Lecture 2 Number Systems ITEC 1000 “Introduction to Information Technology” Lecture 2 Number Systems Hexadecimal Decimal Octal Binary Prof. Peter Khaiter

Lecture Template: Types of number systems Number bases Range of possible numbers Conversion between number bases Common powers Arithmetic in different number bases Shifting a number

Types of Number Systems Additive: Numbers have intrinsic value: e.g.: Roman Numerals: LVIII = 50 + 5 + 1 + 1 + 1 = 58 Positional: Value depends on position: e.g.: Decimal system: 55 = 5 x 10 + 5 x 1 Additive Number Systems are not used much any more: Awkward to use. Prone to errors.

Definitions The Base of a number system – how many different digits (incl. zero) are used in the system. Base 2: 0, 1 Base 5: 0, 1, 2, 3, 4 Base 8: 0, 1, 2, 3, 4, 5, 6, 7 Base 10: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 Base 16: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

Definitions Bit – a cell holding a single binary number (0 or 1) Byte = 8 bits (can hold 28 = 256 different patterns/values) Word – a fixed-sized group of bits that the computer handles together. Typical word sizes: 4, 8, 16, 32, 64, 128 bits 1K = 1024 bytes

Magnetic Core Memory http://en.wikipedia.org/wiki/Computer

Common Number Systems System Base Symbols Used by humans? Used in computers? Decimal 10 0, 1, … 9 Yes No Binary 2 0, 1 Octal 8 0, 1, … 7 Hexa- decimal 16 0, 1, … 9, A, B, … F

Positional decimal system The number 125 means: 1 group of 100 (100 = 102) 2 groups of 10 (10 = 101) 5 groups of 1 (1 = 100)

Place values (1 of 2) In our usual positional number system, the meaning of a digit depends on where it is located in the number Example: 3 7 3 2 3 groups of 1000 7 groups of 100 3 groups of 10 2 groups of 1

Place values (2 of 2) Weight 12510 => 5 x 100 = 5 2 x 101 = 20 1 x 102 = 100 125 Base

Representing in bases: 10, 2, 8, 16 86510 = 8 x 102 + 6 x 101 + 5 x 100 = 800 + 60 + 5 10112 = 1 x 23 + 0 x 22 + 1 x 21 + 1 x 20 = 8 + 2 + 1 = 1110 258 = 2 x 81 + 5 x 80 = 16 + 5 = 2110 A716 = 10 x 161 + 7 x 160 = 160 + 7 = 16710 Note: The subscript naming the base is itself given in base ten (10), by convention. Base

Counting in bases (1 of 3) Decimal Binary Octal Hexa- decimal 1 2 10 3 1 2 10 3 11 4 100 5 101 6 110 7 111

Counting in bases (2 of 3) Decimal Binary Octal Hexa- decimal 8 1000 9 1001 11 1010 12 A 1011 13 B 1100 14 C 1101 15 D 1110 16 E 1111 17 F

Counting in bases (3 of 3) Decimal Binary Octal Hexa- decimal 16 10000 20 10 17 10001 21 11 18 10010 22 12 19 10011 23 13 10100 24 14 10101 25 15 10110 26 10111 27

Estimating magnitude: Binary 1101 01102 = 21410 1101 01102 > 19210 (128 + 64 + additional bits to the right) Place 27 26 25 24 23 22 21 20 Value 128 64 32 16 8 4 2 1 Evaluate 1 x 128 1 x 64 0 x 32 1 x16 0 x 8 1 x 4 1 x 2 0 x 1 Sum for Base 10

Range of possible numbers R = BK where R = range B = base K = number of digits Example #1: Base 10, 2 digits R = 102 = 100 different numbers (0…99) Example #2: Base 2, 16 digits R = 216 = 65,536 or 64K 16-bit PC can store 65,536 different number values

Decimal Range for Bit Widths Bits Digits Range 1 0+ 2 (0 and 1) 4 1+ 16 (0 to 15) 8 2+ 256 10 3 1,024 (1K) 16 4+ 65,536 (64K) 20 6 1,048,576 (1M) 32 9+ 4,294,967,296 (4G) 64 19+ Approx. 1.6 x 1019 128 38+ Approx. 2.6 x 1038 See also Ch. 3, p. 74

