Archimedean Property

Archimedean Property:

Theorem: If x > 0, then for any y \in \mathbb{R} there exist n \in \mathbb{N} such that nx > y.

Proof: When y \leqslant 0, the theorem is evident. For y > 0. Let the theorem be false, so that nx \leqslant y{\text{ }}\forall n \in \mathbb{N}. Thus, S = \left\{ {nx:n \in \mathbb{N}} \right\} is a non-empty set bounded above (fornx \leqslant y ). Therefore, by completeness axiom S has the supremum. Let \alpha = \sup S. Then nx \leqslant \alpha {\text{ }}\forall n \in \mathbb{N} \Rightarrow \left( {n + 1} \right)x \leqslant \alpha {\text{ }}\forall n \in \mathbb{N}  \Rightarrow nx \leqslant \alpha - x{\text{ }}\forall n \in \mathbb{N}. Thus, \alpha - x is also an upper bound of S. But \alpha - x < \alpha . This contradicts that \alpha = \sup S. Hence, the assumption nx \leqslant y{\text{ }}\forall n \in \mathbb{N} is false, and so the statement of the theorem is true.

Corollary 1: For any x \in \mathbb{R}{\text{ }}\exists {\text{ }}n \in \mathbb{N}such thatn > x.
It follows from the theorem, on replacing y by x and taking 1 for x.

Corollary 2: The set \mathbb{N}is bounded below but unbounded above.

Corollary 3: For any x \in {\mathbb{R}^ + } there exist n \in \mathbb{N} such that x > 1/n.

Corollary 4: For any x \in \mathbb{R} there exist n,m \in \mathbb{Z} such that n > x > m.

Corollary 5: For any x \in \mathbb{R} there exist unique n \in \mathbb{Z} such that n + 1 > x \geqslant n.

Proof: By completeness axiom the set \left\{ {m:m < x \wedge m \in \mathbb{Z}} \right\} being bounded above has the suprema, say n,n \in \mathbb{Z}. Thus, x \geqslant n \wedge n + 1 > x, i.e. n + 1 > x \geqslant n, where n \in \mathbb{Z}. Since n is a suprema therefore, it is unique.

The above integer n is usually denoted by \left[ x \right]and is called the integral part of the number x.

Corollary 6: For any x \in \mathbb{R} there exist unique n \in \mathbb{N} such that x \geqslant n > x - 1.

Corollary 7: For any x \in \mathbb{R} there exist unique n \in \mathbb{N} such that x > n \geqslant x - 1.

Example: For any x \in {\mathbb{R}^ + } there exist unique n \in \mathbb{N} such that \frac{{n\left( {n + 1} \right)}}{2} > x \geqslant \frac{{n\left( {n - 1} \right)}}{2}.
Solution: For real number \sqrt {\left( {2x + \frac{1}{4}} \right)} + \frac{1}{2} by corollary 5, there exist unique n \in \mathbb{N} such that

 n + 1 > \sqrt {\left( {2x + \frac{1}{4}} \right)} + \frac{1}{2} \geqslant n

Or

 {\left( {n + \frac{1}{2}} \right)^2} > 2x + \frac{1}{4} \geqslant {\left( {n - \frac{1}{2}} \right)^2}

Or

{n^2} + n > 2x \geqslant {n^2} - n

i.e.

\frac{{n\left( {n + 1} \right)}}{2} > x \geqslant \frac{{n\left( {n - 1} \right)}}{2}

The above example illustrates that every positive integer n can be uniquely expressed as n = \frac{{i\left( {i - 1} \right)}}{2} + j, for unique i,j \in \mathbb{N} \wedge 1 \leqslant j \leqslant i.

Such unique representation of natural members are sometimes very helpful in investigating the enumerability of certain sets.