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Old 2022-05-18, 13:36   #23
retina
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Default The truth about 1+2+3+...=-1_12

Nice vid.

https://www.youtube.com/watch?v=YuIIjLr6vUA

Supersums, Zeta and Eta functions, are also included for some extra spice.
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Old 2022-05-20, 02:14   #24
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Quote:
Originally Posted by retina View Post
Good job.

Now we have two answers.

A > B, and
B > A

Anyone want to speculate now that A = B?
https://en.wikipedia.org/wiki/Schr%C...nstein_theorem
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Old 2022-05-20, 02:40   #25
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Quote:
Originally Posted by paulunderwood View Post
Good job.

Although that says that |A| = |B|. Which is what VBCurtis suggested up thread.

Last fiddled with by retina on 2022-05-20 at 02:41 Reason: Linky
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Old 2022-05-20, 14:42   #26
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Default 2^∞ = 0

The infinite product of 2×2×2×2×2×2×2... = 0

If we set p = 2×2×2×2×2×2×2... we can see that 2×2×2×2×2×2×2... = 2×p

Leading to p = 2×p

Subtract p from both side; 0 = p

Giving 2×2×2×2×2×2×2... = 0



Sadly the Mersenne number 2^∞-1 = -1, and therefore isn't prime.
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Old 2022-05-20, 16:15   #27
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Quote:
Originally Posted by retina View Post
The infinite product of 2×2×2×2×2×2×2... = 0

If we set p = 2×2×2×2×2×2×2... we can see that 2×2×2×2×2×2×2... = 2×p

Leading to p = 2×p

Subtract p from both side; 0 = p

Giving 2×2×2×2×2×2×2... = 0



Sadly the Mersenne number 2^∞-1 = -1, and therefore isn't prime.
\[2^{\aleph_0} - 1 = \aleph_1\] and I am not sure primeness is defined for these.
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Old 2022-05-21, 01:26   #28
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Quote:
Originally Posted by paulunderwood View Post
2^{aleph_0} - 1 = aleph_1 ...
I showed my working to arrive at the real answer of 0.
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