Hi Experts,
I have a problem with this one.
The question is (Data Sufficiency):
Is x>y?
i) ax > ay
ii) a^2 x > a^2 y
The correct answer is B, because an integer square is always positive... but how says that a is an integer?? If it would be a complex number a^2 could be negative.
Thanks in advance!
Is x>y?
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Is x>y?
i) ax > ay => ax-ay > 0, a(x-y) > 0 posiible if a and (x-y) have the same +ve or -ve sign.
a may be -2 and x-y may be 5-7 = -2 but a(x-y) = +4
ii) a^2 x > a^2 y => a²(x-y) > 0 logic similar to (i) but a² will be +ve for real numbers and (x-y) > 0
The eq holds for any real nos. even if it is not a integer but a fraction.
In the OG11 under the heading 4.3 Problem Solving Sample Questions it is stated that
Numbers : All numbers used are real numbers.
In the Official Guide there is no discussion on complex nos, though real nos are explained.
Anyway, better wait for the experts.
i) ax > ay => ax-ay > 0, a(x-y) > 0 posiible if a and (x-y) have the same +ve or -ve sign.
a may be -2 and x-y may be 5-7 = -2 but a(x-y) = +4
ii) a^2 x > a^2 y => a²(x-y) > 0 logic similar to (i) but a² will be +ve for real numbers and (x-y) > 0
The eq holds for any real nos. even if it is not a integer but a fraction.
In the OG11 under the heading 4.3 Problem Solving Sample Questions it is stated that
Numbers : All numbers used are real numbers.
In the Official Guide there is no discussion on complex nos, though real nos are explained.
Anyway, better wait for the experts.
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