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Appendix A

Number systems

A.1. Overview

The efficient use of a microprocessor or microcontroller requires a working knowledge of binary, decimal, and hexadecimal numbering systems. This section provides a background for those who are unfamiliar with these numbering systems and who do not know how to convert from one number system to another one.

Number systems are classified according to their bases. The numbering system used in everyday life is base 10 or the decimal number system. The most commonly used numbering system in microprocessor and microcontroller applications is base 16, or hexadecimal. In addition, base 2 (binary) or base 8 (or octal) number systems are also used.

A.2. Decimal number system

As you all know the numbers in this system are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. We can use the subscript 10 to indicate that a number is in decimal format. For example, we can show decimal number 235 as 235₁₀.

In general, a decimal number is represented as follows:

$a_{n} \times 10^{n} + a_{n - 1} \times 10^{n - 1} + a_{n - 2} \times 10^{n - 2} + \dots + a_{0} \times 100$

For example, decimal number 825₁₀ can be shown as follows:

$825_{10} = 8 \times 10^{2} + 2 \times 10^{1} + 5 \times 10^{0}$

Similarly, decimal number 26₁₀ can be shown as:

$26^{10} = 2 \times 10^{1} + 6 \times 10^{0}$

or,

$3359^{10} = 3 \times 10^{3} + 3 \times 10^{2} + 5 \times 10^{1} + 9 \times 10^{0}$

A.3. Binary number system

In binary number system, there are two numbers: 0 and 1. We can use the subscript 2 to indicate that a number is in binary format. For example, we can show binary number 1011 as 1011₂.

In general, a decimal number is represented as follows:

$a_{n} \times 2^{n} + a_{n - 1} \times 2^{n - 1} + a_{n - 2} \times 2^{n - 2} + \dots + a_{0} \times 2^{0}$

For example, binary number 1110₂ can be given as follows:

$1110_{2} = 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 0 \times 2^{0}$

Similarly, binary number 10001110₂ can be shown as:

$10001110_{2} = 1 \times 2^{7} + 0 \times 2^{6} + 0 \times 2^{5} + 0 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 0 \times 2^{0}$

A.4. Octal number system

In octal number system, the valid numbers are 0, 1, 2, 3, 4, 5, 6, 7. We can use the subscript 8 to indicate that a number is in octal format. For example, we can show octal number 23 as 23₈.

In general, an octal number is represented as:

$a_{n} \times 8^{n} + a_{n - 1} \times 8^{n - 1} + a_{n - 2} \times 8^{n - 2} + \dots + a_{0} \times 8^{0}$

For example, octal number 237₈ can be given as:

$237_{8} = 2 \times 8^{2} + 3 \times 8^{1} + 7 \times 8^{0}$

Similarly, octal number 1777₈ can be given as:

$1777_{8} = 1 \times 8^{3} + 7 \times 8^{2} + 7 \times 8^{1} + 7 \times 8^{0}$

A.5. Hexadecimal number system

In hexadecimal number system, the valid numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. We can use the subscript 16 of H to indicate that a number is in hexadecimal format. For example, we can show hexadecimal number 1F as 1F₁₆ or as 1F_H.

In general, a hexadecimal number is represented as:

$a_{n} \times 16^{n} + a_{n - 1} \times 16^{n - 1} + a_{n - 2} \times 16^{n - 2} + \dots + a_{0} \times 16^{0}$

For example, hexadecimal number 2AC₁₆ can be given as:

$2 A C_{16} = 2 \times 16^{2} + 10 \times 16^{1} + 12 \times 16^{0}$

Similarly, hexadecimal number 3FFE₁₆ can be represented as:

$3 FF E_{16} = 3 \times 16^{3} + 15 \times 16^{2} + 15 \times 16^{1} + 14 \times 16^{0}$

A.6. Converting binary numbers ınto decimal

To convert a binary number into decimal, write the number as the sum of the powers of 2.

Example A.1

Convert binary number 1011₂ into decimal.

