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Tricky Example of a Function

  1. Oct 26, 2004 #1
    I was asked to find a function that is continuous nowhere but its square is continous everywhere but 0. The domain of the function is [0,1].

    I can't come up with an example...
    Any help is appreaciated. Thanks
     
  2. jcsd
  3. Oct 26, 2004 #2

    StatusX

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    if your allowed to define it piecewise:

    f(x)=
    1 for x rational
    -1 for x irrational

    and if it absolutely needs to be discontinuous at 0

    0 for x=0
     
  4. Oct 27, 2004 #3

    Tide

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    Hmm. That raises the question: If p is a number in [0, 1], what is the "next number" after p? Is it rational or irrational? It seems rather difficult to answer for a continuum! :-)
     
  5. Oct 27, 2004 #4
    One of the properties of real numbers is that there is no next number.
     
  6. Oct 27, 2004 #5

    matt grime

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    How on earth did you reach that conclusion?
     
  7. Oct 27, 2004 #6

    HallsofIvy

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    On the contrary, it's very easy to answer: there is NO "next number" in the set of real numbers or the set of rational numbers. If there were, every subset would be "well ordered" and that's a property of integers, not rational or real numbers.
     
  8. Oct 27, 2004 #7

    Tide

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    Actually, it wasn't a conclusion. It was a question for StatusX to ponder in which he proposes a "solution" to Nebula's query by, in effect, making "every other number" change the sign of a function. The answer, of course, is that there is no "next number" in the continuum of reals. No matter how close two numbers are you can always find other numbers in between.
     
  9. Oct 27, 2004 #8

    StatusX

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    he wanted a function that was nowhere continous, so thats what i gave. i never said that rationals and irrationals alternate, if thats what you mean, which doesn't even make sense. all i mean is that the limit of f(x) as x goes to c does not exist for any c, and so does not equal f(c), so the function isnt continuous anywhere. maybe thats not the right way to define continuity, but i still think its obviously discontinous.
     
  10. Oct 28, 2004 #9

    matt grime

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    You did have a conclusion, tide: that the example 'raised the question'... how or why does it make you think of this question? The example does not make any such 'alternating sign every other number' argument. It is just +1 and -1 for two different dense subsets. Any such function is a solution, and there's no need to put quotation marks around the word as if it weren't.

    There's no next number in the rationals either, and that isn't a continuum, but the question in my mind is still, 'what has this got to do with anything?'
     
  11. Oct 28, 2004 #10

    jcsd

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    By the defintion of continuity and the fact that the rationals and irrationals are both dense in R the function is obviously discontinious as at every point.
     
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