Is n an integer? (1) n^2 is an integer (2) n^(1/2) is an

If we take \(\sqrt{3}\) = 1.732050807568877
and square it, we still get 2.999999999999
which is not an integer. That's the reason I marked D

Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 Please read.

\(\sqrt{3}=1.732050807568877293527446341505872366942805253810380628055806...\). It's an irrational number, it goes on forever. Anyway, what's your question?

How can square of non-integer be an integer?
1.73 is a non- integer and so is its square

The square root of 3 is a number which when squared gives 3.

Statement 1 says \(n^2\) is an integer.
According to the solution, \((\sqrt{2})^2\) is an integer.
But \(\sqrt{2}\) is an irrational number which is equal to 1.41421356237 approx.
And when we square it we would get 1.9999999 approx.
We do not get an integer on squaring an irrational number but rather a value close to an integer.
So, if we exclude irrational numbers, we should get integer value for \(\sqrt{n^2}\)
For this reason, i marked D.
Can someone please explain why is my thinking wrong and why are we taking approximate values ?

The square root of 2 is a number (whatever it is) which when squared gives 2.

Statement 1: n^2 = Integer

i.e. n^2 = 1, 2, 3, 4, 5....
i.e. \(n^2 = 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}\)....
NOT SIFFICIENT

Statement 2: \(\sqrt{n}\) = 1, 2, 3, 4, 5....

i.e. n = 1, 4, 9, 16, 25...
SUFFICIENT

Answer: Option B

Actually why you dont ROOT both sides in (1)?

Is this because we dont know if integer is negative or not?

I mean n^2 = int and by ROOTING both sides you get n = sqrt(int) which then implies that since n can be any value is not sufficient (?). I find all this smart number approach unnecessary for this question.

Your explanation is fine if you can easily understand n = sqrt(int) and sqrt(int) may or may NOT be an Integer therefore 1st statement is NOT SUFFICIENT

However it's not equally easy for everyone to understand as the natural bias of our mind makes us think that "if n is an integer then n^2 will also be an Integer therefore if n^2 is an Integer then n also must be an Integer" which would lead to incorrect answers.

Being a trainer I keep asking this question to the students and around 70% students get this question wrong who just begin to learn DS

Also, Smart Number approach is definitely a very good approach however the Smart number approach in isolation isn't as good enough in isolation as the Algebraic approach in isolation isn't good enough

I suggest that use of both Smart Number and Algebraic approach is the best way to perform best in any aptitude test.

Hi, Bunel how could n2 could be 3 in your assumption. I think it has to be 1,2,4,9 which is sufficient.

Why cannot n^2 be 3? In this case \(n=\sqrt{3}\) or \(-\sqrt{3}\), so in this case n is an irrational number.

For more on number theory (for example on irrational numbers) check the following link: https://gmatclub.com/forum/math-number- ... 88376.html

Hope it helps.

We need to determine whether n is an integer.

Statement One Alone:

n^2 is an integer.

If n^2 is an integer, n may or may not be an integer. For instance, if n^2 = 4, then n is an integer (since n = 2 or -2). However, if n^2 = 5, then n is not an integer (since n = √5 or -√5). Statement one is not sufficient to answer the question.

Statement Two Alone:

√n is an integer.

In order for √n to be an integer, n must be an integer. This is because n = (√n)^2, and any integer squared is also an integer. Statement two is sufficient to answer the question.

Answer: B

A proper way to answer any DS question

Check the video solution of this question

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Interesting but cunning question
Statement 1
( \(\sqrt{2} \))^2 = 2 yet, \(\sqrt{2}\) is not an integer
Conversely, 2^2 =4 and 2 is an integer.

Statement 2
is sufficient by its own.

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