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I Where has this proof gone wrong? ∞= 1/0

  1. Sep 16, 2017 #1
    Now I expect that most of you on this forum would be familiar with the equality between point nine reoccurring and one:

    0.999...=1

    If your not familiar please review https://en.wikipedia.org/wiki/0.999...

    Now this equality can be used to imply something else, which is rather heterodox, consider the below:

    Point nine reoccurring is only one infinith smaller than one
    0. 999 ... +(1/∞)= 1

    But as point nine reoccurring and one are equal, point nine reoccurring is one infinith smaller than itself
    0. 999 ... +(1/∞)= 0. 999 ...

    It is only logical then to conclude that one infinith is equal to zero
    0. 999 ... − 0. 999 ... +(1/∞)= 0. 999 ... − 0. 999 ...

    1/∞= 0


    Which can in turn be inverted to reveal that infinity is equal to one divided by zero

    ∞ =1/0

    As the mathematical consensus is that division by zero is undefined, why is the proof incorrect?
     
  2. jcsd
  3. Sep 16, 2017 #2
    1/is not a real number, so you cannot add it to other numbers. is not a number, it's a symbol that means something only in a special context - limits. "one infinith smaller" does not make sense.
     
  4. Sep 16, 2017 #3

    mfb

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    There is no such thing. Whatever (1/∞) is supposed to mean, you are assuming (1/∞)=0 here, and that equation doesn't make sense in the real numbers.
     
  5. Sep 16, 2017 #4

    WWGD

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    And the EDIT: Extended Reals (Standard Reals with a "Point at Infinity" appended to them) are not a ring, i.e., do not have a multiplication defined ( and, I believe, no "reasonable" version definable), so the operation ## \infty \times 0## is not defined. Still, if you had a ring, then ## 0 \times r =0 ## for any r in the ring. So this operation is not definable within the Extended Reals either. Maybe you can try the Surreals or the Hyperreals where you have infinities as numbers. Still, a lot of the work on those numbers is pretty esoteric.
     
    Last edited: Sep 16, 2017
  6. Sep 17, 2017 #5
    Why is infinity not a number, while zero is?
     
  7. Sep 17, 2017 #6
    It's not a real number. Why do you think it should be? Do you understand what "infinity" means in context of limits? Or other contexts (cardinality)? How would you define it as a real number? We have a few constructions of real numbers, in all of them 0 is a precisely defined object.
     
  8. Sep 17, 2017 #7

    jbriggs444

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    If ##\infty## were a member of the field of real numbers and if one agreed that ##\frac{1}{\infty}=0## then one could proceed to prove nonsense. For example...

    By definition, ##\frac{a}{b}## is that [unique] number which, when multiplied by ##b## produces ##a##. So if ##\infty## is a real number and ##\frac{1}{\infty}=0## then, by definition:
    $$0 \cdot \infty = 1$$
    But it is also well known and provable that for all ##a##, ##0 \cdot a = 0##. In particular, if ##\infty## is a member of the field of real numbers then:
    $$0 \cdot \infty = 0$$
    Two things that are equal to the same thing are equal to each other. So it would immediately follow that:
    $$0 = 1$$

    There are some choices for how to deal with this problem. One way to go is to accept that ##\infty## is a legitimate number but to leave ##0 \cdot \infty## undefined. This leads to the extended reals as @WWGD has explained. Another way to go is to allow for multiple infinite numbers, each with its own infinitesimal inverse. This leads to Robinson's Non-standard Analysis and fields such as the "hyperreals" or "surreals" [as @WWGD has also suggested].

    In the standard real numbers, there are no infinite numbers and only one infinitesimal number. Zero counts as the infinitesimal.
    In the non-standard real numbers, there are many infinite numbers and many infinitesimal numbers but still only one zero (which still has no inverse).
     
    Last edited: Sep 17, 2017
  9. Sep 17, 2017 #8

    FactChecker

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    There are many ways for something to be infinite and many different "sizes" of infinity (cardinality). So any time you try to use infinity as a number in a calculation, you are opening a can of worms that may be much more complicated than the original problem.
     
  10. Sep 17, 2017 #9

    fresh_42

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    With this point of view given, my argument will be:
    Zero is not an element of the multiplicative group of the real number filed. Therefore the question, whether it has an inverse or even can be multiplied by something else simply does not exist. It is in a different set!

    More elaborated: The connection between addition, where the zero belongs to, and multiplication, where inverses belong to, in a field is set by the distributive law only. It forces ##0 \cdot a = 0##. As ##\infty## isn't an element of either of the groups, there is no chance to operate with it. You could object "... then let us define it as an element" of one or both of the groups, but then you will run into contradictions of all kind as already instanced in previous posts, including yours. You have to be rigorous from the start, not at the end and wonder why the sand castle breaks down.
     
  11. Sep 17, 2017 #10
    A problem here is the use of ##\infty## without clearly distinguishing it from the real numbers, which have a cardinality equal to ##\aleph_1##, assuming the continuum hypothesis.

