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I've been wondering how you go about proving an impossibility e.g. when I looked up Abel's impossibility theorem it says nothing about the proof and only restates the theorem when I'd like to know how it's done. Is the way to prove an impossibility that you assume the opposite - that it's possible - and derive a contradiction from the possibility and hence you have proven an impossibility? Thank you

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Abel's proof is the grand finale of a 3rd-year undergraduate abstract algebra class. It's a lot of work, understanding how it's done (but well worth it!). No one's going to do it out for you here, but lhf's reference is a good one, if you're ready for it. –  Gerry Myerson Jul 25 '11 at 12:57

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up vote 5 down vote accepted

Proving "not $P$" can be done by showing two things: "if not $Q$, then not $P$" and "not $Q$". But this is equivalent to proving the contrapositive, "if $P$ then $Q$", along with "not $Q$". The latter is what we call "proof by contradiction" (assuming the opposite).

The Wikipedia article gives a proof of Abel-Ruffini which boils down to "If the Galois group of $S_n$ is not solvable, then general $n^{th}$ degree polynomials are not solvable by radicals", together with "$S_5$ is not a solvable group". This particular proof is not technically "proof by contradiction" (but is equivalent to such a proof).

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+1 and accepted answer since you tell how to apply Galois groups –  Niklas rtz Jul 25 '11 at 21:15

Contradiction is one possible way but not the only one. Consider the famous Greek problems. The proof that angle trisection cannot be done with ruler and compass in general relies on an algebraic framing of the problem and then showing that no solution can be achieved using algebraic invariants, in this case the fact that ruler and compass constructions can only happen in extension fields whose degree is a power of 2, while angle trisection leads to extensions of degree 3. That you cannot double the cube follows a similar argument. However, the proof that the circle cannot be squared relies on a proof that $\pi$ is not algebraic and proofs of this fact are usually by contradiction. For the impossibility of solving equations of degree 5 in radicals, one considers permutations of the roots and the properties of subgroups of $S_5$. See Field Theory and its Classical Problems by Hadlock for a nice exposition of this theory.

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+1 for the clarification while I accepted the other answer since it directly mentioned the Abel-Ruffini theorem and Galois groups –  Niklas rtz Jul 25 '11 at 21:13

There is a recent more or less popular level book by Peter Pesic on Abel's proof which might be of interest to you. It has an appendix with a full annotated translation of Abel's original paper. I have not seen this translation, but I have gone through the original paper. That paper is fairly short, and in some parts kind of heavy going. That's almost inevitable, after all it is original research, and the result escaped the great Lagrange. Another problem is that the group-theoretic machinery that we are accustomed to is only there in embryo.

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Thank you for the answer –  Niklas rtz Aug 9 '11 at 22:51

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