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Math Help - Survey of Mathmatics Question

  1. #1
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    Smile Survey of Mathmatics Question

    Hi everyone. I am a newbie here. I am taking an online math class, so I am pretty much on my own. It is Survey of Mathmatics. I hope someone can help me with this one problem.

    Calculate the number of subsets for the set, then calculate the number of proper subsets,
    {x l x e N and x lies between 8 and 12}

    The e should be a sign for element but I can't make that.
    I appreciate any help you can give me.
    Mirage20
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mirage20 View Post
    Hi everyone. I am a newbie here. I am taking an online math class, so I am pretty much on my own. It is Survey of Mathmatics. I hope someone can help me with this one problem.

    Calculate the number of subsets for the set, then calculate the number of proper subsets,
    {x l x e N and x lies between 8 and 12}

    The e should be a sign for element but I can't make that.
    I appreciate any help you can give me.
    Mirage20
    first, let's write out the elements of the set. there's not that much

    the set is {9,10,11}

    now the number of subsets of a set is given by 2^n, where n is the number of elements in the set.

    the number of proper subsets is given by 2^n - 1

    Can you take it from here?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mirage20 View Post
    Hi everyone. I am a newbie here. I am taking an online math class, so I am pretty much on my own. It is Survey of Mathmatics. I hope someone can help me with this one problem.

    Calculate the number of subsets for the set, then calculate the number of proper subsets,
    {x l x e N and x lies between 8 and 12}

    The e should be a sign for element but I can't make that.
    I appreciate any help you can give me.
    Mirage20
    Quote Originally Posted by Jhevon View Post
    first, let's write out the elements of the set. there's not that much

    the set is {9,10,11}

    now the number of subsets of a set is given by 2^n, where n is the number of elements in the set.

    the number of proper subsets is given by 2^n - 1

    Can you take it from here?
    What Jhevon says is correct but he provides no explanation. The reason that there are 2^n subsets of a set with n elements is that each element of the set may be in or out of a sunset, so we may label each element with a 1 if it is in the sunset or 0 if it is out. Thus these are two possible labellings for each element and so 2^n possible different labellings for the elements of the set and so 2^n different subsets.

    There are 2^n-1 proper subsets because all but one of the subsets are proper (only the subset which is the set itself is not a proper subset).

    RonL

    RonL
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  4. #4
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    We can also prove this by induction.
    Say that \{1,2,...,n\} has 2^n subsets. Then we will show use this to show that \{1,2,...,n,n+1\} has 2^{n+1} subsets.
    ----

    Given, \{1,2,...,n,n+1\}. Included in all the subsets of this set are all the subsets of \{1,2,...,n\} which is 2^n by the induction step.
    Now it remains to count the number of subsets having n+1. We can have have the following:
    \{ \{ n+1 \} \} = {n\choose 0}
    \{ \{ n+1, 1 \} , \{ n+1, 2 \} , \{ n+1, 3\} ,,, \{ n+1,n \} \} = {n\choose 1}

    \{ \{n+1,1,2\}, \{n+1,1,3\}, ... ,\{n+1,n-1,n\} = {n\choose 2 }.
    .....
    \{\{1,2,...,n,n+1\}\} = { n\choose n}

    So in total we have,
    {n\choose 0} + {n\choose 1} + .... + {n\choose n} = 2^n

    Thus, in overall total we have,
    2^n + 2^n = 2\cdot 2^n = 2^{n+1}
    And induction is complete.
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