State whether the following statement is true or false. If the statement is false,rewrite it correctly.
If $A = \{1, 2\}$ and $B = \{3, 4\}$,then $A \times \{B \cap \varnothing\} = \varnothing$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) The statement is True.
Step $1$: Evaluate the intersection $B \cap \varnothing$. Since the intersection of any set with the empty set is the empty set,$B \cap \varnothing = \varnothing$.
Step $2$: Substitute this into the expression: $A \times \{B \cap \varnothing\} = A \times \{\varnothing\}$.
Step $3$: The Cartesian product $A \times \{\varnothing\}$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in \{\varnothing\}$. This results in $\{(1, \varnothing), (2, \varnothing)\}$.
Wait,let us re-evaluate the original expression: $A \times \{B \cap \varnothing\}$.
Since $B \cap \varnothing = \varnothing$,the expression becomes $A \times \{\varnothing\}$.
This is $NOT$ equal to $\varnothing$. Therefore,the statement is False.
Corrected statement: If $A = \{1, 2\}$ and $B = \{3, 4\}$,then $A \times \{B \cap \varnothing\} = \{(1, \varnothing), (2, \varnothing)\}$.

Explore More

Similar Questions

If $P = \{a, b, c\}$ and $Q = \{r\}$,find the sets $P \times Q$ and $Q \times P$. Are these two products equal?

Let $A = \{a, b, c\}$ and $B = \{1, 2\}$. Then the relation $R$ defined from set $A$ to set $B$ is a subset of which of the following?

Let $A = \{1, 2, 3\}$,$B = \{3, 4\}$,and $C = \{4, 5, 6\}$. Find $A \times (B \cap C)$.

If $G = \{7, 8\}$ and $H = \{5, 4, 2\}$,find $G \times H$ and $H \times G$.

If $A = \{0, 1\}$ and $B = \{1, 0\}$,then $A \times B$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo