"Stuart Gall" <[EMAIL PROTECTED]> wrote in message
news:<9q297s$hek$[EMAIL PROTECTED]>...
> Yes Dr Fairman,
> I have a DL assignment
>
> Q1 Write a formula for the number of non abelian groups of order N
> Q2 Show that every even number greater than 2 is the sum of two primes
>
> You said my first assignment would be free.
> Could you get the answers to me by Monday please
>
> Thanks
> --
>
> Stuart Gall
> -----------------------
-------------------------
> This message is not provable.
> "Nimish Shah" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > > Dear DL Students,
> > >
> > > I have Ph.D. degree in mathematics, physics, electrical engineering,
> > > computer science.
> > > I render assistance for DL students in performing
> > > assignments, theses, pfojects, courseworks, etc.
> > > with detailed solutions descriptions.
> >
> > He is back again, and yet still no details about his 4 PhDs!
> >
> > Nim.
> >
> > NB. I guess it would help if he uses the pural form of `degrees`,
> > spell-checked his advert (projects), stop using pompous phrases with
redundant words ("help" -> "render assistance"), etc.
> >
> >
> >
Dear Stuart,
I promised to perform only one question for free and you put two.
Below is Q2 solution. If you need Q1 solution, contact me privately
and we shall negotiate on my fee rate.
You need me to prove that a+b= 2*k (1) Where: a,b any primes and k is
some natural integer
Note, that: a) even and odd natural numbers alternate, i.e.
adjacent difference (as absolute value) is equal 1; b)prime number is
always odd number (by its definition);
So, any prime could be represented as (2*n+1) where n is some (not
any) natural integer number.
Hence (1) could be rewritten as a/2+b/2=k or
(2*n+1)/2+(2*m+1)/2=n+m+1 = k
since n and m are natural integers, it is obvious that k is also
natural integer, that was to be proved.
Dr. Fairman.
11 Oct,2001
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