November 2006 - mathfun@mailman.xmission.com

RE: [math-fun] Ambiguously placed cities
by Andy Latto 09 Dec '06

09 Dec '06

> Q2. Are there 10 road lengths that lead to distinct town > configurations? > >> > -------------------------------------------------------------- > ---------------------------- > > Q1. Draw a non-isosceles triangle ABC with rational sides. Let M be > the midpoint of AB, and let DD' be a segment drawn through M that's > perpendicular to CM, such that the distances d(D,M) and > d(D',M) are equal > and rational. > > This implies distinct town configurations A,B,C,D and A,B,C,D' having > the same sest of 6 distances. The fact that DMD' is perpendicular to CM shows that the distances CD and CD' are equal. But I don't see why AD and AD' should be equal in this contstruction. Andy Latto andy.latto(a)pobox.com

10 36

Re: [math-fun] Ambiguously placed cities
by Daniel Asimov 05 Dec '06

05 Dec '06

Ed writes: << Refering to http://www.mathpuzzle.com/AmbiguousTowns.gif . . . Between four towns are roads of length 3, 5, 6, 7, ~3.10, ~5.44 . There are two distinct town configurations. (See picture in link) Q1. Are there 6 integer road lengths that lead to distinct town configurations? Q2. Are there 10 road lengths that lead to distinct town configurations? >> ------------------------------------------------------------------------------------------ Q1. Draw a non-isosceles triangle ABC with rational sides. Let M be the midpoint of AB, and let DD' be a segment drawn through M that's perpendicular to CM, such that the distances d(D,M) and d(D',M) are equal and rational. This implies distinct town configurations A,B,C,D and A,B,C,D' having the same sest of 6 distances. By scaling everything up by the product of all 6 denominators, the distances also become integral. Q2. Given an answer to Q1, add a 5th town on the perpendicular bisector of the segment DD', and far away from the other 4. This guarantees 10 road lengths with two distinct town configurations. It's unclear to me if this can be done with all distances integral. (Apropos this last comment, it's an unsolved problem whether there is a point in the plane at integral distances from all 4 vertices of an NxN square, for any integer N.) --Dan

3 4

[math-fun] [0, 1] to (0, 1)?
by Kerry Mitchell 01 Dec '06

01 Dec '06

More of an intellectual curiosity question than anything useful. Is there a "nice" function that is one-to-one and continuous that maps the closed interval [0, 1] to the open interval (0, 1)? Ideally, such a "nice" function would be not a piecewise function or a power series and something that could be implemented with standard library functions. If such a thing doesn't exist, is there at least a "not nice" function that is one-to-one and maps the intervals? If not, why not? Thanks for any pointers or wisdom, Kerry Mitchell -- lkmitch(a)gmail.com www.fractalus.com/kerry

4 4

RE: [math-fun] Add odd digits, subtract even
by Eric Angelini 01 Dec '06

01 Dec '06

> http://www.cetteadressecomportecinquantesignes.com/134.htm ... Should be ok now ... Best, É.

3 2

Re: [math-fun] Question related to p-adics
by Daniel Asimov 30 Nov '06

30 Nov '06

<< It is also isomorphic to the homomorphisms on the group Q/Z (i.e., fractions with addition modulo 1). >> I take it this means Hom(Q/Z, Q/Z) = End(Q/Z). (If not, what are the homomorphisms into?) Will just free-associate about this group for a bit, since I've always liked it. Unlike Q, it's a countable direct sum of subgroups, namely for each prime p the group of elements G_p := {k/p^n mod 1 : k in Z, n in Z+, (k, p) = 1 } under addition. So a homomorphism h: Q/Z -> Q/Z is determined by its values on all the elements, for each p, of form 1/p^k, that is, h(1/p), h(1/p^2), . . . , h(1/p^n), . . . with the requirement, of course, that h(px) = h(x) + . . . h(x) (p summands) for any x and in particular x = 1/p^n for any n = 1,2,3,.... Say k is the smallest positive integer j such that x= h(1/p^j) is nonzero. Then clearly px = 0 (mod p), so x= r/p for 0 < r < p. Then for y = h(1/p^(k+1)) we must have py = x, so y = s/p^2 where ps = r (mod p^2), so s is any member of the set {r, r+p, . . ., r + (p-1)p} of p possible values. Etc. So it's beginning to look as though this group End(Q/Z) is the direct sum of the groups Hom(G_p, G_p) = End(G_p), which are each like the inverse limit . . . -> Z_p^4 -> Z_p^3 -> Z_p^2 -> Z_p where each map is multiplication by p. Okay, that sheds some light on Franklin's comment. --Dan

1 0

Re: [math-fun] Question related to p-adics
by Michael Reid 30 Nov '06

30 Nov '06

> Is there a name for the ring which is the direct product of the p-adic > numbers for all primes p, and/or for the direct product of the p-adic > integers for all primes p? perhaps this is "notation" rather than a "name", but the latter one is (isomorphic to) Z^hat , which is the profinite completion of Z (both in the category of groups and in the category of rings). mike

1 0

[math-fun] Question related to p-adics
by franktaw＠netscape.net 30 Nov '06

30 Nov '06

Is there a name for the ring which is the direct product of the p-adic numbers for all primes p, and/or for the direct product of the p-adic integers for all primes p? The latter, in particular, has some nice properties. Just as a p-adic integer can be represented as a base p expansion with infinitely many places to the left of the decimal point, such an integer can be represented as an infinite base factorial number. It is also isomorphic to the homomorphisms on the group Q/Z (i.e., fractions with addition modulo 1). Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.

1 0

RE: [math-fun] Add odd digits, subtract even
by Eric Angelini 30 Nov '06

30 Nov '06

Ooooops, thanks Michael -- back to work :-( Best, É. [copy to MathFun] -----Message d'origine----- De : Michael Kleber [mailto:michael.kleber@gmail.com] Envoyé : jeudi 30 novembre 2006 19:30 À : Eric Angelini Objet : Re: [math-fun] Add odd digits, subtract even I think you have a mistake in that data... you show 25->32, but it should be 25->28(->18->11), right? --Michael On 11/30/06, Eric Angelini <Eric.Angelini(a)kntv.be> wrote: > > Hello MathFun, > have a quick look there, please (though this might be old hat): > http://www.cetteadressecomportecinquantesignes.com/134.htm > best, > É. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

1 0

[math-fun] Add odd digits, subtract even
by Eric Angelini 30 Nov '06

30 Nov '06

Hello MathFun, have a quick look there, please (though this might be old hat): http://www.cetteadressecomportecinquantesignes.com/134.htm best, É.

1 0

Re: Re: [math-fun] Ambiguously placed cities
by Daniel Asimov 30 Nov '06

30 Nov '06

This (I think) unsolved problem (that I mentioned recently) is so close to the present topic that I'm curious if anything at all is known about it (or if math-fun folks can solve it with their combined brainpower. QUESTION: Given a unit square in the plane, does there exist a point having its 4 distances to the vertices all rationals? (Or clearing denominators, the equivalent question as mentioned earlier: Q2: Does there exist an integer n such that given a square of side n in the plane, there is a point having its 4 distances to the vertices all integers? ) also, i'd like to understand why mike reid's statement << it is possible to have arbitrarily many points in the plane, no three collinear, such that all distances between them are integers, and no two such distances are equal. >> is true. ------------------------------------------------------------------- And what about the higher-dimensional analogues: Q3: What is the maximum number of points in R^n, no hyperplane (i.e., n-1 dimensional plane) containing n+1 of them, such that all interpoint distances are integers? ? --Dan

1 0