What does this equal to?

[SJ]

(-1) ^ (2/3)

interesting enough, when i asked 2 friends today, they responded simultaneously, one said "1" and one said "undefined".

so is it ( (-1) ^ (2) ) ^ (1/3) or ( (-1) ^ (1/3) ) ^ 2 ?

[OneOffDriveByPoster]

SJ @ Fri Apr 17, 2009 4:16 pm wrote:

so is it ( (-1) ^ (2) ) ^ (1/3) or ( (-1) ^ (1/3) ) ^ 2 ?

It doesn't matter:
( (-1) ^ 2 ) ^ (1/3)
= 1 ^ (1/3)
= 1

( (-1) ^ (1/3) ) ^ 2
= (-1) ^ 2
= 1

[[Gandalf]]

For negative numbers you'll only run into trouble with even roots. Besides (-1)^(1/2) is defined, as i.

[endless]

http://www.google.ca/search?hl=en&safe=off&q=%28-1%29%5E%282%2F3%29&btnG=Search&meta=

[SJ]

ahh ok. thanks, i get it now. when i plug it into my calculator it gives me an error, so i guess it doesnt support imaginary numbers. though, i found out that if i graph y=x^(2/3) on graphmatica it's a cusp graph thats defined on both positive and negative sides.

[saltpro15]

well, we can't argue with our Google overlords, that must be the correct solution!

[Brightguy]

OneOffDriveByPoster @ Fri Apr 17, 2009 4:25 pm wrote:

It doesn't matter:
( (-1) ^ 2 ) ^ (1/3)
= 1 ^ (1/3)
= 1

( (-1) ^ (1/3) ) ^ 2
= (-1) ^ 2
= 1

This isn't using the principal value, though; Google is correct if you want that. This came up two years ago [my post is basically unreadable since LaTeX broke ].

[zero-impact]

On a similiar note. Today I was playing with imaginary numbers and came up with this

Where did I go horribly wrong??

[apomb]

zero-impact @ Fri Apr 17, 2009 6:42 pm wrote:

On a similiar note. Today I was playing with imaginary numbers and came up with this

Where did I go horribly wrong??

because (-1)(-1) ≠ √1

well... maybe i guess it does

[CodeMonkey2000]

√1 =+/- 1

There are 2 answers.

[zero-impact]

does that not still mean that 1 can equal -1 or +1?

[CodeMonkey2000]

What? How did you come to that conclusion?

√1 =+/- 1

You ignored the second root in your series of equations, so the whole thing is flawed.

[zero-impact]

I understand what you are saying now. I knew it was flawed, I was simply asking how. Thank you.

[Brightguy]

CodeMonkey2000 @ Fri Apr 17, 2009 8:35 pm wrote:

There are 2 answers.

This also came up before. Any nonzero number has two distinct square roots, but the standard is √ denotes a function, so it is single-valued and defined to be the principal square root.

The real problem is that (a*b)^c = (a^c)*(b^c) doesn't always hold. It does hold when a, b have positive real part or when c is an integer (of course ab must be nonzero). Again we are using the principal value (otherwise a^b for irrational b would have an infinite number of values, for example).

[endless]

Brightguy @ Sat Apr 18, 2009 12:23 am wrote:

CodeMonkey2000 @ Fri Apr 17, 2009 8:35 pm wrote:

There are 2 answers.

This also came up before.

ha, reading through that thread is hilarious.