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Dividing by Infinity

  1. Jun 11, 2010 #1
    I was just thinking about how division by zero is undefined and began wondering if dividing by infinity is undefined as well. I understand the limits that accompany this idea [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP194219b2befa7461g38e00003i7ih720ai6i1egc?MSPStoreType=image/gif&s=38&w=66&h=36 [Broken] and everywhere I look its defined as 0 but I came up with a little contradiction so to speak as follows
    Assume these to be true:
    A[itex]\neq[/itex]0
    [itex]\frac{A}{\infty}[/itex]=0
    Then,
    [itex]\frac{1}{\infty}=\frac{0}{A}[/itex]
    1=[itex]\frac{\infty*0}{A}[/itex]=0
    1=0

    Can someone explain this to me and why division by ##\infty## is 0?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 11, 2010 #2
    Hmmm, zero times infinity is not zero. That's why you got 1=0 wrongly!!!
     
  4. Jun 11, 2010 #3
    Also note that in "common" mathematics, you cannot use infinity as an object that you plug in into equation. Writing infinity is a concept, but it is not a valid number. When people write [itex]\ldots=\infty[/itex] they mean: "Please translate this statement into the definition of limits with all these [itex]\varepsilon[/itex].

    Maybe someone can point out the extensions of mathematics where you are allowed to use infinity? Maybe it's not possible without violating the known algebra? I mean if we see
    [tex]a\cdot 0[/tex] we know the result is zero, whatever the value of a. But this fundamental general rule might be broken, if you introduce infinities?
     
  5. Jun 11, 2010 #4

    Hurkyl

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    The extended real number system that is commonly used in real analysis, for example. Also, the projective complex numbers which are used extensively in complex analysis.

    Cardinal and ordinal numbers have infinite elements, but poor arithmetic properties.
     
  6. Jun 11, 2010 #5
    And which properties do you lose with that?
    Is it still true that [itex]a\cdot 0=0[/itex] for any [itex]a[/itex]?
    But this would be in conflict with
    [tex]1=\frac{1}{\infty}\cdot\infty=0\cdot\infty=0[/tex]
    ?
     
  7. Jun 11, 2010 #6
  8. Jun 11, 2010 #7
    I'm completely sure:
    Which of the rules isn't valid:
    [tex]a^{-1}\cdot a=1[/tex]
    [tex]0\cdot a=0[/tex]
    ? Does it mention it on the webpage?
     
  9. Jun 11, 2010 #8
    Yes, it does. If a = ±∞, both are undefined, although the latter is sometimes defined for convenience as 0 in real analysis.
     
  10. Jun 11, 2010 #9
    So are you saying the above rules are not generally valid, but demand that the variable isn't infinity?

    That's different to normal calculations, where you can be sure the rules work no matter what the value of the variables are?
     
  11. Jun 11, 2010 #10
    As far as I know, yes. Note also though that infinity is not the only place trouble occurs. The first rule you mentioned doesn't hold when a = 0 either. Of course, there may be other ways to define arithmetic on the extended reals, but everything is a product of how we define them.
     
  12. Jun 11, 2010 #11
    Operations usually don't work on everything, it isn't strange at all. For example you can't use the inner product between vectors of different size.
    "That's different to normal calculations, where you can be sure the rules work no matter what the value of the variables are?"
    In normal calculations is 0/0=1 ? No, not without ruining associativity, so it doesn't hold for normal calculations either.
     
  13. Jun 11, 2010 #12

    Hurkyl

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    There is no such thing as an arithmetic identity that works no matter what the variables are (except for the trivial identity where both sides are identical text).

    e.g. the commutative law you learned in elementary school:
    x * y = y * x​
    comes with the condition that the two letter variables are supposed to be real numbers, and the remaining variable (*) is supposed to be real number multiplication.
     
  14. Jun 11, 2010 #13
    Yep. About an hour after posting I realized that 0*[tex]\infty[/tex] is indeterminate.
     
  15. Feb 4, 2012 #14
    So what will be the correct expression for "Infinity/infinity"? Will it be Undefined? Thanks,
     
  16. Feb 4, 2012 #15

    Deveno

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    analysts perfer the term "indeterminate". ∞ is not a real number, but can crop up in various limit processes. depending on the circumstances, the value of a limit that is ∞/∞ could be anything (there is a special rule called L'Hopital's Rule for dealing with some of these situations).

    in fact, since we might imagine one ∞, why not two (a more infinite infinity), or three, or...an infinite number, each with its own peculiarities? since all of these would be "not-finite", when trying to figure out what ∞/∞ might be, one becomes uncertain as to which infinity is being divided by which other one.

    you see, that's one trouble with ∞. with a finite number, we have a clear idea "which" number it is: that number, and no other. it's hard to make a claim that ∞ is quite so definite, that there is ONLY one, and that it always behaves in a very predictable way. for example: we know that for a finite number x, x and x+1 are not the same. are ∞ and ∞+1 the same? what do you think?
     
  17. Feb 4, 2012 #16

    Hurkyl

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    No, analysis also prefer "undefined" for talking about this arithmetic operation.

    But calculus often talks about forms, and "indeterminate" would be correct term for the limit form written that way.
     
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