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

Ok, i deleted my previous answer and try to summarize all the results here.

The short answer is: NO, it is NOT defined in the real case in a mathematical sense for the following reasons:

1a)

x = √4 means x = 2, because, as said in here, a number is well-defined if it has a _unique_ value. You shouldn´t confuse this with solving polynomials like x^2 = 4, which has the solutions x_1 = +√4 _and_ x_2 = -√4.

1b)

A polynomial of degree n, like a*x^n + b*x^(n-1) + ... = c, has exactly n complex (sometimes some or all of them are real) solutions. Why does a polynomial of the form x^n = c have multiple solutions? That´s because the information about the original _phase_ (meaning the sign in the real number case) is lost in the powering process (->entropy of information) and you need to account for all the phases that lead to the given result.

x = n√(c) can be seen as a polynomial of degree 1 and has only one complex solution, which is sometimes (in case c > 0) real (the principal root, as someone also mentioned on here).

(This is, to be exact, only the case for algebraic/non-transcendental values and finite integer n, otherwise the expression would´t be a polynomial. Other cases make this more complicated, but in principal it stays the same.)

Example:

z = ³√(-8) = 1+ i√3

see: http://www.wolframalpha.com/input/?i=cube+root+of+...

1c)

x = ±c means that x = +c _and_ x = -c, so that x = 0 if c = 0.

x_1/2 = ±c means that x_1 = +c _and_ x_2 = -c.

±d = ±c means that +d = +c _and_ -d = -c so that c = d.

This is according to the definition of the ± - sign, which makes it a one-to-one expression.

2)

x^(a/b) = (x^(1/b))^a = (x^a)^(1/b) has to hold for any given x, meaning the order in which operations are applied to the variable _needs_ to be arbitrary. Why should any order be preferred over any other? I must be able to chose a = 1/c and b = 1/d so that this law still holds.

3)

_Every_ other laws for powers and roots have to hold for _every_ value of x. Meaning that for example ³√(x) = 6√(x^2) has to be well-defined in every case. This is _not_ true if you let roots have negative results as shown here:

-2 = ³√(-8) = 6√((-8)^2) = 6√(64) = 2

-2 = 2

which shows that ³√(-8) is not well-defined.

I still don´t get why everyone in here still ignores this fact. A negative root contradicts mathematical rules!

4)

What i said in my previous answer concerning the complex case was wrong. I thought the above problem wouldn´t occur, but it´s also there if you allow complex answers. That is because roots of complex numbers are only defined on the _sliced_ complex plane! This problem is deeply related to the complex logarithm, which is as well only defined on the sliced plane and therefore does _not_ accept negative values of the real axis (as pointed out by others here before). Don´t believe me? Try plotting the complex cube root. It will give you a plot of the followig form: http://en.wikipedia.org/wiki/File:Complex_cube_roo... . There are singularities on the real axis in the negative direction (different limits for approaching from above and from below the real axis). This is not a null set of singularities, but infinite, so you cannot find a way around this.

(I said this, because for example in integration, if the Riemann integral of a curve is not defined because of some points being displaced to make the curve discontinuous, you can choose the Lebesgue definition of the integral as a "workaround" to this problem if the number of discontinuities is a null set.)

5)

The solution to x^3 = -8 is x_1/2/3 = -³√8. Again, don´t confuse this with x = ³√(-8), which is a different problem. You can´t deduce that ³√(-8) = -2 just because (-2)^3 = -8. It does not work this way. These are different problems and the root is not well-defined for negative x, as shown above.

Maybe some parts of the discussion in here were due to the fact that the solutions to an equation are also referred to as "roots". The roots of an equation and the root-function are different things.

6)

I think it´s important to mention, that it is a matter of convention, which "branch" of the root is used. You can of course let the even roots of positive real values be negative, as for example √4 = -2. But you have to stick to that convention once you chose it. This means, you cannot have an equation like

√4 = √4

and use different conventions on different sides so that

-2 = 2

The roots in an equation have to return either positive _or_ negative values _only_, so that you get a true statement in either case. This redefinition directly influences odd roots. They are now defined for negative real values _only_, not giving well-defined results for positive real numbers.

Note that you would also have to redefine the (complex) logarithm. In case you chose negative answers it is not defined on the positive real axis.

