# Why use radical notation instead of rational exponents?

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 ?

• How do you define the meaning of $x^{1/n}$ in the first place, if not as the $n$th root of $x$? Oct 16, 2012 at 20:06
• @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. Oct 16, 2012 at 20:09
• @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. Oct 16, 2012 at 21:36
• ... 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. Oct 17, 2012 at 10:59
• $1 = 1^{\frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = (-1)^{2\times \frac{1}{2}} = (-1)^1 = -1$ And this is why they are introduced with the radical. Oct 17, 2012 at 12:12

## 2 Answers

(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.

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.