You're not wrong, but you should probably tell the whole story if you're going to post on HN.

Primary Objection: Um, Ohms law gives V=IR. And P=IV. So, P=I^2R. Except that P ALSO can be calculated as P=V^2/R. So, we got squared losses anyway and R should go the other direction. WTF?

The answer is that a copper wire is a distributed system, not a lumped system. So, you derive the power loss from the generated heat across a differential slice of the line and then integrate over the whole length of the line--which gives you Power=RI^2 in the limit (and V->0 for any individual slice as the differential ->0).

It is one of those (un?)fortunate happenstances (like the Bohr hydrogen atom model) where the system in the limit sums to something we already know and leads to a bit of misunderstanding when you confuse two things which appear to be related but really aren't...