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How to proof 0\0=?

  1. Oct 15, 2007 #1
    how to proof 0\0=???

    hi
    im anew member..........
    and i search about an appear proof to 0\0 = undefind
    becuse i want to arrive to scince fact after proof
    please hellp me????
    hashim​
     
  2. jcsd
  3. Oct 15, 2007 #2

    CompuChip

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    Hi, welcome to PF.

    I think this discussion has already been held quite often, so might use the search function. In short: íf it existed, it would be a limit, as we cannot divide by zero. Moreover, íf this limit which we call 0/0 were defined, it would be unique.

    But both
    [tex]\lim_{x \to 0} \frac{0}{x} = 0[/tex]
    and
    [tex]\lim_{x \to 0} \frac{x}{x} = 1[/tex]
    are good candidates, though they are not equal, and therefore there is no limit.

    You can also view it as a multiplication problem, 0/0 would have to be the number you need to multiply 0 with to get 1 (which is the reason division by zero is not allowed for any number)
     
  4. Oct 15, 2007 #3

    Gib Z

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    It's simple to prove any real number divided by zero is undefined, because we define division as the inverse operation to multiplication. Division is defined as such:
    [tex]\frac{a}{b} = c [/tex] If and only if [tex]a=bc[/tex], where a, b and c are real numbers.

    Now 2 cases: For values of a not equal to 0, and when a=0.
    When a is not equal to zero, then the definition states that [tex]\frac{a}{0}=c[/tex] if and only if [tex]a=0\cdot c = 0[/tex]. But we already decided that a is not equal to 0. So the first case of a not being equal to 0 can not hold, it is not defined.

    The second case of when a is also 0 becomes thus:
    [tex]\frac{0}{0} = c[/tex] where c is some real number, if and only if [tex]0 = 0 \cdot c[/tex].

    However, we can see that any value of 0 will satisfy that condition. Hence, if we were to evaluate 0/0, we could say it is equal to any value we like. We can't do that because then many inconsistencies come up when dealing with real numbers, so mathematicians like to leave it undefined.
     
  5. Oct 15, 2007 #4
    mr GIB Z
    please i dont understand......
    the proof is not clearlly.......
    when i was studdy at universty the doctor proof it to our .........but
    i search about it but i dot find it???
    the proof was by contradiction............
    please hellp me
     
  6. Oct 15, 2007 #5

    Gib Z

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    What I did WAS Proof by contradiction.

    There were 2 cases. For the first one, I assume a is not equal to 0 but the definition implies that a IS 0, so that is a contradiction.

    In the second case, The definition shows that you can choose whatever you like for the answer, which is again a contradiction. Read through my post again.
     
  7. Oct 16, 2007 #6
    mr GIb Z

    i dont understand proof???? please hellp me??
    hashim

    Edit: hashim, you have been warned several times to stop with the obnoxious formatting.
     
    Last edited by a moderator: Oct 16, 2007
  8. Oct 16, 2007 #7

    Hurkyl

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    0/0 is undefined because the definition of division says so. (I'm assuming you mean division of real numbers, division of rational numbers, or something 'ordinary' like that)

    There is nothing to prove here; there can only be explanations of why people chose to use that definition.
     
  9. Oct 16, 2007 #8
    So 0/0 can be proven to be defined but the definition states that it must be undefined?
     
  10. Oct 16, 2007 #9

    Hurkyl

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    Incorrect.

    0/0 cannot be proven to be defined, and the definition states that it must be undefined.
     
  11. Oct 16, 2007 #10
    This because 0/0=1 but 0(0)=0?
     
  12. Oct 16, 2007 #11

    Hurkyl

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    No. This is because domain of the division operator requires a nonzero denominator. Mathematically speaking, 0/0 is gibberish.
     
  13. Oct 17, 2007 #12
    N/0 where N != 0 doesn't seem to have any solutions (by definition of division) as Gib Z proved. But 0/0 seems to have infinitely many solutions, and is thus indeterminate. I.e., in the case of 0/0, any number multiplied by 0 yields zero. Thus if division of two functions gives an indeterminate form, it can have a finite value depending on the functions, if it exists. For reference, one can take a look at L'Hospital's rule. Ofcourse, all the above uses concept of limits.

