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# Show that the relation $R$ in $R$ defined as $R = {(a, b) : a \leq b}$, is reflexive and  transitive but not symmetric.

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 in a set of real numbers, the relation $R=\{(a,b):a\leq b\}$:
For any $a \in R$, $a \leq a$. Therefore $(a,a) \in R$. Hence $R$ is reflexive.
For any $a \neq b$, we observe that while $a \leq b$ might be true, $b \leq a$ will not be, unless $a=b$.Hence is $R$ is not symmetric.
We can verify this with a simple subsitution:
Let $a = 2, b = 4$. While $a \leq b \rightarrow 2 \leq 4$ is true, $b \leq a \rightarrow 4 \not \leq 2$. Hence $R$ is not symmetric.
For any $(a,b) \in R$, if $a \leq b$ and for any $(b,c) \in R, b \leq c$:
$\Rightarrow a \leq b \leq c \rightarrow a \leq c$.
$\Rightarrow$ If $a \leq c$ then $R$ is transitive.
edited Mar 8, 2013