Question

Suppose that the relation\(R\)is irreflexive. Is the relation\({R^*}\)necessarily irreflexive?

Short answer

\({R^*}\) is not irreflexive.

Step-by-step solution

Given

\(R\) is irreflexive and \((a,a) \notin R\).

Concept of Irreflexive and Composite

A relation\(R\) on a set\(A\)is irreflexive if\((a,a) \notin R\)for every element\(a \in A\).

The composite\(S^\circ R\)consists of all ordered pairs\((a,c)\)for which there exists an element\(b\)such that\((a,b) \in R\)and\((b,c) \in S\).

Check for Irreflexive

Given that\(R\)is irreflexive and\((a,a) \notin R\).

\({R^*}\)is not necessarily irreflexive. For example, if\((a,b) \in R\)and\((b,a) \in R\), then:

\(\begin{array}{l}(a,a) \in R^\circ R = {R^2}\\(b,b) \in R^\circ R = {R^2}\end{array}\)

Since\({R^*} = {R_1} \cup {R_2} \cup \ldots \cup {R^n}:{R^2} \subseteq {R^*}\)

\(\begin{array}{l}(a,a) \in {R^*}\\(b,b) \in {R^*}\end{array}\)

By the definition of irreflexive, \({R^*}\) is then not irreflexive.