Official solution
Given that S is a non-empty subset of R.
\[\bullet\] P : There is a rational number x \[\in\] S such that x > 0.
Now, we need to find the negation of P. Clearly, P is equivalent to saying that "There is a positive rational number in S.
So, its negation (\[\sim\] P) is "There is no positive rational number in S".
Thus, for \[\sim\] P : There exists no positive rational number in S.
\[\bullet\] \[\Leftrightarrow\] \[\sim\] P : Every rational number x \[\in\] S satisfies x \[\le\] 0.