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anonymous
 4 years ago
More fun:
0 = 0
=0+0+0+0+0+...
=(11)+(11)+(11)+(11)+...
=11+11+11+11+...
= 1+ 1 +1 + 1 +1 + 1 +1 +...
= 1 + (1+1) + (1+1) + (1+1) +...
= 1 + 0 + 0 + 0 + 0 + ...
= 1 + 0
= 1
Therefore 0 = 1, Q.E.D.
Where's the problem?
anonymous
 4 years ago
More fun: 0 = 0 =0+0+0+0+0+... =(11)+(11)+(11)+(11)+... =11+11+11+11+... = 1+ 1 +1 + 1 +1 + 1 +1 +... = 1 + (1+1) + (1+1) + (1+1) +... = 1 + 0 + 0 + 0 + 0 + ... = 1 + 0 = 1 Therefore 0 = 1, Q.E.D. Where's the problem?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't need help, I just like to post fun things while I wait for other stuff. Can you find the problem in this reasoning? (there has to be a problem, or else all numbers equal each other and cease to have meaning!)

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2there will be 1 at last?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am not sure that can done or not because we dont know how many how many terms are there in 0+0+0+0+0... Infinity is unreachable.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let there are n terms in 0+0+0+0+... Then Now, let 0=(11) Then (11)+(11)+... 11+11+11... Now this will have 2n terms........ and 2n will be even numbers. Now, 11+11+1... =1(11)(11)+... If u will group that like that. and to get 1 there must be odd number of terms But we know it has 2n terms which is even. Thus, it cannot be 1

zugzwang
 4 years ago
Best ResponseYou've already chosen the best response.1How about \[0 = \sum_{0}^{\infty}0\]is fine \[0=\sum_{0}^{\infty}(1  1)\]is fine But \[0 = 11+11+11... = \sum_{0}^{\infty}(1)^{n}\] is not fine ...?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(11)+(11)+(11)+(11)+... \ne1 + (1+1) + (1+1) + (1+1) +..\]Evidently. But why? http://en.wikipedia.org/wiki/Grandi's_series http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1error is from line four to line five a term is missing =11+11+11+11+... = 1+ 1 +1 + 1 +1 + 1 +1 + 1 ...
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