Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

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.



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.




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: