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Statement 1 says \(n^2\) is an integer. According to the solution, \((\sqrt{2})^2\) is an integer. But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx. And when we square it we would get 1.9999999 approx. We do not get an integer on squaring an irrational number but rather a value close to an integer. So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\) For this reason, i marked D. Can someone please explain why is my thinking wrong and why are we taking approximate values ?

The square root of 2 is a number (whatever it is) which when squared gives 2.
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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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04 Apr 2016, 13:00

1

This post received KUDOS

Bunuel wrote:

vwjetty wrote:

Is n an integer?

(1) n^2 is an integer

(2) n^(1/2) is an integer

Please explain. Thanks.

Is n an integer?

(1) n^2 is an integer --> not sufficient, as if \(n^2=4\) answer is YES, but if \(n^2=3\) answer is NO. <<<< n will be square root of the value. So it can be an integer or non-integer, depending on the value whether it is a perfect square or not. So, not sufficient.

(2) n^(1/2) is an integer --> \(\sqrt{n}=integer\) --> \(n=integer^2=integer\). Sufficient.

Answer: B.

I have added few more explanation for the 1st option, as i got this question wrong because of that problem. Hope you guys will not repeat the same mistake. #kudos
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stmt 1: says that n^2 is an integer. So n could be 4, which n^16 or n could be 5.477 and n^2 would be 30. This gives us two different options so insuff

stmt 2: says that sqrt n is an integer. that means n is the square of an (or integer x integer) which will always give you an integer. suff

Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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20 Jun 2013, 14:07

Stament 1 : n^2=integer n=+/- sqrt(integer) => Sqrt of integer may or may not be integer depending on whether chose integer is perfect square or not. insufficient

Statement 2: sqrt (n) = integer => Squaring both sides => n= integer^2 = some other integer (as intg*intg=intg) = sufficient

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?
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\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?

How can square of non-integer be an integer? 1.73 is a non- integer and so is its square

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?

How can square of non-integer be an integer? 1.73 is a non- integer and so is its square

The square root of 3 is a number which when squared gives 3.
_________________

Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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21 Jul 2015, 20:29

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Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]

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12 Sep 2015, 07:28

Statement 1 says \(n^2\) is an integer. According to the solution, \((\sqrt{2})^2\) is an integer. But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx. And when we square it we would get 1.9999999 approx. We do not get an integer on squaring an irrational number but rather a value close to an integer. So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\) For this reason, i marked D. Can someone please explain why is my thinking wrong and why are we taking approximate values ?

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