Boolean Algebra

This page contains the CBSE Computer Science with Python class 12 Unit-4 chapter 1 Boolean Algebra. You can find the solutions for chapter 1 of CBSE class 12 Computer Science with Python Exercise. So is the case if you are looking for CBSE class 12 Computer Science with Python related topic Boolean Algebra questions and answers for the Exercise.
Buy Class 12 CBSE Computer Science with Python Books NOW!
Exercise
Question 1
1. The commutative law of Boolean algebra states that
A + B = A . B
i)
True
ii)
False ✔
Answer 1
ii) False. In commutative law, order changes but operator stays same: A + B = B + A and A . B = B . A. The statement A + B = A . B is not generally true.
Question 2
2. Applying Dr. Morgan’s theorem to the expression (ABC)’, we get
i)
A’+B’+C’ ✔
ii)
(A+B+C)’
iii)
A+B’+C.C’
iv)
A (B+C)
Answer 2
i) A’ + B’ + C’. By DeMorgan’s theorem, complement of a product becomes sum of complements: (ABC)' = A' + B' + C'.
Question 3
3. Which Boolean Algebra theorem allow us to group operands in an expression in any order without affecting the results of the operation:
i)
Associative ✔
ii)
Commutative
iii)
Boolean
iv)
Distributive
Answer 3
i) Associative. Associative law allows regrouping without changing result: (A+B)+C = A+(B+C) and (A.B).C = A.(B.C).
Question 4
4. Applying Dr. Morgans theorem to the expression ((x+y)’+z)’, we get
i)
(x+y)z
ii)
(x’+y’)z
iii)
(x+y)z’ ✔
iv)
(x’+y’)z’
Answer 4
iii) (x+y)z’. Since ((x+y)'+z)' = ((x+y)')'.z' = (x+y)z', option (iii) is correct.
Question 5
5. Applying the distributive law to the expression A (B+C’+D), we get_____________
i)
AB+AC+AD
ii)
ABCD
iii)
A+B+C+D
iv)
AB+AC’+AD ✔
Answer 5
iv) AB + AC’ + AD. By distributive law, A(B+C'+D) = AB + AC' + AD.
Question 6
6. A variable is a symbol used to represent a logical quantity that can have a value of 1 or 0
i)
True ✔
ii)
False
Answer 6
i) True. In Boolean algebra, a variable represents a logical quantity and can take only two values: 0 or 1.
Question 7
7. The OR operation is Boolean multiplication and the AND operation is Boolean addition
i)
True
ii)
False ✔
Answer 7
ii) False. In Boolean algebra, OR corresponds to addition (+) and AND corresponds to multiplication (.), not the other way round.
Question 8
8. In Boolean Algebra, A+1 = 1
i)
True ✔
ii)
False
Answer 8
i) True. This is the null (domination) law of OR: A + 1 = 1.
Question 9
9. In the Commutative law, in ORing and ANDing of two variables. The order in which the variables are ORed or ANDed makes no difference
i)
True ✔
ii)
False
Answer 9
i) True. Commutative law says order of variables does not affect result: A+B = B+A and A.B = B.A.
Question 10
10. Verify using truth table, DeMorgan’s theorem for three variables.
Answer 10
For three variables, DeMorgan’s theorem is: (A+B+C)' = A'.B'.C' and (A.B.C)' = A' + B' + C'.
Verification of (A+B+C)' = A'.B'.C'
A
B
C
(A+B+C)’
A’.B’.C’
0
0
0
1
1
0
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
0
0
1
1
0
0
0
1
1
1
0
0
Verification of (A.B.C)' = A'+B'+C'
A
B
C
(A.B.C)’
A’+B’+C’
0
0
0
1
1
0
0
1
1
1
0
1
0
1
1
0
1
1
1
1
1
0
0
1
1
1
0
1
1
1
1
1
0
1
1
1
1
1
0
0
Since LHS and RHS columns match in every row for both identities, DeMorgan’s theorem for three variables is verified.
