Mathematics for Computing
and Informatics 1A (MCI511S)
Lecture Notes
2021
By G. Tapedzesa
2021
TAPEDZESA G., NUST
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CHAPTER 1: NUMBER SYSTEMS
Learning Outcomes
Upon completion of this unit, you will be able to:
• Recognize different types of number systems as they relate to computers
• Describe the decimal, binary, hexadecimal and octal system
• Identify and define a base/radix, power, positional notation, and most and least
significant digits as they relate to decimal, binary, octal, and hexadecimal number
systems
• Convert a number in binary, octal or hexadecimal to a number in the decimal
system.
• Convert a number in the decimal system to a number in binary, octal and
hexadecimal.
• Perform arithmetic operations in binary, octal, and hexadecimal number systems
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NUMBER SYSTEMS: Introduction
• A number system defines how a number can be represented using distinct
symbols.
• A number can be represented differently in different systems represented
differently in different systems
• The number system that you are familiar with, that you use every day, is the
decimal number system, also commonly referred to as the base-10 positional
number system.
• When you perform computations such as
3 + 2 = 5, 𝑜𝑟 21 – 7 = 14
you are using the decimal number system.
• This decimal system is ingrained into your subconscious; it’s the natural way
that you think about numbers.
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• Computers can most readily use two symbols, and therefore a
base-2 system, or binary number system, is most appropriate.
• The base-2 system has exactly two symbols: 0 and 1.
• The base-10 symbols are termed digits. The base-2 symbols are
termed binary digits, or bits for short.
• All base-10 numbers are built as strings of digits (such as 6349).
• All binary numbers are built as strings of bits (such as 1101).
Just as we would say that the decimal number 12890 has five
digits, we would say that the binary number 11001 is a five-bit
number.
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The Decimal Number System
The decimal (base 10) system contains ten digits
denoted by the symbols
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Any positive integer in the decimal system can be
expressed, in expanded notation, as a sum of
powers of 10 multiplied by a digit. For instance,
8253 = 8 × 103 + 2 × 102 + 5 × 101 + 3 × 100
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The Decimal Number System
• Any fractional number, represented in the decimal system by
a string of decimal digits together with an embedded decimal
point may also be expressed in expanded notation by using
negative powers of 10. For example,
𝟖𝟓𝟑. 𝟓𝟐𝟔
= 𝟖 × 𝟏𝟎𝟐 + 𝟓 × 𝟏𝟎𝟏 + 𝟑 × 𝟏𝟎𝟎 + 𝟓 × 𝟏𝟎−𝟏 + 𝟐 × 𝟏𝟎−𝟐
+ 𝟔 × 𝟏𝟎−𝟑
• This decimal fraction is said to have three decimal places,
which is the number of digits to the right of the decimal point
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Face Value and Place Value
• The face value or inherent value of a symbol is the
value of that symbol standing alone
• For example, the face value of 6 in numbers 256, 165, 698 is 6 even if it is used
in different number positions
• The place value or positional value of a numeric
symbol is directly related to the base of a system
• In the case of the decimal system, each position has a
value of 10 times greater than the position to its right
• For example: in the number 423, the symbol 3 represents the units (100), the symbol
2 represents the tens position (101), and the symbol 4 represents the hundreds
position (102).
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Binary System
• All data in a computer is represented in binary. The
pictures stored on your hard drive—it’s all bits. The
YouTube video of the cat falling off the chair that you
saw this morning—bits. Your Facebook page—bits. The
tweet you sent—bits. Everything is bits.
• To understand how computers work, you have to speak
the language. And the language of computers is the
binary number system.
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The Binary Number System
• The binary system is a positional numeration system to
the base 2.
• Those binary numbers that have no fractional part are
called binary integers.
• A string of eight bits (such as 11000110) is termed a
byte.
• The table in the next slide gives some idea of the
correspondence between binary numbers and the
more familiar decimal numbers.
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You should “memorize” the binary representations of
the decimal digits 0 through 15 shown below
• Here we used the 4-bit representation.
• When you are given the size of the storage location, include the leading zeros
to show all bits in the storage location.
• For example, if told to represent decimal 5 as an 8-bit binary number, your
answer should be 00000101
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•The idea of describing numbers using a
positional system, as we have illustrated for
base-10 and base-2, can be extended to any
base.
•To avoid confusion when the base in use if not
clear from the context, or when using multiple
bases in a single expression, we append a
subscript to the number to indicate the base,
and write:
510 = 1012
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You should, henceforth, be able to readily shift between
the binary and decimal number representations.
