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I Zero x infinity?

  1. Dec 29, 2015 #1
    This seems a very simple case to me, yet I have heard it said that the answer is some undefined real number.

    Yet zero times anything means no iterations of whatever the object is; whether that be a real number , an imaginary number or an undefined number.

    Whatever it is I don't see how one can get away from the fact that what is specified is zero iterations of that number! So how can zero times anything not be zero?

    Looked at from the other side, it means an infinite (that is an undetermined number) times nothing and however many times one iterates this process the answer must still be nothing (or zero)!
     
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  3. Dec 29, 2015 #2

    Samy_A

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    Is this about limits?

    ##0.\infty## is undefined in ##\mathbb R##.
    But, if you have two sequences, ##(a_n)## and ##(b_n)##, with ##\displaystyle\ \lim_{n\rightarrow +\infty}a_n=0##, and ##\displaystyle\ \lim_{n\rightarrow +\infty}{b_n}=\infty##, then ##\displaystyle\ \lim_{n\rightarrow +\infty}a_nb_n## could be ##0##, another real number, ##\infty##, or undefined.

    For example, take ##a_n=n, \ b_n=1/n##, then ##\displaystyle\ \lim_{n\rightarrow +\infty}a_n=\infty##, ##\displaystyle\ \lim_{n\rightarrow +\infty}b_n=0##, and ##\displaystyle\ \lim_{n\rightarrow +\infty}a_nb_n=1##.

    Or: ##a_n=n\sin²(n), \ b_n=1/n##, then ##\displaystyle\ \lim_{n\rightarrow +\infty}a_n=\infty##, ##\displaystyle\ \lim_{n\rightarrow +\infty}b_n=0##, and ##\displaystyle\ \lim_{n\rightarrow +\infty}a_nb_n=\lim_{n\rightarrow +\infty}\sin²(n) ## is undefined (limit doesn't exist).
     
    Last edited: Dec 29, 2015
  4. Dec 29, 2015 #3

    HallsofIvy

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    What you say about "0", "Yet zero times anything means no iterations of whatever the object is; whether that be a real number , an imaginary number or an undefined number" simply makes no sense. Multiplication itself is defined as an operation on numbers- we first define multiplication on "natural numbers", then extend that definition to "integers", then "rational numbers", "real numbers", and "imaginary numbers" but it makes no sense to even talk about defining multiplication on "undefined numbers"! We simply can't say anything about what happens when we multiply any number, including 0, by an "undefined number". Now, there are ways of extending the real or complex numbers to include "infinite" (as well as "infinitesimal") numbers so that we have defined "infinity" but how we define multiplication of such numbers, and the result of such a multiplication, depends on exactly how we have defined those numbers- and there several different ways to do that.
     
  5. Dec 29, 2015 #4
    No.It is not about limits nor any particular realm of mathematics; it is purely about the logical meaning of zero x anything. That whatever it is that one multiplies by zero. one is saying that one wants none of whatever it is, therefore the result has to be zero.
     
  6. Dec 29, 2015 #5

    Samy_A

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    You have to define what multiplication by zero means. For a real number ##x##, clearly ##x.0=0##.
    But if you want to look at ##0.\infty##, you have to define what you mean (I gave the example of limits because that is a context where I sometimes see expressions as ##0.\infty## used informally).

    You are of course entitled to define ##0.\infty=0## if you so wish, but the real question is if this is useful.
    For example, with this definition the product rule for limits isn't true anymore (see my previous post for examples).

    The point of course is that you can't do usual arithmetic with ∞.

    EDIT: weird, @HallsofIvy 's post only became visible 25 minutes after he posted it. :smile:
     
    Last edited: Dec 29, 2015
  7. Dec 29, 2015 #6
    Excuse me if this seems a bit basic, ∞ is a symbol representing something, a quantity that is undefinable. but a quantity nonetheless.
    Zero x ∞ therefore means zero times that quantity, zero iterations of that quantity - however big it is -or else there has to be a limit beyond which the quantity ∞ means something tangible beyond being a mere quantity?
    How can zero times any quantity be other than zero?
     
  8. Dec 29, 2015 #7

    Samy_A

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    I'll play along. Let's define ##0.\infty=0##.
    Using similar logic,we should define ##x/\infty=0## for any ##x \in \mathbb R##.
    There is nothing forbidden with these definitions, but the problem is you can't do arithmetic with them.

    For example: ##2/\infty=0##, but ##0.\infty=0##, therefore we get ##(2/\infty).\infty=0##. Simplifying by removing the infinity for numerator and denominator, we get ##2=0##. That's tongue in cheek, but it shows that defining ##0.\infty=0## doesn't lead very far.

    ##\infty## is not simply a "very big number", so applying usual number rules and logic to it doesn't work.
     
    Last edited: Dec 29, 2015
  9. Dec 29, 2015 #8

    Mark44

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    No, ∞ does not represent a quantity. A "quantity that is undefinable" is not a quantity.
    Zero times any specific number is zero, but the expression ##0 \cdot \infty## is one of several indeterminate forms that show up in calculus, in the study of limits. This expression is indeterminate because different expressions having this form can come out to be anything.
     
