Thread: Division Result

  1. #16
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    Divided by zero != divided by a number that APPROACHES zero.

    I was referring to sth like 1/0, without limits.

  2. #17
    C++ Witch laserlight's Avatar
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    Same thing (or at least I meant by undefined what you mean by indeterminate).
    From what I understand, undefined and indeterminate are two different concepts. Undefined means that the symbol or operation has no meaning, but indeterminate means that it can take some arbitrary value, depending on circumstances. In that sense, since 0/0 is indeterminate in mathematics, it is incorrect to say that "anything divided by zero is undefined".
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #18
    Malum in se abachler's Avatar
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    Quote Originally Posted by anirban View Post
    What would be the result of the following divisions?

    Code:
    a. 1/0
    b. 1/INFINITY
    c. INFINITY/0
    d. 0/INFINITY
    e. INFINITY/1
    f. INFINITY/INFINITY
    g. 0/0
    a. 1/0 is unit infinity

    b. zero

    c. infinity/0 is infinity squared

    d. 0/infinity is zero or technically zero squared since '0 divided by any number is zero'

    e. Infinity /1 is still infinity

    f. infinity / infinity is 1 as long as both are positive unit infinity

    g. 0/0 is 1 since 0 / x = 0 * 1/ x and 1/0 = infiinty and multiplying a number by its recipricol is always 1 and also dividing any number by itself is always 1

    remember that 5/0 != 4/0 and 1/0 != -1/0

  4. #19
    C++ Witch laserlight's Avatar
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    g. 0/0 is 1 since 0 / x = 0 * 1/ x and 1/0 = infiinty and multiplying a number by its recipricol is always 1 and also dividing any number by itself is always 1
    By that same reasoning, 0/0 is 2 since 0 / x = 0 * 2/x. Consequently, if you do not call 0/0 indeterminate (or at least undefined), then we have a proof that 1=2, upon which it follows that 1=0. How do you address the statements made earlier in this thread concerning the undefined and indeterminate nature of the various expressions given in anirban's question?
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  5. #20
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    From what I can remember of calculus:
    a. undefined (infinitely large)
    b. undefined (infintely small)
    c. undefined (infinitely large)
    d. 0
    e. infinity
    f. 1
    g. indeterminable

    others of note are:
    0^0 = 0 * 1/0 = indeterminate
    0^infinity = 0
    infinity^0 = infinity * 1/infinity = undefined
    0 * infinity = undefined

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