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Re: If y is a positive integer, is square_root of y [#permalink]
30 Jun 2012, 02:50

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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]
30 Jun 2012, 03:08

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]
30 Jun 2012, 03:16

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]
30 Jun 2012, 17:22

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]
01 Jul 2012, 01:24

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. _________________