Convert +610 to binary using the excess_127 system.

Chapter 2 Data Representation AITT 4300 DIGITAL COMPUTER STRUCTURES

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2.1 Decimal and Binary Representation

o The number system we normally use is the decimal system. o It uses 10 as the base. o But a number system can use any base. o Computers work with the binary system (base 2). o Other systems used with computers are octal (base 8) and hexadecimal (base 16).

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Bases and Exponents Any number squared means that number times itself. Example: 10 is the base and 2 is the exponent:

◦ 102 = 10*10 = 100

When a number is raised to a positive integer power, it is multiplied by itself that number of times. Example: the base is 5 and the exponent is 6:

56 = 5*5*5*5*5*5 = 15,625

Exception 1: When a non‐zero number is raised to the power of 0, the result is always 1. ◦ 5,3450 = 1 ◦ 40 = 1 ◦ (–31)0 = 1

Exception 2: 00 is undefined

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The Decimal System

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The first eight columns of the decimal system

107 106 105 104 103 102 101 100

10,000,000 1,000,000 100,000 10,000 1,000 100 10 1

ten‐millions millions hundred‐ thousands

ten‐thousands thousands hundreds tens ones

Expanded Notation

The ten digits that are used in the decimal system are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Any number in the decimal system can be written as a sum of each digit multiplied by the value of its column. This is called expanded notation. The number 6,825 in the decimal system actually means:

5*100 = 5*1 = 5

+ 2*101 = 2*10 = 20

+ 8*102 = 8*100 = 800

+ 6*103 = 6*1,000 = 6,000

6,825

Therefore, 6,825 can be expressed as: 6*103 + 8*102 + 2*101 + 5*100

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The Binary System

The binary system follows the same rules as the decimal system. The difference is that the binary system uses a base of 2 and has only two digits (0 and 1). The rightmost column of the binary system is the one’s column (20). It can contain a 0 or a 1. The next number after one is two; in binary, a two is represented by a 1 in the two’s column (21) and a 0 in the one’s column (20) ◦ one‐hundred in decimal is represented by a 1 in the one‐hundred’s (102) column and 0s in the ten’s column (101) and the one’s column (100)

◦ in binary, a 1 in the 22’s column represents the number 4; i.e. 100

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The Binary System

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The first eight columns of the binary system

Power of 2 27 26 25 24 23 22 21 20

Decimal value 128 64 32 16 8 4 2 1

Binary representation

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Converting the Decimal Number 2910 to Binary 29 is less than 32 but greater than 16 so put a 1 in the 16’s (24) column.

29 – 16 = 13 13 is less than 16 but greater than 8 so put a 1 in the eight’s (23) column

13 – 8 = 5 5 is less than 8 but greater than 4 so put a 1 in the four’s (22) column

5 – 4 = 1 1 is less than 2 so there is nothing in the two’s (21) column Put a 0 in the two’s column You have 1 left so put a 1 in the one’s (20) column Therefore, 2910 = 111012 Power of 2 25 24 23 22 21 20

Decimal value 32 16 8 4 2 1

Binary representation 0 1 1 1 0 1

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Converting the Decimal Number 17210 to Binary There is one 128 in 172 so put a 1 in the 128’s (27) column

172 – 128 = 44 44 is less than 64 so put a 0 in the 64’s (26) column 44 is less than 64 but greater than 32 so put a 1 in the 32’s (25) column

44 – 32 = 12 12 is less than 16 but greater than 8 so put a 0 in the 16’s (24) column and a 1 in the eight’s (23) column

12 – 8 = 4 Put a 1 in the four’s (22) column

4 – 4 = 0 Put 0s in the last two columns Power of 2 2

7 26 25 24 23 22 21 20

Decimal value 128 64 32 16 8 4 2 1

Binary

representation 1 0 1 0 1 1 0 0

Converting Binary to Decimal

To convert a binary number back to decimal, just add the value of each binary column.

