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Homework2_sol_1_16_10

Homework2_sol_1_16_10 - Homework#2 due January 25 at noon...

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Homework #2 due January 25 at noon ECE 15a Winter 2010 For problems 1-6 supply the reasons for each step. (10p) 1. Write out the proof that a(a+b)=a in Theorem 4 (lecture #3), referring each step to the correct postulate. Proof: a =1·a P2: In Boolean algebra there exist distinct identity elements:1 for (·). =(b+1)·a Theorem 3: b+1=1 =b·a+1·a P3: Each operation is distributive over the other. =b·a+a P2: In Boolean algebra there exist distinct identity elements:1 for (·). =b·a+a·a Theorem 2: a·a=a =a·(a+b) P3: Each operation is distributive over the other. (10p) 2. Prove that in every Boolean algebra every triple of elements a, b,c satisfies the identity ab+bc+ca=(a+b)(b+c)(c+a). Proof: (a+b)(b+c)(c+a) = (ab+ac+b+bc)(c+a) = abc+ac+bc+bc+ab+ac+ab+abc = abc+ac+bc+ab = ab(c+1)+bc+ca = ab+bc+ca (10p) 3. Prove, that if a+x=b+x and a+x’=b+x’, then a=b. (hint: check the proof of Theorem 5 in lecture #3). Proof: (a+x)(a+x’) = a+ax’+ax+xx’ = a+a(x+x’)+0 = a+a = a (b+x)(b+x’) = b+bx’+bx+xx’ = b+b(x+x’)+0 = b+b = b Since a+x=b+x and a+x’=b+x’, we have a = (a+x)(a+x’) = (b+x)(b+x’) = b.
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