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Difference between revisions of "Three Greek problems of antiquity"

(→‎Squaring of a Circle)
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All three constructions have been shown to be impossible.  
 
All three constructions have been shown to be impossible.  
  
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The impossibility of these constructions relies on the characterization of the [[constructible number]]s. It can be shown that, starting with a segment of length <math>1</math>, one can construct a segment of length <math>x</math> iff <math>x</math> can be formed from the rational numbers in a finite number of steps using only the operations addition, subtraction, multiplication, division, and taking square roots. Using [[field theory]], one can then show that <math>x</math> must be an [[algebraic number]] and that the degree of it's [[minimal polynomial]] must be a power of <math>2</math>.
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The impossibility of these constructions relies on the characterization of the [[constructible number]]s. It can be shown that, starting with a segment of length <math>1</math>, one can construct a segment of length <math>x</math> if <math>x</math> can be formed from the rational numbers in a finite number of steps using only the operations addition, subtraction, multiplication, division, and taking square roots. Using [[field theory]], one can then show that <math>x</math> must be an [[algebraic number]] and that the degree of it's [[minimal polynomial]] must be a power of <math>2</math>.
 
==Trisection of the General Angle==
 
==Trisection of the General Angle==
 
Statement: Given an angle, construct by means of [[straight edge]] and [[compass]] only, an angle one-third its measure.  
 
Statement: Given an angle, construct by means of [[straight edge]] and [[compass]] only, an angle one-third its measure.  
  
 
This is impossible because the number <math>\cos 20^{\circ}</math> would have to be constructible in this case, and the minimal polynomial of <math>\cos 20^{\circ}</math> is <math>8x^3-6x-1</math>, which has degree <math>3</math>.
 
This is impossible because the number <math>\cos 20^{\circ}</math> would have to be constructible in this case, and the minimal polynomial of <math>\cos 20^{\circ}</math> is <math>8x^3-6x-1</math>, which has degree <math>3</math>.
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You can also go to this website: http://www.murderousmaths.co.uk/books/MMoE/trisect.htm
  
 
==Doubling of a Cube==
 
==Doubling of a Cube==
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This is impossible because the number <math>\sqrt[3]{2}</math> would have to be constructed, and the minimal polynomial of <math>\sqrt[3]{2}</math> is <math>x^3-2</math>, which has degree <math>3</math>.
 
This is impossible because the number <math>\sqrt[3]{2}</math> would have to be constructed, and the minimal polynomial of <math>\sqrt[3]{2}</math> is <math>x^3-2</math>, which has degree <math>3</math>.
 +
 +
Just buy the book: http://www.murderousmaths.co.uk/books/MMoE.htm
  
 
==Squaring the Circle==
 
==Squaring the Circle==
 
Statement: Given a circle, construct by means of [[straight edge]] and [[compass]] only, a square with area same as that of the circle.
 
Statement: Given a circle, construct by means of [[straight edge]] and [[compass]] only, a square with area same as that of the circle.
  
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This is impossible, as the [[transcendental number]] <math>\sqrt{\pi}</math> would have to be constructible in this case.
+
This is impossible because you will have to construct <math>\sqrt{\pi}</math>, which is a [[transcendental number]].
 +
 
 +
You can go to this website: http://www.murderousmaths.co.uk/books/MMoE/sqcirc.htm
  
 
==See Also==
 
==See Also==

Latest revision as of 16:06, 11 June 2022

The three Greek problems of antiquity were some of the most famous unsolved problems in history. They were first posed by the Greeks but were not settled till the advent of Abstract algebra and Analysis in modern times.

All three constructions have been shown to be impossible.

The impossibility of these constructions relies on the characterization of the constructible numbers. It can be shown that, starting with a segment of length $1$, one can construct a segment of length $x$ if $x$ can be formed from the rational numbers in a finite number of steps using only the operations addition, subtraction, multiplication, division, and taking square roots. Using field theory, one can then show that $x$ must be an algebraic number and that the degree of it's minimal polynomial must be a power of $2$.

Trisection of the General Angle

Statement: Given an angle, construct by means of straight edge and compass only, an angle one-third its measure.

This is impossible because the number $\cos 20^{\circ}$ would have to be constructible in this case, and the minimal polynomial of $\cos 20^{\circ}$ is $8x^3-6x-1$, which has degree $3$.

You can also go to this website: http://www.murderousmaths.co.uk/books/MMoE/trisect.htm

Doubling of a Cube

Statement: Given a cube, construct by means of straight edge and compass only, a cube with double the volume.

This is impossible because the number $\sqrt[3]{2}$ would have to be constructed, and the minimal polynomial of $\sqrt[3]{2}$ is $x^3-2$, which has degree $3$.

Just buy the book: http://www.murderousmaths.co.uk/books/MMoE.htm

Squaring the Circle

Statement: Given a circle, construct by means of straight edge and compass only, a square with area same as that of the circle.

This is impossible because you will have to construct $\sqrt{\pi}$, which is a transcendental number.

You can go to this website: http://www.murderousmaths.co.uk/books/MMoE/sqcirc.htm

See Also

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