Theorem: If, then for any  there exist  such that .

Proof: When, the theorem is evident. For . Let the theorem be false, so that . Thus,  is a non-empty set bounded above (for ). Therefore, by completeness axiom  has the supremum. Let . Then . Thus,  is also an upper bound of. But . This contradicts that. Hence, the assumption  is false, and so the statement of the theorem is true.

Corollary 1: For any such that.
It follows from the theorem, on replacing  by  and taking  for.

Corollary 2: The set is bounded below but unbounded above.

Corollary 3: For any  there exist  such that.

Corollary 4: For any  there exist  such that.

Corollary 5: For any  there exist unique  such that.

Proof: By completeness axiom the set  being bounded above has the suprema, say . Thus, , i.e. , where . Since is a suprema therefore, it is unique.
            The above integer is usually denoted by and is called the integral part of the number.

Corollary 6: For any  there exist unique  such that.

Corollary 7: For any  there exist unique  such that.

Example: For any  there exist unique  such that.
Solution: For real number  by corollary 5, there exist unique  such that 
                                   
Or                            
Or                              
i.e.                            
            The above example illustrates that every positive integer can be uniquely expressed as , for unique .
            Such unique representation of natural members are sometimes very helpful in investigating the enumerability of certain sets.