   Question Answers

# If $\text{A=}\left\{ 1,2,3 \right\},\text{ B=}\left\{ 4 \right\},\text{ C=}\left\{ 5 \right\}$ then, verify that $\text{A}\times \left( \text{B-C} \right)=\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)$

Hint: To solve this question, we will use these formulas. The Cartesian product of two sets P and Q is given by,
$\text{P}\times \text{Q=}\left\{ \left( p,q \right) \right\}:p\in P,q\in Q$
And subtraction of two sets P and Q is given by,
$P-Q=\left\{ x:x\in P\text{ and x}\notin \text{Q} \right\}$

Given that, $\text{A=}\left\{ 1,2,3 \right\},\text{ B=}\left\{ 4 \right\},\text{ C=}\left\{ 5 \right\}$
We will separately calculate $\left( \text{A}\times \text{B} \right),\left( \text{A}\times C \right)\text{ and }\left( \text{B-C} \right)$ first. The Cartesian product of two sets P and Q is given by;
$\text{P}\times \text{Q=}\left\{ \left( p,q \right) \right\}:p\in P,q\in Q$
And subtraction of two sets P and Q is given by,
$P-Q=\left\{ x:x\in P\text{ and x}\notin \text{Q} \right\}$
The possible combination of $A\times B$ using the above formula is given by;
$A\times B=\left\{ \left( 1,4 \right),\left( 2,4 \right),\left( 3,4 \right) \right\}$
The possible combination of $A\times C$ using the above formula is given by;
$A\times C=\left\{ \left( 1,5 \right),\left( 2,5 \right),\left( 3,5 \right) \right\}$
The possible value of B-C using the above formula of difference of two set is given by;
$B-C=\varnothing \text{ (empty set)}$
Now, the value of $A\times \left( B-C \right)$ using the above formula is $\varnothing$ as there is no element in B-C. So, $A\times \left( B-C \right)$ is also an empty set.
Now, consider $\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)$
Using the formula of difference of set stated above, we get:
$\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)=\left\{ \left( 1,4 \right),\left( 2,4 \right),\left( 3,4 \right) \right\}-\left\{ \left( 1,5 \right),\left( 2,5 \right)\left( 3,5 \right) \right\}=\varnothing$
As there is no common element. Hence, the value of $\text{A}\times \left( \text{B-C} \right)=\left( \text{A}\times \text{B} \right)-\left( \text{A}\times C \right)=\varnothing$
Hence verified.

Note: The biggest possibility of mistake which the students make while solving this question is considering $\text{A}\times \left( \text{B-C} \right)=\left\{ \left( 1,0 \right),\left( 2,0 \right),\left( 3,0 \right) \right\}$ This step is wrong, because 0 and $\varnothing$ are different elements, here $B-C=\varnothing \text{ (empty set)}$ and not 0. Therefore, $\text{A}\times \left( \text{B-C} \right)=\left\{ \left( 1,0 \right),\left( 2,0 \right),\left( 3,0 \right) \right\}$ is wrong as $0\notin B-C$