Let A = {x ϵ W : x < 2}, B = {x ϵ N : 1 < x ≤ 4} and C = {3, 5}

Solution:

(i) Given:

$A=\{x \in W: x<2\}$

Here, W denotes the set of whole numbers (non – negative integers).

$\therefore A=\{0,1\}$

$[\because$ It is given that $x<2$ and the whole numbers which are less than 2 are $0 \& 1]$

$B=\{x \in N: 1

Here, N denotes the set of natural numbers.

$\therefore B=\{2,3,4\}$

[ $\because$ It is given that the value of $x$ is greater than 1 and less than or equal to 4 ]

and $C=\{3,5\}$

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

By the definition of the union of two sets, $(B \cup C)=\{2,3,4,5\}$

$=\{0,1\} \times\{2,3,4,5\}$

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.

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

$=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}$

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

Now, $A \times B=\{0,1\} \times\{2,3,4\}$

$=\{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}$

and $A \times C=\{0,1\} \times\{3,5\}$

$=\{(0,3),(0,5),(1,3),(1,5)\}$

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

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

$(A \times B) \cup(A \times C)=\{(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)\}$

= L. H. S

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

(ii) Given:

$A=\{x \in W: x<2\}$

Here, W denotes the set of whole numbers (non – negative integers).

$\therefore \mathrm{A}=\{0,1\}$

$[\because$ It is given that $x<2$ and the whole numbers which are less than 2 are 0,1$]$

$B=\{x \in N: 1

Here, N denotes the set of natural numbers.

$\therefore \mathrm{B}=\{2,3,4\}$

$[\because$ It is given that the value of $x$ is greater than 1 and less than or equal to 4$]$

and $C=\{3,5\}$

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

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

$=\{0,1\} \times\{3\}$

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.

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

$=\{(0,3),(1,3)\}$

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

Now, $A \times B=\{0,1\} \times\{2,3,4\}$

$=\{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\}$

and $A \times C=\{0,1\} \times\{3,5\}$

$=\{(0,3),(0,5),(1,3),(1,5)\}$

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

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

$(A \times B) \cap(A \times C)=\{(0,3),(1,3)\}$

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

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