Exercise 3.1 a.)
A B
3174 0522 b.) 4165 1654
3.1.1 What is the sum of A and B if they represent unsigned 12-bit octal numbers? The result should
be written in octal. Show your work. a.) First is to convert the octal numbers base 8 into binary numbers.
A B
3
011
1
001
7
111
4
100
0
000
5
101
2
010
2
010
So in Binary A: 011 001 111 100 two
and B: is 000 101 010 010 two
A 011 001 111 100
+B
=
000
011
101
111
010
001
010
110
The binary sum of A and B is 011 111 001 110 two
, in octal it is 3716. b.) We’ll do the same thing as in a)
A B
4
100
1
001
6
110
5
101
1
001
6
110
5
101
So the binary representation of A: 100 001 110 101 two
and B: 001 110 101 100 two
.
A 100 001 110 101
+B
=
001
110
110
000
101
100
100
001
4
100
The binary sum of A and B is 110 000 100 001 two
, in octal it is 6041.
To check this and every problem I just used the calculator on my computer and set it to programmer.
3.1.2 What is the sum of A and B if they represent signed 12-bit octal numbers stored in sign-
magnitude format? The result should be written in octal. Show your work.
“In computer science , the sign bit is a bit in a signed number representation that indicates the sign of a number. Only signed numeric data types have a sign bit, and its place is usually the leftmost, where the most significant bit in unsigned numbers resides. Floating point numbers in IEEE format are always signed, with the sign bit in the leftmost position. Typically if the sign bit is 1 then the number is negative (in the case of two’s complement integers) or non-positive (for ones’ complement integers, sign-and-magnitude integers, and floating point numbers), while 0 indicates a non-negative number” ( http://en.wikipedia.org/wiki/Sign_bit ).
We are going to convert the octal numbers into binary as in 3.1.1 a.)
A B
3
011
1
001
7
111
4
100
0
000
5
101
2
010
2
010
Since the most significant bit is 0 for both A and B, we get the same value for signed numbers. The binary representation of A: 011 001 111 100 two
and B: 000 101 010 010 two
.
A 011 001 111 100

+B
=
000
011
101
111
010
001
010
110
The binary sum of A and B is 011 111 001 110 two
, in octal it is 3716. b.) We’ll start off the same way as above by converting the octal into binary.
A B
4
100
1
001
6
110
5
101
1
001
6
110
5
101
4
100
In A the most significant bit is 1, so it is negative. Leave the most significant bit which is the 1, which leaves you with:
00 001 110 101
Now you want to take the 2′s complement, simply switch every 0 to 1 and every 1 to 0, and add one to the result. You will get:
A’2
11 110 001
Take that and add 1 to it:
11 110 001
11 110 001
010
010
+1
011
B is the same as before, since the number is positive.
B
1
001
6
110
5
101
4
100
The binary representation of A is 11 110 001 011 two
, and B is 001 110 101 100 two
. The sum of A and B is:
A’2 11 110 001 011
+B 001 110 101 100
= 01 100 110 111
01 100 110 111 two
= 1467 octal.
3.1.4 What is A – B if they represent unsigned 12-bit octal numbers? The result should be written in
octal. Show your work.
A B a.) b.)
7040
4365
0444
3412 a.) The first step is the same as 3.1.1 and 3.1.2, convert to binary:
A B
7
111
0
000
4
100
0
000
0
000
4
100
A – B:
A
-B
=
111
000
110
000
100
011
100
100
111
110 011 111 100 two
, which is 6374 octal. b.) Again convert the octal into binary:
000
100
100
4
100
4
100

A
4
100
3
011
6
110
5
101
B
3
011
4
100
1
001
2
010
A – B:
A
-B
=
110
011
000
011
100
111
110
001
101
101
010
011
= 000 111 101 011_two, which is 0753 octal.
3.1.5 What is A- B if they represent signed 12-bit octal numbers stored in sign-magnitude format? The
result should be written in octal. Show your work. a.) Since A is negative you are going to do steps similar to problem 3.1.2, drop the most significant bit giving you 11 000 100 000 two
, then take the 2′s complement (simply switch every 0 to 1 and every 1 to 0) and then add one to the result. The table below shows the steps:
A
7 0 4 0
111 000 100 000
Drop the most significant bit
11 000 100
Switch 0′s and 1′s
000
00 111 011
2′s complement
111
00 111 100 000
B is the same as before, since the number is positive.
B
0 4 4 4
000 100
Now do A – B:
A 00
-B
=
000
00
100
111
100
010
100
100
100
111
000
100
100
00 010 111 100 two
is 0274 octal. b.) Since A is negative you are going to do steps similar to problem 3.1.2, drop the most significant bit giving you 00 011 110 101 two
, then take the 2′s complement (simply switch every 0 to 1 and every 1 to 0) and then add one to the result. The table below shows the steps:
A
4
100
3
011
6
110
5
101
Drop the most significant bit
00 011 110
Switch 0′s and 1′s
11 100 001
2′s complement
101
010