Conversion Among Bases Hexadecimal Decimal Octal Binary

Binary to Decimal (1 of 3) Hexadecimal Decimal Octal Binary

Binary to Decimal (2 of 3) Technique Multiply each bit by 2n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results

Binary to Decimal (3 of 3) Bit “0” 1010112 => 1 x 20 = 1 1 x 21 = 2 0 x 22 = 0 1 x 23 = 8 0 x 24 = 0 1 x 25 = 32 4310 Bit “0”

Octal to Decimal (1 of 3) Decimal Octal Binary Hexadecimal

Octal to Decimal (2 of 3) Technique Note: 80 = 1, 81 = 8, Multiply each bit by 8n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results together Note: 80 = 1, 81 = 8, 82 = 64 83 = 512, Etc.

Octal to Decimal (3 of 3) 7248 => 4 x 80 = 4 x 1 = 4 2 x 81 = 2 x 8 = 16 7 x 82 = 7 x 64 = 448 46810

Hexadecimal to Decimal (1 of 3) Octal Binary Hexadecimal

Hexadecimal to Decimal (2 of 3) Technique Multiply each bit by 16n, where n is the “weight” of the bit The weight is the position of the bit, starting from 0 on the right Add the results Note: 160 = 1, 161 = 16, 162 = 256 163 = 4096, Etc.

Hexadecimal to Decimal (3 of 3) ABC16 => C x 160 = 12 x 1 = 12 B x 161 = 11 x 16 = 176 A x 162 = 10 x 256 = 2560 274810

Decimal to Binary (1 of 3) Decimal Octal Binary Hexadecimal

Decimal to Binary (2 of 3) Technique Divide by two, keep track of the remainder First remainder is bit 0 (LSB, least-significant bit) Second remainder is bit 1 Etc.

Decimal to Binary (3 of 3) 12510 = ?2 2 125 62 1 2 31 0 2 15 1 2 7 1 2 125 62 1 2 31 0 2 15 1 2 7 1 2 3 1 2 1 1 2 0 1 12510 = 11111012

Octal to Binary (1 of 3) Decimal Octal Binary Hexadecimal

Octal to Binary (2 of 3) Technique Convert each octal digit to a 3-bit equivalent binary representation See Table, Slides 12-14

Octal to Binary (3 of 3) 7058 = ?2 7 0 5 111 000 101 7058 = 1110001012

Hexadecimal to Binary (1 of 3) Octal Binary Hexadecimal

Hexadecimal to Binary (2 of 3) Technique Convert each hexadecimal digit to a 4-bit equivalent binary representation See Table, Slides 12-14

Hexadecimal to Binary (3 of 3) 10AF16 = ?2 1 0 A F 0001 0000 1010 1111 10AF16 = 00010000101011112

Decimal to Octal (1 of 3) Decimal Octal Binary Hexadecimal

Decimal to Octal (2 of 3) Technique Divide by 8 Keep track of the remainder

Decimal to Octal (3 of 3) 123410 = ?8 8 1234 154 2 8 19 2 8 2 3 8 0 2 123410 = 23228

Decimal to Hexadecimal (1 of 3) Octal Binary Hexadecimal

Decimal to Hexadecimal (2 of 3) Technique Divide by 16 Keep track of the remainder

Decimal to Hexadecimal (3 of 3) 123410 = ?16 16 1234 77 2 16 4 13 = D 0 4 123410 = 4D216

Binary to Octal (1 of 3) Decimal Octal Binary Hexadecimal

Binary to Octal (2 of 3) Technique Group bits in threes, starting on right Convert to octal digits See Table, Slides 12-14

Binary to Octal (3 of 3) 10110101112 = ?8 1 011 010 111 1 3 2 7 10110101112 = 13278

