• Definition Let $A$ be a non empty set and let $R$ be a relation from $A$ to itself.We say that $R$ is symmetric if the pair

$(a, b) \in R \Rightarrow (b, a) \in R$

• Example $A = \{1,2,3,4,5\}$

$R=\{(x, y) \in A \times A |$the difference between x and y is an odd number $\}$

$R= \{(1,2), (1,4), (2,1), (2,3), (2,5), (3,2),$

$(3,4), (4,1), (4,3), (4,5), (5,2), (5,4)\}$

$\therefore$ R is a symmetric relation

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

$R = \{(n, m) \in A \times A |n$ divides $m\}$

$= \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$

$(2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (5,5), (6,6)\}$

$(1,2) \in R$ but $(2,1) \not\in R$

$(2,6) \in R$ but $(6,2) \not\in R$

$\Rightarrow R$ is not a symmetric relation.

Let $A= \{1,2,3,4,5\}$

$x=P(A) =$ Set of all subsets of $A$

Define a relation $R$ on $X$ as follows:

(A₁, A₂) $\in R$ if A₁ $\subseteq$ A₂

{2} $\subseteq$ A & {2,3} $\subseteq$ A

{2} $\subseteq$ {2,3} $\Rightarrow$ ({2}, {2,3}) $\in$ R.

{2,3} $\cancel{\subseteq}$ {2} $\Rightarrow$ ({2,3}, {2}) $\not\in$ R

$\therefore R$ is not symmetric.

Let A be a non empty set and let R be a relation from A to itself. We say that R is reflexive if the pair (a,a)$\in$ R for all a $\in$ R.

• Example A={1,2,3,4,5}

X=P(A)

R={(A₁, A₂)$\in$ P(A)xP(A)}$| A_1 \subseteq A_2$}

If $B \subseteq A$ $\Rightarrow$ (B, B)$\in$ R

$\therefore$ R is reflexive

• Example A={1,2,3,4,5} R={(x,y)$\in$ A $\times$ A | x divide y}.

{(1,1), (2,2), (3,3), (4,4), (5,5)}$\subseteq R$

$\Rightarrow$ R is reflexive

• Example A={1,2,3,4}

R={(n,m) |m=n²}

(1,1)$\in$ R but (2,2)∉ R $\Rightarrow$ R is not reflexive.

(2,4) $\in$ R but (4,2)∉ R

$\Rightarrow$ R is not symmetric.

Let A be a non empty set and let R be a relation from A to itself. We say that R is transitive if the following holds

(a,b) & (b,c)$\in$ R $\Rightarrow$ (a,c) $\in$ R.

A= {1,2,3,4,5,6,7,8}

R={(n,m) $\in$ A $\times$ A|n divides m}.

(2,4) $\in$ R & (4,8) $\in$ R

(2,8) $\in$ R

n divides m and m divides k

$\Rightarrow$ n divides k.

$\therefore$ R is transitive

A={1,2,3,4,5}

X=P(A)

R={(A₁, A₂) $\in$ P(A) $\times$ P(A) |A₁ $\subseteq$ A₂}

(A₁, A₂) $\in$ R and (A₂, A₃) $\in$ R

$\Rightarrow$ A₁ $\subseteq$ A₂ and A₂ $\subseteq$ A₃

A₁ $\subseteq$ A₂ $\subseteq$ A₃

A₁ $\subseteq$ A₃ i.e, (A₁, A₃)$\in$ R

$\therefore$ R is transitive.

A={1,2,3,4,5}

R ={(n,m) |the difference between n and m is an odd number}

(n,m) $\in$ R and (m,k) $\in$ R $\Rightarrow$ (n,k) $\in$ R

(1,2) $\in$ R & (2,3) $\in$ R

But (1,3)$\notin$ R

$\therefore$ R is not transitive

Definition A relation R from a non empty set A to itself is called an equivalence relation if

• R is symmetric

• R is reflexive &

• R is transitive

• A={1,2,3,4,5}

• R= {(n,m)| n divides m}

  • R is not symmetric $\Rightarrow$ R is not an equivalence relation

  • R is reflexive

  • R is transitive

• A= {1,2,3,4,5}

• x=P(A)

• R={(A₁, A₂₎ $\in$ P(A) $\times$ P(A)|A₁ $\subseteq$ A₂}

  • R is not symmetric $\Rightarrow$ R is not an equivalence relation.

  • R is reflexive

  • R is transitive

Let $\Z$ denote the set of all integers. Define R on $\Z$ as follows:

(n,m) $\in$ R if n≡ m (mod 9)

i.e, n-m is divisible by 9

Show that, R is an equivalence relation.

i) Suppose (n,m) $\in$ R

i.e, n≡ m (mod 9)

i.e, n-m is divisible by 9

i.e, there exists an integer k such that n-m = k.9

$\Rightarrow$ m-n=-k.9

$\Rightarrow$ m- n is divisible by 9.

$\Rightarrow$ m≡n (mod 9)

$\Rightarrow$ (m,n)$\in$ R.

