Dividing a number by zero

Discussion in 'Physics & Math' started by chikis, Mar 10, 2013.

  1. origin Heading towards oblivion Valued Senior Member

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    11,876
    Thanks for the tip Pete. I realized it was wrong and meant to find out the right way but I got side tracked...
     
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  3. KitemanSA Registered Senior Member

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    624
    Nope. \( -\infty \times 0 \) is undefined.

    Your error is in not following all the steps in altering the equation.

    You should have asked whether \(\frac{0 \times -1}{0} = 0 \times -\infty \) . Then you would have seen the undefined nature of the first term as well as the second.

    Oh, and yes, they are both equally undefined.

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  5. origin Heading towards oblivion Valued Senior Member

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    11,876
    I think my error was wasting my time on the silly and intuitively obviously wrong concept you have on this.
     
    Last edited: Mar 12, 2013
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  7. rpenner Fully Wired Valued Senior Member

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    4,833
    You are free to disagree with the "mathematical authority" of textbooks, Wikipedia, math professors and people that genuinely know what they are talking about, but that disagreement comes with a price:

    from American Math Monthly vol 105 no 7.

    So if you disagree with 0 having no reciprocal then you are not talking about the same numbers as the rest of the world has been using since antiquity. This means that you aren't allowed to divide by zero and get your new type of number until you explain what rules you are using.

    The two classic ways out of this are infinitesimals, where you never talk about dividing by zero but rather \(\epsilon\) a positive number smaller than any rational number. Then \(1/\epsilon = \infty\) is without contradiction. But it is a different number system and a valid number might by \( - 8 \infty + \pi + \frac{27}{3} \epsilon - \sin \epsilon^2\).

    Another classic way out is with a Riemann sphere, so that \(1/0 = \infty\) but like 0, infinity has no sign. Infinity in this concept is not quite a number, and it cannot be added or subtracted, or divided by itself, nor can it be multiplied by zero.
    \(\begin{array}{r|cccc} + & 0 & 1 & 2 & \infty \\ \hline \\ 0 & 0 & 1 & 2 & \infty \\ 1 & 1 & 2 & 3 & \infty \\ 2 & 2 & 3 & 4 & \infty \\ \infty & \infty & \infty & \infty & {\tiny \textrm{undefined}} \end{array}\) \(\begin{array}{r|cccc} \quad \quad - & 0 & 1 & 2 & \infty \\ \hline \\ 0 & 0 & -1 & -2 & \infty \\ 1 & 1 & 0 & -1 & \infty \\ 2 & 2 & 1 & 0 & \infty \\ \infty & \infty & \infty & \infty & {\tiny \textrm{undefined}} \end{array}\) \(\begin{array}{r|cccc} \quad \quad \times & 0 & 1 & 2 & \infty \\ \hline \\ 0 & 0 & 0 & 0 & {\tiny \textrm{undefined}} \\ 1 & 0 & 1 & 2 & \infty \\ 2 & 0 & 2 & 4 & \infty \\ \infty & {\tiny \textrm{undefined}} & \infty & \infty & \infty \end{array}\) \(\begin{array}{r|cccc} \quad \quad \div & 0 & 1 & 2 & \infty \\ \hline \\ 0 & {\tiny \textrm{undefined}} & 0 & 0 & 0 \\ 1 & \infty & 1 & {\tiny \frac{1}{2} } & 0 \\ 2 & \infty & 2 & 1 & 0 \\ \infty & \infty & \infty & \infty & {\tiny \textrm{undefined}} \end{array}\)

    So one is a dodge in that we invent tiny new numbers to divide by something "close" to zero, but never divide by zero. The other is a dodge in that we divide (anything except zero) by zero to get something which has only some of the properties of a number.
     
