confuted

04-12-2003, 04:33 PM

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?

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?