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Archimedean property (Theorem)

Let $ x$ be any real number. Then there exists a natural number $ n$ such that $ n > x$.

This theorem is known as the Archimedean property of real numbers. It is also sometimes called the axiom of Archimedes, although this name is doubly deceptive: it is neither an axiom (it is rather a consequence of the least upper bound property) nor attributed to Archimedes (in fact, Archimedes credits it to Eudoxus).

Proof. Let $ x$ be a real number, and let $ S = \{ a \in \mathbb{N} : a \leq x \}$. If $ S$ is empty, let $ n = 1$; note that $ x < n$ (otherwise $ 1 \in S$).

Assume $ S$ is nonempty. Since $ S$ has an upper bound, $ S$ must have a least upper bound; call it $ b$. Now consider $ b - 1$. Since $ b$ is the least upper bound, $ b - 1$ cannot be an upper bound of $ S$; therefore, there exists some $ y \in S$ such that $ y > b - 1$. Let $ n = y + 1$; then $ n > b$. But $ y$ is a natural, so $ n$ must also be a natural. Since $ n > b$, we know $ n \not\in S$; since $ n \not\in S$, we know $ n > x$. Thus we have a natural greater than $ x$. $ \qedsymbol$

Corollary 1   If $ x$ and $ y$ are real numbers with $ x > 0$, there exists a natural $ n$ such that $ nx > y$.
Proof. Since $ x$ and $ y$ are reals, and $ x \neq 0$, $ y/x$ is a real. By the Archimedean property, we can choose an $ n \in \mathbb{N}$ such that $ n > y/x$. Then $ nx > y$. $ \qedsymbol$
Corollary 2   If $ w$ is a real number greater than 0, there exists a natural $ n$ such that $ 0 < 1/n < w$.
Proof. Using Corollary 1, choose $ n \in \mathbb{N}$ satisfying $ nw > 1$. Then $ 0 < 1/n < w$. $ \qedsymbol$
Corollary 3   If $ x$ and $ y$ are real numbers with $ x < y$, there exists a rational number $ a$ such that $ x < a < y$.
Proof. First examine the case where $ 0 \leq x$. Using Corollary 2, find a natural $ n$ satisfying $ 0 < 1/n < (y-x)$. Let $ S = \{ m \in \mathbb{N} : m/n \geq y \}$. By Corollary 1 $ S$ is non-empty, so let $ m_0$ be the least element of $ S$ and let $ a = (m_0-1)/n$. Then $ a< y$. Furthermore, since $ y \leq m_0/n$, we have $ y - 1/n < a$; and $ x < y - 1/n < a$. Thus $ a$ satisfies $ x < a < y$.

Now examine the case where $ x < 0 < y$. Take $ a = 0$.

Finally consider the case where $ x < y \leq 0$. Using the first case, let $ b$ be a rational satisfying $ -y < b < -x$. Then let $ a = -b$. $ \qedsymbol$



"Archimedean property" is owned by Daume. [ full author list (3) | owner history (2) ]
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See Also: Archimedean semigroup, existence of square roots of non-negative real numbers

Other names:  axiom of Archimedes, Archimedean principle

Attachments:
Archimedean ordered fields are real (Theorem) by rspuzio
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Cross-references: rational, least element, rational number, least upper bound, upper bound, least upper bound property, consequence, axiom, natural number, real number
There are 17 references to this entry.

This is version 6 of Archimedean property, born on 2002-08-29, modified 2007-01-15.
Object id is 3396, canonical name is ArchimedeanProperty.
Accessed 19790 times total.

Classification:
AMS MSC: 12D99 (Field theory and polynomials :: Real and complex fields :: Miscellaneous)
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Axiom vs. Theorem by rspuzio on 2007-07-30 15:28:21
While the phrase "is neither an axiomn nor attributed to Archimedes"
is fine rhetoric, the first half of it needs to be taken with a
teaspoon of salt. Whether or not a statement is an axiom depends
on how one chooses to axiomatize one's theory. To be sure, if one
chooses the least upper bound property as an axiom, then the
Archimedean property follows as a consequence, hence is a theorem.

However, there is more than one way to axiomatize the theory of
real numbers. For instance, I could also characterize the real
number system as follows:

* The real number system is an Archimedean ordered field.

* No proper extension field of the real number system is Archimedean.

In this case, the roles are reversed --- the Archimedean property is
now an axiom and the least upper bound property is now a theorem.
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