Check the following relations $ R $ and $ S $ for reflexivity, symmetry and transitivity:
(a) $ aRb $ iff $ b $ is divisible by $ a $ , $ a,b \in N $
(ii) $ {l_1}S\,{l_2} $ iff $ {l_1} \bot {l_2} $ , where $ {l_1} $ and $ {l_2} $ are straight lines in a plane.
Answer
618.6k+ views
Hint: To solve this problem we will have to check for the reflexivity, symmetry and transitivity separately. To check reflexivity of a relation $ R $ , we will check whether $ \left( {a,a} \right) \in R $ for every $ a \in R $ .To check symmetry, we will check if $ \left( {a,b} \right) \in R $ then $ \left( {b,a} \right) \in R $ . To check transitivity, we will check if $ \left( {a,b} \right) \in R $ and $ \left( {b,c} \right) \in R $ then $ \left( {a,c} \right) \in R $ .
Complete step-by-step answer:
(i)
Given:
Relation $ R\left( {a,b} \right) $ is such that $ b $ is divisible by $ a $ .
Now, we can write relation $ R $ as,
$ R = \left\{ {\left( {a,b} \right):b\;{\rm{is divisible by }}a} \right\} $
In order to check the reflexivity of the relation, $ \left( {a,a} \right) $ must belong to the relation $ R $ such that $ a $ is divisible by $ a $ . Since $ \left( {a,a} \right) \in R $ .
Therefore $ R $ is reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {a,b} \right) \in R $ then $ \left( {b,a} \right) \in R $ . To check for symmetry, we will have to take an example.
Suppose $ \left( {a,b} \right) $ are $ \left( {3,9} \right) $
Here $ \left( {3,9} \right) \in R $ , as 9 is divisible by 3, but $ \left( {9,3} \right) \in R $ 3 is not divisible by 9.
Therefore, from the example it is clear that $ R $ is not symmetric.
In order to check the transitivity, we will have to assume another value $ c \in R $ such that $ \left( {a,b} \right) \in R $ , $ b $ is divisible by a and $ \left( {b,c} \right) \in R $ , $ c $ is divisible by $ b $ .
If $ b $ is divisible by a and $ c $ is divisible by $ b $ then $ c $ is divisible by $ a $ .
Since, $ \left( {a,b} \right) \in R $ and $ \left( {b,c} \right) \in R $ therefore, $ \left( {a,c} \right) \in R $ .
Hence $ R $ is transitive.
(ii)
Given: Relation $ S\left( {{l_1},{l_2}} \right) $ such that $ {l_1} \bot {l_2} $ , where $ {l_1} $ and $ {l_2} $ are straight lines in a plane.
Relation $ S $ can be expressed as,
$ S = \left\{ {\left( {{l_1},{l_2}} \right):{l_1}\;{\rm{is \text perpendicular to }}{l_2}} \right\} $
In order to check the reflexivity of the relation, $ \left( {{l_1},{l_1}} \right) $ must belong to the relation $ S $ but a line can never be perpendicular to itself. So, $ \left( {{l_1},{l_1}} \right) \notin S $ .
Therefore $ S $ is not reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {{l_1},{l_2}} \right) \in S $ then $ \left( {{l_2},{l_1}} \right) \in S $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ then $ {l_2} $ is also perpendicular to $ {l_1} $ . Hence $ \left( {{l_2},{l_1}} \right) \in S $ .
Therefore, $ S $ is symmetric.
In order to check the transitivity, we will have to assume another value $ {l_3} \in S $ such that $ \left( {{l_1},{l_2}} \right) \in S $ , $ {l_1} $ is perpendicular to $ {l_2} $ and $ \left( {{l_2},{l_3}} \right) \in S $ , $ {l_2} $ is perpendicular to $ {l_3} $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ and $ {l_2} $ is perpendicular to $ {l_3} $ then $ {l_1} $ is not perpendicular to $ {l_3} $ .
Hence $ \left( {{l_1},{l_2}} \right) \notin S $ .
Hence $ S $ is not transitive.
