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    So I have attached the picture.

    I have two questions.

    1) In the green box, is that some rule? Or can you just do that?

    2) I don't understand anything after the red line. can anyone shed some light? Why do they take x to be an element in that? Is it a misprint?
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    (Original post by 2710)
    So I have attached the picture.

    I have two questions.

    1) In the green box, is that some rule? Or can you just do that?

    2) I don't understand anything after the red line. can anyone shed some light? Why do they take x to be an element in that? Is it a misprint?
    1) becomes clear if you write out the definition of each module:

    \frac{A}{A \cap B}+\frac{B}{A \cap B} = \{a+(A\cap B) + b +(A\cap B)|a \in A, b \in B\}=\{a+b+(A\cap B)|a \in A, b \in B\}=\frac{A+B}{A \cap B}

    2) is just a misprint, it should say x \in \frac{A}{A \cap B} \cap \frac{B}{A \cap B}
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    (Original post by Mark13)
    1) becomes clear if you write out the definition of each module:

    \frac{A}{A \cap B}+\frac{B}{A \cap B} = \{a+(A\cap B) + b +(A\cap B)|a \in A, b \in B\}=\{a+b+(A\cap B)|a \in A, b \in B\}=\frac{A+B}{A \cap B}

    2) is just a misprint, it should say x \in \frac{A}{A \cap B} \cap \frac{B}{A \cap B}
    Thank you
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    Sorry to bother again. But Im stuck on the very last bit >_>

    The bit which says (b-a) e AnB ==> a=b-(b-a) e AnB

    I cannot see why this implies that a is an element of AnB. I mean, if b were an element of AnB, then I could conclude it by additive closure, but I dont know that b is in AnB, and if I did, I wouldnt need a anyways >_>

    Thanks
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    (Original post by 2710)
    Sorry to bother again. But Im stuck on the very last bit >_>

    The bit which says (b-a) e AnB ==> a=b-(b-a) e AnB

    I cannot see why this implies that a is an element of AnB. I mean, if b were an element of AnB, then I could conclude it by additive closure, but I dont know that b is in AnB, and if I did, I wouldnt need a anyways >_>

    Thanks
    The notes you've posted are quite brief at that point - basically, they're showing that a is in A and a is in B, so therefore a is in AnB.

    You've got a in A by definition, and since you know b is in B, and you've deduced b-a is in B, you get that a is in B by writing a = b-(b-a).
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    (Original post by Mark13)
    The notes you've posted are quite brief at that point - basically, they're showing that a is in A and a is in B, so therefore a is in AnB.

    You've got a in A by definition, and since you know b is in B, and you've deduced b-a is in B, you get that a is in B by writing a = b-(b-a).
    Ah I see that makes more sense. Yeh the answers are a bit vague!

    Thanks !!
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    Sorry to bring up this old thread. But I was looking back, and I was just wondering why:

    \frac{A}{A \cap B} \cap \frac{B}{A \cap B} has to = 0? It didnt bother me before, so I either understood it and forgot, or I just totally missed it >__>

    Thanks
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    "Alice has saved £10 more than Becky. Becky has saved £3 more than Charlie. Altogether Alice, Becky and Charlie have saved £26.50. How much has Alice saved?"

    EASY QUESTION
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    "Alice has saved £10 more than Becky. Becky has saved £3 more than Charlie. Altogether Alice, Becky and Charlie have saved £26.50. How much has Alice saved?"
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    (Original post by killer maths)
    "Alice has saved £10 more than Becky. Becky has saved £3 more than Charlie. Altogether Alice, Becky and Charlie have saved £26.50. How much has Alice saved?"
    Ah I see, Direct sum of two modules being isomorphic to the original means that the two modules are independent. Thanks very much!
 
 
 
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