The problem deals with the issue of square roots - and how they relate to integers. When you think of integers you have to consider integers, fractions, decimals and square roots.

if n^2 is an integer then n could be an integer or a square root - for instance (SQRT2)^2=2 so you can't be sure that n is an integer.

however if SQRT(n) is an integer that means that n is a perfect square - which means that n is the product of 2 integers therefor you know that n must be an integer.

Answer is B.

[quote="tpr-becky"]The problem deals with the issue of square roots - and how they relate to integers. When you think of integers you have to consider integers, fractions, decimals and square roots.

if n^2 is an integer then n could be an integer or a square root - for instance (SQRT2)^2=2 so you can't be sure that n is an integer.

however if SQRT(n) is an integer that means that n is a perfect square - which means that n is the product of 2 integers therefor you know that n must be an integer.

Answer is D.[/quote]

Thanks for the explanation. I think you mean the answer is B and not D.

Now the question,which I have.

I understand that sqrt(2) is not an integer but I thought in the following way.

value of sqrt(2) in terms of decimal will be

1.4142135623730950488016887242097 in fact it will contain some other digits as well and if we do a square of the same then it will result in 1.999999999999999999,which is not an integer and the same will be true for the sqrt of all the numbers

Am I wrong

you are correct for my answer (typo) but on the other fact you are only wrong because you have rounded off the number - yes the SQRT of 2 is a long decimal but by the very fact of it being the SQRT of 2 when you square it you will get 2. That is why when you are dealing with these type of questions it is important to think about integers, fractions, decimals and square roots - becuase square roots act in a different manner.

Thanks