Equivalence Relation on a given set

Equivalence Relation

            A relation R, defined on a set P is an equivalence relation if 

          and only if

       (i) R is reflexive, i.e. p R p for all p ∈ P.

       (ii) R is symmetric, i.e. p R q ⇒ q R p for all p,    q ∈ P.

        (iii) R is transitive, i.e. p R q and q R r ⇒ p R r    for all p, q, r ∈ P.

        The relation defined by “x is equal to y” on the    set P of Integers  is an equivalence relation.

  Solved examples on equivalence              relation                     

      Q 1.  Let P be a set of triangles in a plane. The 

       relation R is defined as “m is                                                            

       similar to n, m, n ∈ P”.

 Solution:

 We see that R is;

     (i) Reflexive, for, every triangle is similar to  

     itself.

     (ii) Symmetric, for, if m is similar to n, then n

     is also similar to m.

     (iii) Transitive, for, if m be similar to n and n be

     similar to r, then m is also similar to r.

  Hence R is an equivalence relation.

        Q 2.   A relation R is defined on the set Z by “p 

        R q if p – q is divisible by 2” for

        p, q ∈ Z. Examine if R is an equivalence 

        relation on Z.

           Solution:

       (i) Let p ∈ Z. Then p – p is divisible by 2. 

     Therefore p R p holds for all p in Z and R is 

     reflexive.

  (ii) Let p, q ∈ Z and pRq hold. Then p – q is 

  divisible by 2 and therefore

  q – p is also divisible by 2.

  Thus, p R q ⇒ q R p and therefore R is 

  symmetric.

(iii) Let p, q, r ∈ Z and p R q, q R r both hold. Then p – q and q – r are both divisible by 2.

Therefore p – r = (p – q) + (q – r) is divisible by 2.

Thus, p R q and q R r  ⇒ p R r and therefore R is transitive.

Since R is reflexive, symmetric and transitive so, R is an equivalence relation on Z.

      Q.3.   Let x be a positive integer. A relation R is 

      defined on the set Z by “a R b

      if and only if p – q is divisible by x” for p, q ∈ Z. 

      Show that R is an

      equivalence relation on set Z.

Solution:

(i) Let p ∈ Z. Then p – p = 0, which is divisible by x

Therefore, p R p holds for all p ∈ Z.

Hence, R is reflexive.

(ii) Let p, q ∈ Z and p R q holds. Then p – q is divisible by x and therefore, q – p is also divisible by x.

Thus, p R q ⇒ q R p.

Hence, R is symmetric.

(iii) Let p, q, r ∈ Z and p R q, q R r both hold. Then p – q is divisible by x and q – r is also divisible by x. Therefore, p – r = (p – q) + (q – r) is divisible by x.

Thus,  p R q and q R r ⇒ p R r

Therefore, R is transitive.

Hence R is an equivalence relation on set Z

          

      Q.4. Let S be the set of all lines in 3 dimensional 

      space. A relation ρ is defined

      on S by “l ρ m if and only if l lies on the plane of 

       m” for l, m ∈ S.

      Examine if ρ is (i) reflexive, (ii) symmetric, 

    (iii) transitive

Solution:

(i) Reflexive: Let l ∈ S. Then l is coplanar with itself.

Therefore, lρl holds for all l in S.

Hence, ρ is reflexive

(ii)  Symmetric: Let l, m ∈ S and l ρ m holds. Then l lies on the plane of m.

Therefore, m lies on the plane of l. Thus, l ρ m ⇒  m ρ l and therefore ρ is symmetric.

(iii) Transitive: Let l, m, p ∈ S and l ρ m, m ρ p both hold. Then l lies on the plane of m and m lies on the plane of p. This does not always implies that l lies on the plane of p.

That is, l ρ m and m ρ p do not necessarily imply l ρ p.

Therefore, ρ is not transitive.

Since, R is reflexive and symmetric but not transitive so, R is not an equivalence relation on set Z

Transitive relation

TRANSITIVE RELATION

Let P be a set on which the relation R is defined.

R is said to be transitive, if

(p, q) ∈ R and (q, r) ∈ R ⇒ (p, r) ∈ R,

That is pRq and qRr ⇒ pRr where p, q, r ∈ P.

The relation is said to be non-transitive, if

(p, q) ∈ R and (q, r) ∈ R do not imply (p, r ) ∈ R.

For example, in the set P of whole numbers if the relation R be defined by ‘x less than y’ then

p < q and q < r imply p < r, that is, pRq and qRr ⇒ pRr.

Hence this  is a transitive relation.

Solved examples of transitive relation on given set:

1. Let L be given positive integer.

Let R = {(p, q) : p, q  ∈ Z and (p – q) is divisible by L}.

Show that R is transitive relation.

Solution:

Given R = {(p, q) : p, q ∈ Z, and (p – q) is divisible by L}.

Let p, q, r ∈ R such that (p, q) ∈ R and (q, r) ∈ R. Then

   ⇒ (p – q) is divisible by L,and (q – r) is divisible by L.

   ⇒ {(p – q) + (q – r)} is divisible by L.

   ⇒ (p – r) is divisible by L.

   ⇒ (p, r) ∈ R.

Therefore, (p, q) ∈ R and (q, r) ∈ R   ⇒ (p, r) ∈ R.

