

Subject:
Basic Math Proof  Apostol Calculus I 2nd ed  Theorem 1.5 in I3.2 on p18
Category: Science > Math Asked by: entaroadunga List Price: $2.00 
Posted:
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! 

There is no answer at this time. 

Subject:
Re: Basic Math Proof  Apostol Calculus I 2nd ed  Theorem 1.5 in I3.2 on p18
From: berkeleychocolatega 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*(bc) = a* (b+(c)) = a*b + a*(c) = a*b + (  a*c ) = a*b  a*c. 
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