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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. _________________