more on rational exponents

[richardt]

Greetings:

Okay, I really need to put this topic to rest but, there are still a few things that I have trouble with.

1) Do we, or do we not define a^m/n as (a^1/n)^m for all real numbers, a, and natural m, n ?

It seems that we have ambiguity without such definition.

Example: In evaluating (-1)^3/2 via that definition, we have ((-1)^1/2)³ = (i)³ = -i.

By contrast, if we cube first, we have, [(-1)³]^1/2 = [-1]^1/2 = i.

This is problematic (for me) as, supposing all above is correct, we have (-1)^3/2 = -i not= [(-1)³]^1/2, which flies in the face of (a^m)ⁿ = a^mn ; a, m, n real.

Or, does that last property only hold for a^mn real ?

Or, is my problem in thinking that (-1)^m/n is well defined for n even and m, n relatively prime?

That is, is there such a thing as a principle solution of x^n/m = -1 [i.e., if x = (-1)^m/n, then x^n/m = -1]

I am inclined to go with (-1)^3/2 = -i per the proposed definition for two additional reasons as follows:

i) (-1)^3/2 = (-1)^{1 + 1/2} = -1 * (-1)^1/2 = -i.

ii) Solving via De Moivre gives the arguments, 3/2(pi) and 3/2(3 pi), the least being 3pi/2 which corresponds with -i.

And that leads to my final inquiry which goes back to the topic of principle roots. Given that the n-roots of a real number, r, have value |r|e^{i(theta + 2k*pi)/n} for appropriate theta, does the principle root necessarily correspond with k = 0 ?

Does this have a corollary for rational exponents?

Thanks for listening.

Any any all guidance is accepted gratefully.

Rich

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[pka]

1) Do we, or do we not define a^m/n as (a^1/n)^m for all real numbers, a, and natural m, n ?
It seems that we have ambiguity without such definition.

Example: In evaluating (-1)^3/2 via that definition, we have ((-1)^1/2)³ = (i)³ = -i.
By contrast, if we cube first, we have, [(-1)³]^1/2 = [-1]^1/2 = i.

The answer to that is NO!
It is true only (m/n) is in reduced form, and if n is even then a>0.
Therefore, that 'rule' does not apply to complex numbers such as i.

Although the complex numbers include all real numbers, not all complex operations are consistent the corresponding real operation. So when you are thinking about rules for real operations do not bring complex, non-real, numbers into the discussion. As you can see above, it does not work.

 

[Dale10101]

Salute

PKA set you straight, I am just saying that I benefited from your question, and salute the depth of your pursuit ... impressive, keep on keeping on.

 

[richardt]

Two things...

1) Salute: Thank you for your kind words.

2) PKA: Thank you for your response. ...Just a couple of questions for clarification:

a) So what is the value of (-1)^3/2 ? How would you assess this expression?

b) How does this sound for a general proposition ?

Let r be real and m, n relatively prime integers. Then (r^1/n)^m = (r^m)^1/n = r^m/n if and only if r >= 0 or n is odd.

c) Will you please set me straight on the definition of a principle root ?

Thanks, in advance.

Rich

 

[pka]

Just a couple of questions for clarification:
a) So what is the value of (-1)^3/2 ? How would you assess this expression?

In the real number system that is simply undefined.

In the complex number system it is \(\displaystyle \exp \left( {\dfrac{{3\pi i}}{2}} \right) = \cos \left( {\dfrac{{3\pi }}{2}} \right) + i\sin \left( {\dfrac{{3\pi }}{2}} \right)\)

b) How does this sound for a general proposition ?
Let r be real and m, n relatively prime integers. Then (r^1/n)^m = (r^m)^1/n = r^m/n if and only if r >= 0 or n is odd.

That works in the real numbers.

c) Will you please set me straight on the definition of a principle root ?

In the real number system \(\displaystyle \displaystyle\sqrt[n]r\) is the principle nth root if n is odd.

In the real number system \(\displaystyle \displaystyle\sqrt[n]r\) is the principle nth root if n is even & \(\displaystyle r\ge 0~.\)

 

[richardt]

Thank you, PKA. I believe I've got it....for the most part. Now, I really don't wish to beat this to death but, just a bit more if you please. We have discussed the idea of a principal root as pertains to those expressions that map from the reals to the reals - and I'm okay with that now, thanks to you. To expand upon that, however,

In the complex numbers, is there such a thing as a principal root? I have two cases of interest.

1) r^1/n: mapping real to complex: Example: (-1)^1/2 = i. What is it that dictates the solution, i, over its counterpart -i ? Is i taken to be a principal root ? And, by the same token, why, in evaluating (-1)^3/2, do we choose the argument (3/2)pi over the alternative, (3/2)*3pi = 9pi/2 ? Are the resulting roots considered principal ?

