Chapter No.6
BOOLEAN ALGEBRA
Q.6.01 Complete the following statements.
(i)
(ii)
(iii)
(iv)
(v)
Ans:
In Boolean algebra plus sign stands for ________________ operation.
AND operation is used for logical ________________
Double complementation has ________________ effects.
In Boolean algebra 1+1+1 is equal to ________________.
According to the distribution law { A + ( B . C ) } = ________________
(i) OR ii) multiplication
iii) cancellation
iv) 1
v) (A+B).(A+C)
Q.6.02 Which of the following statements is True or False?
(i)
(ii)
(iii)
(iv)
(v)
Ans:
i) False
AND operation is also called as logical addition.
In a NOT gate, the output is negative of input.
Two series switches can represent AND operation.
NOT of ( A + B ) is equal to NOT of A OR NOT of B.
A variable that is same within two adjacent squares of Karnaugh map is dropped.
ii) False iii) True iv) True v) False
Q.6.03 Encircle one Choice A, B, C or D I each case
(i)
(ii)
Boolean algebra is also known as
(A) Logical algebra
(B) Code algebra
(C) Switching algebra
(D) Digital algebra
An OR operation has 4 variables. The possible number of combinations in its truth table
are:
(B) 4
(B) 8
(C) 16
(D) 32
(iii) The output will be one if all inputs except one are zero in case of:
(C) NOT operation
(B) AND operation
(C) OR operation
(D) NOT of OR
(iv) According to Demorgon’s Law ̅̅̅̅̅̅̅ is equal to:
(D) ̅
(v)
i)(C)
(C) ̅ ̅
(B) A . B
(D) A + B
According to absorption law A . ( A + B ) is equal to:
(E) 1 + B
Ans:
̅
ii) (C)
(B) A B
(C) A
(D) A + 1
iii) (C)
iv) (C)
v) (C)
1
Q.6.04 Match the items given in Column I with those given in Column II
Column I
i)
ii)
iii)
iv)
v)
Ans:
AND
OR
A+̅
A+̅
A+(A.B)
i) (e)
ii) (c)
Column II
a)
b)
c)
d)
e)
1
A+B
Union
A
Intersection
iii) (a)
iv) (b)
v) (d)
Q. 6.05 What is Boolean algebra?
Ans:
George Boole, the founder of symbolic logic presented it as mathematical logic, which is now
known as Boolean algebra. Boolean algebra finds its application to problem in the form of statements
having answers either true or false. Boolean algebra uses algebraic notation to express logical
relationship in the same way as in ordinary algebra to express mathematical relationships.
Q.6.06 What do you understand by the logical operations AND and OR.
Ans: Logical operation AND:
In Boolean algebra the operation AND means logical multiplication or it is represent with or without a
dot between the variables as X.Y or XY. The resulting variable will be the value 1 only if all variables are
1, otherwise it will be 0 if one or more variables are 0. The operation NOT is illustrated in the following
table:
X
0
0
1
1
AND Operation
Y
0
1
0
1
X.Y
0
0
0
1
Logical operation OR:
In Boolean algebra OR operation means logical addition and it is represented by plus (+) sign between
two variables. The resulting variable will be the value 0 only if all variables are 0, otherwise it will be 1 if
one or more variables are 1. The operation NOT is illustrated in the following table:
2
OR Operation
Y
0
1
0
1
X
0
0
1
1
X+Y
0
1
1
1
Q.6.07 Find the values of the Boolean expressions
i)
ii)
iii)
Ans:
XY + X̅
when X = 1 and Y = 0
(X+Y) . (XY)
when X = 1 and Y = 0
(X+̅) . (̅+Y) when X = 1 and Y = 1
(i) XY + X̅
when X = 1 and Y = 0
XY + X̅
=(1.0)+(1.1)
= 0+1
=1
Note: Remaining questions are same as above.
Q.6.08 State and prove the two basic Demorgan’s theorems. Find out the complement of the
following Boolean expression:
i) XY + X̅
Ans:
Demorgan’s Laws.
(a) ̅̅̅̅̅̅̅= ̅ . ̅
Ans:
iii) (X+̅) . (̅+Y)
ii) (X+Y) . (XY)
(a)
(b) ̅̅̅̅̅̅ = ̅ + ̅
L.H.S
= ̅̅̅̅̅̅̅
= ̅̅̅
=̅.̅
= R.H.S
(b)
L.H.S = ̅̅̅̅̅̅
= ̅ ̅̅
=̅+̅
= R.H.S
Complements of Boolean expression:
i) XY + X̅
= (X+Y) . (X+̅)
ii) (X+Y) . (XY)
= (X .Y) + (X+Y)
iii) (X+̅) . (̅+Y)
3
= (X.̅) + (̅.Y)
Q.6.09 What is Truth Table? Construct a truth table for AND, OR and NOT operation for three
variables X, Y and Z.
Ans:
A truth table is a table that shows the result of a Boolean expression for all the possible
combinations of the values given to the variables related to the operators in the expression.
