Bunuel wrote:
metallicafan wrote:
If y is a positive integer, is \(\sqrt{y}\) an integer?
(1) \(\sqrt{4y}\) is not an integer.
(2) \(\sqrt{5y}\) is an integer.
If \(y\) is a positive integer is \(\sqrt{y}\) an integer?Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.
(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.
(2) \(\sqrt{5*y}\) is an integer --> \(y\) can not be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.
Answer: D.
Hope it helps.
Thank you Bunuel. I arrived to the same conclusion. However, I have a doubt:
Ok, we know that \(x\sqrt{5}\).
But how can we be so sure that x is not for example a a huge number like 10*10^10000000.... to make \(x\sqrt{5}\) a integer?
Maybe, I am forgeting a concept. Please, your help.
(irrational numbers are not terminating decimals). So, \(integer*irrational\neq{integer}\) (no matter how large x is, \(x\sqrt{5}\) will never be an integer).