Math Problem Statement

prove that composition of binary relation is assosciative

Solution

To prove that the composition of binary relations is associative, let's define some basic terms and then proceed with the proof.

Definitions:

  1. Let AA be a set, and let RR, SS, and TT be binary relations on AA. That is, RR, SS, and TT are subsets of A×AA \times A, the Cartesian product of AA with itself.

  2. The composition of two relations RR and SS, denoted RSR \circ S, is defined as: RS={(x,z)A×AyA such that (x,y)S and (y,z)R}.R \circ S = \{(x, z) \in A \times A \mid \exists y \in A \text{ such that } (x, y) \in S \text{ and } (y, z) \in R \}. In words, the pair (x,z)(x, z) is in RSR \circ S if there exists an intermediate element yAy \in A such that (x,y)S(x, y) \in S and (y,z)R(y, z) \in R.

Goal:

We need to prove that the composition of relations is associative. That is, for any relations RR, SS, and TT on a set AA, we must show: (RS)T=R(ST).(R \circ S) \circ T = R \circ (S \circ T).

Proof:

  1. Start with the left-hand side (RS)T(R \circ S) \circ T:

    • By the definition of composition, (RS)T(R \circ S) \circ T consists of all pairs (x,w)A×A(x, w) \in A \times A such that there exists a zAz \in A where (x,z)T(x, z) \in T and (z,w)RS(z, w) \in R \circ S.
    • From the definition of RSR \circ S, we know that (z,w)RS(z, w) \in R \circ S if there exists a yAy \in A such that (z,y)S(z, y) \in S and (y,w)R(y, w) \in R.
    • So, (x,w)(RS)T(x, w) \in (R \circ S) \circ T if and only if there exist z,yAz, y \in A such that:
      • (x,z)T(x, z) \in T,
      • (z,y)S(z, y) \in S,
      • (y,w)R(y, w) \in R.
  2. Now consider the right-hand side R(ST)R \circ (S \circ T):

    • By the definition of composition, R(ST)R \circ (S \circ T) consists of all pairs (x,w)A×A(x, w) \in A \times A such that there exists a yAy \in A where (x,y)ST(x, y) \in S \circ T and (y,w)R(y, w) \in R.
    • From the definition of STS \circ T, we know that (x,y)ST(x, y) \in S \circ T if there exists a zAz \in A such that (x,z)T(x, z) \in T and (z,y)S(z, y) \in S.
    • So, (x,w)R(ST)(x, w) \in R \circ (S \circ T) if and only if there exist z,yAz, y \in A such that:
      • (x,z)T(x, z) \in T,
      • (z,y)S(z, y) \in S,
      • (y,w)R(y, w) \in R.
  3. Comparison of both sides:

    • From the two definitions above, we can see that both (RS)T(R \circ S) \circ T and R(ST)R \circ (S \circ T) contain exactly the same pairs (x,w)A×A(x, w) \in A \times A, as they rely on the same existence conditions for intermediate elements yy and zz.

Thus, we conclude: (RS)T=R(ST),(R \circ S) \circ T = R \circ (S \circ T), which proves that the composition of binary relations is associative.


Would you like to explore further details about the proof, or ask any questions? Here are some related queries:

  1. How is the composition of relations used in real-world applications?
  2. Can the composition of functions also be shown to be associative?
  3. How does this proof change if we extend to more than two relations?
  4. Are there any types of relations where this property doesn't hold?
  5. Can you illustrate the associativity of relation composition with an example?

Tip: When studying properties of relations, visualizing the set relationships using diagrams can often help make abstract concepts clearer.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Binary Relations
Algebra

Formulas

R ∘ S = {(x, z) | ∃ y ∈ A such that (x, y) ∈ S and (y, z) ∈ R}

Theorems

Associative Property of Relation Composition

Suitable Grade Level

Undergraduate (College Level)