Given, \(f: R \rightarrow R\) defined as \(f(x)=\frac{5 x-3}{4}\)
For one-one Let \(x_1, x_2 \in R\)
such that \(f\left(x_1\right)=f\left(x_2\right)\)
\(\Rightarrow  \frac{5 x_1-3}{4}=\frac{5 x_2-3}{4} \)
\(\Rightarrow  5 x_1-3=5 x_2-3\)
\(\Rightarrow  5 x_1=5 x_2 \Rightarrow x_1=x_2\)
So, f is one-one
For onto
Let \(y \in R\) (codomain) be any arbitrary element.
Then, y=f(x)
\(\Rightarrow  y=\frac{5 x-3}{4} \Rightarrow 4 y=5 x-3\)
\(\Rightarrow  5 x=4 y+3 \Rightarrow x=\frac{4 y+3}{5}\)
Thus, for each \(y \in R\), there exists
\(x=\frac{4 y+3}{5} \in R\)
such that \(f(x)=f\left(\frac{4 y+3}{5}\right)\)
\(=\frac{5\left(\frac{4 y+3}{5}\right)-3}{4}=\frac{4 y+3-3}{4}=y\)
So, f is onto.

Given, relation R is defined on a set of real number \(\mathbf{R}\). such that \(R=\{(x, y): x \cdot y\) is an irrational number }
Reflexive \(\sqrt{2}\) is a real number.
and \(\sqrt{2} \cdot \sqrt{2}=2\) is not an irrational number
\(\therefore \quad(\sqrt{2}, \sqrt{2}) \notin R\)
\(\Rightarrow R\) is not reflexive.
Symmetric consider \(\sqrt{3}\) and \(\sqrt{5}\) are two real numbers clearly, \(\sqrt{3} \cdot \sqrt{5}=\sqrt{15}\) is an irrational number
\(\Rightarrow \sqrt{5} \cdot \sqrt{3}=\sqrt{15}\)
is an irrational number.
\(\Rightarrow R\) is symmetric.
Transitive Consider three real numbers \(2, \sqrt{5}\) and 3.
Clearly, \(2 \cdot \sqrt{5}=2 \sqrt{5}\) is an irrational
\(\sqrt{5} \cdot 3=3 \sqrt{5} \text { is an irrational }\)
But \(2 \cdot 3=6\) is not an irrational number
\(\Rightarrow R\) is not transitive.

Given set A = {5, 6, 7, 8, 9} and relation R = {(a, b) : la - bl is divisible by 2}
For reflexive: For a \(\in\) A,
(a, a) \(\in\) R \(\Rightarrow\) |a - a| = 0 is divisible by 2, true. Hence, reflexive.
For symmetric: For a, b \(\in\) A
Let (a, b) \(\in\) R Rand (b, c) \(\in\) R
\(\Rightarrow\) |a - b| is divisible by 2.
\(\Rightarrow\)  |(b - a)| = |b - c| is divisible by 2
\(\Rightarrow\) (b, a) \(\in\) R. Hence, symmetric.
For transitive: For a, b, c \(\in\) A
Let (a, b) E Rand (b, c) \(\in\) R
\(\Rightarrow\) la - bl, i.e. (a - b) is divisible by 2
and Ib - c], i.e. (b - e) is divisible by 2 ...(i)
:. a - e = (a - b) + (b - e) is divisible by 2 [from (i)]
\(\Rightarrow\) la - el is divisible by 2 \(\Rightarrow\) (a,c) \(\in\) R. Hence, transitive relation R is an equivalence relation as it is reflexive, symmetric and transitive.
All elements related to 6, i.e. la - 61 is divisible by 2

Relation R is defined by (a, b) R (c, d) \(\Leftrightarrow\) ad = be for all (a, b), (c, d) \(\in\) N x N.
For reflexive: (a, b) R (a, b) \(\Leftrightarrow\) ab = ba, which is true in N. Hence, reflexive.
For symmetric: (a, b) R (c, d) \(\Leftrightarrow\) ad = be
\(\Leftrightarrow\) cb = da \(\Leftrightarrow\) (c, d) R (a, b).
Hence, symmetric.
For transitive: Consider (a, b) R (c, d) and
(c, d) R (e,f) \(\Leftrightarrow\) ad = be and
cf = de \(\Leftrightarrow\) ad-cf = be-de \(\Leftrightarrow\) af = be
\(\Leftrightarrow\) (a, b) R (e,f). Hence, transitive.
Since relation R is reflexive, symmetric and transitive.
Hence, relation R is an equivalence relation.

