$\begin{array}{1 1} \text{R is reflexive, symmetric and transitive} \\ \text{R is reflexive and symmetric, not transitive} \\ \text{R is reflexive and transitive, not symmetric} \\ \text{R is transitive and symmetric, not reflexive}\end{array} $

- A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
- A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
- A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$

Given $A = \{1,2,3,4,5,6\}$ and $R=\{(x,y):y \;\; \text {is divisible by } x\}$:

For $R=\{(x,y):y \;\; \text {is divisible by } x\}$ to be reflexive, $(a,a) \in R\;$ for every $\; a\in\;A$

$\Rightarrow\; $If $ \;x=y, \; R=\{(x,y):y \;\; \text {is divisible by } x\} = R=\{(x,x):x \;\; \text {is divisible by } x\}$

Since $x$ is divisible by itself, i.e., by $x$, the above Relation holds true.

Therefore, $(x,x) \in R$ for all $x \in R$. Hence $R$ is reflexive.

We can verify this w/ a simple substitution:

If $x=1, y = 1$ and $(1,1) \in R$. Hence $R$ is reflexive.

For $R=\{(x,y):y \;\; \text {is divisible by } x\}$ to be symmetric, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$

Given $A = \{1,2,3,4,5,6\}$, this need not be true in all cases as we can verify w/ this simple substitution:

Let $x=1, y=2$. We can see that 2 is divisble by 1, but not vice versa, as the number 1 is only divisble by itself. Therefore, $R$ is not symmetric.

For $R=\{(x,y):y \;\; \text {is divisible by } x\}$ to be transitive, if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$

$\Rightarrow, (x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$.

If $y$ is divisible by $x$, and $z$ is divisible by $y$, then we can see that $z$ is divisible by $x$. Hence $R$ is transitive.

We can verify this w/ a simple substitution:

Let us choose $x,y,z$ such that $y$ is divisible by $x$, and $z$ is divisible by $y$.

Let $x=1, y = 3, z = 6 \rightarrow, 3$ is divisible by $1. 6$ is divisible by$ \;3$. And we can see that $6$ is divisible by $1$ Hence $R$ is transitive..

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