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# If x is a positive integr is x^(1/2) (Sq root of x) is an

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If x is a positive integr is x^(1/2) (Sq root of x) is an [#permalink]  22 Aug 2005, 11:27
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If x is a positive integr is x^(1/2) (Sq root of x) is an integer?
1. (4x)^(1/2) is an intger
2. (3x)^(1/2) is not integer.

from 1
let (4x)^(1/2) =k where k is an integer

x^(1/2)=k/2 . so does value of root x depends on odd/even. What wrong am I doing.
What is the way to approach this sort of problem.
S

Source og 222
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Regards, S

Last edited by saurya_s on 22 Aug 2005, 11:43, edited 3 times in total.
Manager
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Could you use more brackets or describe the question more precisely please
Senior Manager
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A ?

1. (4x)^(1/2) is an intger
=> x^(1/2) has to be integer

2. (3x)^(1/2) is not integer.
=> since 3^(1/2) is not integer already, we cannot decide the property of
x
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I choose A too..Inline with qpoo..
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Re: DS IS Root x an integer [#permalink]  22 Aug 2005, 17:28
saurya_s wrote:
If x is a positive integr is x^(1/2) (Sq root of x) is an integer?
1. (4x)^(1/2) is an intger
2. (3x)^(1/2) is not integer.

from i, (4x)^(1/2) = (4)^(1/2) (x)^(1/2) = +or- 2 (x)^(1/2) is an integer. so sqrt (x) is an integer.

from ii, (3x)^(1/2) is not integer means sqrt(x) could or couldnot be an integer because sqrt(3) doesnot give any integer.

so only A is correct.
Director
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Himalaya, can you know what is conceptual error in my method?
Thanks mate
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hmm what if X=1/4

then Sqrt (4X)=Sqrt(1)=1,

or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

I would go with E on this...
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fresinha12 wrote:
hmm what if X=1/4

then Sqrt (4X)=Sqrt(1)=1,

or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

I would go with E on this...

Shud be "A"...X is +ve integer can't be 1/4.

Saurya, u r correct in ur approach except that ur variable K has to be an even integer as the LHS has a factor of 2.
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banerjeea_98 wrote:
fresinha12 wrote:
hmm what if X=1/4

then Sqrt (4X)=Sqrt(1)=1,

or Sqrt(4X)=2Sqrt(x), if this is an integer then Sqrt(x) is either an integer or is equal to 1/2 or Y/2....in that case its not an integer... again I am assuming irrational numbers can be viewed as fractions....there is no precise defintion that they are not...

I would go with E on this...

Shud be "A"...X is +ve integer can't be 1/4.

Saurya, u r correct in ur approach except that ur variable K has to be an even integer as the LHS has a factor of 2.

So baner, what is wrong with my approach. So, answer should be E with my way as we can't say if k is odd or even?
Can you plz explain this.
S
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saurya_s wrote:
Himalaya, can you know what is conceptual error in my method? Thanks mate
saurya_s wrote:
If x is a positive integr is x^(1/2) (Sq root of x) is an integer?
1. (4x)^(1/2) is an intger
2. (3x)^(1/2) is not integer.
from 1, let (4x)^(1/2) =k where k is an integer. x^(1/2)=k/2. so does value of root x depends on odd/even. What wrong am I doing. What is the way to approach this sort of problem.

surya, it doesnot matter. if (4x)^(1/2) is an intger, x^(1/2) must be an intger because sqrt(4) is an integer (+or-2).

you did x^(1/2)=k/2, which still must give you an integer. if you are supposing x as fraction (e.g1/4 or 0.25), which is not possible in this case because x is a positive integer as per the statement. so sqrt(x) must have an integer.

hope that helps.
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