R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b – c) is even (Why?). So, (a – c) is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z.
note that all even integers are related to zero, as (0, ± 2), (0, ± 4) etc., lie in R and no odd integer is related to 0, as (0, ± 1), (0, ± 3) etc., do not lie in R. Similarly, all odd integers are related to one and no even integer is related to one. Therefore, the set E of all even integers and the set O of all odd integers are subsets of Z satisfying following conditions:
(i) All elements of E are related to each other and all elements of O are related to each other.
(ii) No element of E is related to any element of O and vice-versa.
(iii) E and O are disjoint and Z = E \(\cup\) O.
The subset E is called the equivalence class containing zero and is denoted by [0]. Similarly, O is the equivalence class containing 1 and is denoted by [1]. Note that [0] \(\ne\)[1], [0] = [2r] and [1] = [2r + 1], r \(\in\) Z. Infact, what we have seen above is true for an arbitrary equivalence relation R in a set X. Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets A_i called partitions or subdivisions of X satisfying:
(i) all elements of A_i are related to each other, for all i.
(ii) no element of A_i is related to any element of A_j, i \(\ne\) j.
(iii) \(\cup\) A_j = X and A_i \(\cap\) A_j = f, i \(\ne\) j.
The subsets Ai are called equivalence classes. The interesting part of the situation is that we can go reverse also. For example, consider a subdivision of the set Z given by three mutually disjoint subsets A₁, A₂ and A₃ whose union is Z with
A₁ = {x \(\in\) Z : x is a multiple of 3} = {..., – 6, – 3, 0, 3, 6, ...}
A₂ = {x \(\in\) Z : x – 1 is a multiple of 3} = {..., – 5, – 2, 1, 4, 7, ...}
A₃ = {x \(\in\) Z : x – 2 is a multiple of 3} = {..., – 4, – 1, 2, 5, 8, ...}
Define a relation R in Z given by R = {(a, b) : 3 divides a – b}. Following the arguments similar to those , we can show that R is an equivalence relation. Also, A₁ coincides with the set of all integers in Z which are related to zero, A₂ coincides with the set of all integers which are related to 1 and A₃ coincides with the set of all integers in Z which are related to 2. Thus, A₁ = [0], A₂ = [1] and A₃ = [2].
In fact, A₁ = [3r], A₂ = [3r + 1] and A₃ = [3r + 2], for all r \(\in\) Z.

We have, \( y=f(x)=\sqrt{4-x^{2}, 0 \leq x \leq 2,0 \leq y \leq 2} \)
For one-one Let \( x_{1}, x_{2}\) be any two elements of the interval 0 \(\leq \)x \(\leq \)2,
such that \(f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow \sqrt{4-x_{1}^{2}}=\sqrt{4-x_{2}^{2}} \Rightarrow 4-x_{1}^{2}=4-x_{2}^{2} \Rightarrow \)
\(x_{1}^{2}=x_{2}^{2} \Rightarrow x_{1}=\pm x_{2} \Rightarrow x_{1}=x_{2}\)
\(\left[x_{1} \neq-x_{2},\right. since x_{1} and x_{2} are non-negative ] \)
So, f(x) is one-one. For onto Let k \(\in\)[0, 2] be any arbitrary element and let
\(f(x)=k \Rightarrow \sqrt{4-x^{2}}=k \)
On squaring both sides, we get \(4-x^{2}=k^{2}\) 
[\(\because\) x is non-negative, so we take positive]
 Also, for \(0 \leq k \leq 2\), we have \(0 \leq \sqrt{4-k^{3}} \leq 2 \Rightarrow 0 \leq x \leq 2\), which is true.
Thus, for each k \( \in\)[0, 2], there exists \(x=\sqrt{4-k^{2}} \in[0,2] \) such that f(x) = k .
So, f(x) is onto. Hence, f is bijective function.

Given, f : N\( \rightarrow\) N, defined by
\(f(n)=\left\{\begin{array}{ll} \frac{n+1}{2}, & \text { if } n \text { is odd } \\ \frac{n}{2}, & \text { if } n \text { is even } \end{array}\right.\)
Here, \(f(1)=\frac{1+1}{2}=1 and f(2)=\frac{2}{2}=1 \)
So, f is not one-one, i.e. f is many-one. Now, consider a natural number m in codomain N .
Case I : Let \(m=f(n)=\frac{n+1}{2} \Rightarrow 2 m=n+1 \)
\(\Rightarrow n=2 m-1 \), which is odd for each m \(\in \)N .
Thus, for each m \(\in \)N (codomain), there exists , such that \(f(2 m-1)=\frac{2 m-1+1}{2}=m \) 
Case II : Let m = f(n) = n / 2
\(\Rightarrow \)n = 2 m, which is even for each m \(\in\) N
Thus, for each m \(\in\) N (codomain), there exists n = 2 m \(\in\) N, such that \(f(2 m)=\frac{2 m}{2}=m \)
Since, from both cases, we have pre-images. So, we can choose either of the case to get the pre-image. Therefore, f is onto.
Hence, f is many-one and onto function.

