Let $S=\{1,2,3,4\}$. A binary relation $P$ is defined on $S$ such that for any $a, b \in S, a P b$ i

Question

Let $S=\{1,2,3,4\}$. A binary relation $P$ is defined on $S$ such that for any $a, b \in S, a P b$ if and only if $b=a+1$. Let $k$ be the total number of ordered pairs in $P$. Let $p$ be the minimum number of additional ordered pairs required to make $P$ a reflexive relation. Let $q$ be the minimum number of additional ordered pairs required to make $P$ a symmetric relation. Determine the value of $k+p+q$.

Professor PC Thomas Classes · Answer

Let the given set be $S=\{1,2,3,4\}$. A binary relation $P$ is defined on $S$ such that for any $a, b \in S, a P b$ if and only if $b=a+1$. We need to find the ordered pairs in $P$.
For $a=1, b=1+1=2$. Since $2 \in S,(1,2) \in P$.
For $a=2, b=2+1=3$. Since $3 \in S(2,3) \in P$.
For $a=3, b=3+1=4$. Since $4 \in S,(3,4) \in P$.
For $\boldsymbol{a}=\mathbf{4}, \boldsymbol{b}=\mathbf{4}+\mathbf{1}=\mathbf{5}$. Since $\mathbf{5} 
otin S$, there is no pair starting with $\boldsymbol{a}=\mathbf{4}$.
So, the relation $P=\{(1,2),(2,3),(3,4)\}$.
$\boldsymbol{k}$ is the total number of ordered pairs in $\boldsymbol{P}$. From the above, $\boldsymbol{k}=\mathbf{3}$.
A relation $P$ on a set $S$ is reflexive if for every $a \in S,(a, a) \in P$. The set $S=\{1,2,3,4\}$. For $P$ to be reflexive, it must contain the pairs $(1,1),(2,2),(3,3),(4,4)$.
The current relation $P=\{(1,2),(2,3),(3,4)\}$ does not contain any of these pairs.
Therefore, the minimum number of additional ordered pairs required to make $P$ a reflexive relation is $p=4$.
A relation $P$ on a set $S$ is symmetric if for every $(a, b) \in P,(b, a)$ must also be in $P$.
The current relation $P=\{(1,2),(2,3),(3,4)\}$.
- For $(1,2) \in P$, we need $(2,1)$ to be in $P$.
- For $(2,3) \in P$, we need $(3,2)$ to be in $P$.
- For $(3,4) \in P$, we need $(4,3)$ to be in $P$.
None of these required pairs are currently in $P$. Therefore, the minimum number of additional ordered pairs required to make $P$ a symmetric relation is $q=3$. $k+p+q=3+4+3=10$.