Jay,

I had another look at what you were trying to do and I think I can make things a bit clearer for you.

Jayroberts wrote:

So I know this is supposed to be a simple problem, but I keep working it and not getting anywhere. so by n=2 he means find the wedge product of two "vectors" whos "basis" has dimension 2.

This question (in my edition at least) does not ask for the special case n=2. It looks like the numbering of the questions may have changed. In my book the special case for n=2 is asked for in exercise 11.10

Jayroberts wrote:

I set up the following.

a=xi+yk

b=wi+zk

where w,x,y,z run through the reals, and i,k represent my "basis" for the algebra.

so a^b=(xi+yk)(wi+zk)

which gives us:

a^b = xiwi+xizk+ykwi+ykzk

You are on track but it looks you are trying to interpret the elements that have been wedged together as scalars. In fact there is a natural identification between them in the case n=2, but I think it is much clearer if you leave all the wedges in.

With your notation, you should have:

a^b=(xi+yk)^(wi+zk)

= xi^wi+xi^zk+yk^wi+yk^zk (associativity)

= 0 + xz(i^k) + yw(k^i) + 0 (bilinearity)

= (xz-yw)(i^k)

Similarly

b^a=(wi+zk)^(xi+yk)

= (zx-wy)(k^i)

= -(zx-wy)(i^k)

= -a^b

Let me know if this helps.