Animated Solution for Mathematics - Sets and Relations: Let S be a non-empty subset of R. Consider the following statement : P : There is a rational number $x\in S$ such that $x>0$. Which of the following statements is the negation of the statement P ?

The statement provided is $P:\exists x\in (S\cap \mathbb{Q}),x>0$.

This statement asserts that within the intersection of set $S$ and the set of rational numbers $\mathbb{Q}$, there exists at least one element $x$ that is strictly positive. Our objective is to determine the negation, denoted as $ eg P$.

The first step in our journey is to understand the power of the quantifier. The symbol $\exists$ stands for "there exists."

When we negate an existential statement, we are essentially asserting that the existence of such an element is impossible. If it is impossible for "at least one" to exist, then it must be true that for all elements, the condition fails.

Thus, the existential quantifier $\exists$ transforms into the universal quantifier $\forall$. This is the first, and perhaps most critical, shift in our logical perspective.

Now, we turn our attention to the condition $x>0$. If we are negating the claim that $x$ is strictly greater than zero, we must consider the entire real number line.

If a number is not strictly greater than zero, it must be either negative or exactly zero. Therefore, the negation of $x>0$ is $x\le 0$.

By combining these two insights—the transformation of the quantifier and the negation of the condition—we arrive at our final logical destination:

This translates to: "Every rational number $x\in S$ satisfies $x\le 0$."

This is the beauty of logic—it is not about guessing; it is about the systematic, step-by-step dismantling of a statement until only the truth remains. Keep this precision in your toolkit, and you will find that even the most complex problems become clear.