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Possibly a very stupid question.

  1. Jan 26, 2009 #1
    Division by zero is undefined.....but why doesn't somebody just define it?

    Like they did with root (-1).
  2. jcsd
  3. Jan 26, 2009 #2
  4. Jan 26, 2009 #3


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    Arithmetical operations have to result in unique numbers.
    4 multiplied by 6 can result in one and only one number.
    100 divided by 12 results in one and only one number.
    1 divided by zero does not result in a unique number. What it results in is not numerically definable.
  5. Jan 27, 2009 #4
    [itex]\forall x\in \mathbb{R}, x \times 0 = 0[/itex]


    Let be [itex]a, b \in \mathbb{R}, a \neq b[/itex]

    We said before that [itex]a \times 0 = 0[/tex] and also [itex]b \times 0 = 0[/itex].

    So, [itex]a \times 0 = b \times 0[/tex]. If we were able to divide by 0, we will have that [itex]a = b[/itex] which is a contradition with our suppose.
  6. Jan 27, 2009 #5
    Check the archives. There are good discussions about this question.

    The basic problem is there IS no way to define it in such a way that it is the full inverse of multiplication.

    What does that mean? Two operators are inverses if they "undo" each other, regardless of input. Addition and subtraction are the best known example. For any x and y, x + y - y = x. You start with x, you do "+y" then you "undo" the +y with a -y.

    But there is no way to do this with multiplication. You can do it for *almost* every single number. Clearly, if we have 16 * 4 / 4, that equals 16 and division has "undone" the multiplication. In fact, this works for every number except 0. Try it out.

    The proof of this lies in the fact that multiplication is not one-to-one (it does not map unique inputs to unique outputs). Functions which are one-to-one all have inverse functions. Functions which are not one-to-one may only, at best, have partial inverses.

    This contrasts to imaginary numbers. Square root, unlike division, IS one-to-one. No two real numbers have the same square root. However, it happens that some reals don't have square roots. The imaginary number system is a way to extend the real line to preserve the square root function's one-to-one-ness, and at the same time, upgrade square root from a partial function to a total function.
  7. Jan 27, 2009 #6
    This definition is too restrictive, just say that y and it's inverse under the operation gives the identity.
    Here's a counterexample. Define the binary operation on a set S = {a,b,c} as follows
    \mc{1}{l}{*} & \mc{1}{l}{\bfseries a} & \bfseries b & \bfseries c \\ \hline
    a & a & b & c \\
    b & b & a & b \\
    c & c & b & a
    The operation is commutative, [itex]a[/itex] is the identity, and for each x in S, the inverse is also x. Let's do b on c, [itex]cb = a[/itex] let's undo b by it's inverse, then [itex](cb)b = ab = b[/tex] but now the original [itex]c[/itex]
    This is due to the fact that your definition would work only for associative operations!

    Does that mean addition is one-to-one? What is the definition of a one-to-one here?
    Take [itex](7,1) \neq (5,3)[/tex] but [itex]+(7,1) = +(5,3)[/itex], in other words, I can't see how given an output, one can determine the unique ones which produced it (the usual one-to-one definition!)
    Actually "map unique inputs to unique outputs" doesn't suggest one-to-oneness but rather that the operation is well defined, i.e given a unique x, it gets mapped to only one f(x)!
    Last edited: Jan 27, 2009
  8. Jan 27, 2009 #7
    I wasn't giving a full definition. Just the flavor of it.

    You shouldn't really speak of an operator's inverse. You speak about sections of that operator, where only one argument is applied, such as (*2), the doubling function. The inverse function would be (/2), the halving function.

    The argument is that the section (*0), which maps all real numbers to 0, has no inverse.

    When speaking about addition, what I meant was that all functions (+n) have an inverse called (-n). To contrast, for all n EXCEPT 0, (*n) has an inverse (/n). That exception cannot be removed because (*0) is not one-to-one. In fact, it is quite the opposite.
  9. Jan 27, 2009 #8
    Let me clarify that when I said "y and it's inverse under the operation gives the identity.", I was implicitly assuming "if it exists" both for the inverse and for the identity. Obviously an operation doesn't need to have any of these!

    Absolutely! I never argued that!

    I'm sorry but I can't follow, what "all functions (+n)" are you talking about here, do you mean to define a function for each n, why not just a single function that sends each element to it's inverse?

    Surely the exception cannot be removed, but then if we define such function, actually *0 wouldn't be "not one-to-one" but rather undefined!
    This is what my argument is about, I can't understand what do you mean by one-to-one!
  10. Jan 27, 2009 #9
    Tac-Tics, I should have assumed the definition meant to be informal, so I hope my reply didn't offend you. I simply couldn't understand what you meant by one-to-one! But this is off the main argument. I think you already made your point which I definitely agree with.

    Last edited: Jan 27, 2009
  11. Jan 27, 2009 #10
    You did not offend.

    I might as well add one more idea about division. The standard accepted definition is that division isn't even defined at 0. If we're talking about the function f(x) = 1/x (or using "section" notation (1/)), 0 isn't even part of the domain. The advantage of this is that f is actually a continuous function (by the epsilon-delta definition) even though its graph clearly makes a large "jump" from negative infinity to positive infinity.
  12. Jan 27, 2009 #11
    I'm glad to hear that.

    Surely, there's an algebraic explanation why it is undefined, as was elaborated by you and others, but here's an interesting example.
    The same operations of R (+ and *) defined on a set with just one element. We can actually freely divide by 0 since we already know that the multiplicative identity equals the additive identity.

    But I realize also that the original poster didn't question why it's undefined but rather why we can't find a way to come around it, which I believe was put nicely in the fourth post by MiGUi, which if such inverse exists, it would result in equating all the elements (in other words, reduces the set R to a singleton set, which is contradictory).
  13. Jan 27, 2009 #12


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    Simple-minded thought: a number, n is a specific (although unknown, variable) number, and zero is a specific number. n divided by 0 is infinity? But infinity is NOT a number. Division by zero is impossible. Good? Not Good?
  14. Jan 27, 2009 #13
    I wouldn't go with that argument, infinity is a "number" in the extended reals isn't it?
  15. Feb 2, 2009 #14
    Is it possible to construct an algebraic system where zero is not unique, prehaps have a unique zero for every element which would than permit division by zero?

    Or maybe think of the zero as an operater rather than an identity element where the inverse would represent division by zero.

    sorry for the general questions but any response would be greatly appreciated.
  16. Feb 2, 2009 #15
    Algebraicly speaking, there's a structure called wheels, which does something similar to that. Not sure how useful it is though! See http://dx.doi.org/10.1017/S0960129503004110

    Division by zero is still not defined in the extended real line, since you would get both [itex]+\infty[/itex] and [itex]-\infty[/itex]. However, it can be defined on the extended complex plane [itex]C \cup \{\infty\} [/itex] since there's only one infinity, and you would get [itex]\frac{1}{0} = \infty[/itex]. But the structure you get is NOT a field, like R or C. So you won't be treating it as an algebraic structure and get the usual contradictions!
    Last edited: Feb 2, 2009
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