(i) Let
A = (1,2,3)
A x A = (1, 1) (2, 2) (3, 3) (1, 2) (2, 1) (3, 2) (2, 3) (1, 3), (3, 1)
\(\therefore\) R = {(1, 1) (1, 2) (2, 1) (2, 2) (2, 3) (1,3) (3, 3) (3,2) (3, 1)}.
(ii) R = {(1, 1) (1,2) (2,1) (2,2) (2,3), (3, 3) (3,2)}

The given function is f: X → Y and relation on X is R={(a, b): f(a) = f (b)}
Reflexive Since, for every x ∈ X, we have
f'(x) = f(x)
⇒ (xx) ∈ R, ∀ ∈ X Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R
Then, f(x)=f(y)
⇒ f(y)=f(x)
⇒ (y, x) ∈ R
Thus, (x, y) ∈ R ⇒  (y, x)∈ R, ∀x, y∈ X
Therefore, R is symmetric.
Transitive Let x, y, z∈ X such that
(x, y) ∈ R and (y, z) ∈ R
Given a relation 5 in \(N \times N\), defined as
(a, b) S(c, d), if a+d=b+c.
Reflexive Let (a, b) be any arbitrary element of \(N \times N\)
i.e. \((a, b) \in N \times N\), where \(a, b \in N\)
Now, as a+b=b+a
[∴ addition is commutative ]
Therefore \quad(a, b) S(a, b)
So, S is reflexive.
Symmetric \(\operatorname{Let}(a, b),(c, d) \in N \times N\), such that (a, b)
S(c, d). Then, a+d=b+c
\( \Rightarrow  b+c=a+d \Rightarrow c+b=d+a \)
\(\Rightarrow (c, d) S(a, b)\)
So, S is symmetric.
Transitive Let \((a, b),(c, d),(e, f) \in N \times N\) such that (a, b) S(c, d) and (c, d) S(e, f).
Then, a+d=b+c and c+f=d+e
On adding the above equations, we get
a+d+c+f=b+c+d+e 
\( \Rightarrow  a+f=b+e \Rightarrow(a, b) S(e, f)\)
So, S is transitive.
Thus, S is reflexive, symmetric and transitive. Hence, S is an equivalence

A relation R on a non-empty set A is called a transitive relation if (a, b), (b, c) \(\in R\) then (a, c) \(\in R\) , i.e., aRb, bRc implies aRc.
Thus a relation R on a non empty set A is said to be transitive if there exist a, b, c \(\in A\) such that (a, b)(b, c) \(\in R\)  implies (a, c) . \(\in R\)
Example
Let A = (1, 2, 3, 6)
R = (3, 6) (6, 1) (3, 1)
3 R 6 and 6 R 1 \(\Rightarrow \)(3, 1)  \(\in R\)
\(\therefore\) A is transitive.

Symmetric Relation : A relation R on a set A is called symmetric relation if aRb implies bRa, for every a,b \(\in a\) i.,e if (a,b)  \(\in R\) \(\Rightarrow \) (b,a)  \(\in R\)For every a,b \(\in A\)
Example 
A = (1,2,3)
A x A =(1,2) (2,1) (1,1) (2,2) (3,3) (1,3) (2,3)  (3,1) (3,2 )  \(\in R\) 
since (a,b)  \(\in R\)  (b,a)  \(\in R\) for every a, b  \(\in A\) 
Relation is said to be symmetric 

Reflexive Relation : A relation R on a set A is called reflexive relation if aRa for every \(a\in A\) ; if (a, a) \(\in \) R, for every \(a\in A\)
Example let
A = [1, 2, 3]
A x A =(1, 1) (1, 2)(1, 3) (2, 1) (2, 2) (2, 3) (3, 1)(3, 2) (3, 3)  \(\in R\) 
Since (a, a) \(\in R\) for every \(a\in A\)

\(f o f(x)=f(f(x))=F\left\{\left(3-x^3\right)^{1 / 3}\right\}=\left[3-\left\{\left(3-x^3\right)^{1 / 3}\right\}^3\right]^{1 / 3}=\left(x^3\right)^{1 / 3}=x\)

