CSE115Lecture01NumberSystems.pptx

Lecture 01 Number Systems CSE115: Computing Concepts Introduction to Numbering Systems Base: The number of fundamental symbols in a numbering system (e.g. 0, 1, 2 etc.) We are all familiar with the decimal number system (Base 10). Some other number systems that we will work wi...

Lecture 01 Number Systems CSE115: Computing Concepts Introduction to Numbering Systems Base: The number of fundamental symbols in a numbering system (e.g. 0, 1, 2 etc.) We are all familiar with the decimal number system (Base 10). Some other number systems that we will work with are: Binary Base 2 Octal Base 8 Hexadecimal Base 16 2  Common Number Systems Used by Used in System Base Symbols humans? computers? Decimal 10 0, 1, … 9 Yes No Binary 2 0, 1 No Yes Octal 8 0, 1, … 7 No No Hexa- 16 0, 1, … 9, No No decimal A, B, … F Decimal numbers are used by humans Binary numbers are used by computers Octal and Hexa-decimal numbers are useful to represent lon binary numbers in short form Lets do some counting (1 of 3) Hexa- Decimal Binary Octal decimal 0 0 0 0 1 1 1 1 2 10 2 2 3 11 3 3 4 100 4 4 5 101 5 5 6 110 6 6 7 111 7 7 Lets do some counting (2 of 3) Hexa- Decimal Binary Octal decimal 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F Lets do some counting (3 of 3) Hexa- Decimal Binary Octal decimal 16 10000 20 10 17 10001 21 11 18 10010 22 12 19 10011 23 13 20 10100 24 14 21 10101 25 15 22 10110 26 16 Etc. 23 10111 27 17 Bits and Bytes A single binary digit is called a bit. A collection of 8 bits is called a byte. There are 2 fundamental digits in the binary number system. One of them is 0 and the other is 1. Write the base as subscript 2510 = 110012 = 318 = 1916 Base Conversion Among Bases The possibilities: Decimal Octal Binary Hexadecimal A closer look at a decimal number Weight 12510 => 5 x 100 = 5 2 x 101 = 20 1 x 102 = 100 125 Base Binary to Decimal Decimal Octal Binary Hexadecimal Binary to Decimal 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 Example 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 Octal to Decimal Decimal Octal Binary Hexadecimal Octal to Decimal Technique 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 Example 7248 => 4 x 80 = 4 2 x 81 = 16 7 x 82 = 448 46810 Hexadecimal to Decimal Decimal Octal Binary Hexadecimal Hexadecimal to Decimal 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 Example 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 Decimal Octal Binary Hexadecimal Decimal to Binary Technique Divide by two, keep track of the remainder First remainder is bit 0 (LSB, least-significant bit) Second remainder is bit 1 Etc. Example 12510 = ?2 2 125 2 62 1 2 31 0 2 15 1 2 7 1 2 3 1 2 1 1 0 1 12510 = 11111012 Decimal to Octal Decimal Octal Binary Hexadecimal Decimal to Octal Technique Divide by 8 Keep track of the remainder Example 123410 = ?8 8 1234 8 154 2 8 19 2 8 2 3 0 2 123410 = 23228 Decimal to Hexadecimal Decimal Octal Binary Hexadecimal Decimal to Hexadecimal Technique Divide by 16 Keep track of the remainder Example 123410 = ?16 16 1234 16 77 2 16 4 13 = D 0 4 123410 = 4D216 Octal to Binary Decimal Octal Binary Hexadecimal Octal to Binary Technique Convert each octal digit to a 3-bit equivalent binary representation Example 7058 = ?2 7 0 5 111 000 101 7058 = 1110001012 Binary to Octal Decimal Octal Binary Hexadecimal Binary to Octal Technique Group bits in threes, starting on right Convert to octal digits Example 10110101112 = ?8 1 011 010 111 1 3 2 7 10110101112 = 13278 Hexadecimal to Binary Decimal Octal Binary Hexadecimal Hexadecimal to Binary Technique Convert each hexadecimal digit to a 4-bit equivalent binary representation Example 10AF16 = ?2 1 0 A F 0001 0000 1010 1111 10AF16 = 00010000101011112 Binary to