To be clear, the "version" of (0,1) {x = 0} would have a "tail" pointing in the xz plane. That tail would have no scalar value (not even 1), but would retain a directional value in case (0,1) needs to interact with other algebraic terms. In other words, we tend to use scalars without coordinates, but shouldn't we then also be able to have coordinates without scalar values attached? AKA the invisible "arrowhead" of the vector-- sans the "stick". Thus multiplying (0,1) by "j" (one of Hamilton's orthogonal operators) would not change (0,1) at all, but would spin it on it's axis. From most observational directions, a Native American's arrow doesn't look any different when spun 90 degrees (by "j"), but it in fact IS different. One can arrive at THIS level of reasoning by attempting to act on an identity vector with the sqrt(-1) version {say "j" acting on vector (1,0) } which does not change it 90 degrees, but instead spins it. Hamilton says this spin done twice is -1 the scalar. But the (1,0) vector wouldn't really change direction, it would still look like (1,0) to most outside observation. But is it? We only have Hamilton's opinion, that j^2 = -1. This needs peer review! LOL.