B Constructing a Line Segment Equal to a Circle's Circumference?

Trysse
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Is there any way to construct a line segment, that has the lenght of the circumference of a circle using only a ruler and a compass?

My intuition says "no"

Or phrasing the question in another way: given two line segments, can I prove, that the longer line segment has the length of the circumference of a circle to which the shorter line segment is the radius/diameter?

Or In another way:, can I construct a circle, that has a circumference equal to a given line segment?

My intuition for "no" is based in the irrationality of Pi.
 
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Trysse said:
My intuition for "no" is based in the irrationality of Pi.
Not the irrationality. Square root of two is irrational, but you can construct it. It is the diagonal of a square with unit length side.
 
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"Squareing the circle" is known to be impossible with ruler and compass. I suspect this is the problem here, since having squared the circle (ie. constructed a line segment of length \sqrt{\pi}r you can with ruler and compass construct a line segment of length \pi r^2, then from that and a line segment of length r construct a segment of length \pi r, and finally double that to get a line segment of length 2\pi r.
 
Trysse said:
Is there any way to construct a line segment, that has the lenght of the circumference of a circle using only a ruler and a compass?

My intuition says "no"

Or phrasing the question in another way: given two line segments, can I prove, that the longer line segment has the length of the circumference of a circle to which the shorter line segment is the radius/diameter?

Or In another way:, can I construct a circle, that has a circumference equal to a given line segment?

My intuition for "no" is based in the irrationality of Pi.
The answer is "no" due to the transcendency of ##\pi##. This means that ##\pi## cannot be written as
$$
0=a_n\pi^n +a_{n-1}\pi^{n-1}+\ldots+a_2\pi^2+a_1\pi+a_0
$$
with integers ##a_0,\ldots,a_n.##

The technical reason is: ##\pi## is not included in any Galois extension of the rational numbers.

Constructible lengths are included in certain Galois extensions of the rational numbers, namely those of degrees being a power of two, IIRC.
 
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Thanks, that was helpful.
martinbn said:
Not the irrationality. Square root of two is irrational, but you can construct it.
Good point.
pasmith said:
"Squareing the circle" is known to be impossible with ruler and compass.
I knew that "squaring the circle" was impossible, but I did not make the connection.
fresh_42 said:
The answer is "no" due to the transcendency of π.
I knew the term transcendency but was unaware that this had a geometric consequence.
 
fresh_42 said:
Constructible lengths are included in certain Galois extensions of the rational numbers, namely those of degrees being a power of two, IIRC.
It's even stricter than this. Operations with a straightedge and compass can only introduce square roots, so a complex number number is constructible if and only if it is in a field extension of ##\mathbb{Q}## generated by iterated square roots. Not every extension of degree ##2^n## is a tower of quadratic extensions (and I also don't see why the extensions being Galois should be relevant).
 
Infrared said:
It's even stricter than this. Operations with a straightedge and compass can only introduce square roots, so a complex number number is constructible if and only if it is in a field extension of ##\mathbb{Q}## generated by iterated square roots. Not every extension of degree ##2^n## is a tower of quadratic extensions (and I also don't see why the extensions being Galois should be relevant).
IIRC was short for: "I am too lazy to look it up." van der Waerden had it around Galois-theory, so I took what was left in my memory. The statement itself wasn't false.
 
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