# Finding the fallacy in this broken proof

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 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
I think the truth reason is "addition of an infinite number of terms cannot be assumed to be associative or commutative. In particular, rearranging terms in an infinite sum can actually change the sum." says here math.stackexchange.com/questions/427635/… – iMath Apr 19 '15 at 8:39

Let's acknowledge that the sequence $1,0,1,0,1,0,...$ is divergent. The result will follow from this fact. Now, consider the series $$\sum_{n=1}^{\infty}(-1)^{n-1}=1-1+1-1+\ldots\:\:\:\:\:\:\:\:\:\:\:\:(*)$$ which appears on your calculation. By definition, we know that the limit of the sequence of partial sums is the limit of the series. Sequence of the partial sums of our series is as follows:

$$(s_n)=1,0,1,0,1,0,\ldots$$ which is divergent as we know and indicated in advance.

Therefore, treating the RHS of the equation $(*)$ as a number is mathematically incorrect, so calculations may yield wrong results.

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The fallacy is that bracketing isn't necessarily a valid operation for an infinite series. In this case, it doesn't work well at all.

The reasoning goes that if you bracket this way:

   x = (1 - 1) + (1 - 1) + ... then x = 0


and if you bracket this other way

   x = 1 + (-1 + 1) + ... then x = 1


but the problem here is that "bracketing" doesn't quite work correctly in this case.

These equations require you to append two numbers at a time to the series to maintain the 0 or 1 sum. That seems to cause problems. Which sum you get depends on whether you "stop" at an even numbered term or at an odd numbered term. But that's the fallacy, there is no "stopping".

In other words, this is true:

   x = (1 - 1) + (1 - 1) + ... then x = 0


but only if you "stop" when there is a "...(1-1)" at the "end" of the series. But what if you "stopped" when there is a "...(1-1) + 1" at the "end" of the series? Then x = 1.

Conversely,

   x = 1 + (-1 + 1) + (-1 + 1) + ... then x = 1


is only true if you "stop" when there is a "...(-1 + 1)" at the "end" of the series. But what if you "stopped" when there is a "...(-1 + 1) - 1" at the "end of the series? then x = 0.

The fallacy is there is no "end" and you can't "stop" when you are processing an infinite series. There's always one more term.

A better technique is to accept the series are infinite (be one with infinity ahem) and then use addition of two infinite series:

        x =   1 - 1 + 1 - 1 ...
-1 +  x =  -1 + 1 - 1 + 1 ...
-1 + 2x =   0 + 0 + 0 + 0 ....
-1 + 2x =   0
2x = 1
x = 1/2


This is valid because:

• the series "x" and the series "-1 + x" are both infinite but have the same cardinality/length (according to Cantor).
• since they are the same length we can match up each term in the first series to each term in the second.
• they continue to match up nicely to an infinite number of terms
• and so the "0 + 0 ..." continues all the way to infinity
• an infinite sum of 0's is 0
• the rest of the elementary operations are also valid

John

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I'm sorry for posting this late, I just wrote this up as an answer for this (Why is this $0 = 1$ proof wrong?) question before it was marked duplicate. Maybe it has some tiny new bits in it

One possible answer: The proof is wrong because infinite summation is not associative (as is for example shown by this :D )

It is however interesting because it shows that when dealing with infinite series, one has to be careful what one is actually talking about and therefore introduce some notions that might seem technical at a first glance. For example there is usually made a distinction between the sequence of partial sums and it limes.

If the latter one exists (i.e. if the series converge), one sometimes identifies it with it's series, but this identification can lead to problems as in the argument you show: In the first and second line, you are dealing with convergent series, so you can make the identification of limes (left hand side) and sequence of partial sums (right hand side). In the third line, the only way to make sense of the right hand side is a sequence of partial sums, the left hand side is a number however, so we cannot write down equality. The last lines have again a convergent sum on the right, but the sequence of partial sums is different from the one in line two!

I hope this was not too technical, feel free to ask for explanations!

By the way, it is very interesting, that in other contexts, precisely the above argument can be used to show interesting (true) things, for example that you cannot unknot a knot by adding another one to it (check http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle if you are interested)

Also the proofs of the "sum of natural numbers equals minus one over 12" proofs you mention are probably wrong (Is it the youtube videos?) and i would say not accepted fact (and i really mean the proofs, the statement can still be made sense of in a different way) but still a good way to generate interest for these questions i guess :)

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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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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

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

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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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