Archimedean Property of Real Number

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

. ∎

Corollary 1.

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

. ∎

Corollary 2.

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.

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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Archimedean Property of Real Number

Proof.

Corollary 1.

Proof.

Corollary 2.

Proof.

Corollary 3.

Proof.

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