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    Let A = {11,00}. Find A^n for n = 0, 1, and 3
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    (Original post by wdkmwd)
    Let A = {11,00}. Find A^n for n = 0, 1, and 3
    What is A? The matrix \begin{pmatrix}1&1\\0&0    \end{pmatrix}?

    If so consider index laws.
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    (Original post by morgan8002)
    What is A? The matrix \begin{pmatrix}1&1\\0&0    \end{pmatrix}?

    If so consider index laws.


    A is actually a set.


    http://postimg.org/image/pmmnuktrd/
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    (Original post by Mathemagicien)
    So you want the Cartesian product?


    Yes mate.


    I would love that
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    (Original post by wdkmwd)
    Yes mate.


    I would love that
    Can you think of what A^1 would be?
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    (Original post by wdkmwd)
    Yes mate.


    I would love that
    Oh, and for any set S we have S^0 = \{\emptyset\}.
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    (Original post by Zacken)
    Can you think of what A^1 would be?



    Probably the same thing.

    I also think A^0 is 1 isn't it?
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    (Original post by Mathemagicien)
    Well, AFAIK, a Cartesian product of two sets A,B is the set of points (x,y) where x is in A, and y is in B

    E.g. {a,b}x{c,d}={ (a,c), (a,d), (b,c), (b,d) }

    (And thus, as Zacken says, A^0 is the empty set)



    Thanks for that, but could you continue?
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    A^3 doesn't seem to be uniquely defined, since cartesian product isn't associative.
    edit: No, I'm being stupid. Just combine all the elements in 3-tuples.
    (Original post by Zacken)
    Oh, and for any set S we have S^0 = \{\emptyset\}.
    Shouldn't it be \emptyset, since \{1\}\times \{\emptyset\} = \big(1, \emptyset\big).
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    (Original post by morgan8002)
    A^3 doesn't seem to be uniquely defined, since cartesian product isn't associative.
    edit: No, I'm being stupid. Just combine all the elements in 3-tuples.


    Shouldn't it be \emptyset, since \{1\}\times \{\emptyset\} = \big(1, \emptyset\big).
    I'm working off this. :confused:
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    (Original post by Zacken)
    I'm working off this. :confused:
    Apparently \{\emptyset \} is right and \big(S, \emptyset) = S\forall S. The second is required for the first, but I'm not sure why the second is true.
 
 
 
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