11 100 001 011
B is the same as before, since the number is positive.
B
3 4 1 2
011
A – B
A
100 001 010
-B
=
11
011
00
100
100
000
00 000 000 001 two
is 0001 octal.
Exercise 3.2
A B
001
001
000
011
010
001 a. b.
1446
2460
672F
4935
3.2.1 What is the sum of A and B if they represent unsigned 16-bit hexadecimal numbers? The result should be written in hexadecimal. Show your work.
Since the left most bit is the sign bit in the binary number, then the following rule can be used with any binary number in order to figure out either it is a positive number, or a negative number: 1)If the most left bit is 0, and then the binary number is a positive number. 2) If the most left bit is 1, then the binary number is a negative number.
“Hexadecimal is a positional numeral system with a radix , or base, of 16. It uses sixteen distinct symbols, most often the symbols 0–9 to represent values zero to nine, and A, B, C, D, E, F (or alternatively a–f) to represent values ten to fifteen.” (http://en.wikipedia.org/wiki/Hexadecimal) a.) The first step is to convert from hexadecimal to binary.
A B
1
0001
A + B
A
4
0100
0001
4
0100
0100
6
0110
0100 0110
6
0110
7
0111
2
0010
F
1111
+B
=
0110
0111
0111
1011
0010
0111
1111
0101
Now you want to convert 0111 1011 0111 0101 back into hexadecimal.
0111 1011 0111 0101
7 B 7 5
A + B = 7B75 b.) Again convert the hexadecimal numbers into binary.
A B
2
0010
A + B
A
+B
4
0100
0010
0100
6
0110
0100
1001
0
0000
0110
0011
0000
0101
4
0100
9
1001
3
0011
5
0101

= 0110 1101 1001 0101
Now take 0110 1101 1001 0101 and convert it back into hexadecimal.
0110 1101 1001 0101
6 D 9 5
A + B = 6D95
3.2.2 What is the sum of A and B if they represent signed 16-bit hexadecimal numbers stored in sign-
magnitude format? The result should be written in hexadecimal. Show your work
A B a. 1446 672F b. 2460 4935 a.)The first step is to convert from hexadecimal to binary.
A B
1 4 4 6 6 7 2 F
0001 0100 0100 0110 0110 0111 0010 1111
Both A and B have zero at the most left bit, so they are positive numbers. Since both A and B are positive numbers, then the sum of A+B in singed 12-bits will be similar in unsigned 16 -bits.
A in binary
B
0001
0110
0100
0111
0100
0010
0110
1111
A+B (binary) 0111 1011 0111 0101
A+B = 7b75 b.)The first step is to convert from hexadecimal to binary.
A B
2
0010
4
0100
6
0110
0
0000
4
0100
9
1001
3
0011
5
0101
Both A and B have zero at the most left bit, so they are positive numbers. Since both A and B are positive numbers, then the sum of A+B in singed 12-bits will be similar in unsigned 16 -bits.
A in binary
B
0010
0100
0100
1001
0110
0011
0000
0101
0110 1101 1001 0101 A+B (binary)
A+B= 6d95

