# Help debunk a proof that zero equals one (no division)?

Unlike the more common variant of proof that 0=1, this does not use division.

So, the reasoning goes like this:

\begin{align} 0 &= 0 + 0 + 0 + \ldots && \text{not too controversial} \\ &= (1-1) + (1-1) + (1-1) + \ldots && \text{by algebra}\\ &= 1 + (-1 + 1) + (-1 + 1) \ldots && \text{by associative property}\\ &= 1\\ \\ &\therefore 0 =1 \end{align}

I can't help but feel that something went wrong here, specifically with the use of the associative property. However, I can't come up with a mathematically compelling reason.

Where's the error?

• The reason this proof doesn't work is because the associative property doesn't hold for infinite sums. Why doesn't it hold for infinite sums? Precisely because this proof gives a counterexample. ;) – Dustan Levenstein Nov 27 '13 at 6:02
• The second line is incorrect since $\sum_{n=0}^\infty (-1)^n\not\in \mathbb{R}$ – JMag Nov 27 '13 at 6:02
• By distributive property did you reshuffle the parenthesis? If so you aren't allowed to change the order of addition in an infinite sum like that. Unless we have a very nice series. I think J.Maglione's answer is the best. – TheNumber23 Nov 27 '13 at 6:03
• Brain fart, I've edited to change to "associative" now. Dustan, you have an interesting argument, but at the moment it feels like circular reasoning. I'll mull over this now. Thanks! – danmcardle Nov 27 '13 at 6:05
• It's not circular reasoning; the fact of the matter is you technically had no reason to believe that the manipulations were valid in the first place, since the rules for algebra are only given for finite sums and products. In the theory of infinite series, much of the intuition that you've gotten from algebra breaks down. – Dustan Levenstein Nov 27 '13 at 6:11

Not all algebraic rules generalize to infinite series in the way that one might hope. When they fail, it is because something fails to converge. In this case, what fails to converge is the series that should appear between the two lines in the middle of the "proof": $$1-1+1-1+1 \cdots.$$ Indeed, this series fails to converge because the sequence of partial sums $\{1, 1-1, 1-1+1,\ldots\}$ oscillates between $1$ and $0$ and does not converge to any value.
• Maybe to put another nail in the coffin, you can use $\epsilon=1/2$ to show the series does not converge. – DBFdalwayse Nov 27 '13 at 6:10
• @DBFdalwayse True, although I think it's fairly intuitive that the sequence $\{1,0,1,0,\ldots\}$ does not converge. – Trevor Wilson Nov 27 '13 at 6:12