For simplicity and universality, one always chooses the convention of positive answers. This is also the way that feels most natural to us humans.

This is, by the way, the flaw in deducing the -2 = 2 equation in 3). I used a different convention for the root function in the first step returning all negative values as opposed to the last step, in which i chose that root-functions always return positive answers in real numbers.

7)

However, for practical purposes you can define ³√(-8) to be equal to -2. This violates of course all the involved laws and mathematical definitions, some of them which i mentioned above, but, whenever you come across a problem of that form you can carefully(!) say that by ³√(-8) you mean -³√(8) as the solution to a polynomial and give the result -2 as a _practical_ result.

I don´t know where one might come across such a practical problem. I have never seen such an expression in my whole life, as far as i remember.

To emphasize this again: This is for practical purposes, the root-function is still not defined for negative values in the mathematical sense!

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Some further notices

@Randy:

When saying that

√4 = √4

leads to

±2 = ±2

and look at it in the way you did, giving 4 equations, you use different conventions of the root in the middle two, namely +2 = -2 and -2 = +2. Using different conventions in one equation is forbidden. This is also the reason for the laws of ± i gave in 1c).

Also, for what reason do you discard certain results? There is no justification to produce results and throw some of them away just because they seem illogical ;-)

I´d also like to thank you for the discussion, it was fun and fruitful :-)

@ Bhaskar:

In your deduction you used the complex(!) logarithm Ln and seperated it to two real logarithms, what cannot be done without a lot of care.

Ln(r * e^(i*phi)) = Ln(r) + Ln(e^i * phi)

r is real and positive, so the left part reduces to the real ln

= ln(r) + Ln(e^(i * phi))

the right part is defined _only_ for phi ≠ (2k+1)π with integer k. This is because of uniqueness problems, as phi = π returns the same value as phi = -π in the Exp-function. You can see the resulting singularities in this plot: http://en.wikipedia.org/wiki/File:Complex_log.jpg . The complex natural logarithm is, as well as the real one, not defined for negative numbers on the real axis. Therefore, your deduction is wrong.

My Professor asked, if, like the real logarithm, the complex logarithm is also undefined for negative values in my final math examination. I showed him the same deduction you are giving here and he started laughing (i was very nervous and couldn´t concentrate a lot). See the paragraph "Definition of principal value" in this article for a brief explanation: http://en.wikipedia.org/wiki/Complex_logarithm .

Quoting you:

"edit: if a^(p/q) is not defined, how can you guys apply the laws of exponentiation to it? I mean how can you write a^(p/q) = (a^1/q)^p = (a^p)^1/q ?"

That´s the point :-) These laws must hold for every a, for which the expression is well-defined. And by showing that these laws produce false or contradictory results for certain values of a (as for example in 3)), one shows that for these values the initial expression is not well-defined.

We say that a given number is defined when it has a unique value.

In any number of the type, x^(p/q), first the root is to be taken and then power is raised. Thus, x^(p/q) = [(x^(1/q)]^p and not [(x^p)]^(1/q). For example, (-1)^1 = -1 and so also (-1)^(2/2).

But (-1)^(2/2) = [(-1)^(1/2)]^2 = i^2 = -1 and not [(-1)^2]^(1/2) = 1.

In this sense, x^(2/3) = [(x^(1/3)]^2 and x^(1/3) will have one real and two complex roots. So in R, x^(2/3) will be defined. In Z, if principal root is accepted as the value, then even in Z it is defined.

EDIT:

In my above explanation, I have stated that

"x^(2/3) = [x^(1/3)]^2 and that x^(1/3) will have one real and two complex roots."

This is true for x > 0 and x < 0. Hence, the question asked has no special significance for x < 0.

Edit:

Leaving aside all complicated discussions, final short answer to the question

"Is x^(2/3) defined if x < 0" ? which

=> "Does x^(2/3) have a unique real value ?"

and the answer if YES.

Edit:

I had asked opinion of subject expert "Mr. Duke" for the above question. His exemplary explanation is as under.

