Bounds for the index of a radical

[FranzDiCoccio]

Hi,
I know that, at least formally, the index of a radical should be positive and integer. That is if I introduce
\sqrt[x]{2}
I need to assume x\in \mathbb N and x>0.

However, my calculator has no problem in calculating the radical for any x\neq 0, say x=-\pi.
The result it gives is based on the assumption
\sqrt[x]{2} = 2^{\frac{1}{x}}
and on the fact that the exponent of the exponential function can be any number. Therefore
\sqrt[-\pi]{2} = 2^{-\frac{1}{\pi}}\approx 0.802

It seems to me that the above "assumption" that a radical can be replaced by an exponential holds true in any case, provided that the radicand is positive.

I wonder whether I'm overlooking some strange case where the above assumption fails.
I do not see any, so I guess that the bounds x\in \mathbb N and x>0. are just formal.
I think that the idea is that it is not really worth bothering with weird indexes in radicals.
Those really interested in \sqrt[x]{2} for any x, should just stop using radicals and work with exponentials only.

Is that it, or is there more to it?
Thanks a lot for your input
Franz

 

[fresh_42]

This is all true for the real numbers. But both notations: $\sqrt[p]{r}$ and $r^{\frac{1}{p}}$ only point towards one real solution of $x^p=r\; , \;p\in \mathbb{N}$ and neglect all other (i.g. complex) solutions, resp. $p\cdot \log(x)=\log(r)\; , \;p\in \mathbb{R}$. However, one does not write $\sqrt[-p]{r}$ for $r^{-\frac{1}{p}}$. In this case it is common and less confusing to choose $\sqrt[p]{\frac{1}{r}}$ or $(\frac{1}{r})^{\frac{1}{p}}$.

Notation is always a convention. It's purpose is to communicate the same content.

 

[Mark44]

Hi,
I know that, at least formally, the index of a radical should be positive and integer. That is if I introduce
\sqrt[x]{2}
I need to assume x\in \mathbb N and x>0.

All of the radical expressions I've ever seen have an index that is a positive integer greater than 1, such as $\sqrt 2$ (index of 2 is implied), or $\sqrt[3] 8$, and so on.
Of course, you can convert a radical expression to one in exponental form, and then there aren't the same restrictions on what the exponent can be.
So while I've never seen something like this: $\sqrt[\pi] x$, there's no reason you can't write it as $x^{1/\pi}$.

 

[FranzDiCoccio]

This is all true for the real numbers. But both notations: $\sqrt[p]{r}$ and $r^{\frac{1}{p}}$ only point towards one real solution of $x^p=r\; , \;p\in \mathbb{N}$ and neglect all other (i.g. complex) solutions, resp. $p\cdot \log(x)=\log(r)\; , \;p\in \mathbb{R}$.

Ok... right... My "bound" r\geq0 implicitly limits all this to real solutions. What you're saying is that p\in \mathbb{N} allows for p-1 further complex solutions (if you allow them).

However, one does not write $\sqrt[-p]{r}$ for $r^{-\frac{1}{p}}$. In this case it is common and less confusing to choose $\sqrt[p]{\frac{1}{r}}$ or $(\frac{1}{r})^{\frac{1}{p}}$.

Notation is always a convention. It's purpose is to communicate the same content.

Ok, so you sort of agree with me. It's just a matter of notational conventions. Using non positive-integer indexes for radicals is in principle allowed, but it is uselessly complicated, so it is basically never used.
The usual bonds are given in order to avoid a uselessly complicated notation, which might cause unneeded confusion and undue awe.

I remember that as a student I used to feel a little uneasy when I looked at math exercises containing radicals, no matter how simple.
When I realized that a radical is basically just another way of writing a power (or an exponential), I felt much less "threatened".
I think that this is because a "square root" is sort of universally recognized as "mathematically difficult stuff" by the layman.I sometimes think that radicals should be dropped altogether, and replaced by powers... on the other hand, \sqrt[n]{x} with odd n is defined for x\in \mathbb{R}, whereas x^{1/n} only for x\geq 0.

Sorry, I'm always referring to real numbers because that is what is mostly taught in Italian high-school. Complex numbers are taught as well, but not very in depth.

 

[FranzDiCoccio]

All of the radical expressions I've ever seen have an index that is a positive integer greater than 1, such as $\sqrt 2$ (index of 2 is implied), or $\sqrt[3] 8$, and so on.
Of course, you can convert a radical expression to one in exponental form, and then there aren't the same restrictions on what the exponent can be.
So while I've never seen something like this: $\sqrt[\pi] x$, there's no reason you can't write it as $x^{1/\pi}$.

Yes, I agree with what you say. My question was basically inspired by the $\sqrt[y]{x}$ key in a pocket calculator. The calculator won't complain if $y\not\in \mathbb{N}$, provided that $x\geq 0$.

 

[fresh_42]

I sometimes think that radicals should be dropped altogether, and replaced by powers... on the other hand, $\sqrt[n]{x}$ with odd $n$ is defined for $x\in \mathbb{R}$, whereas $x^{1/n}$ only for $x\geq 0$.

No. They both mean the same and whether they are defined for $x\in \mathbb{R}$ or for $x\geq 0$ only depends on the context, i.e. real or complex. You are right in so far, as the root notation should be avoided, and is. As @Mark44 has said, it is normally only used in case of square or cubic roots. But even these are often written as powers. If your calculator has two buttons: $\sqrt[y]{x}$ and $x^y$ then it is a redundancy. The latter will do, esp. as there is probably also the button $1/x$ available.

 

[FranzDiCoccio]

No. They both mean the same and whether they are defined for $x\in \mathbb{R}$ or for $x\geq 0$ only depends on the context, i.e. real or complex.

Sorry, that is what I meant to say with my last remark. In high school, context is usually $\mathbb{R}$, which introduces the constraint $x\geq 0$ (otherwise you might end up proving that $2=-2$).

You are right in so far, as the root notation should be avoided, and is. As @Mark44 has said, it is normally only used in case of square or cubic roots. But even these are often written as powers. If your calculator has two buttons: $\sqrt[y]{x}$ and $x^y$ then it is a redundancy. The latter will do, esp. as there is probably also the button $1/x$ available.

I completely agree. If one needs to choose, $\sqrt[y]{x}$ should go. It is precisely this redundancy that caused my question.
Thanks a lot for your help!

 

[Svein]

If you think that is weird, take a look at the complex version: y=z^{\alpha}, where both z and α are complex. You start out by observing that obviously \log(y)=\alpha\cdot \log(z) which looks correct as long as z stays away from 0. But - not so fast - \log(z) has an infinite amount of values differing by some integer multiply of 2πi: \log(z)=\log(z_{0})+n\cdot 2\pi i, so \log(y)=\alpha\log(z_{0})+\alpha \cdot n\cdot 2\pi i. Taking the exponential on both sides gives y=e^{(\alpha\log(z_{0})+\alpha \cdot n\cdot 2\pi i)}=e^{\alpha\log(z_{0})}\cdot e^{\alpha \cdot n\cdot 2\pi i}. The first factor is trivial, but the second factor can have a finite amount of values (if α is rational) or an infinite amount of values.