Note: In this type of problem, all the values of the relation must satisfy the conditions of reflexivity, symmetry and transitivity. This can be done taking different examples of the values from the same relation.
Complete step-by-step answer:
(i)
Given:
Relation $ R\left( {a,b} \right) $ is such that $ b $ is divisible by $ a $ .
Now, we can write relation $ R $ as,
$ R = \left\{ {\left( {a,b} \right):b\;{\rm{is divisible by }}a} \right\} $
In order to check the reflexivity of the relation, $ \left( {a,a} \right) $ must belong to the relation $ R $ such that $ a $ is divisible by $ a $ . Since $ \left( {a,a} \right) \in R $ .
Therefore $ R $ is reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {a,b} \right) \in R $ then $ \left( {b,a} \right) \in R $ . To check for symmetry, we will have to take an example.
Suppose $ \left( {a,b} \right) $ are $ \left( {3,9} \right) $
Here $ \left( {3,9} \right) \in R $ , as 9 is divisible by 3, but $ \left( {9,3} \right) \in R $ 3 is not divisible by 9.
Therefore, from the example it is clear that $ R $ is not symmetric.
In order to check the transitivity, we will have to assume another value $ c \in R $ such that $ \left( {a,b} \right) \in R $ , $ b $ is divisible by a and $ \left( {b,c} \right) \in R $ , $ c $ is divisible by $ b $ .
If $ b $ is divisible by a and $ c $ is divisible by $ b $ then $ c $ is divisible by $ a $ .
Since, $ \left( {a,b} \right) \in R $ and $ \left( {b,c} \right) \in R $ therefore, $ \left( {a,c} \right) \in R $ .
Hence $ R $ is transitive.
(ii)
Given: Relation $ S\left( {{l_1},{l_2}} \right) $ such that $ {l_1} \bot {l_2} $ , where $ {l_1} $ and $ {l_2} $ are straight lines in a plane.
Relation $ S $ can be expressed as,
$ S = \left\{ {\left( {{l_1},{l_2}} \right):{l_1}\;{\rm{is \text perpendicular to }}{l_2}} \right\} $
In order to check the reflexivity of the relation, $ \left( {{l_1},{l_1}} \right) $ must belong to the relation $ S $ but a line can never be perpendicular to itself. So, $ \left( {{l_1},{l_1}} \right) \notin S $ .
Therefore $ S $ is not reflexive.
In order to check the symmetry of the relation, we will have to check if $ \left( {{l_1},{l_2}} \right) \in S $ then $ \left( {{l_2},{l_1}} \right) \in S $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ then $ {l_2} $ is also perpendicular to $ {l_1} $ . Hence $ \left( {{l_2},{l_1}} \right) \in S $ .
Therefore, $ S $ is symmetric.
In order to check the transitivity, we will have to assume another value $ {l_3} \in S $ such that $ \left( {{l_1},{l_2}} \right) \in S $ , $ {l_1} $ is perpendicular to $ {l_2} $ and $ \left( {{l_2},{l_3}} \right) \in S $ , $ {l_2} $ is perpendicular to $ {l_3} $ .
If $ {l_1} $ is perpendicular to $ {l_2} $ and $ {l_2} $ is perpendicular to $ {l_3} $ then $ {l_1} $ is not perpendicular to $ {l_3} $ .
Hence $ \left( {{l_1},{l_2}} \right) \notin S $ .
Hence $ S $ is not transitive.
Note: In this type of problem, all the values of the relation must satisfy the conditions of reflexivity, symmetry and transitivity. This can be done taking different examples of the values from the same relation.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is the situation called when no party gets the class 9 social science CBSE

Find the sum of series 1 + 2 + 3 + 4 + 5 + + 100 class 9 maths CBSE

Name 10 Living and Non living things class 9 biology CBSE

What is the Full Form of ICSE, CBSE and SSC

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

What is the difference between Atleast and Atmost in class 9 maths CBSE

