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February 16[edit]

Triangle Inequality[edit]

Hello all! I read the article on the Triangle Inequality, and it raised a question. Apologies if this is roundabout; I'm pretty weak with mathematical terminology so it's difficult to find the most concise phrasing. :)

Actual question: In 2-dimensional Euclidean geometry, is satisfying the Triangle Inequality merely a necessary condition to form a triangle? Or is it also sufficient? That is, for every set of three side lengths that satisfies the Triangle Inequality, does there always exist a triangle?

(I think that I'm asking if the "converse" of the Triangle Inequality holds, but I'm not sure if that is the correct logical term. So I thought that whole long question was necessary to clarify my intent.)

Thank you very much for your help!

JonathanHopeThisIsUnique (talk) 17:21, 16 February 2017 (UTC)[reply]

Yes, it is necessary and sufficient. You can convince yourself of this by first trying to draw a triangle with sides a, b, c with a=b+c. The two short sides will exactly overlap the longer side. Now lengthen one of the shorter sides—the two shorter sides will bulge out away from the long side, forming a triangle. Loraof (talk) 17:36, 16 February 2017 (UTC)[reply]

Actually the converse is dealt with in the section Triangle inequality#Converse. Loraof (talk) 17:56, 16 February 2017 (UTC)[reply]

Just a little history: The converse of the triangle inequality is in Euclid (Book 1, Prop 22) but he doesn't give the triangle inequality itself (at least that I could find). It seems likely that Euclid's purpose wasn't so much to prove that a triangle with the given sides exists, but to show how to construct it. In any case, the proof given in the linked article seems a bit modern (perhaps OR), calling in complex numbers to prove what should be a purely geometrical result. (The proof makes heavy use of the Pythagorean theorem as well, which in turns assumes you're in the Euclidean plane, so the question of whether the converse would hold in non-Euclidean geometry is unsettled.) It should be noted also that the converse is for the statement about triangles, but the triangle inequality has been generalized in many ways and taken as an axiom in some cases, so it's hard to say what a converse to these versions might be. --RDBury (talk) 04:53, 17 February 2017 (UTC)[reply]

The section Triangle inequality#Euclidean geometry give a proof of the triangle inequality, and says "This proof appears in Euclid's Elements, Book 1, Proposition 20.[6]"

It looks to me like the proof of the converse in Triangle inequality#Converse starts out in paragraph 2 by assuming the conclusion. Any objection to my removing the proof? Loraof (talk) 16:12, 17 February 2017 (UTC)[reply]

Actually I see how to fix the wording so it doesn't appear to assume the conclusion. Loraof (talk) 16:52, 17 February 2017 (UTC)[reply]

I missed Prop. 20; Euclid tends to keep converses together, but I should have allowed for exceptions. And you're right in that the proof in the article appears more circular than it really is. I guess my biggest objection to it is that it does analytically, needlessly mixing in imaginary numbers, what Euclid does more easily by construction. Euclid's proof is not without its problems though in that it makes an assumption which Euclid neither proves nor takes as an axiom. Namely that if a circle has a point inside and a point outside another circle then the circles intersect. Prop. 1 (construction of an equilateral triangle) is actually a special case and suffers from the same fault, as has been noted by many. But axioms have been added since which resolve such issues so I don't see anything wrong with removing the proof and replacing it with a reference to Euclid. Euclid's proof is straightforward and not very noteworthy in itself, but perhaps the flaw makes it so. --RDBury (talk) 20:01, 17 February 2017 (UTC)[reply]

This raises an interesting line of enquiry: can one construct a counterexample? I'm thinking of a Euclidean plane except over the rational numbers instead of the reals. At a glance it would seem (to my untutored eye) to conform to all Euclid's axioms, yet two circles can overlap without intersecting. I don't see that even Hilbert's axioms change this. What am I missing? —Quondum 22:05, 17 February 2017 (UTC)[reply]

...and there are no equilateral triangles with all their vertices in Q^2. So the construction fails there. --JBL (talk) 22:21, 17 February 2017 (UTC)[reply]

I can see how Hilbert's congruence axiom III.1 rules out Q², and how via construction it ensures that equilateral triangles exist. I guess that this axiom would also rule out nonintersecting overlapping circles (not yet obvious to me). Yet, a geometry over some proper subfield of R should satisfy all the axioms, and this in itself should give pause to question whether arbitrary crossing continuous curves intersect, for example. —Quondum 23:41, 17 February 2017 (UTC)[reply]

Hmm. I see the axiom of completeness, which gets excessively complicated for me. Not to worry. —Quondum 23:55, 17 February 2017 (UTC)[reply]

To top it off I'm pretty sure you have to throw in the Archimedean property to rule out geometry over the hyperreals. Heath's solution is to admit what he calls Dedekind's postulate, which is basically a reformulation of Dedekind cuts applied to a line segment. It is astonishing how deep these issues get. Meanwhile, I'll start a thread on the Triangle inequality talk page on whether the proof should be kept. --RDBury (talk) 03:26, 18 February 2017 (UTC)[reply]

Thanks a lot for your quick response, and your detailed answers. I really appreciate it. :) JonathanHopeThisIsUnique (talk) 00:59, 18 February 2017 (UTC)[reply]