$\begin{array}{1 1} \text{Reflexive, Transitive and Symmetric} \\ \text{Neither Reflexive, Transitive nor Symmetric} \\ \text{Only Reflexive} \\ \text{Only Reflexive and Transitive}\end{array} $

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- 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 set $Z$ of all integers and $R=\{(x,y):x-y \;\; \text {is an integer}\}$:

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

$\Rightarrow$ If $x=y$, then, $x-y = x-x = 0$ which is an integer for all $x$. Therefore R is relfexive.

We can verify via simple subsitution also, where if $x=y=1, x-y = 1-1 = 0$.

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

$\Rightarrow$ For any ordered pair (x,y), $R = x-y $, which is an integer.

$\Rightarrow$ For the ordered paid (y,x), $R = y-x = - (x-y)$ which must also be an integer, since $x-y$ is an integer. Therefore, $R$ is symmetric.

We can verify via simple subsitution:

Let $x=3, y = 5, (x,y) \in R \rightarrow x-y = 3-5 = -2$, which is an integer.

Similarly, $(y,x) \in R \rightarrow y-x = 5-2 = 2$ which is also an integer. Hence $R$ is symmetric.

For $R=\{(x,y):x-y \;\; \text {is an integer}\}$ 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$\; a_1,a_2,a_3 \in A$

$\Rightarrow$ For any ordered pair (x,y), $R = x-y $, which is an integer.

$\Rightarrow$ For any ordered pair (y,z), $R = y-z $, which is an integer.

$\Rightarrow$ It follows that for any ordered pair (x,z), $R = x-z$, which must also be an integer, since $x-y$ and $y-z$ are integers, hence $x-z$ which is nothing but $x-y+y-z$ must also be an integer. Therefore $R$ is transitive.

We can verify via simple subsitution:

Let $x=3, y = 5, (x,y) \in R \rightarrow x-y = 3-5 = -2$, which is an integer.

Let $y=5, z = 10, (y,z) \in R \rightarrow y-z = 5-10 = -5$, which is an integer.

We can see that $(x,z) \in R \rightarrow x-z= 3-10 = -7$, which is also an integer.

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