- #1

Organic

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L1 is a circle with a perimeter's length 1.

/ is "divided by"

n>2

**C is the roughness of some polygon which is closed by L1 circle (all its vertices are ON the circle) and calculated as L1/n.**

*RF*A circle is not a polygon, therefore we are talking about its roughness-magnitude, which is notated as

**A and calculated by L1/2^aleph0 (=0)**

*RF*A 1 dimensional closed geometrical object is not a circle and it is not a polygon. Let us call it SC for semi-circle.

**B 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).**

*RF***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

**A = 1/2^aleph0 = 0 (there are no "holes").**

*RF*aleph0 < c (does not have the power of the continuum) therefore

**B = 1/aleph0 > 0 (there are "holes").**

*RF*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

**B there are aleph0 radii with r=y, but there are 2^aleph0 radii with r<y.**

*Rf*aleph0 < 2^aleph0, therfore r<y.

S2 is the area of a

**A geometrical object (a circle).**

*RF*S1 is the area of a

**B geometrical object (a SC).**

*RF*In case of

**A we know that S2 = 3.14... r^2 (where r=y).**

*Rf*But in case of

**B r<y, therefore S1 = 3.14... r^2 (where r<y).**

*Rf*S2 - S1 = x where x > 0.

A question: can we use x to ask meaningful questions on the CH problem ?

Organic