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# Groups and Sets watch

1. I have no ideas. This is the last question in the Maths homework I have today. The other problems related to groups and sets are all solved, save this difficult one...Again, I have no ideas...

Consider two sets S and T with a mapping a: S --> T. If A and B are subsets of S, show that a (A U B) = a (A ) U a (B )

PS: Don't laugh at me; I'm just 15 and I'm the youngest student in my maths class. One more time, I have absolutely NO ideas on how to solve this problem. :embarasse
2. Suppose that y is a(A u B). Then there is an x in A u B such that a(x) = y. We know that (i) x is in A, or (ii) x is in B.
• If (i) is true then y is in a(A). So y is in a(A) u a(B).
• If (ii) is true then y is in a(B). So y is in a(A) u a(B).

We have proved that y is in a(A) u a(B).

--

Suppose that y is in a(A) u a(B). Then (i) y is in a(A), or (ii) y is in a(B).
• If (i) is true then there is an x in A such that a(x) = y. Since x is in A u B, y is in a(A u B).
• If (ii) is true then there is an x in B such that a(x) = y. Since x is in A u B, y is in a(A u B).

We have proved that y is in a(A u B).

--

We have proved that if y is in a(A u B) then y is in a(A) u a(B), and vice versa. So a(A u B) = a(A) u a(B)
3. (Original post by Jonny W)
Suppose that y is a(A u B). Then there is an x in A u B such that a(x) = y. We know that (i) x is in A, or (ii) x is in B.

* If (i) is true then y is in a(A). So y is in a(A) u a(B).
* If (ii) is true then y is in a(B). So y is in a(A) u a(B).

We have proved that y is in a(A) u a(B).

--

Suppose that y is in a(A) u a(B). Then (i) y is in a(A), or (ii) y is in a(B).

* If (i) is true then there is an x in A such that a(x) = y. Since x is in A u B, y is in a(A u B).
* If (ii) is true then there is an x in B such that a(x) = y. Since x is in A u B, y is in a(A u B).

We have proved that y is in a(A u B).

--

We have proved that if y is in a(A u B) then y is in a(A) u a(B), and vice versa. So a(A u B) = a(A) u a(B)
This makes some sense. Thanks a lot.

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Updated: February 27, 2005
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