Odd/even functions and fractional indices

[dyn]

Hi.
I would like to check that my understanding is correct. For $f(x)=x^{1/n}$ where n is an integer. If n is odd then f(x) is an odd function while if n is even then f(x) is neither odd or even as it involves the square root function which is only defined for non-negative x.
For $f(x) = x^{m/n}$ where n and m are integers then the above rule applies and if m is even then f(x) is even and m odd gives f(x) odd.
Examples $f(x) = x^{1/4}$ and $f(x) = x^{3/2}$ are neither odd or even as they are only defined for non-negative x and $f(x) = x^{2/5}$ is even and $f(x) = x^{3/5}$ is odd.
Have i got all this right ?
Thanks

 

[mfb]

How do you define any of these for negative x? There are ways to define x^c for all complex x and c but that won't work nicely in the real numbers, and the result will not be an odd or even function.

 

[Stephen Tashi]

Hi.
I would like to check that my understanding is correct. For $f(x)=x^{1/n}$ where n is an integer. If n is odd then f(x) is an odd function while if n is even then f(x) is neither odd or even as it involves the square root function which is only defined for non-negative x.

It's hard to formulate a systematic and rigorous treatment of fractional exponents when restricted to the real number system.

In the complex number system, the notation $x^{1/n}$ is ambigous. A non-zero complex number has n distinct n-th roots. (For example, 1 has 3 distinct cube roots.) So the notation $f(x) = x^{1/n}$ does not define a function.

Restricting ourselves to the real number system, a high school algebra course might expect students to evaluate $(-8)^{2/3}$ as $)(-8)^{1/3})^2 =(-2)^2 = 4$. With such a convention, the odd roots of numbers are unique. However we have the problem that $(-8)^{4/6}$ is not defined for $x < 0$ but after we reduce the fraction 4/6 to 2/3, we get something that is defined.

The rule $(x^a)^b = x^{ab}$ does always work when $a$ and $b$ are not integers.
[The above sentence should be corrected to say "does not always work"]
For example , let $x = -1,\ a = 2/3,\ b = 3/2$. With the above convention $((-1)^{2/3}))^{3/2} = (1)^{3/2} = 1$ But $(-1)^{(2/3)(3/2)} = (-1)^1 = -1$.

The high school algebra texts of my youth did not present an exhaustive treatment of fractional exponents. The authors knew that fraction exponents are a minefield when restricted to the real number system. The implied advice was "Don't waste your time pondering fractional exponents until you study complex numbers". Perhaps that's good advice!

 

[dyn]

In my original question I was only referring to real numbers . That's why I stated $x^{1/2}$ is neither odd or even because it only exists for non-negative numbers while the graph of $x^{1/3}$ shows that it is an odd function

 

[Stephen Tashi]

where n and m are integers then the above rule applies and if m is even then f(x) is even and m odd gives f(x) odd.

The set of "integers" includes the negative integers.

 

[Stephen Tashi]

Have i got all this right ?

Yes, in the sense that people can find a way of interpreting your notation and reach the same conclusions. What is lacking is your own explanation of your notation.

In normal mathematical notation, if $a = b$ and $x^a$ exists then $x^b$ exists and $x^a = x^b$. However, apparently in you notation $4/6 = 2/3$ and $(-8)^{2/3}$ exists, but $(-8)^{4/6}$ does not. So your conclusions cannot be explained by properties of numbers alone. Your notation indicates particular algorithms. To deduce your conclusions we must define how notation is interpreted as algorithms. Someone who wished to define $x^{m/n}$ by an algorithm that reduced $m/n$ to its "lowest terms" before proceeding would disagree with your conclusions.

 

[Quasimodo]

The rule $(x^a)^b = x^{ab}$ does always work when $a$ and $b$ are not integers.
[The above sentence should be corrected to say "does not always work"]
For example , let $x = -1,\ a = 2/3,\ b = 3/2$. With the above convention $((-1)^{2/3}))^{3/2} = (1)^{3/2} = 1$ But $(-1)^{(2/3)(3/2)} = (-1)^1 = -1$.

1^(3/2)=-1, as well!

 

[mfb]

In my original question I was only referring to real numbers . That's why I stated $x^{1/2}$ is neither odd or even because it only exists for non-negative numbers while the graph of $x^{1/3}$ shows that it is an odd function

$x^{1/3}$ doesn't have an unambiguous value for negative x. You can define it to be $-(-x)^{1/3}$ but that is not the only option, and it leads to trouble when you want to look at e.g. $x^{2/6}$. Which is the same, as 1/3 = 2/6, but now you could evaluate it completely differently.

 

[dyn]

The rule $(x^a)^b = x^{ab}$ does always work when $a$ and $b$ are not integers.
For example , let $x = -1,\ a = 2/3,\ b = 3/2$. With the above convention $((-1)^{2/3}))^{3/2} = (1)^{3/2} = 1$ But $(-1)^{(2/3)(3/2)} = (-1)^1 = -1$.

