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Maths question

Let A = {11,00}. Find A^n for n = 0, 1, and 3
Original post by wdkmwd
Let A = {11,00}. Find A^n for n = 0, 1, and 3


What is A? The matrix (1100)\begin{pmatrix}1&1\\0&0 \end{pmatrix}?

If so consider index laws.
Reply 2
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wdkmwd
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Original post by morgan8002
What is A? The matrix (1100)\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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wdkmwd
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Original post by Mathemagicien
So you want the Cartesian product?




Yes mate.


I would love that
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Zacken
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Original post by wdkmwd
Yes mate.


I would love that


Can you think of what A1A^1 would be?
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Zacken
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Original post by wdkmwd
Yes mate.


I would love that


Oh, and for any set SS we have S0={∅}S^0 = \{\emptyset\}.
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wdkmwd
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Original post by Zacken
Can you think of what A1A^1 would be?





Probably the same thing.

I also think A^0 is 1 isn't it?
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wdkmwd
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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?
A3A^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 SS we have S0={∅}S^0 = \{\emptyset\}.


Shouldn't it be ∅\emptyset, since {1}×{∅}=(1,∅)\{1\}\times \{\emptyset\} = \big(1, \emptyset\big).
(edited 8 years ago)
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Zacken
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Original post by morgan8002
A3A^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}×{∅}=(1,∅)\{1\}\times \{\emptyset\} = \big(1, \emptyset\big).


I'm working off this. :confused:
Original post by Zacken
I'm working off this. :confused:

Apparently {∅}\{\emptyset \} is right and (S,∅)=S∀S\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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