Now to show that it is impossible to Circle the Sqaure,
Double the Cube, and to Trisect an Arbitrary Angle. We just found
that for any intermediate field between the rationals and C
[F(a1,...,an):Q] = 2mmust be true, but we know that Also,
[Q(3 Last we can use the trig identity: cos(A) = 4 cos3(A/3) - 3 cos (A/3).Setting A=600 and x = cos(A/3) we get 1/2 = 4x3 - 3xwhich tells us that [Q(x):Q] = 3 This concludes our little story on the three classical geometric construction problems. It is amazing how using the power of modern mathematics we can solve a puzzle which perplexed so many, for so long so ... simply! |
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