Question

Check the following relations for being reflexive, symmetric, transitive and thus choose the equivalence relations if any.
(i) $aRb\ if\ \left| a \right|\le b;a,b\in \ set\ of\ real\ numbers $.
(ii) $aRb\ if\ a < b;a,b\in N $.

Answer 1

Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify if each relation holds or not and hence deduce the answer.

Complete step-by-step solution -
Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. $\left( a,a \right)\in R$ while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we have two relations and we have to check them for reflexive, symmetric, transitive and thus equivalence.
Now, in (i) part we have the relation $aRb\ if\ \left| a \right|\le b;a,b\in \ set\ of\ real\ numbers$.
Now, we have been given relation R as \[aRb\Rightarrow \left| a \right|\le b\].
Now, we check for reflexive. So, we have \[aRa\Rightarrow \left| a \right|\le a\] now, if a < 0 then $\left| a \right|>0$ and therefore, $\left| a \right|{\nleq}a$ for a < 0. So, the relation is not reflexive.
Now, for symmetric we have if,
$aRb\Rightarrow \left| a \right|\le b........\left( 1 \right)$
Now, if we take a = 3, b = 5. So,
$\begin{align}
  & aRb\Rightarrow \left| 3 \right|\le 5 \\
 & bRa\Rightarrow \left| 5 \right|\le 3 \\
\end{align}$
Which is not true. Hence, the relation is not symmetric.
Now, for transitive we have if,
$\begin{align}
  & aRb\Rightarrow \left| a \right|\le b........\left( 1 \right) \\
 & bRc\Rightarrow \left| b \right|\le c........\left( 2 \right) \\
\end{align}$
Now, since $\left| a \right|>0\ \ \forall a\in \mathbb{R}$ therefore, b > 0 and since b > 0. We can write equation (2) as b < c as modulus of a positive number is the number itself. So, now we have,
$\begin{align}
  & \left| a \right|\le b\le c \\
 & \Rightarrow \left| a \right|\le c \\
 & \Rightarrow aRb\ and\ bRc\Rightarrow aRc \\
\end{align}$
Hence, the relation is transitive only.
Now, in part (ii) we have,
$aRb\ if\ a < b {;} a,b \in N$
Now, for reflexive we have,
$aRa\ if\ a < a$ which is not true for all $a\in N$ set of natural numbers.
Now, for symmetric we have if,
$aRb\Rightarrow a < b.........\left( 1 \right)$
But $b{\nleq}a$ as from (1) we have a < b. So, the relation is not symmetric.
Now, for transitive we have if,
$\begin{align}
  & aRb\Rightarrow a < b........\left( 1 \right) \\
 & bRc\Rightarrow b < c........\left( 2 \right) \\
\end{align}$
Then from (1) and (2) we have,
 $\begin{align}
  & a < b < c \\
 & \Rightarrow a < c \\
\end{align}$
Hence, $aRc$ holds.
Therefore, if $aRb\ and\ bRc\Rightarrow aRc$.
Hence, the relation is $aRb\ if\ a < b$ transitive only.

Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. Also these types of questions are solved easily by giving examples and counter examples. Also, we have to check the relation for reflexive, symmetric and transitive relation to check it for equivalence relation.