September 2009 - mathfun@mailman.xmission.com

Re: [math-fun] induction from completeness?
by James Propp 04 Sep '09

04 Sep '09

Josh Zucker writes: >I'm not sure exactly what you're taking as your axioms here, so I'm >not sure if I'm on the right track at all. They include all the usual first-order axioms governing N, Z, Q, and R, but no second-order axioms other than the axiom of completeness. >This last line isn't making sense to me. x is a lower bound for E, >and it is an integer, so x is in E, and thus not in S. I fumbled the end of that proof. Plus, the whole thing was too terse to be readable. I should have written: Let S be some set of positive integers that contains 1 and has the property that for all n in S, n+1 is also in S. Suppose there exists some positive integer that is not in S. Let x be the greatest lower bound of the set E of all such exceptional positive integers. If x is not an integer, then we can find x' > x with no integers between x and x', and x' is also a lower bound on E, contradicting the choice of x as the greatest lower bound. Hence x is an integer. It follows that x-1 and x+1 are integers, and we'll make the assumption that there are no integers between x-1 and x, or between x and x+1. Now since x is a lower bound for E (remember, it's the greatest lower bound of E) and since x+1 isn't a lower bound for E (since it's greater than the greatest lower bound of E), there must be some element of E in the interval [x,x+1). But there's only one integer in this interval, namely x, so we must have x in E. That is, the integer x is not in S. Since the set S is inductive, it follows that x-1 is not in S. Hence x-1 is in E. But then x is not a lower bound on E: contradiction. (I grant that it seems a bit odd to derive a fundamental property of Z from a property of R, especially if you accept Kronecker's ontology of number!) Jim Propp

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Re: [math-fun] smooth torus
by Dan Asimov 04 Sep '09

04 Sep '09

As a torus fan, I've been working on this. The answer to the question is Yes. Proof: We seek two C^oo closed curves x,y: R/Z -> R^3 such that F(s,t) := x(s) + y(t) is everywhere non-singular. F is non-singular at (s,t) precisely when x'(s) x y'(t) is not the 0 vector. WLOG we assume that x and y are each parametrized by arclength. (So in fact the domain of F will be R/(cZ) x R/(dZ), which will not affect anything.) Since the only way unit vectors a, b can satisfy a x b = 0 is that a = +-b. So we seek curves x, y such that both x'(s) + y'(t) and x'(s) - y'(t) are nonzero for all s,t. A known theorem* about the *curve of tangent vectors* to a unit-speed closed curve in R^3 states that C is such a curve on S^2 if [either C is planar or else C contains points on both sides of every great circle]. So consider a monkey saddle, the graph of z = x^3 - 3xy^2 in R^3. Now consider its intersection C with the unit sphere x^2 + y^2 + z^2 = 1. It's easy to verify that C intersects both sides of every great circle (and so is the curve of unit tangents to a closed space curve, in fact many of them). Further, note that the antipodal map a: S^2 -> S^2 carries C into itself, so any curve D disjoint from C is also disjoint from a(C). By perturbing C slightly off itself, we get another curve on S^2 that clearly intersects both sides of every great circle, and so it, too, is the curve of unit tangents to a closed space curve. Thus setting {x'(s)} = C and {y'(t)} = D, we can let x(0) and y(0) be arbitrary points of R^3 and recover the closed curves x and y from x' and y', resp., by integration. Hence (s,t) -> x(s) + y(t) is locally non-singular for all (s,t), or technically speaking, an immersion of T^2 into R^3. ⟡ Note that this does *not* prove that the image {x(s)+y(t)} is non-self-intersecting, no less whether it can be, so I suppose that remains as an unsolved problem (unless Veit know the answer). --Dan ___________________________________________________________ * See Werner Fenchel, On the differential geometry of closed space curves, Bull. Amer. Math. Soc. 57, (1951). 44–54. This theorem is due to Vigodsky. (For the converse, note that for every plane in R^3, the two planes parallel to it that intersect the space curve at the greatest distance apart must both be tangent to the curve.) >> But here's something for torus fans: Can the Minkowski sum of two smooth closed curves in R^3 ever be a smooth torus? If you don't know what a Minkowski sum is, let x(s) and y(t) be smooth closed curves in R^3 parameterized on the unit interval. Now consider the set z(s,t) = x(s) + y(t), (s,t) in [0,1]^2. Question: can you find an x(s) and y(t) so z(s,t) is a smooth surface for all (s,t) in [0,1]^2 ? >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Quadruple letter
by bobmayans＠aol.com 04 Sep '09

