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Direct sum, modules question Watch

1. 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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2. (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:

2) is just a misprint, it should say
3. (Original post by Mark13)
1) becomes clear if you write out the definition of each module:

2) is just a misprint, it should say
Thank you
4. 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
5. (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).
6. (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 !!
7. Sorry to bring up this old thread. But I was looking back, and I was just wondering why:

has to = 0? It didnt bother me before, so I either understood it and forgot, or I just totally missed it >__>

Thanks
8. "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
9. "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?"
10. (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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