Composition of Relations is a mathematical concept that links elements across different sets by combining multiple relations. This powerful tool is fundamental in fields like graph theory, function composition, and system modeling, enabling the exploration of complex connections and relationships within various structures.
In this article, we will understand the meaning of relations and its types, composition of relations, properties of composition of relations, its applications in mathematics and solved examples for better understanding.
Definition of Relations
Relation is a relationship between two sets Through diagrams, we can express that one set of parts interacts with another. The relation R is a subset of A x B; ( Here, on each side the sets/of groups involved in the operation are linked by the Cartesian product).
Since relation R from A to B is a subset of A x B this implies that R ⊆ A x B; for (a, b) ∈ R we say "a related to b by the given under consideration" and in symbolic manner write it as (aRb).
For example, we can represent our sets like this: A = {1, 2, 3} and B = {4, 5}. The relation R is returned from A to B, i.e.
R = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}
Now, it can be seen from the relation (R) in this example that set (A) is related to every element of Set (B) by the Relation.
Types of Relations
The primary types of relations are given below:
- Reflexive Relation: When all elements of set (A) are related to themselves through a relation (R), it is a reflexive relation. For any a ∈ A, (a, a) ∈ R. For instance, consider the set (A): {1, 2, 3}. A reflexive relation (R) is defined as {(1, 1), (2, 2), (3, 3)}, where each element of set (A) is related to itself.
- Symmetric Relation: A relation (R) is symmetric for a set (A) if (a R b) implies (b R a). From a mathematical perspective, (a, b) ∈ R implies (b, a) ∈ R for any a, b ∈ A. For instance, with a set (A) of {1, 2}, A relation (R) defined as {(1, 2), (2, 1)} is symmetric if it meets the above criterion.
- Transitive Relation: A relation (R) on a set (A) is transitive if aRb and bRc entail aRc. Mathematically, if a, b, c ∈ A, then (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R. Using A = {1, 2, 3}, the relation R = {(1, 2), (2, 3), (1, 3)} is transitive.
- Antisymmetric Relation: For a relation R on a set A to be antisymmetric, aRb and bRa must entail a = b. Mathematically, a = b for every a, b ∈ A, (a, b) ∈ R, and (b, a) ∈ R. For instance, the relation R = {(1, 1), (2, 2)} is antisymmetric with A = {1, 2}.
Composition of Relations
A composition of relations is a method whereby two sets are combined to produce a new one by use of a combination of every feasible pair of elements from the two sets, so indirectly.
For example, let us consider that we have three sets: A, B, and C. Let R be a relation from A to B and let S be a relation from B to C. The composition S ⚬ R is a relation from A to C that contains all pairs (a, c) such as there is an element b ∈ B such that (a, b) ∈ R and (b, c) ∈ S.
Suppose, we have three sets namely, A={1, 2}, B={3, 4}, and C={5, 6}, with R = {(1, 3), (2, 4)}, and S = {(3, 5), (4, 6)}. Then the composition S ⚬ R would be {(1, 5), (2, 6)}, because 1 is linked to 5 through 3 and 2 is linked to 6 through 4.
Mathematical Representation of Composition of Relations
The mathematical operation of composition of relations S ⚬ R is defined as:
S ⚬ R = (a, c) | ∃b ∈ B such that (a, b) ∈ R and (b, c) ∈ S
Interpreting the definition above:
- (a, b) ∈ R denotes a in A and b in B relate based on the relation R.
- (b, c) ∈ S denotes b in B and c in C relate based on the relation S
- (a, c) ∈ S ⚬ R reveals an indirect relationship between a in A and c in C based on b in B.
Composition of this form allows us to connect elements in multiple sets in a natural way, which underpins several mathematical and practical applications.
Properties of Composition of Relations
The properties of composition of relations are as follows:
- Identity Relation: For any set A, the identity relation IA on A is defined as IA = {(a, a) | a ∈ A}. The identity relation acts as a neutral element in the composition: R ⚬ IA = R and IB ⚬ R = R, where R ⊆ A x B.
- Associativity: Composition of relations is associative. If R ⊆ A x B, S ⊆ B x C and T ⊆ C x D then T ⚬ (S ⚬ R) = (T ⚬ S) ⚬ R.
