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    http://www.campus.manchester.ac.uk/s.../MATH10101.pdf

    B6.(i)
    Prove this
    x \in (A \cup B)-C \Rightarrow x \in (A-C) \cup B(from definition of subsets)


    So
    x \in (A \cup B)-C
    =>
    x \in (A \cup B) and x \not \in C
    =>
    (x \in A or x \in B) and x \not \in C
    =>
    (x \in A and x \not \in C) or (x \in B and x \not \in C)
    =>
    x \in ((A-C) \cup (B-C))

    So
    x \in ((A-C) \cup (B-C)) \Rightarrow x \in(A-C) \cup B

    This is true because if B-C=B then its trivially true. But if
    B-C \subset B
    then
    (A-C) \cup (B-C) \subseteq (A-C) \cup B
    Because when x is in A-C its trivially true. But when x is in B-C then its in B, since  x \in B-C \Rightarrow x \in B. So in both cases
    x \in (A-C) \cup (B-C) \Rightarrow x \in (A-C) \cup B

    (Is this a good explanation, I would draw a truth table but then that would be like 9 lines of working.)

    Is the last bit 75? As it seems like just using the above. Part (ii) is easy.

    Oh yeah. For A2.
    1. True, choose x=3-y.
    2. False, choose y=-x
    3. True, choose x=0
    4. True, choose y= \sqrt{2}. Strange question.
    5. False, choose y=0. Then x^2 \geq 0 then
    -x^2 \leq 0 so -x^2 \not \in Z^{+}

    (how do you make Z look like it is in the paper?)

    P.S. Three hours? More like half an hour. Glad it is long because I have noticed that a level exams can be pain for time. FP2 was the worst as I needed to rush and think quick to make sure I had completed it making me do lots of unlogical stupid things.
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    (Original post by Simplicity)
    I did this
    x \in (A \cup B)-C \Rightarrow x \in (A-C) \cup B
    Which means that you've assumed the result. Game over.
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    (Original post by DFranklin)
    Which means that you've assumed the result. Game over.
    I didn't assume result. That obvious by the definition of subsets. Okay, I will change wording.
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    (Original post by Simplicity)
    This is true as x in (A-C) or x in (B-C). But if B has no common elements with C then its trivially true. If B has some or one common element of C then x is always in (A-C) so thats trivially true.
    False. A = {}, B = {1,2}, C = {2}, x = 1. x \in (A \cup B) - C, B has a common element with C, but x isn't in A-C.

    Note: "B has some or one common element of C" doesn't really make sense, so I'm having to guess at what you mean. But I can't see any reasonable interpretation of your words that is going to work.
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    (Original post by DFranklin)
    False. A = {}, B = {1,2}, C = {2}, x = 1. x \in (A \cup B) - C, B has a common element with C, but x isn't in A-C.

    Note: "B has some or one common element of C" doesn't really make sense, so I'm having to guess at what you mean. But I can't see any reasonable interpretation of your words that is going to work.
    Yeah, I as thinking that.

    Is there anything wrong with this
    (A \cup B)-C = (A-C) \cup (B-C).

    What I meant is this
    B-C \subset B then
    (A-C) \cup (B-C) \subseteq (A-C) \cup B
    Because when x is in A-C its trivially true. But when x is in B-C then its in B. So in both cases
    x \in (A-C) \cup (B-C) \Rightarrow x \in (A-C) \cup B

    Yeah, I changed it so it should make sense now. Thanks.
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    That is basically OK, but in a question like this you can't really ever say "{blah} is trivially true". If there's a reason, give it. If it really is obvious, just state the result. "trivially true" tends to be a red flag that means "I know this must be true but I can't see how to prove it".

    More generally, I think you are trying too hard to keep things "symbolic" and not use words. If I was proving this, I'd go:

    Take x \in (A \cup B) - C. Then x \not \in C, and either x \in A or x \in B (or both).

    Suppose x \in A. Then x \in A - C (since x \not \in C), so x \in (A-C) \cup B.

    And if instead x \in B then x \in B \cup (A-C) (since the RHS contains B).

    So in all cases, x \in (A-C) \cup B and we're done.

    Note that although I've used words, my answer is still much shorter than yours.
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    Woe betide the helpful shout to read alot more before attempting to answer such questions, I dare say you have not headed to such suggestions. I seem to have amassed warnings for promoting the download of copyright books so to promote downloading free share material;

    http://www.peter-dixon.staff.shef.ac...ching/STDN.PDF

    combined with the pdf at the bottom here (which has the same constructs and definitions required to know for the question sheet you are using);

    titled elementary set theory with a universal set

    http://en.wikipedia.org/wiki/New_foundations

    The above coupled with some questions should be a good start to getting you on the way to an organized approach (and the new foundation pdf describes a construction of induction set theoretically which will no doubt be more reason to read).
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    (Original post by DeanK22)
    Woe betide the helpful shout to read alot more before attempting to answer such questions, I dare say you have not headed to such suggestions. I seem to have amassed warnings for promoting the download of copyright books so to promote downloading free share material;

    http://www.peter-dixon.staff.shef.ac...ching/STDN.PDF

    combined with the pdf at the bottom here (which has the same constructs and definitions required to know for the question sheet you are using);

    titled elementary set theory with a universal set

    http://en.wikipedia.org/wiki/New_foundations

    The above coupled with some questions should be a good start to getting you on the way to an organized approach (and the new foundation pdf describes a construction of induction set theoretically which will no doubt be more reason to read).
    I can't read more that would be impossible. I'm still trying to understand a book called An introduction to Mathematical Reasoning; numbers, sets and function. I don't see how learning more advance set theory will help.
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    (Original post by Simplicity)
    I can't read more that would be impossible. I'm still trying to understand a book called An introduction to Mathematical Reasoning; numbers, sets and function. I don't see how learning more advance set theory will help.
    The first one isn't exactly advanced and the second one is certainly for people who have no experience with set theory at all
 
 
 
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