Let a = {x | x = 2k-1, K ∈ Z}, B = {x | x = 2K + 3, K ∈ Z} (2)A={x|x=2k,k∈z｝,B={x|x=-2k,k∈z｝ (2)A={x|x=k+1/4,k∈z｝,B={x|x=k/2-1/4,k∈z｝ | Math enthusiast | Mathematical art

Let a = {x | x = 2k-1, K ∈ Z}, B = {x | x = 2K + 3, K ∈ Z} (2)A={x|x=2k,k∈z｝,B={x|x=-2k,k∈z｝ (2)A={x|x=k+1/4,k∈z｝,B={x|x=k/2-1/4,k∈z｝

Analysis: (1) a is an odd set, B = {x | x = 2K + 3, K ∈ Z} = {x | x = 2 (K + 1) + 1, K ∈ Z} is also an odd set, so a = B
(2) A and B are even sets, so a = B
(3) In a, the elements x = (4K + 1) / 4 are odd but not all, such as - 1, 3, etc., while in B, the elements x = (2k-1) / 4 are all odd, so a is the proper subset of B

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