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According to my resources, $\sqrt x = x^{\frac{1}{2}}$, and $\sqrt[3] x = x^{\frac{1}{3}}$, etc, or to put it simply, $\sqrt[y] x = x^{\frac{1}{y}}$.

However, since I understand an exponent to mean the amount of times a given number (x in this example) is multiplied by itself, I'm not sure if this way of displaying exponents makes sense.

After all, if $x^3$ is shorthand for $x\times x\times x$ and $x^4$ is shorthand for $x\times x\times x\times x$, wouldn't $x^{\frac{1}{2}}$ be the same thing as $x \over 2$? (I say this because if the exponent refers to the quantity of times x is multiplied by itself, at first glance, it seems like $x^{\frac{1}{2}}$ would (maybe?) mean $x\over2$ ... ?)


On the other hand, since $x^y \times x^z = x^{y+z}$ we can see that $x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x$ which is what we would expect. If you think about it this way, it makes perfect sense (yea, it's expected) that $x^{\frac{1}{y}} = \sqrt[y] x$.


I realize that this is a rather basic question, so why would I ask it?

Well, the beauty of mathematics (versus, English's grammar rules, perhaps) is that numbers make sense! Hence, there's no reason why you shouldn't be able to understand why something works in math.
(Thank God numbers aren't arbitrary!)

Hence, I'd like to be able to understand why $\sqrt[y] x$ can be written as $x^{\frac{1}{y}}$. One more thing, according to my calculator $x^{1.5} \times x^{1.5} = x^3 = 125$, why does $x^{1.5} = x \times x^{\frac{1}{2}}$?


I guess my basic question is somewhere along the lines of: "$\sqrt[y] x = x^{1/y}$ feels arbitrary (except for the point about adding exponents made above). Why isn't it arbitrary?"

Is it because since $\sqrt x \times \sqrt x = x$ and we know that we add the exponents when multiplying, $\sqrt x$ must be the same as $x^{\frac{1}{2}}$ since this allows us to multiply the square roots, adding $^{\frac{1}{2}}$ to $^{\frac{1}{2}}$ thus getting $x^1$ which is the correct answer?


PS: Did I just answer my own question in a rather informal, unstandardized way?

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  • $\begingroup$ Fractional exponents always make sense, even when $x$ is negative - provided, of course, you go into complex numbers... and what you ask about follows from the rules of exponents. $\endgroup$ Commented Sep 27, 2016 at 1:41
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    $\begingroup$ I think you not only answered your own question, but that the answer you gave was the standard way. You want the arithmetic with exponents to mimic the arithmetic of nth roots: just as you said when you wrote "on the other hand since..." I am not sure if there is anything else to say, other than this is why we define fractional (and negative) exponents this way. $\endgroup$ Commented Sep 27, 2016 at 1:46
  • $\begingroup$ @User0112358: Thanks. That makes sense. :) $\endgroup$
    – J. Allan
    Commented Sep 27, 2016 at 1:51

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There are two reasons. First, chained exponentiation follows the rule:$$ \left(x^y\right)^z = x^{yz}.$$ So, if $y= 1/z$ then $\left(x^y\right)^z = x$, meaning that exponentiation by the reciprocal obeys the property that defines taking a root.

Second, if you want to know how these things are actually calculated, that's an interesting question, too. We can define two functions $\operatorname{e}^x$ and it's functional inverse $\ln x$ (that is $\operatorname{e}^{\ln x} = \ln [\operatorname{e}^x] = x$) in terms of power series. $\ln x$, in particular, has the property: $$\ln x^y = y\ln x,$$ making it possible to calculate $x^y$ using the identity: $$x^y = \operatorname{e}^{y\ln x}.$$ I don't think that this is how it is actually calculated by computers, because this way is probably pretty slow, but it can be done this way.

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    $\begingroup$ Thanks, Sean. The chained exponentiation was what I was trying to say in my post, but didn't quite know how to say. :) +! $\endgroup$
    – J. Allan
    Commented Sep 27, 2016 at 2:00

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