2010-09-11, 08:00 PM
Zalfor Wrote:Yes, but only solves for Real numbers. I is not counted as a real number in this equation. Only u.
Whereas right that a "i < 3u"-definition does not (given u is a real number), as far as we currently know, make any sense, does not mean that this does not make sense.
Say we define i = sqrt(-1), and say that u = - sqrt(-1), then if we substitute in:
sqrt(-1) < 3 * -(sqrt(-1))
Dividing by sqrt(-1) gives us "1 < -3", which is true.
Basically, any imaginary number lower than 1/3 * i = u satisfies this equation.
Noah