Solution A.1

Write the number as the sum of the powers of 2:

$\begin{matrix} 1011_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0} \\ = 8 + 0 + 2 + 1 \\ = 11 \end{matrix}$

or, 1011₂ = 11₁₀

Example A.2

Convert binary number 11001110₂ into decimal.

Solution A.2

Write the number as the sum of the powers of 2:

$\begin{matrix} 11001110_{2} = 1 \times 2^{7} + 1 \times 2^{6} + 0 \times 2^{5} + 0 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 0 \times 2^{0} \\ = 128 + 64 + 0 + 0 + 8 + 4 + 2 + 0 \\ = 206 \end{matrix}$

or, 11001110₂ = 206₁₀

Table A.1 presents the decimal equivalent of numbers from 0 to 31.

Table A.1

Decimal equivalent of binary numbers.
Binary	Decimal	Binary	Decimal
00000000	0	00010000	16
00000001	1	00010001	17
00000010	2	00010010	18
00000011	3	00010011	19
00000100	4	00010100	20
00000101	5	00010101	21
00000110	6	00010110	22
00000111	7	00010111	23
00001000	8	00011000	24
00001001	9	00011001	25
00001010	10	00011010	26
00001011	11	00011011	27
00001100	12	00011100	28
00001101	13	00011101	29
00001110	14	00011110	30
00001111	15	00011111	31

A.7. Converting decimal numbers ınto binary

To convert a decimal number into binary, divide the number repeatedly by 2 and take the remainders. The first remainder is the least significant digit (LSD), and the last remainder is the most significant digit (MSD).

Example A.3

Convert decimal number 28₁₀ into binary.

Solution A.3

Divide the number into 2 repeatedly and take the remainders:

28/2	→	14	Remainder 0	(LSD)
14/2	→	7	Remainder 0
7/2	→	3	Remainder 1
3/2	→	1	Remainder 1
½	→	0	Remainder 1	(MSD)

The required binary number is 11100₂.

Example A.4

Convert decimal number 65₁₀ into binary.

Solution A.4

Divide the number into 2 repeatedly and take the remainders:

65/2	→	32	Remainder 1	(LSD)
32/2	→	16	Remainder 0
16/2	→	8	Remainder 0
8/2	→	4	Remainder 0
4/2	→	2	Remainder 0
2/2	→	1	Remainder 0
½	→	0	Remainder 1	(MSD)

The required binary number is 1000001₂.

Example A.5

Convert decimal number 122₁₀ into binary.

Solution A.5

Divide the number into 2 repeatedly and take the remainders:

122/2	→	61	Remainder 0	(LSD)
61/2	→	30	Remainder 1
30/2	→	15	Remainder 0
15/2	→	7	Remainder 1
7/2	→	3	Remainder 1
3/2	→	1	Remainder 1
½	→	0	Remainder 1	(MSD)

The required binary number is 1111010₂.

A.8. Converting binary numbers ınto hexadecimal

To convert a binary number into hexadecimal, arrange the number in groups of 4 and find the hexadecimal equivalent of each group. If the number cannot be divided exactly into groups of 4, insert zeroes to the left-hand side of the number.

Example A.6

Convert binary number 10011111₂ into hexadecimal.

Solution A.6

First, divide the number into groups of 4 and then find the hexadecimal equivalent of each group:

$\begin{matrix} 10011111 = 1001 1111 \\ 9 F \end{matrix}$

The required hexadecimal number is 9F₁₆.

Example A.7

Convert binary number 1110111100001110₂ into hexadecimal.

Solution A.7

First, divide the number into groups of 4 and then find the equivalent of each group:

$\begin{array}{l} 1110111100001110 = 1110 1111 0000 1110 \\ E F 0 E \end{array}$

The required hexadecimal number is EF0E₁₆

Example A.8

Convert binary number 111110₂ into hexadecimal.

Solution A.8

Since the number cannot be divided exactly into groups of 4, we have to insert zeroes to the left-hand side of the number:

$\begin{array}{l} 111110 = 0011 1110 \\ 3 E \end{array}$

The required hexadecimal number is 3E₁₆.

