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 gt 0`
The negation of p is

Options

A. There is a rational number `x in S` such that `x le 0`
B. There is no rational number `x in S` such that `x le 0`
C. Every rational number `x in S` such that `x le 0`
D. `x in S` and `x le 0 implies x` is not rational

Answer

To find the negation of the statement \( p \): "There is a rational number \( x \in S \) such that \( x > 0 \)", we will follow these steps:

### Step 1: Understand the Original Statement
The original statement \( p \) can be expressed in logical symbols as:
\[ p: \exists x \in S, \, (x > 0) \]
This reads as "there exists an \( x \) in the set \( S \) such that \( x \) is greater than 0".

### Step 2: Negate the Statement
To negate the statement \( p \), we need to apply the rules of negation to the existential quantifier. The negation of "there exists" (\( \exists \)) is "for all" (\( \forall \)). Therefore, the negation of \( p \) is:
\[ \neg p: \forall x \in S, \, (x \leq 0) \]
This means "for all \( x \) in the set \( S \), \( x \) is less than or equal to 0".

### Step 3: Write the Negation in Words
Now, we can express the negation in a more verbal form:
"The statement \( p \) is negated to say that every rational number \( x \) in the set \( S \) is less than or equal to 0".

### Final Answer
Thus, the negation of the statement \( p \) is:
\[ \forall x \in S, \, (x \leq 0) \]
or in words: "Every rational number \( x \) in \( S \) is less than or equal to 0".

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