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Re: a contradiction worth reading (unless a silly mistake was made)
Posted:
Jul 22, 2011 11:39 PM
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On Jul 22, 11:27 pm, Simon Roberts <2ass...@gmail.com> wrote:
> This is very poorly written but if you can decifer it you will see my
> statement is true. You have been kind patient and spent alot of time
> and effort on this. like I said I'll take up the challenge and I'll
> verify that the statement is false. But I gotta go now and get some
> immediate help with my head. I'm not looking for sympathy and it's not
> a poor excuse but I really must leave you now. thank you. and good
> bye. There are two universities and two hospitals in this small city,
> very convenient:) I feel awefully guilty about my rant and it's
> affecting me. I can't help myself sometimes. I'm sorry.
>
> > >Say the set S = {1,2,3,...,(n-2),(n-1)}.
> > >Then if we squared each element x_i in S, we would get a new
> > >set S' (with (n-1)/2 elements) whose elements would be
> > >(x_i)^2 mod n. Note we are assuming (n-1)/2 is odd, so
> > >n = 3 mod 4. If we square each element in S'
> > >((x_i)^2)^2 mod n we would end up with the same exact set S'
> > >and so on and so on.
>
> Let S be the set {1,2,...,(n-3)/2,(n-1)/2}
>
> let x be any member of set S
> and y be any other member of set S
>
> x^2=y^2 (mod n) implies either:
> (1) x=y (acontradictionif x!=y)
> (2) x = -y (mod n) (acontradiction)
> therefore x^2!=y^2 mod n if x!=y
>
> let x^2 = -y^2 (mod n)
> but x^(n-1) - y^(n-1) = 0 mod n
> or
> (x^2)^(n-1)/2 - (y^2)^(n-1)/2 = 0 mod n
> or
> (x^2)^(n-1)/2 + (x^2)^(n-1)/2 = 0 mod n
> or
> 2x^(n-1)= 0 mod n (acontradiction)
> therefore x^2!=-y^2 mod n for any x and y in S
>
> let x^2 (mod n) make the set S' where x is any member of S
> because of the above arguement S' will have (n-1)/2 distinct members
> (note: the only way it's a smaller subset of S' is if x^2=y^2 (mod n)
> x!=y (a condtradiction))
>
> let y^(2^i) mod n be any a member of the set S''={1,...,n-1} that is
> not a member of set of S'
> (where i>1).
>
> Then because S' has (n-1)/2 distict member with no two adding to n.
> We will have
> an x^2 in S' such that x^2 + y^(2^i) = 0 mod n (acontradiction).
>
> Therefore x^(2^i) mod n will also make up the same set S' or a subset
> of S' (i>1)
> *****************************************************************************************
> (the only way it's a smaller than (a subset) S' is if there is a c and
> d in S
> (c!=d mod n) such that
> c^(2^i)=d^(2^i) (mod n)
>
> that is
> if c^(2^i)=d^(2^i) mod n
>
> then
>
> (c^2-d^2)(c^2+d^2)(c^4+d^4)....(c^(2^(i-1))+d^(2^(i-1))=0 mod n
>
> but
>
> (c^2+d^2)(c^4+d^4)....(c^(2^(i-1))+d^(2^(i-1))!=0 mod n
>
> therefore
>
> (c+d)(c-d)=0 mod n
>
> acontradiction
> ********************************************************************************
> EXPANDING THE ORIGINAL SET S
> let z and g are any member of (the new set) (1,2,...,{n-1}) then z^2
> mod n will form the same set S'and hence z^(2^i) mod n will form the
> same set S' that is
> z!=g and z^(2^i)=g^(2^i) mod n only if z+g=0 mod n (i>=1) ---> z^2=g^2
> mod n only if z+g=0 mod n
"Jeremy" Pearl Jam
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