How should I calculate(-4)^(2/4)? Do I calculate [(-4)^(2)]^(1/4) way
or [(-4)^(1/4)]^2 way? Why do I have to use one way over the other
because normally, (x^a)^b = x^(ab)= x^(ba) but it
doesn't seem to work here.

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Request for Question Clarification by mathtalk-ga on 13 Nov 2004 14:11 PST

Hi,kokomomo-ga:

Raising a nonzero real number (base) to a fractional power (exponent)
only gives a well-defined result (the "principal" root) if the base is
positive.

Another law of exponents is that (xy)^a = (x^a)(y^a).  Using this you
can reduce the problem you posed to considering:

  4^(2/4)  and   (-1)^(2/4)

There's no difficulty in computing the principal value of 4^(2/4)
either way, as (4^2)^(1/4) or (4^(1/4))^2.  Of course the most direct
way is 4^(1/2) = 2, but you get to the same result by taking
intermediate principal (positive) roots in either of the other two
computations.

If you like I can explain in a bit of detail why this doesn't apply to
fractional powers of -1.

regards, mathtalk-ga

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Clarification of Question by kokomomo-ga on 13 Nov 2004 18:27 PST

Are you saying that (-4)^(2/4) = 2i ?
Then, what is the domain of the function x^(2/4)? Is it all real
number or just positive rational numbers?

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Clarification of Question by kokomomo-ga on 13 Nov 2004 18:32 PST

What happend to the rule x^(a/b)=x^(a*1/b)? Do I do x^(1/b) first and
then  raise the result to a power or do I do x^a first and raise the
result by 1/b power second?

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Request for Question Clarification by mathtalk-ga on 14 Nov 2004 06:31 PST

Hi, kokomomo-ga:

The expression x^(2/4) doesn't define a function by itself, so one
cannot say what domain it has without further interpretation.

This is "the" square root function.  As you know, a positive real
number has two square roots.  We speak of the positive square root of
a positive real number as "principal".  See the discussion here:

[Square Root -- Mathworld (A Wolfram Web Resource)]
http://mathworld.wolfram.com/SquareRoot.html

Since a function gives a single value according to the definition of
"function", we must make some choices to interpret a square root
operation as a function.  The choice of a positive root for a positive
real number seems natural enough (and useful, esp. as often extended
to a zero square root for zero).  But once we study the square root
relationship for complex numbers, we see that it's impossible to
extend this function definition to all complex numbers.  Some complex
points, forming a curve extending from zero out to infinity, have to
be excluded in order to define a square root function (these excluded
points are called a "branch cut").  One is free to choose a branch cut
for the square root function in many different ways, but the
conventional choice is to exclude the negative real numbers.

When we write x^(2/4), the parentheses indicate that we must evaluate
2/4 first.  So we are asking to raise x to the power 0.5, ie. to take
a square root.

But if you ask to apply this operation to -4, then you really do have
to explain what you mean by this, because -4 is a negative real number
and excluded by convention from the usual "principal branch"
definition of a square root function.  If the definition is not made,
then it is possible to produce "paradoxes" from the resulting
confusion.

Basically what happens is that in the complex plane, the principal
branch of the square root function will give us a limiting value +2i
if we approach -4 from the upper half plane (where imaginary parts are
positive) and a limiting value -2i if we approach from the lower half
plane (where imaginary parts are negative).  No choice of (-4)^0.5 is
going to agree with both of these limits.

regards, mathtalk-ga

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Clarification of Question by kokomomo-ga on 15 Nov 2004 01:17 PST

Dear Sir:
You wrote ...
When we write x^(2/4), the parentheses indicate that we must evaluate
2/4 first.  So we are asking to raise x to the power 0.5, ie. to take
a square root.

Then
Does 4th root of (x^2) gives the same result as the square of the 4th
root of x ? Including complex numbers, doesn't the values of x can be
all real numbers?
What happens to the rule (x^a)^b = x^(ab)=x^(ba)=(x^b)^a   ?

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Request for Question Clarification by mathtalk-ga on 16 Nov 2004 04:35 PST

The short answer is that if we restrict x to be a positive real
number, the usual laws of exponents hold without restriction, even if
you allow the exponents to be complex numbers:

(x^a)^b = x^(ab)=x^(ba)=(x^b)^a

If you allow base x to be other than a positive real number, the
expressions are only unambiguous for integer exponents.

As I noted before, -4 has two square roots +2i and -2i.  There is no
standard convention for determining which of these is meant by
(-4)^(2/4).  You are certainly free to define the value to be one or
the other, but the "principal branch" of the square root function in
the complex plane by definition excludes -4 as an argument.

regards, mathtalk-ga

Hi there.