Table A.2 presents the hexadecimal equivalent of numbers 0 to 31.

Table A.2

Hexadecimal equivalent of decimal numbers.
Decimal	Hexadecimal	Decimal	Hexadecimal
0	0	16	10
1	1	17	11
2	2	18	12
3	3	19	13
4	4	20	14
5	5	21	15
6	6	22	16
7	7	23	17
8	8	24	18
9	9	25	19
10	A	26	1A
11	B	27	1B
12	C	28	1C
13	D	29	1D
14	E	30	1E
15	F	31	1F

A.9. Converting hexadecimal numbers ınto binary

To convert a hexadecimal number into binary, write the 4-bit binary equivalent of each hexadecimal digit.

Example A.9

Convert hexadecimal number A9₁₆ into binary.

Solution A.9

Writing the binary equivalent of each hexadecimal digit:

$A = 1010_{2} 9 = 1001_{2}$

The required binary number is 10101001₂.

Example A.10

Convert hexadecimal number FE3C₁₆ into binary.

Solution A.10

Writing the binary equivalent of each hexadecimal digit:

$F = 1111_{2} E = 1110_{2} 3 = 0011_{2} C = 1100_{2}$

The required binary number is 1111111000111100₂.

A.10. Converting hexadecimal numbers ınto decimal

To convert a hexadecimal number into decimal, we have to calculate the sum of the powers of 16 of the number.

Example A.11

Convert hexadecimal number 2AC₁₆ into decimal.

Solution A.11

Calculating the sum of the powers of 16 of the number:

$\begin{matrix} 2 A C_{16} = 2 \times 16^{2} + 10 \times 16^{1} + 12 \times 16^{0} \\ = 512 + 160 + 12 \\ = 684 \end{matrix}$

The required decimal number is 684₁₀.

Example A.12

Convert hexadecimal number EE₁₆ into decimal.

Solution A.12

Calculating the sum of the powers of 16 of the number:

$\begin{matrix} E E_{16} = 14 \times 16^{1} + 14 \times 16^{0} \\ = 224 + 14 \\ = 238 \end{matrix}$

The required decimal number is 238₁₀.

A.11. Converting decimal numbers ınto hexadecimal

To convert a decimal number into hexadecimal, divide the number repeatedly into 16 and take the remainders. The first remainder is the LSD, and the last remainder is the MSD.

Example A.13

Convert decimal number 238₁₀ into hexadecimal.

Solution A.13

Dividing the number repeatedly into 16:

238/16	→	14	Remainder 14 (E)	(LSD)
14/16	→	0	Remainder 14 (E)	(MSD)

The required hexadecimal number is EE₁₆.

Example A.14

Convert decimal number 684₁₀ into hexadecimal.

Solution A.14

Dividing the number repeatedly into 16:

684/16	→	42	Remainder 12 (C)	(LSD)
42/16	→	2	Remainder 10 (A)
2/16	→	0	Remainder 2	(MSD)

The required hexadecimal number is 2AC₁₆.

A.12. Converting octal numbers ınto decimal

To convert an octal number into decimal, calculate the sum of the powers of 8 of the number.

Example A.15

Convert octal number 15₈ into decimal.

Solution A.15

Calculating the sum of the powers of 8 of the number:

$\begin{matrix} 15_{8} = 1 \times 8^{1} + 5 \times 8^{0} \\ = 8 + 5 \\ = 13 \end{matrix}$

The required decimal number is 13₁₀.

Example A.16

Convert octal number 237₈ into decimal.

Solution A.16

Calculating the sum of the powers of 8 of the number:

$\begin{matrix} 237_{8} = 2 \times 8^{2} + 3 \times 8^{1} + 7 \times 8^{0} \\ = 128 + 24 + 7 \\ = 159 \end{matrix}$

The required decimal number is 159₁₀.

A.13. Converting decimal numbers ınto octal

To convert a decimal number into octal, divide the number repeatedly into 8 and take the remainders. The first remainder is the LSD, and the last remainder is the MSD.

Example A.17

Convert decimal number 159₁₀ into octal.