Do you mean the rule $(x^a)^b = x^{ab}$does NOT always work when $a$ and $b$ are not integers ? I can only find that $(x^a)^b = x^{ab}$ without much reference to what $a$ and $b$ are

 

[Stephen Tashi]

Do you mean the rule $(x^a)^b = x^{ab}$does NOT always work when $a$ and $b$ are not integers ?

Yes, that's what I meant.

I can only find that $(x^a)^b = x^{ab}$ without much reference to what $a$ and $b$ are

There are many souces on the web state "the laws of exponents" without stating the proper restrictions that must be applied to the laws. For example, the Wolfram page http://mathworld.wolfram.com/ExponentLaws.html hints that the formula on it apply only to integral exponents by using $m$ and $n$ to denote the exponents, but it doesn't state the restriction explicitly.

 

[dyn]

So if I encounter $(x^{m/n})^{p/q}$ in general terms without knowing $x,m.n,p,q$ what do I do ? Is it only defined for certain $x,m,n,p,q$ ?
In the past I'm sure I used $(x^{m/n})^{p/q} = x^{(mp)/(nq)}$ and it always seemed to work

 

[Stephen Tashi]

So if I encounter $(x^{m/n})^{p/q}$ in general terms without knowing $x,m.n,p,q$ what do I do ? Is it only defined for certain $x,m,n,p,q$ ?
In the past I'm sure I used $(x^{m/n})^{p/q} = x^{(mp)/(nq)}$ and it always seemed to work

My rule of thumb is that if $x \ge 0$ the laws of exponents for integer exponents provide patterns that work for rational exponents.

There might be some expert who knows a set of rules for manipulating fractional exponents that works automatically when the base is negative. That expert isn't me! If I encountered $((-8)^{m/n})^{p/q}$ in appyling math to something, I'd think about what the expression means in the particular application and ask myself if writing it as $(-8)^{(mp)/(nq)}$ meant the same thing.

 

[Mark44]

1^(3/2)=-1, as well!

No.
Possibly you are thinking of this as $(1^{1/2})^3$, and mistakenly thinking that $1^{1/2} = \pm 1$.
$1^{1/2} = \sqrt 1$ which is unambiguously 1.

 

[Quasimodo]

No.
Possibly you are thinking of this as $(1^{1/2})^3$, and mistakenly thinking that $1^{1/2} = \pm 1$.
$1^{1/2} = \sqrt 1$ which is unambiguously 1.

No, I am thinking 1^(1/2)=x, implying x^2=1, etc.

 

[Mark44]

No, I am thinking 1^(1/2)=x, implying x^2=1, etc.

But the equations $x = 1^{1/2} = \sqrt 1$ and $x^2 = 1$ are not equivalent. In the first equation, x = 1 only, while in the second equation, $x = \pm 1$.

This is an error a lot of people make, thinking that, for example, $\sqrt 1 = \pm 1$.

 

[SammyS]

Added in Edit:

Never mind.

 

[dyn]

This is an error a lot of people make, thinking that, for example, $\sqrt 1 = \pm 1$.

Surely the square root of +1 is +1 or -1 as both values squared give 1 ? Unless you are taking the square root sign to mean taking only the positive value ?

 

[Mark44]

Surely the square root of +1 is +1 or -1 as both values squared give 1 ?

No. When we say "square root of a number," what we mean is the principal square root of that number.

Unless you are taking the square root sign to mean taking only the positive value ?

Yes, the principal square root is the positive square root. A given positive number n will have two square roots. The notation $\sqrt n$ is the positive number a such that $a^2 = n$.

 

[dyn]

Hi.
I would like to check that my understanding is correct. For $f(x)=x^{1/n}$ where n is an integer. If n is odd then f(x) is an odd function while if n is even then f(x) is neither odd or even as it involves the square root function which is only defined for non-negative x.
For $f(x) = x^{m/n}$ where n and m are integers then the above rule applies and if m is even then f(x) is even and m odd gives f(x) odd.
Examples $f(x) = x^{1/4}$ and $f(x) = x^{3/2}$ are neither odd or even as they are only defined for non-negative x and $f(x) = x^{2/5}$ is even and $f(x) = x^{3/5}$ is odd.
Have i got all this right ?
Thanks

How do you define any of these for negative x? There are ways to define x^c for all complex x and c but that won't work nicely in the real numbers, and the result will not be an odd or even function.

$x^{1/3}$ doesn't have an unambiguous value for negative x. You can define it to be $-(-x)^{1/3}$ but that is not the only option, and it leads to trouble when you want to look at e.g. $x^{2/6}$. Which is the same, as 1/3 = 2/6, but now you could evaluate it completely differently.

I am using Thomas' Calculus and it shows $x^{1/2}$ and $x^{3/2}$ as only existing for non-negative $x$ so they are neither odd or even functions. It also shows graphs showing that $x^{1/3}$ is an odd function and $x^{2/3}$ is an even function. It also states that the cube root function is defined for all real $x$ and that $x^{2/3} = (x^{1/3})^2$

 

[Stephen Tashi]

It looks like you have a typo there.

