...what is.... i²....equal to?
"i" isn't a variable, but stands for "imaginary numbers" if i remember correctly (jeez, that was nerdy!)
i admit, it is a hard concept to accept but it does equal something!
2006-07-10 10:55:18 · 7 answers · asked by ξℓ Çђαηφσ 7 in Science & Mathematics ➔ Mathematics
As the first answer says, i is defined as the square root of -1, so i^2 would be -1. The reason for its existance is that there's no real number that's the square root of -1, or any other negative number for that matter. Any positive or negative number squared will yield a positive number. Certain applications in engineering and such need to deal with square roots for negative numbers, so the imaginary number i was conceived.
2006-07-10 11:04:54 · answer #1 · answered by Kyrix 6 · 0⤊ 0⤋
i ** 2 = -1
2006-07-10 11:12:35 · answer #2 · answered by diogenese19348 6 · 0⤊ 0⤋
Don't have to prove i^2 = -1. That's the definition of i.
2006-07-10 11:13:30 · answer #3 · answered by amolitor99 1 · 0⤊ 0⤋
Can u guys prove that i^2 = -1 ?
2006-07-10 11:05:33 · answer #4 · answered by fwrs 2 · 0⤊ 0⤋
exactly what he said.... i is sort of an imaginary number but when you times it on itslef, like i^2 it equals -1, but I don't remember what i stands for
2006-07-10 11:01:36 · answer #5 · answered by binoxi 4 · 0⤊ 0⤋
-1 ...by definition i^2 is -1
2006-07-10 10:57:10 · answer #6 · answered by lavi_or_lavinia 2 · 0⤊ 0⤋
i ^2 (how do you make that cool superscript "2") =-1
i^3=1
i^4=-1
because i is defined as the square root of -1, when you square it you naturally get -1
2006-07-10 11:01:00 · answer #7 · answered by enginerd 6 · 0⤊ 0⤋