"The definition of the real-valued power function x^n of a real variable x and n=const usually goes in the following steps (each requiring the previous to be defined previously):

1) n - natural: ordinary multiplication n times;

2) n=0: x^0=1 for every x not 0; 0^0 remains undefined;

3) n<0, integer: x^n = 1/x^(-n) for every x not 0;

4) rational n=p/q (our case) where p is integer, q - natural (i.e. if n<0 the minus sign is assigned to the numerator), and p, q are co-prime (this is very important - for example x^1.6=x^(8/5), not as x^(16/10) - or we should use the irreducible representation of n):

x^n = x^(p/q) = q_th_Root(x^p) = (q_th_Root(x))^p

Both expressions above yield the same result for co-prime p and q, as in Your answer.

5) irrational n needs involving limits or exponential and logarithmic functions (out of scope of the question).

To summarize: x^(2/3) = (x^2)^(1/3) = (x^(1/3))^2 is unambiguously defined for all real x, having a single non-negative value. This function is even, it is a standard example in Calculus books for a graph with a cusp point (horn) in the origin as a point of its global minimum; the right derivative in x=0 is infinity, the left is -infinity, i.e. with vertical tangent in the extremal point unlike the usual situation of horizontal tangent according Fermat's Theorem for differentiable functions."

Step 4) of his explanation is in conformity with Randy and proving me incorrect in my initial statements. My mistake in the given example of (-1)^(2/2) is not taking p and q as coprime.

In the final analysis, Randy deserves BA.

I agree with Jake z

a^x is defined as e^(x*ln(a)). ln(a) is not defined for negative a. so, a^x is not defined for negative a.

edit: @ Madhukar Daftary: the definition of a^x is for two numbers 'a' and 'x'. I ought to have said a^b. but it doesn't matter which one is the variable.

edit: if a^(p/q) is not defined, how can you guys apply the laws of exponentiation to it? I mean how can you write a^(p/q) = (a^1/q)^p = (a^p)^1/q ?

the definition of a^x runs like this:

first we define ln(x) as the integral of 1/t from 1 to x

then we define the exp(.) function as the inverse of the ln(.) function.

then we find that e^(lna) = a

finally, we define a^x as (e^lna)^x = e^(x lna) for any a>0 and real x.

if you can't apply the laws of exponentiation, you can't have complex roots and all those things.

its ok to think of x^(2/3) as 'a cube root of x multiplied two times'. but what if you have x^√2 ? so, its good to have a consistent definition like a^x = e^(x lna). of course, if you define x^(p/q) as 'the qth root of x raised to the power p' for integer p and q, then complex roots enter.

by the way, Happy New Year to all of you!!!

edit: ok, I got it finally. its just a matter of definition. if you think of x^(2/3) in a prosy way like 'the cube root of x multiplied two times' then its not defined because there's no single cube root of 1. you should think if x^(2/3) as e^((2/3) lnx). Madhukar Daftary pointed out to me that e^(iπ) = -1.

so let a = -c, where c is positive.

e^(b lna) = e^(b ln(ce^iπ))

= e^(b (lnc + ln(e^iπ)))

= e^(b (lnc + iπ))

= e^(b lnc) * e^(b iπ)

= e^(b lnc) (cos(bπ) + isin(bπ))

phew! we're done.

so the e^(b lna) definition covers all the reals.

really enjoyed this question.

If x<0 then x^(2/3) is NOT defined as a real number. It is a complex number.

Note that x^(2/3) = (cube root of x)^2

The square root function is NOT defined for x <0

Hence, x^(2/3) is not a real number for x <0. It is a complex number.

As an example, (-1)^(2/3) = -0.5 + 0.866025404i, which is a complex number.

obviously...

take x == any negative number....

X^2 == positive

n even x^[1/3] is defined for all x...

example -1,-8,etc..

It is NOT defined for x < 0. No one here knows what they're talking about.

If x<0, then x^(2/3) is a complex number.

Note that x^(2/3) = cube root of x squared

The square root function is only defined for numbers 0 or greater.

If x< 0, it means we're taking the square root of a -ve number, and so the answer is a complex number.

x^(2/3) is NOT defined under the real number system for x<0

x^(2/3) , with x<0 comes out as a complex number (not a real number).

Try any number x<0 using your calculator or some mathematical software, the answer is always a complex number (not a real number)

Absolutely. x^(2/3) can be written as (x^2)^(1/3) or (x^(1/3))^2.

In the first case, x^2 is always greater than 0. The cube root function is defined for all positive numbers; so the whole function is defined.

