A Summer of Maths (ASoM) 2016

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    (Original post by Zacken)
    For any set X we have (trivially) \emptyset \subset X. So the set of all subsets of X has to contain all the subsets of X, one of which is \emptyset.

    Think about what being a subset means. If I claim \emptyset \subset X, I'm saying that for all x \in \emptyset, I also have x \in X, this is a vacuously true statement.

    ETA: can you name me an element in \emptyset that is not in X?
    Oh, I see. I got confused due to the fact that \Omega is a non-empty set, and for some reason I think of non-empty set as an element i.e 0, and in this way I confused myself with the statement that G includes the empty set, haha. Sorry about that, I am sometimes a big idiot.
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    Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.
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    (Original post by EnglishMuon)
    For Q4 of exercise 1.3 (p11) of the Beardon book, is it necessary to look at the product of every element individually e.g. to show associativity/closure? I dont see that the cycles in different elements are disjoint so I dont think I can just conclude that  \alpha \beta = \beta \alpha for alpha beta are any 2 elements in the set.
    Are you referring to the composition of permutations in a symmetry group?

    If so, the fact that disjoint cycles commute should intuitively be obvious. Disjoint cycles operate on different numbers.

    Formally,

    Take disjoint cycles a,b. The permutation ab 'does b first, then does a', and the permutation ba does this in the opposite order. We can show that ab = ba by showing that each element has the same image for ab and for ba.

    Suppose an element x is in cycle b, and b takes x to y. Then y is in b. So y is not in a (a does not 'act' on y) and ab takes x to y. Now suppose u is in cycle a, taken to v. Since u is not in cycle b, overall ab will take u to v.

    Similarly you can show that ba will also take x to y, and u to v. Therefore ab = ba. They are the same function (permutation).


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    (Original post by Ecasx)
    Are you referring to the composition of permutations in a symmetry group?

    If so, the fact that disjoint cycles commute should intuitively be obvious. Disjoint cycles operate on different numbers.

    Formally,

    Take disjoint cycles a,b. The permutation ab 'does b first, then does a', and the permutation ba does this in the opposite order. We can show that ab = ba by showing that each element has the same image for ab and for ba.

    Suppose an element x is in cycle b, and b takes x to y. Then y is in b. So y is not in a (a does not 'act' on y) and ab takes x to y. Now suppose u is in cycle a, taken to v. Since u is not in cycle b, overall ab will take u to v.

    Similarly you can show that ba will also take x to y, and u to v. Therefore ab = ba. They are the same function (permutation).


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    What I meant is that thats what I would like to do but the cycles aren't disjoint in the first place so all I could see is checking each element separatly (even though there were only 4 elements so it wasnt too bad)
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    (Original post by EnglishMuon)
    What I meant is that thats what I would like to do but the cycles aren't disjoint in the first place so all I could see is checking each element separatly (even though there were only 4 elements so it wasnt too bad)
    Ah, yes. A case by case check is really the best way forward.


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    (Original post by Jellymath)
    Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.
    I think what you are looking for is the Introductory sheet A0 in this http://www.damtp.cam.ac.uk/user/examples/
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    (Original post by Jellymath)
    Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.
    It is not summer work, rather it is used in Jesus as an example sheet before they cover enough content in lectures for them to be able to attempt a proper example sheet and have a proper supervision. It can be found here.
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    (Original post by Jellymath)
    Does anyone have the link for the example sheet that cambridge send as summer work? I remember that it had a question on proving the irrationality of pi and the sum of inverse squares.
    I think (not certain) you are right that Cambridge do send out some work to do after offers are confirmed, though I am not certain if you can find it anywhere online easily. It is probably better to wait to receive it anyway.

    At least, I have heard of it before.

    Similarly, this website has some "Introductory Example Sheets" designed for the very beginning of the course - kind of like a zeroth example sheet. I just did the N&S one and isn't really difficult at all, but the Groups one may be better (I haven't looked).
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    There is a booklet but tis really easy.


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    The worksheet referred to above is here: https://www.maths.cam.ac.uk/sites/ww...s/workbook.pdf
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    That makes it even.
    Can't believe I wrote you're. lol.
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    (Original post by EricPiphany)
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    That makes it even.
    Can't believe I wrote you're. lol.
    That moment when uve been looking at the signs of permutations for the past 2 hours and go on tsr to have a break...
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    (Original post by EnglishMuon)
    That moment when uve been looking at the signs of permutations for the past 2 hours and go on tsr to have a break...
    My whole life is a break.

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    (Original post by EricPiphany)
    My whole life is a break.
    pls no moar XD
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    (Original post by EnglishMuon)
    pls no moar XD
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    (Original post by EnglishMuon)
    lol u know wat i mean

    Muons Theorem: Every number = 0
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    I just realised your username is EnglishMuon and not EnglishMoon
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    (Original post by Imperion)
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    I just realised your username is EnglishMuon and not EnglishMoon
    hahaha who knows could be a typo
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    Guys after reading this thread, I went and got Algebra and Geometry by Beardon. Its been interesting so far (although I've just begun chapter 2).I was reading the page on "The Principle of Induction II", and I didn't understand something.When he says: "If not, then we can write m+1=ab and so (obviously) P(m+1)=P(a)P(b)".Shouldn't that be P(m+1)=P(a)+P(b) ?Edit: How the **** do I put paragraphs in?
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    (Original post by Oxmath)
    Guys after reading this thread, I went and got Algebra and Geometry by Beardon. Its been interesting so far (although I've just begun chapter 2).I was reading the page on "The Principle of Induction II", and I didn't understand something.When he says: "If not, then we can write m+1=ab and so (obviously) P(m+1)=P(a)P(b)".Shouldn't that be P(m+1)=P(a)+P(b) ?Edit: How the **** do I put paragraphs in?
    I don't think the alternative you gave is true either, consider m = 3. Also it's used again later in the proof.
    Removing the restriction the the factors counted by P(n) have to be prime seems to solve the problem.
 
 
 
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