Set Theory Proof Homework: Can You Help?

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Homework Statement



Homework Statement are located in the pdf below. I also upload the file onto an online viewer for pdf

for those people who are afraid to download attachments.

Here is the link: http://view.samurajdata.se/psview.php?id=c1f5a372&page=1

Homework Equations



None.

The Attempt at a Solution



The attempt at a solution is located in the pdf below.

Here is the link again for those who do not wish to download the file: http://view.samurajdata.se/psview.php?id=c1f5a372&page=1

Can anyone see if I did anything wrong in the proof? I am not talking about the format of the proof, just the underlying logic. Thanks!
 

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If you can use deMorgan's Laws, this can be done quite a bit shorter.

B - A = B \cap A^C

So
A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)

The latter expression is just U, the universal set.
 
Mark44 said:
If you can use deMorgan's Laws, this can be done quite a bit shorter.

B - A = B \cap A^C

So
A \cup (B - A) = A \cup (B \cap A^C) = (A \cup B) \cap (A \cup A^C)

The latter expression is just U, the universal set.

Yes, I know that. I should have mentioned it before hand that I wanted to use the "choose-an-element method" to prove it rather than algebraic manipulation. Other than that, do you think that there was anything wrong with my proof in the pdf?
 
IMO, you are overcomplicating things by using contradiction proofs. Why don't you just prove the two statements directly?
 
Mark44 said:
IMO, you are overcomplicating things by using contradiction proofs. Why don't you just prove the two statements directly?

To be honest, I do not really know how to prove one of the statements.

That is, A + B is a subset of A + (B - A).

The plus signs means "union."

Attempt:

1) Let x be in A + B.
2) We can say x is in A or x is in B.
3) If x is in A, then x is in A + (B - A).
4) If x is in B, then x is B and x can either be in A or not in A.
5) If x is in B and x is in A, then x is in A + (B - A).
6) If x is in B and x is not in A, then x is in (B - A).

I do not even know if I got it right...
 
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number0 said:
To be honest, I do not really know how to prove one of the statements.

That is, A + B is a subset of A + (B - A).
How about writing the expression on the right as
B \cap A^C

number0 said:
The plus signs means "union."

Attempt:

1) Let x be in A + B.
2) We can say x is in A or x is in B.
x could be in both sets, if it is in their intersection.

Look at three different cases:
1) x is in A but not in B.
2) x is in B but not in A.
3) x is in A \cap B.

For each case, show that x is also in A U (B - A).
number0 said:
3) If x is in A, then x is in A + (B - A).
4) If x is in B, then x is B and x can either be in A or not in A.
5) If x is in B and x is in A, then x is in A + (B - A).
6) If x is in B and x is not in A, then x is in (B - A).

I do not even know if I got it right...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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