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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. :)

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Where's X (Y = √Y)? –  Guess who it is. Aug 28 '10 at 2:25
The square root symbol, as a function on the real numbers, generally indicates only the positive root. –  Qiaochu Yuan Aug 28 '10 at 2:28
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), –  Guess who it is. Aug 28 '10 at 2:55
Read your question again, you said Y = √Y ; where's the X there? –  Guess who it is. Aug 28 '10 at 3:22
@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 4:39

1 Answer 1

up vote 1 down vote accepted

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).

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Thanks a lot :) –  Electrifyings Aug 30 '10 at 2:31

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