50 points ☺
When the relation R will be reflexive and Symmetric.
Give two or more cases :)
Answer according to the points :)
Answers
Relation R is reflexive for every ' a'
belongs to A then (a,a) belongs to R
example:1 element is equal to itself
example:2 R = {(1,1),(2,2),(3,3)} in A =
{ 1,2,3} is reflexive
example:3. A is subset of A
Relation R is symmetric if every
(a,b)belongs to R implies (b,a) belongs
to R then it is symmetric
example:1 R = {(1,1),(2,3),(3,2),(2,2)} is
A= {1,2,3} is symmetric
example:2 A is subset to B then B is the
subset of B
First of all, let us learn about sets. Then we will move to relations.
❈ Definition of sets :
⇒ According to German mathematician G. Cantor- A set is a well-defined collection of distinct objects of our perception or of our thought, to be conceived as a whole.
• If S be a set and a, b, c belong to S, then we can write a, b, c ∈ S
• If S be a set and d doesn't belong to S, then we can write d ∉ S
❈ Writing a set :
A set S is expressed as
S = {x : x has a property p}
❈ A few examples :
1. Set of Natural Numbers
ℕ = {1, 2, 3, 4, ...}
2. Set of Integers
ℤ = {- ∞, ..., - 1, 0, 1, ..., ∞}
3. Set of Rational Numbers
ℚ = {x : x = a/b, a, b ∈ ℤ, b ≠ 0}
4. Set of Real Numbers ℝ is the superset of ℕ, ℤ, ℚ and Irrational numbers. It means that all elements of ℕ, ℤ, ℚ and Irrational numbers belong to the set of Real numbers ℝ.
5. Set of natural numbers less than 20
S = {n ∈ ℕ : n < 20}
6. Set of Complex Numbers is the superset of the set of Real Numbers ℝ.
=====⇒
In order to learn about Relations, we must learn about Cartesian product of sets first.
Let, A and B be two non-empty sets. The Cartesian product of this two sets is denoted by A × B and defined by
A × B = {(a, b) : a ∈ A, b ∈ B}
=====⇒
Since we have learnt about Sets, Cartesian products, we move to Relations.
❈ Relation : Let, A and B are two non-empty sets. A binary relation R between A and B is a subset of A × B.
• If (a, b) ∈ R, we can say that a is related to b and is denoted by aRb.
• If (a, b) ∉ R, we can say that a is not related to b.
❈ Divisions of a relation :
Let, S be a non-empty set and R be a binary relation on S.
1. Reflexive relation : The relation R is said to be reflexive if (a, a) ∈ R for all a in S.
Here, aRa is reflexive.
Examples :
i) aRa iff (if and only if) a - b is divisible by 5, where a, b ∈ ℤ and R is defined on ℤ.
⇒ Let, a ∈ ℤ. Then a - a = 0, divisible by 5. Then aRa holds. R is reflexive.
2. Symmetric relation : The relation R is said to be reflexive if (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b in S.
Examples :
i) aRb iff (if and only if) a - b is divisible by 5, where a, b ∈ ℤ and R is defined on ℤ.
⇒ Let, a, b ∈ ℤ and aRb holds. Then a - b is divisible by 5 and since a, b are Integers, b - a is also divisible by 5.
Thus aRb ⇒ bRa, and therefore R is symmetric.
3. Transitive relation : If for any three elements a, b, c in S, aRb & bRc ⇒ cRa, R is said to be transitive relation.
Example :
i) aRb iff (if and only if) a - b is divisible by 5, where a, b ∈ ℤ and R is defined on ℤ.
⇒ Let, a, b, c ∈ ℤ and aRb, bRc hold.
aRb ⇒ (a - b) = 5h (say) ...(i)
bRc ⇒ (b - c) = 5k (say) ...(ii)
Now, a - c = (a - b) + (b - c)
= 5h - 5k = 5 (h - k), divisible by 5
Thus, aRc holds and therefore R is transitive.
4. Equivalence relation : If the relation R on any set be reflexive, symmetric and transitive altogether, R is called an Equivalence relation.
Example :
i) aRb iff (if and only if) a - b is divisible by 5, where a, b ∈ ℤ and R is defined on ℤ.
⇒ As we have proved earlier that R is reflexive, symmetric and transitive at a time, R is said to be equivalence relation in this case.
❈ Homeworks for you :
• Determine the nature of the following relations R on the set ℤ.
i) aRb if and only if a ∈ ℤ, b ∈ ℤ and a ≤ b,
ii) aRb if and only if a ∈ ℤ, b ∈ ℤ and ab ≥ 0,
iii) aRb if and only if a ∈ ℤ, b ∈ ℤ and a² + b² is a multiple of 2,
iv) aRb if and ony if a ∈ ℤ, b ∈ ℤ and 2a + 3b is divisible by 5.