Exercises 0.3 #2

~ ⊂ ℕ(2) × ℕ(2). If (a1, a2) ∈ ℕ(2) then a1 + a2 = a2 + a1. So ((a1, a2), (a1, a2)) ∈ ~. If ((a1, a2), (b1, b2)) ∈ ℕ(2) × ℕ(2) then a1 + b2 = a2 + b1. By commutativity b1 + a2 = b2 + a1, so ((b1, b2), (a1, a2)) ∈ ℕ(2) × ℕ(2). If ((a1, a2), (b1, b2)), ((b1, b2), (c1, c2)) ∈ ℕ(2) × ℕ(2) then a1 + b2 = a2 + b1 and b1 + c2 = b2 + c1. Subtracting gives a1 – c1 = a2 – c2, so a1 + c2 = a2 + c1. Thus ((a1, a2), (c1, c2)) ∈ ℕ(2) × ℕ(2). Hence ~ is an equivalence relation.

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