Are true means = must be true without any refutation.

We have questions in which we do have "could" and those in which we do have "must"

I worded this way because the question was good but the wording poor

Yes, I, like suhDude was a bit confused about the phrasing of the question only because I found that "C" was always true but incomplete (i.e. x didn't HAVE to be less than 0 when y<0 to produce a true inequality; x could also be greater than 1 when y<0), while "E" was the only expression that is "completely true" given the stated inequality.


Once I figured that a correct answer selection is one that produces a valid (y/x) > y statement (i.e. I'm not looking for answer items that give the full range of valid x-values for the stated y-range), then I figured that the answer is definitely C & E.





We know that if x and y are both real numbers (a valid assumption on the GRE), that they can both either be less than zero, equal to zero, or greater than zero.

Because division by zero is undefined (see ETS math convention #6 under "Mathematical expressions, symbols, and variables"), we know that x cannot equal zero.

So y=0 or y<0 or y>0 and x<0 or x>0

1.) let y = 0:
(0 / x) > 0
0 > 0
-> 0>0 is always FALSE
(doesn't matter what x is in this case)
->-> so (y/x) > y is FALSE when y = 0 and x = anything


2.) let x<0:
-multiply both sides of inequality by x<0:
y<xy

a.) let x<0 & y<0:
-divide both sides of inequality by y<0:
1>x == x<1
-intersection of x<0 and x<1 == x<0,
->-> (y/x)>x is TRUE when x<0 & y<0
(e.g. x = -2, y = -4; (-4/-2 = 2) > -4 == TRUE)

b.) let x<0 & y>0:
-divide both sides of inequality by y>0:
1<x == x>1
-intersection of x<0 and x>1 is CONTRADICTION
->-> (y/x) > y is FALSE when x<0 & y>0


3.) let x>0:
-multiply both sides of inequality by x>0:
y>xy

a.) let x>0 & y<0:
-divide both sides of inequality by y<0:
1<x == x>1
-intersection of x>0 and x>1 == x>1
->-> (y/x)>y is TRUE when x>1 and y<0
(e.g. x=2, y=-4; (-4/2 = -2) > -4 == TRUE)

b.) let x>0 & y>0:
-divide both sides of inequality by y>0:
1>x == x<1
-intersection of x>0 and x<1 == 0<x<1
->-> (y/x)>y is TRUE when 0<x<1 and y>0
(e.g. x=1/2, y=4; (4/(1/2) = 8) > 4 ==TRUE)


---------------
So, selecting statements that are always true:

A = FALSE (demonstrated by case #1)
B = FALSE (demonstrated by case #1)
C = TRUE (demonstrated by case #2a)
*at first I felt that this was FALSE because as demonstrated in case#3a, when y<0, x doesn't HAVE to be less than 0 for the given inequality to be valid. The (y/x) > y inequality is also valid when y<0 and x>1. A set of x and y values however, when picked from the "C" range, will [i]always produce a true (y/x) >y statement, regardless of whether or not other values/combinations/ranges also produce true statements, so "C" is definitely TRUE.[/i]
D = FALSE (demonstrated by cases #2b, #3b)
E = TRUE (demonstrated by case#3b)

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