Hexadecimal Decimal Octal Binary Hexadecimal Binary to Hexadecimal Technique Group bits in fours, starting on right Convert to hexadecimal digits Example 10101110112 = ?16 10 1011 1011 2 B B 10101110112 = 2BB16 Octal to Hexadecimal Decimal Octal Binary Hexadecimal Octal to Hexadecimal Technique Use binary as an intermediary Example 10768 = ?16 1 0 7 6 001 000 111 110 2 3 E 10768 = 23E16 Hexadecimal to Octal Decimal Octal Binary Hexadecimal Hexadecimal to Octal Technique Use binary as an intermediary Example 1F0C16 = ?8 1 F 0 C 0001 1111 0000 1100 1 7 4 1 4 1F0C16 = 174148 Conversion Exercise Hexa- Decimal Binary Octal decimal 33 1110101 703 1AF Try not to use a calculator! Conversion Exercise Answer Hexa- Decimal Binary Octal 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 Value 10-12 pico p.000000000001 10-9 nano n.000000001 10-6 micro .000001 10-3 milli m.001 103 kilo k 1000 106 mega M 1000000 109 giga G 1000000000 1012 tera T 1000000000000  Common Powers (2 of 2) Base 2 Power Preface Symbol Value 210 kilo k 1024 220 mega M 1048576 230 Giga G 1073741824 What is the value of “k”, “M”, and “G”? In computing, particularly w.r.t. memory, the base-2 interpretation generally applies Example In the lab… 1. Double click on My Computer 2. Right click on C: 3. Click on Properties / 230 = Complement Complement is the negative equivalent of a number. If we have a number N then complement of N will give us another number which is equivalent to –N So if complement of N is M, then we can say M = -N So complement of M = -M = -(-N) = N So complement of complement gives the original number Types of Complement For a number of base r, two types of complements can be found  1. r’s complement  2. (r-1)’s complement Definition:  If N is a number of base r having n digits then o r’s complement of N = rn – N and o (r-1)’s complement of N = rn-N-1 Example Suppose N = (3675)10 So we can find two complements of this number. The 10’s complement and the 9’s complement. Here n = 4 10’s complement of (3675) = 104 - 3675 = 6325 9’s complement of (3675) = 104 - 3675 -1 = 6324 Short cut way to find (r-1)’s complement In the previous example we see that 9’s complement of 3675 is 6324. We can get the result by subtracting each digit from 9. Similarly for other base, the (r-1)’s complement can be found by subtracting each digit from r-1 (the highest digit in that system). For binary 1’s complement is even more easy. Just change 1 to 0 and 0 to 1. (Because 1-1=0 and 1-0=1) Example Find the (r-1)’s complement in short cut method.  (620143)8 Ans: 157634  (A4D7E)16 Ans: 5B281  (110100101)2 Ans: 001011010 Short cut way to find r’s complement From the definition we can say,  r’s complement of (N) = (r-1)’s complement +1 So, we can first find the (r-1)’s complement in short cut way then add 1 to get the r’s complement. Example: r’s complement of (620143)8 =157634 +1 = 157635 This method is a two step process. But we can find it in a one step process also. Short cut way to find r’s complement One step process:  Start from rightmost digit to left.  Initial zeros will remain unchanged  Rightmost non-zero digit will be subtracted from r  Rest of the digits will be subtracted from r-1 Example:  Find the 10’s complement of (529400)10  Rightmost 2 zeros will not change, 4 will be subtracted from 10 and rest of the digits 529 will be subtracted from 9  So the result is 470600 Example Find the r’s complement in short cut method. (8210)10 Ans: 1790 (61352)10 Ans: 38648 (6201430)8 Ans: 1576350 (A4D7E0)16 Ans: 5B2820 Example for binary For binary: start from rightmost bit Up to first 1 – no change. For rest of the bits toggle (Change 1 to 0 and 0 to 1)  (11010010100)2 Ans: 00101101100  (01101001011)2 Ans: 10010110101  (10000000)2 Ans: 10000000