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Binary-to-Decimal Conversion
• To convert a binary number to a decimal number, we simply write
the binary number as a sum of powers of 2. For example,
10112 = 1 × 23 + 0 × 22 + 1 × 21 + 1 × 20 = 8 + 2 + 1 = 1110
1101012 = 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20
= 32 + 16 + 4 + 1 = 5310
Exercise: Convert each of the following binary numbers as decimal
numbers
1101102 ,
10000012 ,
101.11012
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Decimal-to-Binary Conversion
• Given a binary number, you can now convert it to the
equivalent decimal number.
• The method of converting a decimal number to a binary
number, which is much more readily adaptable to
programming on a computer, entails repeatedly dividing the
decimal number by 2, keeping track of the remainder at each
step.
• To convert the decimal number, say x, to binary we follow the
steps on the next slide:
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Step 1. Divide 𝑥 by 2 to obtain a quotient and remainder. The remainder will be
0 or 1.
Step 2. If the quotient is zero, proceed to Step 3. Otherwise, go back to Step 1,
assigning 𝑥 to be the value of the most-recent quotient from Step 1.
Step 3. The sequence of remainders from bottom up forms the binary
representation of the number.
Example: Convert the decimal number 53 to its binary equivalent.
53 ÷ 2 = 26 𝑟 1
26 ÷ 2 = 13 𝑟 0
13 ÷ 2 = 6 𝑟 1
6÷2=3𝑟0
3÷2=1𝑟1
1÷2=0𝑟1
Ans = 110101
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• If a decimal number has a fractional part, you convert
the integral part and the fractional part separately as
illustrated with the decimal number 109.78125 below.
• First, convert the integral part 109 to its binary
equivalent: 109 = 11011012 (verify)
• Then convert the fractional part 0.78125 to its binary
equivalent by successively multiplying the fractional
part by 2, noting the integral part of the product, as
follows:
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Decimal-to-Binary Conversion
0.78125 × 2 = 𝟏. 56250
0.5625 × 2 = 𝟏. 1250
0.125 × 2 = 𝟎. 250
0.25 × 2 = 𝟎. 50
0.5 × 2 = 𝟏. 0
You stop when you have a zero fractional part. The sequence of
integral part digits (in red) from top down yields the required
binary representation. That is, 0.78125 = 0.110012
The binary equivalents of the decimal number 109.78125 is the
sum of the binary equivalents of its integral and fractional parts:
109.78125 = 1101101.110012
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Binary Arithmetic
The execution of numerical calculations is essentially the
same in all positional numeration systems.
Binary Addition
The algorithm for the addition of decimal numbers holds
also for the addition of binary numbers; the only addition
facts needed for binary addition are:
0+0=0
0+1=1
1+0=1
1 + 1 = 0, with a carry of 1
1 + 1 + 1 = 1, with a carry of 1
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Binary Addition
Go through examples 1.5 and 1.6 in the prescribed
textbook.
Exercise: In each of the following cases, perform the addition of the
binary numbers:
1.
2.
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Binary Multiplication
The algorithm for decimal multiplication also holds for
binary multiplication. Please note that when performing
the operation of multiplication, it is extremely important
to line up the numbers in the correct columns.
For practice, attempt examples 1.8 and 1.9 in the
prescribed textbook
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Binary Subtraction
The same “borrowing” procedure followed
in the decimal system is applied in the
binary system. The only subtraction tion
facts needed for binary subtraction are:
0−0=0
1−0=1
1−1=0
10 − 1 = 1
0 − 1 = 1, with a borrow of 1 from the next column
Examples 1.12 – 1.14
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Binary Division
• The algorithm for binary division is simply
repeated subtraction of the divisor.
• As in decimal division of integers, a
remainder is possible when one binary
integer is divided by another.
• Also, the division of binary fractions is
handled the same way as the division of
decimal fractions.
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Complements
• All computers represent integers in a manner that
allows both positive and negative numbers
• Remember: (including the negative sign) must be
represented as bits; that is, ones and zeros!
• Many computers store negative numbers in the form
of their arithmetic complements
• Complements are used to reduce subtraction to
addition
• First we discuss these complements in the familiar
decimal system, known as the nines complement
and the tens complement
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Decimal Complements
The nines complement of a decimal number, say A, is
obtained by subtracting each digit of A from 9; and the
tens complement of A is its nines complement plus one.
