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Possible Duplicates:
Reason why the even root of a number always positive
Square roots — positive and negative

I saw the following during a practice exam:

$f(x) = \sqrt x $ for $x ≥ 2$

After the exam, I pointed out to my teacher that this was not actually a function, but instead a mapping, because it can have both positive and negative values. She told me that this didn't matter; you assume the answer of $\sqrt x$ to be positive, unless otherwise stated. However, this contradicts her previous statement that $\sqrt x$ is a mapping. (She gave it as an example of something that was not a function, when defining functions.)

So, is $f(x)$ a function? Does it matter if there is a domain restriction?

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    What exactly do you mean by "it can have both positive and negative values"? The (principal) square root function lives entirely in the first quadrant...2011-07-24
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    @J.M. -2 and +2 are both the square root of 4.2011-07-24
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    That's why one chooses the *principal* square root; when people write $f(x)=\sqrt{x}$ they (usually) mean that the principal value is assumed. Thus, $\sqrt{4}=-2$ is incorrect, but $4=(-2)^2$ is alright.2011-07-24
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    @J.M. I think you should consider expanding your last comment into an answer, as that's all there really is to say.2011-07-24
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    @mix: since I'm trying to answer some other question, could you do that for me instead? ;D2011-07-24
  • 1
    Ah... [here](http://math.stackexchange.com/questions/41878) we are!2011-07-24
  • 0
    @J.M. Aha! good catch. Duplicate.2011-07-24
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    ...and [this puppy](http://math.stackexchange.com/questions/26363) too!2011-07-24
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    I'm just a beginner math student - why the down-votes?2011-07-24

4 Answers 4