By Randall R. Holmes
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Then x ∈ A or x ∈ B. Assume that x ∈ / A. Then x ∈ B and x ∈ / A, so x ∈ B \ A. Hence, x ∈ A ∪ (B \ A). The containment (⊇) follows. Therefore, A ∪ (B \ A) = A ∪ B. Discussion: In the proof of (⊆), after letting x be an arbitrary element of the left-hand set and writing what this means, we turn to the task of showing x is in the right-hand set. 1). 3 Theorem. Let A, B, and C be sets. We have (i) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), (ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Proof. (i) It suffices to show that each side is a subset of the other.
2 Example m + 1 is odd. Solution Assume that m is odd. Then m = 2k + 1 for some k ∈ Z. We have m2 + m + 1 = (2k + 1)2 + (2k + 1) + 1 = (4k 2 + 4k + 1) + 2k + 2 = (4k 2 + 6k + 2) + 1 = 2(2k 2 + 3k + 1) + 1. Now 2k 2 + 3k + 1 ∈ Z, so we conclude that m2 + m + 1 is odd. 2 Divisibility We are accustomed to saying informally that “3 goes evenly into 15” to mean that there is no remainder when 15 is divided by 3. This is the same 63 as saying that “15 is a multiple of 3,” namely, 15 = (5)3. Here is the more formal terminology: Let m and n be integers.
Proof. (i) It suffices to show that each side is a subset of the other. (⊆) Let x ∈ C \ (A ∪ B). Then x ∈ C and x ∈ / A ∪ B. So x ∈ C and also x∈ / A and x ∈ / B. Put another way, x ∈ C and x ∈ / A and also x ∈ C and x∈ / B. This says x ∈ C \ A and x ∈ C \ B, so that x ∈ (C \ A) ∩ (C \ B). The inclusion (⊆) follows. (⊇) Let x ∈ (C \ A) ∩ (C \ B). Then x ∈ C \ A and x ∈ C \ B. So x ∈ C and x∈ / A and also x ∈ C and x ∈ / B. Put another way, x ∈ C and also x ∈ /A and x ∈ / B. This says that x ∈ C and x ∈ / A ∪ B, so that x ∈ C \ (A ∪ B).