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Q58E

Expert-verifiedFound in: Page 583

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**(a) To find Relation\({R^2}\)**

**(b) To find Relation \({R^3}\)**

**(c) To find Relation \({R^4}\)**

**(d) To find Relation\({R^5}\)**

(a)The solution of Relation\({R^2} = \left\{ \begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,4)(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2)(4,3),(4,4),\\(5,1),(5,2),(5,3),(5,4),(5,5)\end{array} \right\}\)

(b) The solution of Relation\({R^3} = \left\{ \begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4),(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2),(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\end{array} \right\}\)

(c) The solution of Relation \({R^4} = \left\{ \begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4)(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2)(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\end{array} \right\}\)

(d) The solution of Relation\({R^4} = \left\{ \begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4)(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2)(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\end{array} \right\}\)

(a) Relation \(R = \{ (1,1),(1,2),(1,3),(2,3),(2,4),(3,1),(3,4),(3,5),(4,2),(4,5),(5,1)\),

\(\qquad (5,2),(5,4)\} \)

On set \(A = \{ 1,2,3,4,5\} \)

(b) Relation\(R = \{ (1,1),(1,2),(1,3),(2,3),(2,4),(3,1),(3,4),(3,5),(4,2),(4,5),(5,1)\)\(\qquad (5,2),(5,4)\} \)

On set \(A = \{ 1,2,3,4,5\} \)**.**

(c) Relation\(R = \{ (1,1),(1,2),(1,3),(2,3),(2,4),(3,1),(3,4),(3,5),(4,2),(4,5),(5,1)\)\(\qquad (5,2),(5,4)\} \)

On set \(A = \{ 1,2,3,4,5\} \)**.**

(d) Relation\(R = \{ (1,1),(1,2),(1,3),(2,3),(2,4),(3,1),(3,4),(3,5),(4,2),(4,5),(5,1)\)\(\qquad (5,2),(5,4)\} \)

On set \(A = \{ 1,2,3,4,5\} \)**.**

**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.**

Consider the relation \(R\)

\({R^2} \Rightarrow \) paths of length 2

\(\begin{array}{c}{R^2} = R \cdot R\\ = \{ (1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,4),(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2),(4,3),(4,4),\\(5,1),(5,2),(5,3),(5,4),(5,5)\} \end{array}\)

Consider the relation \(R\)

\({R^3} \Rightarrow \) paths of length 3

\(\begin{array}{c}{R^3} = {R^2}.R\\ = \{ (1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4)(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2)(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\} \end{array}\)

Consider the relation \(R\)

\({R^4} \Rightarrow \) paths of length 4

\(\begin{array}{c}{R^4} = {R^3} \cdot R\\ = \{ (1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4)(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2)(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\} \end{array}\)

Consider the relation \(R\)

\({R^5} \Rightarrow \) paths of length 5

\(\begin{array}{c}{R^5} = {R^4} \bullet R\\ = \{ (1,1),(1,2),(1,3),(1,4),(1,5),\\(2,1),(2,2),(2,3),(2,4),(2,5),\\(3,1),(3,2),(3,3),(3,4),(3,5),\\(4,1),(4,2),(4,3),(4,4),(4,5),\\(5,1),(5,2),(5,3),(5,4),(5,5)\} \end{array}\)

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