- Inverse Relation: If R ⊆ A x B, the inverse relation R-1 ⊆ B x A is defined as (b, a) ∈ R-1 ⇔ (a, b) ∈ R. The composition of a relation and its inverse has properties similar to functions: R ⚬ R-1 ⊆ IB and R-1 ⚬ R ⊆ IA.
- Distributivity over Union: Composition of relations distributes over the union: R ⚬ (S ⋃ T) = (R ⚬ S) ⋃ (R ⚬ T) and (R ⋃ S) ⚬ T = (R ⚬ T) ⋃ (S ⚬ T).
- Monotonicity: If R1 ⊆ R2 and S1 ⊆ S2 , then: S1 ⚬ R1 ⊆ S2 ⚬ R2.
Applications of Composition of Relations
Some of the common applications of composition of relations are:
- Databases: Instead of discussing table relationships or repeating statements, join is done. When one table links two sets of entities and the other relates one to the third, a join operation (similar to composition) joins the two tables and their relationship.
- Computer Science (State Machines):State machines explicitly mathematically describe transition relations as a set of relations. If a system goes from A to B and then from B to C, these stages illustrate a clear transition.
- Graph Theory: Relations represent graph theory pathways. The composition of graph theory relations concatenates paths. The composition gives a road from A to C if there are paths from A to B and B to C.
- Formal Logic: Predicate logic usually combines relations. The more complicated predicate is the key benefit when examining dictionary relationships.
Conclusion
The composition of relations allows for pairing elements across different sets, revealing indirect connections. This concept is crucial in areas like function composition, graph theory, and system modeling. Key properties such as associativity, identity, and inverses help manage complex structures, aiding in both theoretical and practical problem-solving.
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Solved Examples on Composition of Relations
Example 1: Given the two relations as below:
- Relation (R) as {(1, 20), (24, 38)}
- Relation (S) as {(8, 9), (7, 6)}
Find the composition of this relations i.e. S ⚬ R.
Solution:
Now, let us consider that (a, b) ∈ R, and we need to find the corresponding b in S as (b, c) in order to form the composition of relations.
And finally after combining the pairs we can get S ⚬ R where (a, c) ∈ (S ⚬ R).
But, from the relation R we can find that R doesn't have any b values that matches with the first element of the pair in S. So, we can say that S ⚬ R = ∅.
Example 2: Given the two relations as below:
- Relation (R) as {(1, 20), (25, 35)}
- Relation (S) as {(20, 40), (35, 56)}
Find the composition of this relations i.e. S ⚬ R.
Solution:
Now, let us consider that (a, b) ∈ R, and we need to find the corresponding b in S as (b, c) in order to form the composition of relations.
And finally after combining the pairs we can get S ⚬ R where (a, c) ∈ (S ⚬ R) where. Now, from the given relations:
1. Now if we combine (1, 20) from R and (20, 40) from S it is going to result of (1,40).
2. We can merge (25, 35) from R and (35, 56) from S to get (25,56).
Therefore, we obtain S ⚬ R ={(1,40), (25, 56)}.
Example 3: Given the two matrices R and S
- R = \begin{bmatrix} 0 && 1 && 0 \\ 1 && 0 && 0 \\ 0 && 1 && 1\end{bmatrix}
- S = \begin{bmatrix} 1 && 0 && 1 \\ 0 && 1 && 0 \\ 1 && 0 && 1\end{bmatrix}
Find the matrix representing R ⚬ S.
Solution:
Now we multiply S by R to find the resulting matrix of the expression R ⚬ S.
