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Possible to Multiply or Divide Infinities?

  1. Mar 8, 2008 #1
    Hi all,
    Some people are arguing that it is possible to divide, multiply, add, subtract infinities, but I believe that impossible.

    Am I right or wrong?

    Thank you.
     
  2. jcsd
  3. Mar 8, 2008 #2

    rbj

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    fundamentally, we add, subtract, multiply, and divide numbers. [itex]\infty[/itex] is a mathematical concept, but is not a number.

    but, sometimes it can be meaningful to discuss these operations acting on numbers where one or more numbers are increasing without bound ("going to infinity"). for example, dividing a finite constant by such a number that is increasing without bound would result in something getting closer and closer to zero (to within any given [itex]\epsilon > 0[/itex]).

    [tex] \lim_{n \rightarrow \infty} \frac{1}{n} = 0 [/tex]

    sometimes is stated crudely as

    [tex] \frac{1}{\infty} = 0 [/tex] .

    while this sematic is icky, it does have meaning. but the following don't:

    [tex] (+\infty) + (-\infty) = ?? [/tex]

    or

    [tex] \frac{\infty}{\infty} = ?? [/tex]
     
  4. Mar 8, 2008 #3

    morphism

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    It depends on what you mean by "infinities". Try looking up "cardinal arithmetic", and see if that helps you.
     
  5. Mar 8, 2008 #4
    Yes, I know that you can use limits to know where they are approaching. Explicitly put, there is no answer for:
    [tex]\infty + \infty [/tex]= Undefined
    [tex]\infty - \infty [/tex]= Undefined
    [tex]\infty \times \infty [/tex]= Undefined
    [tex]\infty \div \infty [/tex]= Undefined
     
  6. Mar 9, 2008 #5
    Like it was said arithmetic operations like, +,-, /, *, etc are applied with numbers, but [tex]\infty[/tex] is not a number so you cannot impose these operations when one deals with infinity. however[tex]\infty+\infty=\infty[/tex] this basically is written this way to state the fact that the sum of any two values that increases without bound is still a value that increases without bound. also [tex]\infty*\infty=\infty[/tex] states that the product of two such quantities that increase without bound is still a quantity that increases without bound. THis is what it implies. While [tex]\frac{\infty}{\infty}[/tex] is undefined for the fact that we have one quantity that increases without bound over another quantity that increases without bound, we do not know which one of them grows faster, so it is undefined. Also, the quantity on the denominator might represent say the total number of points in a line, while the quantity in the numerator the total number of points in a plane, so we have two such quantities which quotient we do not actually know. that's why it is undefined. etc....
     
  7. Mar 9, 2008 #6
    If a and b are two infinite cardinals, then ab = max{a,b} and a+b = max{a,b}, assuming the Axiom of Choice.
     
  8. Mar 9, 2008 #7
    I'm pretty sure you can define division on cardinals, though it would be pretty trivial in most cases. a / b = c if c is the smallest cardinal such that a can be written as a disjoint union Ub of c-sized sets. Or something to that effect.
     
  9. Mar 9, 2008 #8

    tiny-tim

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  10. Mar 9, 2008 #9

    HallsofIvy

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    In other words, infinity is not a real number and operations on the real numbers to not apply to infinity. There are other ways to define number systems that include "infinity", such as the cardinals, mentioned above, in which you can do arithmetic on "infinities" but they do not, in general, have the same properties as arithmetic operations on the real numbers.

    tiny-tim, I see no mention of "multiplying by infinity" on the page. And while physicists mgiht think of the delta function as involving "infinity", no mathematician would!
     
  11. Mar 9, 2008 #10

    tiny-tim

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    Hi HI! :smile:
    http://en.wikipedia.org/wiki/Dirac_delta_function:
    Strictly speaking.of course, the delta function is not a function, but something called a distribution.

    Its use is essential in evaluating the very complicated multiple integrals in (perturbative) Quantum Field Theory.

    Whether one regards [tex]\delta(0)[/tex] as being infinite or as being meaningless is a matter of taste - and my taste is to prefer some meaning to no meaning! :redface:

    I admit this is a trick - but I think Raza knows that you can't really multiply by infinity without some trick being involved! :smile:
     
  12. Mar 9, 2008 #11

    Hurkyl

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    In all seriousness, it depends on precisely what meaning you choose for each of the words "divide", "multiply", "add", "subtract", and "infinity".
     
  13. Mar 9, 2008 #12

    HallsofIvy

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    I still see nothing about "multiplying by infinity".
     
  14. Mar 9, 2008 #13

    tiny-tim

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    Ah, but lower down, the delta-function gets multiplied by other functions and integrated.

    And the whole is equal to the sum of its parts, so if you multiply the whole, you must multiply each bit … :smile:

    As Hurkyl sagely says:
     
  15. Mar 10, 2008 #14
    yea that definitely wouldn't work for infinite sets
     
  16. Mar 10, 2008 #15
    Why not?
     
  17. Mar 10, 2008 #16
    meh i guess it would but it would be the trivial a/b = a for all infinite sets
     
  18. Mar 10, 2008 #17
    No, it may be for all regular infinite cardinals. For singular cardinals, it is non-trivial.
     
  19. Mar 10, 2008 #18

    rbj

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    better watch out, tiny. them's maths guys take this pretty seriously and will beat you up. they beat me up (or at least beat on me a bit) for defending the Neanderthal engineering POV of [itex]\delta(t)[/itex]. i can't remember who was involved, maybe wonk was.
     
  20. Mar 10, 2008 #19

    tiny-tim

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    I'm not afraid.

    Physics guys 'n' gals can use force. Maths guys can only rearrange the coordinates. :smile:
     
  21. Mar 10, 2008 #20

    CRGreathouse

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    Of course if you don't have AC that's not well-defined. What about just using the indirect image? Then you can have
    [tex]\aleph_0 / \aleph_0 = \mathbb{Z}^+\cup\{\aleph_0\}[/tex]
    [tex]\aleph_0 / \mathfrak{c} = \emptyset[/tex]
     
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