(4)

Given, $A=\{1,2,3,4,5,6,7\}$

and $R=\left\{\right(x,y)\in A\times A:x+y=7\}$

For Reflexive: Let $y=x$

So, $x+x=7\Rightarrow x=\frac{7}{2}$, which is not possible.

So, the given relation is not reflexive.

For Symmetric: $xRy\Rightarrow x+y=7$

$\Rightarrow$  $y+x=7$  $\therefore$ $xRy\Rightarrow yRx$

Hence, the given relation is symmetric.

For Transitive: Let $xRy$ and $yRz\Rightarrow x+y=7$ and $y+z=7$.

But it does not imply that $x+z=7$. 

So, $R$ is not transitive.

(3)

[IMAGE 2]

$a_1$ divides $b_2$

Each element has 2 choices. 

$\Rightarrow 3\times 2=6$

$a_2$ divides $b_1$ 

Each element has 2 choices. 

$\Rightarrow 3\times 2=6$

$\therefore \text{Total number of relations is }6\times 6=36$

(4)

We have, $A=\{1,3,4,6,9\}$; $B=\{2,4,5,8,10\}$

$R=\left\{\right((a_1,b_1),(a_2,b_2)):a_1\le b_2\text{ and }{b}_1\le a_2\}$

At $a_1=1$, there are 5 choices for $b_2$;

$a_1=3$, there are 4 choices for $b_2$; 

$a_1=4$, there are 4 choices for $b_2$; 

$a_1=6$, there are 2 choices for $b_2$;

$a_1=9$, there is 1 choice for $b_2$;

$\text{∴}$  $\text{Total ways for} (a_1\le b_2)=16$

Now, at $b_1=2$, there are 4 choices for $a_2$; 

$b_1=4$, there are 3 choices for $a_2$;

$b_1=5$, there are 2 choices for $a_2$; 

$b_1=8$, there is 1 choice for $a_2$. 

$\text{Total ways for }(b_1\le a_2)=10$

$\therefore \text{Required number of ways }=16\times 10=160$

(3)

$R=\left\{\right(a,b):3a-3b+\sqrt{7}$ is an irrational number}

Reflexive: For $(a,a)$, we have $3a-3a+\sqrt{7}=\sqrt{7},$ which is an irrational number. $\therefore R$ is reflexive.

Symmetric: Let $(a,b)\in R$, i.e., $3a-3b+\sqrt{7}$ is an irrational number.

Now, we need to check $(b,a)\in R$ or not.

Let $3a=\sqrt{7}$ and $3b=8$

Then $3a-3b+\sqrt{7}=\sqrt{7}-8+\sqrt{7}=2\sqrt{7}-8,$ which is an irrational number.

But $3b-3a+\sqrt{7}=8-\sqrt{7}+\sqrt{7}=8,$ which is not an irrational number.

$\therefore$ For $(a,b)\in R$, $(b,a)\notin R.$

$\therefore R$ is not symmetric.

Transitive: Let $(a,b)$ and $(b,c)\in R.$

Let $3a=8, 3b=2\sqrt{7}, 3c=\sqrt{7}$

Then $3a-3b+\sqrt{7}=8-2\sqrt{7}+\sqrt{7}=8-\sqrt{7},$ which is an irrational number.

Also, $3b-3c+\sqrt{7}=2\sqrt{7}-\sqrt{7}+\sqrt{7}=2\sqrt{7},$ which is an irrational number.

But $3a-3c+\sqrt{7}=8-\sqrt{7}+\sqrt{7}=8,$ which is not an irrational$\Rightarrow (a,c)\notin R.$

$\therefore$  $R$ is not transitive.

(4)

(4)

[IMAGE 3]

Reflexive : $\gcd (a,a)=a\ne 1\text{ for }a\ne 1.$

Symmetric: Let $a=2, b=1$

$\gcd (a,b)=1; b\ne 2a$

Thus, $(2,1)\in R; \gcd (1,2)=1.$ But $2=2\times 1$

$\text{Therefore }(1,2)\notin R.\text{ Hence, }R\text{ is not symmetric.}$

Transitive: $(2,3)\in R$

$(3,8)\in R.$ But $(2,8)\notin R.$

$\text{Hence }R\text{ is not transitive.}$

$\text{R is neither symmetric nor transitive.}$

(3)

If 'R' be a relation defined on $\mathbf{N}$ as $\text{a R b}$ is $2a+3b$ is a multiple of 5, $a,b\in \mathbf{N}$.

