November 2011 - mathfun@mailman.xmission.com

[math-fun] Continuum-cardinality collections of continuum-cardinality subfields of real numbers
by Warren Smith 23 Nov '11

23 Nov '11

Some time ago, I discovered what seemed like a basic new result about the real numbers. The reason I thought it was new was (a) it was basic and had a fairly easy proof, (b) it was not mentioned in the usual "foundations of real analysis" books but due to (a) it should have been. However, I now suspect it is not new, the problem is merely that those books are crap. Anyhow, here are some of "my" results: 1. I can explicitly construct a dense subfield F of the real numbers, having the same cardinality as the real numbers, but omitting all algebraic irrationals. 2. I also can explicitly construct a continuum infinity of such subfields, call them F(x) where x, 0<x<1, is a real parameter, such that the intersection of any two of them is the rational numbers, i.e. they are "almost disjoint" -- and if any finite set of irrationals is chosen, each element from a different F(x), then that set will automatically be algebraically independent. [By "explicitly" I mean (among other things) that I do not require the axiom of choice.] 3. All those subfields each have measure=0. It is impossible for any additive subgroup of the reals to exist, which has positive measure. (The proof of this impossibility is mostly stolen from a recent mathfun post by Fred W. Helenius.) 4. Results similar to the above also work inside the complex numbers, quaternions, octonions, and "16ons." All of these subfields are disconnected. Some literature references, which I have not seen, which apparently prove similar statements, include: Jean Dieudonne: Sur les corps topologiques connexes, Comptes Rendus Acad. Sci. Paris 221 (1945) 396-398. Carsten Elsner: Uber eine effektive Konstruktion grosser Mengen algebraisch unabhangiger Zahlen, Mathematische Semesterberichte 47 (2000) 243-256. Isaac Kapuano: Sur les corps de nombres a une dimension distincts du corps reel, Rev. Fac. Sci. Univ. Istanbul (A) 11 (1946) 30-39. Hellmuth Kneser: Eine Kontinuumsmachtige, algebraisch unabhangige Menge reeler Zahlen, Bull.Soc.Math.Belg. 12 (1960) 23-27. Ernst Steinitz: Algebraische Theorie der Körper, Journal fur.Math 137 (1910) 167-309. Dieudonne and/or Kapuano actually apparently have shown existence of a CONNECTED (indeed, more strongly, "locally connected") dense subfield of the complex numbers (but not "arc connected") including one that is isomorphic with the real numbers. But I think in order to accomplish that they needed to use the Axiom of Choice so their proof of existence is highly nonconstructive. [It is impossible to make a connected additive subgroup of the reals, which is why they were doing complex numbers.] Questions: 1. Do they really require the Axiom of Choice? 2. If the rationals are removed from this connected subfield, does that disconnect it?

