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More fun: 0 = 0 =0+0+0+0+0+... =(1-1)+(1-1)+(1-1)+(1-1)+... =1-1+1-1+1-1+1-1+... = 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?

Mathematics
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I 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!)
there will be 1 at last?
*-1

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Other answers:

I 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.
Sorry for typo
yep u r right !!!
Let there are n terms in 0+0+0+0+... Then Now, let 0=(1-1) Then (1-1)+(1-1)+... 1-1+1-1+1-1... Now this will have 2n terms........ and 2n will be even numbers. Now, 1-1+1-1+-1... =1-(1-1)-(1-1)+... 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
How about \[0 = \sum_{0}^{\infty}0\]is fine \[0=\sum_{0}^{\infty}(1 - 1)\]is fine But \[0 = 1-1+1-1+1-1... = \sum_{0}^{\infty}(-1)^{n}\] is not fine ...?
\[(1-1)+(1-1)+(1-1)+(1-1)+... \ne1 + (-1+1) + (-1+1) + (-1+1) +..\]Evidently. But why? http://en.wikipedia.org/wiki/Grandi's_series http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle
error is from line four to line five a term is missing =1-1+1-1+1-1+1-1+... = 1+ -1 +1 + -1 +1 + -1 +1 + -1 ...

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