Question

(a) To find Relation \({R_1} \cup {R_2}\).

(b) To find Relation \({R_1} \cap {R_2}\).

(c) To find Relation \({R_1} - {R_2}\).

(d) To find Relation \({R_2} - {R_1}\).

(e) To find Relation \({R_1} \oplus {R_2}\).

Short answer

(a)The solution of Relation\({R_1} \cup {R_2} = \left\{ {(a,b)\mid b = ka} \right.\)or\(b = \frac{1}{k}a\),where\(k\)is an integer\(\} \)(b) The solution of Relation\({R_1} \cap {R_2} = \{ (a,b)\mid b = a\} \)

(c) The solution of Relation\({R_1} - {R_2} = \{ (a,b)\mid a\)is a proper divisor of\(b\} \).

(d) The solution of Relation\({R_2} - {R_1} = \{ (a,b)\mid a\)is a multiple of\(b\), greater than\(b\} \).

(e) The solution of Relation\({R_1} \oplus {R_2} = \left\{ {(a,b)\mid b = ka{\rm{ or }}b = \frac{1}{k}a{\rm{, where }}k{\rm{ is a positive integer and }}k \ne 1} \right\}\)

Step-by-step solution

 Given

(a)\({{\rm{R}}_1}\)and\({{\rm{R}}_2}\)be the "divides" and "is a multiple of" relations on the set of all positive integers, respectively.

That is,\({R_1} = \{ (a,b)\mid \)a divide\(b\} \)and\({R_2} = \{ (a,b)\mid a\)is a multiple of\(b\} \).

(b)\({{\rm{R}}_1}\)and\({{\rm{R}}_2}\)be the "divides" and "is a multiple of" relations on the set of all positive integers, respectively.

That is,\({R_1} = \{ (a,b)\mid \)a divide\(b\} \)and\({R_2} = \{ (a,b)\mid a\)is a multiple of\(b\} \).

(c)\({{\rm{R}}_1}\)and\({{\rm{R}}_2}\)be the "divides" and "is a multiple of" relations on the set of all positive integers, respectively.

That is,\({R_1} = \{ (a,b)\mid \)a divide\(b\} \)and\({R_2} = \{ (a,b)\mid a\)is a multiple of\(b\} \).

(d)\({{\rm{R}}_1}\)and\({{\rm{R}}_2}\)be the "divides" and "is a multiple of" relations on the set of all positive integers, respectively.

That is,\({R_1} = \{ (a,b)\mid \)a divide\(b\} \)and\({R_2} = \{ (a,b)\mid a\)is a multiple of\(b\} \).

(e)\({{\rm{R}}_1}\)and\({{\rm{R}}_2}\)be the "divides" and "is a multiple of" relations on the set of all positive integers, respectively.

That is, \({R_1} = \{ (a,b)\mid \) a divide \(b\} \) and \({R_2} = \{ (a,b)\mid a\) is a multiple of \(b\} \).

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 ofn-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.

Determine the value of relation (a)

Let us consider the relation\({R_1}\;\& \;{R_2}\)were,

\({R_1} = \{ (a,b)\mid a\)divides\(b\} \)and\({R_2} = \{ (a,b)\mid a\)multiple of\(b\} \)

\({R_1} \cup {R_2}\)=\(\left\{ {(a,b)\mid (a,b) \in {R_1} \cup {R_2}} \right.\)if and only if\((a,b) \in {R_1}\)or\(\left. {(a,b) \in {R_2}} \right\}\)

=\(\left\{ {(a,b)\mid (a,b) \in {R_1}} \right.\)\( \cup {R_2}\)if and only if\(a\)divides\(b\)or\(a\)is a multiple of\(b\)}

If\(a\)divides\(b\), for some integer\(k\),\(b = ka\)

If\(a\)is a multiple of\(b\), for some integer\(k\),\(b = \frac{1}{k}a\)

Hence,\({R_1} \cup {R_2} = \left\{ {(a,b)\mid b = ka{\rm{ or }}b = \frac{1}{k}a,{\rm{ where }}k{\rm{ is an integer}}{\rm{. }}} \right\}\)

Determine the value of relation (b)

Let us consider the relation\({R_1}\;\& \;{R_2}\)were,

\({R_1}{\rm{ }} \cap \;{R_2} = \)\(\left\{ {(a,b)\mid (a,b) \in {R_1} \cap {R_2}} \right.\)if and only if\((a,b) \in {R_1}\)and\((a,b)\)\(\left. { \in {R_2}} \right\}\)

\( = \left\{ {(a,b)\mid (a,b) \in {R_1}} \right.\)\( \cap {R_2}\)if and only if a divide\(b\)and\(a\)is a multiple of\(b\)\(\} \)

This is possible only if\(a = b\).

Hence,\({R_1} \cap {R_2} = \{ (a,b)\mid b = a\} \)

Determine the value of relation (c)

Let us consider the relation\({R_1}\;\& \;{R_2}\)were,

\({R_1} - {R_2}{\rm{ }} = \)\(\left\{ {(a,b)\mid (a,b) \in {R_1} - {R_2}} \right.\)if and only if\((a,b) \in {R_1}\)and\((a,b)\)\(\left. { \notin {R_2}} \right\}\)

\( = \left\{ {(a,b)\mid (a,b) \in {R_1}} \right.\)\( - {R_2}\)if and only if a divide\(b\)or\(a\)is not a multiple of\(b\)\(\} \)

This can be equated as\(a\)is a proper divisor of\(b\).

Hence,\({R_1} - {R_2} = \{ (a,b)\mid a{\rm{ is a proper divisor of }}b\} \)

Determine the value of relation (d)

Let us consider the relation\({R_1}\;\& \;{R_2}\)were,

\({R_2} - {R_1} = \)\(\left\{ {(a,b)\mid (a,b) \in {R_2} - {R_1}} \right.\)if and only if\((a,b) \in {R_1}\)and\((a,b)\)\(\left. { \notin {R_2}} \right\}\)

=\(\left\{ {(a,b)\mid (a,b) \in {R_2} - {R_1}} \right.{\rm{ if and only if a doesn't divide b or a is a multiple of b}}\} \)

This can be equated as\(a\)is a multiple of, greater than\(b\).

Hence,\({R_2} - {R_1} = \{ (a,b)\mid a\) is a multiple of \(b\), greater than \(b\} \)

Determine the value of relation (e)

Let us consider the relation\({R_1}\;\& \;{R_2}\)were,

\({R_1} \oplus {R_2}{\rm{ }} = \)\(\left\{ {(a,b)\mid (a,b) \in {R_1} \oplus {R_2}} \right.\)if and only if\((a,b) \in {R_1}\)\(\left. { \cup {R_2}{\rm{ and }}(a,b) \notin {R_1} \cap {R_2}} \right\}\)

\( = \)\(\left\{ {(a,b)\mid (a,b) \in {R_1} \cup {R_2}} \right.\)if and only if for some integer\(k\)\({\rm{either }}\left. {b = ka{\rm{ or }}b = \frac{1}{k}a} \right\}{\rm{ }}\)

\( = \left\{ {(a,b)\mid (a,b) \in {R_1} \cap {R_2}{\rm{ if and only if }}a = b} \right\}\)

Hence,\({R_1} \oplus {R_2} = \left\{ {(a,b)\mid b = {\rm{ ka or }}b = \frac{1}{k}a{\rm{, where }}k{\rm{ is a positive integer and }}k \ne 1} \right\}\)