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If x and y are integers, is x > y?

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Re: If x and y are integers, is x > y? [#permalink]
Bunuel wrote:
If x and y are integers, is x > y?

(1) 0 < x^y < 1
(2) y is odd

1) if x^y is less than 1, it does'nt matter if y is odd or even. To satisfy this condition, x has to be positive and y has to be negative. Hence x > y. Suff
2) No information about x. x can be anything here. Not Suff.
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Re: If x and y are integers, is x > y? [#permalink]
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This one can be tricky if you rush through it.

As some users have posted, at first glance, it appears that x must be positive however it is important to remember that even exponents can mask signs or make negative numbers into positive.

Consider x = -2;
Is there any power we can raise -2 to to make it a positive fraction less than 1?
Well to make it positive we have to raise it to an even exponent and to make it a fraction we have to raise it to a negative exponent; Thus raising it to a negative even exponent could satisfy the requirements.
Both y = -2 and y = -4 work, however they give different answer to the question "is x>y?" thus more information is needed.

To finish this off quickly - the 2nd statement tells us very little on its own so we'll need to combine it with statement I;
y is odd means that x cannot be negative as there is no way we can make it a positive number;
Together this information is sufficient
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Re: If x and y are integers, is x > y? [#permalink]
Bunuel wrote:
If x and y are integers, is x > y?

(1) 0 < x^y < 1
(2) y is odd

(x,y)=integers

(1) 0 < x^y < 1 insufic

y=negative

x=-3, y=-2: x^y=1/(-3)^2=1/9, x<y
x=3, y=-1: x^y=1/(3)^1=1/3, x>y

(2) y is odd insufic

(1 and 2) sufic

y=odd_negative
x=positive
x>y

Ans (C)
Re: If x and y are integers, is x > y? [#permalink]
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