A Summer of Maths (ASoM) 2016

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    (Original post by krishdesai7)
    Thanks I wasn't thinkijg of it in terms of orbit decompositions, so that's where I got stuck
    np
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    Anyone got some hints for Q3 p51 of Beardon? aside from faffing about with Vieta's/ triangle inequalities Ive not got too far...
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    (Original post by EnglishMuon)
    Anyone got some hints for Q3 p51 of Beardon? aside from faffing about with Vieta's/ triangle inequalities Ive not got too far...
    For some real x = 1 + \frac{1}{n-1}
    Spoiler:
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    we have x^n = (1 + (x-1))^n > 1 + n(x-1) > \cdots,
    basically follow that till you get to x^n > x+1, then it's just

    Then:
    Spoiler:
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    \displaystyle z^n + z + 1 = 0 \Rightarrow |z|^n  = |z+1| \leq |z| + 1, now set x = |z|


    I've tried to spoiler it as much as possible. :-)
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    (Original post by Zacken)
    For some real x = 1 + \frac{1}{n-1}
    Spoiler:
    Show
    we have x^n = (1 + (x-1))^n > 1 + n(x-1) > \cdots,
    basically follow that till you get to x^n > x+1, then it's just

    Then:
    Spoiler:
    Show
    \displaystyle z^n + z + 1 = 0 \Rightarrow |z|^n  = |z+1| \leq |z| + 1, now set x = |z|

    I've tried to spoiler it as much as possible. :-)
    ah yes, thanks. That was suprisingly simple after trying the
    x^n = (1 + (x-1))^n trick
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    (Original post by EnglishMuon)
    ah yes, thanks. That was suprisingly simple after trying the
    x^n = (1 + (x-1))^n trick
    Oh, and this is a good time to tell you that part (ii) of that question is incorrect. I think it's meant to be 1 + \mathbf{2}z + z^n = 0 and not nz.
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    (Original post by Zacken)
    Oh, and this is a good time to tell you that part (ii) of that question is incorrect. I think it's meant to be 1 + \mathbf{2}z + z^n = 0 and not nz.
    lool sorry I took so long to check my notifs. Spent a good hour or so trying to solve the incorrect one XD Thanks anyway!
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    (Original post by Zacken)
    For some real x = 1 + \frac{1}{n-1}
    Spoiler:
    Show
    we have x^n = (1 + (x-1))^n > 1 + n(x-1) > \cdots,
    basically follow that till you get to x^n > x+1, then it's just

    Then:
    Spoiler:
    Show
    \displaystyle z^n + z + 1 = 0 \Rightarrow |z|^n  = |z+1| \leq |z| + 1, now set x = |z|


    I've tried to spoiler it as much as possible. :-)
    Bernoullis inequality. Nice.
    Weird how its a think even though it is very very triviail init. I guess he goes deeper though.


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    Sorry I must be being really dumb, but how do I show an inverse exists for all elements of our Set (soon to be proved group)
     G = [ \mathrm{ all \ 2x2 \ matrices} \begin{pmatrix} a & b \\c & d \end{pmatrix} : a,b,c,d \in \mathbb{Z}_{p} ] ,p is prime and all elements are non-singular? where the multiplication/ addition of any elements of the matrices is mod p.
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    Will it be impossible to apply for Cambridge with only three As levels
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    (Original post by EnglishMuon)
    Sorry I must be being really dumb, but how do I show an inverse exists for all elements of our Set (soon to be proved group)
     G = [ \mathrm{ all \ 2x2 \ matrices} \begin{pmatrix} a & b \\c & d \end{pmatrix} : a,b,c,d \in \mathbb{Z}_{p} ] ,p is prime and all elements are non-singular? where the multiplication/ addition of any elements of the matrices is mod p.
    If p is prime then we know all possible matrices must be invertible, as the determinant will never be 0, lost my train of thought from here, but I think just some mod arithmetic stuff from there?


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    (Original post by EnglishMuon)
    Sorry I must be being really dumb, but how do I show an inverse exists for all elements of our Set (soon to be proved group)
     G = [ \mathrm{ all \ 2x2 \ matrices} \begin{pmatrix} a & b \\c & d \end{pmatrix} : a,b,c,d \in \mathbb{Z}_{p} ] ,p is prime and all elements are non-singular? where the multiplication/ addition of any elements of the matrices is mod p.
    note that  \mathbb{Z}_{p} is a field.
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    (Original post by drandy76)
    If p is prime then we know all possible matrices must be invertible, as the determinant will never be 0, lost my train of thought from here, but I think just some mod arithmetic stuff from there?


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    hmm maybe but my problem is that just because it is invertible doesnt mean its 'classical' inverse is in G. For example, the 'classical' inverse of our abcd matrix is  \dfrac{1}{ad-bc} \begin{pmatrix} d & {-b} \\{-c} & a \end{pmatrix} 

but this is not  \in G if each of the matrix elements /(ad-bc) is not an integer. So I was thinking we need to rely on the mod. arithmetic to show there is always solutions e,f,g,h to the equations ae+bg=1=cf+dh, ce+dg=0=af+bh where each multiplication and addition is mod p. This though seems really fiddly (if indeed works at all) so Ive probably misunderstood something...
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    (Original post by 13 1 20 8 42)
    note that  \mathbb{Z}_{p} is a field.
    ah yes lol
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    (Original post by EnglishMuon)
    hmm maybe but my problem is that just because it is invertible doesnt mean its 'classical' inverse is in G. For example, the 'classical' inverse of our abcd matrix is  \dfrac{1}{ad-bc} \begin{pmatrix} d & {-b} \\{-c} & a \end{pmatrix} 

but this is not  \in G if each of the matrix elements /(ad-bc) is not an integer. So I was thinking we need to rely on the mod. arithmetic to show there is always solutions e,f,g,h to the equations ae+bg=1=cf+dh, ce+dg=0=af+bh where each multiplication and addition is mod p. This though seems really fiddly (if indeed works at all) so Ive probably misunderstood something...
    Bit distracted by yugioh atm but what about considering what would occur for it not to be a member of G, and showing why that can't occur?(disclaimer: probably fiddly af)


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    (Original post by drandy76)
    Bit distracted by yugioh atm but what about considering what would occur for it not to be a member of G, and showing why that can't occur?(disclaimer: probably fiddly af)


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    yea lol dw, 13 1 20 8 42 said the obvious that Zp is a field (if and only if p is prime). So Zp is a group wrt addition and multiplication mod p so there must always be solutions to the stuff above.
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    Anyone want some nice number theory questions.


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    Pretty easy but a nice lemma.
    Find all positive integers n such that for all odd integers a, if a^2<=n then a|n


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    Extension if you solve it:
    What happens when a is even? Is there a similar result? Justify.


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    (Original post by EnglishMuon)
    yea lol dw, 13 1 20 8 42 said the obvious that Zp is a field (if and only if p is prime). So Zp is a group wrt addition and multiplication mod p so there must always be solutions to the stuff above.
    Ah fair enough, haven't really covered fields yet so it didn't really cross my mind


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    (Original post by physicsmaths)
    Pretty easy but a nice lemma.
    Find all positive integers n such that for all odd integers a, if a^2<=n then a|n


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    Always get this mixed up, is that a divides n, or n divides a?


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