Question

Which of the following operation is not a partial order relation?

• A. "less than or equal ($\leq$)" on $\mathbb{R}$
• B. "subset ($\subseteq$)" on power set of set $A$
• C. "Divides" on $\mathbb{N}$
• D. "less than ($<$)" on $\mathbb{R}$

Hint:

A Partial Order Relation must be Reflexive. If the relation is "strict" (like $<$ or $\subset$), it is automatically not a partial order because an element cannot be strictly less than or a strict subset of itself.

Answer 1

The Correct Option is D

To determine if a relation is a Partial Order Relation, it must satisfy three specific properties: Reflexivity, Antisymmetry, and Transitivity. 1. Defining the Requirements:
• Reflexivity: Every element must be related to itself ($aRa$ for all $a$).
• Antisymmetry: If $aRb$ and $bRa$, then $a = b$.
• Transitivity: If $aRb$ and $bRc$, then $aRc$. 2. Analyzing the Options:
• $\leq$ on $\mathbb{R$:} $a \leq a$ (Reflexive); if $a \leq b$ and $b \leq a$ then $a=b$ (Antisymmetric); if $a \leq b \leq c$ then $a \leq c$ (Transitive). This is a partial order.
• $\subseteq$ on $\mathcal{P(A)$:} $X \subseteq X$ (Reflexive); if $X \subseteq Y$ and $Y \subseteq X$ then $X=Y$ (Antisymmetric); Transitivity holds. This is a partial order.
• "Divides" on $\mathbb{N$:} $a | a$ (Reflexive); if $a | b$ and $b | a$ then $a=b$ (Antisymmetric for positive integers); Transitivity holds. This is a partial order.
• "$<$" on $\mathbb{R$:} A number is never less than itself ($a \nless a$). Therefore, the relation is irreflexive. 3. Conclusion: Since "less than ($<$)" fails the reflexivity requirement, it cannot be a partial order relation. It is instead classified as a strict partial order.