Note: The first table verifies complement of OR: (A+B+C)' = A'.B'.C', while the second table verifies complement of AND: (A.B.C)' = A'+B'+C'.
Question 11
11. Construct a truth table for the following functions
i)
F1(A,B,C) = A + BC’
ii)
F2(A,B,C) = AC + BC + AB’
iii)
F3(w,x,y,z) = y’z + wxy’ + wxz’ + wx’z
iv)
(x + y) . ( y + z) . (z + x)
Answer 11
Constructed truth tables (in binary count order):
Take input combinations as ABC: 000 to 111 and wxyz: 0000 to 1111, then evaluate each expression row-wise.
i) F1(A,B,C)=A+BC'
A
B
C
F1
0
0
0
0
0
0
1
0
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
ii) F2(A,B,C)=AC+BC+AB'
A
B
C
F2
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
0
1
1
1
1
iii) F3(w,x,y,z)=y'z+wxy'+wxz'+wx'z
w
x
y
z
F3
0
0
0
0
0
0
0
0
1
1
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
0
1
0
0
1
1
1
0
1
0
0
1
0
1
1
1
1
1
0
0
1
1
1
0
1
1
1
1
1
0
1
1
1
1
1
0
iv) (x+y).(y+z).(z+x)
x
y
z
F
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1
Question 12
12. Expand the following expressions using De Morgan’s theorem
i)
F1(A,B,C) = (AB)'(ABC)'(A’C)’
ii)
F2(A,B,C) = ((AB + B’C).(AC + A’C’))’
Answer 12
i) F1(A,B,C)
= (AB)'(ABC)'(A'C)'
= (A'+B')(A'+B'+C')(A+C')
Note that while solving the above, we used the following
(AB)'
= A'+B'
(ABC)'
= A'+B'+C'
(A'C)'
= (A')'+(C)'
= A+C'
So, (AB)'(ABC)'(A'C)' = (A'+B')(A'+B'+C')(A+C').
ii) F2(A,B,C)
= ((AB + B'C).(AC + A'C'))'
= (AB + B'C)' + (AC + A'C')'
= (AB)'(B'C)' + (AC)'(A'C')'
= (A'+B')(B+C') + (A'+C')(A+C)
Note that while solving the above, we used the following
(X + Y)'
= X'Y'
(XY)'
= X'+Y'
and double complement (A')' = A
Question 13
13. State the fundamental axioms of Boolean algebra.
Answer 13
Fundamental axioms/laws used in Boolean algebra are:
Identity
A+0=A
A.1=A
Null
A+1=1
A.0=0
Idempotent
A+A=A
A.A=A
Complement
A+A'=1
A.A'=0
Commutative
A+B=B+A
A.B=B.A
Associative
(A+B)+C=A+(B+C)
(A.B).C=A.(B.C)
Distributive
A.(B+C)=A.B+A.C
A+(B.C)=(A+B).(A+C)
These axioms form the base for simplifying and proving Boolean expressions.
Question 14
14. Explain principle of Duality of Boolean algebra.
Answer 14
Duality principle states that if a Boolean identity is true, then its dual is also true.
Meaning of dual: the corresponding Boolean expression obtained by interchanging OR and AND operations, and interchanging 0 and 1.
To form the dual, interchange + and ., and interchange 0 and 1; variables and their complements remain unchanged.
Example 1: dual of A+0=A is A.1=A.
Example 2: dual of A+A'=1 is A.A'=0.
Question 15
15. What do you mean by literal?
Answer 15
A literal is a Boolean variable written either in direct form or complemented form, such as A or A'.
So, each variable occurrence in an expression is a literal.
Example: in AB' + C, the literals are A, B', and C.
Boolean terms (like minterms and maxterms) are formed using one or more literals.