  8. Prof.Layman totally internally reflected Registered Senior Member

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    982
    The secret is that you just write the words "lim" in front of the equation and then you can just cancel zero's right out of the equation. For instance, the tangent line to a curve at exactly the tangent line "h" the height between two points would become zero, so you just write the word "lim" in front of it and then you can cancel out the "h" out of the equation. Then you are left with an equation that is exactly the tangent line to a curve, where "h" would be zero. Then by definition the tangent line of a curve is a line that intersects that curve at only exactly one point, not two points some real distance away from each other.
     
  9. arfa brane call me arf Valued Senior Member

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    7,832
    Is 0 a number? What kind of number is it?

    The natural numbers exclude 0 (naturally), or include it depending on the context (in the context of algorithms there is usually a 'zeroth' step).

    But 0 is neither positive or negative (or it's both, if you like), it isn't odd or even (or again, it's both). So if a number is defined as having sign and being odd or even (up to some approximation of an integer), then 0 appears to lie outside the set of numbers (or at least, the integers or natural numbers).
    Infinity isn't a number either, for the same reasons as well as having no definite value (at least 0 has a definite value).
     
  10. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    You are a danger to children. The limit of an expression as a variable takes on a value (like zero) has nothing to do with canceling or arithmetic.
    \(\lim_{h\to 0} \frac{(x + h)^{\tiny n} \; - \; x^{\tiny n}}{h} = n x^{\tiny n-1}\) has nothing to with the statement about numbers that division by zero cannot result in a number with all the properties of other numbers.

    It's the number with the smallest magnitude. It's the number that is the identity element for addition. It's the number that is its own additive inverse. It's the only number of a field without a multiplicative inverse.
    That is a feature of the choice of definition of the set of natural numbers. 0 is always included in the integers or rationals or reals or complex numbers.
    No. Zero is even, not odd. And zero is an integer, because \(0 = 1 \, + \, -1\). And zero is a real number because it is the least upper bound of the set of negative numbers, which makes it a real number if you know anything about real numbers.
    Infinity is a concept, not a number. But in some number systems (the hyperreal numbers, the ordinal numbers, the surreal numbers), a symbol \(\infty\) (or \(\omega\)) can take on the properties of a number with a magnitude larger than any finite number.
     
  11. Undefined Banned Banned

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    1,695
    My understanding is that zero and infinity do not axiomatically belong to number line, but are undefined singularity or boundary symbols without defined axiomatic status in number theory.

    There are arithmetic conventions using zero as symbolic placeholders, but that is not a number but a convention.

    Similarly, infinity is conventionally used in limits and arguments where unidentified scale or value is used for boundary condition rather than an axiomatically specific number or value.

    That is my personal understanding only.
     
  12. rpenner Fully Wired Valued Senior Member

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    4,833
    Zero is on the number line, halfway between \(-1\) and \(1\).

    //Edit: Like here. ( http://en.wikipedia.org/wiki/Number_line // Edit 2 -- whoops, forgot the Number-Line link )

    And how can you have an "understanding" of any system "axiomatically" unless you can either quote the axioms or point to a citation for them?

    I can point to a machine-checked, browseable proof of the statement "0 is a real number" or "0 ∈ ℝ" here: http://us.metamath.org/mpegif/0re.html

    And I can list the axioms from which that statement follows axiomatically, http://us.metamath.org/mpegif/mmcomplex.html#axioms
     
    Last edited: Mar 12, 2013
  13. Prof.Layman totally internally reflected Registered Senior Member

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    982
    You could say,\( \lim_{a\to 0} \frac {a}{a} = 1 \), so then you could say that as the limit of "a" approuches zero, zero divided by zero is equal to one.
     
  14. rpenner Fully Wired Valued Senior Member

    Messages:
    4,833
    But \(\lim_{a \to 0} \frac{0}{a} = 0\) and \(\lim_{a \to 0, \, b \to 0} \frac{b}{a}\) can take on any value depending on how one approaches the limit. Thus \(\lim_{a \to 0} \frac{a}{a} = 1\) does not cause the expression \(\frac{0}{0}\) to cease to be undefined.
     
