Hey guys!

[macdude425]

"i" is the square root of -1, therefore, i^2=-1, and so on...

 

[Nubius]

"i" is the square root of -1, therefore, i^2=-1, and so on...

So then I was correct in some fashion or another

 

[C0RR0SIVE]

I get enough math at school guys.....

 

[Nubius]

hell, I'm having to relearn math since I basically haven't done any in the last 4 years and I gotta knock these college math classes out of the way here since the local college is cheap compared to the institute up in seattle....I feel like I'm back in 8th grade again

 

[gimli of gloin]

i stands for imaginary number

 

[macdude425]

Exactly. This is good practice for me...I have a test over this stuff on Wednesday

 

[Mr. Switch]

whoops, double posted!

 

[Mr. Switch]

ok i'll take a stab at this,

from my days of Algebra 2,

'^' applies to exponents, ie. x^2 = x squared

i = the square root of -1
i^2 = -1
i^3 = -i because (i^2)(i) = (-1)(i) = -i!
i^4 = 1 because (i^2)( i^2) = (-1)(-1) = 1
i^5 = i because(i)(i^4) = (i)(1) = i!
i^6 = -1 because (i^2)(i^4) = (1)(-1) = -1!

and so on and so on.......

also the square root of a negative number is imaginary, hence the 'i', so the square root of -17 equals just 17i!

these can also be written out so if you have the square root of negative pi, and the square root of negative pi equals the square root of negative 1 times the square root of pi, and the square root of -1 is i, than it can be written out as i root pi! its really hard to explain without pictures.

 

[j4ckaL]

Welcome to tha forums doubledanger.

 

[doubledanger]

Cool, thanks guys for answering that question for me and saving a headache.

Oh and by the way dude sqrt(-17) = sqrt(17)i