History of the three "impossible" compass-and-straightedge problems

Question

I'm preparing a presentation about constructible numbers and I wanted to know some of the history about it to motivate the topic.

I wanted to know if the classical Greek problems (doubling the cube, squaring the circle and the angle trisection) were first proven to be impossible using Field Theory or there is another method that was developed earlier to prove this.

Answer 1

Sites around the internet give Pierre Laurent Wantzel credit for proving these problems as impossible in the paper "Recherches sur les moyens de reconnaître si un problème de Géométrie peut se résoudre avec la règle et le compas" in "Liouville's Journal" (1837).

"In Wantzel's paper he proved the impossibility of the solution under Euclidean restrictions. Wantzel regarded the magnitudes involved not as geometric segments, but as numerical lengths, via analytic geometry. This let him use algebra and arithmetic rather than pure geometry [Dunham, p. 245]." (source)

For details about the proof check out page 7:
http://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/Suzuki.pdf

Ferdinand von Lindemann's proof (1882) that pi is transcendental (non-algebraic and therefore non-constructible) was also important, because squaring the circle by compass and straightedge would require constructing the square root of pi.

A useful chart: http://en.wikipedia.org/wiki/Constructible_number#Impossible_constructions

A useful explanation: http://www.uwgb.edu/dutchs/PSEUDOSC/trisect.HTM

History of the three:

http://www-history.mcs.st-and.ac.uk/Indexes/Greeks.html

http://www.docstoc.com/docs/73739647/History-topic--Squaring-the-circle

Answer 2

To put is simply, Pierre Wantzel used field theory. Building up on the work of Descartes and Gauss, he demonstrated that doubling the cube and trisecting the angle requires solving irreducible cubic equations over the rationals. However, the equations that describe intersections of lines and circles (the shapes made by using a compass and straightedge) are quadratic, and so none of them can ever provide solutions to cubic equations irreducible over the rationals (and by extension, the constructibles). (This is not true with respect to all quartics irreducible over the rationals; $x^4+1=0$ has all four solutions constructible with a compass and straightedge, despite its obvious irreducibility over the rationals)

Ferdinand von Lindemann proved that the sum of a finite set of integer multiples of distinct algebraic powers of $e$, if $e$ is the base of the natural logarithm, cam never equal zero.

If the circle can be squared with a compass and straighedge, then $\sqrt{}$ is constructible, because that would be the ratio of the side of the square with the diameter of the circle. If $\sqrt{}$ is constructible, then it is algebraic. If $\sqrt{}$ is algebraic, then is algebraic, and if $i^2=-1$, the $i$ is algebraic as well, and $e^{i}+1\not =0$, which contradicts Euler's identity. So the circle can not be squared with a compass and a straighedge.