Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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 ?

share|improve this question
How do you define the meaning of $x^{1/n}$ in the first place, if not as the $n$th root of $x$? – Henning Makholm Oct 16 '12 at 20:06
3  
@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. – Henning Makholm Oct 16 '12 at 20:14
1  
@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
2  
$1 = 1^{\frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = (-1)^{2\times \frac{1}{2}} = (-1)^1 = -1$ And this is why they are introduced with the radical. – xavierm02 Oct 17 '12 at 12:12
show 18 more comments

1 Answer

up vote 3 down vote accepted

(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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.