Sets, Relations & Functions
Subsets
Grade 11
Question:
<p>Let \(S\) be a non-empty subset of \(\mathbb{R}\). 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 \(P\)?</p>
<p>There is no rational number \(x \in S\) such that \(x \leq 0\).</p>
<p>Every rational number \(x \in S\) satisfies \(x \leq 0\).</p>
<p>\(x \in S\) and \(x \leq 0 \Rightarrow x\) is not rational.</p>
<p>There is a rational number \(x \in S\) such that \(x \leq 0\).</p>
Step-by-Step Solution
Key Concept: The negation of an existential statement 'there exists x such that P(x)' is the universal statement 'for all x, not P(x)'. Here, negate both the existence quantifier AND the condition (x ∈ S AND x is rational AND x > 0).
<p><strong>Step 1:</strong> Identify the original statement P: ∃x ∈ S (x is rational ∧ x > 0)</p><p><strong>Step 2:</strong> Apply negation rules: ¬(∃x, P(x)) ≡ ∀x, ¬P(x)</p><p><strong>Step 3:</strong> The negation becomes: ∀x ∈ S, if x is rational then x ≤ 0</p><p><strong>Step 4:</strong> In words: "For every rational number x in S, we have x ≤ 0" or equivalently "All rational numbers in S are non-positive" or "There is no positive rational number in S"</p><p><strong>Step 5:</strong> This is equivalent to: "Every rational number in S is either zero or negative"</p><p>∴ <strong>Answer: B</strong> (The negation states that all rational numbers in S are non-positive, i.e., ≤ 0)</p>
Correct Answer: B