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I'm helping my younger sister for her math class. She has recently been taught integer exponents, and has starteed studying radicals (mainly square roots). The next topic will be rational exponents, which she has already read a bit up on.

It's been a long time since I've learnt all this and it has become second nature to me. In particular I've never been fond of radical notation and often end up writing $(expr)^\frac{1}{2}$ instead of $\sqrt{expr}$. This means that while I can give her a good amount of help, I am not always able to justify why things work how they do and what's the motivation behind what she's taught.

Right now she understands integer exponents well; adding, multiplying them, even when rational bases are involved. Roots however confuse her, which means that while she is able to solve e.g. $y^{5/2} (\frac{x}{2})^2 \frac{x}{x^{1/2} y^{7/4}} = \frac{1}{4} x^{5/2} y^{3/4}$, the same thing written with roots $\sqrt{y^5} (\frac{x}{2})^2 \frac{x}{\sqrt{x} \sqrt[4]{y^7}}$ is unclear to her. She now has to relearn every formula she knows like $x^n y^n = (xy)^n$ written with roots $\sqrt{x}\sqrt{y} = \sqrt{xy}$.

I've explained to her that roots and rational exponents are roughly the same and how to rewrite roots as rational exponents ($\sqrt[q]{x^p} = x^\frac{p}{q}$). She isn't allowed to do that in class however, as learning about roots is mandatory, and I'm not actually trying to help her skip the subject.

But since the exponent notation is so much simpler to her (and to me) she asked me what was the reason to use roots in the first place, and I wasn't able to answer. So here's the question: what are the reasons to use roots ? Are there significant cases where the last equation I wrote above does not hold ? Are nth-roots and rational exponents actually different beasts ?

