Let a = {x x = 2k-1, K ∈ Z}, B = {x x = 2K, K ∈ Z}, find a ∩ B, find a ∪ B

A = {x ∣ x = 2K, K ∈ Z}, then the set a is the set of all even numbers,
B = {x ∣ x = 2K + 1, K ∈ Z}, then set B is the set of all odd numbers,
A ∩ B = empty set
A∪B={x|x≠0 ,x∈Z}

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