Check whether the relation $R$ defined in the set $\{1,2,3,4,5,6\}$ as $R =\{(a, b): b=a+1\}$ is reflexive, symmetric or transitive.
Let $A =\{1,2,3,4,5,6\}$.
A relation $R$ is defined on set $A$ as: $R=\{(a, b): b=a+1\}$
$\therefore R =\{(1,2),(2,3),(3,4),(4,5),(5,6)\}$
we can find $(a, a) \notin R,$ where $a \in A$
For instance,
$(1,1),\,(2,2),\,(3,3),\,(4,4),\,(0,5),\,(0,6) \notin R$
$\therefore R$ is not reflexive.
It can be observed that $(1,2) \in R ,$ but $(2,1)\notin R$
$\therefore R$ is not symmetric.
Now, $(1,2),\,(2,3) \in R$
But, $(1,3)\notin R$
$\therefore R$ is not transitive
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.
Determine whether each of the following relations are reflexive, symmetric and transitive:
Relation $\mathrm{R}$ in the set $\mathrm{N}$ of natural numbers defined as
$\mathrm{R}=\{(x, y): y=x+5 $ and $ x<4\}$
Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y= 10\}$ and $R_2 = \{(x,y) \in N\times N : x+ 2y= 10\} $. Then
Let $\mathrm{A}=\{1,2,3,4,5\}$. Let $\mathrm{R}$ be a relation on $\mathrm{A}$ defined by $x R y$ if and only if $4 x \leq 5 y$. Let $m$ be the number of elements in $\mathrm{R}$ and $\mathrm{n}$ be the minimum number of elements from $\mathrm{A} \times \mathrm{A}$ that are required to be added to $\mathrm{R}$ to make it a symmetric relation. Then $m+n$ is equal to:
If $R$ is a relation on the set $N$, defined by $\left\{ {\left( {x,y} \right);3x + 3y = 10} \right\}$
Statement $-1$ : $R$ is symmetric
Statement $-2$ : $R$ is reflexive
Statement $-3$ : $R$ is transitive, then thecorrect sequence of given statements is
(where $T$ means true and $F$ means false)
Determine whether each of the following relations are reflexive, symmetric and transitive :
Relation $\mathrm{R}$ in the set $\mathrm{A}=\{1,2,3, \ldots, 13,14\}$ defined as $\mathrm{R}=\{(x, y): 3 x-y=0\}$