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Square root ?? Doubt

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Square root ?? Doubt [#permalink] New post 29 Jul 2012, 05:52
If Sqrt(x) is an Integer.can x be Non integer ??
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Re: Square root ?? Doubt [#permalink] New post 29 Jul 2012, 06:07
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riteshv wrote:
If Sqrt(x) is an Integer.can x be Non integer ??


The answer is NO.

Given: \sqrt{x}=integer --> square this equation: x=integer^2. Now, since integer^2 is always an integer then x=integer^2=integer. So, \sqrt{x}=integer means that x must be an integer.

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Re: Square root ?? Doubt [#permalink] New post 30 Jul 2012, 16:07
Simply put - an integer multiplied by another integer will always give you an integer. If you want to get something that's not an integer, you're going to have to add a factor that's not an integer - something that is a fraction, or is irrational, and such.

Therefore, if rt(x) is an integer, then x is an integer.
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Re: Square root ?? Doubt   [#permalink] 30 Jul 2012, 16:07
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