  15. Pete It's not rocket surgery Registered Senior Member

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    10,167
    No, that's not how you say it.
    You would say that "The limit of 'a' divided by 'a', as 'a' approaches zero, is equal to one."

    I think you're intended meaning is correct - i.e. you can use limits to resolve some situations where it seems like to have to divide by zero (finding the instantaneous slope of a curve is the first one people learn) - but you need to be careful, and very precise with your language.
     
  16. Undefined Banned Banned

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    1,695
    Sorry, had to cut your links. New here, so can't post links yet.

    How can a line be negative? A line is a line. Arbitrary conventions splitting a line into a positive and a negative half is irrational and axiomatically invalid. If you wish to use invalid arbitrary axioms than you can prove anything you like using your own rules.

    My understanding is that a line is a line. What you make of that line is not what that line is.

    If you arbitrarily introduce a boundary or singularity condition of zero to arbitrarily split your line into plus and minus halves, then you automatically admitting the singulaity and boundary condition of zero is a common origin for both sides. That means zero is both a positive and a negative starting point according to your arbitrary introduction of it like that.

    Similarly, if you introduce a singularity or limit condition of infinity to end your lines into a positive and a negative undefined boundary value, then you are also arbitrarily ending those already arbitrary halves which your arbitrary zero arbitrarily created according to rules you arbitrarily made up as you went along.

    That way of argument, by arbitrarily inventing and introducing symbols for singularities and boundary conditions to suit your prepared definitions is not valid. It is sloppy way of doing number theory line analysis.

    I can see that my naive understanding is not according to your own rules by which you create your own understanding by introducing undefined singularity and boundary limit conditions and symbols to patch up your preferred axiomatic logic which would otherwise fail all by itself without any fudges of these introduced ex-axiomatic understandings and logics that are invalid as understood and used by you.

    Sorry, but you can prove anything the way you are doing it. And all your references are similarly tainted by that same arbitrary introduction of zero and infinity on the number line. They do not belong there. They are singularity and boundary conditions, not values.

    Sorry if my naive understanding doesn't suit you expert opinion.
     
  17. Pete It's not rocket surgery Registered Senior Member

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    10,167
    According to what axioms?

    It seems like you are using your own arbitrary axioms to prove what you like using your own rules?

    It seems like you're offended by learning.
     
  18. Undefined Banned Banned

    Messages:
    1,695
    No. The existing arguments have the axioms which I question with my naive statement of the nature of zero and infinity as singularity and boundary conditions not numbers on a number line.

    If you have any counter arguments to the singularity and boundary condition nature of the two symbols I will listen intently. But saying I am offended by learning is attacking me for pointing out the singularity and boundary condition nature of the two symbols. Not my fault if that is what they seem when closely examined on their own terms and not as fudge understandings to patch prepared axiomatic definitions and number line analysis which falls on its own without those fudges.

    Unless you and what you are propounding yourself is perfect in its own right, then learning works two ways. Maybe you should listen and not attack so quickly before even arguing your own case against my naive understanding which I stated simply.
     
  19. Prof.Layman totally internally reflected Registered Senior Member

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    982
    The tangent line of a curve has been defined, and it assumes that \(\lim_{a \to 0} \frac{a}{a} = 1\). In the case of \(\lim_{a \to 0, \, b \to 0} \frac{b}{a}\), there is just not enough information in order to determine a certain value. It would be like saying that \(\frac{x}{y}\) is not always equal to one, but that does not mean that \(\lim_{a \to 0} \frac{a}{a}\) does not equal one. It just means that you have too many variables and not enough constants to hone in on a correct value. That is the purpose of limits, to find an answer to a problem that cannot normally be solved. It finds answers to problems closer and closer to where the point is that your looking for an answer, and then gives you the answer as to what value those problems are pointing you too. So then if you wanted to find the answer to \(\frac{0}{0}\) then you would have to take the limit, and then after taking the limit you would find that the answer is one, that is the location that has the hole in the graph. That is how you solve for holes in equations.
     
