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"ADH" <adoug48@xxxxxxxxxxx> writes:
>Hi,
>I'm trying to solve this problem on group theory. Most of the problems
>I was able to handle were Zn and U(n) (integers under addition and
>multiplication) but now I'm faced with rationals. I have no idea where
>to start.
I would say, start with the definitions.
>Problem: Let G= {a+b*sqrt(2)},where a and b are rational numbers not
>both 0. Prove that G is a group under ordinary multiplication.
So, the first part of the definition of a group says you are given
a set--and you are, it's G. The next part says you are given a
"binary operation" on G: according to the terms of your problem,
that is supposed to be "ordinary multiplication". I suppose you
are capable of using "ordinary multiplication" to work out the
product (a+b*sqrt(2))(c+d*sqrt(2)): so, do it. The *next* part
of the definition (I may not be breaking the definition up into
parts the way you've seen before, but I hope you can see how
what I'm saying accords with what you've seen before) says that
the set is supposed to be "closed" by this binary operation:
in other words, no matter which two elements of G you subject
to the operation, the result is again an element of G. Well,
if you worked out (a+b*sqrt(2))(c+d*sqrt(2)) like I told you
to [if you didn't, GO BACK AND DO IT NOW, eh?], you ought to
be able to look at the result and decide whether or not it
belongs to G in case both (a+b*sqrt(2)) and (c+d*sqrt(2))
belong to G (in other words, at least one of a and b is non-zero,
and at least one of c and d is non-zero). This is the first
place where you actually have to give a *proof* of anything,
but it's not much of a proof.
Now go on as before, checking one part of the definition at
a time until you're all done.
Later you will have other ways to do this and similar problems,
but it's fairly clear that you're close enough to the beginning
of your study of groups that checking the definition line by line
is the best thing you could be doing.
Lee Rudolph
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