First slide

Relations XII

Question

. Let R be a relation on the set N of natural numbers denoted $by n R m \Leftrightarrow n is a factor of m (i.e., n ∣ m) . Then, R is$

Moderate

A

Reflexive and symmetric
B

Transitive and symmetric
C

Equivalence
D

Reflexive, transitive but not symmetric

Solution

$As n ∣ n for all n \in N, R is reflexive. As 2 | 6 but 6 ∣ 2, R is not symmetric$
$Let n R m and m R p . As n ∣ m and m | p \Rightarrow n | p \Rightarrow n R p .$
So, R is transitive.

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