In the second case, let x = -n, where n is positive. So, x^(1/3) = (-n)^(1/3) = (-1)^(1/3) * n^(1/3)

The cube root of -1 is defined to be -1, so this becomes -n^(1/3).

As n is positive, the cube root function for it is defined. So, -n(1/3) is defined, which means that x^(1/3) is defined.

In general, any function of the form x^(a/b), where a and b are coprime integers or a/b is an irreducible fraction, is defined for all real values of x if and only if b is odd. If the denominator is even, then it will not be defined for any negative values of x. For example, the function x^(3/4) is defined only for positive values of x, while the function x^(4/5) will be defined for all values of x.

Take special note of the fact that a/b has to be an IRREDUCIBLE fraction. For example, the function x^(4/6) may seem as if it is undefined for negative x looking at the even denominator, but it can be reduced to x^(2/3), which makes it defined for all x.

Hope this helps!

@Wycabwin : Hope this answers your question too!

Addendum : Here's an extra link for another slightly more complicated question on domains of functions. The best answer explains it all:

http://answers.yahoo.com/question/index;_ylt=As4l1...

@Wycabwin : Okay, here's a way of seeing if the function is defined at a point or not. Take all the roots of the function, real or otherwise, at that point. If atleast one of the roots is real, then the function is defined at that point. If all the roots are not real, then the function is undefined at that point. I really don't see why the cube root of a negative number isn't real. If we look at it in that way, even positive numbers shouldn't have cube roots. For example, x^(1/3) at x = 1 has three roots, one real and the other two not real. However, we accept the cube root of 1 to be real. Similarly, the function also has three roots at x = -1, of which again, one is real and the other two are not. Aren't the two situations the same? So if the cube root of 1 is real, why isn't the cube root of -1 real? What makes positive numbers special?

Check the link for more details.

@sdfsdf : The function x^(1/3) returns three values when x = -1. These will be e^(iπ/3), e^(2iπ/3) and e^(iπ). These will return values of √3i/2 - 1/2, √3i/2 + 1/2 and -1. Calculators obviously cannot display all three of these at the same time, so it will display only one of them. So which one will it be? Calculators are programmed to display what is called the principal root of the function, which is the root of the function with the lowest argument, or angle. The angles are π/3, 2π/3 and π; of which π/3 is the smallest. So, the calculator will display √3i/2 - 1/2. Ask yourself this : does this necessarily mean that the other two roots are wrong?

Again, look at the link.

@Wycabwin: The flaw here is the assumption that the even root of a number is a single number. I'll elaborate on what the problem is speaking only in terms of real numbers:

A function is basically a one-to-one mapping between two sets. Now, if we consider the square root 'function', it maps each element in the domain to two elements in the range. So, strictly speaking, the square root 'function' isn't a function at all.

Now, lets take the statement (-2)^6 = 2^6. Perfectly true, unquestionable. But if we take the sixth root, we get -2 = 2, which is obviously false. So where is the flaw?

The flaw lies in the fact that all root functions give two outputs.

Lets take again the equality:

(-2)^6 = 2^6

Now, we take the sixth root of both sides. This is the crucial step. The sixth root of (-2)^6 is not -2, but is rather ±(-2). Similarly, the sixth rooth of 2 is ±2. Now we have:

±(-2) = ±2

Which gives us four cases:

-2 = 2

2 = 2

2 = -2

-2 = -2

Out of these four, the first and the fourth are correct roots and the second and third are extraneous. Discarding the extraneous roots, we have 2 =2 or -2 = -2, which is correct.

Similarly, for the cube root case, we have:

-2 = 6√64

Now, following the same logic again, 6√64 is ±2, which gives us:

-2 = ±2

which is equiavalent to:

-2 = 2

-2 = -2

of which the second is true and the first is extraneous. So, we have -2 = -2, which is true.

What I am trying to say is this : both the positive and negative outputs have to be considered when even roots are concerned; you cannot take one and disregard the other.

Now, let me finally get to cube roots. Unlike square roots, cube roots give only a single output for every number. That makes the cube root function a well defined, one-to-one function. This means that if b^3 = a, then ³√a = b always, because it is a one-to-one function. I hope I make the point very clear on this. The square root is NOT a function while the cube root is. Therefore, as (-2)^3 = -8, ³√(-8) = -2.

Again, this is only because the cube root function is one to one. Once again, I have been talking solely about real numbers in this entire argument and haven't introduced complex numbers at all, so all the arguments are compatible and correct with real numbers.

