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If y is a positive integer, is square_root of y
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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.
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Re: If y is a positive integer, is square_root of y
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Re: If y is a positive integer, is square_root of y
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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
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30 Jun 2012, 03:16
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
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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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Re: If y is a positive integer, is square_root of y
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01 Jul 2012, 01:24



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



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Re: If y is a positive integer, is square_root of y
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13 Aug 2015, 14:04
BunuelWhat if y=125? 125 is an integer. sqrt of 5*125 = sqrt of 625 = 25 This would make B insufficient. Am I missing something here?



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Re: If y is a positive integer, is square_root of y
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14 Aug 2015, 16:11
Mascarfi wrote: BunuelWhat if y=125? 125 is an integer. sqrt of 5*125 = sqrt of 625 = 25 This would make B insufficient. Am I missing something here? 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 \(\sqrt{y}\neq Integer\) Hope this helps.



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If y is a positive integer, is square_root of y
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14 Aug 2015, 16:48
It's clear now. Completely forgot what the question was asking while looking at number 2. thanks



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Re: If y is a positive integer, is square_root of y
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25 Aug 2018, 10:00
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