>>9574403

I have a square

I cut it in half and fill half of what i just cut

I now have 1/2 of a square filled and 1/2 of a square unaccounted for.

I cut the unfilled half in half and fill half of what i just cut

I now have 1/2 + 1/4 of a square filled, and 1/4 unaccounted for.

I cut the unfilled half in half and fill half of what i just cut.

I now have 1/2 + 1/4 + 1/8 of a square filled, and 1/8 of a square unaccounted for.

After every cut, you are that cut's value away from the full square

If you did this up to 1/2 + 1/4 + 1/8 + ... + 1/562,949,953,421,312 then there would still be 1/562,949,953,421,312 of the square left unaccounted for. Another cut will double this significand and HALVE the previous value to 1/1.12589991E+15, leaving 1/1.12589991E+15 of the square left unaccounted for. There needs to exist a value [math]\frac{\frac{1}{2^n}}{2} =

0 [/math] in order to no longer have any part of the square left unaccounted.

The problem of this square is the same as [math]\sum_{n=1}^{\infty} \frac{1}{2^n}[/math], which the sum shows an issue. If we assume Sum1/2^n is equal to 1 and leave the limit undefined and wish to know what the limit is, we can compare it to a problem [math]\sum_{n=1}^{x} \frac{9}{10^n} = 0.999 [/math]

n1: 0.9

n2: 0.99 (+0.09)

n3: 0.999 (+0.009)

Thus x=3 is our limit to solve this problem

Take note at x=3, the value added is not insignificant, which is why x=3

Back to [math]\sum_{n=1}^{x} \frac{1}{2^n} = 1[/math] where we wish to define x, if we set the limit to infinity, we have that the infinite'th partial sum is [math]\frac{1}{2^{\infty}} = \frac{1}{\infty} = 0[/math], which is an insignificant value. Whether this 0 is accounted for or not doesn't change the total sum, so this means that the end of significant summing did not occur at n=infinity, it occurred before at a value less than infinity. Since any real number is less than infinity, we are redirected that x= any R, but x= any R is obviously not 1.

Infinity is broken.