Binary to Hexadecimal (1 of 3) Octal Binary Hexadecimal

Binary to Hexadecimal (2 of 3) Technique Group bits in fours, starting on right Convert to hexadecimal digits See Table, Slides 12-14

Binary to Hexadecimal (3 of 3) 10101110112 = ?16 10 1011 1011 2 B B 10101110112 = 2BB16

Octal to Hexadecimal (1 of 3) Binary Hexadecimal

Octal to Hexadecimal (2 of 3) Technique Use binary as an intermediary See Table, Slides 12-14

Octal to Hexadecimal (3 of 3) 10768 = ?16 1 0 7 6 001 000 111 110 2 3 E 10768 = 23E16

Hexadecimal to Octal (1 of 3) Binary Hexadecimal

Hexadecimal to Octal (2 of 3) Technique Use binary as an intermediary See Table, Slides 12-14

Hexadecimal to Octal (3 of 3) 1F0C16 = ?8 1 F 0 C 0001 1111 0000 1100 1 7 4 1 4 1F0C16 = 174148

Exercise – Convert ... Decimal Binary Octal Hexa- decimal 33 1110101 703 1AF Don’t use a calculator! Skip answer Answer

Exercise – Convert (answers) Decimal Binary Octal Hexa- decimal 33 100001 41 21 117 1110101 165 75 451 111000011 703 1C3 431 110101111 657 1AF

Common Powers (1 of 2) Base 10 Power Preface Symbol pico p nano n 10-12 pico p 10-9 nano n 10-6 micro  10-3 milli m 103 kilo k 106 mega M 109 giga G 1012 tera T Value .000000000001 .000000001 .000001 .001 1000 1000000 1000000000 1000000000000

Common Powers (2 of 2) Base 2 What is the value of “k”, “M”, and “G”? Preface Symbol 210 kilo k 220 mega M 230 Giga G Value 1024 1048576 1073741824 What is the value of “k”, “M”, and “G”? In computing, particularly w.r.t. memory, the base-2 interpretation generally applies

Example 1. Double click on My Computer 2. Right click on C: 3. Click on Properties / 230 =

For common bases, add powers Multiplying powers For common bases, add powers ab  ac = ab+c 26  210 = 216 = 65,536 or… 26  210 = 64  210 = 64k

Table of powers Power Base 8 7 6 5 4 3 2 1 256 128 64 32 16 32,768 256 128 64 32 16 32,768 4,096 512 65,536

Number point or radix point Fractions Number point or radix point Decimal point in base 10 Binary point in base 2 No exact relationship between fractional numbers in different number bases Exact conversion may be impossible

Decimal fractions Move the number point one place to the right Effect: multiplies the number by the base number Example: 139.010 139010 Move the number point one place to the left Effect: divides the number by the base number Example: 139.010 13.910

Fractions: Base 10 .258910 10-1 10-2 10-3 10-4 Place 10-1 10-2 10-3 10-1 10-2 10-3 10-4 .258910 Place 10-1 10-2 10-3 10-4 Value 1/10 1/100 1/1000 1/10000 Evaluate 2 x 1/10 5 x 1/100 8 x 1/1000 9 x1/10000 Sum .2 .05 .008 .0009

Fractions: Base 2 .1010112 = 0.67187510 2-1 2-2 2-3 2-4 2-5 2-6 Place 2-1 2-2 2-3 2-4 2-5 2-6 .1010112 = 0.67187510 Place 2-1 2-2 2-3 2-4 2-5 2-6 Value 1/2 1/4 1/8 1/16 1/32 1/64 Evaluate 1 x 1/2 0 x 1/4 1x 1/8 0 x 1/16 1 x 1/32 1 x 1/64 Sum .5 0.125 0.03125 0.015625

Fractions: Base 10 and Base 2 No general relationship between fractions of types 1/10k and 1/2k Therefore a number representable in base 10 may not be representable in base 2 But: the converse is true: all fractions of the form 1/2k can be represented in base 10 Fractional conversions from one base to another are stopped If there is a rational solution, or When the desired accuracy is attained