Question 16
16. List the theorems of Boolean Algebra.
Answer 16
Important theorems/laws include:
•
Commutative, associative, distributive (e.g., A+B=B+A, (A+B)+C=A+(B+C), A(B+C)=AB+AC).
•
Identity, null, idempotent, complement, involution (e.g., A+0=A, A+1=1, A+A=A, A+A'=1, (A')'=A).
•
Absorption, DeMorgan’s and consensus theorems (e.g., A+AB=A, (A+B)'=A'B', AB+A'C+BC=AB+A'C).
These theorems are used to simplify Boolean expressions and logic circuits.
Question 17
17. What is the difference between operator in mathematics and logic?
Answer 17
Aspect
Mathematics
Logic (Boolean)
Values used
Numbers (0, 1, 2, …)
Only binary values 0 and 1
Operators
+, -, *, /
OR +, AND ., NOT '
Meaning of result
Arithmetic quantity
Logical truth value
Example
1+1=2
1+1=1 (OR), 1.1=1 (AND)
So, mathematical operators compute numeric results, while Boolean operators evaluate logical conditions.
Question 18
18. Write Dual of
i)
A + 1 = 1
ii)
x + x’y = x + y
iii)
x + x’ = 1
vi)
x'(y + z) + x'(y’ + z’)
Answer 18
Using the principle of duality, interchange + and ., and interchange 1 and 0, while keeping variables and complements unchanged.
Applying this rule to each given expression:
S No.
Expression
Dual
i)
A + 1 = 1
A . 0 = 0
ii)
x + x'y = x + y
x . (x' + y) = x . y
iii)
x + x' = 1
x . x' = 0
vi)
x'(y + z) + x'(y' + z')
(x' + y.z) . (x' + y'.z')
Question 19
19. How is Boolean Algebra different from Arithmetic algebra?
Answer 19
Aspect
Boolean Algebra
Arithmetic Algebra
Values used
Only two values: 0 and 1
Many numeric values (0, 1, 2, …)
Purpose
Represents logical conditions and switching states
Represents numerical quantities and calculations
Main operations
OR +, AND ., NOT '
Addition, subtraction, multiplication, division
Carry/borrow
Not used in Boolean simplification
Carry and borrow are applicable
Example
1+1=1 (OR), 1.1=1
1+1=2
Hence, Boolean algebra is logic-oriented, while arithmetic algebra is quantity-oriented.
Question 20
20. Prove following
i)
C + A (C + B) + BC = C + AB
ii)
(A + C)A + AC + C = A + C
iii)
BC + A(B+C) = AB + BC + AC
iv)
B + A (B + C) + BC = B + AC
v)
x+ (x. y) = x
vi)
(a’ + b’) . (a’ + b) . (a + b’) = a’ . b’
Answer 20
i) C + A(C + B) + BC
= C + (AC + AB) + BC
= C + AC + AB + BC
= C + AB + BC (since C+AC=C)
= C + AB (since C+BC=C)
ii) (A + C)A + AC + C
= (AA + AC) + AC + C
= A + AC + AC + C
= A + AC + C
= A + C (since A+AC=A)
iii) BC + A(B + C)
= BC + AB + AC
= AB + BC + AC
= AB + BC + AC
iv) B + A(B + C) + BC
= B + (AB + AC) + BC
= B + AB + AC + BC
= B + AC + BC (since B+AB=B)
= B + AC (since B+BC=B)
v) x + (x.y)
= x + xy
= x(1 + y)
= x.1
= x
vi) (a' + b').(a' + b).(a + b')
= a'.(a + b') (using (X+Y)(X+Z)=X+YZ)
= a'a + a'b'
= 0 + a'b'
= a'.b'
Hence all the given identities are proved.
