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# Is square root of x*y an integer? 1). Root x is not an

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Is square root of x*y an integer? 1). Root x is not an [#permalink]

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07 Jan 2006, 16:58
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Is square root of x*y an integer?
1). Root x is not an integer
2). Root y is not an integer
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07 Jan 2006, 17:32
For sqrt(xy) to be an integer:

xy must be an integer AND
either xy must be a perfect square or x and y must each be perfect squares.

(1) and (2) don't give all this information.

I'm not sure about my explanation though .. does anyone agree?

Thanks
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07 Jan 2006, 19:45
I would say D.

sqrt(x*y) = sqrt(x)*sqrt(y)

From statement 1, we know that root x is not an integer. Therefore, no matter what sqrt(y) turns out to be, when it gets multiplied by the value of sqrt(x), the product will not be an integer in any case.

Same logic for statement 2.

Does anyone see a flaw in this logic ? Thanks !
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07 Jan 2006, 20:10
pmenon wrote:
I would say D.

sqrt(x*y) = sqrt(x)*sqrt(y)

From statement 1, we know that root x is not an integer. Therefore, no matter what sqrt(y) turns out to be, when it gets multiplied by the value of sqrt(x), the product will not be an integer in any case.

Same logic for statement 2.

Does anyone see a flaw in this logic ? Thanks !

Your logic will not be valid, since we don't know about "sqrt(y)". It can be an integer or non-integer that will multiply non-integer "sqrt(x)" & make result an integer or non-integer! Say for example...

what if,
sqrt(x) = 1.5 & sqrt(y)=2 ? YES, result is integer
AND
sqrt(x) = 1.5 & sqrt(y)=1.5 ? NO, result is not an integer

It should be 'E'!
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07 Jan 2006, 20:16
vivek123 wrote:
pmenon wrote:
I would say D.

sqrt(x*y) = sqrt(x)*sqrt(y)

From statement 1, we know that root x is not an integer. Therefore, no matter what sqrt(y) turns out to be, when it gets multiplied by the value of sqrt(x), the product will not be an integer in any case.

Same logic for statement 2.

Does anyone see a flaw in this logic ? Thanks !

Your logic will not be valid, since we don't know about "sqrt(y)". It can be an integer or non-integer that will multiply non-integer "sqrt(x)" & make result an integer or non-integer! Say for example...

what if,
sqrt(x) = 1.5 & sqrt(y)=2 ? YES, result is integer
AND
sqrt(x) = 1.5 & sqrt(y)=1.5 ? NO, result is not an integer

It should be 'E'!

Vivek, in your first example, sqrt of (1.5 x 2) is still not an integer. What I am saying is that if we know already that ONE of the roots is not an integer, it wont matter if the other root comes out to an integer or not, since the root of the PRODUCT wont be an integer, and thats what the question is asking. Right ?
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07 Jan 2006, 20:27
let me give an example. say x=3 and y=4

sqrt(xy)=sqrt(12)=sqrt(3)*sqrt(4).

Sqrt(4) is an integer, but sqrt(3) is not. Therefore, there is no way that sqrt(xy) will be an integer itself.

This comes from statement 1 when we are told that root x is not an integer.

I hope this makes sense, I really dont see it any other way
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07 Jan 2006, 20:30
pmenon wrote:
vivek123 wrote:
pmenon wrote:
I would say D.

sqrt(x*y) = sqrt(x)*sqrt(y)

From statement 1, we know that root x is not an integer. Therefore, no matter what sqrt(y) turns out to be, when it gets multiplied by the value of sqrt(x), the product will not be an integer in any case.

Same logic for statement 2.

Does anyone see a flaw in this logic ? Thanks !

Your logic will not be valid, since we don't know about "sqrt(y)". It can be an integer or non-integer that will multiply non-integer "sqrt(x)" & make result an integer or non-integer! Say for example...

what if,
sqrt(x) = 1.5 & sqrt(y)=2 ? YES, result is integer
AND
sqrt(x) = 1.5 & sqrt(y)=1.5 ? NO, result is not an integer

It should be 'E'!

Vivek, in your first example, sqrt of (1.5 x 2) is still not an integer. What I am saying is that if we know already that ONE of the roots is not an integer, it wont matter if the other root comes out to an integer or not, since the root of the PRODUCT wont be an integer, and thats what the question is asking. Right ?

pmenon, I think you are mistaking it...
You are taking sqrt of (1.5 x 2) again?

I said, sqrt(x) = 1.5 & sqrt(y)=2
&
sqrt(x*y) = sqrt(x)*sqrt(y) = 1.5x2 = 3.
is this correct?
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07 Jan 2006, 20:45
AAAAAAAAAAHHH

Sorry for the confusion folks. Vivek, youre right, for some reason i was taking the root of 3 again !!!

I change my answer to E
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07 Jan 2006, 20:49
joemama142000 wrote:
Is square root of x*y an integer?
1). Root x is not an integer
2). Root y is not an integer

should be E.
from i, if x = 3 and y = 2, sqrt(xy) = is not an integer.
if x = 3 and y = 3, sqrt(xy) = is an integer.

from ii, if y = 3 and x = 2, sqrt(xy) = is not an integer.
if y = 3 and x = 3, sqrt(xy) = is an integer.

from i and ii, if x = 2, y = 3, no. if x=y=3, yes. so insuff.
Re: DS square rts   [#permalink] 07 Jan 2006, 20:49
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