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Disagree with OA if x^2 is a positive integer, is x a [#permalink ]
29 Jul 2005, 17:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum. Disagree with OA
if x^2 is a positive integer, is x a positive integer?
1. sqrt x^2 is an integer
2. sqrt x^2 =x
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Re: positive integer [#permalink ]
29 Jul 2005, 19:20
mirhaque wrote:
Disagree with OA if x^2 is a positive integer, is x a positive integer? 1. sqrt x^2 is an integer 2. sqrt x^2 =x
Let us take x^2 = 16 so x = +/- 4
1) +/- 4 both are integers so A does not help
2) agaiin +/- 4 does not help
both together does not help either. So E? THis is too easy or i am missing some thing..
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Agree with E
From stem X can be +/-
From A) same case
From B) identical to A)
When combined it doesn't help much cause in both X is squared.
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I'l go with C...
X^2=integer.... X=integer?
1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff
2) Sqrt(X^2)=X...
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff....
combining them together...Sufficient...
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fresinha12 wrote:
I'l go with C... X^2=integer.... X=integer? 1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff 2) Sqrt(X^2)=X... OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff.... combining them together...Sufficient...
if 2) Sqrt(X^2)=X..., How does it mean X is positive? can you eloborate on that..
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it is a question of denotaion of square root
it seems to me: sqrt(4) = +/- 2
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I'm also getting (E) on this one. What's the OA?
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Quote:
sqrt(x^2)=integer....i.e sqrt (16)= 4, x could be +- 4...Insuff
the above expression is not correct. Sqrt(X^2) is always +ve.
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OA is C. But I think it's E
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read my explanation...it should be C...
sqrt(x^2) is always positive....
mirhaque wrote:
OA is C. But I think it's E
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fresinha12 wrote:
read my explanation...it should be C...
sqrt(x^2) is always positive....
mirhaque wrote:
OA is C. But I think it's E
Hi Freisinha12,
Can you eloborate why sqrt(x^2) is always positive, i am not getting it...
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fresinha12 wrote:
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff.... combining them together...Sufficient...
Agree with fresinha12.
sqrt(x^2) = |x|.
and |x | = x only when x is positive or 0.But from the stem x^2 is positive so x can't be 0. So x is definitely positive.
and from (1) x is an integer.
So using (1) and (2) x is a positive integer.
HMTG.
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i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?
HMTG, how do you know that sqrt x^2=|x| ? x can denote a negative number as well.
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christoph wrote:
i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?
mirhaque wrote:
OA is C. But I think it's E
C cannot be OA because the question asking here is whether x is a positive integer or not?
statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.
from ii, sqrt x^2 =x. here x could be a positive integer or a positive nuber (fraction). so not suff.
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HIMALAYA wrote:
statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.
sqrt() always refers to the nonnegative root.
Let x = -3. x^2 = 9, sqrt(x^2) = 3, which is an integer. But x is not a positive integer.
This is why we need S2 as well.
C is correct.
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I would pick C.
B tells us that X is positive:
B tells us "sqrt x^2 =x" therefore "x= sqrt x^2 "
we know that sqrt can be only positive
therefore x is positive
A tells us that X is integer
both concludes that X is positive and integer