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Let R be a relation on the set N of natural numbers defined by nRm \u21Longleftrightarrow n is a factor of m. Then R is

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1. **Reflexive**: Every $n \in \mathbb{N}$ is a factor of itself ($n \cdot 1 = n$). (Reflexive)
2. **Symmetric**: If $2$ is a factor of $6$ ($2|6$), does $6$ have to be a factor of $2$? No. So **not symmetric**.
3. **Transitive**: If $n$ is a factor of $m$ ($n|m$) and $m$ is a factor of $p$ ($m|p$), does $n$ divide $p$?
   - $m = n \cdot k_1$, $p = m \cdot k_2$
   - $p = (n \cdot k_1) \cdot k_2 = n \cdot (k_1 \cdot k_2)$.
   - Yes, $n$ is a factor of $p$. (Transitive)
4. **Result**: Reflexive and transitive. Because it fails symmetry, it is a 'partial order' rather than an equivalence relation.

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