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Q35E

Expert-verifiedFound in: Page 582

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 \({R_3} \oplus {R_5} = {R_4}\).

The given for all parts are as follows:

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

**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.**

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.

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

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 \({\rm{a}} \ne {\rm{b}}\).

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

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 \({{\rm{R}}_3}\).

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

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 \({R_3}\).

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

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.

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

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 \({{\rm{R}}_3}\).

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

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.

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

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.

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

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