Antonio Vargas
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ARRG
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  • How do you define the meaning of $x^{1/n}$ in the first place, if not as the $n$th root of $x$? – hmakholm left over Monica Oct 16 '12 at 20:06
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    @Henning: It’s the same definition whether you call it $x^{1/n}$ or $\sqrt[n]{x}$. It seems to me that the question is really as much about notation as about the mathematics involved. – Brian M. Scott Oct 16 '12 at 20:09
  • @Brian: My point was more that you do need a distinct definition for this case, as well as a distinct proof that the power laws apply to that case. So its not so much about "learning about roots" that can be skipped but merely the conventional notation for them. – hmakholm left over Monica Oct 16 '12 at 20:14
  • I'm with Henning - the fact that $n$th roots are really exponentiation by $\frac{1}{n}$ is something not worth skipping. In particular, $x^n$ is always defined when $n$ is an integer, and $x^q$ is not always defined. We spend a lot of time on division as a kind of "inverse multiplication" before we start thinking of division as a multiplication by an inverse. – Thomas Andrews Oct 16 '12 at 20:15
  • @Henning: Why do you think that you need a distinct definition? Your second sentence is precisely the point that I was making. – Brian M. Scott Oct 16 '12 at 20:28
  • @Thomas: Where on earth did you get the idea that I advocated skipping the fact that $n$-th roots are exponentiation by $\frac1n$?! – Brian M. Scott Oct 16 '12 at 20:29
  • Yes to above responses (ignoring the back and forth). To ARRG I'd say it might ber historical. The symbols for radicals were presumably in use before fractional exponent versions of the same. And it seems there are lots of examples where some textbook writer prefers to use an archaic symbology rather than the latest more streamlined versions. – coffeemath Oct 16 '12 at 20:49
  • @Brian: You need a distinct definition of what $a^b$ means when $b$ is not an integer. One cannot get by with saying that we already know the power laws and then, hocus-pocus, putting a fraction in the subscript yields interesting things. The root concept needs its own definition and proofs, no matter whether one notates them $\sqrt[n]{a}$ or $a^{1/n}$, and no matter whether one hides the scary word "root" or not. What I'm arguing is that sticking to $a^{1/n}$ notation doesn't really save any work -- all of the understanding needs to be done separately for $a^{1/n}$ anyway. – hmakholm left over Monica Oct 16 '12 at 21:30
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    @Henning: This is a response to something that I never said. Of course you need a distinct definition of $a^b$ when $b$ is not an integer; so what? As I said, for the case $b=\frac1n$ it’s the same definition whether you call it the $n$-th root or the $\frac1n$-th power. – Brian M. Scott Oct 16 '12 at 21:36
  • @Brian: I think you asked me "Why do you think that you need a distinct definition". I responded with an explanation of why I think we need a distinct definition. To repeat my original point again: because this definition is not the same definition as the one we already know for integer powers, there is no real conceptual savings in calling the thing we define a "fractional power" rather than a "root". So if the OP thinks there's a conceptual saving here, it must be because they haven't noticed that calling it $a^{1/n}$ won't let them avoid the work of understanding a new definition. – hmakholm left over Monica Oct 16 '12 at 21:42
  • @Henning: But you misunderstood what I was asking. I was asking why we need distinct definitions for $x^{1/n}$ and $\sqrt[n]{x}$, which is what you appeared to be claiming. We don’t. And from the information in the question it seems likely to me that the OP’s daughter’s problem is at least as much with the notation as with the concept. – Brian M. Scott Oct 16 '12 at 22:01
  • @Brian: Of course we don't need distinct definitions for $x^{1/n}$ and $\sqrt[n]{x}$ -- they are the same thing! If you're asking why I think we do, the answser is that I don't! My point is (now repeated umpteen times) that if the OP and/or their daughter thinks $x^{1/n}$ is easier to understand than $\sqrt[n]{x}$, it must be because they fail to understand that they still have to grasp that $x^{1/n}$ needs a different definition than $x^n$ and that to understand $x^{1/n}$ properly one has to do exactly the same work as to understand $\sqrt[n]{x}$, because they are the same thing! – hmakholm left over Monica Oct 17 '12 at 10:56
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    ... The only reason I can see for thinking that $x^{1/n}$ is any easier to understand than $\sqrt[n]{x}$ is that one falsely believes that everything one knows about power laws and so forth for the integer-exponent case can be transferred formally to the fractional-exponent case and this is not true. It is a new thing that needs new definition, new proofs, and new insight, and any possible "oh, I already knows this" feeling that comes from the apparently-familiar $x^{1/n}$ notation is a false sense of safety that just deceives them into thinking there's nothing new to learn. – hmakholm left over Monica Oct 17 '12 at 10:59
  • @Henning: I think that you underestimate the difficulty notation can cause independently of the conceptual content if the notation seems awkward. There’s probably some conceptual confusion as well, but it still seems entirely possible to me that the bulk of the girl’s difficulty in dealing with the radical notation is purely notational. (As for the rest, I realized long ago that in your initial comment you weren’t claiming what you seemed to me at the time to be claiming; I just wish that you’d understood a little sooner what comment in response was saying $-$ we could have reached ... – Brian M. Scott Oct 17 '12 at 11:06
  • ... the actual point of disagreement a lot quicker!) – Brian M. Scott Oct 17 '12 at 11:06
  • @Brian: I'm afraid I still don't understand "what comment in response was saying". If you did understand what I was saying ($x^n$ and $x^{1/n}$ needs different definition), then why did you keep asking me about something I never claimed ($x^{1/n}$ and $\sqrt[n]{x}$ needs different definitions)? I have tried hard all the way with boldface and italics and everything, to explain what it is I claim. Do you still think I misunderstand anything? – hmakholm left over Monica Oct 17 '12 at 11:31
  • @Henning: "believes [...] the integer-exponent case can be transferred formally to the fractional-exponent case and this is not true.", so what is actually different ? Note that the scope of the course my sister (not daughter ;)) follows doesn't encompass proofs of theorems (or historical notions). It's really an applied course. – ARRG Oct 17 '12 at 11:39
  • @Henning: You didn’t make it clear until well into the thread, in your comment just after coffeemath’s that you were talking about $x^n$ and $x^{1/n}$, not about $x^{1/n}$ and $\sqrt[n]{x}$. Once you finally did make that clear, I realized that you had completely misunderstood my comments up to that point. I understood what you were saying (and explicitly agreed with it), but I couldn’t understand how you had managed to misunderstand my comments so thoroughly that you thought that you were arguing with me when in fact you were agreeing with my original point! Let’s just drop it. – Brian M. Scott Oct 17 '12 at 11:40
  • The main thing that bothers me is that while she knows for example that $x^a x^b = x^{a+b}$ (for integer and rational exponents), she has to learn once more that e.g. $x \sqrt{x} = \sqrt{x^3}$ which is actually the same thing. Worse, they only teach her about square roots, which means that even with the above formula, she wouldn't be able to simplify $x \sqrt[3]{x}$ either. She's doing more work, but learns to express less things... That's what I'm trying to justify the topic to her. – ARRG Oct 17 '12 at 11:43
  • @ARRG: $23^7$ is defined as $\underbrace{23\times23\times\cdots\times23}_{7\text{ times}}$. You can't use that definition to compute that $23^{1/7}$ is because "multiplying $23$ by itself one-seventh of a time" does not make sense. You can punch it into a calculator, but if you have two calculators that give different results, how would you know which of them is broken? There's a genuinely new concept to be learned when you move beyond integral exponents, no matter whether you notate the extension as $23^{1/7}$ or $\sqrt[7]{23}$. – hmakholm left over Monica Oct 17 '12 at 11:43
  • @Brian: But when I did explain (as clearly and plainly as I could) that you had misunderstood my comment, why did you say that my explanation was "a response to something that [you] never said"? I'm completely confused as to what you expect from me at this point. – hmakholm left over Monica Oct 17 '12 at 11:45
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    $1 = 1^{\frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = (-1)^{2\times \frac{1}{2}} = (-1)^1 = -1$ And this is why they are introduced with the radical. – xavierm02 Oct 17 '12 at 12:12
  • @Xaviermo2 True, but we also define $(a^b)^c=a^{bc} for positive $a$ for that reason, so we could still do without roots recalling that – Jean-Sébastien Oct 17 '12 at 12:59
  • If only there was a better notation for exponents, roots, and logs. – Zaz Feb 27 '15 at 11:41

