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A Summer of Maths (ASoM) 2016

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    (Original post by A Slice of Pi)
    For those who would like a refresher on number theory, I've just finished Q4 of this year's BMO2 and it's very nice indeed.

    Here's what I did for this one
    Spoiler:
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    p^2 = \frac{u^2 + v^2}{2} \\ \\ \therefore 2p^2 = u^2 + v^2 = (u+v)^2-2uv \\

    Multiply by two

    4p^2 = 2(u+v)^2-4uv \\ \\ 4p^2 - (u+v)^2 = (u + v)^2 - 4uv \\ \\ \therefore (2p - u - v)(2p + u + v) = (u + v)^2 - 4uv = u^2 - 2uv + v^2

    Now  u^2 - 2uv + v^2 can be written in two ways. Either as  (u - v)^2 or as 2(p^2 - uv).

    This tells us that (2p - u - v)(2p + u + v) is
    1.) Even
    2.) A perfect square

    Thus
    2p - u - v =2^{k}n^2
    for some n.
    When k is odd, this expression is twice a square, and when k is even it is a square, as desired.
    I remember doing that question, it was nice. I did miss that I can directly conclude  2p - u - v =2^{k}n^2 though and ended up doing a slightly longer method of looking at prime factorisation.
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    Hey could someone give me the breakdown on how you can use CFE's to solve Pell's equation?
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    Im timing from when i post this message
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    (Original post by BinaryJava)
    Hey could someone give me the breakdown on how you can use CFE's to solve Pell's equation?
    You might like to take a look at this thread (and the links off it) where I go through using CFs to solve the Pell equation.
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    (Original post by Krollo)
    I spent like 2 hours staring at this **** in the exam. I spent much time finding numerical values to check it was correct because it sounded so unlikely... sadly my impressive arithmetic got no marks.

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    It was certainly challenging! It isn't too long a question really, but the trouble is figuring out how to start. I'm not sure how difficult/long the BMO questions are usually because this is the first one of these I've done. I just did STEP in year 12 and didn't look at BMO.
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    (Original post by Gregorius)
    You might like to take a look at this thread (and the links off it) where I go through using CFs to solve the Pell equation.
    Yes, thanks
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    (Original post by EnglishMuon)
    I remember doing that question, it was nice. I did miss that I can directly conclude  2p - u - v =2^{k}n^2 though and ended up doing a slightly longer method of looking at prime factorisation.
    I tried to find the official solution after I uploaded mine, but couldn't. One I came across used the prime factorisation method and this was quite a long way of doing it. In my method, I used two different expressions for the RHS and one showed that the expression was even, and the conclusion followed quickly from that. I haven't seen anyone else use that method.
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    Can someone please recommend books to learn the Numbers and Sets and Differential Equations module for Part IA?
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    (Original post by LondonGamer)
    Can someone please recommend books to learn the Numbers and Sets and Differential Equations module for Part IA?
    Not a book but MIT OpenCourseWare have a series on solving Differential Equations if you can't find anything more focused


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    (Original post by LondonGamer)
    Can someone please recommend books to learn the Numbers and Sets and Differential Equations module for Part IA?
    Looking at the syllabus for each of them, they are probably the courses you will least need a book for. Differential equations especially looks really non-rigorous (you'll need some analysis to do differential equations properly). You might benefit slightly from getting a book for numbers and sets like "Naive Set Theory", but it shouldn't be necessary.
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    (Original post by EnglishMuon)
    Im timing from when i post this message
    legend says that you're still timing away

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    (Original post by Zacken)
    legend says that you're still timing away

    XD i actually am. still waiting for acknowledgement ...
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    (Original post by BinaryJava)
    Yes, thanks
    Do you know if my previous threads are saved somewhere?
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    Anyone know the significance of this question? :

    Basically it was showing that for any groups

     G_{i} : i \in \{ 1, ..., n \}, G_{1} \times ... \times G_{n} \approx (G_{1} \times ... G_{i} ) \times (G_{i+1} \times ... ) ... ( ... \times G_{n} ) i.e. we can bracket it any way and the group formed is isomorphic to the standard external product of these n groups. But to me this just seems like removing/adding extra brackets from the elements and is straight forwards to prove. so is this a mindless exercise or is there some relevance to other areas? The only thing I can think of is writing elements of  \mathbb{R}^{3} as  ( a,b ) : a \in \mathbb{C} , b \in \mathbb{R} or something
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    (Original post by EnglishMuon)
    Anyone know the significance of this question? :

