Question

Let R be the relation on the set R of all real numbers defined by $aRb$iff $\left| {a - b} \right| \leqslant 1$. Then R is,
$A)$Reflexive and symmetric
$B)$Symmetric only
$C)$Transitive only
$D)$Anti-symmetric only

Answer 1

Hint : First, we will need to know about the relations on reflexive, transitive, symmetric, and anti-symmetric.
After that, we can easily solve the given problem as R is defined as the set of all real numbers and expressed as the $aRb$.
We will use the definitions of the given functions to check whether the options hold or not and hence we deduce the result.

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 which is also known as the equivalent relation definition.
Starting with the reflexive relation, every element in the reflexive is mapped to itself.
That is, $(a,a) \in R$.
Hence reflexive can be expressed as $aRa$(mapped to itself) and from the given that, we can able to rewrite the equation as $aRa \Leftrightarrow \left| {a - a} \right| = 0$(where the symbol $\Leftrightarrow$is represented as if and only if, that means both the parts are equivalent to one another)
Thus, we get, $aRa \Leftrightarrow \left| {a - a} \right| = 0 < 1$(strictly $1$)
Hence the given relation holds on reflexively.
Symmetric relations are those for which if $aRb$then $bRa$
This can be expressed as $aRb \Leftrightarrow \left| {a - b} \right| \leqslant 1$and now taking the negative sign common and thus we get,
$aRb \Leftrightarrow \left| {a - b} \right| \leqslant 1 \Rightarrow \left| { - (b - a)} \right| \leqslant 1$, since the equation is in the modulus so taking out the values, we get hence we get the symmetric relation as $aRb \Leftrightarrow \left| {a - b} \right| \leqslant 1 \Rightarrow \left| {(b - a)} \right| \leqslant 1 \Leftrightarrow bRa$
Thus, symmetric relations hold$aRb \Leftrightarrow bRa$.
Transitive are those elements in which if $aRb$and $bRc$then $aRc$must be held.
Take an example; $aRb \Leftrightarrow \left| {a - b} \right| \leqslant 1$ $a = 1.5,b = 2,c = 2.8$substituting the values we get,
$aRb \Leftrightarrow \left| {1.5 - 2} \right| = 0.5 \leqslant 1$
$bRc \Leftrightarrow \left| {2 - 2.8} \right| = 0.8 \leqslant 1$
But for, $aRc \Leftrightarrow \left| {1.5 - 2.8} \right| = 1.3$(which is not less than or equals to $1$)
Thus, the transitive relation does not hold.
Since R is symmetric, so R will not the antisymmetric because antisymmetric is absolute cannot go on both ways like the symmetric which is $(a,b) \notin R$or $(b,a) \notin R$whenever $x \ne y$
Hence the relation holds on reflexive and symmetric only.
Thus option $A)$Reflexive and symmetric is correct.

Note: It is important to remember that $aRb$ which means a is related to b with the help of the given real number R.
These types of questions are solved using examples and counterexamples to verify the relation is correct or not.
Since if the relation satisfies the symmetric, transitive, and also reflexive then it is called the equivalence relation.
Taking the values out in the modulus, only gives the positive terms.