Solution A.17

Dividing the number repeatedly into 8:

159/8	→	19	Remainder 7	(LSD)
19/8	→	2	Remainder 3
2/8	→	0	Remainder 2	(MSD)

The required octal number is 237₈.

Example A.18

Convert decimal number 460₁₀ into octal.

Solution A.18

Dividing the number repeatedly into 8:

460/8	→	57	Remainder 4	(LSD)
57/8	→	7	Remainder 1
7/8	→	0	Remainder 7	(MSD)

The required octal number is 714₈.

Table A.3 presents the octal equivalent of decimal numbers 0 to 31.

Table A.3

Octal equivalent of decimal numbers.
Decimal	Octal	Decimal	Octal
0	0	16	20
1	1	17	21
2	2	18	22
3	3	19	23
4	4	20	24
5	5	21	25
6	6	22	26
7	7	23	27
8	10	24	30
9	11	25	31
10	12	26	32
11	13	27	33
12	14	28	34
13	15	29	35
14	16	30	36
15	17	31	37

A.14. Converting octal numbers ınto binary

To convert an octal number into binary, write the 3-bit binary equivalent of each octal digit.

Example A.19

Convert octal number 177₈ into binary.

Solution A.19

Write the binary equivalent of each octal digit:

$1 = 001_{2} 7 = 111_{2} 7 = 111_{2}$

The required binary number is 001111111₂.

Solution A.20

Convert octal number 75₈ into binary.

Solution A.20

Write the binary equivalent of each octal digit:

$7 = 111_{2} 5 = 101_{2}$

The required binary number is 111101₂.

A.15. Converting binary numbers ınto octal

To convert a binary number into octal, arrange the number in groups of 3 and write the octal equivalent of each digit.

Example A.21

Convert binary number 110111001₂ into octal.

Solution A.21

Arranging in groups of 3:

$\begin{array}{l} 110111001 = 110 111 001 \\ 6 7 1 \end{array}$

The required octal number is 671₈.

A.16. Negative numbers

The most significant bit of a binary number is usually used as the sign bit. By convention, for positive numbers this bit is 0, and for negative numbers this bit is 1. Fig. A.1 shows the 4-bit positive and negative numbers. The largest positive and negative numbers are +7 and −8, respectively.

Figure A.1 4-bit positive and negative numbers.

To convert a positive number into negative, take the complement of the number and add 1. This process is also called the 2’s complement of the number.

Example A.22

Write decimal number −6 as a 4-bit number.

Solution A.22

First, write the number as a positive number, then find the complement and add 1:

0110	+6
1001	compliment
1	add 1
-------
1010	which is −6

Example A.23

Write decimal number −25 as a 8-bit number.

Solution A.23

First, write the number as a positive number, then find the complement and add 1:

00011001	+25
11100110	compliment
1	add 1
-------------
11100111	which is −25

A.17. Adding binary numbers

The addition of binary numbers is similar to the addition of decimal numbers. Numbers in each column are added together with a possible carry from a previous column. The primitive addition operations are:

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10	generate a carry bit
1 + 1 + 1 = 11	generate a carry bit

Some examples are given below.

Example A.24

Find the sum of binary numbers 011 and 110.

Solution A.24

We can add these numbers as in the addition of decimal numbers:

011	First column	1 + 0 = 1
+ 110	Second column	1 + 1 = 0 and a carry bit
--------	Third column	1 + 1 = 10
1001

Example A.25

Find the sum of binary numbers 01000011 and 00100010.

Solution A.25

We can add these numbers as in the addition of decimal numbers:

01000011	First column	1 + 0 = 1
+ 00100010	Second column	1 + 1 = 10
----------------	Third column	0 + carry = 1
01100101	Fourth column:	0 + 0 = 0
	Fifth column:	0 + 0 = 0
	Sixth column:	0 + 1 = 1
	Seventh column:	1 + 0 = 1
	Eighth column:	0 + 0 = 0

A.18. Subtracting binary numbers

To subtract two numbers, convert the number to be subtracted into negative and then add the two numbers.

