MGMAT: Modules
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- Senior | Next Rank: 100 Posts
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- Joined: Fri Aug 14, 2009 12:03 am
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- Senior | Next Rank: 100 Posts
- Posts: 59
- Joined: Fri Aug 14, 2009 12:03 am
I don't understand why statement (1) is insufficient!!
Since module X is always >0, we can change the equation into x<x*module x
Form here we consider three possible variants:
a) x<0
then x>x*module x
(for instance, -2>-2*2)
but this contradicts our initial condition, so we drop it
b) 0<x<1
then x>x*module x
(for example, 1/2>1/2*1/2)
we see contradiction again, hence, this variant is not workable
c) x>1
then x<x*module x
(e.g. 2<2*2)
this is the only variant that satisfies our problem, we conlcude that module x>1, hence statement (1) should be sufficient.
Could anyone tell me what is the flaw in my reasoning? Will greatly appreciate your input.
Since module X is always >0, we can change the equation into x<x*module x
Form here we consider three possible variants:
a) x<0
then x>x*module x
(for instance, -2>-2*2)
but this contradicts our initial condition, so we drop it
b) 0<x<1
then x>x*module x
(for example, 1/2>1/2*1/2)
we see contradiction again, hence, this variant is not workable
c) x>1
then x<x*module x
(e.g. 2<2*2)
this is the only variant that satisfies our problem, we conlcude that module x>1, hence statement (1) should be sufficient.
Could anyone tell me what is the flaw in my reasoning? Will greatly appreciate your input.
-
- Senior | Next Rank: 100 Posts
- Posts: 59
- Joined: Fri Aug 14, 2009 12:03 am
I got it!
Instead of variant a), we should consider
a.1) -1<x<0
then x<x*module x
(see, -1/2<-1/2*1/2)
satisfies our statement
a.2) x<-1
x>x*module x
(-2>-2*2)
violates statement
So, we have two possible options when x<x*module x, in particular, option a.1) and c). And no guidance as which one to chose. Hence, statement (1) is insufficient.
Instead of variant a), we should consider
a.1) -1<x<0
then x<x*module x
(see, -1/2<-1/2*1/2)
satisfies our statement
a.2) x<-1
x>x*module x
(-2>-2*2)
violates statement
So, we have two possible options when x<x*module x, in particular, option a.1) and c). And no guidance as which one to chose. Hence, statement (1) is insufficient.
For statement 1
case i. if x >0, |x| > 1 therefore x > 1.
case ii. if x < 0, |x| < 1 therefore -1< x <0.
so we have x between -1 and 0 or x > 1.
stmt 1 is insufficient.
from stmt 2 we get x <0. insufficient.
combining stmt 1 and 2 we get -1< x<0 so |x| <1.
case i. if x >0, |x| > 1 therefore x > 1.
case ii. if x < 0, |x| < 1 therefore -1< x <0.
so we have x between -1 and 0 or x > 1.
stmt 1 is insufficient.
from stmt 2 we get x <0. insufficient.
combining stmt 1 and 2 we get -1< x<0 so |x| <1.