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
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.