• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q52E

Expert-verified
Discrete Mathematics and its Applications
Found in: Page 583
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Show that the relation \(R\) on a set \(A\) is antisymmetric if and only if \(R \cap {R^{ - 1}}\) is a subset of the diagonal relation \(\Delta = \{ (a,a)\mid a \in A\} \).

\(R\) is antisymmetric if and only if \(R \cap {R^{ - 1}} \subseteq \Delta \)

See the step by step solution

Step by Step Solution

Step 1: Given Data

Given: Relation \(R\) on a set \(A\) and \({R^{ - 1}}\) is the inverse relation of \(R\).

\(\Delta = \{ (a,a)\mid a \in A\} \)

Step 2: Concept of the antisymmetric, inverse relation and Intersection

Antisymmetric: A relation \(R\) on a set \(A\)is antisymmetric if \((b,a) \in R\) and \((a,b) \in R\) implies \(a = b\)

The inverse relation: \({R^{ - 1}}\) is the set \(\{ (b,a)\mid (a,b) \in R\} \)

Intersection \(A \cap B\): All elements that are both in \(A\) AND in \(B\).

Step 3: Proof that \(R \cap {R^{ - 1}} \subseteq \Delta \)

First part Let us assume that \(R\) is antisymmetric.

If \(R \cap {R^{ - 1}}\) is the empty set, then \(R \cap {R^{ - 1}}\) is a subset of \(\Delta \) (since the empty set is a subset of all sets).

Let us assume that \(R \cap {R^{ - 1}}\) is not empty and let \((a,b) \in R \cap {R^{ - 1}}\).

An element is in the intersection, if the element is in both sets:

\(\begin{array}{l}(a,b) \in R\\(a,b) \in {R^{ - 1}}\end{array}\)

By the definition of the inverse relation

\((b,a) \in R\)

We then know that \((a,b) \in R\)and \((b,a) \in R\). Since \(R\) is antisymmetric:

\(a = b\)

We then know that \(R \cap {R^{ - 1}}\) only contains elements of the form \((a,a)\). Since \((a,a) \in \Delta \) :

\(R \cap {R^{ - 1}} \subseteq \Delta \)

Step 4: Proof that \(R\) is antisymmetric

Second part Let us assume that \(R \cap {R^{ - 1}} \subseteq \Delta \). Let \((a,b) \in R\) and \((b,a) \in R\).

Since \({R^{ - 1}} = \{ (b,a)\mid (a,b) \in R\} \)

\(\begin{array}{l}(a,b) \in {R^{ - 1}}\\(b,a) \in {R^{ - 1}}\end{array}\)

An element is in the intersection if it is in both subsets:

\(\begin{array}{l}(a,b) \in R \cap {R^{ - 1}}\\(b,a) \in R \cap {R^{ - 1}}\end{array}\)

Since \(R \cap {R^{ - 1}} \subseteq \Delta \)

\(\begin{array}{l}(a,b) \in {\rm{ Delta }}\\(b,a) \in {\rm{ Delta }}\end{array}\)

Since \(\Delta = \{ (a,a)\mid a \in A\} \)

\(a = b\)

\(R\) is antisymmetric.

If \(R\) is antisymmetric, then \(R \cap {R^{ - 1}} \subseteq \Delta \) and If \(R \cap {R^{ - 1}} \subseteq \Delta \), then \(R\) is antisymmetric.

This is equivalent with \(R\) is antisymmetric if and only if \(R \cap {R^{ - 1}} \subseteq \Delta \)

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.