@sdfsdf : Ahem, we are taking about cube roots, not square roots...

Edit : This question gets a star for giving my brain a workout, :)

@Wycabwin: Quoting you:

"Even if you choose the definition √(x^2) = +/- x, your calculation would only yield 2 equations (being -2 = 2 and 2 = -2). You cannot cross the results and pick the ones that seem the most logic to you ;-)"

What do you mean 'crossing the results'? I'm not doing anything of that sort. Look at the steps involved again:

-2 = 6√64

As 6√64 = ±2:

-2 = ±2

which gives

-2 = 2

-2 = -2

The first equation is extraneous and invalid, and this leaves us with the second equation, which is true.

As for the ± thing, finding the value of √4 is the same as finding the roots of the equation x = √4. This will have the roots x = ±2, and so by the transitive property of equality √4 = ±2. Note that the ± means EITHER + OR -, and not both simultaneously.

@Wycabwin: Okay, I'm gonna try your approach for a while. Lets take for example the equation of the unit circle:

x^2 + y^2 = 1

Does the point (1/√2, -1/√2) satisfy this? Well, yes, because (1/√2)^2 + (-1/√2)^2 = 1. Now, I'm going to manipulate the function a bit. Pay close heed:

y^2 = 1 - x^2

y = √(1 - x^2)

Have I changed the function in any way? So, the point (1/√2, -1/√2) should still satisfy it, right? Therefore, the following equality should hold true:

-1/√2 = √(1 - (1/√2)^2)

= √(1 - 1/2)

= √(1/2)

Now, according to your definition of square roots, √(1/2) = 1/√2 only. Then, we have:

-1/√2 = 1/√2

which is absurd. The flaw? We failed to consider the negative root. If we did that too, we would have:

-1/√2 = ±1/√2

which gives:

-1/√2 = 1/√2

-1/√2 = -1/√2

You know the drill; the first root is extraneous, so we are left with the second root, which leaves a true equation.

@Wycabwin.... again, :D

First of all, Happy New Year! Short year, good times, a lot of math!

Anyway, back to the topic. You ask about the crossing? As I have mentioned before, a = ±b does not mean that a = b and a = -b simultaneously, that is obviously not true. It means that EITHER a = b OR a = -b. Let me explain why it works this way. When I say that √4 = ±2, it doesn't mean that √4 is both 2 and -2 at the same time; it will be only one of them. Which one it is will be decided by the context and nature of the problem, while the other will be extraneous. Similarly, if there are n roots, then all the n must be considered.

Why do you say that if ±a = ±b, then a = b or -a = -b? It is definitely untrue to assume that we must choose the first sign with the first and the second with the second. What we are doing is performing a function on both sides of a function and equating the outputs. Its like saying that a set {a, -a} is equal to a set {b, -b}. The set is not ordered, so we can't only match corresponding elements. It is necessary to find all the possible maps between the two sets, which results in the four equations a = b, a = -b, -a = b and -a = -b. Maybe my convention is different from the one you are using, I don't know. Anyway, there is no 'order' in the outputs of the square root function, so we can equate them in any way we want.

Anyway, I sort of find the notation ±√n illogical, because √n already returns two values. However, the significance of using this notation for showing the roots of polynomials is different; it is to emphasize the fact that there are two roots. For example, saying that x = ±√2 makes it clear that there are two roots, rather than saying x = √2.

As for the unit circle: I am saying exactly what you are saying. I was confused when you said n√((x)^n) = |x|; you clarified it later by talking about polynomials. No opposition there.

Basically, the spirit of my argument is that I don't see how we can give one root precedence over another, I feel that all roots must be treated equally (wow, that makes me sound like a feminist, XD). I really don't see the point in removing the information that a function has and adding it back later (i.e., saying ±√2; making √2 positive and then adding the negative root too, when it already includes both positive and negative roots. So, why throw away information that you already have?

Okay, my answer reached its character limit. What do I do?

yes it is defined . why not ?

whats the problem?

if x = -8 you get 4

its simple

x^(a/b) has a real answer (and b-1 other complex answers) whenever b is odd. Take x^(1/b), which is always real for b odd, then raise it to the "a" power.

-1000^(2/3)=1000000^1/3=100,yes

God bless you.