Unformatted text preview: I(B n C) ,AI since (A, B) n (B ,A) = 0 = IA n CI IA n B n CI + IB n CI IA n B n CI = IA n CI + IB n CI 21A n B n CI (b) IA EB B EB CI = I(A EB B) EB CI = IA EB BI + ICI21(A EB B) n CI = IAI + IBI 21A n BI + ICI21(A EB B) n CI =IAI+IBI+ICI 21A n BI 21A n CI 21B n CI + 41A n B n CI by Proposition 6.1.1 again using Proposition 6.1.1 by part (a) 23. The weak form of the Principle of Mathematical Induction can be used, but we shall also need the special case n = 2 derived in the text. For n = 1 we have to prove that I All = E I Ai I. Since, in this case, E IAil = IAll, the result holds. Now assume the result is true for n = k. Then given k + 1 sets Al ,A2, . .. , Ak+l> IAI U A2 U··· U Ak+ll = I(AI U··· U Ak) U Ak+ll = IAI U··· U Akl + IAk+llI(Al U··· U Ak) n Ak+ll using the result for n = 2 = IAI U A2 U··· U Akl + IAk+ll I(AI n Ak+l) U (A2 n Ak+l) U··· U (Ak n Ak+l)1...
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 Summer '10
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 Graph Theory, Integers, Natural number, IA EB BI, IA n CI

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