For example truth table for operators AND, OR, and NOT given below for three variables:
X
0
0
0
0
1
1
1
1
Truth Table for AND Operator
Y
Z
0
0
0
1
1
0
1
1
0
0
0
1
1
0
1
1
X.Y.Z
0
0
0
0
0
0
0
1
X
0
0
0
0
1
Truth Table for OR Operator
Y
Z
0
0
0
1
1
0
1
1
0
0
X+Y+Z
0
1
1
1
1
1
1
1
0
1
1
1
0
1
Truth Table for NOT Operator
̅
Z
0
1
1
1
0
1
1
1
0
0
1
0
X
0
0
0
0
1
1
Y
0
0
1
1
0
0
1
1
0
1
1
1
1
1
1
̅
̅
1
1
1
0
0
1
1
0
1
0
1
0
0
0
1
0
0
0
Q.6.10 State and Prove the following laws:
(i) Idempotent Law
(ii) Involution Law
4
(iii) Absorption Law
Ans:
(i) Idempotent Laws:
(a)
A+A=A
a) L.H.S = A + A
(b) L.H.S
(b)
A.A=A
= (A + A) .1
= (A + A) . (A + ̅ )
= A + (A .̅ )
=A+0
=A
= R.H.S
=A.A
= (A . A) + 0
= (A . A) + (A .̅ )
= A . (A + ̅ )
=A.1
=A
= R.H.S
(ii) Involution Law:
It states that double complement has cancellation effect. This can be proved by the method of perfect
induction as shown in truth table.
A
̅
̿
0
1
0
1
0
1
(iii) Absorption Laws:
(a)
A + (A . B) = A
(b)
(a) L.H.S
= A + (A . B)
= (A . 1) + (A . B)
= A . (1 + B)
=A.1
=A
= R.H.S
(b) L.H.S
= A . (A + B)
= (A + 0) . (A + B)
5
A . (A + B) = A
= A + (0 . B)
=A+0
=A
= R.H.S
Q.6.11 Construct a truth table for the following Boolean expression.
(i) XY + ̅Z + YZ
(iii) X̅+XZ + Y ̅
(ii) (X+Y) . (XY)
Ans: Part (i) XY + ̅Z + YZ
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
̅
1
1
1
1
0
0
0
0
Z
0
1
0
1
0
1
0
1
̅Z
0
1
0
1
0
0
0
1
XY
0
0
0
0
0
0
1
1
F = XY + ̅Z + YZ
0
1
0
1
0
0
1
1
YZ
0
0
0
1
0
0
0
1
Note: Remaining part (ii) and (iii) are same as above.
Q.6.13 Simplify the following functions using karnaugh map.
Ans:
(i) A ̅ C +ABC + ̅ ̅
(ii) ABC + A ̅ C + AB ̅
(iii) A ̅ C + ̅B ̅ + ̅ ̅ C
(iv) A ̅ + ̅C + B ̅
Solution:
(i) A ̅ C +ABC + ̅ ̅
= A ̅ C +ABC + A ̅ ̅ + ̅ ̅ ̅
̅
̅
B
B
̅
̅̅ ̅
A
A̅ ̅
A ̅C
ABC
̅
C
C
̅
̅
̅
B
B
6
̅
1
0
0
0
A
1
1
1
0
̅
C
C
̅
̅
B
B
A ̅C
ABC
AB ̅
̅
C
C
̅
̅
̅
B
B
̅
0
0
0
0
A
0
1
1
1
̅
C
C
̅
Identify adjacent groups
F = ( ̅ ̅ ̅ + A ̅ ̅ )+ ( A ̅ C + ABC )
= ̅ ̅ ( ̅+A) + AC ( ̅ +B)
= ̅ ̅ (1) + AC (1)
= ̅ ̅ + AC
(ii) ABC + A ̅ C + AB ̅
̅
̅
A
Identify adjacent groups
F = ( A ̅ C + ABC ) + (ABC + AB ̅ )
=
( ̅ +B) + AB ( ̅ +C)
=
(1) + AB (1)
= AC + AB
(iii) A ̅ C + ̅B ̅ + ̅ ̅ C
7
̅
̅
̅
̅ ̅C
A
A ̅C
B
B
̅B ̅
̅
C
C
̅
̅
̅
B
B
̅
0
A
0
1
0
0
̅
C
C
̅
̅
B
B
̅ ̅C
̅BC
̅B ̅
0
Identify adjacent groups
F = ( ̅ ̅ C + A ̅ C ) + ̅B ̅
= ̅ C ( ̅+A) + ̅B ̅
= ̅ C (1) + ̅B ̅
= ̅ C + ̅B ̅
(iv) A ̅ + ̅C + B ̅
= A ̅C +
̅ ̅ + ̅BC + ̅ ̅ C + AB ̅ + ̅B ̅
̅
̅
A
A̅ ̅
A ̅C
̅
C
8
AB ̅
C
̅
̅
̅
B
B
̅
0
A
1
1
0
1
̅
C
C
̅
Identify adjacent groups
F = ( ̅ ̅ + A ̅ C ) + ( ̅ ̅ C + ̅BC ) + ( ̅
= A ̅ ( ̅ +C) + ̅C ( ̅ +B) + B ̅ ( ̅+A)
= A ̅ (1) + ̅C (1) + B ̅ (1)
=A ̅ + ̅C + B ̅
9
̅ + AB ̅ )