2) z^1/n : mapping complex to complex: Does the idea of a principal root apply there ? For instance, consider the set of 3^rd roots of 3 + 4i. Assuming my work is correct, that set is given by {5^1/3[cos(u_k) + i sin(u_k)] : u_k = (t + 2k*pi)/3 ; t = cos^-1(3/5)} k = 0, 1, 2. Is one of these roots to be dubbed a principal root ? I have read various internet articles that label as principle, that root which corresponds with k = 0. Is there any truth to this?

Okay, I think that should do it for me. Again, pka, thanks for bearing with me on this issue.

Regards,

Rich

 

[pka]

In the complex numbers, is there such a thing as a principal root? I have two cases of interest.
1) r^1/n: mapping real to complex: Example: (-1)^1/2 = i. What is it that dictates the solution, i, over its counterpart -i ? Is i taken to be a principal root ? And, by the same token, why, in evaluating (-1)^3/2, do we choose the argument (3/2)pi over the alternative, (3/2)*3pi = 9pi/2 ? Are the resulting roots considered principal ?

First, there is no universal agreement on this topic.
But although I am old enough to be their father, I do agree with the younger complex analysis on several points.
First definition: i is a symbol that is the solution to the equation \(\displaystyle {x^2} + 1 = 0\).
It is not defined as \(\displaystyle \sqrt{-1}\).

Second, if \(\displaystyle z\) is a non-real complex number then \(\displaystyle \sqrt{z}\) has no meaning.

Now define what is meant as the principal argument of a complex number:
Suppose that \(\displaystyle x\cdot y\ne 0 \) then
\(\displaystyle Arg(x + yi) = \left\{ {\begin{array}{{ll}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \),
please note the capital A .

Now note that using standard notation: \(\displaystyle \exp(i\theta)=\cos(\theta)+i\sin(\theta)\), we can define what is meant by the principal nth root.

If \(\displaystyle \theta=\text{Arg}(z)\) then the the principal nth root of \(\displaystyle z\) is is ​\(\displaystyle \sqrt[n]{|z|}\exp \left( {\dfrac{{\theta i}}{n}} \right)\)

The other nth roots are \(\displaystyle \sqrt[n]{|z|}\exp \left( {\dfrac{{k\pi+\theta i}}{n}} \right),~k=1,2,\cdots,n\)

 

[daon2]

Thank you, PKA. I believe I've got it....for the most part. Now, I really don't wish to beat this to death but, just a bit more if you please. We have discussed the idea of a principal root as pertains to those expressions that map from the reals to the reals - and I'm okay with that now, thanks to you. To expand upon that, however,

In the complex numbers, is there such a thing as a principal root? I have two cases of interest.

1) r^1/n: mapping real to complex: Example: (-1)^1/2 = i. What is it that dictates the solution, i, over its counterpart -i ? Is i taken to be a principal root ? And, by the same token, why, in evaluating (-1)^3/2, do we choose the argument (3/2)pi over the alternative, (3/2)*3pi = 9pi/2 ? Are the resulting roots considered principal ?

2) z^1/n : mapping complex to complex: Does the idea of a principal root apply there ? For instance, consider the set of 3^rd roots of 3 + 4i. Assuming my work is correct, that set is given by {5^1/3[cos(u_k) + i sin(u_k)] : u_k = (t + 2k*pi)/3 ; t = cos^-1(3/5)} k = 0, 1, 2. Is one of these roots to be dubbed a principal root ? I have read various internet articles that label as principle, that root which corresponds with k = 0. Is there any truth to this?

Okay, I think that should do it for me. Again, pka, thanks for bearing with me on this issue.

Regards,

Rich

The principle solution to \(\displaystyle z^n=w\), \(\displaystyle w\in \mathbb{C}\) is the complex number \(\displaystyle z=re^{i\theta / n}\) where \(\displaystyle r=\sqrt[n]{|w|}\) and \(\displaystyle -\pi < \theta=\text{Arg(w)} \le \pi\)

Solving \(\displaystyle z^2=-1\) yields \(\displaystyle \theta=\pi, r=1\), and so if it is agreed on beforehand that \(\displaystyle w^{1/2}\) means the principle solution to \(\displaystyle z^2=w\), then you may write \(\displaystyle z = (-1)^{1/2} = e^{i\pi/2}\), which is of course \(\displaystyle i\).

 

[richardt]

Dear PKA and Daon2:

Thank you so much for your patience and thorough explanations. While it took two days to sink in, I do believe I've got it.

Best Regards,

Rich