Solution Manual For Discrete Mathematics, 7th Edition

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Chapter 1Solutions to Selected ExercisesSection 1.12.{2,4}3.{7,10}5.{2,3,5,6,8,9}6.{1,3,5,7,9,10}8.A9.11.B12.{1,4}14.{1}15.{2,3,4,5,6,7,8,9,10}18. 119. 322. We find thatB={2,3}. SinceAandBhave the same elements, they are equal.23. LetxA. Thenx= 1,2,3. Ifx= 1, since 1Z+and 12<10, thenxB. Ifx= 2, since2Z+and 22<10, thenxB. Ifx= 3, since 3Z+and 32<10, thenxB. Thus ifxA, thenxB.Now suppose thatxB. ThenxZ+andx2<10. Ifx4, thenx2>10 and, for thesevalues ofx,x /B. Thereforex= 1,2,3. For each of these values,x2<10 andxis indeed inB. Also, for each of the valuesx= 1,2,3,xA. Thus ifxB, thenxA. ThereforeA=B.26. Since (1)32(1)2(1) + 2 = 0,1B. Since1/A,A=B.27. Since 321>3, 3/B. Since 3A,A=B.30. Equal31. Not equal34. LetxA. Thenx= 1,2. Ifx= 1,x36x2+ 11x= 136·12+ 11·1 = 6.ThusxB. Ifx= 2,x36x2+ 11x= 236·22+ 11·2 = 6.AgainxB. ThereforeAB.35. LetxA. Thenx= (1,1) orx= (1,2). In either case,xB. ThereforeAB.38. Since (1)32(1)2(1) + 2 = 0,1A. However,1/B. ThereforeAis not a subsetofB.39. Consider 4, which is inA. If 4B, then 4Aand 4 +m= 8 for somemC. However, theonly value ofmfor which 4 +m= 8 ism= 4 and 4/C. Therefore 4/B. Since 4Aand4/B,Ais not a subset ofB.1

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