# Problem related with equation consisting square root

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)? –  Ｊ. Ｍ. 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), –  Ｊ. Ｍ. Aug 28 '10 at 2:55
Read your question again, you said Y = √Y ; where's the X there? –  Ｊ. Ｍ. 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