Question

 Test whether the relation \[{{R}_{1}}\] is
(i) Reflexive (ii) Symmetric and (iii) Transitive
where \[{{R}_{1}}\] on \[{{Q}_{0}}\] defined by \[(a,b)\in {{R}_{1}}\Leftrightarrow a=\dfrac{1}{b}\]

Answer 1

Hint: We will use the definitions of reflexive, symmetric and transitive relations to solve this question. A relation is a reflexive relation If every element of set A maps to itself. A relation in a set A is a symmetric relation if \[({{a}_{1}},{{a}_{2}})\in R\] implies that \[({{a}_{2}},{{a}_{1}})\in R\], for all \[{{a}_{1}},{{a}_{2}}\in A\]. A relation in a set A is a transitive relation if \[({{a}_{1}},{{a}_{2}})\in R\] and \[({{a}_{2}},{{a}_{1}})\in R\] implies that \[({{a}_{1}},{{a}_{3}})\in R\] for all \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\in A\].

Complete step-by-step answer:
 Before proceeding with the question we should know about the concept of relations and different types of relations that are reflexive, symmetric and transitive relations.
A relation in set A is a subset of \[A\times A\]. Thus, \[A\times A\] is two extreme relations.
A relation in a set A is a reflexive relation if \[(a,a)\in R\], for every \[a\in A\].
A relation in a set A is a symmetric relation if \[({{a}_{1}},{{a}_{2}})\in R\] implies that \[({{a}_{2}},{{a}_{1}})\in R\], for all \[{{a}_{1}},{{a}_{2}}\in A\].
A relation in a set A is a transitive relation if \[({{a}_{1}},{{a}_{2}})\in R\] and \[({{a}_{2}},{{a}_{1}})\in R\] implies that \[({{a}_{1}},{{a}_{3}})\in R\] for all \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\in A\].
A relation in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
Now let’s check for reflexivity. Assuming a to be an arbitrary element of \[{{R}_{1}}\].
Then, \[\Rightarrow a\ne \dfrac{1}{a}\] for all \[a\in {{Q}_{0}}\]. Clearly it is not reflexive.
Now moving on to symmetry. Let \[(a,b)\in {{R}_{1}}\Leftrightarrow a=\dfrac{1}{b}\]. But we can also write this as \[\Rightarrow b=\dfrac{1}{a}\]. Hence \[(b,a)\in {{R}_{1}}\]. Thus it is symmetric.
Finally, checking for transitivity. Let \[(a,b)\in {{R}_{1}}\] and \[(b,c)\in {{R}_{1}}\]. So substituting \[(a,b)\in {{R}_{1}}\] we get \[a=\dfrac{1}{b}\] and by substituting \[(b,c)\in {{R}_{3}}\] we get \[b=\dfrac{1}{c}\]. And from these two relations we get \[a=c\]. Hence it is not transitive.
Thus \[(a,b)\in {{R}_{1}}\Leftrightarrow a=\dfrac{1}{b}\]is only symmetric.

Note: Remembering the definition of relations and the types of relations is the key here. We in a hurry can make a mistake in thinking it as a transitive set but we don’t get a reciprocal relationship between a and c and hence it does not belong to \[{{R}_{1}}\].