Question

(a)Which of the 16 relations on \(\{ 0,1\} \), which you listed are reflexive?

(b)Which of the 16 relations on \(\{ 0,1\} \), which you listed are irreflexive.

(c)Which of the 16 relations on \(\{ 0,1\} \), which you listed are symmetric.

(d)Which of the 16 relations on \(\{ 0,1\} \), which you listed are anti-symmetric.(e)Which of the 16 relations on \(\{ 0,1\} \), which you listed are asymmetric.

(f)Which of the 16 relations on \(\{ 0,1\} \), which you listed are transitive?

Short answer

(a)\({R_{\rm{8}}},{R_{13}},{R_{14}},{R_{16}}\)are reflexive in nature.

(b)\({R_1},{R_3},{R_9},{R_4}\)are irreflexive in nature.

(c)\({R_1},{R_2},{R_5},{R_8},{R_9},{R_{12}},{R_{15}},{R_{16}}\)are symmetric in nature.

(d)\({R_1},{R_2},{R_3},{R_4},{R_5},{R_6},{R_7},{R_8},{R_9},{R_{10}},{R_{11}},{R_{12}},{R_{13}},{R_{14}}\)are antisymmetric in nature.

(e)\({R_3},{R_4}\)are asymmetric in nature.

(f)\({R_1},{R_2},{R_3},{R_4},{R_5},{R_6},{R_7},{R_8},{R_9},{R_{10}},{R_{11}},{R_{12}},{R_{13}},{R_{14}},{R_{16}}\) are transitive.

Step-by-step solution

 Given

(a) The given set\(A = \{ 0,1\} \)

(b)The given set\(A = \{ 0,1\} \)

(c)The given set\(A = \{ 0,1\} \)

(d)The given set\(A = \{ 0,1\} \)

(e)The given set\(A = \{ 0,1\} \)

(f)The given set\(A = \{ 0,1\} \)

The Concept of relation

A relation from a set A to a set B is a subset of A×B. Hence, a relation R consists of ordered pairs (a, b), where a∈A and b∈B.

Determine the relation (a)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation \({\rm{R}}\) on a set \({\rm{A}}\) is reflexive if \((a,a) \in R\) for every element \(a \in A\).

Thus, \({R_8},{R_{13}},{R_{14}},{R_{16}}\) are reflexive in nature.

Determine the relation (b)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation \({\rm{R}}\) on a set \({\rm{A}}\) is irreflexive if \((a,a) \notin R\) for every element \(a \in A\).

Thus,\({R_1},{R_3},{R_9},{R_4}\) are irreflexive in nature.

Determine the relation (c)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation \({\rm{R}}\) on a set A is symmetric if \((b,a) \in R\) whenever \((a,b) \in R\).

Thus,\({R_1},{R_2},{R_5},{R_8},{R_9},{R_{12}},{R_{15}},{R_{16}}\) are symmetric in nature.

Determine the relation (d)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation \({\rm{R}}\)on set \({\rm{A}}\) is antisymmetric if \((b,a) \in R\) whenever \((a,b) \in R\) impliesa \( = {\rm{b}}\)Thus,\({R_1},{R_2},{R_3},{R_4},{R_5},{R_6},{R_7},{R_8},{R_9},{R_{10}},{R_{11}},{R_{12}},{R_{13}},{R_{14}}\) are antisymmetric in nature.

Determine the relation (e)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation\({\rm{R}}\)on a set\({\rm{A}}\)is asymmetric if\((b,a) \in R\)implies\((a,b) \notin R\)implies\(a = b\).

Thus,\({R_3},{R_4}\) are asymmetric in nature.

Determine the relation (f)

The set\(\{ 0,1\} \)and need to list 16 different relations.

Let us consider\(A = \{ 0,1\} \). Then,\(A \times A = \{ (0,0),(0,1),(1,0),(1,1)\} \). So, the subsets of\(A \times A\)are precisely the relations on\(A\).

Therefore, the relation on\(A = \{ 0,1\} \)are:

\(\begin{array}{l}{R_1} = \phi \\{R_2} = \{ (0,0)\} \\{R_3} = \{ (0,1)\} \\{R_4} = \{ (1,0)\} \\{R_5} = \{ (1,1)\} \\{R_6} = \{ (0,0),(0,1)\} \\{R_7} = \{ (0,0),(1,0)\} \\{R_8} = \{ (0,0),(1,1)\} \\{R_9} = \{ (0,1),(1,0)\} \\{R_{10}} = \{ (0,1),(1,1)\} \\{R_{11}} = \{ (1,0),(1,1)\} \\{R_{12}} = \{ (0,0),(0,1),(1,0)\} \\{R_{13}} = \{ (0,0),(0,1),(1,1)\} \\{R_{14}} = \{ (0,0),(1,0),(1,1)\} \\{R_{15}} = \{ (0,1),(1,0),(1,1)\} \\{R_{16}} = \{ (0,0),(0,1),(1,0),(1,1)\} \end{array}\)

A relation \({\rm{R}}\) on a set \({\rm{A}}\) is transitive if \((a,b) \in R\) and \((b,c) \in R\) implies \((a,c) \in R\). Thus, \({R_1},{R_2},{R_3},{R_4},{R_5},{R_6},{R_7},{R_8},{R_9},{R_{10}},{R_{11}},{R_{12}},{R_{13}},{R_{14}},{R_{16}}\) are transitive.