Convert the binary number 10112 to decimal o There is a 1 in the one’s column o There is a 1 in the two’s column so the value of that column is 2 o There is a 0 in the four’s column so the value of that is 0 o There is a 1 in the eight’s column so the value of that column is 8

1 + 2 + 0 + 8 = 11 o Therefore, 10112 = 1110

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Convert the Binary Number 101010102 to Decimal

There is a 0 in the one’s column There is a 1 in the two’s column so the value of that column is 2 There is a 0 in the four’s column so the value of that is 0 There is a 1 in the eight’s column so the value of that column is 8 There is a 0 in the 16’s column There is a 1 in the 32’s column so the value of that column is 32 There is a 0 in the 64’s column There is a 1 in the 128’s column so the value of that column is 128

0 + 2 + 0 + 8 + 0 + 32 + 0 + 128 = 170

Therefore, 101010102 = 17010

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2.2 The Hexadecimal System The hexadecimal system uses a base of 16. ◦ there is a one’s column (160)

◦ a 16’s column (161)

◦ a 256’s column (162)

◦ a 4,096’s column (163)

◦ a 65,536’s column (164)

◦ and so forth

Rarely need to deal with anything larger than the 164’s column. The hexadecimal system makes it easier for humans to read binary notation.

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The Hexadecimal System

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The first five columns of the hexadecimal system

Hexadecimal

column 16 4 163 162 161 160

16*16*16*16 16*16*16 16*16 16 1

Decimal

equivalent 65,536 4,096 256 16 1

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Hexadecimal Digits o The decimal system uses 10 digits (0 through 9) in each column (base 10) o The binary system uses two digits (0 and 1) in each column (base 2) o The hexadecimal system uses 16 digits in each column (base 16) o How can you represent 10 in the one’s column in base 16?

o no way to distinguish “ten” (written as 10) from “sixteen” (also written as 10  a one in the 16’s column and a zero in the one’s column)

o Use uppercase letters to represent the digits 10 through 15 o hexadecimal digits are 0 through 9 and A through F

◦ 1010 is represented as A16 ◦ 1110 is represented as B16 ◦ 1210 is represented as C16 ◦ 1310 is represented as D16 ◦ 1410 is represented as E16 ◦ 1510 is represented as F16

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Converting the Decimal Number 2310 to Hexadecimal There is one 16 in 2310 so put a 1 in the 16’s column

 23 – 16 = 7 so put a 7 in the 1’s column

 Therefore, 2310 = 1716

Power of 16 163 162 161 160

Decimal value 4096 256 16 1

Hexadecimal representation 1 7

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Converting the Decimal Number 87510 to Hexadecimal 875 is less than 4,096 but greater than 256 so there is nothing in the 4,096’s (163) column Divide 875 by 256 to see how many 256s there are 875 ÷ 256 = 3 with a remainder of 107 Put a 3 in the 256’s column 107 ÷ 16 = 6 with a remainder of 11 Put a 6 in the 16’s column 11 in decimal notation = B in hexadecimal notation Put a B in the one’s column Therefore, 87510 = 36B16 Power of 16 16

3 162 161 160

Decimal value 4096 256 16 1

Hexadecimal representation 3 6 B

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Converting the Hexadecimal Number 123D16 to Decimal In expanded notation, this hexadecimal number is:

(1*4096) + (2*256) + (3*16) + (D*1)

D in hexadecimal is 13 in decimal, so:

4096 + 512 + 48 + 13 = 4669

Therefore, 123D16 = 466910

Power of 16 163 162 161 160

Decimal value 4096 256 16 1

Hexadecimal representation 1 2 3 D

Using Hexadecimal Notation

Notice:

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Decimal Binary Hexadecimal Decimal Binary Hexadecimal

0 0000 0 8 1000 8

1 0001 1 9 1001 9

2 0010 2 10 1010 A

3 0011 3 11 1011 B

4 0100 4 12 1100 C

5 0101 5 13 1101 D

6 0110 6 14 1110 E

7 0111 7 15 1111 F

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Using Hexadecimal Notation It is common to write a long binary number in hexadecimal notation. The 15 hexadecimal digits represent all combinations of a 4‐bit binary number. Convert the following binary number to hexadecimal notation:

11101010000011112 1. Separate the binary number into sets of 4 digits:

1110 1010 0000 1111 2. Refer to the table, if necessary, to make the conversions

11102 = E16 10102 = A16 00002 = 016 11112 = F16

Therefore, 11101010000011112 is EA0F16

2.3 Integer Representation  How computers process numbers depends on each number’s type.  Integers are stored and processed in quite a different manner from floating point numbers.  Even within the broad categories of integers and floating point numbers, there are more distinctions.  Integers can be stored as unsigned numbers (all nonnegative) or as signed numbers (positive, negative, and zero).  Floating point numbers also have several variations.

