1. **State the problem:** We need to show that the relation $R = \{(a,b) \mid a \leq b\}$ defined on the set $S = \{2,4,6,8,10\}$ is a partial order relation. 2. **Definition of partial order:** A relation $R$ on a set $S$ is a partial order if it is reflexive, antisymmetric, and transitive. 3. **Check reflexivity:** A relation $R$ is reflexive if for every $a \in S$, $(a,a) \in R$. In our case, since $a \leq a$ is true for all $a$, each $(a,a)$ is in $R$. Example: $(2,2), (4,4), (6,6), (8,8), (10,10) \in R$. Hence, $R$ is reflexive. 4. **Check antisymmetry:** $R$ is antisymmetric if for all $a,b \in S$, if $(a,b) \in R$ and $(b,a) \in R$, then $a = b$. Because $a \leq b$ and $b \leq a$ implies $a = b$, antisymmetry holds. 5. **Check transitivity:** $R$ is transitive if for all $a,b,c \in S$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$. Since $a \leq b$ and $b \leq c$ imply $a \leq c$, transitivity is satisfied. 6. **Conclusion:** Since $R$ on $S$ is reflexive, antisymmetric, and transitive, $R$ is a partial order relation on $S$. **Final answer:** The relation $R = \{(a,b) \mid a \leq b \}$ defined on $S = \{2,4,6,8,10\}$ is a partial order relation because it is reflexive, antisymmetric, and transitive.