So, R is a transitive relation.

2. A relation ρ on the set N is given by “ρ = {(x, y) ∈ N × N : x is divisor of y}”. Examine whether ρ is transitive or not transitive relation on set N.

Solution:

Given ρ = {(x, y) ∈ N × N : x is divisor of y}.

Let p, q, r ∈ N and (p, q) ∈ ρ and  (q, r ) ∈ ρ. Then

  (p, q) ∈ ρ and  (q, r ) ∈ ρ  ⇒ p is divisor of q and q is divisor of r.

⇒ p is divisor of r.

 ⇒ (p, r) ∈ ρ

Therefore, (p, q) ∈ ρ and q, r) ∈ ρ ⇒ (p, r) ∈ ρ.

Hence, R is a transitive relation.

Symmetric Relation

SYMMETRIC RELATION

Let P be a set on which the relation R is defined. Then R is said to be a symmetric relation, if (p, q) ∈ R ⇒ (q, p) ∈ R, that is, pRq ⇒ qRp for all (p, q) ∈ R.

e.g- the set P of natural numbers. If a relation P be defined by “x + y = 5”, then this relation is symmetric in P, for

p + q = 5 ⇒ q + p = 5

But in the set P of natural numbers if the relation R be defined as ‘x is a divisor of y’, then the relation R is not symmetric as 4R8 does not imply 8R4  for, 4 divides 8 but 8 does not divide 4.

Solved example on symmetric relation on set:

1. A relation R is defined on the set Z by {p R q if p – q is divisible by 5} for p, q ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let p, q ∈ Z such that pRq hold. Then p – q is divisible by 5, i.e. p – q =5z and therefore q – p =5z hence q - p is divisible by 5.

Thus, pRq ⇒ qRp and therefore R is symmetric.

2. A relation R is defined on the set Z (set of all integers) by {pRq if and only if 2p + 3q is divisible by 5}, for all p, q ∈ Z. Examine if R is a symmetric relation on Z.

Solution:

Let p, q ∈ Z such that pRq holds i.e., 2p + 3q = 5z, which is divisible by 5. Now, 2q + 3p = 5p – 2p + 5q – 3q = 5(p + q) – (2p + 3q) is also divisible by 5.

Therefore pRq implies qRp for all p,q in Z i.e. R is symmetric.

3. Let R be a relation on Q, defined by R = {(p, q) : p, q ∈ Q and p – q ∈ Z}. Show that R is Symmetric relation.

Solution:

Given R = {(p, q) : p, q ∈ Q, and p – q ∈ Z}.

Let (p,q) ∈ R ⇒ (p – q) ∈ Z, i.e. (p – q) is an integer.

               ⇒ -(p – q) is an integer

               ⇒ (q – p) is an integer

               ⇒ (q, p) ∈ R

Thus, (p, q) ∈ R ⇒ (q, p) ∈ R

Therefore, R is symmetric.

4. Let m be given fixed positive integer.

Let R = {(p, q) : p, q  ∈ Z and (p – q) is divisible by m}.

Show that R is symmetric relation.

Solution:

Given R = {(p, q) : p, q ∈ Z, and (p – q) is divisible by m}.

Let (p, q) ∈ R . Then,

                ⇒ (p – q) is divisible by m

               ⇒ -(p – q) is divisible by m

               ⇒ (q – p) is divisible by m

               ⇒ (q, p) ∈ R

Thus, (p, q) ∈ R ⇒ (q, p) ∈ R

Therefore, R is symmetric relation on set Z.

Reflexive Relation

Reflexive Relation

It is a binary element in which every element is related to itself.

Let A be a set and R be the relation defined on it.

R is set to be reflexive, if (p, p) ∈ R for all p ∈ A that is, every element of A is R-related to itself, in other words pRp for every p ∈ A.

A relation R on a set A is not reflexive if there is at least one element p ∈ A such that (p, p) ∉ R.

Consider, for example, a set A = {l,m,n,o}.

The relation R1

But the relation R2

Solved example of reflexive relation on set:

1. A relation R is defined on the set Z (set of all integers) by “aRb if and only if 2a + 3b is divisible by 5”, for all a, b ∈ Z. Examine if R is a reflexive relation on Z.

Solution:

Let b ∈ Z. Now 2b + 3b = 5b, which is divisible by 5. Therefore bRb holds for all b in Z i.e. R is reflexive.

2. A relation R is defined on the set Z by “aRb if a – b is divisible by 5” for a, b ∈ Z. Examine if R is a reflexive relation on Z.

Solution:

Let p ∈ Z. Then p – p is divisible by 5. Therefore pRp holds for all p in Z i.e. R is reflexive.

3. Consider the set Z in which a relation R is defined by ‘aRb if and only if a + 3b is divisible by 4, for a, b ∈ Z. Show that R is a reflexive relation on on setZ.

Solution:

Let a ∈ Z. Now a + 3a = 4a, which is divisible by 4. Therefore aRa holds for all a in Z i.e. R is reflexive.

4. A relation ρ is defined on the set of all real numbers R by ‘xρy’ if and only if |x – y| ≤ y, for x, y ∈ R. Show that the ρ is not reflexive relation.

Solution:

The relation ρ is not reflexive as x = -2 ∈ R but |x – x| = 0 which is not less than -2(= x).