Example
2021
Decimal number
Nine complement
Tens complement
4308
5691
5692
123 123
876 876
876 877
9672
0327
0328
751 620
248 379
248 380
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• To illustrate the use of complements in
subtraction, let A and B be two decimal integers
with the same number of digits, say 4 digits, and
suppose that A is less than B
• We can simplify the difference 𝐵 − 𝐴 as
𝐵 − 𝐴 = 𝐵 − 𝐴 + 10 000 − 10 000
= 𝐵 − 𝐴 + (9999 + 1 − 10 000)
= 𝐵 + (9999 − 𝐴 + 1) − 10 000
= 𝐵 + 9999 − 𝐴 + 1 − 10 000
Thus, we can calculate B − 𝐴 either by adding the
tens complement of A to B and then subtracting
10 000, or by adding the nines complement of A
and then adding 1 and subtracting 10 000.
Attempt example 1.18
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• Suppose now that A is greater than B. For
instance, let us consider A=5872 and B=2148.
Then the difference A-B is negative:
• We can simplify the difference 𝐵 − 𝐴 as
𝐵 − 𝐴 = 2148 + (9999 − 5872 + 1) − 10 000
= 2148 + 4128 − 10 000
= −3724
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Binary Complements
• The terminology and principle of complements in the
decimal system can be easily translated into the binary
system
• Specifically, if A is a binary number, the ones complement
of A is obtained by subtracting each digit of A from 1, and
the twos complement of A is its ones complement plus 1.
Example
2021
Binary number
Ones complement
Twos complement
11011100
00100011
00100100
00100111
11011000
11011001
00011000
11100111
11101000
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Binary Complements
• Observe that taking the ones complement simply
inverts each digit, that is 0 is replaced by 1 and vice
versa.
• Specifically, if A is a binary number, the ones
complement of A is obtained by subtracting each
digit of A from 1, and the twos complement of A is
its ones complement plus 1.
Example
Binary number
Ones complement
Twos complement
11011100
00100011
00100100
00100111
11011000
11011001
Attempt examples
1.20 and
1.21 in the11101000
prescribed
00011000
11100111
textbook
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So, to find the representation of a negative
decimal number, do the following:
• First find the binary representation of the
number without the negative sign
• Then take the two’s complement. The result is
the representation of the negative number.
Every computer today uses two’s
complement notation for the storage of
integers.
Note: a binary number, added to its two’s
complement, results in zero, as expected.
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Tutorial Exercises
1. Find the face value and the place value of the digit 4 in
the following decimal numbers.
(a) 7425
(b) 146 723 (c) 305.54 (d) 0.012345
2. Rewrite in expanded notation.
a) 2468
(b) 54.321
3. Give the face value and the place value of each of the
underlined bit:
(a) 101102 (b) 10110012 (c) 101.110102 (d) 110.00102
4. Rewrite in expanded notation.
a) 1101102
(b) 11.011012
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5. Convert each binary number to its decimal form:
(a) 11100112
(b) 110.10112
6. Convert the decimal number 437.206 251 to its
binary equivalent.
7. Evaluate the following binary sums:
a) 11011 + 1010
(b) 110.1101 + 1011.011
8. Evaluate the following binary products:
a) 110111 × 101
(b) 110.001 × 1.11
9. Evaluate the following binary differences:
a) 1110001 − 111011 (b) 1101.0011 −
110.11011
10. Evaluate 111001 ÷ 1001 to two binary places.
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11. Determine the nines and tens complements of the following decimal
numbers:
(a) 3268
(b) 479 200 (c) 99 132 756
12. Find the following differences using tens complements:
(a) 53 726 − 14 503
(b) 2658 − 4321
13. Determine the ones and twos complements of the binary numbers:
a) 11011 + 1010
(b) 110.1101 + 1011.011
14. Evaluate the following binary differences using twos complements:
a) 11011 − 1010
b) 111000 − 110100
c) 101010011 − 111011001
d) 110.1101 + 1011.011
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• Computer engineers must oftentimes look at the
contents of a specific item in computer memory.
• You might, for instance, have to look at a variable
that is stored at an address:
00000000000100101111111101111100
• Surely, such long binary strings would be cumbersome
to transcribe or to read off to a colleague.
• Even if you have come to love the binary number
system, you would still likely agree that these long
strings are too much of a good thing.
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• Fortunately, large binary numbers can be made much more
compact, and hence easier to work with, if represented in other
number systems, the octal (base 8) and the hexadecimal (base 16)
systems. 8 = 23 , 16 = 24
• Since 8 and 16 are powers of 2, there is almost instant
interconversion between the octal and hexadecimal systems and
the binary system.
• Although the decimal system is more compact than the binary
system, the octal and hexadecimal systems are in fact comparable
in compactness with the decimal system
• Converting between binary and hexadecimal is exceedingly easy—
much easier than converting between binary and decimal.
