Answer:
Explanation:
a) To determine whether R is an equivalence relation on N, we need to check if it satisfies three properties: reflexive, symmetric, and transitive.
Reflexive: For any a ∈ N, aRb holds if and only if a and b end in the same digit. Since a and a both end in the same digit, aRa is true for any a ∈ N. Therefore, R is reflexive.
Symmetric: For any a, b ∈ N, if aRb, then a and b end in the same digit. This also means that b and a end in the same digit, so bRa is also true. Therefore, R is symmetric.
Transitive: For any a, b, c ∈ N, if aRb and bRc, then a and b end in the same digit, and b and c end in the same digit. This means that a and c also end in the same digit, so aRc is true. Therefore, R is transitive.
Since R satisfies all three properties (reflexive, symmetric, and transitive), it is an equivalence relation on N.
b) To determine whether R is an equivalence relation on Z, we need to check if it satisfies the same three properties: reflexive, symmetric, and transitive.
Reflexive: For any a ∈ Z, aRb holds if and only if ab < 0. Since a * a = a^2 > 0 for any nonzero integer a, aRa is false for all nonzero a. However, aR0 is true for any nonzero a since a * 0 = 0 < 0. Therefore, R is reflexive for the element 0.
Symmetric: For any a, b ∈ Z, if aRb, then ab < 0. This means that bR a is also true since b * a = a * b < 0. Therefore, R is symmetric.
Transitive: For any a, b, c ∈ Z, if aRb and bRc, then ab < 0 and bc < 0. However, this does not necessarily imply that ac < 0, since b could be 0. For example, let a = 2, b = 0, and c = -3. Then aRb and bRc are both true, but aRc is false since ac = 2 * -3 = -6 > 0. Therefore, R is not transitive.
Since R does not satisfy the transitive property, it is not an equivalence relation on Z.