R \circ S = \begin{bmatrix}1 && 0 && 1 \\ 0 && 1 && 0 \\ 1 && 0 && 1\end{bmatrix} \times \begin{bmatrix}0 && 1 && 0 \\ 1 && 0 && 0 \\ 0 && 1 && 1\end{bmatrix}
Now, index from 1 for both rows and column
1. For (1,1) entry: (1 x 0) + (0 x 1) + (1 x 0) = 0
2. For (1,2) entry: (1 x 1) + (0 x 0) + (1 x 1) = 2
3. For (1,3) entry: (1 x 0) + (0 x 0) + (1 x 1) = 1
4. For (2,1) entry: (0 x 0) + (1 x 1) + (0 x 0) = 1
5. For (2,2) entry: (0 x 1) + (1 x 0) + (0 x 1) = 0
6. For (2,3) entry: (0 x 0) + (1 x 0) + (0 x 1) = 0
7. For (3,1) entry: (1 x 0) + (0 x 1) + (1 x 0) = 0
8. For (3,2) entry: (1 x 1) + (0 x 0) + (1 x 1) = 2
9. For (3,3) entry: (1 x 0) + (0 x 0) + (1 x 1) = 1
Therefore the resulting matrix is the same as under:
R \circ S = \begin{bmatrix}0 && 2 && 1 \\ 1 && 0 && 0 \\ 0 && 2 && 1\end{bmatrix}
Example 4: Given the two relations as below:
- Relation (R) as {(a, a), (b, b)}
- Relation (S) as {(a, a), (b, b)}
Find the composition of this relations i.e. S ⚬ R.
Solution:
Now, let us consider that (x, y) ∈ R, and we need to find the corresponding y in S as (y, z) in order to form the composition of relations.
And finally after combining the pairs we can get S ⚬ R where (x, z) ∈ (S ⚬ R). Now, from the given relations:
1. Pairing (a, a) from R with (a, a) from S to get (a, a).
2. Pairing (b, b) from R with (b, b) from S to get (b, b).
So, from this we get, S ⚬ R is given as: {(a, a), (b, b)}.
Note: The relation resulted here reflects the reflexive property.
Example 5: Given the two functions as described below:
- f(x) is given as: 2x + 1
- g(x) is given as: 3x - 2
Find f(g(x)) and g(f(x)).
Solution: Now, in order to find f(g(x)) initially we need to substitute g(x) into f(x) and the final expression is given as follows:
f(g(x)) = 2*( 3x - 2) + 1 = 6x - 4 + 1 = 6x - 3
Now, in order to find g(f(x)) initially we need to substitute f(x) into g(x) and the final expression is given as follows:
g(f(x)) = 3*( 2x + 1) - 2 = 6x + 3 - 2 = 6x + 1
Note: The relation resulted here represents the composition in a function.
Example 6: Given the two relations as below:
- Relation (R) is given as: {(1, 2), (2, 3)}
- Relation (S = R-1 ) is given as: {(2, 1), (3, 2)}
Find the composition of this relations i.e. S ⚬ R.
Solution:
Now, let us consider that (a, b) ∈ R, and we need to find the corresponding b in S as (b, c) in order to form the composition of relations.
And finally after combining the pairs we can get S ⚬ R where (a, c) ∈ (S ⚬ R). Now, from the given relations:
1. Now if we combine (1, 2) from R and (2, 1) from S it is going to result of (1,1).
2. We can merge (2, 3) from R and (3, 2) from S to get (2,2).
Therefore, we obtain S ⚬ R ={(1, 1), (2, 2)}.
Note: As we can observe that S represents the inverse of R, so this type of relations are known as composition of inverse relations.
Example 7: Given the two relations as below:
- Relation (R) is given as: { (1, 2), (2, 3) }
- Relation (S) is given as: { (2, 1), (3, 2) }
Find S ⚬ R and R ⚬ S.
Solution:
Now, let us consider that (a, b) ∈ R, and we need to find the corresponding b in S as (b, c) in order to form the composition of relations.
And finally after combining the pairs we can get S ⚬ R where (a, c) ∈ (S ⚬ R). Now, from the given relations:
1. Now if we combine (1, 2) from R and (2, 1) from S it is going to result of (1,1).
2. We can merge (2, 3) from R and (3, 2) from S to get (2,2).
Therefore, we obtain S ⚬ R ={(1, 1), (2, 2)}.
Now, let us consider that (a, b) ∈ S, and we need to find the corresponding b in R as (b, c) in order to form the composition of relations.
And finally after combining the pairs we can get R ⚬ S where (a, c) ∈ (R ⚬ S). Now, from the given relations:
1. Now if we combine (2, 1) from R and (1, 2) from S it is going to result of (2,2).
2. We can merge (3, 2) from R and (2, 3) from S to get (3,3).
Therefore, we obtain R ⚬ S ={(2, 2), (3, 3)}.
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