Konstantin Lynov wrote:
An old one:
Is X>Y ?
(1) square root X > square root Y
(2) X^2 > Y^2
Through the discussion people agreed that the answer is A, since on the GMAT we are dealing only with arithmetic radicals and, therefore, sqrt(X) and sqrt(Y) are non negative.
Disagree on that answer is:
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.