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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter Relations and Functions. This page offers a step-by-step solution to the specific question from Exercise 1, Question 8: Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
A = {1, 2, 3, 4, 5}
R = { (a,b) ; |a – b| is even}
It is clear that for any element a ∈A, we have |a -a| = 0(which is even).
∴R is reflexive.
Let (a, b) ∈ R.
=> |a –b| is even.
=> |- (a –b)| = |b - a| is also even.
=> (b, a) ∈ R is even.
A = {1, 2, 3, 4, 5}
R = { (a, b) : | a – b| is even}
It is clear that for any element a ∈A, we have |a - a | = 0 (which is even).
∴R is reflexive.
Let (a, b) ∈ R.
⇒ |a –b| is even.
⇒ |- (a –b)| = |b - a| is also even.
⇒ (b, a) ∈ R is even.
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (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.
∴R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ |a –b| is even and |(b –c)| is even.
⇒ (a – b) is even and (b –c ) is even.
⇒ (a –c ) = (a – b) + (b – c ) is even. [ Sum of two even integers is even]
⇒ |a – c | is even.
⇒ (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.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is ....
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