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What is limit below this?

  1. Jun 13, 2015 #1
    We know,
    1/0.1=10;
    1/0.01=100;
    1/0.001=1000; 1/0.0001=10000;
    .
    .
    .
    1/0.00000000001=100000000000
    .
    .
    .
    1/0=infinity

    Why is it not correct?
     
  2. jcsd
  3. Jun 13, 2015 #2

    mfb

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    Infinity is not a real number.

    Also consider:
    1/(-0.1)=-10;
    1/(-0.01)=-100;
    ...
    1/0 = -infinity?

    1/(-0.1)=-10;
    1/(0.01)=100;
    1/(-0.001)=-1000;
    1/(0.0001)=10000;
    1/0 = ???
     
  4. Jun 13, 2015 #3

    disregardthat

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    There's a reason why mathematicians say 1/0 is undefined. It just doesn't fit in with the general framework. You are free to define it to be infinity, or whatever you want, but beware that you cannot use the standard rules of arithmetic to manipulate fractions containing 0 as a denominator. You will quickly run into consistencies.
     
  5. Jun 13, 2015 #4

    DaveC426913

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    Here's why you will run into trouble.

    The operation of division is defined as the converse of multiplication.
    Note here, how the first thing they do to define division is to start with multiplication and invert it:
    https://en.wikipedia.org/wiki/Division_(mathematics)


    So, while it is true that
    10 * 0.1 = 1
    100 * 0.01 = 1
    1000 * .001 = 1

    it is not true that
    infinity * 0 = 1.

    Thus it's converse (1/0=infinity) is also not true.


    That bears reiteration: School taught you the shortcut for division, but once you start pushing at the leaky corners of numbers, you've got to use more formal maths.
     
    Last edited: Jun 13, 2015
  6. Jun 13, 2015 #5
    if i devided 10 apple between no one then every one get infinity? thats not correct. so you devide something by zero does not make infinite. its just undefined.

    you can get infinite if 1/ x with x tens to zero. but exactly zero will give you undefined sol.
     
  7. Jun 13, 2015 #6

    DaveC426913

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    No, you will just get a very large number.
     
  8. Jun 13, 2015 #7
    very very large number is not infinity?
     
  9. Jun 14, 2015 #8

    Svein

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    No. Since nothing (in this sense) is larger than infinity, a very large number is not infinity because:
    • You give me an extremely large number (call it P). I answer "but P+1 is larger, so P cannot be infinity"
     
  10. Jun 14, 2015 #9

    micromass

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    Yes in the sense that the limit of ##1/x## when positive ##x## tends to ##0## is infinity. What you are describing in your OP is exactly what limits do, they don't tell us what ##1/0## is exactly, but they do tell how small numbers ##\varepsilon>0## behave when we do the division ##1/\varepsilon##. In this case, if ##\varepsilon## get very close to ##0##, then ##1/\varepsilon## will get bigger and bigger, and thus will be infinity in the limit.

    However, you can only do this when talking about limits. Doing arithmetic with infinity is way more technical when not doing limits.
     
  11. Jun 14, 2015 #10
    Talking about apples. If you divide 10 apples between zero people, you didn't even divide them. You did nothing, therefore your action makes no sense and it's not defined. Both infinity and nothing are very interesting/complicated terms that probably no one can understand.

    @micromass : Talking about limes, I always thought of it as something not so... clean. For example, if you have 1 + 1/2 + 1/4 + 1/8 + ... you cannot just simply say that it all combined gives 2. It will never give 2 whatever you do. I actually had an argument with my math teacher in school about this. I mean, you can say "Yeah, this sum is leaning towards 2", but in fact it will never be 2.
     
    Last edited: Jun 14, 2015
  12. Jun 14, 2015 #11

    jbriggs444

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    In the context of limits, infinity is not a number. It is more of a boundary. No number is larger than infinity in this sense. No number is even equal to infinity.

    When one sees "x-> oo" used in the bounds of a limit, it is an abuse of notation that should be read "as x increases without bound".

    When one sees "oo" used as the value of a limit, it is an abuse of notation that should be read as a statement that the limit does not exist and that it does not exist in a particular way.


