If $Y=\left\{x \mid x\right.$ is a positive factor of the number $2^{p-1}\left(2^p-1\right)$, where $2^p-1$ is a prime number\}. Write $Y$ in the roaster form.
$Y=\left\{x \mid x\right.$ is a positive factor of the number $2^{p-1}\left(2^p-1\right)$, where $2^p-1$ is a prime number $\}$.
So, the factor of $2^{p-1}$ are $1,2,2^2, 2^3, \ldots, 2^{p-1}$.
$$\therefore \quad Y=\left\{1,2,2^2, 2^3, \ldots, 2^{p-1}, 2^p-1\right\}$$
State which of the following statements are true and which are false. Justify your answer.
(i) $35 \in\{x \mid x$ has exactly four positive factors $\}$.
(ii) $128 \in\{y \mid$ the sum of all the positive factors of $y$ is $2 y\}$.
(iii) $3 \notin\left\{x \mid x^4-5 x^3+2 x^2-112 x+6=0\right\}$.
(iv) $496 \notin\{y \mid$ the sum of all the positive factors of $y$ is $2 y\}$.
(i) Since, the factors of 35 are 1, 5, 7 and 35 . So, statement (i) is true.
(ii) Since, the factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128 .
$$\begin{aligned} \therefore \quad \text { Sum of factors } & =1+2+4+8+16+32+64+128 \\ & =255 \neq 2 \times 128 \end{aligned}$$
So, statement (ii) is false.
(iii) We have,
$$x^4-5 x^3+2 x^2-112 x+6=0$$
$\therefore$ For $x=3$,
$$\begin{aligned} & (3)^4-5(3)^3+2(3)^2-112(3)+6 =0 \\ & \Rightarrow 81-135+18-336+6 =0 \\ & \Rightarrow -346 =0 \end{aligned}$$
which is not true.
Hence, statement (iii) is true.
(iv) $\because \quad 496=2^4 \times 31$
So, the factors of 496 are $1,2,4,8,16,31,62,124,248$ and 496 .
$$\begin{aligned} \therefore \quad \text { Sum of factors } & =1+2+4+8+16+31+62+124+248+496 \\ & =992=2(496) \end{aligned}$$
So, $496 \in\{y \mid$ the sum of all the positive factor of $y$ is $2 y\}$.
Hence, statement (iv) is false.
If $L=\{1,2,3,4\}, M=\{3,4,5,6\}$ and $N=\{1,3,5\}$, then verify that $L-(M \cup N)=(L-M) \cap(L-N)$.
Given, $\quad L=\{1,2,3,4\}, M=\{3,4,5,6\}$ and $N=\{1,3,5\}$
$\therefore \quad M \cup N=\{1,3,4,5,6\}$
$\begin{array}{lrrl} & & L-(M \cup N) & =\{2\} \\ & \text { Now, } & L-M & =\{1,2\}, L-N=\{2,4\} \\ & \therefore & (L-M) \cap(L-N) & =\{2\} \\ & \text { Hence, } & L-(M \cup N) & =(L-M) \cap(L-N) .\end{array}$
If $A$ and $B$ are subsets of the universal set $U$, then show that
(i) $A \subset A \cup B$
(ii) $A \subset B \Leftrightarrow A \cup B=B$
(iii) $(A \cap B) \subset A$
$$\begin{array}{ll} \text {(i) Let } & x \in A \\ \Rightarrow & x \in A \text { or } x \in B \Rightarrow x \in A \cup B \\ \text { Hence, } & \subset A \cup B \end{array}$$
$$\begin{aligned} &\text { (ii) If }\\ &\begin{array}{lr} & A \subset B \\ \text { Let } & x \in A \cup B \end{array} \end{aligned}$$
$$\begin{aligned} &\begin{array}{lcr} \Rightarrow & x \in A \text { or } x \in B \Rightarrow x \in B & {[\because A \subset B]} \\ \Rightarrow & A \cup B \subset B & \ldots \text { (i) } \\ \text { But } & B \subset A \cup B & \ldots \text { (ii) } \end{array}\\ \end{aligned}$$
From Eqs. (i) and (ii),
$\begin{aligned} & A \cup B=B \\ & \text{If}\quad A \cup B=B\end{aligned}$
Let $y \in A$
$\Rightarrow \quad y \in A \cup B \Rightarrow y \in B \quad[\because A \cup B=B]$
$\begin{array}{ll}\Rightarrow & A \subset B \\ \text { Hence, } & A \subset B \quad \Leftrightarrow \quad A \cup B=B\end{array}$
(iii) Let $$x \in A \cap B$$
$\Rightarrow \quad x \in A$ and $x \in B \quad \Rightarrow \quad x \in A$
Hence, $$A \cap B \subset A$$
Given that $N=\{1,2,3, \ldots, 100\}$. Then, write
(i) the subset of $N$ whose elements are even numbers.
(ii) the subset of $N$ whose elements are perfect square numbers.
We have, $\quad N=\{1,2,3,4, \ldots ., 100\}$
(i) Required subset $=\{2,4,6,8, \ldots, 100\}$
(ii) Required subset $=\{1,4,9,16,25,36,49,64,81,100\}$