$\therefore$ R is symmetric.

ii) Let n $\in$ $\Z$

n-n=0=0.9

$\Rightarrow$ n≡n (mod 9)

$\Rightarrow$ (n,n) $\in$ R

i.e, R is reflexive

iii) Let (n, m) $\in$ R and (m, r)$\in$ R

(n, m)$\in$ R $\Rightarrow$ n ≡ m (mod 9)

$\Rightarrow$ n - m is divisible by 9

$\Rightarrow$ There exists k $\in$ $\Z$ such that n - m = k.9 $\longrightarrow$(1)

(m, r)$\in$ R $\Rightarrow$ m ≡ r (mod 9)

$\Rightarrow$ m - r is divisible by 9

$\Rightarrow$ There exists p $\in$ Z such that m - r =pq $\longrightarrow$(2)

n - m = k 9 $\quad$ m - r = p 9

n - r = n - m + m - r

= (n - m ) + (m - r)

= k 9 + p 9 =(k + p). 9

$\Rightarrow$

n - r = (k + p) 9

$\Rightarrow$ n - r is divisible by 9

i.e, n ≡ r (mod 9) $\Rightarrow$ (n, r) $\in$ R

$\therefore$ R is an equivalence relation

• Let A={Δ’s |Δ is a triangle in R²}

• R = {(Δ₁, Δ₂) $\in$ A$\times$A |Δ₁ is congruent to Δ₂}

Show that, R is an equivalence relation.

• i) R is symmetric Δ₁ is congruent to Δ₂ <$\Rightarrow$ Δ₂ is congruent to Δ₁

ii) Every triangle is congruent to itself R is reflexive.

iii) Suppose (Δ₁, Δ₂) $\in$ R and (Δ₂, Δ₃)$\in$ R

• (Δ₁, Δ₂) $\in$ R $\Rightarrow$ Δ₁ is congruent to Δ₂ $\longrightarrow$ (i)

• Similarly (Δ₂, Δ₃) $\in$ R $\Rightarrow$ Δ₂ is congruent to Δ₃ $\longrightarrow$ (ii)

• (i) & (ii) $\Rightarrow$ Δ₁ is congruent to Δ₃ $\Rightarrow$ (Δ₁, Δ₃) $\in$ R

$\therefore$ R is an equivalence relation.

• A= {Δ |Δ is a 2 - dimensional triangle}

• R ={(Δ₁, Δ₂)$\in$ A$\times$A |Δ₁ is similar to Δ₂)}

Here, R is an equivalence relation (same proof as in the previous example).

Let

• A = {A|A is a finite set}

• Define R on A as follows:

• (A, B) $\in$ R ,If the number of elements in A is equal to the number of elements in B.

• R = {(A, B)$\in$ A$\times$A | #(A) = #(B)}

Show that R is an equivalence relation.

i) Let (A, B)$\in$ R $\Rightarrow$ #(A) = #(B)

$\Rightarrow$ #(B) = #(A) $\Rightarrow$ (B, A)$\in$ R

$\therefore$ R is symmetric.

ii) Let A be a finite set

#(A)=#(A) $\Rightarrow$ (A, A)$\in$ R

$\therefore$ R is reflexive

iii) Let (A, B)$\in$ R and (B, C)$\in$ R

(A, B)$\in$ R $\Rightarrow$ #(A) = #(B) $\longrightarrow$(1)

similarly (B, C)$\in$ R $\Rightarrow$ #(B) = #(C) $\longrightarrow$(2)

By (1) & (2) we have #(A)= #(C)

$\therefore$ R is transitive.

Thus, R is an equivalence relation:

Let A = $\mathbb{R}^2$ \ {(0, 0)}-punctured plane

R = {((x₁, y₁), (x₂, y₂)) $\in$ A$\times$A |there exists a non-zero scaler λ$\in$R such that (x₁, y₁) = λ (x₂, y₂)}

Show that R is an equivalence relation.

i) Let ((x₁, y₁), (x₂, y₂))$\in$ R

$\Rightarrow$ there exists 0 ≠ λ $\in$ R such that (x₁, y₁) = λ(x₂, y₂)

$\Rightarrow$ $\frac{1}{\lambda}$ (x₁, y₁) = (x₂, y₂)

$\Rightarrow$ (x₂, y₂) = $\frac{1}{\lambda}$ (x₁, y₁)

$\Rightarrow$ ((x₂, y₂), (x₁, y₁))$\in$ R

$\therefore$ R is symmetric.

ii) Let (x₁, y₁)$\in$ A

(x₁, y₁) = 1.(x₁, y₁)

$\Rightarrow$ ((x₁, y₁), (x₁, y₁)) $\in$ R

$\therefore$ R is reflexive.

iii) Let ((x₁, y₁), (x₂, y₂)) $\in$ R & ((x₂, y₂), (x₃, y₃)) $\in$ R

((x₁, y₁), (x₂, y₂)) $\in$ R

$\Rightarrow$ there exists 0 ≠ λ $\in$ $\R$ such that (x₁, y₁) = λ (x₂, y₂) $\longrightarrow$ (1)

similarly ((x₂, y₂), (x₃, y₃)) $\in$R

$\Rightarrow$ there exists 0≠β $\in$ $\R$ such that (x₂, y₂) = β (x₃, y₃) $\longrightarrow$ (2)

Substituting (2) in (1), we get

(x₁, y₁) = λ (x₂, y₂)

= λ.β (x₃, y₃)

= (λβ) (x₃, y₃)

$\Rightarrow$ ((x₁, y₁), (x₃, y₃)) $\in$R

∴ R is transitive

Thus, R is an equivalence relation.

For any two distinct elements x & y in R, one & only one of the following holds:

i) [x]_R = [y]_R

ii) [x]_R ∩ [y]_R = $\phi$

Thank you