Example A.26

Subtract binary number 0010 from 0110.

Solution A.26

First, let’s convert the number to be subtracted into negative:

0010	number to be subtracted
1101	compliment
1	add 1
-------
1110

Now, add the two numbers:

0110

+ 1110

----------

0100

Since we are using 4-bits only, we cannot show the carry bit.

A.19. Multiplication of binary numbers

Multiplication of two binary numbers is same as the multiplication of decimal numbers. The four possibilities are:

0 × 0 = 0

0 × 1 = 0

1 × 0 = 0

1 × 1 = 1

Some examples are given below.

Example A.27

Multiply the two binary numbers 0110 and 0010.

Solution A.27

Multiplying the numbers:


0110
0010
-------
0000
0110
0000
0000
-------
001100 or 1100

In this example, 4-bits are needed to show the final result.

Example A.28

Multiply binary numbers 1001 and 1010.

Solution A.28

Multiplying the numbers:


1001
1010
-------
0000
1001
0000
1001
------------
1011010

In this example, 7-bits are required to show the final result.

A.20. Division of binary numbers

The division of binary numbers is similar to the division of decimal numbers. An example is given below.

Example A.29

Divide binary number 1110 into binary number 10.

Solution A.29

Dividing the numbers:


111

----------
10
---
11
10
---
10
10
---
00

giving the result 111₂.

A.21. Floating point numbers

Floating point numbers are used to represent non-integer fractional numbers. For example: 3.256, 2.1, 0.0036, and so forth. Floating point numbers are used in most engineering and technical calculations. The most commonly used floating point standard is the IEEE standard. According to this standard, floating point numbers are represented with 32-bit (single precision) or 64-bit (double precision).

In this section, we are looking at the format of 32-bit floating point numbers only and see how mathematical operations can be performed with such numbers.

According to the IEEE standard, 32-bit floating point numbers are represented as:

31	30	23	22	0
X	XXXXXXXX		XXXXXXXXXXXXXXXXXXXXXXX
↑	↑		↑
sign	exponent		mantissa

The most significant bit indicates sign of the number, where 0 indicates positive and 1 indicates that the number is negative.

The 8-bit exponent shows the power of the number. To make the calculations easy, the sign of the exponent is not shown, but instead excess 128 numbering system is used. Thus, to find the real exponent, we have to subtract 127 from the given exponent. For example, if the mantissa is “10000000”, the real value of the mantissa is 128 − 127 = 1.

The mantissa is 23-bits wide and represents the increasing negative powers of 2. For example, if we assume that the mantissa is: “1110000000000000000000”, the value of this mantissa is calculated as: 2⁻¹ + 2⁻² + 2⁻³ = 7/8.

The decimal equivalent of a floating point number can be calculated using the formula:

$Number = {(- 1)}^{s} 2^{e - 127} 1. f$

where
s = 0 for positive numbers, 1 for negative numbers
e = exponent (between 0 and 255)
f = mantissa

As shown in the above formula, there is a hidden “1” in-front of the mantissa. That is, mantissa is shown as “1.f.”

The largest and the smallest numbers in 32-bit floating point format are:

The largest number

$0 11111110 11111111111111111111111$

This number is: (2 − 2⁻²³) 2¹²⁷ or decimal 3.403 × 10³⁸. The numbers keep their precision up to 6 digits after the decimal point.

The smallest number

$0 00000001 00000000000000000000000$

This number is: 2⁻¹²⁶ or decimal 1.175 × 10⁻³⁸.

A.22. Converting a floating point number ınto decimal

To convert a given floating point number into decimal, we have to find the mantissa and the exponent of the number and then convert into decimal as shown above.

Some examples are given here.

