You double the voltage and halve the resistance. With longer cables you can invest more in more expensive stuff at the ends to deal with the high voltage.
Resistance stays the same, loss due to resistance goes down. I’m not sure it halves either, it might be better than halving but I’m not sure myself.
Edit:
Basic power loss formula is P=I^2R, so yes power loss is divided by 4 for a 2x increase in voltage assuming the target power delivered is held constant.
Usually the resistance does not stay the same, because it is preferred to use a thinner cable, to reduce its cost.
At a given power, double voltage means half current. If the resistance is kept the same, that means 4 times lower losses. If the resistance is doubled by using a thinner cable, that still results in two times lower losses.
Yeah I agree, I was just pointing out that a wire won’t change resistance due to voltage going up. Of course notwithstanding the wire heating up or something.
