Is x > y ? (1) x^(1/2) > y^(1/2) (2) x^2 > y^2

USEFUL TO KNOW
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Is x>y?

(1) \(\sqrt{x}>\sqrt{y}\) --> as both parts of the inequality are non-negative then according to A we can square them --> \(x>y\). Sufficient.

(2) \(x^2>y^2\) --> \(|x|>|y|\) --> \(x\) is farther from zero than \(y\), but this info is insufficient to say whether \(x>y\) (if \(x=2\) and \(y=1\) - YES but \(x=-2\) and \(y=1\) - NO). Not sufficient.

Answer: A.