# 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.