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B 0/0 = 2 Proof Explanation

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  1. Sep 18, 2016 #1
    Recently I can across this proof:
    0/0= (100-100)/(100-100)
    = (10^2-10^2)/10(10-10)
    =(10-10)(10+10)/10(10-10)
    = (10+10)/10
    =20/10
    = 2
    But this is obviously wrong, as 0/0 is infinity, but which line in this proof is actually wrong?
     
  2. jcsd
  3. Sep 18, 2016 #2

    QuantumQuest

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    ##\frac{0}{0}## is not infinity. It is just not determined i.e it can be any number.

    You can write 0 in whichever way you want, as is done in the proof you post, e.g as a difference of whatever numbers and factorize, write as square difference but watch carefully what is the operation that is the cheat (what I write gives you already a hint).
     
  4. Sep 18, 2016 #3
    I'm not sure, but to apply the a^2-b^2 formula, does a and b have to be distinct (a=/=b)? Is that where it goes wrong?
     
  5. Sep 18, 2016 #4

    QuantumQuest

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    No. Check again carefully

     
  6. Sep 18, 2016 #5

    Ssnow

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    Hi,the form ##\frac{0}{0}## is an indeterminate form so for example I can say that ##\frac{0}{0}=\frac{27-27}{27-27}## and with the an analogue procedure I can conclude that ## \frac{0}{0}=3## ...

    From false assumptions you can deduce everything ...
     
  7. Sep 18, 2016 #6
    So the problem with the proof is that we cannot just rewrite 0 as 100-100? But while factorising algebraic expressions, we split up middle terms to simplify factorisation, which is sort of similar to this, isn't it?
     
  8. Sep 18, 2016 #7

    QuantumQuest

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    Why not? Look for one non-legitimate operation.
     
  9. Sep 18, 2016 #8

    Ssnow

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    not exactly, the problem is when you simplify the factor ##(10-10)## ..., reflect on this...
     
  10. Sep 18, 2016 #9

    Ssnow

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    sorry ##10-10## ...
     
  11. Sep 18, 2016 #10
    Is it because (10+10)*0/10*0 and you can't just cancel just cancel the zeros and assume that they are equal to one? Because that can only be rewritten as (10+10)/10 if I cancel the zeros and say that becomes one?
     
  12. Sep 18, 2016 #11

    QuantumQuest

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    Did you decude that yourself? Meaning do you understand that?
     
  13. Sep 18, 2016 #12
    yeah...
     
  14. Sep 18, 2016 #13

    Stephen Tashi

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    The proof uses the idea that ##\frac{ac}{bc} = \frac{a}{b} ##.
     
  15. Sep 18, 2016 #14
    Thanks for your help QuantumQuest and Ssnow!
     
  16. Oct 3, 2016 #15
    First, you can't cancel zero from both numerator and denominator. When you're cancelling both 10-10 and 10-10, you're actually cancelling zero.
    Second, 0/0 isn't infinity, it can be proved to be equal to any number by cancelling zero's. I myself used to do it this way:
    0*5=0
    or 5=0/0, similarly if division by zero on both sides is allowed, then every number can be proved to be equal to 0/0.
    I also used this way to get an absurd result:
    0=0
    5*0=3*0 since both sides are equal to zero
    now cancelling zero's from both sides:
    5=3
     
  17. Oct 18, 2016 #16
    You can conclude anything if you're allowed to break math rules. In particular, division by 0 is undefined. Actually 0/0 has no value at all, because it is not defined. It is not "infinity" -- it doesn't have any value. And your calculation is actually part of the reason why it is not defined: no definition can be given that makes the other math rules still behave consistently -- you have just shown that if you try to treat it as a real defined quantity, you get weird inconsistent results. So we take it as a math rule that division by 0 has no meaning, like how a square root sign with nothing under it has no meaning.

    Note also -- there are some settings where one can take division by 0 as defined, and you are probably thinking of that when you called "infinity". But if you do that, you then end up sacrificing other math rules (in the most common treatments, taking n/0 as infinity leads to the creation of new undefined expressions -- in fact, 0/0 remains undefined in such treatments.). It's a tradeoff. Division by 0 and grade school math rules together are a logical contradiction. Your experiments are the way you prove that. If you want one side of the contradiction, you have to sacrifice from the other. Since usually we want to keep all the math rules, we are forced to leave division by 0 undefined.

    (This, incidentally, is what makes division by 0 different from taking square roots of negative numbers. In the latter case, all the rules of arithmetic fortunately still work if you allow that. And we get a very beautiful new mathematical system from it. But division by 0 doesn't work the same way. The math rules don't like to be broken in that way. You could say they are more resistant to breaking in one way than in another.)

    So what step is wrong? The very beginning -- where you assumed that 0/0 even had a value at all and then started manipulating it. That is, the moment you said "0/0 = ...". Or, on the other hand, you can say the argument has no flaw -- you are just drawing the wrong conclusions from it. You can consider your argument to be effectively a (or part of a) proof by contradiction of the falsehood of the assumption "0/0 has a value" in the usual system of grade school math rules.

    (Incidentally, other posters bringing up indeterminate forms are also not quite right. Talk of indeterminate forms only makes sense when taking limits. In that sense 0/0 is indeterminate in that if it shows up when taking a limit, it tells you nothing about the value or existence of the limit. On the other hand, if you get 1/0 (say), you know the limit DNE. But we aren't taking limits here, we are doing algebra, so the topic of indeterminate forms is not really relevant here. 0/0 is undefined in grade school algebra. That's the endpoint.)
     
    Last edited: Oct 18, 2016
  18. Oct 18, 2016 #17

    QuantumQuest

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    I see that you're trying to make a rigorous point here, but did you really understand and follow the context and the purpose of the OP question?
     
  19. Oct 18, 2016 #18

    disregardthat

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    I don't know what your point is. He's talking about the way "indeterminate form" is not a relevant topic in this discussion (since it had been brought up), and in particular how "0/0-forms" in the context of limits should not be confused with the algebraic expression 0/0 as they have nothing to do with each-other.
     
  20. Oct 19, 2016 #19

    QuantumQuest

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    @disregardthat
    My point is that OP posted an exercise, trying to figure out where is the error. So, I and Ssnow, provided OP hints to find that point. That essentially, as 0 is the absorbent element for (usual) multiplication, cancelling zeros is a non-legitimate operation. Now, we all know about what is indeterminate and what is not, as well as what holds for limits. But in the context of the present exercise, what sshai45 essentially says, is telling the OP not to do the exercise at all. That was the point of my objection. OP cannot know in advance, things that he/she will understand later.
     
    Last edited: Oct 19, 2016
  21. Oct 19, 2016 #20

    disregardthat

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    That's all well and good, but we should generally aim to provide precise explanations using correct terminology so that the discussion won't risk leaving other potential readers confused.
     
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