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Re: I cant understand how the OA is??? [#permalink]
21 Sep 2010, 21:35

1

This post received KUDOS

Expert's post

If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or an irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Re: I cant understand how the OA is??? [#permalink]
21 Sep 2010, 22:50

OK now i understood the fundamentals about this question. I was thinking what if n=141/100? in that case n^2 will be 2 but n will not be integer. Thanks bunuel

Re: I cant understand how the OA is??? [#permalink]
22 Sep 2010, 17:31

Bunuel wrote:

If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n. _________________

Re: I cant understand how the OA is??? [#permalink]
22 Sep 2010, 22:37

2

This post received KUDOS

Expert's post

BalakumaranP wrote:

Bunuel wrote:

If n=p/q (p and q are nonzero integers), is n an integer?

(1) n^2 is an integer --> \(n^2\) to be an integer \(n\) must be either an integer or irrational number (for example: \(\sqrt{3}\)), (note that \(n\) can not be reduced fraction, for example \(\frac{2}{3}\) or \(\frac{11}{3}\) as in this case \(n^2\) won't be an integer). But as \(n\) can be expressed as the ratio of 2 integers, \(n=\frac{p}{q}\), then it can not be irrational number (definition of irrational number: an irrational number is any real number which cannot be expressed as a fraction a/b, where a and b are integers), so only one option is left: \(n\) is an integer. Sufficient.

(2) (2n+4)/2 is an integer --> \(\frac{2n+4}{2}=n+2=integer\) --> \(n=integer\). Sufficient.

Answer: D.

Hi Bunuel,

for (1), what if p and q are 1732 and 1000, in that casem n^2 will be an integer but not n. Isnt it possible.. 1.732^2 doesnt exactly lead to 3 but we can have such integers giving exact value of sqrt of an integer for n.

That's not true. No reduced fraction when squared can equal to an integer.

Or consider this: \(n^2=\frac{p^2}{q^2}=integer\) --> \(n=\frac{p}{q}=\sqrt{integer}\) --> now, square root of an integer is either an integer or an irrational number. But it can not be an irrational number as we have that \(\frac{p}{q}=\frac{integer}{integer}=\sqrt{integer}\) and we know that an irrational number cannot be expressed as a fraction of two integers, so \(\sqrt{integer}=n=integer\).

Re: If n=(p/q) (p and q are nonzero integers), is an integer? [#permalink]
10 Jul 2014, 06:59

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