2 Answers2

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(Sorry about the generation error!)

To attempt to answer the actual question, I think that it’s a largely two-fold consequence of the historical inertia mentioned in the comments by @coffeemath. On the one hand, it’s simple classroom inertia: it’s ‘always’ been done this way, so we do it this way. On the other hand it’s the practical consideration that since the radical notation does survive in real-world use, students need to learn how to deal with it. None of this, however, justifies the requirement that students deal with it directly, rather than by translating it into a less cumbersome, more easily manipulated notation. Indeed, in my view this is a good occasion to make the point that well-chosen notation makes our mathematical lives easier.

Brian M. Scott
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Aesthetically, I think in typesetting and writing $$\sqrt n \quad \text{and} \quad 2 \pi \sqrt{\frac l g} $$ look better than $$n^{1/2} = n^{\frac 1 2} \quad \text{and} \quad 2 \pi \left(\frac l g\right)^{1/2} = 2 \pi \left(\frac l g\right)^{\frac 1 2}$$

Especially inline when $n^{1/2}$ and $n^{\frac 1 2}$ are squished while $\sqrt n$ is easier to interpret at a glance. Compare

$$\sum_{k=1}^{\sqrt n}k \quad \text{and} \quad \lim_{k \to \sqrt n} k$$ to $$\sum_{k=1}^{n^{1/2}}k \quad \text{and} \quad \lim_{k \to n^{1/2}}k$$

This isn't a strong justification and may be due to the bias of using the squareroot notation in the first place, but it is definitely why I use it.

qwr
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