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Math Help - Set problem

  1. #1
    Gaz
    Gaz is offline
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    Set problem

    If A = \{0, {1\over2} , 1, {3\over2} , 2, {5\over2} , 3, {7\over2} , 4, {9\over2} , 5\}, write down the following set;

    \lbrace2n + 1 \in A : n \in A\rbrace

    The solution is \{1,2,3,4,5\} but I can't get my head round why, could someone please explain this to me?

    Cheers.
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  2. #2
    Senior Member
    Joined
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    Hi.

    Quote Originally Posted by Gaz View Post
    If A = \{0, {1\over2} , 1, {3\over2} , 2, {5\over2} , 3, {7\over2} , 4, {9\over2} , 5\}, write down the following set;

    \lbrace2n + 1 \in A : n \in A\rbrace

    The solution is \{1,2,3,4,5\} but I can't get my head round why, could someone please explain this to me?

    Cheers.
    It is n \in A ,
    what is n? n = 0 or n = 1/2 or n = 1 or n = 3/2 or n = 2 or ... or n = 5

    It should be  2n+1 \in A

    Check it for all n

    n= 0:

    2*0+1 = 1, is 1 \in A? Yes it is! Therefor our solution contains 2*0+1 = 1

    n = 1/2 :

    2*1/2 + 1 = 2,
    is 2 in A? Yes it is, therefor 2 is a solution, too
    ...
    ...
    ...
    n = 5:

    2*5 + 1 = 10+1 =11
    Is 11 in A? No, its not! Therefore 11 is not a solution.

    Do you think you find the other solutions by yourself?

    Regards, Rapha
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  3. #3
    Gaz
    Gaz is offline
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    Right. I've got it now. I know what my problem was!!

    Thanks!
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