Solveeit Logo
Login

Question

Question: Let R be a relation from A = \[\left\\{ {1,2,3,4} \right\\}\]to B = \[\left\\{ {1,3,5} \right\\}\]i....

Let R be a relation from A = \left\\{ {1,2,3,4} \right\\}to B = \left\\{ {1,3,5} \right\\}i.e. (a,b)∈R\left( {a,b} \right) \in R, if a<ba{\rm{ }} < {\rm{ }}b then find ROR−1RO{R^{ - 1}}

Explanation

Solution

According the question find out the relation R from A and B if a<ba{\rm{ }} < {\rm{ }}b and also find out inverse using relation R. Then calculate ROR−1RO{R^{ - 1}}.

Complete step-by-step answer:
Firstly, here we will calculate the relation R that is (a,b)(a,b) that is to be formed by using the condition a<ba{\rm{ }} < {\rm{ }}b .
It is given that A = \left\\{ {1,2,3,4} \right\\}and B = \left\\{ {1,3,5} \right\\}.
So, relation R = \left\\{ {\left( {1,3} \right),\left( {1,5} \right),\left( {2,3} \right),\left( {2,5} \right),\left( {3,5} \right),\left( {4,5} \right)} \right\\}
Now, we will calculate R−1{R^{ - 1}} that is (b,a)(b,a) by reversing all the set values in relation R.
So, {R^{ - 1}} = \left\\{ {\left( {3,1} \right),\left( {5,1} \right),\left( {3,2} \right),\left( {5,2} \right),\left( {3,5} \right),\left( {5,4} \right)} \right\\}
Here, taking one by one all the values of relation R−1{R^{ - 1}} that is (a,b)(a,b)and then find out in relation R which is starting from b that is (b,c)(b,c) . Through which we can calculate the relation ROR−1RO{R^{ - 1}}that is (a,c)(a,c) .
As, ROR−1=(3,1)∈R−1RO{R^{ - 1}} = \left( {3,1} \right) \in {R^{ - 1}} and (1,5)∈R\left( {1,5} \right) \in R
Then, (3,5)∈ROR−1\left( {3,5} \right) \in RO{R^{ - 1}}
As, ROR−1=(3,1)∈R−1RO{R^{ - 1}} = \left( {3,1} \right) \in {R^{ - 1}} and (1,3)∈R\left( {1,3} \right) \in R
Then, (3,3)∈ROR−1\left( {3,3} \right) \in RO{R^{ - 1}}
As, ROR−1=(5,1)∈R−1RO{R^{ - 1}} = \left( {5,1} \right) \in {R^{ - 1}} and (1,3)∈R\left( {1,3} \right) \in R
Then, (5,3)∈ROR−1\left( {5,3} \right) \in RO{R^{ - 1}}
As, ROR−1=(5,1)∈R−1RO{R^{ - 1}} = \left( {5,1} \right) \in {R^{ - 1}} and (1,5)∈R\left( {1,5} \right) \in R
Then, (5,5)∈ROR−1\left( {5,5} \right) \in RO{R^{ - 1}}
As, ROR−1=(3,2)∈R−1RO{R^{ - 1}} = \left( {3,2} \right) \in {R^{ - 1}} and (2,3)∈R\left( {2,3} \right) \in R
Then, (3,3)∈ROR−1\left( {3,3} \right) \in RO{R^{ - 1}}
As, ROR−1=(3,2)∈R−1RO{R^{ - 1}} = \left( {3,2} \right) \in {R^{ - 1}} and (2,5)∈R\left( {2,5} \right) \in R
Then, (3,5)∈ROR−1\left( {3,5} \right) \in RO{R^{ - 1}}
As, ROR−1=(5,2)∈R−1RO{R^{ - 1}} = \left( {5,2} \right) \in {R^{ - 1}} and (2,3)∈R\left( {2,3} \right) \in R
Then, (5,3)∈ROR−1\left( {5,3} \right) \in RO{R^{ - 1}}
As, ROR−1=(5,2)∈R−1RO{R^{ - 1}} = \left( {5,2} \right) \in {R^{ - 1}} and (2,5)∈R\left( {2,5} \right) \in R
Then, (5,5)∈ROR−1\left( {5,5} \right) \in RO{R^{ - 1}}
As, ROR−1=(3,5)∈R−1RO{R^{ - 1}} = \left( {3,5} \right) \in {R^{ - 1}}but there is not any set that starts from 5 in relation R. So, ROR−1RO{R^{ - 1}} cannot be formed.
As, ROR−1=(5,4)∈R−1RO{R^{ - 1}} = \left( {5,4} \right) \in {R^{ - 1}} and (4,5)∈R\left( {4,5} \right) \in R
Then, (5,5)∈ROR−1\left( {5,5} \right) \in RO{R^{ - 1}}
Therefore as of now, we will take all the values of ROR−1RO{R^{ - 1}}without repeating and put them in a relation function.
Hence, RO{R^{ - 1}} = \left\\{ {\left( {3,3} \right),\left( {3,5} \right),\left( {5,3} \right),\left( {5,5} \right)} \right\\}

Note: To solve these types of questions, you need to calculate relation R using the given condition. As, in the above question it is required to calculate ROR−1RO{R^{ - 1}} from which we also need to calculate R−1{R^{ - 1}} .
As, it important to see first the value of R−1{R^{ - 1}} that is (a,b)\left( {a,b} \right) then use the values from R that is (b,c)\left( {b,c} \right)
And hence ROR−1RO{R^{ - 1}}is calculated (a,c)\left( {a,c} \right). So, by following the above method we can calculate any required value.