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$\sqrt{9} = 3$

9 has 2 square roots: 3 and -3.

What is $9^\frac12$? Is $9^\frac12 = \sqrt{9} = 3$ or is $9^\frac12 = \pm3$?

at01
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4 Answers4

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This previous answer of mine talks about the fact there are actually multiple exponentiation operators. The meaning of $9^{\frac{1}{2}}$ depends on which one you are actually using.

Community
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To tell you the truth, it all depends on the given context, because the $n^{th}$ root of a certain number can be any of the n complex solutions to the equation $x^n=a$, so each radical is a choice. If your context is $\mathbb{R}$, then you look for the single real solution if n is odd, or for the positive one if n is even. But if you dabble into something a bit more serious, like $\mathbb{C}$, cubics, quartics, or Galois theory, then you have to be very careful each step of the way that the values you ascribe to each of these roots fulfill certain criteria. For instance, all the various formulae for the zeroes of the cubic or quartic equations rely heavily on which choices one makes for each radical, and if these choices fail to meet certain demands, then the formulae yield false results. E.g., at a certain point, when $\Delta_0=0$, one has to choose the value of $\sqrt{\Delta_1^2\!-\!4\Delta_0^3}=\!\sqrt{\Delta_1^2}=\!\Delta_1$ , even if the quantity in question is negative, otherwise the formula simply doesn't work !

Lucian
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$a^b$ is always positive for $a\in\Bbb R^+$ and $b\in\Bbb R$. This is because $f(x)=a^x$ satisfies the functional equation $f(x+y)=f(x)f(y)$ (i.e. $a^{x+y}=a^xa^y$). It follows that $f(2x)=f(x)^2,$ or $$a^x=f(x)=f\left(\frac x2\right)^2\ge 0$$ Consequently, $9^\frac12\ge 0\implies 9^\frac12=3$

Tim Ratigan
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  • I guess it would have to be that way because $9^1 = 9^\frac12 \cdot 9^\frac12$ and the result of that can't be negative. – at01 Dec 22 '13 at 09:07
  • No, you'll see that $(-3)(-3)=9$, so in that respect $9^\frac12$ would work. But $9^\frac12=(9^\frac14)^2$, so $9^\frac12$ can't be negative assuming $9^\frac14$ is real-valued, which it is. – Tim Ratigan Dec 22 '13 at 09:09
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    What I meant was if $9^\frac12$ could be negative or positive, then one of them could be negative and the other positive making $9^1$ negative. – at01 Dec 22 '13 at 09:10
  • We're assuming that $f(x)=9^x$ is well-defined, that is $f(x)$ is only one value for any $x$. – Tim Ratigan Dec 22 '13 at 09:12
  • I just stumbled upon this question I asked a couple months ago and am reviewing your answer that I accepted. I'm not as sure about your answer anymore. Basically, $9^\frac12$ can be $(9^\frac14)^2$, but that doesn't, by itself, preclude it from also being equal to -3. How do we get away from the fact that $9=(-3)^2$ and so we can replace 9 with $(-3)^2$ in $9^\frac12$? – at01 Feb 25 '14 at 00:52
  • $9^{1/4}$ is, itself, real and nonzero, right? Any real number that is not zero, when squared, is greater than zero. Therefore $9^{1/2}=(9^{1/4})^2>0\implies 9^{1/2}=3$. – Tim Ratigan Feb 25 '14 at 09:39
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$9^\frac{1}{2}=3.$ not $-3$. This is because $9^{\frac{1}{2}}$ is positive.

mathlove
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