- #1
Organic
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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, therefore 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 ?
Organic
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, therefore 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 ?
Organic