## SectionB.17Obtaining 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.