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

Concept:
• A relation \(R\) on a set \(S\) is a Partial Order Relation if it satisfies three properties:
• Reflexivity: \(aRa\) for all \(a \in S\).
• Antisymmetry: If \(aRb\) and \(bRa\), then \(a = b\).
• Transitivity: If \(aRb\) and \(bRc\), then \(aRc\).

Step 1: Check Option (A): \(\le\) on \(\mathbb{R}\)
\(x \le x\) (Reflexive). If \(x \le y\) and \(y \le x\), then \(x = y\) (Antisymmetric). If \(x \le y\) and \(y \le z\), then \(x \le z\) (Transitive). This is a partial order.

Step 2: Check Option (B): \(\subseteq\) on Power Set
\(X \subseteq X\) (Reflexive). If \(X \subseteq Y\) and \(Y \subseteq X\), then \(X = Y\) (Antisymmetric). Inclusion is transitive. This is a partial order.

Step 3: Check Option (C): "Divides" on \(\mathbb{N}\)
\(n|n\) (Reflexive). If \(a|b\) and \(b|a\), then \(a = b\) in natural numbers (Antisymmetric). If \(a|b\) and \(b|c\), then \(a|c\) (Transitive). This is a partial order.

Step 4: Check Option (D): \(<\) on \(\mathbb{R}\)
For a relation to be reflexive, \(x < x\) must be true for all \(x\). However, \(5 < 5\) is false.
Since it fails reflexivity, it cannot be a partial order. It is actually a "Strict Partial Order".