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.