Question ID: #1163
Let $A=\{-2,-1,0,1,2,3\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $y=\max\{x,1\}$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in $R$ to make it reflexive and symmetric relations, respectively. Then $l+m+n$ is equal to
- (1) 12
- (2) 11
- (3) 13
- (4) 14
Solution:
$$A = \{-2, -1, 0, 1, 2, 3\}$$
$$xRy \iff y = \max\{x, 1\}$$
$$x = -2 \Rightarrow y = \max\{-2, 1\} = 1 \Rightarrow (-2, 1) \in R$$
$$x = -1 \Rightarrow y = \max\{-1, 1\} = 1 \Rightarrow (-1, 1) \in R$$
$$x = 0 \Rightarrow y = \max\{0, 1\} = 1 \Rightarrow (0, 1) \in R$$
$$x = 1 \Rightarrow y = \max\{1, 1\} = 1 \Rightarrow (1, 1) \in R$$
$$x = 2 \Rightarrow y = \max\{2, 1\} = 2 \Rightarrow (2, 2) \in R$$
$$x = 3 \Rightarrow y = \max\{3, 1\} = 3 \Rightarrow (3, 3) \in R$$
$$R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$$
$$l = \text{number of elements in } R$$
$$l = 6$$
To make $R$ reflexive, it must contain $(a, a)$ for all $a \in A$.
Elements needed for reflexivity: $(-2, -2), (-1, -1), (0, 0)$
$$m = 3$$
To make $R$ symmetric, if $(a, b) \in R$, then $(b, a)$ must be in $R$.
Elements needed for symmetry: $(1, -2), (1, -1), (1, 0)$
$$n = 3$$
$$l + m + n = 6 + 3 + 3$$
$$l + m + n = 12$$
Ans. (1)
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