Question 21
21. For the given truth table, write Boolean expression for function G1, G2, G3, G4, G5

X
Y
Z
G1
G2
G3
G4
G5
0
0
0
0
0
0
0
1
0
0
1
0
1
0
1
0
0
1
0
1
0
0
1
0
0
1
1
1
1
0
1
0
1
0
0
1
0
0
1
1
1
0
1
0
0
1
0
1
1
1
0
0
1
1
0
0
1
1
1
0
0
1
0
1
Answer 21
Method (SOP form): pick only those rows where the output is 1. For each such row, write its minterm and add all minterms with OR.
Note: truth-table rows are numbered here from 1 to 8, while minterm indices are m0 to m7.
Row-to-minterm example: XYZ=010 gives X'YZ' (0 means complemented variable, 1 means direct variable).
Note: In the 5th column, minterm is written only for rows where that function value is 1. Otherwise - is shown (including XYZ=000 by convention here).
G1’s Truth Table to SOP Mapping
X
Y
Z
G1
Row to Minterm value
0
0
0
0
–
0
0
1
0
–
0
1
0
1
X'YZ'
0
1
1
1
X'YZ
1
0
0
1
XY'Z'
1
0
1
0
–
1
1
0
0
–
1
1
1
0
–
G1=1 at rows 3,4,5 i.e. m(2,3,4). So, Boolean Expression for the function G1 = X'YZ' + X'YZ + XY'Z'.
G2’s Truth Table to SOP Mapping
X
Y
Z
G2
Row to Minterm value
0
0
0
0
–
0
0
1
1
X'Y'Z
0
1
0
0
–
0
1
1
1
X'YZ
1
0
0
0
–
1
0
1
0
–
1
1
0
1
XYZ'
1
1
1
0
–
G2=1 at rows 2,4,7 i.e. m(1,3,6). So, Boolean Expression for the function G2 = X'Y'Z + X'YZ + XYZ'.
G3’s Truth Table to SOP Mapping
X
Y
Z
G3
Row to Minterm value
0
0
0
0
–
0
0
1
0
–
0
1
0
0
–
0
1
1
0
–
1
0
0
0
–
1
0
1
1
XY'Z
1
1
0
1
XYZ'
1
1
1
1
XYZ
G3=1 at rows 6,7,8 i.e. m(5,6,7). So, Boolean Expression for the function G3 = XY'Z + XYZ' + XYZ.
G4’s Truth Table to SOP Mapping
X
Y
Z
G4
Row to Minterm value
0
0
0
0
–
0
0
1
1
X'Y'Z
0
1
0
1
X'YZ'
0
1
1
1
X'YZ
1
0
0
1
XY'Z'
1
0
1
0
–
1
1
0
0
–
1
1
1
0
–
G4=1 at rows 2,3,4,5 i.e. m(1,2,3,4). So, Boolean Expression for the function G4 = X'Y'Z + X'YZ' + X'YZ + XY'Z'.
G5’s Truth Table to SOP Mapping
X
Y
Z
G5
Row to Minterm value
0
0
0
1
–
0
0
1
0
–
0
1
0
0
–
0
1
1
0
–
1
0
0
1
XY'Z'
1
0
1
1
XY'Z
1
1
0
0
–
1
1
1
1
XYZ
G5=1 at rows 1,5,6,8 i.e. m(0,4,5,7). So, Boolean Expression for the function G5 = X'Y'Z' + XY'Z' + XY'Z + XYZ.
Question 22
22. Express the OR operator in the terms of AND and NOT operator.
Answer 22
Using DeMorgan’s theorem, OR can be written using only AND and NOT operations.
Start with (A'.B')'. By DeMorgan, (X.Y)' = X' + Y'.
So, (A'.B')' = (A')' + (B')' = A + B.
Hence, A + B = (A'.B')'.
Question 23
23. List the reasons used for algebraic proof of Associative theorm.
Answer 23
The algebraic proof of associative theorem is justified using standard Boolean laws at each step, such as:
•
Commutative law
•
Distributive law
•
Identity law
•
Idempotent law
•
Complement law
•
Absorption law
These laws provide the reason for every transformation in the algebraic proof.