We all know about rational numbers. Informally, these are ratios of integers, i.e., numbers of the form where and are integers (with of course). Any two rational numbers have a sum a product and there is an order relation such that or These operations and the relation obey certain properties, which makes the set of rational numbers an ordered field.
Using this system we can solve most of the elementary problems in mathematics and related disciplines. But there are issues this system can’t address. One such problem is the classical one which the Greeks encountered. Measure the length of the hypotenuse of a right-angled triangle whose other two sides are of unit length. Taking as the required length, the Pythagorean theorem gives us Now a simple and well-known argument shows that Let us place the hypotenuse on the -axis in such a way that one end coincides with the origin and the other end lies on the positive -axis. Then does not correspond to a rational number; it corresponds to something else.
So we see that the rational numbers are not sufficient for the purpose of measuring lengths of straight line segments; we need new numbers for this. As you have probably already guessed, these new numbers are the irrational numbers. Together with the rational numbers they form the set of real numbers. The operations and the relation on can be extended to and the resulting system is said to be the system of real numbers. We shall construct this system in two different ways: by Dedekind cuts, and by Cauchy sequences (to be disussed in a subsequent post).
We shall now construct the set of real numbers using what are called Dedekind Cuts. This construction is named after the German mathematician Richard Dedekind. We see that the point does not correspond to a rational number. This means each rational number either lies on the left of or on the right of “cuts” the set of rational numbers into two halves: the set of those rational numbers that lie on the left of and the set of those rational numbers that lie on the right of . The partition of is said to be a Dedekind cut.
The set satisfies the following properties.
- (1) is a nonempty proper subset of
- (2) If and such that then
- (3) For every there exists such that
The second property says that every rational number less than some element of is also an element of The third property says that has no largest element. We let denote the set of those subsets of that satisfy properties (1), (2) and (3). Observe that each element corresponds to the Dedekind cut and conversely. Since we can intuitively see that each Dedekind cut corresponds to a unique point on a straight line and conversely, the elements in corresponds to the points on a straight line. Observe that every element satisfies for all in other words, every element of is an upper bound of But there is no least upper bound because has no smallest element.
We shall extend the operations and the relation from to to create a system of real numbers. Note that these extensions will really be extensions only if Strictly speaking, we don’t have because the element in are subsets of whereas the elements of are not. To get around this we identify with the set With this identification we have Now first we extend the order relation from to For we define iff (recall that are the subsets of ). Using the properties (1), (2) and (3) one can prove that the relation is reflexive, anti-symmetric and transitive, and that or Thus is a legitimate order relation on But why is it an extension of the relation on If then (because of our identification) A moment of reflection convinces one that iff Hence is an extension of There is one property that has which does not have: The least upper bound property. It says that every non-empty subset of that has some upper bound has a smallest upper bound. This removes the difficulty we had with and in the previous paragraph. It can be easily proved. Consider a non-empty subset Let have an upper bound This means for every and hence for every To complete the proof, i.e., to show that is the least upper bound of one has to prove that That is easy. This relatively easy proof is an advantage of this construction over the other construction involving Cauchy sequences.
Extension of the operations from to are less trivial and are done as follows. For we define
- If then ; and, if , or , if , or , if .
- If then ; if , then .
It can be proved that is an ordered field; because of the least upper bound property is a complete ordered field. Further, it has as a subsystem.