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, x1, x2∈Z. If m, x1, x2satisfy the equations m+x1= 0 and
m+x2= 0, then x1=x2.
Proof. Let m, x1, x2∈Z, and let m+x1= 0 and m+x2= 0. Since = is both symmetric
and transitive, m+x1=m+x2. Adding (−m) to each side of the equation gives
(−m)+(m+x1)=(−m)+(m+x2)
By axiom 1.1.(ii),
((−m) + m) + x1= ((−m) + m) + x2
Then Proposition 1.8 can be applied to both sides of the equation to get 0 + x1= 0 + x2.
Applying axiom 1.1.(i) gives x1+ 0 = x2+ 0. It then remains to use axiom 1.2 to get
x1=x2.
Revised
Proposition 1.14. For all m∈Z,m·0=0=0·m.
Proof. Let m∈Z. By axiom 1.2, 0 ∈Zand 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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