    ##\infty## is defined here as "an abstract concept describing something without any bound or larger than any number." i.e. two different definitions, and later
    "In real analysis, the symbol ##\infty##, called "infinity", is used to denote an unbounded limit. ##{n \rightarrow \infty}## means that x grows without bound..."

    i.e. ##\infty## as normally used refers to a large but not infinite number.

    So ## \lim_{x \rightarrow+\infty} {\frac {x+1} {2x+1}}=0.5## should really be read as "as x increases without limit the expression becomes arbitrarily close to, but is never equal to ##0.5##.

    Using the '=' sign in this way is very convenient but also misleading.

    Using x equal to the smallest infinity ##\aleph_0## you get ##{\frac {\aleph_0+1} {2\aleph_0+1}}## which is undefined.

    Using '=' in a different way ...

    Given that ##0.999999....=1##
    ## \lim_{x \rightarrow+\infty}({0.999999....+\frac 1 x})>1## since 1/(a positive number which can increase without limit, but is always finite) is always >0.
     
  12. Sep 17, 2017 #11

    I like Serena

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    As said before, infinity is not defined for the real numbers, and neither is division by zero.

    However, we can extend the real numbers to the Real Projective Line or to the Extended Real Number Line.
    In both cases your proof is valid, and we have indeed ##\frac 1\infty = 0##.
    It's only on the Real Projective Line that we have ##\frac 1 0=\infty## though.
    Have to be careful since not all of the usual operations are well defined.

    For the record, the use of infinity in Projective Geometry allows us full equivalency between ellipses, parabolas, and hyperboles.
     
  13. Sep 18, 2017 #12
    Hi All,

    You've made quite clear to me that infinity isn't a real number because it hasn't been defined as one. However, what I really want to understand is why it wasn't included.

    Obviously if something nonsensical results of the inclusion then there is a reason to exclude it. jbriggs444 follows this logic in his explanation:

    That being said, the 0=1 equality is only arrived at because it is assumed that:

    If a=b and a=c then b=c

    This is not always the case, consider Yx0=0

    Y can be 1, 0 or 524 as all numbers work for Y. Just because Y=1 and Y=0 doesn't mean that 1=0
     
  14. Sep 18, 2017 #13

    jbriggs444

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  15. Sep 18, 2017 #14
  16. Sep 18, 2017 #15

    jbriggs444

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    Let me try to translate to more coherent prose...

    Let Y be the solution the equation, "(Y-2)(Y-3)=0"...
    That's a violation of the rules of mathematical discourse. The word "the" carries with it an assertion of existence and uniqueness. You are not allowed to use the word unless you are prepared to provide those guarantees. In this case there is no guarantee of uniqueness.

    Let Y be a solution the equation, "(Y-2)(Y-3)=0". Clearly then, either Y=2 or Y=3. We can conclude that both Y=2 and Y=3 are true.
    That's just wrong. We can conclude one of the two possibilities (Y=2 or Y=3) must hold. But we do not know which. We most certainly know that both cannot simultaneously be true.

    Edit: Let me try to put it in an entirely different way...

    If we are given the equation "(Y-2)(Y-3)=0", that is an assertion about some state of affairs. We might have a state of affairs where Y=2, the sky is blue and water is wet. We might have a different state of affairs where Y=3, Pluto is not a planet and Elvis has left the building. Those are both consistent with the truth of the equation. [The equation rules out every state of affairs where Y is anything other than 2 or 3]

    The transitive property: "if a=b and b=c then a=c" holds within a single state of affairs. You have to have decided on specific values for a, b and c. You are not allowed to equivocate, picking one state of affairs where a=b holds and a different state of affairs where b=c holds.
     
    Last edited: Sep 18, 2017
  17. Sep 18, 2017 #16
    How would you approach plotting it on a Cartesian plane?

    Flip a coin, if heads plot (0,2), if tails plot (0,3)?

    That type of thinking seems very pre-Schrödinger's cat.
     
  18. Sep 18, 2017 #17

    jbriggs444

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    Plotting what on a Cartesian plane.

    You can plot the graph of f(Y) = (Y-2)(Y-3) and get a parabola which intersects the Y axis at Y=2 and Y=3.
    You can plot the solution set to (Y-2)(Y-3)=0 on the number line by putting dots at 2 and 3.

    None of that is relevant to the fallacy of equivocation.
     
  19. Sep 18, 2017 #18

    FactChecker

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    It's not always true that ∞ - ∞ = 0. So including ∞ as a real number means that some fundamental rules of arithmetic no longer hold for the reals. That is too much to give up.
     
    Last edited: Sep 18, 2017
  20. Sep 18, 2017 #19

    Mark44

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    No. The correct conclusion is that EITHER Y = 2 OR Y = 3. It is incorrect to conclude that Y can be both numbers simultaneously, which is what the conjunction "and" implies.
     
  21. Sep 18, 2017 #20

    WWGD

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    Y=2 _Or_ Y=3 for product to be 0. Y=2 _and_ Y=3 does not follow.

    EDIT: Still, per@ILikeSerena's last post, we can do arithmetic on ## \infty ## in ##P_1(K)## ; arithmetic rules can be extended from K to work with the projective line.
     
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