lostNfound said:
I'm going through a abstract algebra book I found and am trying to learn more about group theory by going through some of the proofs and practice sets, but am having trouble with the following:
Prove that G={a+b*sqrt(2) | a,b E R; a,b not both 0} is a group under ordinary multiplication.
Any help would be awesome!
first, we need to show closure. so we need to show the product of two elements of G is again in G. now:
(a + b√2)(c + d√2) = ac + 2bd + (ad + bc)√2, however:
we need to show that either
a) ac + 2bd ≠ 0 or
b) ad + bc ≠ 0.
so let's suppose that ac + 2bd = 0 (because if it isn't, there's nothing to prove). we need to show that whenever this is true, we cannot have ad + bc = 0.
now ac + 2bd = 0 means:
ac = -2bd. can both a and c be 0? no, because that would force either b or d to be 0, so that one of our original numbers wasn't in G. similarly we can't have b = d = 0. so if ac = 0, the only way that could happen is
1) b = c = 0, a,d ≠ 0 or
2) a = d = 0, b,c ≠ 0
suppose it was the second one. then bc ≠ 0, so ad + bc = bc ≠ 0. and if it was the first one, ad ≠ 0, so ad + bc = ad ≠ 0.
so, the only cases left are where NONE of a,b,c or d are 0. in this case, we have a = -2bd/c, and so
ad + bc = -2bd
2/c + bc. now suppose that somehow this was 0 (we are going to derive a contradiction). then:
-2bd
2 + bc
2 = 0 dividing by b (which is not 0):
-2d
2 + c
2 = 0, so
c
2 = 2d
2, and thus (c/d)
2 = 2. but this imples that c/d = ±√2, and since c/d are rational, that √2 is rational. this cannot be, so ad + bc cannot be 0.
thus, we have closure (this is the hard part).
multiplication is obviously associative (considering these as real numbers), or you can prove it directly (the computation is tedious, but not challenging).
that leaves 2 things to verify:
G has an identity, and G possesses inverses.
verify that 1 = 1 + 0√2 is an identity, and that:
1/(a + b√2) = a/(a
2 + 2b
2) - [b/(a
2 + 2b
2)]√2.
