Let\[R\] be the relation on the set\[R\]of all real numbers defined by\[aRb\]if \[\left| {a - b} \right| \le 1\]. What type of relation\[R\]?
A. Reflexive and symmetric
B. Symmetric only
C. Transitive only
D. Anti-symmetric only

Hint: First we will check whether \[R\] is reflexive. We will put \[a\] in place \[b\]in the inequality \[\left| {a - b} \right| \le 1\] and check whether it satisfies the inequality. To check the symmetric property we will swap the positions of \[a\] and \[b\] and check whether it holds or not. To check transitive, we will assume that \[a = 1\], \[b = 2\], and \[c = 3\] and check whether the values satisfy the inequality \[\left| {a - b} \right| \le 1\], \[\left| {b - c} \right| \le 1\], and \[\left| {a - c} \right| \le 1\]. If it does not satisfy, then the relation is not transitive.
Formula Used:
We will use the definition of the relations, i.e., A relation in a set A is called reflexive relation if \[\left( {a,a} \right) \in R\] for every element \[a \in A\]. While symmetric relations are those for which if\[aRb\]then\[bRa\]and transitive are those relations in which if \[aRb\] and \[bRc\] then \[aRc\] must be held.
Complete step by step solution:
Given condition is \[\left| {a - b} \right| \le 1\]
(i) First we will check the reflexive relation:
\[ \Rightarrow aRa\]
Now we will use the given condition, we will get,
\[ \Rightarrow \left| {a - a} \right| = 0 < 1\]
Which holds true,
So, the given relation is reflexive relation.
(ii) Second we will check the symmetric relation,
\[ \Rightarrow aRb\]
Now we will use the given relation, we will get,
\[\left| {a - b} \right| \le 1\]
Now we will swap the values of \[a\]and\[b\], we will get,
\[\left| {b - a} \right| \le 1\]
Which states that, \[bRa\],
So the given relation is symmetric relation.
So, from the above we can conclude that \[R\] is not anti-symmetric
(iii) Now we will check for Transitive relation:
For transitive relation, we will substitute some values for \[a\]and\[b\],
For example take \[a = 1\], \[b = 2\]and\[c = 3\],
We will substitute the values in the relation, we will get,
\[ \Rightarrow 1R2\]
Now using the condition given,
\[ \Rightarrow \left| {1 - 2} \right| \le 1\]
Now we will simplify, we will get,
\[ \Rightarrow \left| { - 1} \right| \le 1\]
Now we will further simplify, we will get,
\[ \Rightarrow 1 = 1\]
Now we will take
\[ \Rightarrow 2R3\]
Now using the condition given,
\[ \Rightarrow \left| {2 - 3} \right| \le 1\]
Now we will simplify, we will get,
\[ \Rightarrow \left| { - 1} \right| \le 1\]
Now we will further simplify, we will get,
\[ \Rightarrow 1 = 1\]
Now we will take
\[ \Rightarrow 1R3\]
Now using the condition given,
\[ \Rightarrow \left| {1 - 3} \right| \le 1\]
Now we will simplify, we will get,
\[ \Rightarrow \left| { - 2} \right| \le 1\]
Now we will further simplify, we will get,
\[ \Rightarrow 2 \le 1\]
Which cannot be true,
So, the given relation is not transitive.
The correct option is A.
Note: Students often do a common mistake. They check the transitive property by using inequality without assuming the value of the variables. The check the inequality \[\left| {a - c} \right| \le 1\] by using the inequalities \[\left| {a - b} \right| \le 1\] and \[\left| {b - c} \right| \le 1\] . But it is the wrong procedure. The correct procedure is first we will assume the value of the variables \[a,b.c\] and check the inequality \[\left| {a - c} \right| \le 1\].