classical problems of constructibility

classical problems of constructibility

There are at least three

classical problems of constructibility

:

1.

Trisecting the angle

: Can an arbitrary angle be trisected?

2.

Doubling the cube

: Given an arbitrary cube, can a cube with double the volume be constructed?

3.

Squaring

the circle

: Given a circle (

http://

planetmath

.org/Circle

), can a square with the same area as the circle be constructed?

The ancient Greeks knew that these constructions were possible using various tools:

•

Archimedes trisected the angle using a

compass

and a ruler with one mark on it. His construction is discussed in the entry

trisection of angle

.

•

Menaechmus constructed a

line segment

of length

2

3

by determining the

intersections

of the

parabolas

y

2

=

2

⁢

x

and

x

2

=

y

. This is an amazing feat considering that

analytic

was not known at the time.

•

A way of squaring the circle is discussed in the entry

variants on compass and straightedge constructions

.

Since the ancient Greeks were interested in performing constructions with the

minimal

amount of tools, they wanted to know whether these constructions were possible using only compass and straightedge. With this constraint in place, answers were elusive until the advent of abstract algebra. The problem was that, working in

geometry

alone, there is really no way to

prove

that a construction is impossible. By using abstract algebra, people could finally prove that certain compass and straightedge constructions were impossible.

The discovery that trisecting the angle using only compass and straightedge is impossible is attributed to Pierre Wantzel. He actually proved a sharper result from which the result about trisecting the angle immediately follows.

Theorem 1

(Wantzel).

It is impossible to trisect a

60

∘

angle using only compass and straightedge.

Proof.

It should first be noted that

60

∘

is a

constructible angle

(

http://planetmath.org/Constructible2

) since

cos

⁡

60

∘

=

1

2

is a

constructible number

. (See the

theorem on constructible angles

for more details.) Thus, we are working in the field of constructible numbers.

Suppose that

20

∘

is a constructible angle. Then

cos

⁡

20

∘

is also a constructible number. Using the triple angle

formulas

(

http://planetmath.org/TrigonometricIdentities

), we have that

cos

⁡

60

∘

=

4

⁢

cos

3

⁡

20

∘

-

3

⁢

cos

⁡

20

∘

. Thus,

1

2

=

4

⁢

cos

3

⁡

20

∘

-

3

⁢

cos

⁡

20

∘

. Therefore,

8

⁢

cos

3

⁡

20

∘

-

6

⁢

cos

⁡

20

∘

=

1

. Hence,

(

2

⁢

cos

⁡

20

∘

)

3

-

3

⁢

(

2

⁢

cos

⁡

20

∘

)

-

1

=

0

.

Let

α

=

2

⁢

cos

⁡

20

∘

. Then

α

is a constructible number and

α

3

-

3

⁢

α

-

1

=

0

. Since

1

and

-

1

are not roots of

x

3

-

3

⁢

x

-

1

, the

polynomial

is

irreducible

(

http://planetmath.org/IrreduciblePolynomial

) over

ℚ

by the

rational root theorem

. Thus,

x

3

-

3

⁢

x

-

1

is the

minimal polynomial

for

α

over

ℚ

. Hence,

[

ℚ

(

α

)

:

ℚ

]

=

3

, contradicting the

theorem on constructible numbers

. The result follows.
∎

The discovery that doubling the cube using only compass and straightedge is impossible is also attributed to Pierre Wantzel.

Theorem 2

(Wantzel).

Doubling the cube is impossible using only compass and straightedge.

Proof.

By

scaling

so that the sides of the original cube are of length

1

, the possibility of this construction is

equivalent

(

http://planetmath.org/Equivalent3

) to

2

3

being a constructible number. Since

[

ℚ

(

2

3

)

:

ℚ

]

=

3

, the theorem on constructible numbers yields that

2

3

is not a constructible number.
∎

The discovery that squaring the circle is impossible is attributed to Ferdinand von Lindemann.

Theorem 3

(Lindemann).

Squaring the circle is impossible using only compass and straightedge.

Proof.

By scaling so that the radius (

http://planetmath.org/Radius2

) of the circle is of length

1

, the possibility of this construction is equivalent to

π

being a constructible number. Note that

π

is

transcendental

. (See this result (

http://planetmath.org/ProofOfLindemannWeierstrassTheoremAndThatEAndPiAreTranscendental2

) for more details.) Thus,

π

is also transcendental. Therefore,

[

ℚ

⁢

(

π

)

:

ℚ

]

is not even finite, let alone a power of

2

. The theorem on constructible numbers yields that

π

is not a constructible number.
∎

References

1

Rotman, Joseph J.

A First Course in Abstract Algebra

. Upper Saddle River, NJ: Prentice-Hall, 1996.

Title

classical problems of constructibility

Canonical name

ClassicalProblemsOfConstructibility

Date of creation

2013-03-22 17:18:37

Last modified on

2013-03-22 17:18:37

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

14

Author

Wkbj79 (1863)

Entry type

Topic

Classification

msc 01A20

Classification

msc 51M15

Classification

msc 12D15

Related topic

TheoremOnConstructibleNumbers

Related topic

TheoremOnConstructibleAngles

Related topic

CompassAndStraightedgeConstruction

Related topic

ConstructibleAnglesWithIntegerValuesInDegrees

Defines

trisecting the angle

Defines

doubling the cube

Defines

squaring the circle

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