Let a,b, and c be real numbers. if a + b = a + c, then b = c. Suggestion: Use Ax

Question

Let a,b, and c be real numbers. if a + b = a + c, then b = c. Suggestion: Use Axiom A3, Axiom Al, and Axiom A2. 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

we know there exists a number like -a for which (-a)+a=0 (A3). Now add (-a) to both sides: LHS: (-a)+(a+b) = ((-a)+a) + b (A1) = 0 + b (A3) = b (A2) in a similar way for RHS we get c, therefore b=c proof ends.

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