I'm rather sure you meant that to say:

The rule $(x^a)^b = x^{ab}$ does not always work when $a$ and $b$ are not integers.

That was acknowledged in post #10

 

[Quasimodo]

No. When we say "square root of a number," what we mean is the principal square root of that number.

NO! We certainly don't mean that!
Why don't you just admit that Stephen Tashi has made a mistake and come to his defence with such nonsense? The n-th roots of unity are known since 1600's.

 

[Mark44]

No. When we say "square root of a number," what we mean is the principal square root of that number.

NO! We certainly don't mean that!

You took me out of context here. My reply, quoted above, was a response to what dyn wrote in post #17, where he said that $\sqrt 1 = \pm 1$.

Why don't you just admit that Stephen Tashi has made a mistake and come to his defence with such nonsense? The n-th roots of unity are known since 1600's.

Sure, but that's not what is being discussed here. The notation $\sqrt n$ or $\sqrt[4] n$ signifies the principal square root or fourth root of a nonnegative real number n. The fact that n has two square roots and four fourth roots is not relevant in this discussion.

 

[Quasimodo]

Sure, but that's not what is being discussed here. The notation √nn\sqrt n or 4√nn4\sqrt[4] n signifies the principal square root or fourth root of a nonnegative real number n. The fact that n has two square roots and four fourth roots is not relevant in this discussion.

If it is a matter of notation, refer me where it is written that $1^{1/2} =1$ as Mr. Tashi claims. (Because, I do not see any square root symbol...

 

[Mark44]

If it is a matter of notation, refer me where it is written that $1^{1/2} =1$ as Mr. Tashi claims. (Because, I do not see any square root symbol...

Check any precalculus textbook and it will show that $x^{1/2}$ and $\sqrt x$, with $x \ge 0$ are different notations that mean the same thing

 

[dyn]

I am using Thomas' Calculus and it shows $x^{1/2}$ and $x^{3/2}$ as only existing for non-negative $x$ so they are neither odd or even functions. It also shows graphs showing that $x^{1/3}$ is an odd function and $x^{2/3}$ is an even function. It also states that the cube root function is defined for all real $x$ and that $x^{2/3} = (x^{1/3})^2$

Any thoughts on the above found in Thomas' Calculus ? If $x^{1/3}$ is defined for all real $x$ then surely $x^{2/6}$ must be equivalent to it as 2/6=1/3 or we are in serious trouble !

$x^{1/3}$ doesn't have an unambiguous value for negative x. You can define it to be $-(-x)^{1/3}$ but that is not the only option, and it leads to trouble when you want to look at e.g. $x^{2/6}$. Which is the same, as 1/3 = 2/6, but now you could evaluate it completely differently.

 

[Mark44]

$x^{1/3} = x^{2/6}$, obviously. The problem comes with the latter expression if you try to evaluate it as $\left( x^{1/6}\right)^2$ for negative x. What this shows, in my view, is that the usual rules for exponents don't carry through when you raise negative numbers to some fractional power.

The rule that you cited earlier, $x^{ab} = (x^a)^b$ doesn't apply in general if x < 0.

 

[dyn]

It seems that fractional powers of negative numbers is a minefield ! I have seen no mention of it any books. Stephen Tashi asked a question looking for a book covering it in the Textbook forum. I'm amazed that it is not covered anywhere.

 

[Stephen Tashi]

Any thoughts on the above found in Thomas' Calculus ? If $x^{1/3}$ is defined for all real $x$ then surely $x^{2/6}$ must be equivalent to it as 2/6=1/3 or we are in serious trouble !

Math education is indeed in serious trouble when it tries to sort out exponentiation.

If we define $x^{2/6}$ to be the square of the sixth root of $x$ then, in the real number system, $(-8)^{2/6}$ is undefined. This agrees with the statements you made in your original post. However, it has the awkware consequence that for a rational number $q$, the notation "$x^q$" is not uniquely defined because the above definition depends on how we choose to write $q$ as a fraction.

If we consider complex numbers then $-8$ has a principal sixth root and we may square it and, if all goes well, get an answer of -2. However, according to https://en.wikipedia.org/wiki/Nth_root the principal cube root of $-8$ , as usually defined in complex analysis, is $1 + i \sqrt{3}$ so applying this definition to $(-8)^{2/3}$ we get $(-8)^{2/3} = (1 + i\sqrt{3})^2 = -2 + 2i\sqrt{3}$

In the real number system, one way to make "$x^q$" unambiguous is to declare that for a rational number $q$ and a real number $x$, the notation $x^q$ is to be interpreted by expressing $q$ as a fraction in "lowest terms" as $m/n$ and then, if possible, taking the $n$th principal root of $x$ in the real numbers system and raising it to the $m$th power.

With that convention $( (-8)^{1/6})^2$ does not exist in the real number system since it requires finding the sixth root of (-8) However $(-8)^{2/6}$ does exist. So we still have trouble with $(x^a)^b = x^{ab}$.

 

[dyn]

This has been a very illuminating thread but what amazes me is that this issue seems to be never mentioned in calculus of real numbers !