Answer to Question #198868 in Discrete Mathematics for Vina

Question

Let T be a relation from A = {0, 1, 2, 3} to B = {0, 1, 2, 3, 4} such that (a, b) T iff b2 – a2 is an odd number. (A, B  U = Z.) (Hint: Write down all the elements of T. For example, if 4  B and 1  A then 42 – 12 = 16 – 1 = 15 which is an odd number, thus (1, 4)  T.) Answer questions 5 and 6 by using the defined relation T. Question 5 Which one of the following alternatives provides only elements belonging to T? 1. (3, 1), (4, 1), (3, 2) 2. (0, 1), (2, 4), (2, 3) 3. (3, 0), (1, 2), (3, 4) 4. (1, 0), (1, 2), (1, 3)   Question 6 Which one of the following statements regarding the relation T is true? 1. T is transitive. 2. T is symmetric. 3. T is antisymmetric. 4. T is irreflexive.  

Answer 1

let T be a relation from A ={0,1,2,3} to B={0,1,2,3,4}

such that 

(a,b)∈T iff b²-a² is an odd number. 

(A,B)⊆U=Z)

Question5

which one of the following alternatives provides only elements belonging to T?

1. (3,1),(4,1),(3,2)

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

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

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

Question6

which one of the following statements regarding the relation T is true?

1. T is transitive

2. T is symmetric

3. T is antisymmetric

4. T is irreflexive

solution:-

firs of all we write all (a,b)=b²-a² ,and select element of T,which b²-a²=odd.

(0,0)=0 (0,1)=1 (0,2)=4 (0,3)=9 (0,4)=16

(1,0)=-1 (1,1)=0 (1,2)=3 (1,3)=8 (1,4)=15

(2,0)=-4 (2,1)=-3 (2,2)=0 (2,3)=5 (2,4)=12

(3,0)=-9 (3,1)=-8 (3,2)=-5 (3,3)=0 (3,4)=7

we create a set of element which show b²-a²=odd.

T={(0,1) (0,3),(1,0),(1,2),(1,4),(2,1),(2,3),(3,0),(3,2),(3,4)}

answer5

elements of option 3 contains in the T set.

so option 3 is correct answer.

answer6

1. T is transitive:-

definition of transitive:-

if (a,b) where b²-a²=odd and (b,c) where c²-b²=odd

then (a,c) where c²-a²=odd.

it property called transitive.

check:-for (a,c)

c²-a²=(c²-b²)+(b²-a²)

     =odd+odd

    =even {we know that sum of two odd is even always}

so T is not transitive

2. T is symmetric

definition of symmetric

if (a,b) where b²-a²=odd 

then (b,a) where a2-b2=odd.

it property called symmetric

check:-for (b,a)

a2-b2=-(b²-a²)

     =-(odd)

     =odd {negative odd is also odd}

so T is symmetric

3. T is antisymmetric

option 2 is correct hence option 3 is incorrect.

4. T is irreflexive

definition of reflexive

if (a,a) where a2-a2=odd 

it property called reflexive

check:-for (a,a)

a2-a2=0 {0 is not odd}

so T is not reflexive

hence T is irreflexive

here option 2 and 4 are correct option