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I recently saw in the Thread on Mathematical Misconceptions a post which implied that the following equation has no known solutions: $$\frac{\sqrt{x}}{2} = -1$$ Why does this not have any solutions? It seems to me that $\sqrt{x} = -2$, so $x = (-2)(-2) = 4$. Does it not have any solutions because $\sqrt{x}$ is explicitly defined as $-2$, and if $x$ were $4$, then $\sqrt{x}$ would have to equal $\pm 2$?

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The symbol $\sqrt{x}$ refers to the unique non-negative number whose square is $x$. There is no ambiguity in terms of signs for the symbol $\sqrt{x}$.

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    Finally someone who gives the only right answer. No, it's not context dependent. – Git Gud Dec 19 '13 at 23:01
  • Wonderfully definitive. Cheers – Newb Dec 19 '13 at 23:04
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    Is it widely known that the symbol $\sqrt x$ is so narrowly defined? I wasn't aware of that, and I'll bet many others aren't either. Is there a good reason for limiting the definition of this symbol? – Betty Mock Dec 19 '13 at 23:05
  • @BettyMock Without this restriction, $x \mapsto \sqrt{x}$ isn't even a function, since it's not single-valued. In lower level courses, many people skim over this fact and simply emphasize that the equation $x^2 = ...$ will always have two solutions. –  Dec 19 '13 at 23:07
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    I don't think I ever had the need to name $-2$ has $\sqrt 4$, in fact $-\sqrt 4$ suffices. – Git Gud Dec 19 '13 at 23:07
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    My algebra book went to great lengths to ensure that people knew $\sqrt{x}\ge0$... I'm sort of the opposite of Betty--I thought that everyone knew $\sqrt{x}$ was the principal 2nd root... – apnorton Dec 19 '13 at 23:08
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    A good reason on why the square root is defined as positive: http://math.stackexchange.com/questions/158566/is-it-wrong-to-say-sqrtx-times-sqrtx-pm-x-forall-x-in-mathbbr?rq=1 – apnorton Dec 19 '13 at 23:17
  • Just read the very lengthy discussion on the link above provided by anorton; the length alone tells me that there is a lot of ambiguity floating around. That discussion mostly focuses on $\sqrt x$ as a function; T Bongers discusses it as a symbol. I think we all agree that a function must map each element of its domain to a unique value. Now if we say the domain is the complex numbers surely f(z) = $\sqrt z$ is defined, but it has "branches", thus dealing with the multiple value issue. (continued) – Betty Mock Dec 20 '13 at 00:07
  • Would it not be consistent to say that when the domain is real x > 0 the function has two branches ? This allows us to be complete, to avoid arbitrary definitions, and to introduce the idea of "branch" to less advanced students in an easy to understand example. Those that go on to complex variables will understand "branch" right away. – Betty Mock Dec 20 '13 at 00:13
  • @anorton You REMEMBER your algebra book? Are you a young person? I don't remember mine at all, let alone what it said about square roots. Although I do remember the names of my pet dinosaurs, Slinky and Blinky. – Betty Mock Dec 20 '13 at 00:16
  • @BettyMock haha... yeah, I am a fairly young user. ;) my age is in my profile. – apnorton Dec 20 '13 at 01:34
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    @BettyMock : there's no good reason not to limit the definition of $\sqrt{x}$. If you want both roots, then write $\pm \sqrt{x}$. No need to mess with confusing "branches" until you're dealing with complex numbers. – Stefan Smith Dec 20 '13 at 03:48
  • @stephan smith -- of course whatever the community thinks is the right definition will prevail. But sometimes we should ask questions. Re branches, I at first found them confusing in complex analysis; if someone had just pointed out the real sqrt function, all would have been clear. – Betty Mock Dec 21 '13 at 01:25
  • @anorton glad you found it funny; I was a little worried I might have offended you. I've spent much of my life programming, including currently. Use this site to take breaks from it. I know for sure I never had a textbook. – Betty Mock Dec 21 '13 at 01:32