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Section B.17 Obtaining the Reals from the Rationals

A full discussion of this would take us far away from a discrete math class, but let's at least provide the basic definitions. A subset \(S\subset \rats\) of the rationals is called a cut (also, a Dedekind cut), if it satisfies the following properties:

  1. \(\emptyset\neq S\neq \rats\text{,}\) i.e, \(S\) is a proper non-empty subset of \(\rats\text{.}\)

  2. \(x\in S\) and \(y\lt x\) in \(\rats\) implies \(y\in S\text{,}\) for all \(x,y\in \rats\text{.}\)

  3. For every \(x\in S\text{,}\) there exists \(y\in S\) with \(x\lt y\text{,}\) i.e., \(S\) has no greatest element.

Cuts are also called real numbers, so a real number is a particular kind of set of rational numbers. For every rational number \(q\text{,}\) the set \(\bar{q}= \{p\in \rats: p\lt q\}\) is a cut. Such cuts are called rational cuts. Inside the reals, the rational cuts behave just like the rational numbers and via the map \(h(q)=\bar{q}\text{,}\) we abuse notation again (we are getting used to this) and say that the rational numbers are a subset of the real numbers.

But there are cuts which are not rational. Here is one: \(\{p\in \rats: p\le 0\}\cup \{p\in \rats: p^2\lt 2\}\text{.}\) The fact that this cut is not rational depends on the familiar proof that there is no rational \(q\) for which \(q^2=2\text{.}\)

The operation of addition on cuts is defined in the natural way. If \(S\) and \(T\) are cuts, set \(S+T=\{s+t:s\in S, t\in T\}\text{.}\) Order on cuts is defined in terms of inclusion, i.e., \(S\lt T\) if and only if \(S\subsetneq T\text{.}\) A cut is positive if it is greater than \(\bar{0}\text{.}\) When \(S\) and \(T\) are positive cuts, the product \(ST\) is defined by

\begin{equation*} ST= \bar{0}\cup\{st:s\in S, t\in T, s\ge0, t\ge 0\}. \end{equation*}

One can easily show that there is a real number \(r\) so that \(r^2=\bar{2}\text{.}\) You may be surprised, but perhaps not, to learn that this real number is denoted \(\sqrt2\text{.}\)

There are many other wonders to this story, but enough for one day.