PLease show details For every a and b in R. -(a + b) = (-a) + (-b). Suggestion:

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For every a and b in R. -(a + b) = (-a) + (-b). Suggestion: Add the two expressions together and work carefully through equalities to arrive at 0. Be sure to watch your parentheses, use of axioms, and previous propositions. For all a,b,c R, (a + b) + c = a + (b + c). There exists a unique number 0 R such that a+0= 0+a = a for every a R. For all a R, there exists a unique number -a R such that a + (-a) = (-a) + a = 0. For all a, b R, a + b = b + a. For all a,b,c R, (a . b) . c = a . (b . c). There exists a unique number 1 R such that a.1 = 1.a = a for every a R. For all nonzero a R, there exists a unique number a-1 R such that a .a-1 = a-1.a = 1. For all a, b R, a . b = b . a. For all a,b,c R, a.(b + c) = a.b + a.c. 1 0. For all a R, exactly one of the following statements is true: 0

Explanation / Answer

add a+b to both sides: LHS: (a+b)+(-(a+b))=0 (A3) RHS: (a+b)+(-a)+(-b) = (a+(-a))+b+(-b) (A1,A4) =0+0 (A3) =0 (A2) Hence (a+b)+(-(a+b)) = (a+b)+(-a)+(-b) add -(a+b) to both sides: -(a+b) = (-a)+(-b)

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