The negation of a statement '$x \in A \cap B \rightarrow (x \in A \text{ and } x \in B)$' is
Show Hint
Remember that the negation of an "If P, then Q" statement is always "P happened, and NOT Q occurred". This pattern instantly tells you that the first part must remain unchanged, which easily leaves option (B) as the correct choice!
$x \in A \cap B \rightarrow (x \in A \text{ or } x \in B)$
$x \in A \cap B \text{ and } (x \notin A \text{ or } x \notin B)$
$x \in A \cap B \text{ or } (x \in A \text{ or } x \in B)$
$x \notin A \cap B \text{ and } (x \in A \text{ and } x \in B)$
Show Solution
The Correct Option isB
Solution and Explanation
Step 1: Name the parts.
Let $p$ be $x \in A \cap B$ and $q$ be ($x \in A$ and $x \in B$). The statement is $p \rightarrow q$.
Step 2: Negate the implication.
The negation of $p \rightarrow q$ is $p \wedge \sim q$.
Step 3: Negate q.
By De Morgan, $\sim q$ is ($x \notin A$ or $x \notin B$). So the answer is $x \in A \cap B$ and ($x \notin A$ or $x \notin B$).
\[ \boxed{\text{Option 2}} \]