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Unsigned Integer Format

 A computer processes information in the form of bytes.  Bytes are normally 8 to 16 bits.  To store 112 and 1011012 both must have the same length as a byte.  Do this by adding 0s to the left of the number to fill up as many places as needed for a byte.  This is called the unsigned form of an integer.

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Unsigned Binary Integers Store the decimal integer 510 in an 8‐ bit memory location:

Convert 510 to binary: 1012 Add five 0s to the left to make 8 bits:

000001012

Store the decimal integer 92810 in a 16‐bit memory location:

Convert 92810 to binary: 11101000002 Add six 0s to the left to make 16 bits:

00000011101000002

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Overflow

If you try to store an unsigned integer that is bigger than the maximum unsigned value that can be handled by that computer, you get a condition called overflow.

Store the decimal integer 2310 in a 4‐bit memory location:  range of integers available in 4‐bit location is 010 through 1510

Therefore, attempting to store 2310 in a 4‐bit location gives an overflow.

Store the decimal integer 65,53710 in a 16‐bit memory location: range of integers available in 16‐bit location is 010 through 6553510

Therefore, attempting to store this number in a 16‐bit location gives an overflow.

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Range of Unsigned Integers

Number of Bits Range

8 0...255

16 0...65,535

32 0...4,294,967,295

64 0...18,446,740,000,000,000,000 (approximate)

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Sign‐and‐Magnitude Format The simple method to convert a decimal integer to binary works well to represent positive integers and zero.

We need a way to represent negative integers.

The sign‐and‐magnitude format is one way.

The leftmost bit is reserved to represent the sign.

The other bits represent the magnitude (or the absolute value) of the integer.

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Store the decimal integer +2310 in an 8‐bit memory location using sign‐ and‐magnitude format

Convert 2310 to binary: 101112 Since this is an 8‐bit memory location, 7 bits are used to store the magnitude of the number.

101112 uses 5 bits so add two 0s to the left to make up 7 bits: 00101112 Finally, look at the sign. This number is positive so add a 0 in the leftmost place to show the positive sign.

Therefore, +2310 in sign‐and‐magnitude format in an 8‐bit location is 000101112

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Store the decimal integer -1910 in a 16‐bit memory location using sign‐ and‐magnitude format

Convert 1910 to binary: 100112 Since this is a 16‐bit memory location, 15 bits are used to store the magnitude of the number.

100112 uses 5 bits so add ten 0s to the left to make up 15 bits: 0000000000100112 Finally, look at the sign. This number is negative so add a 1 in the leftmost place to show the negative sign.

Therefore, ‐1910 in sign‐and‐magnitude format in an 8‐bit location is 10000000000100112

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The Problem of the Zero (a) Store the decimal integer 010 in an 8‐bit memory location using sign‐and‐magnitude format: Convert 010 to binary: 02 Since this is an 8‐bit memory location, 7 bits are used to store the magnitude of the number. The number 02 uses 1 bit so add six 0s to the left to make up 7 bits: 00000002 Look at the sign. Zero is considered a non‐negative number so you should add a 0 in the leftmost place to show that it is not negative. Therefore, 010 in sign‐and‐magnitude in an 8‐bit location is: 000000002 (b) ... but...given that 100000002 is an 8‐bit binary integer in sign‐and‐magnitude form, find its decimal value: First convert the rightmost 7 bits to decimal to get 010 Look at the leftmost bit; it is a 1. So the number is negative. Therefore, 100000002 represents the decimal integer –010

One’s Complement Format

The fact that 0 has two possible representations in sign‐and‐magnitude format is one of the main reasons why computers usually use a different method to represent signed integers. There are two other formats that may be used to store signed integers. The one’s complement method is not often used, but it is explained here because it helps in understanding the most common format: two’s complement. To complement a binary digit, you simply change a 1 to a 0 or a 0 to a 1.

In the one’s complementmethod, positive integers are represented as they would be in sign‐and‐ magnitude format. The leftmost bit is still reserved as the sign bit.

+610, in a 4‐bit allocation, is still 01102 In one’s complement, –610 is just the complement of +610

–610 becomes 10012 The range of one’s complement integers is the same as the range of sign‐and‐magnitude integers. BUT… there are still two ways to represent the zero.

