Comprehensive Solutions for Boolean Algebra Homework Assignment

This document presents comprehensive solutions to various Boolean Algebra problems. It covers simplifying Boolean expressions, truth table analysis, binary and hexadecimal conversions, and 2's complement representation. The solutions are detailed and step-by-step, demonstrating the application of key concepts in discrete mathematics and computer science. The assignment addresses topics such as hexadecimal to ASCII conversion, binary to decimal conversion, and bitwise operations. The document includes examples of logical equivalencies and number system conversions, providing a valuable resource for students studying theoretical computer science and related fields. This is a valuable resource for students seeking to improve their understanding of Boolean Algebra and related concepts.

Running head: BOOLEAN ALGEBRA
BOOLEAN ALGEBRA
Name of the Student
Name of the University
Author Note

1BOOLEAN ALGEBRA
Solution 1:
AB(A+B)(C+C)
= AB(A+B)C
= AB(A+A’B)C
=(ABA+ABA’)C
= (AB+0B)C
= ABC
Solution 2:
(p,*r)+q
Solution 3:
[(p^q’)v(rvq)]^s

2BOOLEAN ALGEBRA
Solution 4:
P Q R P’ (Q v R) P’ ^ (Q v R)
0 0 0 1 0 0
0 0 1 1 1 1
0 1 0 1 1 1
0 1 1 1 1 1
1 0 0 0 0 0
1 0 1 0 1 0
1 1 0 0 1 0
1 1 1 0 1 0
Solution 5:
Simplifying the first expression, we get,
(A+C) (AD + AD’) +AC +C
=(A+C)A(D+D’)+AC+C
=(A+C)A+AC+C
=(AA+AC)+AC+C
=(A+AC)+AC+C

3BOOLEAN ALGEBRA
Making the truth table of the normalized form
A C AC A+AC AC+C
(A+AC)+
(AC+C)
0 0 0 0 0 0
0 1 0 0 1 1
1 0 0 1 0 1
1 1 1 1 1 1
Making truth table of A+C
A C A+C
0 0 0
0 1 1
1 0 1
1 1 1
From the above two tables we can conclude that the given two expressions are equivalent
Solution 6:
Since the characters are in hexadecimal form, we group them in two to decode the in ascii
values.
44= D
69= i
73= s

4BOOLEAN ALGEBRA
63= c
72= r
65= e
74= t
65= e
20= [space]
4D= M
61= a
74= t
68= h
65= e
6D= m
61= a
74= t
69= i
63= c
7 = since this number is single, therefore discarded
On successful decoding, the message is: Discrete Mathematic

5BOOLEAN ALGEBRA
Solution 7:
(00010001)₂ = (0 × 2⁷) + (0 × 2 ) + (0 × 2⁵) + (1 × 2⁴) + (0 × 2³) + (0 × 2²) + (0 × 2¹) + (1 ×⁶
2 )⁰
= 0+0+0+16+0+0+0+1
= (17)₁₀
(010111100)₂ = (0 × 2⁸) + (1 × 2⁷) + (0 × 2 ) + (1 × 2⁵) + (1 × 2⁴) + (1 × 2³) + (1 × 2²) + (0 ×⁶
2¹) + (0 × 2 )⁰
= 0+128+0+32+16+8+4+0+0
= (188)₁₀
(1111010)₂ = (1 × 2 ) + (1 × 2⁵) + (1 × 2⁴) + (1 × 2³) + (0 × 2²) + (1 × 2¹) + (0 ×⁶
2 )⁰
= 64+32+16+8+0+2+0
= (122)₁₀
We Know that,
(1011100)₂ = (1 × 2 ) + (0 × 2⁵) + (1 × 2⁴) + (1 × 2³) + (1 × 2²) + (0 × 2¹) + (0 × 2 )⁶ ⁰
=64+0+16+8+4+0+0
= (92)₁₀
(1110101)₂ = (1 × 2 ) + (1 × 2⁵) + (1 × 2⁴) + (0 × 2³) + (1 × 2²) + (0 × 2¹) + (1 ×⁶
2 )⁰
= 64+32+16+0+4+0+1

6BOOLEAN ALGEBRA
= (117)₁₀
So,
92 + 117 = (209)10
1011100 + 1110101 = (209)10
Solution 8:
We know that, 12 = 0001100 in 7 bits
1’s complement = 1110011
2’s complement = 1110011+1
=1110100
We know that, -2 = 0000010 in 7 bits
1’s complement = 1111101
2’s complement = 1111101+1
=1111110
We know that, -8 = 0001000 in 7 bits
1’s complement = 1110111
2’s complement = 1110111+1
=1111000
We know that, 0 = 0000000 in 7 bits
1’s complement = 1111111

7BOOLEAN ALGEBRA
2’s complement = 1111111+1
=10000000
Since the requirement in the problem statement is for 7 bits, therefore, 2’s complement in 7
bit is 0000000
The highest possible integer in seven bit is 63
In the problem, it is mentioned that the highest bit will be 7 bits. If we do a 2s complement of
all 1s then the seventh bit will increase a bit by 1 since for binary addition, 1+1 = 10.
Therefore we will take (0111111) in order to receive highest 2s complement and the value of
(0111111) is 63.
For the lowest integer, the number will be zero.
Solution 9:
9A88
Converting the hexadecimal number to binary, we get,
(9A88) = (1001101010001000)₁₆ 2
Therefore, from the above conversion we can conclude that this is a 16 bits notation.
4AF6
Converting the hexadecimal number to binary, we get,
(4AF6) = (0100101011110110)₁₆ 2
Therefore, from the above conversion we can conclude that this is a 16 bits notation.
DA
Converting the hexadecimal number to binary, we get,
(DA) = (11011010)₁₆ 2
Therefore, from the above conversion we can conclude that this is a 8 bits notation.

8BOOLEAN ALGEBRA
Decimal sum of 9A88 and 4AF6 are as follows:
(9A88 ) = (9 × 16⁴) + (10 × 16³) + (8 × 16²) + (8 × 16¹) + (-16 × 16 ) = (39560)₁₆ ⁰ ₁₀
(4AF6 ) = (4 × 16⁴) + (10 × 16³) + (15 × 16²) + (6 × 16¹) + (-16 × 16 ) = (19190)₁₆ ⁰ ₁₀
39560 + 19190
= (58750)10
Hexadecimal sum of 9A88 and 4AF6 is (9A88 + 4AF6) = (E57E)16 = (58750)10