Re:Why don't we define “imaginary” numbers for every “imposs...
by devit on Tuesday April 11, 2023 @09:20AM
(#35524745)
You can do that, but there's a tradeoff of losing properties that otherwise hold.
For example, by adding the imaginary numbers, there is no longer an ordering compatible with addition and multiplication (ordering compatible with multiplication means that z > 0 and x > y implies x * z > y * z: assuming that, if 0 < i, then 0 = 0 * i < i * i = -1, absurd, or if 0 > i and thus 0 < -i, then 0 = 0 * -i < -i * -i = -1, absurd).
You can certainly add a number x such that x = x + 1 (e.g. what is commonly called an infinity or NaN), but that implies no longer having additive left inverses assuming you keep associativity of addition and 0 != 1 (since otherwise 0 = -x + x = -x + (x + 1) = (-x + x) + 1 = 0 + 1 = 1).
For example, by adding the imaginary numbers, there is no longer an ordering compatible with addition and multiplication (ordering compatible with multiplication means that z > 0 and x > y implies x * z > y * z: assuming that, if 0 < i, then 0 = 0 * i < i * i = -1, absurd, or if 0 > i and thus 0 < -i, then 0 = 0 * -i < -i * -i = -1, absurd).
You can certainly add a number x such that x = x + 1 (e.g. what is commonly called an infinity or NaN), but that implies no longer having additive left inverses assuming you keep associativity of addition and 0 != 1 (since otherwise 0 = -x + x = -x + (x + 1) = (-x + x) + 1 = 0 + 1 = 1).