Solution to Let T be a relation from A = {0, 1, 2, 3} to B = … - Sikademy
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Archangel Macsika

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.

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Here's the Solution to this 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)


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


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