Pons asinorum: what's wrong with this proof?
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From: Ken Pledger on 8 Aug 2010 18:29 In article <i3hreu$3u4$1(a)reader1.panix.com>, kj <no.email(a)please.post> wrote: > According to Wikipedia, the popular explanation for the nickname > "Pons asinorum" given to one of Euclid's theorem is that it was > considered "the first real test ... of the intelligence of the > reader.... Alternatively, the nickname may have been copied from scholastic logic. By about 1500, logicians had developed a diagram called "Pons Asinorum" as a mnemonic for constructing syllogisms (cf. I.M. Bochenski, trans. I. Thomas, "A History of Formal Logic" p.220). Perhaps Euclid's criss-cross diagram for I.5 resembled it enough to suggest the transfer of nickname. > > This proof seems to me insanely complex when compared to the > following proof. > > Let ABC be an isosceles triangle, with BA = BC. We want to show > that angles BAC and BCA are equal. > > Proof: Produce a second triangle C'B'A' by reflecting ABC relative > to the line passing through B and perpendicular to AC (actually, > any line will do). Since BA = BC = B'C' = B'A', and AC = C'A', it > follows that ABC and C'B'A' are congruent.... You have just used side-side-side congruence, which is Euclid I.8. Its proof uses I.7, which uses I.5 which you are trying to prove. In article <ab50b583-cbed-4141-95d4-b711fd931afd(a)x21g2000yqa.googlegroups.com>, bert <bert.hutchings(a)btinternet.com> wrote: > .... > Euclid seems to have been happy with invariance > under reflection, at least to the extent that > his rules allow an arbitrary triangle to be > proved congruent to its own mirror image.... In article <28771833-b506-4e4f-8d26-0cf23d19a4af(a)v41g2000yqv.googlegroups.com>, bert <bert.hutchings(a)btinternet.com> wrote: > .... > Well, Pappus's proof looks exactly the same as > that anecdotal automatically-generated proof, > and it looks like a pretty "shortcut" to me.... Pappus's idea is easy and attractive if you take care to list triangles' vertices in corresponding order; for example saying that trianges ABC and DEF are congruent only if A corresponds to D, B to E and C to F. Apparently Euclid hadn't thought of that useful convention, and usually just lists his triangle vertices in alphabetical order. With the convention about order, Pappus proves that (in your notation) triangles ABC and CBA are congruent (side-angle-side, which is Euclid I.4), so the interior angles at A and C are equal. In article <i3m425$qim$1(a)reader1.panix.com>, kj <no.email(a)please.post> wrote: > In <1ffd4e45-31d8-4af8-80ed-b882edf9f1c8(a)f6g2000yqa.googlegroups.com> Chip > Eastham <hardmath(a)gmail.com> writes: > > .... It seems obvious that if > any two angles are equal (and, in particular, if the two base angles > are equal) their complements (relative to the straight line) are > also equal. IOW, if a + a' = b + b' = Pi radians, and we have > already proven that a = b, then it follows that a' = b'.... You assume Euclid I.13, that adjacent angles on a straight line add up to two right angles. Euclid's proof inserts a perpendicular to make two actual right angles, and then his proof looks something like the associatve law of addition. After that, your argument follows from Common Notion 3 (one of his axioms): "If equals be subtracted form equals, the remainders are equal." It's easy to make lots of assumptions when we look at diagrams; but the Greeks from Hippocrates onwards were trying hard to tie up all the logical loose ends. We know that they didn't succeed completely, but it still impresses me that a small number of people reached the standard of Euclid's "Elements" in just over a century. Ken Pledger. |