1 0

[math-fun] Heronian triangle lattice poses
by Fred lunnon 23 Nov '11

23 Nov '11

I finally caught up with Michael Reid's proof, and at least understand HOW his algorithm works --- even if I still have only the faintest notion WHY ... Essentially it computes a magic complex number via Gaussian integer GCD, representing a rotation about a chosen vertex at the origin, through which a chosen edge along the axis is rotated to the desired lattice pose. There are 6 ways to choose the vertex and edge, 4 Gaussian units to vary the GCD, and 2 conjugations of the rotation (yes or no), yielding in general 48 different possible results. If the triangle possesses one axial and one non-axial lattice pose, each can be translated, rotated and rotated trivially in 24 variations, yielding again 48 results. While I can't quite put my finger on exactly why, I think the proof must accommodate the following stronger result. Conjecture: (modulo trivial symmetries) right, axial non-right, non-axial Heronian triangles have respectively just one axial, one axial plus one non-axial, one non-axial pose. On 11/18/11, Michael Reid <reid(a)gauss.math.ucf.edu> wrote: > > Theorem. Any Heronian triangle can be embedded in the plane so its > vertices have integer coordinates. > Proof: First position the triangle so that A is at (0, 0) , B is > at (a, 0) . It is easy to show that vertex C has rational coordinates. > Consider the points as complex numbers; the vertices are all in Q(i) , > the fraction field of Z[i] . It is well-known that Z[i] , which is > the ring of Gaussian integers, is a PID, and even a Euclidean domain. > > Write the coordinate(s) of C as alpha / beta , in lowest terms. > We have |AC|^2 = N(alpha / beta) = alpha alpha' / (beta beta') > (where ' denotes complex conjugation, and N(z) is the norm > of z , i.e. z z' ). Since |AC|^2 is (the square of) an integer, > we have beta divides alpha alpha' . As it is relatively prime to > alpha , we have beta divides alpha' , and thus beta' divides alpha , > i.e. alpha = beta' gamma , for some gamma in Z[i] . Since beta > is relatively prime to alpha , it follows that beta is also > relatively prime to beta' . > > Also, |BC|^2 = N(a - alpha / beta) = > a^2 - a (alpha / beta + alpha' / beta') + alpha alpha' / (beta beta') > is an integer, so it follows that a (alpha / beta + alpha' / beta') = > a (alpha beta' + alpha' beta) / (beta beta') is an integer. Note that > beta is relatively prime to alpha beta' and therefore to > alpha beta' + alpha' beta . Take complex conjugates to see that beta' > is also relatively prime to alpha beta' + alpha' beta , and therefore > beta beta' is relatively prime to alpha beta' + alpha' beta . > However, beta beta' divides a (alpha beta' + alpha' beta) . This > now implies that beta beta' divides a , so we may write > a = b beta beta' , where b is a rational integer. > > Finally, rotate the triangle by multiplying by beta / beta' , which > has magnitude 1 . The vertices of the triangle rotate to 0 , > a beta / beta' = b beta^2 and alpha / beta' = gamma , which are > Gaussian integers. Therefore the vertices of the rotated triangle > have integer coordinates. QED > > Note that this gives a constructive way to find such an embedding. > > Michael Reid

1 0

[math-fun] Galois theory before abstract algebra?
by Henry Baker 22 Nov '11

22 Nov '11

Can anyone point me to a decent exposition (in English) available on the web of Galois theory prior to abstract algebra? I would imagine that this would have to be something written in the late 19th century. I want something written from the point of view of equations & roots, not rings, fields, extension fields, etc. Thanks in advance for any links.

3 2

[math-fun] Galois theory for dummies
by Warren Smith 22 Nov '11

22 Nov '11

http://books.google.com/books/about/Theory_of_equations.html?id=FfwjAAAAMAAJ Theory of equations James Víctor Uspensky 1948 -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 0

[math-fun] dragon triple points
by Bill Gosper 21 Nov '11

21 Nov '11

ACW> What about thirds and sevenths? http://gosper.org/drag3rds7ths.png The palindromic nature of the dragon will make the end segments congruent, but only steps of 1/5/2^n will coincide with the single-point "necks" connecting the lobes. Is this dragon-subdivision all old news? You don't need a magic, exact Drag function to predict that the lobes have area 1/5/2^n, n>-2, nor Julian's triple point {13/30, 21/30, 23/30}. But it's sure nice to be able to check! I would be interested to hear if anyone else has bothered to implement an exact rational dragon function. And especially an inverse. One mystery that bears investigating: Divide the domain [0,1] into 2^n equal subintervals, chaining 2^n little subdragons with necks at <integer>/5/2^(n+k). Many of these necks will lie in the interior of the big [0,1] dragon. Why aren't they triple points? --rwg On Thu, Nov 17, 2011 at 6:00 AM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: > In[135]:= $RecursionLimit = 9999; Clear[Drag];> Drag[t_, a1_: 1, a0_: 0] := Drag[t, b1_: 1,> b0_: 0] = (Drag[t, s1_: 1, s0_: 0] = ((a0 - s0)/(s1 - a1));> Module[{t2 = 2*t, n}, n = Floor[t2]; t2 -= n; Switch[n,> 0, (I + 1)/2*Drag[t2, a1*(I + 1)/2, a0],> 1, 1 + (I - 1)/2*Drag[1 - t2, (I - 1)/2*a1, a1 + a0], 2, 1]])> E.g.,> In[136]:= Drag[7/22]>> Out[136]= 23/82 + 39 I/82>> In[137]:= % - Drag[113/355]>> Out[137]= 517517964757026532787/96808512898827726880686 +> 650828352727980554837 I/96808512898827726880686>> In[138]:= N[%]>> Out[138]= 0.00534579 + 0.00672284 I> (The Dragon function is continuous.)>> Dividing the domain [0,1] into equal fifths,> ListLinePlot[> Partition[Table[{Re@#, Im@#} &@Drag[t/4/2048], {t, 8190}],1638],> AspectRatio -> Automatic, Axes -> None]>> may surprise you:> http://gosper.org/dragon5ths.png>> The obligatory continuous texture drift:> ListLinePlot[Table[{Re@#, Im@#} &@Drag[t], {t, 0, 1, 1/513}],> AspectRatio -> Automatic, Axes -> None]>> and> ListLinePlot[Table[{Re@#, Im@#} &@Drag[t], {t, 1/1024, 1, 1/513}],> AspectRatio -> Automatic, Axes -> None]>> http://gosper.org/driftdrags.png> --rwg

