Let R be the relation on N (set of Natural numbers) defined by R = {(a, b): a, b ∈ N and b is divisible by a}. Then the relation R is

The correct option is Reflexive, Transitive but not symmetric.

Let us analyze the given relation $R$ defined on the set of natural numbers $\mathbb{N}$:
$R=\left\{\right(a,b):a,b\in \mathbb{N}\text{and}b\text{is divisible by}a\}$

To determine the nature of the relation $R$, we test for three properties: reflexivity, symmetry, and transitivity.

1. Reflexivity:
A relation $R$ on $\mathbb{N}$ is reflexive if for every $a\in \mathbb{N}$, the pair $(a,a)\in R$.
Since any natural number $a$ is always divisible by itself (i.e., $a/a=1$, which is an integer), the condition "$a$ is divisible by $a$" holds true for all $a��\mathbb{N}$.
Therefore, $(a,a)\in R$ for all $a\in \mathbb{N}$, meaning the relation $R$ is reflexive.

2. Symmetry:
A relation $R$ on $\mathbb{N}$ is symmetric if $(a,b)\in R$ implies $(b,a)\in R$.
Let us test this with a counterexample. Consider $a=2$ and $b=4$.
Since 4 is divisible by 2, we have $(2,4)\in R$.
However, 2 is not divisible by 4 (since $2/4=0.5$, which is not a natural number). Thus, $(4,2)\notin R$.
Since $(a,b)\in R$ does not guarantee that $(b,a)\in R$, the relation $R$ is not symmetric.

3. Transitivity:
A relation $R$ on $\mathbb{N}$ is transitive if whenever $(a,b)\in R$ and $(b,c)\in R$, it must be that $(a,c)\in R$.
Let $(a,b)\in R$ and $(b,c)\in R$.
This means:
- $b$ is divisible by $a$, so there exists some integer $k$ such that $b=ka$.
- $c$ is divisible by $b$, so there exists some integer $m$ such that $c=mb$.
Substituting $b=ka$ into the equation for $c$ gives:
$c=m\left(ka\right)=\left(mk\right)a$
Since $m$ and $k$ are integers, their product $mk$ is also an integer. This shows that $c$ is divisible by $a$, which implies $(a,c)\in R$.
Therefore, the relation $R$ is transitive.

Conclusion:
Since $R$ is reflexive and transitive, but not symmetric, it is classified as Reflexive, Transitive but not symmetric.