There is a bit more to this question that you might think because the
answer is actually what we call an 'imaginary number'. You get an
imaginary number whenever you square-root a negative number, and it is
indicated by the symbol 'i'. You probably don't need to know a lot of
detail about imaginary numbers are unless you're going into rocket
science or hard-core engineering. For now we just have to believe in
them in the same way that some people believe in ghosts - ie.
something that's there - only not there.

First of all express the stuff in brackets in as simple terms as
possible. In this case (2/4) reduces to (1/2) quite nicely.

As you probably know x^(1/2) is the same as ?x and x^(1/3) is the same as ³?x etc.

So to recap:

(-4)^(2/4) is the same as (-4)^(1/2)
(-4)^(1/2) is the same as ?(-4)

The best way to deal with imaginary numbers is to simply remember that:

?(-1) = i

... if we identify 'i' at the earliest opportunity it will allow us to
deal wiht the rest of the equation so it's often useful to isolate it
at the earliest opportunity.

We know that ?(ab) is equal to ?a x ?b
and that -4 is equal to (-1 x 4)

so using this rule:

?(-4) is equal to [?4] x [?(-1)]

and taking into account the imaginary number part:

[?4] x [?(-1)] = 2 x i

This, as you might guess, is expressed as 2i (the 'i' is normally
written after the number) and therefore 2i is the answer to the
question!

Interestingly enough:
i^0 = 1
i^1 = i
i²  = -1
i^3 = -i
i^4 = 1
i^5 = i
i^6 = -1
i^7 = -i
i^8 = 1
i^9 = i
i^10 = -1
i^11 = -i
i^12 = 1
i^13 = i
i^14 = -1
i^15 = -i
etc etc...

intresting eh?

By the way if you're wondering I studied Engineering at degree level
where I used lots of imaginary numbers; now I work in a bank where I
don't.

Alex

Are you saying that (-4)^(2/4) = 2i ?
Then, what is the domain of the function x^(2/4)? Is it all real
number or just positive rational numbers?

yes, I think everyone is saying -4^(2/4)=2i... 

Based upon an excellent definition of domain at
http://mathforum.org/library/drmath/view/54551.html
  - The domain of a function is the set of all the stuff you can 
    plug into the function.

The short answer is, for f(x)=x^(2/4), x is any number: real,
imaginary, or complex, although you may wish to specify if you don't
want a result to be imaginary numbers. If you don't want imaginary
numbers, your domain is all positive real numbers, and zero.

with respect to 
What happend to the rule x^(a/b)=x^(a*1/b)? Do I do x^(1/b) first and
then  raise the result to a power or do I do x^a first and raise the
result by 1/b power second?

http://oakroadsystems.com/math/expolaws.htm#RationalExponents shows
your answer quite frankly.

a) [(-4)^2]^(1/4)
=  [16]^(1/4)
= +2, -2, +2i, -2i
"Raised to 1/4" means "fourth root of", so, 4 roots.

b) [(-4)^(1/4)]^2
= [(4)^(1/4) *(-1)^(1/4)]^2
= [{+,-sqrt(2)}*{+,-sqrt(i)}]^2
= 2*i
= 2i

Which is correct?

2i, or +,-2i, are ok, but what about +,-2?

Here is where extraneous roots come in. If you raise a number to an
even exponent, you are going to introduce extraneous roots. Here, +,-2
are extraneous roots.
When you did (-4)^2 to get 16, the extraneous roots went in.

There is nothing wrong with
(x^a)^b = x^(ab) = x^(ba).
x^(a/b) = x^(a *1/b) = (x^(1/b))^a.
But if a is even, or if a/b is even, then extraneous roots might or will come in.

If you are familiar with the quadratic formula then you are familiar
with extraneous roots. Because the formula has a square root of
something, you get two roots always, a plus and a minus.
More often than not, one of the roots, the negative root, is
extraneous. It does not check with the original equation. It is not a
solution of the problem. It is rejected. It is extraneous.

---------
y = x^(2/4) is not a function. For any value of x, there are more than
one value of y. So, no function, no domain.

Corrections:
The quadratic formula does not always give a positive root and a
negative root. Sometimes the two roots are both positive, one being
lesser than the other. Even here, though, sometimes one of the two
positive roots is extraneous.

The correct answer is the following:

Given the following problem:
(-4)^(2/4)

You must remember to do order of opperations. The (2/4) is the first
thing that you must do. Therefore the problem becomes:

(-4)^(1/2)

Which is simply 2i. Remeber, however that when you take a square root
you are given two roots, so the answer to the initial problem is
+/-2i. Hope this helps.