A Summer of Maths (ASoM) 2016

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    (Original post by drandy76)
    Natural numbers are the counting numbers, ( the non negative and non-zero integers ) whole numbers are just integers AFAIK


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    The natural numbers are the positive integers. Some people include 0, others don't. Whole numbers are the same thing. I don't think you can include negatives in the whole numbers so that's where the whole numbers differ from the integers.
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    (Original post by krishdesai7)
    Or are both conventions followed by different people?
    This. Some people think that whole numbers are numbers without a fractional part, hence making them equivalent to the integers. Others think they are non-negative integers. Others think they are positive integers. Mathematicians tend not to use the word "whole numbers" for that very precise reason, and instead use phrases like "non-negative integers", "non-positive integers", "integers", "positive integers", etc...

    Some of the notation/terminology in maths just makes me shake my head, like the whole domain/co-domain/range thing and the whole f^{-1}(A) being the pre-image, but using inverse notation. Fun tymes.
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    I propose that mathmos must carry a pen and pad everywhere, so that any verbally ambiguous/ awkward terms can be expressed using notation( thereby abandoning, whole/natural numbers etc)


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    (Original post by Zacken)

    Some of the notation/terminology in maths just makes me shake my head, like the whole domain/co-domain/range thing and the whole f^{-1}(A) being the pre-image, but using inverse notation. Fun tymes.
    Yeah lol. I spend so much time trying to figure out what exactly the difference between co-domain and range is before I finally gave up.
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    What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,  3^{m} = 1 in  \mathbb{Z}_{7} . Deduce that 7 divides  1+3^{2001} ".

    I just checked a couple of m values and pretty easy to see it works for m=6. So then I said  3^{2001} = 3^{2001-333(6)}=3^{3} (mod 7) so  3^{2001} +1 = 27+1 = 0 mod 7 . Is there a more general way without having to specifically work out m?
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    (Original post by krishdesai7)
    Yeah lol. I spend so much time trying to figure out what exactly the difference between co-domain and range is before I finally gave up.
    Ooooh, get used to it, there's a lot of that coming your way Got to have coimages and kernels and cokernels yet!
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    (Original post by EnglishMuon)
    What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,  3^{m} = 1 in  \mathbb{Z}_{7} . Deduce that 7 divides  1+3^{2001} ".

    I just checked a couple of m values and pretty easy to see it works for m=6. So then I said  3^{2001} = 3^{2001-333(6)}=3^{3} (mod 7) so  3^{2001} +1 = 27+1 = 0 mod 7 . Is there a more general way without having to specifically work out m?
    Just FLT.



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    (Original post by physicsmaths)
    Just FLT.



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    oh yea lol for the first bit
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    (Original post by EnglishMuon)
    What's the expected method for Q8 pg 29 in the beardon book? : " Show that for some integer m,  3^{m} = 1 in  \mathbb{Z}_{7} . Deduce that 7 divides  1+3^{2001} ".

    I just checked a couple of m values and pretty easy to see it works for m=6. So then I said  3^{2001} = 3^{2001-333(6)}=3^{3} (mod 7) so  3^{2001} +1 = 27+1 = 0 mod 7 . Is there a more general way without having to specifically work out m?
    I don't see this as being too different from any expected method tbh


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    (Original post by drandy76)
    I don't see this as being too different from any expected method tbh


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    kk, thanks. Sometimes I just have no idea if im being really dumb as its feels strange not to check every solution you do against some ms
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    (Original post by EnglishMuon)
    kk, thanks. Sometimes I just have no idea if im being really dumb as its feels strange not to check every solution you do against some ms
    I've started to implement a mental checklist, if at every stage I can feasibly follow my working without any undue assumptions, I can be reasonably convinced that my solution is correct


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    Is that a typo on q6 page 41 of Beardon? In mine it says az+ (b bar) z but i think its meant to say az + b (z bar)
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    (Original post by Gregorius)
    Ooooh, get used to it, there's a lot of that coming your way Got to have coimages and kernels and cokernels yet!
    Oh god no. Why you do this to me. This is downright torture. Aren't kernels bad enough that one must throw cokernals at me too.
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    Aren't kernels the popcorn thing.
    One thing i learnt from edexcel physics. Qed.


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    (Original post by physicsmaths)
    Aren't kernels the popcorn thing.
    One thing i learnt from edexcel physics. Qed.


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    nah m8, ur thinking of potato batteries. kernels are wat k is for in kfc.
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    (Original post by physicsmaths)
    Aren't kernels the popcorn thing.
    One thing i learnt from edexcel physics. Qed.


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    Yeah kernels are the popcorn thing. But they're the nice kernels. (Don't judge me but I actually like eating them from the bottom of the popcorn bowl) But no; the kernals I've been talking about are the members of the input (domain?) that map to zero under a given transformation. At least that's what I've understood. It's also very plausible that I'm really stupid and Dr Cowley was in fact talking about the popcorn thing
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    (Original post by EnglishMuon)
    nah m8, ur thinking of potato batteries. kernels are wat k is for in kfc.
    Close but not quite, Kernel is the guy who started Kfc, the K stands for Kool


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    Um guys, could I have a bit of help with this question?

    Beardon book page 11, EX 1.3, Q5
    Suppose that the permutation ρ of {1,...,n} satisfies  \rho^3 = I . Show that ρ is a product of 3-cycles, and deduce that if n is not divisible by 3 then ρ fixes some k in {1,...,n}
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    (Original post by krishdesai7)
    Um guys, could I have a bit of help with this question?

    Beardon book page 11, EX 1.3, Q5
    Suppose that the permutation ρ of {1,...,n} satisfies  \rho^3 = I . Show that ρ is a product of 3-cycles, and deduce that if n is not divisible by 3 then ρ fixes some k in {1,...,n}
    Think about the orbit decomposition of any permutation, e.g. why can't we have 2 cycles or 4+ cycles in this case? Also remember that 1-cycles fix elements so by definition any power of one cycles is the identity so they don't effect anything.

    Further hint:
    Spoiler:
    Show
    Recall that an n-cycle,  \sigma , to the power of n is the identity
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    (Original post by EnglishMuon)
    Think about the orbit decomposition of any permutation, e.g. why can't we have 2 cycles or 4+ cycles in this case? Also remember that 1-cycles fix elements so by definition any power of one cycles is the identity so they don't effect anything.

    Further hint:
    Spoiler:
    Show
    Recall that an n-cycle,  \sigma , to the power of n is the identity
    Thanks I wasn't thinkijg of it in terms of orbit decompositions, so that's where I got stuck
 
 
 
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