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Converting from any base to base 10
• We can convert a base 𝑏 number, 𝑁𝑏 , into its decimal
representation by writing 𝑁𝑏 in expanded notation and
calculating by decimal arithmetic
• This conversion is accomplished by the following algorithm,
which distinguishes between the integral part and the
fractional part of the number:
Integral part: Multiply the leftmost digit by the base 𝑏 and add
the next digit to the right. Multiply the sum by the base 𝑏 and
add the next digit. Repeat the process until the rightmost digit
is added. The final answer is the required decimal integral part.
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Converting from any base to base 10
Examples
1. Convert the octal number 1753 to its base 10 equivalent
17538 = 1 × 83 + 7 × 82 + 5 × 81 + 3 × 80 = 100310
2. Convert the quintary (base-5) number 142 to its base- 10
equivalent
1425 = 1 × 52 + 4 × 5 + 2 = 47
3. Convert the base-4 number 132.12 to its decimal equivalent
132.124 = 1 × 42 + 3 × 4 + 2 + 1 × 4−1 + 2 × 4−2
= 20.555555556
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Converting from any base 𝒃 to base 10
Fractional part: Multiply the rightmost digit
by the base 1Τ𝑏 and add the next digit to the
right. Multiply the sum by the base 1Τ𝑏 and
add the next digit. Repeat the process until
the leftmost digit is added and the sum is
1
multiplied by Τ𝑏. The final answer is the
required decimal fractional part.
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Examples:
a) Convert the quintary (base-5) number 3412 to its decimal
equivalent.
34125 = 3 × 53 + 4 × 52 + 1 × 51 + 2 × 50 = 48210
b) Convert the number 56237 to its base-10 equivalent
56237 = 5 × 73 + 6 × 72 + 2 × 71 + 3 × 70 = 202610
c) Convert the number 231.20123 to its decimal equivalent
231.20123
= 2 × 32 + 3 × 31 + 1 × 30 + 2 × 3−1 + 1 × 3−3 + 2 × 3−4
= 28.72 … …10
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Example: Consider the quintary number 2401.23145 .
2401.23145 =
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Decimal-to-base-b Conversion
We can convert a decimal number to its base-b representation using
the following algorithm, which also distinguishes between the integral
part of the number and its fractional part.
Integral part: Divide the decimal number and each succeeding
quotient by b until a zero quotient is obtained. The sequence of
remainders, from bottom up, yields the required base-b
representation.
Fractional part: Multiply the decimal number and the fractional part of
each succeeding product by b until a zero fractional part is obtained.
Then the finite (or infinite repeating) sequence of integral parts of the
products gives the base-b digit representation, from top down, yields
the required base-b representation.
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Decimal-to-base-b Conversion
Example
1. Convert the decimal number 1589 to its base-7 equivalent
1589 ÷ 7 = 227 𝑟 0
227 ÷ 7 = 32 𝑟 3
32 ÷ 7 = 4 𝑟 4
4÷7=0 𝑟 4
Therefore 158910 = 44307
2. Convert the decimal number 2546 to its base-5 equivalent
2546 ÷ 5 = 509 𝑟 1
509 ÷ 5 = 101 𝑟 4
101 ÷ 5 = 20 𝑟 1
20 ÷ 5 = 4 𝑟 0
4÷5=0 𝑟4
Therefore 25465 = 40141
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Decimal-to-base-b Conversion
Example
1. Convert the decimal number 15.89 to its base-7 equivalent
Integer part = 15:
15 ÷ 7 = 2 𝑟 1
2÷7=0
𝑟 2
15 = 217
fractional part = 0.89:
0.89 × 7 = 6.23
0.23 × 7 = 1.61
0.61 × 7 = 4.27
0.27 × 7 = 1.89
0.89 × 7 = 6.23
0.89 = 0. 6ሶ 1ሶሶ 4ሶ 1ሶ
ሶ ሶ 4ሶ 1ሶ
Ultimately, 15.89 =21. 61
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Example 1: Convert the decimal number 684 to its quintal (base 5)
representation.
Solution: We divide 684, and each subsequent quotient, by 5, noting all
remainders.
Divisions
Quotients
Remainders
684 ÷ 5
136
4
136 ÷ 5
27
1
27 ÷ 5
5
2
5÷5
1
0
1÷5
0
1
The sequence of remainders from bottom up gives the quintal form for
68410 = 102145 .
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Example 2: Convert the decimal number 0.4704 to its
quintal (base 5) representation.
Solution: We multiply 0.4704, and each subsequent
fractional part, by 5, noting the integral part of each
product.
