Exercises 0.3 #1

i) If m, n ∈ ℕ then m × n ∈ ℕ and m + n ∈ ℕ. So [X = {2k | k ∈ ℕ} ∪ {2k + 1 | k ∈ ℕ}] ⊂ ℕ. Let n ∈ ℕ. Since ℕ is not bounded above, there is a largest k ∈ ℕ such that 2k ≤ n. So 2k ≤ n < 2(k + 1) = 2k + 2. Because there is no natural number between a natural number and its successor, n = 2k or n = 2k + 1. So ℕ ⊂ X. Thus X = ℕ. Let m ∈ {2k | k ∈ ℕ} ∩ {2k + 1 | k ∈ ℕ}. Then m = 2k1 = 2k2 + 1 for k1, k2 ∈ ℕ. If k1 = k2 then 2k2 + 1 > 2k1. If k1 < k2 then 2k2 + 1 > 2k1. If k1 > k2 then k1 ≥ k2 + 1 > k2 ⇒ 2k1 > 2k2 + 1. Contradiction. Thus {2k | k ∈ ℕ} and {2k + 1 | k ∈ ℕ} are disjoint.

ii) Skipped (similar to (i)).

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