Question

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

Answer 1

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\\}

Complete step-by-step answer:
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 $0\notin B-C$