If A = {1, 3, 5) B = {3, 4} and C = {2, 3}, verify that:
Question:

If A = {1, 3, 5) B = {3, 4} and C = {2, 3}, verify that:

(i) $A \times(B \cup C)=(A \times B) \cup(A \times C)$

(ii) $A \times(B \cap C)=(A \times B) \cap(A \times C)$

 

Solution:

(i) Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3}

L. H. S $=A \times(B \cup C)$

By the definition of the union of two sets, $(B \cup C)=\{2,3,4\}$

$=\{1,3,5\} \times\{2,3,4\}$

Now, by the definition of the Cartesian product,

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e.

$P \times Q=\{(p, q): p \in P, q \in Q\}$

$=\{(1,2),(1,3),(1,4),(3,2),(3,3),(3,4),(5,2),(5,3),(5,4)\}$

R. H. S $=(A \times B) \cup(A \times C)$

Now, $A \times B=\{1,3,5\} \times\{3,4\}$

$=\{(1,3),(1,4),(3,3),(3,4),(5,3),(5,4)\}$

and $A \times C=\{1,3,5\} \times\{2,3\}$

$=\{(1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\}$

Now, we have to find $(A \times B) \cup(A \times C)$

So, by the definition of the union of two sets

$(A \times B) \cup(A \times C)=\{(1,2),(1,3),(1,4),(3,2),(3,3),(3,4),(5,2),(5,3),(5,4)\}$

= L. H. S

$\therefore$ L. H. S $=$ R. H. S is verified

(ii) Given: $A=\{1,3,5\}, B=\{3,4\}$ and $C=\{2,3\}$

L. H. S = A × (B \cap C)

By the definition of the intersection of two sets, $(B \cap C)=\{3\}$

$=\{1,3,5\} \times\{3\}$

Now, by the definition of the Cartesian product,

Given two non – empty sets $P$ and $Q$. The Cartesian product $P \times Q$ is the set of all ordered pairs of elements from $\mathrm{P}$ and $\mathrm{Q}$, . .i.e.

$P \times Q=\{(p, q): p \in P, q \in Q\}$

$=\{(1,3),(3,3),(5,3)\}$

R. $H . S=(A \times B) \cap(A \times C)$

Now, $A \times B=\{1,3,5\} \times\{3,4\}$

$=\{(1,3),(1,4),(3,3),(3,4),(5,3),(5,4)\}$

and $A \times C=\{1,3,5\} \times\{2,3\}$

$=\{(1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\}$

Now, we have to find $(A \times B) \cap(A \times C)$

So, by the definition of the intersection of two sets,

$(A \times B) \cap(A \times C)=\{(1,3),(3,3),(5,3)\}$

$=\mathrm{L} . \mathrm{H} . \mathrm{S}$

$\therefore \mathrm{L} . \mathrm{H} . \mathrm{S}=\mathrm{R} . \mathrm{H} . \mathrm{S}$ is verified

 

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