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If y is a positive integer, is square_root of y

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If y is a positive integer, is square_root of y [#permalink] New post 30 Jun 2012, 02:46
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If y is a positive integer, is \sqrt{y} an integer?

(1) \sqrt{4y} is not an integer.
(2) \sqrt{5y} is an integer.
[Reveal] Spoiler: OA

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

Answer: D.

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Re: If y is a positive integer, is square_root of y [#permalink] New post 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.
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Re: If y is a positive integer, is square_root of y [#permalink] New post 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).

Hope it's clear.
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Re: If y is a positive integer, is square_root of y [#permalink] New post 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!
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Expert Post
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Re: If y is a positive integer, is square_root of y [#permalink] New post 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.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If y is a positive integer, is square_root of y [#permalink] New post 12 Jun 2013, 03:27
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If y is a positive integer, is square_root of y   [#permalink] 12 Jun 2013, 03:27
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