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Chapter 9: Problem 58

Let \(R\) be the relation on the set \(\\{1,2,3,4,5\\}\) containing the ordered pairs \((1,1),(1,2),(1,3),(2,3),(2,4),(3,1),\) \((3,4),(3,5),(4,2),(4,5),(5,1),(5,2),\) and \((5,4) .\) Find \(\begin{array}{llll}{\text { a) } R^{2}} & {\text { b) } R^{3} .} & {\text { c) } R^{4}} & {\text { d) } R^{5}}\end{array}\)

Short Answer

Expert verified
a) R^2 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,3), (2,4), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,2), (4,4), (4,5), (5,1), (5,2), (5,4) }b) R^3 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }c) R^4 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }d) R^5 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }

Step by step solution

01

Understanding the Problem

The problem requires finding the powers of a given relation R on a set. Specifically, we need to find R^2, R^3, R^4, and R^5.
02

Definition Review

By definition, if R is a relation on set A, then R^2 is the relation that includes all pairs (a, c) such that there exists some b in A for which (a, b) and (b, c) are in R. Similarly, R^3 includes all pairs (a, d) such that there exist b and c in A for which (a, b), (b, c), and (c, d) are in R, and so on.
03

Finding R^2

To find R^2, look for pairs (x, z) such that there exists an element y for which (x, y) and (y, z) are both in R.Given R = { (1,1), (1,2), (1,3), (2,3), (2,4), (3,1), (3,4), (3,5), (4,2), (4,5), (5,1), (5,2), (5,4) }:Pairs are: (1,3),(1,4),(1,4), (1,5),(2,1),(2,5),(3,2),(3,3),(3,4),(3,5),(4,1),(4,3),(4,4),(4,5),(5,1),(5,3),(5,4),(5,5)R^2 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,3), (2,4), (2,5), (3,1), (3,2), (3,4), (3,5), (4,1), (4,2), (4,4), (4,5), (5,1), (5,2), (5,4) }
04

Finding R^3

To find R^3, look for pairs (x, w) such that there exist elements y and z for which (x, y), (y, z), and (z, w) are in R.Continuing from R^2:Pairs: (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(3,5),(4,1),(4,2),(4,3),(4,4),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)R^3 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }
05

Finding R^4

To find R^4, look for pairs (x, v) such that there exist elements y, z, and w for which (x, y), (y, z), (z, w), and (w, v) are in R.Continuing from R^3:R^4 = R^3 (every connection is preserved)R^4 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }
06

Finding R^5

To find R^5, look for pairs (x, u) such that there exist elements y, z, w, and v for which (x, y), (y, z), (z, w), (w, v), and (v, u) are in R.Continuing from R^4:R^5 = R^4 (every connection is preserved)R^5 = { (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5) }

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relation Composition
In discrete mathematics, a relation on a set is a collection of ordered pairs of elements from the set. The composition of relations is a fundamental concept that helps us understand how two relations can be combined to form a new relation. This concept is closely tied with the idea of chaining operations.
Matrix Multiplication
Another crucial concept related to relation powers is matrix multiplication. It provides a more formalized approach to solving relation compositions, especially in discrete mathematics.
Discrete Mathematics
Discrete mathematics is a branch of mathematics that deals with discrete elements that use distinct and separate values. It encompasses various topics critical for theoretical computer science and combinatorial mathematics.

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Most popular questions from this chapter

Show that the poset of rational numbers with the usual less than or equal to relation, \((\mathbf{Q}, \leq),\) is a dense poset.

List the ordered pairs in the equivalence relations produced by these partitions of \(\\{0,1,2,3,4,5\\}\) a) \(\\{0\\},\\{1,2\\},\\{3,4,5\\}\) b) \(\\{0,1\\},\\{2,3\\},\\{4,5\\}\) c) \(\\{0,1,2\\},\\{3,4,5\\}\) d) \(\\{0\\},\\{1\\},\\{2\\},\\{3\\},\\{4\\},\\{5\\}\)

Which of these collections of subsets are partitions of the set of bit strings of length 8? a) the set of bit strings that begin with 1, the set of bit strings that begin with 00, and the set of bit strings that begin with 01 b) the set of bit strings that contain the string 00, the set of bit strings that contain the string 01, the set of bit strings that contain the string 10, and the set of bit strings that contain the string 11 c) the set of bit strings that end with 00, the set of bit strings that end with 01, the set of bit strings that end with 10, and the set of bit strings that end with 11 d) the set of bit strings that end with 111, the set of bit strings that end with 011, and the set of bit strings that end with 00 e) the set of bit strings that contain 3k ones for some nonnegative integer k, the set of bit strings that contain 3k + 1 ones for some nonnegative integer k, and the set of bit strings that contain 3k + 2 ones for some nonnegative integer k.

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Show that the partition of the set of all identifiers in C formed by the equivalence classes of identifiers with respect to the equivalence relation \(R 31\) is a refinement of the partition formed by equivalence classes of identifiers with respect to the equivalence relation \(R 8\) . (Compilers for "old" C consider identifiers the same when their names agree in their first eight characters, while compilers in standard C consider identifiers the same when their names agree in their first 31 characters.)
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