Relations

posted on: 04 Apr, 2012 | updated on: 15 Jun, 2012

Relation Algebra is the algebra of Sets that deals with a Set of finite relations that is closed under some operators. These operators work upon one or more relations.
In mathematics a relation is a set of ordered pairs and the symbol for the set is denoted by “”. There are some examples of relations,
(0, 2), (55, 21), (3, 50),
(0, 2), (2, 5), (3, 9),
A relation is a set of ordered pairs as said above and there is no matter about the Numbers that are in a relation. In short, any set of numbers would be called a relation so long as these numbers appears in pairs.
The Domain of a algebra relation is the set of all the first numbers arranged in the ordered pairs. In short the domain is all of the x values while The Range of a relation is the set of the second numbers that are arranged in the each pair or in other words it is the set of the ‘y’ values.
For example if we have a set of a relation as
S = (0,1), (2, 33), (40, 25),
Here domain of this relation is 0, 2, 40 and the range of the relation is 1, 33, 25.
Like the usual formula for the relation of the Position and time of a ball can be notices and the general formula that shows this relation is,
g (time) = position,
The term relation shows the relationship between the sets of information like consider about all the people in the class and their height. If the name of the students in the class and the height of the students are paired together then this is called a relation. In relations the pairs of names and heights are ordered that means names comes first and the other that is height comes second.
A function is a well mannered relation. All Functions are relations but not all relations are functions.
Relation algebra is an algebraic structure equipped with the Boolean operations of conjunction, dis-conjunction and negation, the Boolean constants, the relational operations and the converse in such a way that these operations and constants satisfy certain equations that constitutes algebra relations.
Relation algebra is equal to expressive power to relational Calculus and this result is named as Codd's theorem.

If there are three relations R₁, R₂, R₃ are defined on set A = a, b, c as follows.
R₁= (a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c) ,
R₂ = (a, b), (b, a), (a, c), (c, a),
R₃ = (a, b), (b, c), (c, a)
Find whether R₁, R₂, R₃is reflexive, symmetric and transitive?

Given that (1) R₁= (a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c),
                                       (2) R₂ = (a, b), (b, a), (a, c), (c, a) ,
                                       (3) R₃ = (a, b), (b, c), (c, a)
If we check the condition for reflexive i.e. a relation ‘R’ on Set ‘A’ is said to be reflexive if every element of ‘A’ is related to itself.

(1) Reflexive: - (a, a), (b, b), (c, c) ∊ R₁. So we can say that relation R₁ is reflexive.

Symmetric: - Here we observe that (a, b) ∊ R₁ but (b, a) ∉ R₁. So relation R₁is not symmetric on relation ‘A’.

Transitive: - Here we observe that (b, c) ∊ R₁ and (c, a) ∊ R₁but (b, a) ∉ R₁. So relation ‘R₁’ is not transitive on relation ‘A’.

(2) Reflexive: - (a, a), (b, b), (c, c) are not present in R₂. So we can say that relation R₁ is not reflexive.

Symmetric: - Here we observe that the order pair obtained by interchanging the components of ordered pairs in R₂ is also in R₂. So relation R₁is symmetric on relation on ‘A’.

Transitive: - Here we observe that (a, b) ∊ R₂ and (b, a) ∊ R₂but (a, a) ∉ R₂. So relation R₂is not transitive on relation ‘A’.

(3) Reflexive: - (a, a), (b, b), (c, c) are not present in R₃. So we can say that relation
           R₃ is not reflexive.

Symmetric: - We observe that the (b, c) ∊ R₃ but (c, a) ∉ R₃. So relation R₃is symmetric on relation on ‘A’.

Transitive: - We find that (a, b) ∊ R₃ and (b, c) ∊ R₃but (a, c) ∉ R₃. So relation R₃is not transitive on relation ‘A’.

Show that the relation R on set A = 2, 3, 4 is given by R = (2, 2), (3, 3), (4, 4), (2, 3), (3, 4). Is reflexive but neither symmetric nor transitive?

