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„The solution of the problem of squaring the circle by compass and straightedge requires the construction of the number √π. If √π is constructible, it follows from standard constructions that π would also be constructible.“
„In 1882, the task was proven to be impossible, as a consequence of the Lindemann –Weierstrass theorem which proves that pi (π) is a transcendental, rather than an algebraic irrational number; that is, it is not the root of any polynomial with rational coefficients.“
From Wikipedia
This Paper was in the study cases category for the first three days:
https://drive.google.com/file/d/1AruLaitcjlSgIUQOUZ8AQlFk8nZun1zn/view?usp=sharing
It is now difficult to find it among hundreds of other…
related to the topic: How people find out pi? ; Area and circumference of a circle
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„Study case“ or solution ?
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Re: „Study case“ or solution ?
by Guest » Thu Feb 25, 2021 11:24 am
Archimedes value Pi: 22/7
[tex]r=\frac{22}{7}[/tex]
[tex]a=r \sqrt {3,1640625}=\frac{99}{28}\sqrt{\frac{5}{2}}[/tex]
[tex]A=3,1640625(\frac{22}{7})^{2}=31\frac{397}{1568}[/tex]
[tex]A=(\frac{99}{28}\sqrt{\frac{5}{2}})^{2}=31\frac{397}{1568}[/tex]
[tex]r=\frac{22}{7}[/tex]
[tex]a=r \sqrt {3,1640625}=\frac{99}{28}\sqrt{\frac{5}{2}}[/tex]
[tex]A=3,1640625(\frac{22}{7})^{2}=31\frac{397}{1568}[/tex]
[tex]A=(\frac{99}{28}\sqrt{\frac{5}{2}})^{2}=31\frac{397}{1568}[/tex]
- Guest
Re: „Study case“ or solution ?
by Guest » Sat Feb 27, 2021 3:37 pm
What do you mean by "find out pi"?
I believe it was Archimedes who first determined (or he copied what some other person had said- ancient Greece didn't have copyright laws) that the circumference of any circle is a constant multiple of the diameter. We call that constant [tex]\pi[/tex]. Typically we use an approximation such as "3.14" or "22/7" but those are NOT equal to [tex]pi[/tex]. (My favorite mnemonic, "May I have a large container of coffee" gives [tex]\pi[/tex] , 3.1415926, to 7 decimal places) [tex]\pi[/tex] is a "transcendental number". it has infinitely many decimal places and cannot be written as a fraction nor even as simple calculation, like [tex]\sqrt{2}[/tex]- there are complicated formulas for [tex]\pi[/tex] to any desired number of decimal places.
I believe it was Archimedes who first determined (or he copied what some other person had said- ancient Greece didn't have copyright laws) that the circumference of any circle is a constant multiple of the diameter. We call that constant [tex]\pi[/tex]. Typically we use an approximation such as "3.14" or "22/7" but those are NOT equal to [tex]pi[/tex]. (My favorite mnemonic, "May I have a large container of coffee" gives [tex]\pi[/tex] , 3.1415926, to 7 decimal places) [tex]\pi[/tex] is a "transcendental number". it has infinitely many decimal places and cannot be written as a fraction nor even as simple calculation, like [tex]\sqrt{2}[/tex]- there are complicated formulas for [tex]\pi[/tex] to any desired number of decimal places.
- Guest
Re: „Study case“ or solution ?
by Guest » Sun Feb 28, 2021 11:56 am
What do you mean by "find out pi"?
simply make a segment of length Pi
https://youtu.be/Z0UIshe0Fx8?t=408
- Guest
Re: „Study case“ or solution ?
by Guest » Mon Dec 27, 2021 3:14 pm
I don't speak that language so I don't know what he was saying, whether this was a joke or he was serious.
He appears to be talking about three "impossible constructions", "trisecting an arbitrary angle" (given an angle, using only the straight edge and compasses, divide it into three equal angles), "duplicating the cube" (given a cube, using only the three dimensional analogues of a straight edge and compasses, construct a cube with twice the volume), and "squaring the circle".
Each of those was proven to be impossible long ago using the concept of "constructable numbers". A number, x, is constructible if and only if, given a line segment of length 1, it is possible using only straight edge and compasses, to construct a line segment of length x. It has been proven that the only "constructible numbers" are those that are "algebraic of order a power of 2". For example, if we have a line segment of length 1, we can use compasses and straight edge to construct a perpendicular at one end of that segment, strike a length 1 on that perpendicular, then connect those two points constructing a line segment of length [tex]\sqrt{2}[/tex]. Of course, [tex]\sqrt{2}[/tex], is "algebraic of order 2" since it satisfies [tex]x^2= 2[/tex].
IF it were possible to trisect an arbitrary angle then it would be possible to construct a line segment satisfying a cubic equation- a number algebraic of order 3, NOT a power of two.
If it were possible to "duplicate the cube" taking, say, a cube of side length 1 so volume 1, we would have constructed a cube of volume 2, so side length [tex]\sqrt[3]{2}[/tex], a number algebraic of order 3, not a power of 2.
If it were possible to "square the circle", taking a circle of radius 1, so area [tex]\pi[/tex], you would construct a square of area [tex]\pi[/tex] so side length [tex]\sqrt{\pi}[/tex], a transcendental number, not algebraic of any order!
He appears to be talking about three "impossible constructions", "trisecting an arbitrary angle" (given an angle, using only the straight edge and compasses, divide it into three equal angles), "duplicating the cube" (given a cube, using only the three dimensional analogues of a straight edge and compasses, construct a cube with twice the volume), and "squaring the circle".
Each of those was proven to be impossible long ago using the concept of "constructable numbers". A number, x, is constructible if and only if, given a line segment of length 1, it is possible using only straight edge and compasses, to construct a line segment of length x. It has been proven that the only "constructible numbers" are those that are "algebraic of order a power of 2". For example, if we have a line segment of length 1, we can use compasses and straight edge to construct a perpendicular at one end of that segment, strike a length 1 on that perpendicular, then connect those two points constructing a line segment of length [tex]\sqrt{2}[/tex]. Of course, [tex]\sqrt{2}[/tex], is "algebraic of order 2" since it satisfies [tex]x^2= 2[/tex].
IF it were possible to trisect an arbitrary angle then it would be possible to construct a line segment satisfying a cubic equation- a number algebraic of order 3, NOT a power of two.
If it were possible to "duplicate the cube" taking, say, a cube of side length 1 so volume 1, we would have constructed a cube of volume 2, so side length [tex]\sqrt[3]{2}[/tex], a number algebraic of order 3, not a power of 2.
If it were possible to "square the circle", taking a circle of radius 1, so area [tex]\pi[/tex], you would construct a square of area [tex]\pi[/tex] so side length [tex]\sqrt{\pi}[/tex], a transcendental number, not algebraic of any order!
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