    Regards,
    Sleek.
     
  14. Oct 17, 2007 #13

    Gib Z

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    Ok say we have what you are talking about: [tex]\lim_{a\to 0, b\to 0} \frac{a}{b}[/tex].
    This is actually the ratio of two infinitesimals, or differentials. And you know that can have a finite value, like you said. But that is different to the actual, 0/0, rather than this limit. Basically given values of a and b, arbitrarily small, there will still be a finite value that it evaluates to. ie In the definition of division i proposed, there WILL be a solution to that equation. However when b is EXACTLY 0, there no unique number will work, and assigning some principal value will ALWAYS lead to inconsistencies in arithmetic.
     
  15. Oct 17, 2007 #14

    CompuChip

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    OK, one final attempt to convince people:
    There is also a theorem, I think, that says that
    [tex]\lim_{(a, b) \to (0, 0)} f(a, b) [/tex] (1)
    exists iff
    [tex]\lim_{a \to 0} \left( \lim_{b \to 0} f(a, b) \right)[/tex] (2)
    and
    [tex]\lim_{b \to 0} \left( \lim_{a \to 0} f(a, b) \right)[/tex] (3)
    both exist, in which case all the limits are equal.
    Now take [itex]f(a, b) = a / b[/itex] and you will see that the limit in (2) does not exist.
     
    Last edited: Oct 17, 2007
  16. Oct 17, 2007 #15

    Gib Z

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    I think you meant that the first limit does not exist, but either way that theorem works =]
     
  17. Oct 20, 2007 #16
    It makes no sense to say 0/0 is defined as undefinable.

    If we define division between the numbers x and y as: x/y =df (the z: x=y*z),
    then neither 0/0 or x/0 are unique.
    That is to say neither 0/0 nor 1/0 exist, even though they are defined!

    There is no unique number that 0/0 is.
    There is no unique number that 1/0 is.

    Note: within hypercomplex numbers x/(a zero divisor) does not exist either.
     
  18. Oct 20, 2007 #17

    Hurkyl

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    Whether or not that's true, it's not quite what I said. I said that / is defined so that 0/0 is an undefined expression.


    That's not a definition. It fails two key semantic criteria: a function must have a value at each point in its domain, and that value must be unique.


    But in the spirit of what you were trying to do, you can define a ternary relation, such as
    "z is a quotient of x and y" iff x = y * z.​
    In this case, the expressions
    z is a quotient of 0 and 0​
    and
    z is a quotient of 1 and 0​
    are, in fact, well-defined. (and, of course, the former is identically true and the latter is identically false)
     
  19. Oct 20, 2007 #18
    You said "0/0 is undefined because the definition of division says so."
    Presumably then, your defiition of 0/0 says that 0/0 is undefined??

    What exactly does your definition of division say??

    Quote:
    If we define division between the numbers x and y as: x/y =df (the z: x=y*z),
    then neither 0/0 or x/0 are unique.

    What the hell are you talking about?

    Where do you get your information?
     
  20. Oct 20, 2007 #19

    CompuChip

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    Here, for example, which corresponds to my definition by the way. Note the part where it says
    or
    This is also what I learned in my first year Mathematics course. What's your definition?
     
  21. Oct 20, 2007 #20

    Hurkyl

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    Definition A real number x is said to be invertible iff there exists a real number y such that xy = 1.

    Definition [itex]\mathbb{R}^\times[/itex] is the set of invertible real numbers.

    Definition [itex](\cdot)^{-1}[/itex] is the function whose domain and range is [itex]\mathbb{R}^\times[/itex] and satisfies the equation [itex]x x^{-1} = 1[/itex] for all invertible x.

    Definition / is the binary function whose domain is [itex]\mathbb{R} \times \mathbb{R}^\times[/itex], whose range is [itex]\mathbb{R}[/itex] and is given by [itex]x/y = xy^{-1}[/itex].


    Sorry for the verbosity, but the typical presentation really does go in stages like this. You could combine the definitions to get something that looks similar to yours, but with one key difference: it requires the denominator to be an invertible number.

    Since 0 is not invertible, (1,0) and (0,0) are not in the domain of /, and so 1/0 and 0/0 are undefined expressions.
     
    Last edited: Oct 20, 2007
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