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Math Help - cardinality

  1. #1
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    cardinality

    Let O be the set of odd integers and E be the set of even integers.

    Prove that |Z| = |E|

    Any advice?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    What about finding a bijection from Z into E ?
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  3. #3
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    I did figure that part out but I'm having trouble proving 1-1
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    I did figure that part out but I'm having trouble proving 1-1
    what was the bijection you found?

    also, did you learn that |\mathbb{Z}| = |\mathbb{N}| and |E| = |\mathbb{N}|?
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  5. #5
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    The way I was taught to solve a problem like this: |Z| = |E| was to find a bijection. Finding a bijection means proving that the function is 1-1 and onto. Right? I just don't know how to prove that it's 1-1.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    The way I was taught to solve a problem like this: |Z| = |E| was to find a bijection. Finding a bijection means proving that the function is 1-1 and onto. Right? I just don't know how to prove that it's 1-1.
    you have to have the bijection first in order to prove it is 1-1. how can you show a function is 1-1 if you don't know what the function is? That is why I asked you to tell us your bijection, we can then help you to show it is 1-1
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  7. #7
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    I guess I'm going at the wrong way then. Can you show me how to find the bijection?
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  8. #8
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    Quote Originally Posted by jzellt View Post
    I guess I'm going at the wrong way then. Can you show me how to find the bijection?
    f(x) = 2x. Show that is a bijection.
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  9. #9
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    I did get that part then.

    Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.

    To show 1-1, I said for every x plugged into the function, there would be exactly one output, thus 1-1.

    How does that sound?
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  10. #10
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    Quote Originally Posted by jzellt View Post
    I did get that part then.

    Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.
    Not quite. You have to show that for any even number e(that is for any element in E) , there exists an x such that e = 2x.
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jzellt View Post
    I did get that part then.

    Is this right? To show it is onto I said that any x plugged into the function would come out even. Then the image of f is E and thus onto.

    To show 1-1, I said for every x plugged into the function, there would be exactly one output, thus 1-1.

    How does that sound?
    1-1 means, f(x_1) = f(x_2) \implies x_1 = x_2 for all x_1,x_2 \in \mathbb{Z}.

    So assume f(x_1) = f(x_2) \implies 2x_1 = 2x_2 \implies x_1 = x_2 by dividing through by 2. hence f(x) is one-to-one

    Follow Isomorphisms directions to show f(x) is onto
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