**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 = 2k_{1} = 2k_{2} + 1 for k_{1}, k_{2} ∈ ℕ. If k_{1} = k_{2} then 2k_{2} + 1 > 2k_{1}. If k_{1} < k_{2} then 2k_{2} + 1 > 2k_{1}. If k_{1} > k_{2} then k_{1} ≥ k_{2} + 1 > k_{2} ⇒ 2k_{1} > 2k_{2} + 1. Contradiction. Thus {2k | k ∈ ℕ} and {2k + 1 | k ∈ ℕ} are disjoint.

**ii)** *Skipped* (similar to **(i)**).

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