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Math Help - one to one correspondence

  1. #1
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    one to one correspondence

    How can I determine wether this function f(x) = x^2 is a one-to-one correspondence. For Z -> Z that is.
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  2. #2
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    Quote Originally Posted by Madspeter View Post
    How can I determine wether this function f(x) = x^2 is a one-to-one correspondence. For Z -> Z that is.
    Consider f(-1)~\&~f(1). Are they equal?
    So is f a bijection?
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  3. #3
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    both -1 and 1 give the same value. So my guess would be that it's not a bijection.

    sounds more like it's surjective?
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  4. #4
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    Quote Originally Posted by Madspeter View Post
    both -1 and 1 give the same value. So my guess would be that it's not a bijection.

    sounds more like it's surjective?
    OH? f(?)=2.
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  5. #5
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    Quote Originally Posted by Plato View Post
    OH? f(?)=2.
    madspeter, after you answer this question, ask yourself this.

    If the original question were from R -> R, would it be injective, surjective, or neither?
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  6. #6
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    neither.

    right?
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  7. #7
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    Quote Originally Posted by Madspeter View Post
    neither.

    right?
    yup.

    what if we change it to [0, \infty) -> [0, \infty)?
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  8. #8
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    I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.
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  9. #9
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    Quote Originally Posted by Madspeter View Post
    I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.
    The function maps integers to integers.
    1.4xxxx is not an integer.
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  10. #10
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    even with [0, ) -> [0, )?

    excuse me for not knowing basics, I'm trying.
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  11. #11
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    Quote Originally Posted by Madspeter View Post
    I suppose that could solve F(?) = 2 with an ugly number like 1.4xxxx? Not really sure.
    if it were from R -> R, but in your question its from Z -> Z.
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  12. #12
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    Quote Originally Posted by Madspeter View Post
    even with [0, ) -> [0, )?

    excuse me for not knowing basics, I'm trying.

    with that, x^2 would be bijective. because we restricted the domain and range.
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  13. #13
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    allright. Thanks to both of you.

    now if you have some ekstra spare time.
    I have the answers lying around for these questions..

    F(x) = -x is bijective
    F(x) = 2x is not bijective

    however, I fail to see why 2x is not bijective.
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  14. #14
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    Quote Originally Posted by Madspeter View Post
    allright. Thanks to both of you.

    now if you have some ekstra spare time.
    I have the answers lying around for these questions..

    F(x) = -x is bijective
    F(x) = 2x is not bijective

    however, I fail to see why 2x is not bijective.

    from what to what?

    R -> R?
    Z -> Z?
    N -> N?
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  15. #15
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    Z -> z
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