The relation ‘R’ on Set A = 2, 3, 4 where,
R = (2, 2), (3, 3), (4, 4), (2, 3), (3, 4)
Since 2, 3, 4 ∊ A and (2, 2), (3, 3), (4, 4) ∊ R i.e. for each a ∊ A, (a, a) ∊ R. So relation ‘R’ is reflexive.
And we find that (2, 3) ∊ R but (3, 2) ∉ R.
So relation ‘R’ is not symmetric.
Also, (2, 3) ∊ R and (3, 4) ∊ R but (2, 4) ∉ R. So ‘R’ is not transitive.

If there are three relations R₁, R₂, R₃ are defined on set A = 1, 2, 3 as follows.
R₁= (1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),
R₂ = (1, 2), (2, 1), (1, 3), (3, 1),
R₃ = (1, 2), (2, 3), (3, 1)
Find whether R₁, R₂, R₃is reflexive, symmetric and transitive?

Given that (1) R₁=(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),
(2) R₂ = (1, 2), (2, 1), (1, 3), (3, 1),
(3) R₃ = (1, 2), (2, 3), (3, 1).
If we check the condition for reflexive i.e. a relation ‘R’ on Set ‘A’ is said to be reflexive if every element of ‘A’ is related to itself.
(1) Reflexive: - (1, 1), (2, 2), (3, 3) ∊ R₁. So we can say that relation ‘R₁’ is reflexive.
Symmetric: - Here we observe that (1, 2) ∊ R₁ but (2, 1) ∉ R₁. So relation ‘R₁’ is not symmetric on relation ‘A’.
Transitive: - Here we observe that (2, 3) ∊ R₁ and (3, 1) ∊ R₁but (2, 1) ∉ R₁. So relation R₁is not transitive on relation ‘A’.

(2) Reflexive: - (1, 1), (2, 2), (3, 3) are not present in R₂. So we can say that relation ‘R₁’ is not reflexive.
Symmetric: - Here we observe that the order pair obtained by interchanging the components of ordered pairs in R₂ is also in R₂. So relation R₁is symmetric on relation on ‘A’.
Transitive: - Here we observe that (1, 2) ∊ R₂ and (2, 1) ∊ R₂but (1, 1) ∉ R₂. So relation R₂ is not transitive on relation A.

(3) Reflexive :-( 1, 1), (2, 2), (3, 3) are not present in R₃. So we can say that relation R₃ is not reflexive.
Symmetric: - We observe that the (2, 3) ∊ R₃ but (3, 1) ∉ R₃. So relation R₃is symmetric on relation on ‘A’.
Transitive: -We find that (1, 2) ∊ R₃ and (2, 3) ∊ R₃but (1, 3) ∉ R₃. So relation R₃is not transitive on relation A.

Show that the relation R on R defined as R = (p, q): p ≤ q, Is reflexive and transitive but not symmetric?

Given the relation R on R defined as R = (p, q): p ≤ q,
Here we have R = (p, q): p ≤ q, where p, q ∊
First we check reflexive condition:
Reflexivity: - For any a ∊ R, we have
               P ≤ P
     ⇒ (P, P) ∊ R for all P ∊ R.
     ⇒ So, ‘R’ is reflexive.
Symmetry: We find that (2, 3) ∊ R but (3, 2) ∉ R. so R is not symmetric.
Transitivity: let (P, q) ∊ R and (q, r) ∊ R. then,
         ⇒      P ≤ q and q ≤ r
         ⇒      p ≤ r
         ⇒    (p, r) ∊ R, so ‘R’ is transitive.

Find x and y, if (x + 3, 5) = (6, 2x + y)?

x + 3 = 6 and 2x + y = 5
Therefore x = 3 and 6 + y = 5
Therefore y = -1.

Let f: RR be given by f(x) = x² + 3 Find
(i) {x : f(x) = 28}
(ii) The pre-images of 39 and 2 under ‘f’.

(i) Given f(x) = 28 = x² + 3
        x² = 28 – 3 = 25
x = «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mn»25«/mn»«/msqrt»«/math» = 5
(ii)  To find f^-1 (39) and f^-1 (2)
f^-1 (39) = x  f (x) = 39 = x² + 3
x² = 39 – 3 = 36
x = = 6
f^-1 (2) = x f (x) = 2 = x² + 3
x² = 2 – 3 = -1
x = «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/msqrt»«/math» .