Problem 3 - Applied Discrete Structures

Question

Let \(A=\\{1,2,3,4,5\\}\) and define \(r\) on \(A\) by \(x r y\) iff \(x+1=y .\) We define \(r^{2}=r r\) and \(r^{3}=r^{2} r .\) Find: (a) \(r\) (b) \(r^{2}\) (c) \(r^{3}\)

Step-by-Step Solution

Answer

(a) \( r = \{(1, 2), (2, 3), (3, 4), (4, 5)\} \), (b) \( r^2 = \{(1, 3), (2, 4), (3, 5)\} \), (c) \( r^3 = \{(1, 4), (2, 5)\} \).

Step 1: Understanding the Relation r

The relation "\( x \, r \, y \)" is defined such that \( x+1=y \). This means for each element in set \( A=\{1,2,3,4,5\} \), it relates \( x \) to \( y \) if \( y \) is the immediate successor of \( x \). We need to list down such pairs from set \( A \).

Step 2: Listing Pairs for r

Looking at set \( A \) and the relation criterion \( x+1=y \), we find the following pairs: for \( x = 1 \), \( y = 2 \); for \( x = 2 \), \( y = 3 \); for \( x = 3 \), \( y = 4 \); for \( x = 4 \), \( y = 5 \). Thus, \( r = \{(1, 2), (2, 3), (3, 4), (4, 5)\} \).

Step 3: Finding r squared (r^2)

The relation \( r^2 \) is the composition of \( r \) with itself, meaning \( x \, r^2 \, z \) if there exists a \( y \) such that \( x \, r \, y \) and \( y \, r \, z \). Therefore, for the pairs in \( r \), we compose: \( (1, 3) \) because \((1, 2)\) and \((2, 3)\); \( (2, 4) \) because \((2, 3)\) and \((3, 4)\); and \( (3, 5) \) because \((3, 4)\) and \((4, 5)\). Thus, \( r^2 = \{(1, 3), (2, 4), (3, 5)\} \).

Step 4: Finding r cubed (r^3)

The relation \( r^3 \) is calculated by composing \( r^2 \) with \( r \), meaning \( x \, r^3 \, w \) if there exists a \( z \) such that \( x \, r^2 \, z \) and \( z \, r \, w \). Applying this to our relations: \( (1, 4) \) because \((1, 3)\) and \((3, 4)\); and \( (2, 5) \) because \((2, 4)\) and \((4, 5)\). Hence, \( r^3 = \{(1, 4), (2, 5)\} \).

Key Concepts

Set theory

Set theory is a foundational mathematical theory that studies collections of objects, known as sets. A set is a well-defined collection of distinct elements. In this context, the set \(A=\{1,2,3,4,5\}\) is an example of a finite set containing five elements. In set theory, elements are unique and listed without repetition.

In our exercise, we deal with set \(A\), and utilize its elements to explore relationships or binary relations between the members. Understanding set theory is crucial because it helps us understand how to define collections and manipulate these collections in various mathematical frameworks.

A few fundamental concepts in set theory include:

• **Elements:** Basic objects or members within a set, e.g., 1, 2, 3, 4, and 5 in our example.
• **Subset:** A set whose elements are all contained within another set. If set \(B\) contains only elements from \(A\), then \(B\) is a subset of \(A\).
• **Cardinality:** The number of elements in a set, for set \(A\), its cardinality is 5.
• **Union, Intersection, and Difference:** Operations allowing formation of new sets by combining or comparing two sets. While not directly used in this problem, these operations are fundamental in set explorations.

Through manipulating sets like \(A\), we can delve into different mathematical relations, gaining insights into how separate elements can be interlinked.

Binary relations

Binary relations describe how elements from one set relate to elements from another set or the same set. In this exercise, we define a binary relation \(r\) on set \(A\) with the condition \(x \ r \ y\) if \(x+1=y\). This creates ordered pairs, revealing specific relationships between numbers in the set.

A binary relation on a set can be thought of as a subset of the Cartesian product of the set with itself. For our set \(A\), the pairs satisfying condition \(x+1=y\) form the relation \(r\), i.e., \(\{(1,2), (2,3), (3,4), (4,5)\}\).

Binary relations encompass various properties:

• **Reflexive:** Every element relates to itself. \(r\) does not satisfy this because \(x+1 eq x\).
• **Symmetric:** If \(x \, r \, y\), then \(y \ r \ x\). \(r\) is not symmetric because if \(1 \, r \, 2\), \(2 \, r \, 1\) doesn't hold.
• **Transitive:** If \(x \, r \, y\) and \(y \, r \, z\), then \(x \, r \, z\). In itself, \(r\) requires further compositions for transitive relationships.
• **Antisymmetric:** If \(x \, r \, y\) and \(y \, r \, x\), then \(x = y\). As with symmetry, \(r\) is not antisymmetric here.

Understanding these properties helps identify whether we need simple or compound operations to achieve desired outcomes like finding \(r^2\) or \(r^3\).

Relation composition

Relation composition is a method to combine two relations to form a new one. In our exercise, we perform this by composing the relation \(r\) with itself to find \(r^2\) and then composing that result with \(r\) again to determine \(r^3\).

Here's how relation composition works:

• **Composition for \(r^2\):** We combine \(r\) with itself. This means finding pairs where an intermediate element connects consecutive relations such that if \((x, y)\) and \((y, z)\) exist in \(r\), then \((x, z)\) is part of \(r^2\). From our example, \(r^2 = \{(1, 3), (2, 4), (3, 5)\}\).
• **Composition for \(r^3\):** We take \(r^2\) and find its composition with \(r\). So we find such \((x, w)\) pairs where the intermediate steps include \((x, z)\) in \(r^2\) and \((z, w)\) in \(r\). For instance, \((1, 4)\) and \((2, 5)\) are included in \(r^3\).

Using such compositions, relations transform into more complex relationships, illustrating how initial binary relations can cascade into broader connections across elements in a set. This process is fundamental in understanding dynamic relations and higher-order connections in discrete structures.