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by neelgandham » Thu Dec 01, 2011 8:17 am
sqrt(z)>0, Is sqrt(z)>z ? The question can be rephrased to Is 0<z<1 ?
1) z>0
z can be 0.5 or 8.So Insufficient!
2) sqrt(z) < 1
Given: sqrt(z)>0(question) and sqrt(z) < 1 (option).So, 0<sqrt(z)<1.
Implies 0<z<1 Sufficient!

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by pemdas » Thu Dec 01, 2011 8:27 am
the knowledge of number properties should hint us here, that sqrt(z) can be > z if and only if 0<z<1, so aiding your question we're asked: 0<z<1?
st(1) z>0 this is Not Sufficient
st(2) sqrt(z)<1 is possible only if z<1, hence Sufficient and we answer Yes

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sohrabkalra wrote:sqrt(z) > 0 , is sqrt(z) >z ?

1) z>0
2) sqrt(z) < 1

Although the question seems fairly simple , i have a doubt !
Will raise that and provide OA after a few replies !
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by sohrabkalra » Thu Dec 01, 2011 8:35 am
My question is that Question stem does not (gmat will i guess) tell us that Z>0 which is a necessary condition for statement 2 to be sufficient(as it is in the sqrt) so should we not be going with C? or we just assume that z>0 is implied !
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by neelgandham » Thu Dec 01, 2011 8:39 am
sohrabkalra wrote:My question is that Question stem does not (gmat will i guess) tell us that Z>0 which is a necessary condition for statement 2 to be sufficient(as it is in the sqrt) so should we not be going with C? or we just assume that z>0 is implied !
sqrt(z) > 0 , is sqrt(z) >z ?
The question quotes that the value of sqrt(z)>0, implies sqrt(z) is defined and is only possible if z > 0
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by pemdas » Thu Dec 01, 2011 8:41 am
per GMAT math conventions the sqroot of negative numbers is not defined
it's poossible to measure negative number roots in math but GMAT avoids it by setting its rules in Math conventions. Hence z>0 is must for sqroot(z)>0
sohrabkalra wrote:My question is that Question stem does not (gmat will i guess) tell us that Z>0 which is a necessary condition for statement 2 to be sufficient(as it is in the sqrt) so should we not be going with C? or we just assume that z>0 is implied !
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by pemdas » Thu Dec 01, 2011 8:50 am
also note and don't confuse sqroot(z) can be negative too. Here z is considered [sqroot(z)]^2 and it can be only positive, whereas sqroot(z) can be negative.

z=a^2
sqroot(z)=a

|a|= -a or +a
a^2 is always positive

z can be negative in sqroot(z), but then z is complex number and GMAT relies only on real numbers
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by AbhiJ » Fri Dec 02, 2011 12:48 am
neelgandham wrote:
sohrabkalra wrote:My question is that Question stem does not (gmat will i guess) tell us that Z>0 which is a necessary condition for statement 2 to be sufficient(as it is in the sqrt) so should we not be going with C? or we just assume that z>0 is implied !
sqrt(z) > 0 , is sqrt(z) >z ?
The question quotes that the value of sqrt(z)>0, implies sqrt(z) is defined and is only possible if z > 0
Adding further even if in the q-stem we are not mentioned that sqrt(z) > 0 , the result will be the same.

From (2) we know that sqrt(z) < 1 , now z has to be > 0, because if z < 0, then sqrt(z) will not be a real number but an imaginary number. The imaginary cannot be compared with real number 1.
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