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Re: OG 11th edition [#permalink]
12 Dec 2011, 12:31
Answer is indeed B.
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]
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? _________________
\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). 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]
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]
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 ?
Re: Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an [#permalink]
13 Sep 2015, 02:14
Expert's post
kunal555 wrote:
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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