    Basically it was showing that for any groups

     G_{i} : i \in \{ 1, ..., n \}, G_{1} \times ... \times G_{n} \approx (G_{1} \times ... G_{i} ) \times (G_{i+1} \times ... ) ... ( ... \times G_{n} ) i.e. we can bracket it any way and the group formed is isomorphic to the standard external product of these n groups. But to me this just seems like removing/adding extra brackets from the elements and is straight forwards to prove. so is this a mindless exercise or is there some relevance to other areas? The only thing I can think of is writing elements of  \mathbb{R}^{3} as  ( a,b ) : a \in \mathbb{C} , b \in \mathbb{R} or something
    It should have the same properties as the multiplication of two groups, which is easier to work with. The LHS consist of various n-tuples of elements - one element taken from each group and then the multiplication is done 'component wise' if I remember my group theory. Why would it have to be in  \mathbb{R}^{3} ?
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    (Original post by A Slice of Pi)
    It should have the same properties as the multiplication of two groups, which is easier to work with. The LHS consist of various n-tuples of elements - one element taken from each group and then the multiplication is done 'component wise' if I remember my group theory. Why would it have to be in  \mathbb{R}^{3} ?
    yeah thats just the definition of direct products, but my question was what relevance does this exercise have? So as an example I gave the example of rewriting  \mathbb{R} ^{3} as effectively '2-tuples' as  ((a,b) , c) instead of the usual  (a,b,c) . But still i don't really see the significance of this result overall.
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    (Original post by EnglishMuon)
    yeah thats just the definition of direct products, but my question was what relevance does this exercise have? So as an example I gave the example of rewriting  \mathbb{R} ^{3} as effectively '2-tuples' as  ((a,b) , c) instead of the usual  (a,b,c) . But still i don't really see the significance of this result overall.
    If you're looking for an example where this might be useful or relevant then I can't really think of one. It's still a result that is good to know though.
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    (Original post by EnglishMuon)
    yeah thats just the definition of direct products, but my question was what relevance does this exercise have? So as an example I gave the example of rewriting  \mathbb{R} ^{3} as effectively '2-tuples' as  ((a,b) , c) instead of the usual  (a,b,c) . But still i don't really see the significance of this result overall.
    I can think of two lines of answer to this question. The first is purely didactic: you are given the definition of a product of groups and showing that the re-bracketing property hold shows that you really have understood the definitions. Learning rigour, it's good for the soul!

    The second is to think about where such a re-bracketing property is not true. In the category of sets  A \times (B \times C) \ne (A \times B) \times C, for example. So, why is it true for groups? What has happened? Why has replacing equals with isomorphism made such a difference? Can you come up with non-trivial algebraic examples where it fails even for isomorphism?
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    (Original post by Gregorius)
    I can think of two lines of answer to this question. The first is purely didactic: you are given the definition of a product of groups and showing that the re-bracketing property hold shows that you really have understood the definitions. Learning rigour, it's good for the soul!

    The second is to think about where such a re-bracketing property is not true. In the category of sets  A \times (B \times C) \ne (A \times B) \times C, for example. So, why is it true for groups? What has happened? Why has replacing equals with isomorphism made such a difference? Can you come up with non-trivial algebraic examples where it fails even for isomorphism?
    Well in the case  A, B, C are groups, algebraically its clear to show it holds by looking at the mapping  \phi ( (a,b),c) ) = (a,(b,c)) and showing its one to one and onto. If the product of n-tuples is done element wise,  \phi ( (a,b),c)((d,e),f ))= \phi((ad,be),cf)=(ad,(be,cf))=(a  ,(b,c))(d,(e,f))= \phi( (a,b),c)) \phi( (d,e),f)) so the two are isomorphic. Since the composition of two isomorphisms is an isomorphism we can apply this multiple times to any number of groups to bracket it how u want. So I think the isomorphism makes the difference as it distinguishes between the elements of each set. e.g. if  (a,(b,c))=((a,b),c) then  a=(a,b), (b,c)=c which clearly can't hold, as the elements on either side of the equals are in different sets. Isomorphisms are usually described as 'structure preserving mappings' informally so if we had a plain old set (i.e. with no structure) we'd just be talking about bijection id imagine so if that is the case  A \times (B \times C) \approx (A \times B) \times C still makes sense if we're dealing with sets.
    When you say "Can you come up with non-trivial algebraic examples where it fails even for isomorphism?" do you mean in reference to the bracketing of these algebraic structures being isomorphic (or not in this case)?
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    (Original post by EnglishMuon)
    When you say "Can you come up with non-trivial algebraic examples where it fails even for isomorphism?" do you mean in reference to the bracketing of these algebraic structures being isomorphic (or not in this case)?
    Yup.
 
 
 
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