MHT CET · Maths · Sets and Relations
Let \(S\) be a non-empty subset of \(\mathbb{R}\). Consider the following statement:
p : There is a rational number \(x \in \mathrm{~S}\) such that \(x\gt0\).
Which of the following statements is the negation of the statement p?
- A There is a rational number \(x \in \mathrm{~S}\) such that \(x \leq 0\).
- B There is no rational number \(x \in \mathrm{~S}\) such that \(x \leq 0\).
- C Every rational number \(x \in \mathrm{~S}\) satisfies \(x \leq 0\).
- D \(x \in \mathrm{~S}\) and \(x \leq 0 \Rightarrow x\) is not a rational number.
Answer & Solution
Correct Answer
(C) Every rational number \(x \in \mathrm{~S}\) satisfies \(x \leq 0\).
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Given statement is } \\
& \\
& \exists x \in \mathrm{~S} \text {, such that } x\gt0 \\
& \therefore \quad \sim(\exists x \in \mathrm{~S}, \text { such that } x\gt0) \\
& \equiv \forall x \in \mathrm{~S}, x \leq 0
\end{aligned}\)
i.e., Every rational number \(x \in S\) satisfies \(x \leq 0\).
& \text { Given statement is } \\
& \\
& \exists x \in \mathrm{~S} \text {, such that } x\gt0 \\
& \therefore \quad \sim(\exists x \in \mathrm{~S}, \text { such that } x\gt0) \\
& \equiv \forall x \in \mathrm{~S}, x \leq 0
\end{aligned}\)
i.e., Every rational number \(x \in S\) satisfies \(x \leq 0\).
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