Where is the flaw in this proof?

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I walked into the printer room at my place of work and saw this proof written on the white board. I can't really find anything wrong with it, but I figured you guys have probably seen it a thousand times. I don't know what it's called, or I would google for it.

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

Somewhere at the end there's another -1 tacked on, except that it goes on forever. >.<
 

Answers and Replies

  • #2
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Theres a -1 missing, if you see you had 6 numbers in each step, but in the last one 7, so you're missing a -1
 
  • #3
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The equations

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

and

1 = 1 + (-1 + 1) + (-1 + 1) + ...

are correct, so it is simple thing to conclude that the equation

(1 - 1) + (1 - 1) + ... = 1 + (-1 + 1) + (-1 + 1) + ...

is incorrect, since it is equivalent with 0 = 1. That is where the mistake is.

Your notation

1 - 1 + 1 - 1 + ...

is merely a point where you lose track of what your own notation means. Do you know what you mean by this series? Is it supposed to be a number? The series doesn't converge in standard sense at least.
 
  • #4
Hurkyl
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First off -- do you realize that the expression
0 + 0 + 0...​
doesn't involve the addition operation you learned as a kid?

It couldn't possibly be, because the addition operation you learned as a kid is a binary operation -- it works on two arguments. You can iterate it, but you can only make finite expressions out of it, not infinite ones.

Instead, this is an 'infinite sum' operation, which comes with its own properties and rules for being manipulated.


(Okay, technically, the expression is a little ambiguous. It could be interpreted as an infinite sum, or it could be interpreted as a few applications of elementary-school addition followed by an infinite sum -- either way, the expression has the same value)
 
  • #5
mathman
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The infinite series 1 -1 +1 -1 .... does not converge. It has 2 limit points 1 and 0 and oscillates between them.
 
  • #6
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I thought I studied this sort of thing in college, this is called a conditionally convergent series. If we rearrange terms, we can get different results.

For example since there are infinite positive and negative terms, we could rearrange the series as -1 + (1-1) +(1-1).....=-1.
 
  • #7
HallsofIvy
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Yes, but "rearranging" is not what is going on here. Rearranging means changing the order of the terms. Here you are just shifting parentheses. It is more a matter of the "associative law" not being true for infinite series.
 
  • #8
By rearranging you could get indeed any integer value as the "sum" of the series. Just pick as many 1's or -1's as you need (luckily there are infinitely many) and then have the rest in the form (1-1) in order not to flaw your result:smile:
 
  • #9
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Every time a man has tried to sum an infinite series, he has died before ever completing it.

Like HallsofIvy said, you can only sum a finite series of numbers. Once you have an infinite series, you need to resort to analysis. What you *actually* do for these problems is take the LIMIT of the partial sums. Because there are many series which do converge and are very useful to us, we often write [tex]\Sigma_k^\infty s_k[/tex] instead of [tex]\lim_{n->\infty}\Sigma^n_k s_k[/tex].

Of course, with the series 1 - 1 + 1 - 1 + 1 - 1 + ..., we aren't getting closer to any final answer. The infinite sum isn't 0, because you can never "approximate" 0 within an error of less than 1. And neither is it equal to 1 or -1, because at no point does the sequence start to favor one over the other.
 
  • #10
Hurkyl
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Like HallsofIvy said, you can only sum a finite series of numbers.
Of course you can sum infinite series of numbers. For example,
What you *actually* do for these problems is take the LIMIT of the partial sums.
is exactly what "summing an infinite series" means! (for the elementary calculus definition of "sum")

The thing that you cannot do is iteratively apply the binary + operation an infinite number of times.
 
  • #11
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Of course you can sum infinite series of numbers. For example,

is exactly what "summing an infinite series" means! (for the elementary calculus definition of "sum")

The thing that you cannot do is iteratively apply the binary + operation an infinite number of times.

"Sum" means two different things in the two contexts. When I say you can't sum an infinite series, I mean precisely that standard iterative methods fail to terminate. It's convenient to talk about it as a simple application of addition, but it's not, and so we need calculus to solve it!

I just like saying it that way, because I think a lot of non-math people get this idea that a mathematician is someone who knows how to plug 1 - 1 + 1 - 1 .... into a calculator faster than a normal person and somehow manager to get an answer out of it!
 
  • #12
Gib Z
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I thought I studied this sort of thing in college, this is called a conditionally convergent series. If we rearrange terms, we can get different results.

For example since there are infinite positive and negative terms, we could rearrange the series as -1 + (1-1) +(1-1).....=-1.
Not exactly. Conditionally Convergent means that a series, as it is with its terms of alternating sign, converges, but the sum of the absolute value of each term diverges. A good example may be [tex]\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}[/tex], which converges to ln 2, but when we take the absolute value of each term, it diverges.

Riemann proved that any conditionally convergent series may be rearranged to give any sum wanted, or +/- infinity.

The reason this isn't a conditionally convergent series is that, although it does exhibit that last property, it doesn't conditionally converge to begin with.
 
  • #13
mathman
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A necessary (but not sufficient) for a series to converge is that the terms in the sequence tend to 0. For this series, it is obviously not so.
 
  • #14
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There is a theorem due to Riemann that a conditionally convergent series can be made to sum to any required value by rearrangement, the same holds for oscillating 'divergent' series through asymptotic considerations.

The real flaw with that 'proof' becomes evident if you use a more general expansion using power series or Dirichlet series, the problem of it being inconsistent is due to an extra term being introduced to the power series on one side and not the other, whence the equality ceases to hold. But this is the most general case when it comes to resummations because they can be consistent.
 

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