Multiplications
Products
Integral parts
0.4704 × 5
2.352
2
0.352 × 5
1.76
1
0.76 × 5
3.8
3
0.8 × 5
4.0
4
The sequence of integral parts from top down yields the
required quintal form for 0.470410 = 0.21345 .
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Octal and Hexadecimal System
In the base-8 number system we use the symbols
0
1
2
3
4
5
6
7
Binary to Octal conversions
Exercise: Convert the base-6 number 1542 to its base-8 equivalent
Holiday exercise
1. Convert the binary number 001 011 010 to its octal equivalent (8 = 23 )
1011010 = 1328
1. Convert the binary number 011 011 110 to its octal equivalent (8 = 23 )
11011110 = 3368
In the hexadecimal (base-16) number system we use the symbols
0
1
C
2020
2
D
3
E
4
F
5
6
7
8
9
A
B
𝐴𝐹234 + 100𝐵𝐶 =
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Octal and Hexadecimal System
In the base-8 number system we use the symbols
0
1
2
3
4
5
6
7
Octal to Binary conversions
1. Convert the octal number 567 to its binary equivalent
5678 = 1011101112
2. Convert the binary number 01110001.1100 to its hexadecimal equivalent
1110001.11 = 71. 𝐶16
3. Convert the binary number 10011101.1111 to its hexadecimal equivalent
10011101.1111 = 9𝐷. 𝐹16
In the hexadecimal (base-16) number system we use the symbols
0
1
C
2
D
3
E
4
F
5
6
7
8
9
A
B
𝑨𝑭𝟐𝟑𝟒 + 𝟏𝟎𝟎𝑩𝑪 =
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Octal and Hexadecimal System
The base-8 and base-16 systems are special: 8 = 23 , 16 = 24
Converting from binary to octal
1. Convert the binary number 011 010 001 to its octal equivalent
110100012 = 3218
2. Convert the binary number 011 011.110 to its octal equivalent
11011.112 = 33.68
Converting from octal to binary
1. Convert the octal number 751 to its binary equivalent
7518 = 1111010012
2. Convert the octal number 6540.102 to its binary equivalent
6540.1028 = 110101100000.0010000102
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Octal and Hexadecimal System
The base-8 and base-16 systems are special: 8 = 23 , 16 = 24
Converting from binary to hexadecimal
1. Convert the binary number 0010 1111 1011 to its hexadecimal equivalent
10111110112 = 2𝐹𝐵16
2. Convert the binary number 0010 1101 0101.1011 1000 to its hexadecimal
equivalent
1011010101.101112 = 2𝐷5. 𝐵816
Converting from hexadecimal to binary
1. Convert the hexadecimal number BED0.5B to its binary equivalent
𝐵𝐸𝐷0.5𝐵 = 1011111011010000.010110112
2
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General Arithmetic
Octal: 0 1 2 3 4 5 6 7 10 11 12 13 14 15 16 17 20 21 22…
Decimal: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18…
1.Evaluate the octal addition: 7512+4532
71 512
+4 532
14 244
Therefore, 7512+4532=14244 in base-8
2. Evaluate the octal addition: 157.230 + 254.011
11 51 7.230
+254.011
433.241
Therefore, 157.230 + 254.011 = 433.241 in base-8
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General Arithmetic
Base-7: 0 1 2 3 4 5 6 10 11 12 13 14 15 16 20 21 22 23 24…
Base-10: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18…
1.Evaluate the base-7 subtraction: 512-453=26 in base-7
1
54 10 2
−4 5 3
26
Base-3: 0 1 2 10 11 12 20 21 22 100 101 102 110 111 112 120 121 122 200…
Base-10: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 …
2. Evaluate the base-3 subtraction: 121.2 - 102.11=12.02 in base-3
121 1.21 0
−102.11
12.02
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General Arithmetic
Hex: 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F…
Dec: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31…
1. Evaluate the base-16 addition F3A9+2D0E
𝐹3𝐴9
+2𝐷0𝐸
120𝐵7
Subtraction
1.Evaluate the hexadecimal subtraction: F3A9.B2-2D0E.FF
𝐸1 391 8. 𝐴1 2
−2𝐷0𝐸. 𝐹𝐹
𝐶69𝐴. 𝐵3
2020
TAPEDZESA G., NUST
51
General Arithmetic
1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
1. Evaluate the base-16 addition F3A9+2D0E
𝐹3𝐴9
+2𝐷0𝐸
120𝐵7
Subtraction
1.Evaluate the hexadecimal subtraction: F3A9.B2-2D0E.FF
𝐸1 391 8. 𝐴1 2
−2𝐷0𝐸. 𝐹𝐹
𝐶69𝐴. 𝐵3
2020
TAPEDZESA G., NUST
52