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Thread: Square Root = Two Answers

  1. #16
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    Re: Square Root = Two Answers

    For all non-negative  \ \ x, \ \ x^{1/2} \ = \ \sqrt{x}.

    f(x) \ = \ \sqrt{x}  is a function. As such, it gives one value.
    Last edited by greg1313; Dec 29th 2016 at 09:18 AM.
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  2. #17
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    Re: Square Root = Two Answers

    Quote Originally Posted by greg1313 View Post
    For all non-negative  \ \ x, \ \ x^{1/2} \ = \ \sqrt{x}.

    f(x) \ = \ \sqrt{x}  is a function. As such, it gives one value.
    that's fine and I agree, but the reason for this is not the reasoning as given on quickmath.com, it's because it is defined this way.
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  3. #18
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    Re: Square Root = Two Answers

    Quote Originally Posted by romsek View Post
    that's fine and I agree, but the reason for this is not the reasoning as given on quickmath.com, it's because it is defined this way.
    That post didn't state it was from the reasoning as given on quickmath.com. It's a new stand-alone post.
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    Re: Square Root = Two Answers

    Quote Originally Posted by greg1313 View Post
    That post didn't state it was from the reasoning as given on quickmath.com. It's a new stand-alone post.
    I'm not going to argue with you but that is an interesting interpretation of post #14
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    Re: Square Root = Two Answers

    Quote Originally Posted by zzephod View Post
    It depends on the context. Without qualification I would take it to denote the (two) distinct values of $3e^{n\pi i};\ n\in \mathbb{Z}$, because that is the interpretation in the contexts that I normally work with.
    Quote Originally Posted by greg1313 View Post
    There is no context to consider. You are being asked straight-up for value(s) of  \ \  9^{1/2}.
    Quote Originally Posted by romsek View Post
    I'm not going to argue with you but that is an interesting interpretation of post #14
    Given the fact that the original post is is the pre-university elementary algebra forum I find the post #14 spot-on and I find those of zzephod off-putting in that they are totally inappropriate for the level of discussion.
    Suppose $x\in\mathbb{R}$ then
    \text{square root(s)}(x)=\begin{cases}\text{does not exist } &: x<0 \\ 0 &: x=0\\ \pm\sqrt x&:x>0\end{cases}
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