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Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 03:50

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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.

Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 04:08

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. _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 04:16

Expert's post

metallicafan wrote:

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.

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (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).

Re: If y is a positive integer, is square_root of y [#permalink]

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30 Jun 2012, 18:22

Bunuel wrote:

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (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).

Hope it's clear.

Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational? Thanks! _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: If y is a positive integer, is square_root of y [#permalink]

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01 Jul 2012, 02:24

Expert's post

metallicafan wrote:

Bunuel wrote:

The point is that \(\sqrt{5}\) is an irrational number, and the decimal representation of an irrational number never repeats or terminates (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).

Hope it's clear.

Thank you Bunuel! An additional question: all the square roots (or roots in general) of prime numbers are irrational? Thanks!

Since no prime can be a perfect square (or perfect cube, ...), then \(\sqrt{prime}=irrational\). _________________

Re: If y is a positive integer, is square_root of y [#permalink]

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12 Aug 2015, 19:52

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No, your calculations are fine and are consistent with the question. If you have y = 125, \(\sqrt{125}\) \(\neq\) Integer and thus get an unambiguous "no" making statement 2 sufficient. The original question is

"\(\sqrt{y}\) an integer?" So getting an unambiguous yes or no will be sufficient. Per Bunuel's solution, it is shown that y can not be a perfect square leading to \(\sqrt{y}\neq Integer\)

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