2) Let a, b , and c be real numbers. PROVE If a + b = a + c , then b = c . Sugge

Question

2) Let a, b, and c be real numbers. PROVE If a + b = a + c, then b = c.

Suggestion: Use Axiom A3, Axiom A1, and Axiom A2

A1 For all a, b, c  , (a + b) + c = a + (b + c)

A2 There exists a unique number   0 such that a+0 = 0+a = a

such that a + (-a) = (-a) + a = 0.

THANK YOU!!!

1) Let a and b be real numbers. PROVE If ab = 0, then a = 0 or b = 0. Suggestion: Consider the following axiom which says For all nonzero a , there exists a unique number a-1 , such that 2) Let a, b, and c be real numbers. PROVE If a + b = a + c, then b = c. Suggestion: Use Axiom A3, Axiom A1, and Axiom A2 A1 For all a, b, c , (a + b) + c = a + (b + c) A2 There exists a unique number 0 such that a+0 = 0+a = a for every a . A3 For all a , there exists a unique number -a such that a + (-a) = (-a) + a = 0. THANK YOU!!!

Explanation / Answer

It helps when you post the statements that we're allowed to take as axiomatic. Here goes:

1. Suppose not: suppose there exist a,b in R with a0b s.t. a*b=0. By definition of inverse: since a and b are non-zero, there must exist a-1 and b-1 in R s.t. a-1*a=1 and b*b-1 = 1. Since a*b=0, then a-1*(ab)*b-1 = a-1*(0)*b-1. On the left-hand side of this equation, we have 1*1; on the right-hand size, we have a-1*0 = 0, since zero times and real number is zero. Thus, 1=0. Contradiction.

2. Let a,b,c be in R, and suppose that a + b = a + c, and let (-a) denote the unique additive inverse of a. Then a+b + (-a) = a+c + (-a). By commutativity of addition, this is the same as a+(-a)+b = a+(-a)+c. By associativity (A1), this is equivalent to (a+(-a))+b = (a+(-a))+c. By A3, a+(-a)=0, so this is equivalent to 0+b = 0+c. By A2, 0+b=b on the left-hand side and 0+c=c on the right-hand side, so we have b=c. The end.

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