Let S be non empty subset of R. consider the following statement:
P: There is rational number $x$ such that $x > 0$
Which of the following statement is negation of statement P:
A) There is no rational number $x \in S$ such that $x \leqslant 0$
B) Every rational number $x \in S$ satisfies $x \leqslant 0$
C) $x \in S$ and $x \leqslant 0$ $x$ is not a rational number
D) There is a rational number $x \in S$ such that $x \leqslant 0$

Hint:At first read the statement carefully that is given in the question. Then do its negation such that the original statement is also considered. Match with the options given and choose the correct one.

Complete step-by-step answer:
First let us understand the question. The question says that S is a non-empty subset of R, which means S contains at least one real number.
Now the statement P says there is a rational number such that $x > 0$ which means $x$ is a positive rational number belonging to S.
Now we have to take its negation considering statement P.
The opposite of the given mathematical statement is said to be negation.
Hence we will take the opposite of P.
While taking P into consideration.
This means we can’t change $x$ is a rational number to $x$ is not a rational number.
But we can change $x > 0$ to $x \leqslant 0$ i.e. negation.
Since means we can’t change $x$ is a rational number to $x$ is not a rational number
Hence option A and C are eliminated but we have to change it
Hence we will change it to “every rational number $x \in S$”
The negation of statement P is
Every rational number such that $x \in S$ such that $x \leqslant 0$.

So, the correct answer is “Option B”.

Note:In such questions involving sets, negation and numbers, we need to be very careful as a small mistake can be committed at any point. As in the above equation any one can change “$x$ is a rational number to $x$ is not a rational number”. Hence we need to take care of such mistakes.