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Q35E

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Found in: Page 582

### Discrete Mathematics and its Applications

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

# Exercises 34–37 deal with these relations on the set of real numbers:$${R_1} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a > b} \right\},$$ the “greater than” relation,$${R_2} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a \ge b} \right\},$$ the “greater than or equal to” relation,$${R_3} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a < b} \right\},$$ the “less than” relation,$${R_4} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a \le b} \right\},$$ the “less than or equal to” relation,$${R_5} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a = b} \right\},$$ the “equal to” relation,$${R_6} = \left\{ {\left( {a,\;b} \right) \in {R^2}|a \ne b} \right\},$$ the “unequal to” relation.35. Find(a) $${R_2} \cup {R_4}$$.(b) $${R_3} \cup {R_6}$$.(c) $${R_3} \cap {R_6}$$.(d) $${R_4} \cap {R_6}$$.(e) $${R_3} - {R_6}$$.(f) $${R_6} - {R_3}$$.(g) $${R_2} \oplus {R_6}$$.(h) $${R_3} \oplus {R_5}$$.

(a)The solution of Relation $${R_2} \cup {R_4} = {R^2}.$$

(b) The solution of Relation $${R_3} \cup {R_6} = {R_6}$$

(c) The solution of Relation $${R_3} \cap {R_6} = {R_3}.$$

(d) The solution of Relation $${R_4} \cap {R_6} = {R_3}$$.

(e) The solution of Relation $${R_3} - {R_6} = f$$.

(f) The solution of Relation $${R_6} - {R_3} = {R_5}$$.

(g) The solution of Relation $${R_2} \oplus {R_6} = {R_4}$$.

(h) The solution of Relation $${R_3} \oplus {R_5} = {R_4}$$.

See the step by step solution

## Step 1: Given

The given for all parts are as follows:

$$\begin{array}{l}{R_1} = \left\{ {(a,b) \in {R^2}/a > b} \right\}\\{R_2} = \left\{ {(a,b) \in {R^2}/a \ge b} \right\}\\{R_3} = \left\{ {(a,b) \in {R^2}/a < b} \right\},\\{R_4} = \left\{ {(a,b) \in {R^2}/a \le b} \right\}\\{R_5} = \left\{ {(a,b) \in {R^2}/a = b} \right\},\\{R_6} = \left\{ {(a,b) \in {R^2}/a \ne b} \right\}\end{array}$$

## Step 2: The Concept of relation

An n-array relation on n sets, is any subset of Cartesian product of the n sets (i.e., a collection of n-tuples), with the most common one being a binary relation, a collection of order pairs from two sets containing an object from each set. The relation is homogeneous when it is formed with one set.

## Step 3: (a) Determine the value of relation

The solution of Relation

The union of two relations is the union of these sets.

Thus $${R_2} \cup {R_4}$$ holds between two real numbers if $${R_2}$$ holds or $${R_4}$$ holds (or both, it goes without saying).

Since it is always true that $${\rm{a}} \le {\rm{b}}$$ or $${\rm{b}} \le {\rm{a}},{R_2} \cup {R_4}$$ is all of $${{\rm{R}}^2}$$, i.e., the relation that always holds.

$$\begin{array}{c}{R_2} \cup {R_4} = \{ (a,b)\left. { \in {R^2}\mid a \ge b{\rm{ or }}a \le b} \right\}\\ = {R^2}\end{array}$$

## Step 4: (b) Determine the value of relation

The solution of Relation

For $$({\rm{a}},{\rm{b}})$$ to be in $${R_3} \cup {R_6}$$, we must have $${\rm{a}} < {\rm{b}}$$ or $${\rm{a}} \ne {\rm{b}}$$.

$$\begin{array}{c}{R_3} \cup {R_6} = \left\{ {(a,b)\mid (a,b) \in {R_3}{\rm{ or }}(a,b) \in {R_6}} \right\}\\ = \{ (a,b)\mid a < b{\rm{ or }}a \ne b\} \\ = \{ (a,b)\mid a < b\} \\ = {R_3}\end{array}$$

## Step 5: (c) Determine the value of relation

The solution of Relation

The intersection of two relations is the intersection of these sets.

Thus $${R_3} \cap {R_6}$$ holds between two real numbers if $${{\rm{R}}_3}$$ holds and $${{\rm{R}}_6}$$ holds as well.

