Lazybum wrote:
Bunuel wrote:
Is |x| > |y|?
(1) x^2 > y^2. Since both sides are non-negative, then we can safely take the square root: |x| > |y|. Sufficient.
Or: "is |x| > |y|?" can be rewritten as: is x^2 > y^2? (we can safely square the whole inequality since both sides are non-negative). This statement directly answers the question. Sufficient.
(2) x > y. Clearly insufficient: consider x=1 and y=0 for an YES answer and x=1 and y=-2 for a NO answer. Not sufficient.
Answer: A.
Hi Bunuel --
For S1 -- can you please elaborate the theory re: why
if x^2 > y^2 -- that implies |x| > |y|
I can memorize this but i would like to understand logically the theory behind something like. How would one make this assumption logically without memorizing this information ?
I have tested a few test cases and this is true -- but logically or on a number line -- how does one make sense of this theory ? [ x^2 > y^2 -- implies |x| > |y| ]
Since both side of the inequality are non-negative, then we can safely take the square root to get \(\sqrt{x^2} > \sqrt{y^2}\). Next, since \(\sqrt{a^2}=|a|\), then \(\sqrt{x^2} > \sqrt{y^2}\) gives |x| > |y|.
does anyone know why this is "not negative" as bunuel pointed out? oh actually nvm, its a fundamental identity that if you square a number, it always is greater than or equal to 0.
i tried to solve this question differently but it didnt work, does anyone understand why?
I used the "critical method" cited by bunuel, here:
https://gmatclub.com/forum/absolute-mod ... 10849.htmli listed out all the scenarios
x > y when x>0 and y>0
-x > y when x<0 and y>0
x > -y when x>0 and y<0
-x > -y when x<0 and y<0
this just doesn't work and i'm confused why.[/quote]
Dear
Lazybumthe values are non-negative because you cannot take square root from the negative value.
Even if the X = -2 then the (-2)^2= 4 and \(\sqrt{4} \)= 2 not the - 2
The question is asking is the absolute value of X grate then the absolute value of Y
2d statement gives you variables X and Y in lieu of Absolute variables.