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Another fallacy (please help)

  1. Dec 11, 2013 #1
    hey guys this day my friend showed me a falacy that stunned my mind, it was like this:

    let's say we add powers of 2 til the infinity,
    so 1+2+4+8+16+32......
    we don't know the answer, right? but we know we can multiply the whole thing by one, so
    1*(1+2+4+8+16+32....) = 1*1+1*2+1*4...... same thing
    now, we know that 1 = 2-1,
    so
    1*(1+2+4+8+16+32....) = (2-1)*(1+2+4+8+16+32....)
    1+2+4+8+16+32.... = 2+4+8+16+32+64...... -1-2-4-8-16-32
    almost every numbers and it's negative counterpart cancel each other
    so
    1+2+4+8+16+32.... = [STRIKE]2+4+8+16+32+64[/STRIKE]...-1[STRIKE]-2-4-8-16-32[/STRIKE]
    so
    1+2+4+8+16+32.....= -1

    i want a convincing reason why this is false, a very convincing one (e.g.: not logic)
     
  2. jcsd
  3. Dec 11, 2013 #2
    The falacy is the fact that they math you did assumes that the 2 sets are of the same size. When they are infact not.
    the 2+4+8.... set is larger than the -1-2-4 etc one by 1 term

    Since they both go to infinity that extra term has to also be infinity and the answer would therefore be infinty.

    The falacy is exploiting the fact that infinity can be of different sizes
     
  4. Dec 11, 2013 #3

    jbriggs444

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  5. Dec 11, 2013 #4

    PeroK

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    You can also do it this way:

    1 + 2 + 4 + 8 + ... = 1 + (-1 + 3) + (-3 + 7) + (-7 + 15) + ...

    = (1 - 1) + (3 - 3) + (7 - 7) + ...

    = 0

    The real answer is that an infinite sum is inherently not well defined. You cannot add an infinite set of numbers together. So, you have to come up with a formal definition of what you mean by an infinite sum. See, for example:

    http://en.wikipedia.org/wiki/Series_(mathematics)
     
  6. Dec 11, 2013 #5
    let's keep adding these number til infinity...

    every number of the set 2+4+8.... is 2*|a| it's counter part from the set -1-2-4.... let's say it is its double.

    now, pretend that we reached a VERY large number in the first set, it will be double it's counterpart, right?
    now since these series are powers of two (with the second being negative ones) even if we reach a very high number, in the negative series, we'll find one right AFTER it's counterpart that cancels it, up until infinity, so why bother since we will always find a number that will cancel it out?

    so if all numbers cancel out each other, none of them will stay, except for -1, thus, you didn't convince me enough.
     
  7. Dec 11, 2013 #6
    HAHA good point
    Well frame it in another way then
    You have 2 sets of number
    2,4,8 etc call this A
    and -1,-2,-4 etc call this B

    In your original argument you say that A[N] + B[N-1] = 0 leaving you with just B[0] = -1
    Well what if instead you did A[N] + B[N] (which is in my mind more correct cause the size of both sets are same size)
    That leaves you with 1,2,4... the sum of which would be infinty

    in the first method as I tried to state before you'll always have 1 left over in A that you didnt account for even if you extend it to infinty, but I think this way illistrates it better :)
     
  8. Dec 11, 2013 #7

    PeroK

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    This is wrong, as (1 + 2 + 4 + ...) is NOT a number. So, you can't multiply it by 1.
     
  9. Dec 11, 2013 #8

    D H

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  10. Dec 11, 2013 #9
    This makes absolutely no sense, two countable infinities cannot be of different sizes.
     
  11. Dec 11, 2013 #10

    jbunniii

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    It is possible to enlarge the set of real numbers to include ##\infty## and ##-\infty##, resulting in the extended real number system. In this system, ##1 + 2 + 4 + \ldots = \infty##, and it is perfectly valid to multiply this by ##1## to obtain ##1 \cdot \infty = \infty##. The problem isn't really with that step. The next step, however, is problematic. Expressions of the form ##\infty - \infty## are undefined, even in the extended real number system. So this is undefined:
    $$(2 + 4 + 8 + \ldots) - (1 + 2 + 4 + \ldots)$$
    because both of the parenthesized items are ##\infty##.

    The reason ##\infty - \infty## is undefined is because there is no way to define it consistently. To see this, suppose we assigned some definite value, ##\infty - \infty = x##, where ##x## is some real number. Now ##1 + \infty = \infty##, so we may write
    $$x = \infty - \infty = (1 + \infty) - \infty = 1 + (\infty - \infty) = 1 + x$$
    But ##x = 1 + x## is impossible for any real number ##x##: subtracting ##x## from both sides, we get ##0 = 1##.
     
