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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9$\begingroup$ It's a definition. $\endgroup$– user223391Jan 10, 2017 at 23:08
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$\begingroup$ $x^{a/b}=\sqrt[b/a]{x}=\sqrt[b]{x^a}$ (if $x>0$, etc). $\endgroup$– YoTengoUnLCDJan 10, 2017 at 23:11
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$\begingroup$ At least for $x>0$ and $b\ne0$, $$x^{a/b}=\left(x^a\right)^{1/b}=\left(x^{1/b}\right)^a=\left(\sqrt[b]{x}\right)^a=\sqrt[b]{x^a}$$ $\endgroup$– DidJan 10, 2017 at 23:11
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4 Answers
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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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1$\begingroup$ Yes we do, the imaginary number system. $\sqrt{-1} = i$, $\sqrt{-2} = \pm i\sqrt{2}$ $\endgroup$– TravisJan 10, 2017 at 23:22
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$\begingroup$ @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. $\endgroup$– RGSJan 10, 2017 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? $\endgroup$– RGSJan 10, 2017 at 23:14
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3$\begingroup$ @RSerrao Nah, yours is fine too. Leave them both. $\endgroup$– mweissJan 10, 2017 at 23:24
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$\begingroup$ Sorry to accept @RSerrao's, even though you answered first, his answer just seemed more inclusive and exploitative to me $\endgroup$– TravisJan 10, 2017 at 23:27
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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 $y$).
answered Jan 10, 2017 at 23:15
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$\begingroup$ ... works out fairly nicely for positive $x$ $\endgroup$– GEdgarJan 10, 2017 at 23:20
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$\begingroup$ @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 $\ddot{\smile}$.) $\endgroup$ Jan 10, 2017 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 $\sqrt[2]{2}$
answered Jan 10, 2017 at 23:20
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2$\begingroup$ I think the question was, "Why is this the definition?" Definitions may be stipulated but they are not arbitrary. $\endgroup$– mweissJan 10, 2017 at 23:25