3.2.3 Convert A into a decimal number, assuming it is unsigned.
Repeat assuming it stored in sign-magnitude format. Show your work.
A a. b.
1446
2460
Unsigned Format: a.) The value of A which is 1446 (base 16) that can be represented in a binary form (base 2) by
A 1 4 4 6
0001 0100 0100 then converting the binary number to decimal number:
0110
(0* 2^0) + (1* 2^1)+(1* 2^2)+(0* 2^3)+(0* 2^4)+(0* 2^5)+(1* 2^6)+(0* 2^7)+(0* 2^8)+(0* 2^9)+(1*
2^10)+(0* 2^11) +(1* 2^12)+(0* 2^13)+(0* 2^14)+(0* 2^15) = 2+4+64+1024++4096 =
5190 (base 10). b.)The value of A is 2460 (base 16) which can be represented in a binary form (base 2) by
A 2 4 6 0
0010 0100 0110 0000
(0* 2^0) + (0* 2^1)+(0* 2^2)+(0* 2^3)+(0* 2^4)+(1* 2^5)+(1* 2^6)+(0* 2^7)+(0* 2^8)+(0* 2^9)+(1*
2^10)+(0* 2^11)+(0* 2^12)+(1* 2^13)+(0* 2^14)+(0* 2^15) = 32+64+1024+8192 =
9312 (base 10).
Sign-Magnitude Format: a.)The value of A which is 1446 (base 16) that can be represented in a binary form (base 2) by
A 1 4 4 6
0001 0100 0100 0110 then converting the binary number to decimal number by the following way:
(0* 2^0) + (1* 2^1)+(1* 2^2)+(0* 2^3)+(0* 2^4)+(0* 2^5)+(1* 2^6)+(0* 2^7)+(0* 2^8)+(0* 2^9)+(1*
2^10)+(0* 2^11) +(1* 2^12)+(0* 2^13)+(0* 2^14)+(0*(- 2^15)) = 2+4+64+1024++4096 =
5190 (base 10). b.)
The value of A is 2460 (base 16) which can be represented in a binary form (base 2) by
A 2 4 6 0
0010 0100 0110 0000
(0* 2^0) + (0* 2^1)+(0* 2^2)+(0* 2^3)+(0* 2^4)+(1* 2^5)+(1* 2^6)+(0* 2^7)+(0* 2^8)+(0* 2^9)+(1*
2^10)+(0* 2^11)+(0* 2^12)+(1* 2^13)+(0* 2^14)+(0* *(-2^15)) = 32+64+1024+8192 =
9312 (base 10).

3.2.4 What is A – B if they represent unsigned 16-bit hexadecimal numbers? The result should be
written in hexadecimal. Show your work.
A B a. b.
C352
5ED4
36AE
07A4 a.) The first step is to convert from hexadecimal to binary. Then subtract A to B
A C 3 5 2 B 3 6
1100
1100
0011
0011
0101
0101
0010
0010
A E
0011 0110 1010 1110
A-B =
0011
1000
0110
1100
1010
1010
1110
0100
A-B= 8CA4 b.) The first step is to convert from hexadecimal to binary. Then subtract A to B
A 5 E D 4 B 0 7
0101
0101
1110
1110
1101
1101
0100
0100
A 4
0000 0111 1010 0100
A-B =
0000
0101
0111
0111
1010
0011
0100
0000
A-B=5730
3.2.5 What is A – B if they represent signed 16-bit hexadecimal numbers stored in sign-magnitude
format? The result should be written in hexadecimal. Show your work.
A B a. b.
C352
5ED4
36AE
07A4 a.) The first step is to convert from hexadecimal to binary
A C 3 5 2 B 3 6 A E
1100 0011 0101 0010 0011 0110 1010 1110
Since A is negative you are going to take the 2′s complement (simply switch every 0 to 1 and every 1 to
0) and then add one to the result
0011 1100 1010 1101
0011 1100 1010
+1
1110
B is the same as before, since the number is positive.

A
-
B
0011
0011
1100
0110
1010
1010
1110
1110
A-B =
A-B=0600
0000 0110 0000 0000 b.) The first step is to convert from hexadecimal to binary
A 5 E D 4 B 0
0101 1110 1101 0100 0000
7
0111
A
1010
4
0100
Both A and B have zero at the most left bit, so they are positive numbers. Since both A and B are positive numbers, then the sum of A-B in singed 12-bits will be similar in unsigned 16 -bits.
A
-
B
A-B =
0101
0000
0101
1110
0111
0111
1101
1010
0011
0100
0100
0000
A-B= 5730
3.2.6 Convert A into a binary number. What makes base 16 (hexadecimal) an attractive numbering
system for representing values in computers?
A a. b.
C352
5ED4 a.)The value of A which is C352 (base 16) that can be represented in a binary form (base 2) by
A C 3 5 2
1100 0011 0101 then converting the binary number to decimal number:
0010
(0* 2^0) + (1* 2^1)+(0* 2^2)+(0* 2^3)+(1* 2^4)+(0* 2^5)+(1* 2^6)+(0* 2^7)+(1* 2^8)+(1* 2^9)+(0*
2^10)+(0* 2^11) +(0* 2^12)+(0* 2^13)+(1* 2^14)+(1* 2^15)
= 50002(base 10). b.)The value of A is 5ED4 (base 16) which can be represented in a binary form (base 2) by
A 5
0101
E
1110
D
1101
4
0100
(0* 2^0) + (0* 2^1)+(1* 2^2)+(0* 2^3)+(1* 2^4)+(0* 2^5)+(1* 2^6)+(1* 2^7)+(0* 2^8)+(1* 2^9)+(1*
2^10)+(1* 2^11)+(1* 2^12)+(0* 2^13)+(1* 2^14)+(0* 2^15)
= 4+16+64+128+512+1024+2048+4096+16384
= 24276(base 10).

