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Suppose that the relation \(R\) is irreflexive. Is \(R^{2}\) necessarily irreflexive? Give a reason for your answer.

Short Answer

Expert verified
No, \(R^2\) is not necessarily irreflexive because the existence of elements \((a, c)\) and \((c, a)\) in \(R\) can imply \((a, a)\) in \(R^2\).

Step by step solution

01

Understand Irreflexive Relation

A relation \(R\) on a set \(A\) is irreflexive if for all elements \(a\) in \(A\), the pair \((a, a)\) is not in \(R\). In other words, no element is related to itself.
02

Define \(R^2\)

The composition \(R^2\) of \(R\) is defined such that for elements \(a, b\) in set \(A\), \((a, b)\) is in \(R^2\) if there exists an element \(c\) in \(A\) such that \((a, c)\) is in \(R\) and \((c, b)\) is also in \(R\).
03

Check \(R^2\) for Irreflexivity

To determine if \(R^2\) is irreflexive, check if \((a, a)\) is not in \(R^2\) for all \(a\) in \(A\). This happens if and only if there does not exist any \(c\) such that both \((a, c)\) and \((c, a)\) are in \(R\).
04

Analyze Relation

Since \(R\) is irreflexive, \((a, a)\) is not in \(R\) for any \(a\). However, this does not guarantee that there is no \(c\) such that both \((a, c)\) and \((c, a)\) are in \(R\). The presence of such a \(c\) would mean \((a, a)\) is in \(R^2\).
05

Provide Conclusion

Therefore, \(R^2\) is not necessarily irreflexive even if \(R\) is irreflexive, as there might exist elements \(a\) and \(c\) where \( (a, c) \) and \( (c, a) \) are in \( R \), leading to \( (a, a) \) being in \( R^2 \).

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Key Concepts

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

Irreflexive Relation
Let's dive into the concept of irreflexive relations. An irreflexive relation is a type of relation where no element relates to itself. Formally, for a set \( A \), a relation \( R \) on \( A \) is irreflexive if for every element \( a \) in \( A \), the pair (a, a) is not in \( R \). This means if you pick any element from the set, it should not be connected to itself through the relation \( R \).

For example, consider a set \( A = \{1, 2, 3\} \). An irreflexive relation \( R \) could be \( \{(1, 2), (2, 3), (3, 1)\} \). In this case, none of the elements 1, 2, or 3 have pairs such as (1, 1), (2, 2), or (3, 3) in the relation, maintaining the irreflexive property.
Relation Composition
Understanding relation composition helps to determine more complex properties of relations, such as whether composing a relation with itself retains certain properties. When you compose a relation \( R \) with itself, denoted as \( R^2 \), you are essentially linking the pairs in \( R \) through an intermediate step.

Formally, for elements \( a \) and \( b \) in a set \( A \), (a, b) is in \( R^2 \) if there exists an element \( c \) in \( A \) such that (a, c) is in \( R \) and (c, b) is also in \( R \). This means you find a connection from \( a \) to \( c \) and then from \( c \) to \( b \).
For example, if \( R = \{(1, 2), (2, 3)\} \), then \( R^2 \) includes the pair (1, 3) because you can go from 1 to 2, and then 2 to 3. Logical stitching of pairs this way builds the composite relation \( R^2 \).
Discrete Mathematics
Discrete Mathematics is a branch of mathematics dealing with countable, distinct elements. It covers a wide range of topics, including relations, functions, graphs, and algorithms. The study of relations, particularly irreflexive relations and their compositions, forms a core part of this field.
In discrete mathematics, a relation on a set is simply a collection of ordered pairs of elements from that set. Handling properties like reflexivity, symmetry, and transitivity helps in understanding and solving various problems.
Discrete mathematics emphasizes understanding finite systems. For example, when working with sets and relations, you might be managing database relationships, function mappings, or connections in a network. These concepts are foundational for computer science, as they directly apply to algorithms, data structures, and programming logic.

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Most popular questions from this chapter

Let \(R\) be a relation from a set \(A\) to a set \(B\) . The inverse relation from \(B\) to \(A,\) denoted by \(R^{-1}\) , is the set of ordered pairs \(\\{(b, a) |(a, b) \in R\\} .\) The complementary relation \(\overline{R}\) is the set of ordered pairs \(\\{(a, b) |(a, b) \notin R\\}\). Let \(R\) be the relation \(R=\\{(a, b) | a \text { divides } b\\}\) on the set of positive integers. Find \(\begin{array}{ll}{\text { a) } R^{-1} .} & {\text { b) } \overline{R}}\end{array}\)

Let \(R\) be a relation on a set \(A\) with \(n\) elements. If there are \(k\) nonzero entries in \(\mathbf{M}_{R},\) the matrix representing \(R,\) how many nonzero entries are there in \(\mathbf{M}_{R^{-1}},\) the matrix representing \(R^{-1}\) , the inverse of \(R ?\)

How can the directed graph representing the symmetric closure of a relation on a finite set be constructed from the directed graph for this relation?

A partition \(P_{1}\) is called a refinement of the partition \(P_{2}\) if every set in \(P_{1}\) is a subset of one of the sets in \(P_{2}\) . Show that the partition of the set of people living in the United States consisting of subsets of people living in the same county (or parish) and same state is a refinement of the partition consisting of subsets of people living in the same state.

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 \(\begin{array}{llll}{\text { a) } R^{2}} & {\text { b) } R^{3} .} & {\text { c) } R^{4}} & {\text { d) } R^{5}}\end{array}\)

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