Let $A=\{0,3,4,6,7,8,9,10\}$ and $R$ be the relation defined on $A$ such that $R=\{(x, y) \in A \times A: x-y$ is odd positive integer or $x-y=2\}$. The minimum number of elements that must be added to the relation $R$, so that it is a symmetric relation, is equal to ____________.
Answer (integer)
19
Solution
We have, $A=\{0,3,4,6,7,8,9,10\}$
<br/><br/>Case I : $x-y$ is odd, if one is odd and one is even and $x>y$.
<br/><br/>$\therefore$ Possibilites are $\{(3,0),(4,3),(6,3),(7,6),(7,4)$, $(7,0),(8,7),(8,3),(9,8),(9,6),(9,4),(9,0),(10,9),(10$, $7),(10,3)\}$
<br/><br/>No. of cases $=15$
<br/><br/>Case II : $x-y=2$
<br/><br/>$\therefore$ Possibilities are $\{(6,4),(8,6),(9,7),(10,8)\}$
<br/><br/>$\therefore$ No. of cases $=4$
<br/><br/>So, minimum ordered pair to be added $=15+4=19$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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