Problem 5 - A Gentle Introduction to the Art of Mathematics

Question

We will use the symbol \(\mathbb{Z}^{*}\) to refer to the set of all integers except \(0 .\) Define a relation \(Q\) on the set of all pairs in \(\mathbb{Z} \times \mathbb{Z}^{*}\) (pairs of integers where the second coordinate is non- zero) by \((a, b) \mathbf{Q}(c, d) \Longleftrightarrow a d=\) \(b c .\) Show that \(Q\) is an equivalence relation.

Step-by-Step Solution

Answer

Relation Q is reflexive, symmetric, and transitive, thus is an equivalence relation.

Step 1: Define Equality under Relation Q

Given the relation \(Q\) on pairs \((a, b)\) and \((c, d)\) in \(\mathbb{Z} \times \mathbb{Z}^{*}\), define \(Q\) by \( (a, b) \mathbf{Q}(c, d) \Longleftrightarrow a d = b c \). This means that pairs \((a, b)\) and \((c, d)\) are related if and only if \(a d = b c\).

Step 2: Prove Reflexivity

For any pair \((a, b) \in \mathbb{Z} \times \mathbb{Z}^{*}\), we need to show that \((a, b) \mathbf{Q} (a, b)\). This implies \(a b = b a\), which is always true. Therefore, the relation \(Q\) is reflexive.

Step 3: Prove Symmetry

If \((a, b) \mathbf{Q} (c, d)\), then by definition \(a d = b c\). To show symmetry, we need to prove that \( (c, d) \mathbf{Q} (a, b) \). Given \(a d = b c\), it's clear that \(c b = d a\), which means \( (c, d) \mathbf{Q} (a, b)\). Thus, the relation \(Q\) is symmetric.

Step 4: Prove Transitivity

Assume \((a, b) \mathbf{Q} (c, d)\) and \((c, d) \mathbf{Q} (e, f)\). This means \(a d = b c\) and \(c f = d e\). To prove transitivity, show \((a, b) \mathbf{Q} (e, f)\), which means \(a f = b e\). Starting with \(a d = b c\), multiply both sides by \(f\) to get \(a d f = b c f\). Similarly, from \(c f = d e\), multiply both sides by \(b\) to obtain \(b c f = b d e\). Since \(a d f = b c f\) and \(b c f = b d e\), we conclude \(a f = b e\). Therefore, \( (a, b) \mathbf{Q} (e, f)\), proving transitivity.

Key Concepts

Understanding Reflexivity

Reflexivity is a fundamental property of equivalence relations. To say a relation is reflexive means that every element is related to itself. Consider the relation \( Q \) defined on pairs \((a, b)\) in \( \mathbb{Z} \times \mathbb{Z}^{*} \) by \( (a, b) \mathbf{Q} (a, b) \Longleftrightarrow a d = b c \). For reflexivity, we check if any pair \((a, b)\) is indeed relateable to itself. This requires \( a b = b a \), which holds true as multiplication is commutative. Thus, reflexivity of the relation \( Q \) is established since \( a b = b a \) is always satisfied.

Exploring Symmetry

Symmetry is another key aspect of equivalence relations. A relation \( Q \) is symmetric if whenever \( (a, b) \mathbf{Q} (c, d) \), it follows that \( (c, d) \mathbf{Q} (a, b) \). Given the definition of \( Q \), if \( a d = b c \), we need to show the reverse is also true. By rearranging \( a d = b c \), we obtain \( c b = d a \). Therefore, \( (c, d) \mathbf{Q} (a, b) \) holds true, confirming the symmetry of the relation \( Q \). In other words, if \((a, b)\) is related to \((c, d)\), then \((c, d)\) is indeed related to \((a, b)\) as well.

Grasping Transitivity

Transitivity forms the third pillar of equivalence relations. For \( Q \) to be transitive, if \( (a, b) \mathbf{Q} (c, d) \) and \( (c, d) \mathbf{Q} (e, f) \), then it must follow that \( (a, b) \mathbf{Q} (e, f) \). With \( a d = b c \) and \( c f = d e \), we need to establish \( a f = b e \). Transforming the first equality by multiplying both sides by \( f \): \( a d f = b c f \). Doing similarly with the second: \( c f = d e \) results in \( b c f = b d e \). Upon seeing that \( a d f = b c f \) and \( b c f = b d e \), the transitive property finalizes as \( a f = b e \). Thus, transitivity is proven, ensuring the relation \( Q \) is indeed transitive.