Question

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?

Answer 1

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!)

Answer 2

there will be 1 at last?

Answer 3

*-1

Answer 4

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.

Answer 5

Sorry for typo

Answer 6

yep u r right !!!

Answer 7

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

Answer 8

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 ...?

Answer 9

\[(1-1)+(1-1)+(1-1)+(1-1)+... \ne1 + (-1+1) + (-1+1) + (-1+1) +..\]Evidently. But why?

http://en.wikipedia.org/wiki/Grandi's_series
http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle

Answer 10

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 ...