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Store the decimal integer -3710 in an 8‐bit memory location using one’s complement format

Convert 3710 to binary: 1001012 Since this is an 8‐bit memory location, 7 bits are used to store the magnitude

The number 1001012 uses 6 bits so add one 0 to the left to make up 7 bits:

01001012

This number is negative. Complement all the digits by changing all the 0s to 1s and all the 1s to 0s

Add a 1 in the 8th bit location because the number is negative

Therefore, –3710 in one’s complement in an 8‐bit location is 110110102

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Converting One’s Complement to Decimal To convert a one’s complement number back to decimal: Look at the leftmost bit to determine the sign. If the leftmost bit is 0, the number is positive and can be converted back to decimal immediately. If the leftmost bit is a 1, the number is negative. ◦ Un‐complement the other bits (change all the 0s to 1s and all the 1s to 0s) ◦ then convert the binary bits back to decimal ◦ Remember to include the negative sign when displaying the result!

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The Problem of the Zero Again a) Store the decimal integer 010 in an 8‐bit memory location using one’s complement format:

Convert 010 to binary: 02 Since this is an 8‐bit memory location, 7 bits are used to store the magnitude The number 02 uses 1 bit so add six 0’s to the left to make up 7 bits: 00000002 Zero is considered non‐negative so add a 0 in the leftmost place Therefore, 010 in one’s complement in an 8‐bit location is 000000002 (b) but... given that 111111112 is a binary number in one’s complement form, find its decimal value:

Look at the leftmost bit. It is a 1 so you know the number is negative Since the leftmost bit is 1, all the other bits have been complemented. “un‐complement” them to find the magnitude of the number. When you un‐complement 11111112, you get 00000002 Therefore, 111111112 in one’s complement represents the decimal integer –010

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Why the Fuss About Nothing? Why is there so much fuss about the zero? Why not just define zero in binary as 00002 (or 000000002 or 00000000000000002) and be done with it? In a 4‐bit allocation, the bit‐pattern 11112 still exists. Unless the computer knows what to do with it, the program will get an error. It might even not work at all. One possible scenario: If the result of a calculation using one’s complement was 11112, the computer would read this as –0. If you then tried to add 1 to it, what would the answer be? ◦ The number that follows 11112 in a 4‐bit allocation is 00002. ◦ That would mean, using one’s complement, that –0 + 1 = +0. This certainly would not be an irrelevant issue!

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Two’s Complement Integers To find the two’s complement of an X‐bit number: 1. If the number is positive, just convert the decimal integer to binary

and you are finished. 2. If the number is negative, convert the number to binary and find

the one’s complement. 3. Add a binary 1 to the one’s complement of the number. 4. If this results in an extra bit (more than X bits), discard the leftmost

bit.

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Rules of Binary Addition

Rule 1 + 0 = 1 1 + 1 = 10

Example 1

1 0

+ 1

1 1

1

+ 1

1 0

Example 2

1 0 1

+ 1 0

1 1 1

1 1

+ 1

1 0 0

Example 3

1 0 0

+ 1

1 0 1

1 0 1

+ 1

1 1 0

Finding the Two’s Complement of 8‐bit Binary Integers

Find the two’s complement of +4310 as an 8‐bit binary integer: Convert 4310 to binary: 1010112 Add zeros to the left to complete 8 bits: 00101011

Since this is already a positive integer, you are finished.

Find the twos complement of –4310 as an 8‐bit binary integer: Convert 4310 to binary: 1010112 Add zero’s to the left to complete 8 bits: 00101011

Since the number is negative, do the one’s complement to get: 11010100

Now add binary 1 to this number: 11010100

+ 1

11010101

Therefore, –4310 in two’s complement in an 8‐bit location is 11010101

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Carrying the 1With Binary Addition Find the two’s complement of –2410 as an 8‐bit binary integer: Convert 2410 to binary: 110002 Add zeros to the left to complete 8 bits: 00011000 Since the number is negative, do the one’s complement to get:

11100111 Now add binary 1 to this number:

11100111 + 1 11101000

Therefore, –2410 in two’s complement in an 8‐bit location is 11101000

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When the Two’s Complement Cannot Be Done Find the two’s complement of –15910 as an 8‐bit binary integer: Convert 15910 to binary: 100111112 10011111 already takes up 8 bits so there is nothing left for the sign bit Therefore, –15910 cannot be represented as a two’s complement binary number in an 8‐bit location.

