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

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 cannot 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 \(\sqrt{integer}=n=integer\).

Hope it's clear.

Originally answered B, but D's clear now.

Thanks Bunuel!

Tricky I picked B but makes perfect sense now. Thanks!

Hi
I am having difficulty understanding statement1.
If n^2=100 then n=10
If n^2=3 then n=sqrt(3) --> why cant we express it as p/q
sorry It may be a dumbest question, but need to clarify..Thanks

We can not represent irrational numbers as p/q as the irrational numbers are non repeating non terminating decimals and there is no way we can write p/q for such expression.

One more thing: To check if a square of a number is irrational or not, we need to check if its prime factors have "even powers"

You are concluding from statement 1 that N is an integer, but it is also given that N is of the form P/Q, a fraction. How can an Integer be a fraction?

For example, if n = 2/1, or 6/3, or 10/10, ....

By defining ‘n’ as a ratio of integers, the question data eliminates the possibility of ‘n’ being an irrational number (in simpler terms, a root). That’s valuable information, especially when you are interpreting information like the one given in statement 1.

From statement 1 alone, \(n^2\) is an integer.

This is where the question data is valuable. Had it not been told to us that n=\(\frac{p}{q}\), p and q being non-zero integers, interpreting statement 1 would take a different turn.

Because, \(n^2\) being an integer does not automatically make ‘n’ an integer; ‘n’ could be a root also.

However, in this question, we know that ‘n’ is a rational number and hence cannot be an irrational root. Therefore, if \(n^2\) is an integer, it necessarily means that n is an integer.
Statement 1 alone is sufficient to answer the question with a definite YES. Answer options B, C and E can be eliminated. Possible answer options are A or D.


From statement 2 alone, \(\frac{(2n+4)}{2}\) is an integer.

Upon simplification, we obtain n+2 as an integer. This is possible only if n is an integer itself.
Statement 2 alone is sufficient to answer the question with a definite YES. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!

Regarding statement 2:
is it possible that n = 1/3 and the equation becomes (2/3+4)/2? in this case too the result will be an integer but n is a fraction
confused here!

14/6 is not an integer, hence, it is not a valid example

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