Example A.30

Find the decimal equivalent of the floating point number given below:

$0 10000001 10000000000000000000000$

Solution A.30

Here,

sign = positive
exponent = 129 − 127 = 2
mantissa = 2⁻¹ = 0.5

The decimal equivalent of these number is +1.5 × 2² = +6.0

Example A.31

Find the decimal equivalent of the floating point number given below:

$0 10000010 11000000000000000000$

Solution A.31

In this example,

sign = positive
exponent = 130 − 127 = 3
mantissa = 2⁻¹ + 2⁻² = 0.75

The decimal equivalent of the number is +1.75 × 2³ = 14.0

A.22.1. Normalizing the floating point numbers

Floating point numbers are usually shown in normalized form. A normalized number has only one digit before the decimal point (A hidden number 1 is assumed before the decimal point).

To normalize a given floating point number, we have to move the decimal point repetitively one digit to the left and then increase the exponent after each move.

Some examples are given below.

Example A.32

Normalize the floating point number 123.56

Solution A.32

If we write the number with a single digit before the decimal point, we get:

$1.2356 \times 10^{2}$

Example A.33

Normalize the binary number 1011.1₂

Solution A.33

If we write the number with a single digit before the decimal point, we get:

${1.111}^{3}$

A.22.2. Converting a decimal number into floating point

To convert a given decimal number into floating point, we have to carry out the following steps:

• Write the number in binary
• Normalize the number
• Find the mantissa and the exponent
• Write the number as a floating point number

Some examples are given below:

Example A.34

Convert decimal number 2.25₁₀ into floating point.

Solution A.34

Writing the number in binary:

${2.25}_{10} = {10.01}_{2}$

Normalizing the number,

${10.01}_{2} = {1.001}_{2} \times 2^{1}$

Here, s = 0, e − 127 = 1 or e = 128, and f = 00100000000000000000000

(Remember that a number 1 is assumed on the left-hand side, even though it is not shown in the calculation). We can now write the required floating point number as:

s	e	f
0	10000000	(1)001 0000 0000 0000 0000 0000

or, the required 32-bit floating point number is:

$01000000000100000000000000000000$

Example A.35

Convert the decimal number 134.0625₁₀ into floating point.

Solution A.35

Writing the number in binary:

${134.0625}_{10} = 10000110.0001$

Normalizing the number,

$10000110.0001 = 1.00001100001 \times 2^{7}$

Here, s = 0, e − 127 = 7 or e = 134, and f = 00001100001000000000000

We can now write the required floating point number as:

s	e	f
0	10000110	(1)00001100001000000000000

or, the required 32-bit floating point number is:

$01000011000001100001000000000000$

A.22.3. Multiplication and division of floating point numbers

The multiplication and division of floating point numbers is rather easy and the steps are given below:

• Add (or subtract) the exponents of the numbers
• Multiply (or divide) the mantissa of the numbers
• Correct the exponent
• Normalize the number
• The sign of the result is the EXOR of the signs of the two numbers

Since the exponent is processed twice in the calculations, we have to subtract 127 from the exponent.

An example is given below to show the multiplication of two floating point numbers.

Example A.36

Show the decimal numbers 0.5₁₀ and 0.75₁₀ in floating point and then calculate the multiplication of these numbers.

Solution A.36

We can convert the numbers into floating point as:

${0.5}_{10} = 1.0000 \times 2^{- 1}$

Here, s = 0, e − 127 = −1 or e = 126, and f = 0000

or,

${0.5}_{10} = 0 01110110 (1) 000 0000 0000 0000 0000 0000$

Similarly,

${0.75}_{10} = 1.1000 \times 2^{- 1}$

Here, s = 0, e = 126, and f = 1000

or,

${0.75}_{10} = 0 01110110 (1) 100 0000 0000 0000 0000 0000$

Multiplying the mantissas, we get “(1)100 0000 0000 0000 0000 0000.” The sum of the exponents is 126 + 126 = 252. Subtracting 127 from the mantissa, we obtain, 252 − 127 = 125. The EXOR of the signs of the numbers is 0. Thus, the result can be shown in floating point as:

$0 01111101 (1) 100 0000 0000 0000 0000 0000$

The above number is equivalent to decimal 0.375 (0.5 × 0.75 = 0.375), which is the correct result.