04 Sep '09

The Oxford English Dictionary has, under "ashes", a very old spelling: esssse Bob Mayans

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Re: [math-fun] Legitimate word with a triple letter
by Dan Asimov 04 Sep '09

04 Sep '09

Using the National Puzzlers' League word search facility (which accepts regular expressions on any or all of a huge collections of huge wordlists), these came up among non-proper words in Merriam-Webster's Second International Unabridged Dictionary: ------------------------ bossship demigoddessship goddessship headmistressship patronessship wallless whenceeer ------------------------ The last one is probably a typo (for whencever?) or else one of those bogus plants inserted to detect copyright violations. --Dan Gene wrote: << >From the amazon.fr web site: › Visitez la Page Que Vous Avez Créée >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Legitimate word with a triple letter
by Dan Asimov 04 Sep '09

04 Sep '09

Borgmann's book is filled with fascinating word lore, but much of it with words so obscure that they won't be found in any standard arbiter for word games. But in fact zzz appears as a lowercase word in two very respectable dictionaries: Encarta and Webster's New World. (See < www.onelook.com >.) --Dan Hans writes: << [rwg wrote:] << Doesn't someone have a copy of Borgmann's fabled Language on Vacation? Dmitri doesn't do multiple letters per se but does reference AA- AATAMAD (an unidentified town in Palestine) as the first word in English "exhibiting a consonant" and ZZZZ (to snore) as the last. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Legitimate word with a triple letter
by Eugene Salamin 04 Sep '09

04 Sep '09

>From the amazon.fr web site: › Visitez la Page Que Vous Avez Créée -- Gene

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Re: [math-fun] induction from completeness?
by N. J. A. Sloane 04 Sep '09

04 Sep '09

Jim, See the book by Van der Poorten, Notes on Fermat's Last Theorem, 1st ed., paragraph 1.7, page 7. It might not be quite what you want, but it is close. Neil > Or maybe this is a standard derivation.

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Re: [math-fun] induction from completeness?
by Dan Asimov 04 Sep '09

04 Sep '09

Of course, for this to make any sense at all, we would need to know what axiom system this is describing a possible model of -- and whether it's a first-order or second-order theory. --Dan << . . . if the positive integers are closed under + and *, then if 1 is not the smallest positive integer there are infinitely many between 0 and 1. I'd be hard pressed to still call them "integers", but I guess that might be your nonstandard model -- say, all fractions whose denominators are powers of 2 or something like that? >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] induction from completeness?
by James Propp 03 Sep '09

03 Sep '09

Here's my purported proof of the principle of induction, modulo a couple of tacit pesky details that can all be settled by an appeal to the assumption that 1 is the smallest positive integer: Let S be some set of positive integers that contains 1 and has the property that for all n in S, n+1 is also in S. Suppose there exists some positive integer m that is not in S. Let x be the greatest lower bound of the set E of all such exceptional positive integers. If x is not an integer, then we can find x' > x with no integers between x and x', and x' is also a lower bound on E, contradicting the choice of x as the greatest lower bound. Hence x is an integer. Now since x is a lower bound for E and x+1 isn't, we must have x in S and x+1 not in S, which is also a contradiction. Jim

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[math-fun] induction from completeness?
by James Propp 03 Sep '09

03 Sep '09

I think I can derive the principle of induction for |N from the Dedekind-completeness of |R, via the supplementary assumption that 1 is the smallest positive integer, but I'm wondering if there's a subtle fallacy. Or maybe this is a standard derivation. Also, I'm wondering if one can derive induction from completeness without using the assumption that 1 is the smallest positive integer. My guess is "no", but I don't see off-hand a non-standard model that makes this clear. I can supply more details of my derivation tomorrow, if anyone likes. Jim

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