Fractions: From Base B To Base 10 Determine the appropriate weight for each fractional digit as a negative power of the base Multiply each digit by its weight Add the values Example: .A116 = 10x16-1 + 1x16-2 = 10x0.0625 + 1x0.00390625 = 0.6289062510

Fractions: From Base 10 To Base B Multiply the fraction by the base value B (i.e., 2, 8 or 16) Record the values that move to the left of the radix point and drop them Repeat the process until the value being multiplied is zero, or the desired number of digits of accuracy is attained

Fractions: From Base 10 To Base B 0.67312510 = ?2 0.673125 x 2 1.346250 0.692500 1.385000 0.770000 0.1010112 1.540000 1.080000

Fractions: From Base 2 To Base 8, 16 Group digits from left to right in groups of 3 (base 8) or 4 (base 16) Supplement the right-most group with 0’s, if necessary Convert each group to the desired base. See Table, Slides 12-14

Fractions: From Base 8, 16 To Base 2 Convert each octal (base 8) or hexadecimal (base 16) digit to its 3-bit or 4-bit representation See Table, Slides 12-14

Fractions: between Base 8 and Base 16 Use binary conversion as an intermediary Example: 0.C816 = ?8 1100 10002 6 2 0 0.C816 = 0.628

Mixed number conversion Convert whole part and fraction part separately See Table, Slides 12-14

Arithmetic operations (1 of 14) Binary Addition Two 1-bit values A B A + B 1 10 0 and carry 1 to the next more significant bit, i.e. “two”

Arithmetic operations (2 of 14) Two n-bit values Add individual bits (see Table) Propagate carries 1 1 10101 21 + 11001 + 25 101110 46 Note: superscripts are carried amounts.

Addition (different bases) (3 of 14) Problem Largest Single Digit Decimal 6 +3 9 Octal +1 7 Hexadecimal +9 F Binary 1 +0

Addition (different bases) (4 of 14) Problem Carry Answer Decimal 6 +4 Carry the 10 10 Octal +2 Carry the 8 Hexadecimal +A Carry the 16 Binary 1 +1 Carry the 2

Addition Table: Base 10 (5 of 14) 310 + 610 = 910 + 1 2 3 4 5 6 7 8 9 10 11 12 8 etc 13

Addition Table: Base 8 (6 of 14) 38 + 68 = 118 + 1 2 3 4 5 6 7 10 11 12 13 14 15 16

Arithmetic operations (7 of 14) Binary Subtraction Two 1-bit values A B A - B 1 and borrow 1 from the next more significant bit

Arithmetic operations (8 of 14) Two n-bit values Subtract individual bits (see Table) Keep track of the borrowings 00100101 – 00010001 = 00010100 00100101 3710 - 00010001 - 1710 00010100 2010 10

Arithmetic operations (9 of 14) Multiplication Decimal (just for fun) 35 x 105 175 000 35 3675

Arithmetic operations (10 of 14) Binary multiplication, two 1-bit values A B A  B 1

Arithmetic operations (11 of 14) Binary multiplication, two n-bit values As with decimal values 1110 x 1011 1110 1110 0000 1110 10011010

Multiplication Table: Base 10 (12 of 14) 310 x 610 = 1810 x 1 2 3 4 5 6 7 8 9 10 12 14 16 18 15 21 24 27 20 28 32 36 25 30 35 40 45 42 48 54 49 56 63 etc.

Multiplication Table: Base 8 (13 of 14) 38 x 68 = 228 x 1 2 3 4 5 6 7 10 12 14 16 11 17 22 25 20 24 30 34 31 36 43 44 52 61

Arithmetic operations (14 of 14) Binary division, two n-bit values As with decimal values 100001/11 = 1011 11 ) 100001 quotient 1011 11 dividend 00100 divisor 11 11 11

Shifting a number Shifting a decimal number to the left by one position is equivalent to multiplying by 10 Shifting a binary number to the left by one position is equivalent to multiplying by 2 General rule: shifting a number in any base left one digit multiplies its value by the base; shifting one digit right divides its value by the base

Thank you! Reading: Lecture slides and notes, Chapter 3