Archimedean property

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 $n\ot\\in S$ ; since $n\ot\\in S$ , we know $n>x$ . Thus we have a natural greater than $x$ .∎

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 $x\eq 0$ , $y/x$ is a real. By the Archimedean property, we can choose an $n\\in\\mathbb{N}$ such that $n>y/x$ . Then $nx>y$ .∎

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 $0<1/n<w$ .∎

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 $a=-b$ .∎

单词	ArchimedeanProperty
释义	Archimedean property 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 $n\ot\\in S$ ; since $n\ot\\in S$ , we know $n>x$ . Thus we have a natural greater than $x$ .∎ 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 $x\eq 0$ , $y/x$ is a real. By the Archimedean property, we can choose an $n\\in\\mathbb{N}$ such that $n>y/x$ . Then $nx>y$ .∎ 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 $0<1/n<w$ .∎ 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 $a=-b$ .∎
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