Doron Posted October 17, 2003 Report Share Posted October 17, 2003 Hi, L1 is a circle with a perimeter's length 1. / is "divided by" n > 2 RFC is the roughness of some polygon which is closed by L1 circle (all its vertices are ON the circle) and calculated as L1/n. A circle is not a polygon, therefore we are talking about its roughness-magnitude, which is notated as RFA and calculated by L1/2^aleph0 (=0) A 1 dimensional closed geometrical object is not a circle and it is not a polygon. Let us call it SC for semi-circle. RFB is the roughness-magnitude of a SC which is closed by L1 circle (all its points are ON the circle) and calculated as L1/aleph0 (>0). I think that SC must exist if the circle exists as a geometrical object. More to the point: Let us say that [oo] = 2^aleph0 = c (has the power of the continuum). Therefore: 1/[oo] = 0 1/0 = [oo] The first known transfinite cardinals are aleph0 and 2^aleph0. 2^aleph0 = c (has the power of the continuum) therefore RFA = 1/2^aleph0 = 0 (there are no "holes"). aleph0 < c (does not have the power of the continuum) therefore RFB = 1/aleph0 > 0 (there are "holes"). y = the size of the radius of L1 circle The magnitude of 2^aleph0 is greater than the the magnitude of aleph0. In the case of RfB there are aleph0 radii with r=y, but there are 2^aleph0 radii with r < y. aleph0 < 2^aleph0, therfore r < y. S2 is the area of a RFA geometrical object (a circle). S1 is the area of a RFB geometrical object (a SC). In case of RfA we know that S2 = 3.14... r^2 (where r=y). But in case of RfB r < y, therefore S1 = 3.14... r^2 (where r < y). S2 - S1 = x where x > 0. A question: can we use x to ask meaningful questions on the CH problem ? Link to comment Share on other sites More sharing options...
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