Basic Math Proof - Apostol Calculus I 2nd ed - Theorem 1.5 in I3.2 on p18
Category: Science > Math
Asked by: entaroadun-ga
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11 Nov 2006 08:15 PST
Expires: 11 Dec 2006 08:15 PST
Question ID: 781871
Does anyone have a solution for the theorem in the title? a(b - c) = ab - ac I can come up with two separate approaches, but not one that uses only the 6 axioms given and 4 theorems already proven. One approach I have uses the fact that "a x 0 = 0", and the other uses the fact that "-(a(-c)) = ac". The text doesn't provide a solution. Thanks!
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Re: Basic Math Proof - Apostol Calculus I 2nd ed - Theorem 1.5 in I3.2 on p18
From: berkeleychocolate-ga on 13 Nov 2006 10:53 PST
First prove a*0=0 as follows: a*0 = a*(0+0)=a*0 + a*0. Then subtract a*0 from both sides to get that a*0 = 0. Secondly prove that a*(-c) = - a*c by a*(-c) + a*c = a*(-c+c) = a*0 = 0. From this equation it follows that a*(-c) = - a*c from the definition of - a*c. Finally use these two facts as follows: a*(b-c) = a* (b+(-c)) = a*b + a*(-c) = a*b + ( - a*c ) = a*b - a*c.
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