www.ConcurrentInverse.com 2005 October 11 (c) 2005 Wm.C.Corwin So far it appears with the pentagon, dodecahedron and icosahedron that all dimensions can be expressed in terms of any other dimension with a factor of the form [i/j][k+l[5]] . i,j,k,and l are integers. Dividing or multiplying one such form with another results in the same form. It is not immediately obvious if such numbers can be added. There are occasions in the derivation of the factors in summary.txt where they can be added. Adding one to the ratio is equivalent to adding two numbers. So we will explore adding one to numbers of the form [i/j][k+l[5]] ---------------------------------------------------------------------------- -------------------------------------------------------------------------- The general case will have to be considered: [k+l[5]] = 1 + [h+f[5]] where k,l,h, and f are rationals. It the above is possible we will separate the rational and irrational parts after eliminating the radicals. k + l[5] = 1 + 2[h+f[5]] + h + f[5] k-1-h + (l-f)[5] = 2[h+f[5]] {k-1-h} + 5{l-f} + 2(k-1-h)(l-f)[5] = 4h + 4f[5] {k-1-h} -4h + 5{l-f} = 0 2(k-1-h)(l-f) -4f =0 {k-h} -2(k+h) +1 +5{l-f} =0 2(k-h-1)(l-f) -4f =0 Separating out the rational and irrational parts: {k-h}-2(k+h)+1+5{l-f}= 0 and 2(k-h)(l-f)-2l-2f =0 (k-h)(l-f) -(l+f) =0 kl -hl -kf +hf -l -f =0 kl -l(h+1) -f(k+1) +hf =0 kl -l(h+1) = f(k+1) -hf kl -(k+1)f = l(h+1) -hf k(l+f) =f = l +h(l-f) -(l+f) = (l-f)(h-k) h = k - (l+f)/(l-f) h -k = (f+l)/(f-l) h = k + (f+l)/(f-l) k = h + (l+f)/(l-f) Fortunately we can choose f and l to be rationals and then k-h is a rational making the otherwise quadratic equation linear in h, and hence k is also rational. choose l f then get k h 2 1 21/4 9/4 k = h + 3 k -h = 3 k +h = 2h + 3 9 -2(2h+3) +1 +5 =0 h = 9/4 k = 21/4 [21/4 + 2[5]] = 1 + [9/4 + [5]] [21+8[5]] = 2 + [9+4[5]] = 6.23606797749978969640 choose l f then get k h 3 1 29/4 21/4 [29+12[5]] = 2 + [21+4[5]] = 7.47213595499957939 = 7.47213595 choose l f then get k h 3 2 41/4 21/4 [41+12[5]] = 2 + [21+8[5]] = 8.236067977499 = 8.236067977 The relations are denumerable in number. If l-f is 1 then k-h is an integer and the identity will have a one in four chance of using the 1 in the identity, otherwise k-h will be fractional making the rational constant term other than one, unless by chance x/{n} + y/n + z turns out to be an integer. If a particular h and f are desired as the independent variables the procedure could be used iteratively making a numerical method to see how close an identity could be found. Maybe tables would be useful. Since it is awkward for h to be an independent variable we can take an approach of exhausting the possibilities. For a given h what are the limits on k and l for a fixed given f ? {k-h}-2(k+h)+1+5{l-f}= 0 k - h = (l+f)/(l-f) k + h = 2h + (l+f)/(l-f) {(l+f)/(l-f)} -4h -2(l+f)/(l-f) +1 +5{l-f} =0 4h = {(l+f)/(l-f)} -2(l+f)/(l-f) +1 +5{l-f} k = h + (l+f)/(l-f) l f k h 2 1 21/4 9/4 = 2.25 3 1 29/4 21/4 = 5.25 3 2 41/4 21/4 = 5.25 4 1 4 2 7/2 13/2 = 6.5 4 3 41/4 69/4 =17.25 5 1 5 2 5 3 35/4 19/4 = 4.75 5 4 6 1 6 2 6 3 31/4 19/4 = 4.75 6 4 23/2 13/2 = 6.5 6 5 9 3 39/4 31/4 = 7.75 29 23 45/4 29/4 ... 45/2 15/2 ? ? 461/4 381/4 21 19 981/4 861/4 31 29 ... ----------------------------------------------------------------------------- The well known relationship of Phi = 1 + 1/Phi where Phi = (1+[5])/2 gives [6+2[5]] = 2 + [6-[5]] [3/2 + [5]/2] = 1 + [3/2 - [5]/2] for l f k h 1/2 -1/2 3/2 3/2 in agreement with the above formulas. So Phi corresponds to one, with low number coefficint s, in the series of the above identities . l f k h 3/2 1/2 7/2 3/2 5/2 3/2 15/2 7/2 =1+2.6180339887 ... ---------------------------------------------------------------------------- What integer values of l,f,k,h are possible? {k-h}-2(k+h)+1+5{l-f}= 0 k - h = (l+f)/(l-f) k + h = (1 + 5{l-f} + {(l+f)/(l-f)})/2 54, 83 ? 137, 29 ? Adding 1 to the form [h+f[5]] at best results in half integer k,l,h,f so [2] and 2 is considered at http://www.issi1.com/corwin/calculator/add_2.txt and demonstrated at http://www.issi1.com/corwin/calculator/add_2.html ----------------------------------------------------------------------------- Speculation: Besides things like the distance between inscribed and circumscribed circles and spheres, these identities may have other uses. Also, maybe these identities are properties of space in three or maybe even five dimensions corresponding to the five regular solids and may serve as eigen like values for the existance of solutions of spatial relationships. ---------------------------------------------------------------------------- In example 7.2 of Sacred Geometry, Philosophy and Practice by Robert Lawlor, the author marvles over the seeming coincidence of points lining up. It may correspond to other addition identities. related http://www.issi1.com/corwin/platonic.txt http://www.issi1.com/corwin/calculator/summary.txt http://www.issi1.com/corwin/calculator/pentagon.txt other related references: Pythagorean triangles: 3,4,5 ... etc http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/Pythag.html#mnformula 360deg/n trigonometric values by Gauss http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/simpleTrig.html#solved oscillating spheres, natural occurrances, but mystic http://ascension2000.com/DivineCosmos/03.htm fine structure constant alpha(n1,n2) = n2*cos(pi/n1)*tan(pi/(n1*n2))/pi http://www.fine-structure-constant.org http://www.btinternet.com/~ugah174/strong4.gif ---------------------------------------------------------------------------- APPENDIX Furtive preliminary oversimplified attempts: A possible addition could be reduced to the form [k+l[5]] = 1 + [h+[5]] , where h,k,and l are rational and it would cover the cases where i,j,k,and l of the two added numbers are integers or even rationals. For the rational and irrational parts to separate {k-(h+1)} + 5{l-1} = 4h where {x} denotes x**2 l(1+h) -lk + k = 1-h For h = 1 there are no solutions ------- h=3 has solutions l=1+[2] k=4+[2] For h=5 6l-lk+k=4 and {k-6}+{l-1}=20 k=10 and l=3/2 gives 16+5/4=17+1/4 < 20 so there are solutions This is way overdetermined so the more general case will have to be considered. However the above special case for f=1 will serve to check further work.