[Nevermark]

The proof is supposed to be showing that ij = -ji. Not that both ij = ji = 0.

In this case, i and j are two dimensional units. So yes, ij is a bivector (unit scalar quantity of two dimension units), so ij does not equal zero.

I think the ij = -ij requires another axiom. Or another constraint on how the space operates which results in the same result.

[ndriscoll]

I didn't show ij = ji = 0. I'm assuming i⋅j = j⋅i = 0 (an orthogonal basis), but i⋅j != ij and j⋅i != ji.

(i+j)^2 = (i+j)⋅(i+j) because v^2=v⋅v for any vector v. When you expand RHS, you get i⋅i + i⋅j + j⋅i + j⋅j = 1 + 0 + 0 + 1 = 2. When you expand the LHS (i+j)^2 as Clifford multiplication, you get 2 + ij + ji. Since RHS and LHS are equal, ij+ji=0.

[Nevermark]

Ah, thanks! I will have to go through those steps more carefully.

> Or another constraint on how the space operates which results in the same result.

I see that would be the orthonormality of i and j.