• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q8E

Expert-verified Found in: Page 581 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Show that the relation ${\mathbit{R}}{\mathbf{=}}{\mathbit{\varphi }}$ on a non-empty set ${\mathbf{\text{S}}}$is symmetric and transitive, but not reflexive.

Hence $\text{R}$ is symmetric, transitive and not reflexive.

See the step by step solution

## Step 1: Given data

The relation $R=\varphi$ on a non-empty set is given.

## Step 2: Concept used of relation

A relation${\mathbf{\text{R}}}$on a set ${\mathbf{\text{A}}}$is called reflexive if ${\mathbf{\left(}}{\mathbit{a}}{\mathbf{,}}{\mathbit{a}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$for every element ${\mathbit{a}}{\mathbf{\in }}{\mathbit{A}}$.

A relation ${\mathbf{\text{R}}}$ on a set ${\mathbf{\text{A}}}$is called symmetric if ${\mathbf{\left(}}{\mathbit{b}}{\mathbf{,}}{\mathbit{a}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$ whenever ${\mathbf{\left(}}{\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$, for all ${\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{\in }}{\mathbit{A}}$.

A relation ${\mathbf{\text{R}}}$ on a set ${\mathbf{\text{A}}}$ such that for all ${\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{\in }}{\mathbit{A}}$, if ${\mathbf{\left(}}{\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$ and ${\mathbf{\left(}}{\mathbit{b}}{\mathbf{,}}{\mathbit{a}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$ then ${\mathbit{a}}{\mathbf{=}}{\mathbit{b}}$ is called anti symmetric.

A relation${\mathbf{\text{R}}}$ on a set${\mathbf{\text{A}}}$ is called transitive if whenever${\mathbf{\left(}}{\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$and${\mathbf{\left(}}{\mathbit{b}}{\mathbf{,}}{\mathbit{c}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$ then ${\mathbf{\left(}}{\mathbit{a}}{\mathbf{,}}{\mathbit{c}}{\mathbf{\right)}}{\mathbf{\in }}{\mathbit{R}}$ for all${\mathbit{a}}{\mathbf{,}}{\mathbit{b}}{\mathbf{,}}{\mathbit{c}}{\mathbf{\in }}{\mathbit{A}}$

## Step 3: Solve for relation

Symmetric Since $\left(a,b\right)\in R$ is always false (as $\text{R}$is the empty set), the conditional statement $\left(\left(a,b\right)\in R\right)\to B$is true for any statement B.

Let B be the statement " $\left(b,a\right)\in R$ ". The conditional statement

"If $\left(a,b\right)\in R$, then $\left(b,a\right)\in R$ " is then always true and thus the relation $\text{R}$is symmetric by the definition of symmetric. Transitive Since $\left(a,b\right)\in R$ is always false (as $\text{R}$is the empty set) and since $\left(b,a\right)\in R$ is always false, the statement " $\left(a,b\right)\in R$, and $\left(b,a\right)\in R$ " is also always false. Then the conditional statement $\left(\left(a,b\right)\in R$ and $\left(b,a\right)\in R\right)\to B$ is true for any statement B.

Let B the statement " $a=b$ ". The conditional statement

" If $\left(a,b\right)\in R$ and $\left(b,a\right)\in R$, then $a=b$ is then always true and thus the relation $\text{R}$ is transitive by the definition of transitive.

Not reflexive For any element $a\in S:\left(a,a\right)\notin R$ because $\text{R}$ is the empty set. By the definition of reflexive, $R$ is then not reflexive.

Hence $\text{R}$ is symmetric, transitive and not reflexive. ### Want to see more solutions like these? 