OT: mathematical conundrum

[Cursitor Doom]

Ok, let's try this:

If 'i' is the square root of minus 1, what is 'i' squared?

[Chris Hogg]

-1

[Cursitor Doom]

Bugger. You got it right straight away. I was hoping for a lengthy, argumentative thread on this!

[Bob Eager]

Anyway, it's j really.

[charles]

that depends on whether you are a mathematicial or an engineer. The latter tend to use "i" for current.

[Robin]

I thought the whole point about "i" was that it ain't real.

[Andy Cap]

[Cursitor Doom]

Actually, it is. The label "imaginary numbers" is a misnomer.

[Chris Hogg]

But what is i^4?

[Chris Hogg]

Since when? I've always known it as 'i'. See also

[F]

-1: the same as (i^2 x i^2)?

[Tim Streater]

Whether it is or not, its square is still -1.

[Chris Hogg]

Is the wrong answer. No imaginary coconut for you, then!

[Albert Zweistein]

I beg to differ: -1 x -1 = +1.

[Cursitor Doom]

+1

(that's giving the answer AND agreeing with Tim. ;-)

[Cursitor Doom]

i is j; they're the same thing. It's just that i already reserved in electronics and electrical engineering as the symbol for current, so in these disciplines we use j instead of i to avoid confusion.

[GB]

Isn't that a bit like say: Pegasus without his wings is just an ordinary horse. It requires that Pegasus is there in the first place, surely?

[The Natural Philosopher]

1

[dr.s.lartius]

One can IMHO reconcile the mathematical and engineering notations by defining j as equal to -i resulting in i * j = 1 - not to be confused with quaternions

For further amusement, see

and the article containing it.

[Tim Streater]

This thread was entitled "mathematical conundrum". I've yet to understand what the conundrum is, perhaps I missed it, if so could someone point it out?