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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.)Which one of the following statements regarding the relation T is true?a.T is transitive.b.T is symmetric.c.T is antisymmetric.d.T is irreflexive.

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.)Which one of the following statements regarding the relation T is true?a.T is transitive.b.T is symmetric.c.T is antisymmetric.d.T is irreflexive.

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Solution

To answer this question, we first need to find all the elements of the relation T.

The elements of T are as follows:

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

Now, let's check each of the given options:

a. T is transitive: A relation is transitive if for every (a,b) and (b,c) in T, there is also (a,c) in T. In this case, we can see that (0,1) and (1,2) are in T, but (0,2) is not in T. So, T is not transitive.

b. T is symmetric: A relation is symmetric if for every (a,b) in T, there is also (b,a) in T. In this case, we can see that (0,1) is in T, but (1,0) is not in T. So, T is not symmetric.

c. T is antisymmetric: A relation is antisymmetric if for every (a,b) and (b,a) in T, a = b. In this case, we can see that (1,0) and (0,1) are in T, but 1 ≠ 0. So, T is not antisymmetric.

d. T is irreflexive: A relation is irreflexive if for no a in A, (a,a) is in T. In this case, we can see that there is no (a,a) in T for any a in A. So, T is irreflexive.

Therefore, the correct answer is option d. T is irreflexive.

This problem has been solved

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