Exercise 3.3
Overflow occurs when a result is too large to be represented accurately given a finite word size. Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow by many, but in this textbook, that is considered an overflow.) The following table shows pairs of decimal numbers.
A B a. 216 255 b. 185 122
A
3.3.1 Assume A and B are unsigned 8-bit decimal integers. Calculate
A – B. Is there overflow, underflow, or neither? a.) The first step is to convert from decimal to binary, then subtract
A 216 B 255
-B
=
11011000
11011000
1111111
1100111
1111111
A
-B
=
A
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow by many, but in this textbook, that is considered an overflow. " (given by the book)
There is underflow b.) The first step is to convert from decimal to binary, then subtract
185
10111001
B 122
1111010
10111001
1111010
00111111
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow " (given by the book)
There is underflow, but in this textbook, it is considered as an overflow. Because two negative numbers give a positive number.

3.3.2 Assume A and B are signed 8-bit decimal integers stored in sign magnitude format. Calculate A + B. Is there overflow, underflow, or neither?
A B a. 216 255 b. 185 122 a.) The first step is to convert from decimal to binary, then Add A to B
A 216
11011000
B 255
1111111
A
+B
=
11011000
1111111
101010111
A
-B
=
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow by many, but in this textbook, that is considered an overflow. " (given by the book)
There is underflow b.) The first step is to convert from decimal to binary, then subtract
A 185
10111001
B 122
01111010
10111001
01111010
100110011
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow " (given by the book)
There is overflow
3.3.3 Assume A and B are signed 8-bit decimal integers stored in sign magnitude
format. Calculate A – B. Is there overflow, underflow, or neither?
A B a. 216 b. 185
255
122

A
-B
= a.) The first step is to convert from decimal to binary, then Add A to B
A 216
11011000
B 255
01111111
11011000
01111111
101010111
A
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow by many, but in this textbook, that is considered an overflow. ”(given by the book)
There is neither overflow, nor underflow b.) The first step is to convert from decimal to binary, then subtract
A 185 B 122
-B
=
10111001
10111001
01111010
00111111
01111010
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow " (given by the book)
There is neither overflow, nor underflow
3.3.4 Assume A and B are signed 8-bit decimal integers stored in two’s complement format. Calculate A + B using saturating arithmetic. The result should be written in decimal. Show your work.
A B a. 15 139 b. 151 214 a.) The first step is to convert from decimal to binary, Then take the 2’s complement and Add 1,
A 15
00001111
B 139
10001011
2’s complement 11110000
+ 1 11110001
01110100
01110101

Then A + B
A
+B
11110001
01110101
= 01100110
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow" (given by the book)
There is neither overflow, nor underflow b.) The first step is to convert from decimal to binary, Then take the 2’s complement and Add 1,
A 151 B 214
10010111
2’s complement 01101000
+ 1 01101001
11010110
00101001
00101010
Then A + B
A
+B
01101001
00101010
= 10010010
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow "(given by the book)
There is overflow because the sum of two positive numbers gives a negative number.
3.3.5 Assume A and B are signed 8-bit decimal integers stored in two’s complement format. Calculate A – B using saturating arithmetic. The result should be written in decimal. Show your work. a.) The first step is to convert from decimal to binary, Then take the 2’s complement and Add 1,
A 15 B 139
A
-B
=
00001111
2’s complement 11110000
+ 1 11110001
Then A - B
11110001
01110101
01111100
10001011
01110100
01110101
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow" (given by the book)
There is overflow

b.) The first step is to convert from decimal to binary, Then take the 2’s complement and Add 1,
A 151
10010111
B 214
11010110
2’s complement 01101000
+ 1
Then A - B
01101001
00101001
00101010
A
-B
=
01101001
00101010
00111110
"Underflow occurs when a number is too small to be represented correctly—a negative result when doing unsigned arithmetic, for example. (The case when a positive result is generated by the addition of two negative integers is also referred to as underflow "(given by the book)
There is underflow