rational numbers are real numbers

rational numbers are real numbers

Let us first show that the

natural numbers

0

,

1

,

2

,

…

are

contained

in the

real numbers

as constructed above.
Heuristically, this should be clear. We start with

0

.
By adding

1

repeatedly we obtain the natural numbers

0

,

0

+

1

,

(

0

+

1

)

+

1

,

(

(

0

+

1

)

+

1

)

+

1

,

…

,

To make this precise, let

ℕ

be the natural numbers.
(We assume that these exist. For example, all the usual constructions
of

ℝ

rely on the

existence

of the natural numbers.)
Then we can define a map

f

:

ℕ

→

ℝ

as

1.

f

⁢

(

0

)

=

0

, or more precisely,

f

⁢

(

0

ℕ

)

=

0

ℝ

,

2.

f

⁢

(

a

+

1

)

=

f

⁢

(

a

)

+

1

for

a

∈

ℕ

.

By

induction

on

a

one can prove that

f

⁢

(

a

+

b

)

=

f

⁢

(

a

)

+

f

⁢

(

b

)

,

f

⁢

(

a

⁢

b

)

=

f

⁢

(

a

)

⁢

f

⁢

(

b

)

,

a

,

b

∈

ℕ

and

f

⁢

(

a

)

≥

0

,

a

∈

ℕ

⁢

with equality only when

⁢

a

=

0

.

The last claim follows since

f

⁢

(

a

)

>

0

for

a

=

1

,

2

,

…

(by induction),
and

f

⁢

(

0

)

=

0

.
It follows that

f

is an

injection

: If

a

≤

b

, then

f

⁢

(

a

)

=

f

⁢

(

b

)

implies
that

f

⁢

(

a

)

=

f

⁢

(

a

)

+

f

⁢

(

b

-

a

)

, so

a

=

b

.

To conclude, let us show that

f

⁢

(

ℕ

)

⊂

ℝ

satisfies

the

Peano axioms

with

zero element

f

⁢

(

0

)

and
sucessor

operator

S

:

f

⁢

(

ℕ

)

→

f

⁢

(

ℕ

)

k

↦

f

⁢

(

f

-

1

⁢

(

k

)

+

1

)

First, as

f

is a

bijection

,

x

=

y

if and only if

S

⁢

(

x

)

=

S

⁢

(

y

)

is clear.
Second, if

S

⁢

(

k

)

=

0

for some

k

=

f

⁢

(

a

)

∈

f

⁢

(

ℕ

)

, then

a

+

1

=

0

; a

contradiction

.
Lastly, the

axiom of induction

follows since

ℕ

satisfies this axiom.
We have shown that

f

⁢

(

ℕ

)

are a subset of the real numbers that
behave as the natural numbers.

From the natural numbers, the integers and

rationals

can
be defined as

ℤ

=

ℕ

∪

{

-

z

∈

ℝ

:

z

∈

ℕ

}

,

ℚ

=

{

a

b

:

a

∈

ℤ

,

b

∈

ℕ

∖

{

0

}

}

.

Mathematically,

ℤ

and

ℚ

are subrings of

ℝ

that are
ring

isomorphic

to the integers and rationals, respectively.

Other constructions

The above construction follows

[

1

]

. However, there are also
other constructions. For example, in

[

2

]

, natural numbers in

ℝ

are defined as follows. First, a set

L

⊆

ℝ

is

inductive

if

1.

0

∈

L

,

2.

if

a

∈

L

, then

a

+

1

∈

L

.

Then the natural numbers are defined as real numbers that are contained in all

inductive sets

.
A third approach is to explicitly exhibit the natural numbers when
constructing the real numbers. For example, in

[

3

]

,
it is shown that the rational numbers form a

subfield

of

ℝ

using explicit

Dedekind cuts

.

References

1

H.L. Royden,

Real analysis

, Prentice Hall, 1988.

2

M. Spivak,

Calculus

, Publish or Perish.

3

W. Rudin,

Principles of

mathematical analysis

,
McGraw-Hill, 1976.

Title

rational numbers are real numbers

Canonical name

RationalNumbersAreRealNumbers

Date of creation

2013-03-22 15:45:49

Last modified on

2013-03-22 15:45:49

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

6

Author

matte (1858)

Entry type

Result

Classification

msc 54C30

Classification

msc 26-00

Classification

msc 12D99

Related topic

GroundField

Generated on Sat Feb 10 11:19:13 2018 by

LaTeXML