  20. rpenner Fully Wired Valued Senior Member

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    4,833
    In the case of \(\lim_{a \to 23, \, b \to -12} \frac{b}{a} = \frac{-12}{23}\) no matter the rest of the tapdancing you are doing because \(\frac{-12}{23}\) is a number.
    In the case of \(\lim_{a \to 0, \, b \to 0} \frac{b - 12}{23 + a} = \frac{-12}{23}\) because \(\frac{-12}{23}\) is a number and \(\frac{b - 12}{23 + a} = \frac{-12}{23} + \frac{12 a + 23 b}{23 a + 529}\) (when \(a + 23 \neq 0\))
    The expression \(\frac{0}{0}\) is meaningless by itself because there is no curve defined, therefore no concept of tangent applies and concept of an expression where a limit concept can be applied.

    The expression \(\lim_{a \to 0} \frac{\sinh(a) - \sin(a)}{ a^2 \; \sin(a) }\) is not evaluated by saying it "looks like \(\frac{0}{0}\)" and therefore must be 1.
     
  21. Prof.Layman totally internally reflected Registered Senior Member

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    982
    I don't know what kind of tap dancing this is, but \(\lim_{a \to 0} \frac{a}{a} = 1\). Say you picked any number, and then divided it by itself, you would get one. Then pick another number closer to zero and then divide it by itself, you get one. Do it again and again as much as you like, and you will still get one. So then the closer you get to dividing a number by itself increasingly closer to zero, you always get one. So then the limit of \(\frac{0}{0}\) is one. So then if you wanted to find a solution to a problem and use the closer answer you could possibly find then you would replace \(\frac{0}{0}\) with one.

    I don't think \(\frac{a}{b}\) is really said to have to be equal to anything and that is why it is not a fundemental mathmatical principle. \(\frac{a}{b}\) is just equal to \(\frac{a}{b}\) if "a" can be anything and "b" can be anything then \(\frac{a}{b}\) can be anything.

    Say you had no apples, and you didn't divide them any number of times. How many times would you have not divided any apples? Once.

    The point here I am trying to make is that \(\lim_{a \to 0} \frac{a}{a} = 1\) has already been used in mathmatics and it has been proven to work to provide correct answers for the tangent line of a curve. I think there is a slim possibility that it is not true, but the only way we could ever discover that is by using it and applying it to different areas of science and then finding out for sure if it is really always the case or not. Otherwise, if it was not true then limits would be incorrect mathmatics. You wouldn't carry a limit over to a tangent line of a curve just because that is how it is derived. I think it would be a good idea, so then if a problem did come up then you would know that it could possibly be because the answer was found using a limit.
     
  22. Pete It's not rocket surgery Registered Senior Member

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    10,167
    Yes, \(\lim_{a \to 0} \frac{a}{a} = 1\)

    No, it doesn't follow.
    "the limit of \(\frac{0}{0}\)" is not a correct description of \(\lim_{a \to 0} \frac{a}{a}\)..
    The only correct way to say \(\lim_{a \to 0} \frac{a}{a}\) is "The limit of a/a as a approaches zero"

    Consider these limits. Would you say that they are also "the limit of \(\frac{0}{0}\)"?
    \(\lim_{a \to 0} \frac{0}{a}\)
    \(\lim_{a \to 0} \frac{a}{0}\)
    \(\lim_{a \to 0} \frac{2a}{a}\)


    The phrase "the limit of \(\frac{0}{0}\)" doesn't really mean anything. It's ambiguous, not well defined, and should be avoided.
     
  23. arfa brane call me arf Valued Senior Member

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    7,832
    It depends how closely you want to define parity; in the arithmetic sense, if 0 was odd then 1 + 0 would be even, making it even lets 1 + 0 be odd. With subtraction which isn't commutative the arithmetic rule is the same as for addition, since subtraction is addition of a negative number. Making 0 even or "giving it even parity" just makes arithmetic work except for division.

    Zero is a complex number too, it's also the 0-matrix.
     

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