If $Y=\{1,2,3, \ldots, 10\}$ and $a$ represents any element of $Y$, write the following sets, containing all the elements satisfying the given conditions.
(i) $a \in Y$ but $a^2 \notin Y$
(ii) $a+1=6, a \in Y$
(iii) $a$ is less than 6 and $a \in Y$
Given, $$Y=\{1,2,3, \ldots, 10\}$$
(i) $\left\{a: a \in Y\right.$ and $\left.a^2 \notin Y\right\}=\{4,5,6,7,8,9,10\}$
(ii) $\{a: a+1=6, a \in Y\}=\{5\}$
(iii) is less than 6 and $a \in Y\}=\{1,2,3,4,5$, $\}$
$A, B$ and $C$ are subsets of universal set $U$. If $A=\{2,4,6,8,12,20\}$, $B=\{3,6,9,12,15\}, C=\{5,10,15,20\}$ and $U$ is the set of all whole numbers, draw a Venn diagram showing the relation of $U, A, B$ and $C$.
Let $U$ be the set of all boys and girls in a school, $G$ be the set of all girls in the school, $B$ be the set of all boys in the school and $S$ be the set of all students in the school who take swimming. Some but not all, students in the school take swimming. Draw a Venn diagram showing one of the possible interrelationship among sets $U, G, B$ and $S$.
For all sets $A, B$ and $C$, show that $(A-B) \cap(A-C)=A-(B \cup C)$.
$$\begin{aligned} & \text { Let } \quad x \in(A-B) \cap(A-C) \\ & \Rightarrow \quad x \in(A-B) \text { and } x \in(A-C) \\ & \Rightarrow \quad(x \in A \text { and } x \notin B) \text { and }(x \in A \text { and } x \notin C) \\ & \Rightarrow \quad x \in A \text { and }(x \notin B \text { and } x \notin C) \\ & \Rightarrow \quad x \in A \text { and } x \notin(B \cup C) \\ & \Rightarrow \quad x \in A-(B \cup C) \\ & \Rightarrow \quad(A-B) \cap(A-C) \subset A-(B \cup C) \quad \text{.... (i)}\\ & \text { Now, let } \quad y \in A-(B \cup C) \\ & \Rightarrow \quad y \in A \text { and } y \notin(B \cup C) \\ & \Rightarrow \quad y \in A \text { and }(y \notin B \text { and } y \notin C) \\ & \Rightarrow \quad(y \in A \text { and } y \notin B) \text { and }(y \in A \text { and } y \notin C) \\ & \Rightarrow \quad y \in(A-B) \text { and } y \in(A-C) \\ & \Rightarrow \quad y \in(A-B) \cap(A-C) \\ & \Rightarrow \quad A-(B \cup C) \subset(A-B) \cap(A-C) \quad \text{.... (ii)}\\ & \text { From Eqs. (i) and (ii), } \\ & A-(B \cup C)=(A-B) \cap(A-C) \end{aligned}$$
For all sets $A$ and $B,(A-B) \cup(A \cap B)=A$.
$\begin{aligned} \mathrm{LHS} & =(A-B) \cup(A \cap B) \\ & =[(A-B) \cup A] \cap[(A-B) \cup B] \\ & =A \cap(A \cup B)=A=\mathrm{RHS}\end{aligned}$
Hence, given statement is true.