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The Zero Solution a. Find the two’s complement of +010 as an 8‐bit binary integer:

Convert 010 to 8‐bit binary: 000000002 The number is positive so nothing more needs to be done Therefore, +0 in two’s complement in an 8‐bit location is 00000000

b. Find the two’s complement of –010 as an 8‐bit binary integer:

Convert 010 to 8‐bit binary: 000000002 Since the number is negative, do the one’s complement to get: 11111111

Now add binary 1 to this number: 11111111

+ 1

100000000

Recall that Step 4 in the rules for converting to two’s complement states that, after the addition of 1 to the one’s complement, any digits to the left of the maximum number of bits (here, 8 bits) should be discarded. Discard the leftmost 1

Therefore, –010 in two’s complement in an 8‐bit location is 000000002 which is exactly the same as +010

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Why the Method Works How in the world does flipping digits and then adding 1 somehow end up with the negative of the number you started with? Example: using a 4‐bit allocation, since it is easy to manage. • A 4‐bit allocation allows for 16 binary numbers ranging from 0000 to 1111, or 010 to 1510. • Define the “flip side” of any number between 0 and 16 to be 16minus that number. ◦ using 4 bits, there are 16 possible numbers (010 to 1510), so the flip side of a number between 0 and 16 would be 16minus that number.

◦ the flip side of 4 is 16 – 4 = 12.

◦ in two’s complement, the negative of a number is represented as the flip side of its positive value. ◦ using two’s complement notation, a –310 is represented as the flip side of +310. ◦ In a 4‐bit location, this would be 16 – 3 = 13. In an 8‐bit location, this would be 256 – 3 = 253 because 28 = 256.

In mathematical terms, this can be expressed as follows (assuming an X‐bit memory allocation):

◦ For a number, N, the two’s complement is:

2X – |N| where |N| denotes the absolute value of N

2.4 Floating Point Representation All floating point numbers have both an integer part and a fractional part, even if that fractional part is 0. ◦ Note: 6 is an integer but 6.0 is a floating point number

To represent a floating point number in binary: ◦ divide the number into its parts: ◦ the sign (positive or negative) ◦ the whole number (integer) part ◦ the fractional part

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The Integer Part A specific bit is set aside to denote the sign.

To convert the integer part to binary, simply convert the same way you convert positive integers to binary.

The integer part of a floating point binary number is separated from the fractional part.

The dot (or period) between the integer and fractional parts of a binary number will be referred to as a point.

The point is, in effect, a binary point ◦ it does the same thing as a decimal point in the decimal system.

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The Fractional Part We know that the columns in the integer part of a binary number represent powers of 2. ◦ The first column, 20 is the one’s column; the second column, 21 is the two’s column; and so on.

We can think of the fractional part in similar terms.

◦ The decimal number 0.1 represents , the decimal number 0.01 represents and so on.

◦ As the denominators get smaller, each decimal place is 10 raised to the next power.

◦ In decimal notation: , , , etc.

◦ Also can be represented as 10-1, 10-2, 10-3, etc.

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Fractional Part of the Binary System

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The first six columns of the fractional part of a number in the binary system

0.1 0.01 0.001 0.0001 0.00001 0.000001

= 2–1 = 2–2 = 2–3 = 2–4 = 2–5 = 2–6

0.5 0.25 0.125 0.0625 0.03125 0.015625

halves fourths eighths sixteenths thirty‐seconds sixty‐fourths

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Converting a Decimal Fraction to Binary 1. How many bits are allowed for the fractional part of a given number? 2. Create a chart:

3. As you work, you can fill in the boxes in the third row.

4. If the number is equal to or greater than 0.5, put a 1 in the 2–1 column. Otherwise put a 0 in the 2–1 column. Then subtract 0.5 from the decimal number. If the result is 0, you are done.

5. If the result is equal to or greater than 0.25, put a 1 in the 2–2 column. Then subtract 0.25 from the decimal number. If the result is 0, you are done.

6. If the result of your subtraction is less than 0.25, put a 0 in the 2–2 column. Look at the next column. If your number is less than 0.125, put a 0 in the 2–3 column

7. Repeat with each subsequent column until the subtraction either gives a result of 0 or until you reach the end of the bits required.

