prove that composition of binary relation is assosciative

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 \( A \) be a set, and let \( R \), \( S \), and \( T \) be binary relations on \( A \). That is, \( R \), \( S \), and \( T \) are subsets of \( A \times A \), the Cartesian product of \( A \) with itself.
   
2. The composition of two relations \( R \) and \( S \), denoted \( R \circ S \), is defined as:
   \[
   R \circ S = \{(x, z) \in A \times A \mid \exists y \in A \text{ such that } (x, y) \in S \text{ and } (y, z) \in R \}.
   \]
   In words, the pair \( (x, z) \) is in \( R \circ S \) if there exists an intermediate element \( y \in A \) such that \( (x, y) \in S \) and \( (y, z) \in R \).

### Goal:
We need to prove that the composition of relations is associative. That is, for any relations \( R \), \( S \), and \( T \) on a set \( A \), we must show:
\[
(R \circ S) \circ T = R \circ (S \circ T).
\]

### Proof:

1. **Start with the left-hand side** \( (R \circ S) \circ T \):
   - By the definition of composition, \( (R \circ S) \circ T \) consists of all pairs \( (x, w) \in A \times A \) such that there exists a \( z \in A \) where \( (x, z) \in T \) and \( (z, w) \in R \circ S \).
   - From the definition of \( R \circ S \), we know that \( (z, w) \in R \circ S \) if there exists a \( y \in A \) such that \( (z, y) \in S \) and \( (y, w) \in R \).
   - So, \( (x, w) \in (R \circ S) \circ T \) if and only if there exist \( z, y \in A \) such that:
     - \( (x, z) \in T \),
     - \( (z, y) \in S \),
     - \( (y, w) \in R \).

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

3. **Comparison of both sides**:
   - From the two definitions above, we can see that both \( (R \circ S) \circ T \) and \( R \circ (S \circ T) \) contain exactly the same pairs \( (x, w) \in A \times A \), as they rely on the same existence conditions for intermediate elements \( y \) and \( z \).

Thus, we conclude:
\[
(R \circ S) \circ T = R \circ (S \circ T),
\]
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.

prove that composition of binary relation is associative

Explore the proof of why the composition of binary relations is associative, using set theory and algebraic concepts. Understand the core definitions and step-by-step process involved in proving this fundamental property in mathematics. Ideal for undergraduate students studying set theory and algebra.

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