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

[Trysse]

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.

 

[martinbn]

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.

 

[pasmith]

"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 $2\pi r$.

 

[fresh_42]

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.

 

[Trysse]

Thanks, that was helpful.

Not the irrationality. Square root of two is irrational, but you can construct it.

Good point.

"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.

The answer is "no" due to the transcendency of π.

I knew the term transcendency but was unaware that this had a geometric consequence.

 

[Infrared]

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).

 

[fresh_42]

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.