Binary 2–1 2–2 2–3 2–4 2–5 2–6

Decimal 0.5 0.25 0.125 0.0625 0.03125 0.015625

Conversion

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Convert the Decimal Number 0.4 to a 6‐bit Binary Number This number is less than 0.5, so put a 0 in the 2–1 column The number is greater than 0.25, so put a 1 in the 2–2 column, then subtract: 0.4 – 0.25 = 0.15 0.15 is greater than 0.125, so put a 1 in the 2–3 column and subtract: 0.15 – 0.125 = 0.025

0.025 is less than the next column, 0.0625, so put a 0 in the 2–4 column

0.025 is less than the next column, 0.03125, so put a 0 in the 2–5 column

0.025 is greater than the next column, 0.015625, so put a 1 in the 2–6 column

Even though there is a remainder when you subtract 0.025 – 0.015625 = 0.009375, you do not need to do anything more because the problem specified that only 6 bits are needed 0.4 in decimal = 0.011001 in a 6‐bit binary representation

Binary 2–1 2–2 2–3 2–4 2–5 2–6

Decimal 0.5 0.25 0.125 0.0625 0.03125 0.015625

Conversion 0 1 1 0 0 1

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Putting the Two Parts Together: Store the Decimal Number 75.804 as a Binary Number

1. Convert 75 to binary: 1001011 2. Convert 0.804 to binary:

oPut a 1 in the 2–1 column and subtract: 0.804 – 0.5 = 0.304 oPut a 1 in the 2–2 column and subtract: 0.304 – 0.25 = 0.054 oPut a 0 in the 2–3 column. oPut a 0 in the 2–4 column. oYou do not need to do anything more because the problem specified that only 4 bits are needed for the fractional part.

Therefore, 75.80410 = 101011.11002

Binary 2–1 2–2 2–3 2–4

Decimal 0.5 0.25 0.125 0.0625

Conversion 1 1 0 0

2.5 Putting It All Together Just converting a decimal floating point number to binary representation isn’t enough.

There are several concepts to understand before seeing how a floating point number is actually represented inside the computer.

While you will probably use a calculator for conversions, it is valuable to understand the process and will prove helpful when writing programs.

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Scientific Notation

Computers are used for many scientific applications which often use very large or very

small numbers.

Example: the distance from Earth to our nearest star is 24,698,100,000,000 miles. We

would need a 49‐digit binary number to represent this in a computer.

Example: The diameter of an atom would require at least a 30‐digit binary number.

Humans deal with these almost‐impossible‐to‐read numbers with scientific notation.

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Examples of Scientific Notation In scientific notation, a given number is written as a number between 1 and 9 multiplied by the appropriate power of 10.

Examples:

680,000 = 6.8×105

1,502,000,000 = 1.502×109

8,938,000,000,000 = 8.938×1012

0.068 = 6.8×10–2

0.00001502 = 1.502×10–5

0.000000000008938 = 8.938×10–12

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Exponential Notation In programming, instead of writing 10power, we use the letter E followed by the given power. This is called

exponential notation. Notice, you must include the sign of the exponent.

Examples:

680,000 = 6.8E+5

1,502,000,000 = 1.502E+9

8,938,000,000,000 = 8.938E+12

0.068 = 6.8E-2

0.00001502 = 1.502E-5

0.000000000008938 = 8.938E-12

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Converting a Number from Exponential Notation to Ordinary Notation

Move the decimal point the number of places indicated by the integer following E Examples: Given 1.67E–4 ◦ write 1.67 and move the decimal point 4 places to the left, filling in 3 zeros before 1

◦ This gives 0.000167 Given 4.2E+6 ◦ move the decimal point 6 places to the right, filling in 5 zeros to the right of 2 ◦ This gives 4200000, or as it is usually written, 4,200,000

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Base 10 Normalization

Normalized form is similar to scientific notation. Each normalized number has two parts: the scaled portion and the exponential portion. In scientific notation, the decimal point was moved so the first non‐zero digit was immediately to the left of it. In normalized form, the decimal is moved so the first non‐zero digit is immediately to the right of it. The value of the number is always maintained. To normalize a decimal number, after moving the decimal point, the number is multiplied by 10 raised to whatever power is necessary to return the number to its original value.

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Normalized Decimal Numbers

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Number Scaled Portion Exponential Portion Normalized Form

371.2 0.3712 103 0.3712  103

40.0 0.4 102 0.4  102

0.000038754 0.38754 10–4 0.38754  10–4

–52389.37 –0.5238937 105 –0.5238937  105

Normalizing Binary Floating Point Numbers

The IEEE Standard is the most widely accepted standard for representation of floating point numbers in a computer and uses normalized binary numbers.

