Reciprical of i

[confuted]

i is defined as the square root of -1, (sqr(-1) or (-1)^(.5) )

A friend just asked me what 1/i is, so I punched it into the TI-89 and got -i. This would be great, except I can't get the algebra to do that.

x=1/i
x^2=1/(i^2)
x^2=1/(-1)
x^2=-1
x=sqr(-1)
x=i

i!=-i, so I decided to try it with 2/i

x=2/i
x^2=4/(i^2)
x^2=4/-1
x^2=-4
x=sqr(-4)
x=sqr(-1*4)
x=2 * sqr(-1)
x=2i

Same result...I tried again for 3 and ended up with 3i. I don't see any errors in my algebra, but as far as I know, the TI-89 is infallible. I once saw it described as "a massive beast of knowledge" on these boards, and I agree. Would someone please enlighten me as to the nature of my error?

[Clyde]

When you solve for a square root you get two answers plus and minus hence:

x=1/i
x^2=1/(i^2)
x^2=1/(-1)
x^2=-1
x=sqr(-1)
x=i

becomes

x = 1/i
x^2 = 1/(i^2)
x^2 = 1/(-1)
x^2 = -1
x = +/- sqr(-1)
x = +/- i

Since x cannot be +i:

1/i = i
1 = i^2
1 = -1 ..... bzzzt

it must be - i:

1/i = -i
1 = i * -i
1 = -(i*i)
1 = -(-1)
1 = 1

[confuted]

Thank you, I feel dumb now As I said, the TI-89 knows all...

[Polymorphic OOP]

no need to square at all, just rationalize

1/i

Multiply by i / i

i/(-1)

which is

-i