can anyone show me how to prove -i=1/i. thanx
2007-07-23 16:04:14 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
RHS = 1 / i = (1 / i) (i / i) = i / (i²) = i / (-1) = - i
LHS = - i
LHS = RHS
2007-07-23 22:06:17 · answer #1 · answered by Como 7 · 0⤊ 1⤋
-i
= -i^2/i
= 1/i
2007-07-23 16:06:39 · answer #2 · answered by sahsjing 7 · 0⤊ 1⤋
-i=1/i
-1*i=1/i
-1*i*i=1
-1*-1=1
1=1
2007-07-23 16:16:22 · answer #3 · answered by ? 5 · 0⤊ 2⤋
You are given -i = 1/i
Multiply both sides by i:
-i(i) = (1/i)(i)
(-1)(i)(i) = (1/i)(i)
(-1)*(i^2) = 1
(-1) * (-1) = 1
1 = 1
2007-07-23 16:16:47 · answer #4 · answered by Anonymous · 0⤊ 2⤋
1 / i = i / i^2 = i / -1 = -i
The most correct proof begins with one side of the expression and manipulates & simplifies it until you get to the other side of the equality, like in this response and the one above.
2007-07-23 16:08:25 · answer #5 · answered by triplea 3 · 0⤊ 1⤋
i = (-1)^0.5
i^2 = -1
- i^2 = 1
- i * i = 1
1/ i = -i
2007-07-23 16:11:44 · answer #6 · answered by Anonymous · 0⤊ 2⤋