show tht the R on the set A={1,2,3,4,5},given by

R={(x,y):mod of a-b is evn},is an equivalence relation.

show that all the elements of[1,3,5} are related to each other al the elemnts of{2,4} are related to each other.But,no element of{1,3,5}related to any element of {2,4}.

A = {1, 2, 3, 4, 5}

It is clear that for any element a A, we have (which is even).

∴R is reflexive.

Let (a, b) ∈ R.

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ (a, c) ∈ R

∴R is transitive.

Hence, R is an equivalence relation.

Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

  • 4

the ans is given in ncert exercise solutions, ex 1.1 question 8

  • 0

We all know that a relation R in a set A is said to be an equivalence if it is reflexive ,symmetric ,transitive.First consider its reflexivity:

Let a ∈A if (a,a) ∈R implies Ia-aI is even implies 0 which is even .so R is reflexive.

symmetric:

If (a,b) ∈R implies Ia-bI is even .implies I-(a-b)I is even .implies Ib-aI is even.implies (b,a) ∈ R.so r is symmetric.

Transitive:

if (a,b) ∈ R and (b,a) ∈ R implies Ia-bI is even and Ib-aI is even .

Now,

  Ia-bI + Ib-cI is also even(sum of two even number is even).implies

  Ia-b+b-cI implies

  Ia-cI is even.implies

  (a,c) ∈ R.so R is transitive. 

TO SHOW (1,3,5) ∈ R,

(1,1)=I1-1I is even

(1,3)=I1-3I is even

(1,5)= i1-5I is even

(3,1)=I3-1I is even

(3,3)= I3-3I is even

(3,5)= I3-5I is even

(5,1)=I5-1I is even

(5,3)=I5-3I is even

(5,5)= i5-5I is even.

TO SHOW (2,4) ∈ R,

(2,2)= I2-2-I s even.

(2,4)=I2-4I is even

(4,2)=I4-2I is even

(4-4)= I4-4I is even

but no element  of  (1,3,5) is R =(2,4) there exist a ∈A and b ∈ B.Ia-bI is odd as I1-2I is odd .I3-2I is odd,I5-4I is odd.

SOLVED

  • -1
What are you looking for?