# Check whether the relation R defined in the set ${1, 2, 3, 4, 5, 6}$ as $R= {(a,b) : b= a+1}$ is reflexive, symmetric or transitive.

$\begin{array}{1 1} \text{Reflexive, Symmetric and Transitive} \\ \text{Not Reflexive, Only Symmetric and Transitive} \\ \text{Only Transitive} \\ \text{Not Reflexive, Not Symmetric and Not Transitive}\end{array}$

Toolbox:
• A relation R in a set A is called reflexive. if $(a,a) \in R\;for\; all\; a\in A$
• A relation R in a set A is called symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
• A relation R in a set A is called transitive. if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
Given the set {1, 2, 3, 4, 5, 6} and $R= \{(a,b) : b= a+1\}$:
$R= \{(a,b) : b= a+1\}$ is reflexive if $(a,a) \in R\;for\; all\; a\in A$.
$\Rightarrow$ If $a=b, a = a+1 \rightarrow 0 = 1$ which is not correct; hence since $(a,a) \not \in R$, $R$ is not reflexive
We can verify this w/ a simple substitution:
Let $a = b = 5$, $5 \neq (5 + 1 = 6)$. Hence $R$ is not reflexive.
$R= \{(a,b) : b= a+1\}$ is symmetric. if $(a_1,a_2) \in R\;\Rightarrow \; (a_2,a_1)\in R \;$ for $\;a_1,a_2 \in A$
$\Rightarrow$ If $(a,b) \in R \rightarrow b= a+1$
However, if $(b,a) \in R \rightarrow a = b+1 \rightarrow b = a-1$
Since $a+1 \neq a-1$, $R$ cannot be symmetric.
We can verify this w/ a simple substitution:
$a = 1, b = 2 \;$and if$\; (a,b) \in R \Rightarrow b = a+1 \rightarrow 2 = 1 + 1$
However, for $(b,a) \in R \Rightarrow a = b+1 \rightarrow 1 = 2+ 1 =3$, which is not true. Hence $R$ is not symmetric.
$R= \{(a,b) : b= a+1\}$ is transitive if $(a_1,a_2) \in\; R$ and $(a_2,a_3)\in R \Rightarrow \;(a_1,a_3)\in R\;$for all $\; a_1,a_2,a_3 \in A$
$\Rightarrow$ If $(a,b) \in R \rightarrow b= a+1$
$\Rightarrow$ If $(b,c) \in R \rightarrow c= b+1$
$\Rightarrow$ If $(a,c) \in R \rightarrow c= a+1$ which implies that $b = c$, which is not true. Hence $R$ is not transitive.
We can verify this w/ a simple substitution:
Let $a = 1, b = 2, c = 3$ If $\; (a,b) \in R \Rightarrow b = a+1 \rightarrow 2 = 1 + 1$
If$\; (a,c) \in R \Rightarrow c = a+1 \rightarrow 3$ must be equal to $1 + 1 = 2$ which is not the case. Hence $R$ is not reflexive.
edited Mar 8, 2013