Provided x,y,z <>0.
If z.y < x.y
And if z<x , Does it necessarily mean that y >0.
Also, if y > 0 , Does it necessarily mean z<x
Conceptual question
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Can you elaborate ?. Let me post the questionrobbie523 wrote:only why Y is positive, then Z can be greater than X.
zy<xy Is y > 0
1. z<x
zy < xy . Is z<x
1. y >0
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for the first one, stm 1 is insufficient to prove y>0, since nothing about y is given, only know Z<X, doesn't assure wether the inequality will be reversed or not. Y still could be negative and reverse the inequality signzy<xy Is y > 0
1. z<x
zy < xy . Is z<x
1. y >0
for the second one, stm 1 is sufficient, cuz Y is the key factor in reversing the inequality. if Y>0, that mean with or without why, the inequality won't reverse, therefore, assured Z<X
to prove something, you need to use the premise that's already given, and plug in the other factors later. to see it is yes or no!
to prove something right is very difficult, but to prove something wrong, is very easy, you only need one example!
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yes.
If z.y < x.y
And if z<x ,
then Y must be >0
Also, if y > 0 ,
then z must be <x
If z.y < x.y
And if z<x ,
then Y must be >0
Also, if y > 0 ,
then z must be <x
The powers of two are bloody impolite!!
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madhur_ahuja wrote:Can you elaborate ?. Let me post the questionrobbie523 wrote:only why Y is positive, then Z can be greater than X.
zy<xy Is y > 0
1. z<x
from given
y<x-z is y>0
from 1
z-x<0 ie: x-z>0, using this info and plugging in the given
y<+ve doesnt imply that y is +ve...insuff
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If z < x, we can multiply on both sides by y. If y is negative, we'd need to reverse the inequality, so we'd get zy > xy. We know that's not true, however, because we're told in the stem that zy < xy. So y cannot be negative, and since y cannot be zero, y must be positive. So yes, your conclusion is correct, and Statement 1 is sufficient.madhur_ahuja wrote:
Can you elaborate ?. Let me post the question
zy<xy Is y > 0
1. z<x
Since y is positive, we can divide by y on both sides of the inequality zy < xy without needing to reverse the inequality, so we find that z < x, and Statement 1 is sufficient.madhur_ahuja wrote: zy < xy . Is z<x
1. y >0
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