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I've drawn two random points on a two dimensional plane. If the distance between these two points is one unit, can a line of π units be drawn using a compass and straight edge?

Or, is there a proof that shows this is impossible?

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Yes, there's a proof that it is impossible. Galois theory answers the question of which lengths (relative to a unit length) are constructible, and they have to be roots of a polynomial which has a solvable Galois group (among other contraints). However, $\pi$ is transcendental, and satisfies no polynomial equation over the rationals. – bgins Mar 17 '12 at 20:06
also refer to Lindemann–Weierstrass theorem which proves that $\pi$ is transcendental – quartz Mar 17 '12 at 20:09
This is called "squaring the circle" and is one of three impossible problems that have been around from the times of the Ancient Greeks. The best you can do is construct an approximation to $\pi$: have a look at – Théophile Mar 17 '12 at 20:10
up vote 4 down vote accepted

No (to your first question). A constructable number must be algebraic. However $\pi$ is not an algebraic number ($\pi$ is a transcendental number). Also of interest with regards to this is the famous "squaring the circle" problem.

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A constructible number is more specialised than this (duplicating the cube and trisecting the angle are possible with algebraic numbers). Constructible numbers are those attainable through a series of quadratic extensions of the rationals - hence the possibility of inscribing 3-sided ($2+1$), 5-sided ($2^2+1$), 17-sided (=$2^4+1$), 257-sided (=$2^{16}+1$) and 65537-sided ($2^{256}+1$) polygons in a circle - these numbers being prime. But not 7-sided or 41-sided ones, for example. – Mark Bennet Mar 17 '12 at 21:30
This is not quite circle-squaring, is it? That would involve constructing a line segment of length $\sqrt{\pi}$. – Nate Eldredge Mar 17 '12 at 21:50
@NateEldredge If $a$ can be constructed, then so can $\sqrt a$. I mentioned the "squaring the circle link" mainly because this is the only place I saw in Wiki where it was explicitly stated that $\pi$ is not constructable. – David Mitra Mar 17 '12 at 22:09
@DavidMitra: Thanks, I didn't know that. (For future reference, Wikipedia has a construction.) – Nate Eldredge Mar 17 '12 at 22:36

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