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 (for$$nx \leqslant y$$ ). 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 that $$n > x$$.
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 the 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 and 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 is sometimes very helpful in investigating the enumerability of certain sets.