Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0
Yeah. I cheated. You have to either deliberately misunderstand how to measure vectors or else drop a minus sign for it to work my way.
(Or, from my previous example, you could just frame it as you’re getting the hypotenuse by measuring between |AB| and -|AC|𝑖 instead of the way I framed it – but that makes it more obvious that you’re fishing for a particular answer.)
i= √(-1) = imaginary number (1^2) + (√(-1))^2 = 1 - 1 = 0 7
At least, I thought that was the idea in the OP.
Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0
Yeah. I cheated. You have to either deliberately misunderstand how to measure vectors or else drop a minus sign for it to work my way.
(Or, from my previous example, you could just frame it as you’re getting the hypotenuse by measuring between |AB| and -|AC|𝑖 instead of the way I framed it – but that makes it more obvious that you’re fishing for a particular answer.)