  10. Dec 30, 2015 #9
    I find your thought processes interesting - and illuminating! That it is in applying rules that deal with more advanced realms of mathematics that we find the difficulties.
    Surely what we are dealing with here, at its very basic level is the very foundation of maths and that is counting. When we are counting zero of anything has a very definite meaning, whether it is numbers or apples; for in counting we are not necessarily defining what it is we are counting. In this case it is how many instances of the term ∞ we have.
    And as I said in the OP, if one takes it the other way round (again from the point of view of counting, it doesn't matter how many times one adds zero to zero the answer is and always will be zero! So adding zero an infinite number of times can still be nothing (excuse the pun) but zero.

    When you, Samy_A, argue: " For example: 2/∞=0, but 0.∞=0, therefore we get (2/∞).∞=0. Simplifying by removing the infinity for numerator and denominator, we get 2=0. That's tongue in cheek, but it shows that defining 0.∞=0 doesn't lead very far". - (sorry I can't see how to do quotes...)
    you are rounding your result - for 2 divided by any number will always be a +ve real number however small. Yes it is undetermined yet will always be >0.

    As you can see I am not a mathematician (pretty obvious isn't it?) yet I would say (using my logic) that if one wrote ∞.2/∞ the result would not be 2, because ∞ is indeterminate and the two terms of ∞ are not the same number.
    In maths does ∞ = ∞ ? It surely is undetermined yet has certain definite properties? It is very large, positive, and a real though undefined number.

    (Is this more of a philosophical question? About the nature and meaning of terms? I admit my approach is that of a determined Ockhamite!)


    Reference https://www.physicsforums.com/threads/zero-x-infinity.849936/

    Reference https://www.physicsforums.com/threads/zero-x-infinity.849936/[/SUB]
     
    Last edited by a moderator: Dec 30, 2015
  11. Dec 30, 2015 #10

    Samy_A

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    When you apply your counting logic to set ##0.\infty=0##, what you really do is apply a limit (even if you don't use the term).
    Yes 0 times 7 is 0. 0 times 700000000000000 is still 0, and 0 times 700000000000000000000000000 is also just 0.
    But then you jump from large numbers to infinity, and claim that ##0.\infty## must be 0.

    Mathematically, what you do is claim:
    If ##0.a_n=0##, and ##\displaystyle\ \lim_{n\rightarrow +\infty}{a_n}=\infty##, then ##0.\displaystyle\ \lim_{n\rightarrow +\infty}{a_n}=0##.
    But as I showed in my first post, you can use other sequences, and "show" likewise that ##0.\infty=1## or any other value (or undefined).
    So what should 2/∞ be as a non zero real number?
    If you set 2/∞=a>0, then you get ∞=2/a. Even if a is very small, 2/a will be big but finite.
     
    Last edited: Dec 30, 2015
  12. Dec 30, 2015 #11

    WWGD

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    In some cases, like in measure theory, you have loosely that the product is 0 (countably infinity union of sets of measure zero has measure zero). You can see this by having the set ##S_i## have measure ##\epsilon/2^i ## , then letting ##\epsilon \rightarrow 0 ##. But this is cheating.
     
  13. Dec 30, 2015 #12

    Mark44

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    If you are counting specific things, then zero of them means that you have no (= 0) of those things. However, and as I said before, the expression ##0 \cdot \infty## is indeterminate, and does not fall into the category of "counting things."
    There is no rounding going on here. All that Samy_A did was multiply both sides of the equation by ##\infty##.
    No, ##\infty## is not indeterminate. The point is that it makes no sense at all to do arithmetic with ##\infty##.
    Not necessarily. If we're talking about the sizes of sets (or cardinality), the set {1, 2, 3, ... } would seem to be larger than the set {2, 4, 6, ...}, but both sets have the same cardinality. This means that there is a one-to-one mapping between the two sets so that each number in the first set gets paired with a number in the second set, and vice versa.
     
  14. Dec 30, 2015 #13

    WWGD

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    Maybe, OP, you mean to say: ## lim_{x \rightarrow \infty} x .0 =0 ## . Then this is correct.
     
  15. Dec 30, 2015 #14

    Mark44

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    Based on what the OP wrote earlier --
    -- I don't think this is the case.
     
  16. Dec 30, 2015 #15

    WWGD

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    I meant to say that the sequence : ## 1.0=0, 2.0=0,......, n.0=0 ......## converges to ##0##.
     
  17. Dec 30, 2015 #16

    Mark44

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    You confused me at first, but I think this is what you meant: ##1 \cdot 0=0, 2 \cdot 0=0,......, n \cdot 0=0 ......##
     
  18. Dec 30, 2015 #17

    WWGD

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    Yes, sorry, did not know how to tex the ##a \cdot b ##.
     
  19. Dec 30, 2015 #18

    Samy_A

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    More generally, Grimble is trying to make sense of ∞. Nothing wrong with trying, but it so happens that infinity (even countable infinity) behaves very differently than finite numbers, even very large finite numbers.

    @Grimble , have you heard about Hilbert's hotel? It's a rather amusing illustration of how infinity is different from finite numbers. Take a look at it.
     
  20. Dec 30, 2015 #19

    WWGD

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    Sorry, I may have been inadvertently throwing the discussion off track.
     
  21. Dec 30, 2015 #20

    Samy_A

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    Oh, that's not what I meant.

    I was just rebounding on this post:
     
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