A.22.4. Addition and subtraction of floating point numbers

The exponents of floating point numbers must be the same before they can be added or subtracted. The steps to add or subtract floating point numbers are:

• Shift the smaller number to the right until the exponents of both numbers are the same. Increment the exponent of the smaller number after each shift.
• Add (or subtract) the mantissa of each number as an integer calculation, without considering the decimal points.
• Normalize the obtained result.

An example is given below.

Example A.37

Show decimal numbers 0.5₁₀ and 0.75₁₀ in floating point and then calculate the sum of these numbers.

Solution A.37

As shown in Example A.36, we can convert the numbers into floating point as:

${0.5}_{10} = 0 01110110 (1) 000 0000 0000 0000 0000 0000$

Similarly,

${0.75}_{10} = 0 01110110 (1) 100 0000 0000 0000 0000 0000$

Since the exponents of both numbers are the same, there is no need to shift the smaller number. If we add the mantissa of the numbers without considering the decimal points, we get:

(1)000 0000 0000 0000 0000 0000

(1)100 0000 0000 0000 0000 0000

____________________________+

(10)100 0000 0000 0000 0000 0000

To normalize the number, we can shift it right by one digit and then increment its exponent. The resulting number is:

$0 01111111 (1) 010 0000 0000 0000 0000 0000$

The above floating point number is equal to decimal number 1.25, which is the sum of decimal numbers 0.5 and 0.75.

To convert floating point numbers into decimal, and decimal numbers into floating point, the freely available program given in the following website can be used:

http://babbage.cs.qc.edu/courses/cs341/IEEE-754.html

A.23. BCD numbers

BCD (Binary Coded Decimal) numbers are usually used in display systems such as LCDs and 7-segment displays to show numeric values. In BCD, each digit is a 4-bit number from 0 to 9. As an example, Table A.4 presents the BCD numbers between 0 and 20.

Table A.4

BCD numbers between 0 and 20.
Decimal	BCD	Binary
0	0000	0000
1	0001	0001
2	0010	0010
3	0011	0011
4	0100	0100
5	0101	0101
6	0110	0110
7	0111	0111
8	1000	1000
9	1001	1001
10	0001 0000	1010
11	0001 0001	1011
12	0001 0010	1100
13	0001 0011	1101
14	0001 0100	1110
15	0001 0101	1111
16	0001 0110	1 0000
17	0001 0111	1 0001
18	0001 1000	1 0010
19	0001 1001	1 0011
20	0010 0000	1 0100

Example A.38

Write the decimal number 295 as a BCD number

Solution A.38

Writing the 4-bit binary equivalent of each digit,

$2 = 0010_{2} 9 = 1001_{2} 5 = 0101_{2}$

The required BCD number is: 0010 1001 0101₂

Example A.39

Write the decimal equivalent of BCD number 1001 1001 0110 0001₂

Solution A.39

Writing the decimal equivalent of each group of 4-bit, we get the decimal number,

$9961$

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Table of Contents for Appendix A. Number systems

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A.1. Overview

A.2. Decimal number system

A.3. Binary number system

A.4. Octal number system

A.5. Hexadecimal number system

A.6. Converting binary numbers ınto decimal

A.7. Converting decimal numbers ınto binary

A.8. Converting binary numbers ınto hexadecimal

A.9. Converting hexadecimal numbers ınto binary

A.10. Converting hexadecimal numbers ınto decimal

A.11. Converting decimal numbers ınto hexadecimal

A.12. Converting octal numbers ınto decimal

A.13. Converting decimal numbers ınto octal

A.14. Converting octal numbers ınto binary

A.15. Converting binary numbers ınto octal

A.16. Negative numbers

A.17. Adding binary numbers

A.18. Subtracting binary numbers

A.19. Multiplication of binary numbers

A.20. Division of binary numbers

A.21. Floating point numbers

A.22. Converting a floating point number ınto decimal

A.22.1. Normalizing the floating point numbers

A.22.2. Converting a decimal number into floating point

A.22.3. Multiplication and division of floating point numbers

A.22.4. Addition and subtraction of floating point numbers

A.23. BCD numbers

Table of Contents for
Appendix A. Number systems