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Math 301 Homework 1 Charlie Lane McMenomy
Proposition 1.8. If m is an integer, then (−m) + m = 0. Proof. Let m ∈ Z. Then by axiom 1.4, m + (−m) = 0. Applying axiom 1.1(i) gives (−m) + m = 0.
Proposition 1.10. Let m, x 1 , x 2 ∈ Z. If m, x 1 , x 2 satisfy the equations m + x 1 = 0 and m + x 2 = 0, then x 1 = x 2. Proof. Let m, x 1 , x 2 ∈ Z, and let m + x 1 = 0 and m + x 2 = 0. Since = is both symmetric and transitive, m + x 1 = m + x 2. Adding (−m) to each side of the equation gives (−m) + (m + x 1 ) = (−m) + (m + x 2 ) By axiom 1.1.(ii), ((−m) + m) + x 1 = ((−m) + m) + x 2 Then Proposition 1.8 can be applied to both sides of the equation to get 0 + x 1 = 0 + x 2. Applying axiom 1.1.(i) gives x 1 + 0 = x 2 + 0. It then remains to use axiom 1.2 to get x 1 = x 2.
Revised Proposition 1.14. For all m ∈ Z, m · 0 = 0 = 0 · m. Proof. Let m ∈ Z. By axiom 1.2, 0 ∈ Z and 0 + 0 = 0. By replacement, m · 0 = m · (0 + 0). Using axiom 1.1.(iii), m · (0 + 0) = m · 0 + m · 0. Returning to axiom 1.2, m · 0 + 0 = m · 0. Since = is transitive, m · 0 + 0 = m · 0 + m · 0. Then by proposition 1.9, 0 = m · 0. Applying axiom 1.1.(iv) gives m · 0 = 0. Therefore, m · 0 = 0 = m · 0.
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