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Interesting Infinite Powers Paradox

  1. Jul 15, 2013 #1
    Recently, an AT&T commercial has been running on TV where a moderator asks some children about the largest number they could think of.

    At the end, one kid replies “∞ times ∞”, which of course is simply ∞2.

    Natually, one can instantly think of a larger number: ∞. But then, that got me to thinking about ∞.

    Now, by non-standard, but logical, mathematical rules of conventions regarding reciprocal and powers notation, it can be argued that:

    ∞ = 1/0 = 0-1


    -∞ = -1/0 = -0-1

    This naturally leads to the formula:

    = 0(0-1)

    However, due to those rules governing mixtures of powers, I quickly realized that:

    ≠ 0(0-1)


    0(0-1) = 0 = 0 ! (& no, I’m not intending that to represent a factorial) :tongue:

    Therefore, to properly notate the equation, it must go like this:

    = 0-(0-1) = 0-∞ !! (& no, I’m not intending that to represent a double-factorial) :tongue2:

    Which results in the surprising conclusion:

    ∞ = -∞ :bugeye:


    1/0 = -1/0


    1 = -1:surprised

    (Of course, there is an error in the above—the negative in the exponent of 0 is the act of making the reciprocal of 0, NOT ∞.

    It’s essentially the same error that folks often make when proving 1=0. Besides, even if there was no math error in my original calculation, there’s still an error in logic, as the entire exercise is merely a matter of mathematical notation, which is not the same as actual numbers.

    But, ironically, it is quite interesting that the graphing of nearly any function that involves the value of ∞, there is a corresponding value to those functions that implies that ∞ does indeed equal -∞.)
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Jul 15, 2013 #2


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    First, we would like to define what ##\infty## actually is. Contrary to what most laymen believe, there is no unique form of infinity.

    You use the relation ##\infty = \frac{1}{0}##. This relation is actually false for most definitions of ##\infty## that you use. Except for the projective real line.

    In the projective real line, you add one symbol ##\infty## to ##\mathbb{R}##. And this symbol is both larger than each real number and smaller (in fact, the usual ordering relation breaks down). In this case, we can indeed define division by ##0## by ##\infty##. And interestingly enough, ##\infty = -\infty## holds in the projective real line.

    So, what you described, is no more and no less that the projective real line.
  4. Jul 15, 2013 #3


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    And by the way, in the projective real line, it doesn't hold that ##\frac{\infty}{\infty} = 1##. So you can't go from ##\infty = - \infty## to ##1=-1##. So there is no contradiction.
  5. Jul 15, 2013 #4
    Thanks, Micromass.

    I've known about the concept for a long time, but I've never known its formal name.

    Although I see that the first 3 lines of the Arithmatic Operations defined conflict with the first 2 lines of those undefined, if one sets a = ∞
  6. Jul 15, 2013 #5


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    It doesn't conflict, since it says that ##a\in \mathbb{R}##. And ##\infty## is not an element of ##\mathbb{R}##.
  7. Jul 15, 2013 #6
    Oh, okay. I'm not as up on some of the current heiroglyphics used nowadays.

    a is a subset of the Regular Reals, but ∞ is beyond in the Regular. (Sorry, but it also seems I'm not having much fun with the Latex Reference symbols--I have to use IE at my office)
  8. Jul 15, 2013 #7


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    Not "subset", but "element". That is: ##a## is an element of the real numbers, but ##\infty## is not a real number.
  9. Jul 15, 2013 #8


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    a^2-a^2 = a^2-a^2

    a(a-a) = (a+a)(a-a)

    a(a-a) divided by a-a = (a+a)(a-a) divided by a-a



    .... but I divided by zero
  10. Jul 15, 2013 #9


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    ∞ = -∞ is a very natural result if you extend real numbers, as micromass mentioned. It gets even better with complex numbers.
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