\( a * b=2 a+b-a b \\ 3 * 4=2 \times 3+4-3 \times 4 \\ 3 * 4=-2 \)

22 * 4 = HCF (22,4) = 2

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

$fog(x)=f[g(x)]$
$=f(x-7)$  ...................[ Since, $g(x)=x-7$ ]
$=x-7+7$ ..................[ Since, $f(x)=x+7$ ]           
$\Rightarrow$  $f$$o$$g(x)=x\Rightarrow$  $f$$o$$g(7)=7$

Let a, b \(\in A\), a " b = a + b + ab
Commutatively : for all a, b \(\in A\)
.a * b = a + b + ab
= b + a + ba
= b * a
Associatively: Let a, b, c \(\in A\)
(a * b) " c = (a + b + ab) " c
= (a + b + ab) + c + (a + b + ab)c
(a * b) " c = a + b + c + ab + bc + ac + abc
a * (b * c) = a * (b + c + bc)
= a + b + c + ab + bc + ac + abc
Clearly
(a * b) * c = a * (b "c)  \(\forall \) a, b, c  \(\in A\)}
 * is associative.
Identity: Let e \(\in A\)}such that
a * c = a
c + a + ea = a
c = 0
Identity element of A is e = O.
Inverse: Let b \(\in A\)such that
a*b = b*a = e
\(\Rightarrow \) a + b + ab = 0 and b + a + ba = 0
\(\Rightarrow \)  a = -b-ab
\(\Rightarrow \)  a = -b(l + a)
\(\Rightarrow \) b = \(\frac { -a }{ 1+a } \) \([\because \quad a\in A\therefore a\neq -1]\)
Invertible element of A is \(\frac { -a }{ 1+a } \) for all \(a\in A\)

Here \(a*b=\frac { 3ab }{ 5 } ;a,b,\in Q\) is a binary operation:
(I) Commutativity.
For \(a,b,\in Q\),
\(a*b=\frac { 3ab }{ 5 } =\frac { 3ba }{ 5 } .\)
[\(\because \) Rational numbers are commutative under multiplication]
= b * a
Hence,'* is commutative on Q.
(II) Associativity.
For \(a,b,c\in Q,\)
\((a*b)*c=\frac { 3ab }{ 5 } *c\)
\(=\frac { 3\frac { 3ab }{ 5 } c }{ 5 } =\frac { 9abc }{ 25 } \)
And  \(=\frac { 3a\frac { 3bc }{ 5 } }{ 5 } =\frac { 9abc }{ 25 } \)
Thus \(a*(b*c)=a*(b*c)\).
Hence, ' * ' is associative on Q.
(III) Let 'e' be the identity element.
Then \(a*e=a\Rightarrow e*a\Rightarrow \frac { 3ae }{ 5 } \)
\(\Rightarrow \) \(e=\frac { 5 }{ 3 } \).
Hence, the identity element = \(\frac { 5 }{ 3 } \). 

Let 'x' kg of food I and 'y' kg of food II be mixed. We have the table:

Thus LLP problem is as below:
Minimize:  Z=50x+70y ....(1)
Subject to:\(2x+y\ge 8\)   ...(2)
\(x+2y\ge 10\)  ...(3)
and  \(x,y\ge 0\) ....(4)

First of all, we locate the region represented by (2)-(4).
The shaded region, as shown above, is feasible region.
Applying Corner Point Method, we have:

Hence, the minimum cost = Rs. 380 when 2 kg of food I and 4kg of food II are mixed.

The system of constraints is:
\(x+y\ge 8\)  ....(1)
\(3x+5y\le 15\)....(2)
and \(x,y\ge 0\)......(3)

It is observed that there is no point, which satisfies all (1)-(3) simultaneously.
Thus there is no feasible region.
Hence, there is no feasible solution.

\(2*3=\frac { 2 }{ 2 } +\frac { 3 }{ 2 } =1+1=2\)

3*4 = 3(3) + 4(4) = 9 + 16 = 25