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# Boolean Algebra Questions and Answers

Boolean Algebra:

Boolean logic, originally developed by George Boole in the mid 1800s. The outcomes of the Boolean quantities are either true (1) or false (0).

Boolean Algebra Questions and Answers :

Boolean Algebra is a digital electronic subjects, involves simplification of logical expression. By practicing a sets of problems on Boolean helps to understand the logical relationship between the variables. The equation is combination of sum of product and product of sum's

Laws of Boolean Algebraic Expression are:

 Sl.No Law Expression 1 Commutative Law A + B = B + AA B = B A 2 Associate Law (A + B) + C = A + (B + C) (A B) C = A (B C) 3 Distributive Law A (B + C) = A B + A C A + (B C) = (A + B) (A + C) 4 Identity Law A + A = A   A A = A 5 Redundant Law A + A B = A A (A + B) = A 0 + A = A  0 A = 0  1 + A = 1  1 A = A  $\bar{A}$ + A = 1 $\bar{A}$.A =0 A + AB = A + B A($\bar{A}$+ B) = AB 6 De Morgan's Theorem $(\overline{A+B})=\bar{A}\bar{B}$ $(\overline{AB})=\bar{A}+\bar{B}$

We have given some of the Boolean algebra question answer for self study and understanding of  the laws.

Simplify  the Equation:

(A + C)A(D + D) + AC + C          // Distributive rule  //
(A + C)A + AC + C    //  Complementary  Identity. //
A((A + C) + C) + C     //  Commutative & Distributive //
A(A + C) + C
AA + AC + C          // simplify  //
A+(A+1)C           //  A+1 = 1 //
A+C

2. Prove that LHS = RHS

$\bar{A}$BC + A$\bar{B}$C + AB$\bar{C}$+ABC = AB+BC+AC

Take the LHS ABC + ABC + ABC + ABC

= $\bar{A}$ BC+A$\bar{B}$C+AB$\bar{C}$+ABC   // TABE   AB as common term  //
= $\bar{A}$ BC+A$\bar{B}$C+AB($\bar{C}$+C )    //  $\bar{C}$+C  = 1  //
= $\bar{A}$ BC+A$\bar{B}$C+AB                     // take A as common term  //
= $\bar{A}$ BC+A($\bar{B}$C+B)   //  B+BC = B+C  //
= $\bar{A}$BC+A($\bar{B}$+C)
= $\bar{A}$BC+AB+AC
= C(AB+A)+AB+AC
=  C(B+A)+AB+AC
= CB+AC+AB+AC   // AC+AC = AC  //
=  AB+BC+AC
=  RHS
LHS = RHS

3. Give an example for sum of products form of equation.
AB + CD

4. Applying DeMorgan's theorem to the expression $\overline{\overline{(X+Y)}+\overline{Z}}$

$\overline{\overline{(X+Y)}+\overline{Z}}$    \\ $\overline{(\bar{A})}$ = A //

=  (X + Y)+Z