{R}$ is the completion of $\mathbb{Q}$ and is substantially larger. Since $\mathbb{R}$ is uncountable, it's not an extension of finite degree, meaning that you cannot write $\mathbb{R} = \mathbb{Q}(a_1, a_2, ..., a_n)$ for some finite sequence of symbols $a_i$ (nor even a countable sequence). You have to adjoin uncountably many symbols.

If you're interested in a way to construct reals from rationals, take a look at Dedekind cuts.

Solution 3:

You are confusing field extension with an algebraic field extension. In the former case, you only need a non trivial field homomorphism from the rationals to the reals, which will be injective.

Solution 4:

The field extension is of a quite high degree. You need to adjoin infinitely many elements (more precisely continuum many). Nobody can give you an explicit list, at least not a non-redundant one.

A field $L$ being an extention of the field $K$ just means that $K \subset L$ and the operations on elements of $K$ are the same when considered in $K$ and in $L$.

So it really just says the rational are a subset of the reals, and it does not matter whether I add and multiply two rationals thinking of them as rationals or as reals.

Solution 5:

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* $\mathbb{C}_p$.