Thus for $$({\rm{a}},{\rm{b}})$$ to be in $${R_3} \cap {R_6}$$, we must have $${\rm{a}} < {\rm{b}}$$ and $${\rm{a}} \ne {\rm{b}}$$.

Since this happens precisely when $${\rm{a}} < {\rm{b}}$$, we see that the answer is $${{\rm{R}}_3}$$.

$$\begin{array}{c}{R_3} \cap {R_6} = \left\{ {(a,b)\mid (a,b) \in {R_3}{\rm{ and }}(a,b) \in {R_6}} \right\} = \{ (a,b)\mid a < b{\rm{ and }}a \ne b\} \\ = \{ (a,b)\mid a < b\} \\ = {R_3}\end{array}$$

## Step 6: (d) Determine the value of relation

The solution of Relation

For $$({\rm{a}},{\rm{b}})$$ to be in $${R_4} \cap {R_6}$$, we must have $${\rm{a}} \le {\rm{b}}$$ and $${\rm{a}} \ne {\rm{b}}$$

Since this happens precisely when $$a < b$$, we see that the answer is $${R_3}$$.

$$\begin{array}{c}{R_4} \cap {R_6} = \left\{ {(a,b)\mid (a,b) \in {R_4}{\rm{ and }}(a,b) \in {R_6}} \right\}\\ = \{ (a,b)\mid a \le b{\rm{ and }}a < b\} \\ = \{ (a,b)\mid a < b\} \\ = {R_3}\end{array}$$

## Step 7: (e) Determine the value of relation

The solution of Relation

Recall that $${R_3} - {R_6} = {R_3} \cap {\bar R_6}$$.

But $${\bar R_6} = {R_5}$$, so we are asked for $${R_3} \cap {R_5}$$.

It is impossible for $$a < b$$ and $$a = b$$ to hold at the same time, so the answer is $$f$$, i.e., the relation that never holds.

$$\begin{array}{c}{R_3} - {R_6} = \left\{ {(a,b)\mid (a,b) \in {R_3}{\rm{ and }}(a,b) \notin {R_6}} \right\}\\ = \{ (a,b)/a < b{\rm{ and not }}a \ne b\} \\ = f\end{array}$$

## Step 8: (f) Determine the value of relation

The solution of Relation

$${R_6} - {R_3}$$ contains all ordered pairs that are in the relation $${R_6}$$ that do not occur in the relation $${{\rm{R}}_3}$$.

$$\begin{array}{c}{R_6} - {R_3} = \left\{ {(a,b)\mid (a,b) \in {R_6}{\rm{ and }}(a,b) \notin {R_3}} \right\}\\ = \{ (a,b)\mid a \ne b{\rm{ and }}a{\rm{ is not }} < b\} \\ = \{ (a,b)\mid a > b\} \\ = {R_5}\end{array}$$

## Step 9: (g) Determine the value of relation

The solution of Relation

$${R_2} \oplus {R_6}$$ contains all ordered pairs that are in the relation $${{\rm{R}}_2}$$ or in the relation $${{\rm{R}}_5}$$ but not in both.

$$\begin{array}{c}{R_2} \oplus {R_6} = \left\{ {\begin{array}{*{20}{c}}{(a,b)\mid (a,b) \in {R_2}{\rm{ or }}(a,b) \notin {R_6}{\rm{ or }}}\\{(a,b) \notin {R_2}{\rm{ or }}(a,b) \in {R_6}}\end{array}} \right\}\\ = \{ (a,b)/(a \ge b,{\rm{ or }}a \ne b){\rm{ and not }}(a \ge b{\rm{ and }}a \ne b)\} \\ = \{ (a,b)/a \le b\} \\ = {R_4}\end{array}$$

## Step 10: (h) Determine the value of relation

The solution of Relation

$${R_3} \oplus {R_5}$$ contains all ordered pairs that are in the relation $${R_3}$$ or in the relation $${R_5}$$ but not in both.

$$\begin{array}{c}{R_3} \oplus {R_5} = \left\{ {\begin{array}{*{20}{l}}{(a,b)\mid (a,b) \in {R_3}{\rm{ or }}(a,b) \notin {R_5}{\rm{ or }}}\\{(a,b) \notin {R_3}{\rm{ or }}(a,b) \in {R_5}}\end{array}} \right\}\\ = \{ (a,b)/(a < b,{\rm{ or }}a = b){\rm{ and }}not(a < b{\rm{ and }}a = b)\} \\ = \{ (a,b)/a \le b\} \\ = {R_4}\end{array}$$