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Here's the question I've been thinking for $2$ days and couldn't find an answer why.

If $Y = \sqrt{X}$, where both $X$ and $Y$ are real numbers then which of the following is true?

a) $X\geq 0$; $Y\geq 0$

b) $X\leq 0$; $Y\geq 0$

c) $X\leq 0$; $Y\leq 0$

d) $X\geq 0$; $Y\leq 0$

e) Either a) or b)

Now, the answer is a) but I don't know why. I know a value of $Y$ for which root of $Y$ is negative.

$\sqrt{4} = \pm2$

$\sqrt{9} = \pm3$

So the answer should be $X\leq0$; $Y\geq0$, but its not. :(

Can anyone explain me why? Thanks in advance. :)

Stahl
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Electrifyings
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    Where's X (Y = √Y)? – J. M. ain't a mathematician Aug 28 '10 at 02:25
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    The square root symbol, as a function on the real numbers, generally indicates only the positive root. – Qiaochu Yuan Aug 28 '10 at 02:28
  • If $Y = \sqrt{Y}$ where $Y$ is real then $Y \geq 0$. –  Aug 28 '10 at 02:31
  • So whenever a variable is associated with Square root of any number and its in the form of equation, we MUST take only positive values. But when you're supposed to FIND OUT the square root of any number, you should take both positive and negative values. That's what I understood, correct me if I'm wrong. :\ – Electrifyings Aug 28 '10 at 02:45
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    Not much as "must" as "it's the accepted convention". A number can have two square roots, but when we talk about something like √2, we are concerned here only with the nonnegative square root (more precisely, the square root with nonnegative real part to cover the complex case), – J. M. ain't a mathematician Aug 28 '10 at 02:55
  • Understood partially. But didn't got your first question. Where's X (Y = √Y)? I don't know what's that. :\ – Electrifyings Aug 28 '10 at 03:02
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    Read your question again, you said Y = √Y ; where's the X there? – J. M. ain't a mathematician Aug 28 '10 at 03:22
  • If it is not a typo, as J. Mangaldan points at, then you have Y>=0 and X real that is e) is the answer. – AD - Stop Putin - Aug 28 '10 at 04:17
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    @J. Mangaldan: the "nonnegative real part to cover the complex case" is not entirely agreed upon--the two competing definitions for principal square root that I've seen are equivalent to the argument being in $[0,\pi)$ or in $(-\frac{\pi}{2},\frac{\pi}{2}]$, though most calculators/software seem to implement the latter, which would agree with "nonnegative real part." – Isaac Aug 28 '10 at 04:39
  • Isaac: Fair point, though to be honest I've yet to see a computing environment that returns results consistent with your first case. Anybody here got an example handy? – J. M. ain't a mathematician Aug 28 '10 at 04:47
  • Okay, sorry guys. That was a typo. I corrected the question now. :) – Electrifyings Aug 29 '10 at 05:09
  • @J. Mangaldan: I do not know of any computing environment that uses my first case, but I have seen more texts using the first case than the second and the several mathematicians (all of whom predate such computing systems) that I've asked have initially said that the first case was the proper definition of principal square root, though subsequent surveys of textbooks turned up both definitions. This is especially problematic when trying to compute roots of cubic/quartic polynomials on a calculator using explicit formulae from texts, as the choice of principal root seems to affect the formula. – Isaac Aug 29 '10 at 05:16
  • Isaac: Yes, I'm aware of the choice of principal values for the nth root being problematic for solving polynomial equations explicitly; I remember spending a fair bit of time debugging a TI 83+ program for solving a cubic equation. – J. M. ain't a mathematician Aug 29 '10 at 05:23
  • @J. Mangaldan: I had to rewrite half a page of a textbook twice in a 48-hour span (during which I was supposedly on vacation) to go from a botched copy of a cubic formula from somewhere unknown to a fixed version that turned out not to work on any of the devices/software we had at hand, to a more carefully-constructed formula based on what Mathematica spit out when asked to solve the general cubic. This answer of mine has a similarly-careful quartic formula. – Isaac Aug 29 '10 at 08:07
  • Isaac: FWIW, the algorithm I finally had turned out to be consistent with this formulation: http://books.google.com/books?id=w47cyQLVQowC&pg=PA228 Man, multivalued functions are a pain sometimes... :) – J. M. ain't a mathematician Aug 29 '10 at 08:27
  • OMG, GEEK TALK! :D – Electrifyings Aug 30 '10 at 02:35

1 Answers1

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As has been indicated in the comments, the radical symbol is generally taken to be the principal square root function, which for nonnegative real inputs gives nonnegative real outputs. For negative real inputs, this function gives non-real outputs, so X must be nonnegative, so Y must be nonnegative, giving (a).

Isaac
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