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I was attempting to answer this question, and came across the fact that $x$ to a non integer power, $x^\frac{1}{b}$ is equal to $\sqrt[b]{x}$. Why does this work?

Similarly, does $x^{\frac{a}{b}} = a\sqrt[b]{x}$ ?

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    $\begingroup$ It's a definition. $\endgroup$
    – user223391
    Jan 10 '17 at 23:08
  • $\begingroup$ $x^{a/b}=\sqrt[b/a]{x}=\sqrt[b]{x^a}$ (if $x>0$, etc). $\endgroup$ Jan 10 '17 at 23:11
  • $\begingroup$ At least for $x>0$ and $b\ne0$, $$x^{a/b}=\left(x^a\right)^{1/b}=\left(x^{1/b}\right)^a=\left(\sqrt[b]{x}\right)^a=\sqrt[b]{x^a}$$ $\endgroup$
    – Did
    Jan 10 '17 at 23:11
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As noted in the comments, it is so by definition. Let me try to give you an intuitive explanation though:

Remember the rule of exponents that states:

$$(x^a)^b = x^{ab}$$

Now set $b = \frac1a$:

$$(x^a)^\frac1a = x^{a\cdot\frac1a} = x^1 = x$$

Therefore raising a number to the $\frac1a$ is the inverse operation of raising it to $a$. But we know that the inverse function of raising a number to $a$ is just finding the $a$-th root of it, $\sqrt[a]{.}$. Then they should be equal.

Please beware that even though this may seem really nice, one must be careful when raising numbers to fractions, because the apparently innocent expression

$$(-2)^{\frac12}$$

is actually quite dangerous, since it is

$$\sqrt{-2}$$

and we don't know how to find square roots of negative numbers (we actually have but kost of the times the context does not allow you to use imaginary numbers). These problems arise whenever the denominator of the exponent is even and the base is negative.

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    $\begingroup$ Yes we do, the imaginary number system. $\sqrt{-1} = i$, $\sqrt{-2} = \pm i\sqrt{2}$ $\endgroup$
    – Travis
    Jan 10 '17 at 23:22
  • $\begingroup$ @Travis i know, but when these things are first introduced, we know not of that. Plus, many places where those fractions may pop up we just can't have imaginary numbers. $\endgroup$
    – RGS
    Jan 10 '17 at 23:29
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It's because however we define fractional exponents, we want to make sure they behave according to the same algebraic rules as integer exponents.

For example, for integer exponents we have the property $$(x^a)^b = x^{ab}$$ and we want to extend that property to make sure it also holds when one of the exponents is a fraction. That means we want $$(x^{1/a})^b = x^{a/b}$$ for all $a,b$ (with $a \ne 0$). In particular, we need $$(x^{1/b})^b = x^{b/b} = x^1 = x$$ which means (by definition) that $x^{1/b}$ is a $b$th root of $x$, i.e. that $x^{1/b} = \sqrt[b]{x}$.

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  • $\begingroup$ oh wow, our answers are quite similar but you posted yours first; should I delete mine? $\endgroup$
    – RGS
    Jan 10 '17 at 23:14
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    $\begingroup$ @RSerrao Nah, yours is fine too. Leave them both. $\endgroup$
    – mweiss
    Jan 10 '17 at 23:24
  • $\begingroup$ Sorry to accept @RSerrao's, even though you answered first, his answer just seemed more inclusive and exploitative to me $\endgroup$
    – Travis
    Jan 10 '17 at 23:27
  • $\begingroup$ @Travis: "Exploitative"? $\endgroup$
    – mweiss
    Jan 10 '17 at 23:29
  • $\begingroup$ Explanative* Whoops $\endgroup$
    – Travis
    Jan 10 '17 at 23:32
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The intuition comes from the formula $(x^a)^b) = x^{ab}$ (which certainly makes sense if $a$ and $b$ are positive integers). Generalising this, we might expect to have $(x^{\frac{1}{a}})^a = x^{\frac{1}{a}a} = x^1 = x$, so that $x^{\frac{1}{a}}$ should be the $a$-th root of $x$, $\sqrt[a]{x}$. And it turns out that this all works out fairly nicely (e.g., when we use $\log$ and $\exp$ to define $x^y$ for arbitrary $y$).

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  • $\begingroup$ ... works out fairly nicely for positive $x$ $\endgroup$
    – GEdgar
    Jan 10 '17 at 23:20
  • $\begingroup$ @GEdgar: I was being deliberately vague about the extent to which the generalisation works. And I don't see anything wrong with that in the context of this question. (It actually continues to work out fairly well for negative or even complex numbers, although something like $\mathbf{i}^{\frac{1}{\mathbf{i}}}$ needs a bit more care in the working out $\ddot{\smile}$.) $\endgroup$
    – Rob Arthan
    Jan 10 '17 at 23:24
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It basically is a definition;

Consider $\sqrt{2}$, this can be written as $2^{\frac{1}{2}}$ which is same as $\sqrt[2]{2}$

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    $\begingroup$ I think the question was, "Why is this the definition?" Definitions may be stipulated but they are not arbitrary. $\endgroup$
    – mweiss
    Jan 10 '17 at 23:25

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