(i) $\text{a R a}\Rightarrow 5a$ is a multiple of 5

    R is a reflexive relation.

(ii) $\text{a R b},2a+3b=5\alpha$ (let)

Now, $\text{b R a}\Rightarrow 2b+3a=2b+\left(\frac{5\alpha -3b}{2}\right)·3$

$=\frac{15}{2}\alpha -\frac{5}{2}b=\frac{5}{2}(3\alpha -b)=\frac{5}{2}(2a+2b-2\alpha )=5(a+b-\alpha )$

So, R is a symmetric relation.

(iii) $\text{a R b}\Rightarrow 2a+3b=5\alpha$; $\text{b R c}\Rightarrow 2b+3c=5\beta$

Now, $2a+5b+3c=5(\alpha +\beta )$

$\Rightarrow 2a+5b+3c=5(\alpha +\beta ) \mathrm{or} 2a+3c=5(\alpha +\beta -b)$

$\Rightarrow \text{a R c}$

So, R is a transitive relation.

Hence Relation 'R' is an equivalence relation.

(2)

Given relation, $R=\left\{\right(a,b),(b,c\left)\right\}$ and set = $\{a,b,c\}$

For symmetric, since $(a,b),(b,c)\in R$

So, $(b,a),(c,b)$ be in R. For transitive, since $(a,b),(b,c)\in R$

So, $(a,c)$ should be in R. Then $(c,a)$ should also be in R since $(a,c)$ lies in R.

Since $(a,b),(b,a)\in R$, so $(a,a)\in R$ should also lie in R.

Since $(c,b),(b,c)\in R$, so $(c,c)$ should also lie in R.

Since $(b,c),(c,b)\in R$, so $(b,b)$ should also lie in R.

$\therefore$  Elements to be added are:

$(b,a),(c,b),(a,c),(c,a),(a,a),(c,c),(b,b)$.

Total number of elements to be added = 7.

(2)

We have, $(a,b)R(c,d)\iff ad(b-c)=bc(a-d)$

$\Rightarrow \frac{b-c}{bc}=\frac{a-d}{ad} \text{and} \frac{1}{c}-\frac{1}{b}=\frac{1}{d}-\frac{1}{a}\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

For reflexive: $(a,b)R(a,b)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{b}-\frac{1}{a}$ which is false.

Hence, it is not reflexive.

For symmetric: $(a,b)R(c,d)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

$\Rightarrow \frac{1}{c}-\frac{1}{d}=\frac{1}{b}-\frac{1}{a}\Rightarrow (c,d)R(a,b)$

Hence, it is symmetric.

For transitive: $(a,b)R(c,d)\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{d}-\frac{1}{c}$

$\text{and }(c,d)R(e,f)\Rightarrow \frac{1}{c}-\frac{1}{d}=\frac{1}{f}-\frac{1}{e}$

$\Rightarrow \frac{1}{a}-\frac{1}{b}=\frac{1}{e}-\frac{1}{f}\ne (a,b)R(e,f)$

Hence, it is not transitive.

(4)

For relation $T=a^2-b^2\in I$

Then, $(b,a)$ on relation $R\Rightarrow b^2-a^2\in I$

$\therefore T\text{ is symmetric,}$ $S=\left\{\right(a,b):a,b\in R-\{0\}, 2+\frac{{\displaystyle a}}{{\displaystyle b}}>0\}$

$2+\frac{a}{b}>0\Rightarrow \frac{a}{b}>-2\Rightarrow \frac{b}{a}>-\frac{1}{2}$

If $(b,a)\in S$, then $2+\frac{b}{a}$ is not necessarily positive.

$\therefore$   $S\text{ is not symmetric.}$