1 2

[math-fun] Heronian Tetrahedra have integer volumes
by Warren Smith 21 Nov '11

21 Nov '11

My tetrahedron-volume code had a bug, plus my brain was muddled. Upon repairing those, I find this: Computer explored all possible patterns of the 6 edge-lengths s01, s02.., s23 of the tetrahedron MOD 72 such that the 4 triangle squared-areas all were squares MOD 144 and such that gcd(144, s01, s02, ..., s23)=1. It computed for each pattern the determinant formula giving the alleged value of 144*TetVol. The result was always either a nonsquare mod 144, or 0 mod 144. This constitutes a corrected proof of the desired THEOREM: Every tetrahedron with integer edge lengths and triangle-areas, has either whole-integer or irrational volume. (This proof is probably overly brutal and I expect simpler proofs are possible by working mod 2,3,4,8,9 and/or 16 and doing chinese remaindering, but it'll do.)

1 1

[math-fun] backwards calculus-type problem
by Henry Baker 21 Nov '11

21 Nov '11

We all know that f'(x)=df(x)/dx is the limit of [f(x+deltax)-f(x)]/deltax as deltax approaches zero. However, what if I'm interested in a situation where deltax>0 doesn't approach zero. Suppose I have a variable y=f(x) and I want to consider a _fixed_ deltay. Is there any straightforward way to find an x such that deltax is minimized, yet deltay = f(x+deltax)-f(x) ??? There may be no such pair (x,deltax), or there may be a multiplicity of x's with the same minimal deltax. I'm primarily interested in finding one, but finding them all would be even better.

2 1

Re: [math-fun] Fast Divisionless Determinant and Characteristic Polynomial of Matrix
by Warren Smith 21 Nov '11

21 Nov '11

RCS pointed out some lecture slides showing that E.Kaltofen+G.Villard 2001-2002 knew my result, plus showed more strongly that you can get determinant without division in O(n^2.698) operations, which is quite impressive. I'm not completely sure, but I think the way they did it (maybe?), is equivalent to the following: basically the same as my approach, except you regard it as a block matrix, i.e. each matrix entry in my approach is really a bXb block, and they choose the blocksize b to optimize the runtime. What I actually want: an actual practical program to find det(matrix) without divisions reasonably efficiently.

2 1

[math-fun] Exponential tower parenthesization counts
by Warren Smith 21 Nov '11

21 Nov '11

(Was "A nice sequence" but that title sucked) Perhaps (A^B)^C = (A^C)^B is the only identity that one needs to worry about generically. If so, then the count for an n-high tower is going to obey the recurrence T(n) = sum(a,b,c with a+b+c=n and a,b,c>0)of 11 * T(a) * T(b) * T(c) with 11 not 12. Except actually, this is not quite the right recurrence. But anyway this one would be soluble using generating functions.

2 1

Re: [math-fun] A Nice Sequence
by N. J. A. Sloane 21 Nov '11

21 Nov '11

T he Guy-Selfridge paper cited in A198683 is relevant (for Warren's question) Neil

1 0