What are the values of $n$ that satisfy the condition $1/|n| > n$?
(a) $0<n<1$ (and) $ -\infty < n < 0$
(b) $0 < n < \infty$ (or) $-\infty < n < -1$
(c) $0 < n < 1$ (and) $-1 < n < 0$
(d) $-\infty < n < 0$ (or) $0 < n < 1$
(e) $0<n<1$ (or) $-\infty < n < 0$
My problem is not: why the answer is option-a. My question is why we should choose an option that has AND.
To me, option e is correct. In option e, we find OR instead of AND. Option e seems correct to me because $0<n<1$ alone can satisfy the $1/|n| > n$. And $-\infty<n<0$ alone can satisfy the condition as well. To me AND seems appropriate when neither $0<n<1$ nor $-\infty<n<0$ alone can satisfy the condition. If in order to satisfy the condition, both must be applied at a time, only then usage of AND seems correct to me.
The fact why AND are appropriate here was discussed in the site from which I have collected this question. But their explanations don't make any sense to me. I cannot understand their explanation.
Another question is: what are the differences between option d and option e?
The question asks us to consider the set $A =\{n:\frac{1}{|n|} > n\}$. It would have been helpful for the question to state that $n \neq 0$ but we will assume this.
By considering the cases $n>0$ (which leads to $n^2 < 1$ so $0 < n < 1$) and $n<0$ (which leads to $n^2 > -1$ which is true for all $n$), we can see that:
$A = B \cup C$ where $B=\{n: 0 < n <1 \}$ and $C=\{n:n < 0 \}$
It seems slightly odd to write "$-\infty < n < 0$". If we are assuming $n$ is real that is normally just written $n <0$.
Set $A$ is made up of set $B$ and set $C$. So, to consider the options:
The values that $n$ can take are those in set $B$ and in set $C$. So that is option (a).
If the question had said "what is the possible value of $n$" then we would say
$n \in B \cup C$ so $n \in B$ or $n \in C$, and (e) would be the better option.
But I feel it's a bit of nit-picking distinction!