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heshamelaziry
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If x and y are integers ans xy doesn't = zero, is x-y > 0 ?
(1) x/y <1/2
(2) sqrt x^2 = x and sqrt of y^2 = y
(1) this tells us a couple of things depending on the values of x and y.
if x and y are both (+), then 0<2x<y
If x and y are both (-), then 0>x>.5y
If x is (-), but y is (+), then y>0>2x
if y is (+), but x is (+), then 2x>0>y
All this is unnecessary to write out, I only wrote it out to give a greater understanding of the problem. as you may be able to see, (1) is insufficient on it's own.
(2) sqrt x^2 = x and sqrt of y^2 = y
-looking at this, I would say insufficient as well.
So now we get either C or E. Looking at (1), lets see which options would
if x and y are both (+), then 0<2x<y
If x and y are both (-), then 0>2x>y
If x is (-), but y is (+), then y>0>2x
if y is (+), but x is (+), then 2x>0>y
Now, this is where I'm fuzzy on the exact rules of square roots that someone else could help out with, but I believe when you're taking the square root of a square, you taking it in as an absolute value. So if x was negative, then sqrt x^2 would equal a positive number. Now, with the equations in (2), we know that the solution for sqrt x^2 and sqrt y^2, are x and y respectively. because the sqrt of a square will be positive, then both x and y should both be positive, leaving us with only one possibility, 0<2x<y.
so if x =1, then y>2 (3 for example). Let's plug these in.
x-y>0
1-3>0
-2>0 is wrong.
This means that C is sufficient to answer the questions of is x-y>0.
This is a great example, because we disprove the question. Often times people get caught up in trying to answer yes, not yes or no. HOWEVER, i am not completely sure about the absolute value thing regarding the sqrt of squares rule, so take this with a few grains of salt
