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Simplify the Boolean expression F = C(B + C)(A + B + C).
Ans. Simplification of the given Boolean Expression F = C (B +C) (A+B+C) given asF = C (B+C) (A+B+C)= CB + CC [(A+B+C)]= CB + C [(A+B+C)] (Q CC = C)= CBA + CBB + CBC + CA + CB + CC= ABC + CB + CB + CA + CB + CC (Q CBB =CB & CBC = CB)= ABC + CB + CA + C (Q CB+CB+CB = CB; CC = C)= ABC + BC + C (1+A)= ABC + BC + C (Q 1+A = 1)= ABC + C (1+B)= ABC + C (Q 1+B = 1)= C (1+AB) = C {Q (1+AB)=1}
Subtraction of 01000-01001 using 2's complement method. Ans. Firstly 1's complement of 01001 is 10110 and 2's complement is 10110+ 1 =10111. Thus 01000 = 01000 - 01001
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