Question

is x^(2/3) defined for x<0 ?

i think it is .. but i found some serious debate on the net about it .. some suggest it is not

Answer 1

No its not.

Answer 2

why?

Answer 3

Im not sure why but there is the graph to prove it. Let me see what i can find.

Answer 4

why not? this is either (x^2)^(1/3) so there are no negatives being cube-rooted, or (x^(1/3))^2, and you can take the cube root of a negative anyway. Totally exists.
http://www.wolframalpha.com/input/?i=y%20%3D%20x%5E(2%2F3)&t=crmtb01

Answer 5

Well, (-2)^3 = -8

hence why would we not say that -2 = (-8)^(1/3)?

Thus there is no reason why we cannot have a domain for the function f(x) = x^(1/3) of the entire real line. Which is just a fancy way of saying that x^(1/3) makes sense for negative numbers x.

That being the case, then
\[\left( x^{1/3} \right)^2 = x^{2/3}\]
also makes complete sense.

For example, if x = -8, then
\[ x^{2/3} = (-8)^{2/3} = ( (-8)^{1/3} )^2 = (-2)^2 = 4 \]

That's one side of the story.

Answer 6

Now, here's the other side.

We expect exponents to follow certain rules, such as
\[a^x a ^y = a^{x+y}\]
and
\[ (a^x)^y = a^{xy} \]

The trouble is if we allow x^(2/3) then these laws don't quite work ....

Answer 7

for example, if
\[(-8)^{2/3} = 4\]
then
\[ (-8)^{1/3} = -2\]
but
\[(-8)^{1/3} = ((-8)^{2/3})^{1/2} = 4^{1/2} = 2\]
which is a different value.

Because of this problem, people will often just say no, it's problematic to allow x^{2/3}.

Answer 8

That is like saying that i^2 = 1 because sqrt(1^4) = 1 so we can't allow squaring,

Answer 9

*sqrt(i)^4 = 1

Answer 10

Yes. In short, you have to be careful with fractional exponents because you can run into all sorts of problems. But there's no reason not to define x^{2/3} for negative provided you treat it carefully and don't apply exponent laws without thinking carefully about what those operations are doing.

Answer 11

The problem with your counterexample is that the square root function is designed to produce only the positive value so that it remains a function, however, when solving for a solution, ALL values that satisfy a condition are in the solution set. So for these purposes, 4^(1/2) =±2

Answer 12

Yes, I agree. So one just has to be careful with fractional exponents.

Answer 13

since you brought up complex numbers, I'll just comment that in complex analysis this problem is solved by adding Riemann sheets and creating a Riemann surface on which the functions are well defined.

But most of the time, that is not what we're doing. So to avoid these problems of functions not being well defined is to restrict domains of those functions or to restrict the operations we do with those functions.

Answer 14

thank you all.

but, i dont really understand so what i actually do with negative values now ..