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Discrete Mathematics with Graph Theory (3rd Edition) 173

# Discrete Mathematics with Graph Theory (3rd Edition) 173 -...

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Section 6.1 171 18. [BB] Write a = qb + r with 0:::; r < b. Then l~J = q. Clearly the positive integers b, 2b, ... , qb are all less than or equal to a and are all divisible by b. On the other hand, if sb :::; a, then s must belong to the set {I, 2, ... , q}. Hence, there are exactly q such natural numbers, as required. 19. Since IA U BI = IAI + IBI - IA n BI < IAI + IBI, IA n BI > O. Thus An B ¥= 0. 20. [BB] Recall thatAxB = {(a, b) I a E A, bE B}. If A = {a!, a2, ... , an} and B = {bI, b2, ... , b m }, the ordered pairs of Ax B can be enumerated with n rows of m elements each (aI,b l ) (al,b2) (al,b 3 ) (a2, b l ) (a2, b2) (a2, b 3 ) giving nm = IAI x IBI elements in all. 21. IA U BCI = I(AC n B)CI by one of the laws of De Morgan. So IA U BCI = lUI-lAc n BI = lUI -IB , AI = lUI - (IBI -IB n AI) = lUI - IBI + IA n BI 22. (a) [BB] I (A EB B) n CI = I ((A, B) U (B , A)) n CI = 1((A,B)nC) U ((B,A)nC)1 = I(A n C)
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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 + ICI-21(A EB B) n CI = IAI + IBI- 21A n BI + ICI-21(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+ll-I(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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