Problem 3 - A Book of Abstract Algebra

Question

If \((a, b) \sim(c, d)\) means \(a d=b c\), prove that \(\sim\) is an equivalence relation on \(S\).

Step-by-Step Solution

Answer

The relation \( \sim \) is reflexive, symmetric, and transitive, proving it is an equivalence relation.

Step 1: Define Equivalence Relation Properties

To prove that \( \sim \) is an equivalence relation on a set \( S \), we have to show that it is reflexive, symmetric, and transitive. We will verify each of these properties in the following steps.

Step 2: Prove Reflexivity

For a relation to be reflexive, every element must be related to itself. Consider any element \((a, b)\) in \( S \). According to the relation definition, we need \( a \cdot b = b \cdot a \), which is always true. Hence, the relation is reflexive because \( (a, b) \sim (a, b) \).

Step 3: Prove Symmetry

For the relation to be symmetric, if \( (a, b) \sim (c, d) \), then \( (c, d) \sim (a, b) \) should also hold. Since \( (a, b) \sim (c, d) \) implies \( a \cdot d = b \cdot c \), reversing the equation, we get \( c \cdot b = d \cdot a \). Therefore, \( (c, d) \sim (a, b) \) holds. Thus, the relation is symmetric.

Step 4: Prove Transitivity

For transitivity, if \( (a, b) \sim (c, d) \) and \( (c, d) \sim (e, f) \), then \( (a, b) \sim (e, f) \) should hold. We have: \( a \cdot d = b \cdot c \) and \( c \cdot f = d \cdot e \). From these, eliminate \( c \) and \( d \) to solve: \( a \cdot f = b \cdot e \). Therefore, \( (a, b) \sim (e, f) \) is valid, demonstrating that the relation is transitive.

Key Concepts

Reflexivity

Reflexivity is one of the essential properties of an equivalence relation. To say that a relation is reflexive means every element is related to itself. Let's consider our exercise where we examine a set of pairs \((a, b)\). The relation given is such that \((a, b) \sim (c, d)\) when \(a * d = b * c\). In order to show reflexivity, select any pair \((a, b)\). We check whether it relates to itself using the given relation. Here, you need to validate that \(a * b = b * a\). Clearly, this equation holds as both sides are identical. Therefore, each element relates to itself, confirming the reflexive property. Remember, reflexivity must hold for every single element in your set. It’s a critical foundational concept for establishing the results in equivalence relations.

Symmetry

For an equivalence relation, symmetry requires that if one element is related to another, then the reverse also holds true. In our exercise, for two pairs \((a, b)\) and \((c, d)\), we are given that \((a, b) \sim (c, d)\) when \(a * d = b * c\). Let's interpret symmetry: if \((a, b) \sim (c, d)\), then we must also have \((c, d) \sim (a, b)\). From \(a * d = b * c\), you can simply reverse it to get \(c * b = d * a\), which means \((c, d)\) is related to \((a, b)\) as well. This demonstrates symmetry in action, showing that relations are reciprocal. Understanding symmetry helps structure your reasoning about how elements within sets interact or relate in ways that can go back and forth.

Transitivity

Transitivity in equivalence relations can be thought of as a bridge connecting two related pairs through a common element. If an equivalence relation is transitive, it implies that relatedness can be extended across multiple elements in a sequence. In the provided exercise, the transitive property needs to be proven using pairs \((a, b)\), \((c, d)\), and \((e, f)\). Assume \((a, b) \sim (c, d)\) and \((c, d) \sim (e, f)\). This means:

• \(a * d = b * c\) (first relation)
• \(c * f = d * e\) (second relation)

To establish transitivity, you need to reach \(a * f = b * e\). Utilizing the above equalities, you can eliminate \(c\) and \(d\) effectively. This demonstrates the bridge built by transitivity, where the relation between the first and last pair is established. Transitivity aids in linking elements comprehensively, forming a coherent chain that maintains the relationship across the whole set.