  12. Dec 12, 2013 #11
    yes they can,
    don't believe me?
    watch this vid, until the end, it proves that some infinities are bigger than other infinities
    http://www.youtube.com/watch?v=A-QoutHCu4o
     
  13. Dec 12, 2013 #12

    D H

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    Yes, there are different sizes of infinities. The reals are "bigger" than the integers, for example.

    However, 1MileCrash is absolutely correct. Two countably infinite sets cannot be of different sizes because one can construct a one-to-one, onto mapping between them. All countably infinite sets have the same cardinality, ##\aleph_0##, or aleph-null. Read about Hilbert's hotel.
     
    Last edited: Dec 12, 2013
  14. Dec 12, 2013 #13

    pwsnafu

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    To be countable means that it is the same size as the naturals (i.e. a bijection to the naturals). That's the definition of countable.
    If something is larger then it is not countable.
     
  15. Dec 12, 2013 #14

    PeroK

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    The laws of arithmetic cover only finite additions and multiplications. There is no rule of arithmetic to tell you how to add an infinite number of numbers together.

    So, until you've defined what is meant by 1 + 2 + 3 ... you have a recipe for fallacies, contradictions and error.

    Maybe it's too austere to say that 1 + 2 + 3 ... is simply not defined, as that takes all the fun out of proving and disproving these wacky results!
     
  16. Dec 12, 2013 #15

    pwsnafu

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    It's defined as an ordered formal sum. Whether it is convergent is a completely different question.
     
  17. Dec 12, 2013 #16

    D H

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    Nonsense. Adding an infinite number of numbers together is what infinite series, or for that matter, Riemann integration, are all about. 1/2+1/4+1/8+… might have been confusing to Zeno, but it should not be confusing to those living in the 21st century. It is one.


    Infinite sums can result in some apparently weird results. One is the loss of associativity for conditionally convergent series. An absolutely convergent series will always add to the same number no matter how the terms are rearranged. Rearrange the terms of 1/2+1/4+1/8+… and you will always get one. That is no longer the case with conditionally convergent series. Consider the series 1-1/2+1/3-1/4+…, or ##\sum_{n=1}^{\infty} (-1)^{n+1}/n##. This is log(2). Rearrange the terms in the series and you'll get another result. You can get any result you want! So don't rearrange the terms. Addition is not necessarily associative when it comes to infinite series.


    What about non-convergent series such as 1+2+4+8+…, 1+2+3+4+…, or 1+1+1+1+…? One obvious answer is that these are divergent series. The sum is infinite for each of these series, end of story. These obvious answers did not stop some of the best minds in mathematics such as Euler, Abel, and Hardy from toying with such series. These can rigorously be given finite values.

    Divergent series such as these occasionally appear in advanced physics. The regularization techniques used to get rid of those infinities were crucial in developing quantum field theory. "As everyone knows, 1+1+1+1+…=-1/2."

    One key problem with aligning quantum mechanics and general relativity is that these infinities keep popping up and the regularization techniques that worked so well with QED don't work in this regard.
     
  18. Dec 12, 2013 #17

    PeroK

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    Hey, steady on Mentor. There's no need to get insulting. When did infinite series and Riemann integration become part of "arithmetic"?

    a(b + c) = ab + ac (is the distributive law)

    But:

    [itex]a\sum_{i=1}^{\infty}b_i = \sum_{i=1}^{\infty}ab_i[/itex]

    Can only be given meaning through defining limits and mathematical analysis. In particular, it cannot be deduced from the distributive law.

    So, unless you are using the limit of partial sums, then high-school students are going to run into these paradoxes - essentially by assuming that there is nothing untoward with infinite sums.

    My point is that for those without a grasp of mathematical analysis and the rigorous treatment of infinite series, an infinite sum is (at that point in their mathematical development) undefined.

    In the same way, once someone has mastered infinite countable sums, they may run into problems with, say:

    [itex]\sum_{i \in R}b_i[/itex]

    Until they realise that this is not defined by "infinite series" and a new definition of what this sum might mean is required.
     
  19. Dec 12, 2013 #18
    Not to "echo" the others, but in both series, we have a list of numbers. There is a first place, second place, third place, and so on.

    The only difference between the two series is what symbol "occupies" the places, and that surely means nothing for its cardinality of the series.

    In other words, you can't have two infinite "lists" and have them be of different sizes. Being bigger than this type of infinity means (essentially) that it cannot be put into a list in that way.
     
  20. Dec 12, 2013 #19
    (2-1)*(1+2+4+8+16+32....) does not equal 2+4+8+16+32+64...... -1-2-4-8-16-32
     
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