    One can go a step further and think of numbers that are larger and larger as being closer and closer to "infinity". This involves modifying our usual notion of "closeness" where we decide how close two numbers are by looking at the absolute value of their difference. [Since infinity is not a number, that difference would be undefined]. This process amounts to defining a "topology" on an extended set that includes two new members: +oo and -oo.

    Defining this topology effectively gives you two new positions on an extended number line. The old number line had no end points. The new one has a new right hand endpoint and a new left hand endpoint. This process can be called "the two point compactification of the reals". Doing this just defines the new positions. It does not necessarily define any arithmetic operations on those positions.

    With this two point compactification in hand, one can go back and re-interpret the limit notation

    When one sees "x -> oo" used in the bounds of limit, it is correct notation that can be read as "as x approaches positive infinity"
    When one sees "oo" used as the value of the limit, it is correct notation that indicates that the limit is infinity.

    Note that although it is tempting to go ahead and define arithmetic operations on +oo and -oo, it is not possible to do so in a way that preserves all of the axioms of the real numbers as an ordered field. Accordingly, although one can use +oo and -oo as bounds for limits or as the result of a limit, one cannot use them in ordinary arithmetic operations or produce them as the result of an ordinary arithmetic operation.
     
  13. Jun 14, 2015 #12

    Mark44

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    "limits" are what we're talking about, not limes, which are citrus fruits.
    This sum, 1 + 1/2 + 1/4 + 1/8 + ... is equal to 2. The ... (an ellipsis) indicates that we are talking about the sum of an infinite number of terms, with each term following a certain pattern. It's easy to show that the partial sum Sn, 1 + 1/2 + 1/4 + 1/8 + ... + 1/2n - 1, can be made arbitrarily close to 2. This shows that the limit of the partial sums, ##\lim_{n \to \infty} S_n = 2##.
    Presumably your teacher won the argument.

    What you say would be correct if you were talking about the sum of a finite number of terms, which isn't the case here.
     
  14. Jun 14, 2015 #13

    micromass

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    Note that the sum ##1+\frac{1}{2} + \frac{1}{4} + ...## is defined as the limit of the finite sums! So the equality is by definition. If you want to use another interpretation of the infinite sum, be my guest, but the mathematicaly community has agreed on this definition.

    In any case, your argument sounds a lot like Zeno's paradox...
     
  15. Jun 14, 2015 #14
    @Mark44 and @micromass Yes, I am aware of that and I am aware of the definitions regarding geometric sum in this case. I'm just saying that to me it just doesn't seem right to simply say that the sum I was talking about is equal to 2 without specifically stating that you are looking for the limit of that sum. Yes, that is just my interpretation and when I'm re-reading it, I've noticed that I wasn't able to really explain what I meant as English is not my native language. Also, could someone please explain me how to write fractions and stuff like that here in the posts? Zeno's paradox as in Achilles and the tortoise?
     
  16. Jun 14, 2015 #15

    micromass

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    https://en.wikipedia.org/wiki/Zeno's_paradoxes#Dichotomy_paradox

    If you say that the sum can't equal ##2##, then you must first provide a definition of the infinite sum.
     
  17. Jun 14, 2015 #16
  18. Jun 14, 2015 #17

    Mark44

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    There are a couple of sums here.
    1. Partial sum:
    Sn = 1 + 1/2 + 1/4 + ... + 1/2n - 1, the sum of the first n terms. Another way to write this is
    ##\sum_{i = 0}^{n - 1} \frac 1 {2^i}##
    2. Infinite sum:
    S = 1 + 1/2 + 1/4 + ... + 1/2n - 1 + 1/2n + ...
    The latter sum is exactly equal to 2. The way this is proved is by showing that the sequence of partial sums, {Sn}, gets arbitrarily close to 2.

    When you say it doesn't seem right that the latter sum equals 2, all it means is that you don't quite understand how an infinite series (such as the one you posted) can converge to a number.
    LaTeX fractions:
    # # \frac{1}{2^3}# # (omit the spaces between the # characters)
     
  19. Jun 14, 2015 #18

    DaveC426913

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    Of course not.
     
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