MTH202 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || DISCRETE MATHEMATICS || BY VuTech

MTH202 ASSIGNMENT NO. 2 FALL 2022

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MTH202 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || DISCRETE MATHEMATICS || BY VuTech

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Assignment # 2                                                                     MTH202(Fall2022)

 

Total Marks: 10                                                                                      

Due Date: 14 -12-2022

 

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Question No.1:                                                                                                    Marks=5

Let A={1,2,3,4,5} is a set and R is a relation on A having the following ordered pairs

R={(1,1),(1,2),(2,1).(2,2),(2,3),(3,2),(3,3),(4,4),(5,5)}.

Determine whether R is an Equivalence relation? Justify your answer with complete steps.

Solution

For R to be an Equivalence relation, it must be reflexive, symmetric, and transitive.

Reflexive Relation

R is Reflexive if only if, for all `a \in A, (a,a) \in R`

R={(1,1),(2,2),(3,3),(4,4),(5,5)}

Hence R is a reflexive relation.

Symmetric Relation

R is Symmetric if only if, for all `a,b \in A, if (a,b) \in R, then (b,a) \in R`

R={(1,2),(2,1),(2,3),(3,2)}

Since (1,2) is in R and then we have (2,1) in our relation R. Same as (2,3) in R and (3,2) in R.

So, R is a symmetric relation.

Transitive Relation

R is Reflexive if only if, for all `a,b,c \in A, if (a,b) \in R, (b,c) \in R`, `then (a,c) \in R`

In the given relation, we have (1,2) and (2,1) so for transitive relation, there must be (1,1) which is present in R.

So, R is a transitive relation.

 

Here, all three conditions for equivalence relation all true,

 

Hence, R is an equivalence relation.

 

 

Question No.2:                                                                                                    Marks=5

 

Let A = {0, 1, 2, 3, 4, 5} and R be a relation from A to A defined by

`R = {(x,y) : y = x + 1}`. 

Find R and determine whether R is a function or not. Justify your answer.

Also find the domain of R, range of R and inverse of R.

Solution:

Here

First we will calculate A × A

A × A = {(0,0),(0,1),(0,2),(0,3),(0,4)(0,5),

(1,0),(1,1),(1,2),(1,3),(1,4),(1,5),(2,0),

(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),

(3,3),(3,4),(3,5),(4,0),(4,1),(4,2),(4,3),

(4,4),(4,5),(5,0),(5,1),(5,2),(5,3),(5,4),(5,5)}

 

Now,

            R={(0,1),(1,2),(2,3),(3,4),(4,5)}

For R to be a function,

1)      Every element of A must be the first element of some ordered pair

2)      There is no pair that has the same first-order pair but different second order pair.

 

So,

            R is not a function as it does not satisfy first condition of the function.

 

Now,

           

            Domain of R = {0,1,2,3,4}

            Range of R = (1,2,3,4,5)

And

            Inverse of R = R^-1 = {(1,0),(2,1),(3,2),(4,3),(5,4)}

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