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A Summer of Maths (ASoM) 2016

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Zacken
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Original post by EnglishMuon
ah yes, thanks. That was suprisingly simple after trying the
xn=(1+(x−1))nx^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+2z+zn=01 + \mathbf{2}z + z^n = 0 and not nznz.
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+2z+zn=01 + \mathbf{2}z + z^n = 0 and not nznz.


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!
Original post by Zacken
For some real x=1+1n−1x = 1 + \frac{1}{n-1}

Spoiler

basically follow that till you get to xn>x+1x^n > x+1, then it's just

Then:

Spoiler



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=[all 2x2 matrices(abcd):a,b,c,d∈Zp] 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.
Will it be impossible to apply for Cambridge with only three As levels
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=[all 2x2 matrices(abcd):a,b,c,d∈Zp] 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=[all 2x2 matrices(abcd):a,b,c,d∈Zp] 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 Zp \mathbb{Z}_{p} is a field.
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 1ad−bc(d−b−ca)[br] \dfrac{1}{ad-bc} \begin{pmatrix} d & {-b} \\{-c} & a \end{pmatrix} [br] but this is not ∈G \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...
Original post by 1 8 13 20 42
note that Zp \mathbb{Z}_{p} is a field.


ah yes lol
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 1ad−bc(d−b−ca)[br] \dfrac{1}{ad-bc} \begin{pmatrix} d & {-b} \\{-c} & a \end{pmatrix} [br] but this is not ∈G \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.
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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Original post by drandy76
Always get this mixed up, is that a divides n, or n divides a?


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a divides n so n/a=k EZ


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Original post by drandy76
Always get this mixed up, is that a divides n, or n divides a?


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a divides n
Dunno how to double quote so cheers lads


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ImageUploadedByStudent Room1468610947.215133.jpg
Was looking at this and noticed something odd when I looked at the middle two/one, with the outer sides you can 'follow' the trail from top to bottom with no problems, but if you try the same for either of the middle pillar thingies it can't be done, does anyone know how they managed this?


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