We .have a relation R on N x N defined by (a, b)R(c, d), if
ad(b + c) = bc(a + d) .
Reflexive Let (a, b) ∈ N x N be any arbitrary element.
We have to show that (a, b) R (a, b), i.e.
ab(b + a) = ba(a + b), which is always true, as natural numbers are commutative under usual multiplication and addition .•
Since, (a, b) ∈ N x N was arbitrary, so R is reflexive.
Symmetric Let (a, b), (c, d) ∈ N x N
such that (a, b) R (c, d), i.e. ad(b + c) = bc(a + d)
We have to show that (c, d) R (a, b),
i.e. cb(d + a) = da(c + b)
From Equation (i), we have ad( b + c) = bc( a + d)
⇒ da(c + b) = eb(d + a)
[natural numbers are commutative under
usual addition and multiplication] [1]
⇒ cb(d+a) = da(c+b)
⇒ (c, d)R(a, b)
So, R is symmetric. 
Transitive Let (a, b), (c, d) and (e, f) ∈N x N such that (a, b) R (c, d) and (c, d) R (e, f).
⇒ (a, b)R(c, d) ⇒ ad(b+c) = be(a+d)
\( \Rightarrow \frac{b+c}{b c}=\frac{a+d}{a d} \)
\(\Rightarrow \frac{1}{c}+\frac{1}{b}=\frac{1}{d}+\frac{1}{a}\) 
and \((c, d) R(e, f) \Rightarrow c f(d+e)=d e(c+f)\) 
\(\Rightarrow \frac{d+e}{d e}=\frac{c+f}{c f} \)
\( \Rightarrow \frac{1}{e}+\frac{1}{d}=\frac{1}{f}+\frac{1}{c}\) 
Now, adding Equation (ii) and (iii), we get
\(\left(\frac{1}{c}+\frac{1}{b}\right)+\left(\frac{1}{e}+\frac{1}{d}\right)=\left(\frac{1}{d}+\frac{1}{a}\right)+\left(\frac{1}{f}+\frac{1}{c}\right) \) 
\(\Rightarrow \frac{1}{b}+\frac{1}{e}=\frac{1}{a}+\frac{1}{f}\) 
\(\Rightarrow \frac{e+b}{b e}=\frac{f+a}{a f}\) 
\(\Rightarrow a f(e+b)=b e(f+a)\) 
\(\Rightarrow a f(b+e)=b e(a+f) \)
\(\Rightarrow (a, b) R(e, f)\) 
So, R is transitive.
Hence, R is an equivalence relation.

Given, A = Set of all points in a plane.
R = {(P, Q): Distance between P and Q is less than 2 units}
Reflexive Let P∈A be any arbitrary point. Then,
distance between P and P is 0, which is less than 2 units.
⇒ (P, P)∈ R
Thus, (P, P) ∈ R for all P∈A
So, R is reflexive. [1]
Symmetric Let P, Q ∈ A such that (P, Q) ∈ R, i.e. distance between P and Q is less than 2 units.
⇒ Distance between Q and P is less than 2 units.
⇒(Q, P)∈ R
So, R is symmetric. [1)
Transitive Consider the points P, Q and Shaving coordinates (0,0), (1, 0) and (2, 0), respectively. We can observe that, the distance between P and Q is 1 unit, whichis less than 2 units and also the distance between Q and S is 1 unit.
But the distance between P and S is 2 units, which is not less than 2 units. Thus, (P, Q) ∈ R,(Q, S) ∈ R,but (P,S)∉R. So, R is not transitive.

Given, R = {(x, y) : y is divisible by x} and A = {1, 2, 3, 4, 5, 6}
Reflexive Let x ε A be any arbitrary element. We know that, x is divisible by x.
[ ஃ every real number except zero is divisible by itself ] 
\( \Rightarrow \big(x,x\big) \in R\) 
Since,\(x \in A\) was an arbitrary element, therefore 
\(\big(x,x\big) \in R, \vee A\),So, R is reflexive
Symmetric Clearly, 2, 4 e A and 4 is divisible by 2, but 2 is not divisible by 4.
∴ \((2,4) \in R\) but \((4,2) \in R\) 
Transitive Let \(x,y,z \in A\) such that \((x,y) \in R\) and  \((y,z) \in R\) 
Now, as \((x,y) \in R\) therefore y is divisible by x.
i.e. \(\frac{y}{k} = k_{1}(say)\) ...(i)
where, k₁ is a natural number
and as \((y,z) \in R\) therefore z is divisible by y.
i.e.\( \frac{z}{y}= k_{2}(say)\) (ii)
where, k₂ is a natural number.
On multiplying Eqs. (i) and .(ii), we get
\(\frac{y}{x} \times \frac{z}{y}= k_{1} k_{2} \Rightarrow \frac{z}{k} = k_{1} k_{2} \) 
where, k₁, k₂ is a natural number
ஃ z is divisible by x.
Thus,\((x,y) \in R\) for \((x,y),(y,z) \in R\) 
i.e.\((x,y) \in R,(y,z) \in R \Rightarrow (x,z) \in R\) 
Hence, R is transitive

