** Theorem:** If and are two non-empty subsets of such that (i) , (ii) , then either has the greatest member or has the least member.

** Proof:** By (ii), the non-empty set is bounded above. If has the greatest member, it establishes the theorem. If has no greatest member, then by completeness axiom, in , being the set of upper bounds of it has a least member. Hence the theorem is proved.

** Theorem:** Dedekind’s Property is equivalent to completeness axiom in .

** Proof:** In the previous theorem it is shown that the completeness axiom in implies Dedekind’s property. It remains to show that the Dedekind’s property implies the completeness axiom.

Let be a non-empty set bounded above and sets and be defined by

It evidently follows that and are non-empty, disjoint and (i) , (ii) . Therefore, by Dedekind’s property either has the greatest member or has the least member. If possible, suppose has then greatest member, say. Then, such that , since is an upper bound of . This contradiction leads to the fact that has no greatest member. And so, has the least member. Hence, the set of upper bounds of a non-empty set bounded above has a least member, which is the completeness axiom in . Hence the theorem is proved.

** Theorem:** Between any two distinct real numbers there exist infinitely many rational numbers.

** Theorem:** Between any two distinct real numbers there exist infinitely many irrational numbers.