A normalized binary number consists of three parts: ◦ the sign part ◦ the exponential part ◦ the mantissa.

The mantissa is the binary equivalent to the scaled portion (as in previous slide)

The Excess_127 system is used to represent the exponential portion

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

The Excess_127 system Is used to store the exponential value of a normalized binary number. To represent an 8‐bit number in the Excess_127 system:

oAdd 127 to the number oChange the result to binary oAdd zeros to the left to make up 8 bits

Examples: (a) To represent +910 (b) To represent –1310  add 9 + 127 = 136  add (–13) + 127 = 114  Convert 136 to binary: 10001000  Convert 114 to binary: 01110010  +910 in Excess_127 is 10001000  –1310 in Excess_127 is 0111001

Excess_127

Base 2 Normalization

The process is similar to the one followed to normalize a decimal number.

The point is moved but it is moved so that the first non‐zero number (a 1) is immediately to the left of the point.

Then multiply the number by the power of 2 needed to express the original value of the number.

Not necessary to worry about the sign of the number since, in normalized form, the sign bit takes care of this.

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Examples

Normalize the Binary Number +10110 ◦ Move the point to the left 4 places to get 1.0110 ◦ Since the point was moved 4 places to the left, the number is then multiplied by 24 to

get the original number ◦ +10110 in normalized form is 24×1.0110

Normalize the Binary Number +0.11110011 ◦ Move the point to the right 1 place to get 1.1110011 ◦ Since the point was moved 1 place to the right, the number needs to be multiplied by

2–1 to get the original number ◦ +0.11110011 in normalized form is 2–1×1.1110011

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Single Precision Floating Point Numbers IEEE has defined standards for storing floating point numbers. The most common standard is single precision floating point. In single precision format, a normalized floating point number has three parts. ◦ The sign is stored as a single bit ◦ The exponent is stored in 8 bits ◦ The mantissa is stored in the rest of the bits (23 bits) ◦ Single precision uses 32 bits in total to store one floating point number

There is also a double precision representation which allows for a much larger range of numbers. ◦ The sign of the number still uses one bit ◦ The exponent uses 11 bits ◦ The mantissa uses 52 bits. ◦ An 11‐bit exponent uses the Excess_1023 system and can handle exponents up to ±1023 ◦ Double precision uses 64 bits in total to store one floating point number

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Converting a Decimal Number to Single Precision Floating Point Binary 1.The sign bit:

If the number is positive, put a 0 in the leftmost bit. If it is negative, put a 1 in the leftmost bit.

2. Convert the number to binary.

If there is an integer and a fractional part, convert the whole number to binary

3. Normalize the number.

Move the point so it is directly to the right of the first non‐zero number.

4. Count the number of places you moved the point. This is your exponent.

If you moved the point to the right, your exponent is negative. If you moved the point to the left, your exponent is positive.

5. The exponent part:

Convert your exponent to a binary number, using the Excess_127 system. Store this number in the 8 bits to the right of the sign bit.

6. The mantissa: Use the number from Step 3 is used to find the mantissa. When storing the normalized part of the number, the 1 to the left of the point is discarded. Everything to the right of the point is now called the mantissa.

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE

Represent in single precision floating point the normalized number: –2–9 × 1.00001011 The sign is negative so the leftmost bit is a 1 The exponent is –9. Convert this to Excess_127: ◦ Add: (–9) + 127 = 118

◦ Convert 118 to binary: 01110110 ◦ Store this number in the next 8 bits

The rest of the number is 1.00001011 ◦ Discard the leftmost 1 (the one to the left of the point) and store the remainder of the number in the last 23 bits

This number takes up 8 bits while, in single‐precision floating point, the mantissa is 23 bits long. You must add 15 0’s at the end to complete 23 bits. Therefore, –2–9 × 1.00001011 as a single‐precision floating point number is

1 01110110 00001011000000000000000

Hexadecimal Representation

It is much easier to read a hexadecimal number than to read a long string of 0s and 1s. Single precision floating point numbers are often changed to hexadecimal. It’s easy to convert binary to hexadecimal: ◦ Divide the binary number into groups of four digits ◦ Convert each group to a single hexadecimal number Example (from previous slide):

1 01110110 00001011000000000000000 is

BB05800016

PRELUDE TO PROGRAMMING, 6TH EDITION BY ELIZABETH DRAKE