Given, A = Set of real numbers and R = {(a, b) : 1 + ab > O}.
Reflexive Let a be any real number.
Then,
\(1+a a=1+a^{2}>0 \) \({\left[\because a^{2} \geq 0, \forall a \in A\right]} \) 
\((a, a) \in R, \forall a \in A \) 
So, R is reflexive.
Symmetric Let (a, b) ∈ R, then
\((1+a b)>0 \Rightarrow 1+b a>0\) 
\(\ [\because a b=b a, \forall a, b \in A]\) 
\(\Rightarrow \ (b, a) \in R\) 
Transitive We observe that \(\left(1, \frac{1}{2}\right) \in R\) 
and \(1+1 \cdot \frac{1}{2}=1+\frac{1}{2}>0\) 
But, \((1,-1) \notin R\) 
Because 1 + 1 x (-1) = 0 ⊁ 0 
So, R is not transitive.

Given, A = Set of real numbers and R = {(a, b) : a \(\leq \) b}.
Reflexive Let a ∈ A be any arbitrary real number.
We know that, every real number is equal to itself,
i.e. a = a.
So, we can write a\(\leq \)a => (a, a) ∈ R
So, R is reflexive.
Symmetric Let (a, b) e R, then a < b or a = b
If a = b, then b = a
But if we consider a < b, then b 1. a ⇒ (b, a) ∉ R.
e.g. 4 < 5 but 5 ⊀ 4
So, R is not symmetric.
Transitive Let (a, b), (b, c) ∈ R, then
(a,b) ∈ R ⇒ a ≤ b
(b, c) ∈ R ⇒ b:S;c
From Equation (i) and (ii), we get a ≤ c ⇒ (a, c) ∈ R
So, R is transitive.

Given, A = Set of real numbers and R = {(a, b) : a \(\leq \) b}.
Reflexive Let a ∈ A be any arbitrary real number.
We know that, every real number is equal to itself,
i.e. a = a.
So, we can write a \(\leq \)a => (a, a) ∈ R
So, R is reflexive.
Symmetric Let (a, b) e R, then a < b or a = b
If a = b, then b = a
But if we consider a < b, then b 1. a ⇒ (b, a) ∉ R.
e.g. 4 < 5 but 5 ⊀ 4
So, R is not symmetric.
Transitive Let (a, b), (b, c) ∈ R, then
(a,b) ∈ R⇒ a ≤ b
(b, c) ∈ R ⇒ b:S;c
From Equation (i) and (ii), we get a≤c => (a, c) ∈ R
 So, R is transitive.

Let x,y \(\in W\)
If x and y both are even f(x) = f(y)
x + 1 = y + 1
x = y
If x and y both are odd, f(x) = f(y)
x - 1 = y - 1
x = y
If s is odd and y is even i.e., \(x\neq y\) (x-1) is even (y+1) is odd
\(\Rightarrow x\neq y\Rightarrow f(x)\neq f(y)\)
Similarly for X is even and y is odd f is one-one
Range of f = [f(0) ,f(1) ,f(2) .....}
= [1, 0, 3, 2....] = W = co-domain
f is onto
Hence f is invertible
\(f^{ 2 }:W\rightarrow W.f^{ -1 }(x)=\begin{cases} x-1,xis\quad odd \\ x+1,xis\quad even \end{cases}\)

For \(a\in Z,a-a=0,\) which is divisible by 5.
\(\therefore \) \((a,a)\in R\forall a\in Z\).
Thus R is reflexive.
Now let \((a,b)\in R\)  \(\Rightarrow \) a-b is divisible by 5
\(\Rightarrow \) b-a is divisible by 5 \(\Rightarrow \) \((b,a)\in R\).
Thus R is symmetric.
Again let \((a,b)\in R,\ (b,c)\in R\)
\(\Rightarrow \) a-b and b-c are divisible by 5
\(\Rightarrow \) (a-b) + (b-c) = a-c is divisible by 5
\(\Rightarrow \) \((a,c)\in R\).
Thus R is transitive.
Hence, R is an equivalence relation.