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Today, a friend gave me a "proof" of $1=2$ and challenged me to find the fallacy.

$1 = 1$

$1 = 1 + 0 + 0 + 0 ...$

$1 = 1 + 1 - 1 + 1 - 1 + 1 - 1 ...$

$1 = 2 - 1 + 1 - 1 + 1 - 1 ...$

$1 = 2 + 0 + 0 ...$

$1 = 2$

My answer was that once you turn the initial $1 + 1$ into a 2, everything is offset so a $-1$ is always left at the end no matter how many times it is repeated. This negative one balances out the $2$ at the beginning so $1=1$ still holds true. I.e.

$$1 = 1 + 0 + 0 + 0 ... = 1 + (1 - 1) + (1 - 1) + (1 - 1) = 2 + (-1 + 1) + (-1 + 1) - 1$$

However, my friend claimed that my answer only applies if the $+ 1 - 1$ repeats for a finite number of times. He argues that because the sequence repeats infinitely and things work differently when working with infinity, my answer is not valid.

Can anyone enlighten me to the true fallacy in this proof?

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I don't have the time to give an answer right now and I'm sure someone will, but there's one thing you should think about first: what does the symbol $1+1-1+1-1+1-1\cdots$ really mean? There's a rigorous way to define it and that's exactly where the problem lies. Infinity takes a huge role on this matter. So that people can better help you, are you familiar with the concept of series? –  Git Gud Jun 20 '13 at 7:18
I do have a basic understanding of series. –  tangrs Jun 20 '13 at 7:24
I had similar question about that: math.stackexchange.com/questions/417280/… –  Mher Safaryan Jun 20 '13 at 7:24
Thanks for that, I think my question may actually be a duplicate of that. –  tangrs Jun 20 '13 at 7:26
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4 Answers

up vote 5 down vote accepted

Some users pointed me to read up on converging and diverging series.

As I currently understand it, equating $1+0+0+0... = 1+1−1+1−1+1−1...$ is the fallacy because the series on the right does not converge (much less to 1) - therefore, they are not equal.

To prove that this is the fallacy, we can use convergent tests to show that the two sides of the equation are not equal.

Am I correct in my deduction?

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Pretty much so. The symbol $1+1−1+1−1+1−1\ldots$ is just a different way of writing $\sum \limits_{n=0}^{+\infty}a_n$, where $a_0=1$ and $a_{n+1}=(-1)^n$, for all $n\in \Bbb N_0$. And now what does $\sum \limits_{n=0}^{+\infty}a_n$ mean? Well, that's just $$\sum \limits_{n=0}^{+\infty}a_n=\lim \limits_{m\to+\infty}\left(\sum \limits_{n=0}^{m}a_n\right)$$ if the limit exists, but this limit doesn't exist, so it doesn't make sense to talk about the series. A simple way to see the series doesn't converge is to note that $(a_n)_{n\in \Bbb N_0}$ doesn't converge to $1$, (it does not converge). –  Git Gud Jun 20 '13 at 7:58
Thanks, that explains it perfectly –  tangrs Jun 20 '13 at 8:09
I meant doesn't converge to $0$, above. –  Git Gud Jun 20 '13 at 14:30
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The "proof" uses Grandi's series. The sum of a divergent series depends on how it is bracketed. Without brackets:

$1-1+1-...=x\\ 1-x=x\\ x=\frac{1}{2}$

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The third equation your friend presented to you is nonsense. The sum on the right does not converge.

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Look carefully at the original "proof" again. The fifth step conveniently leaves out an item in the fourth step, namely the final -1 It should read 1 = 2 + 0 + 0 - 1, leading to the conclusion 1 = 1.

I think you noticed this too. Some people, like your friend, seem to think that just because you can go on repeating for all eternity, or stop repeating whenever you like, an expression like +1-1 without making any difference, then somehow it's equally OK to break off in the middle of that expression if it suits you. That's the fallacy.

I also agree with Angela Richardson that a lot depends on how you bracket, and I make that point in more detail in answer to question 427635 about Grandi.

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