Number System - Final 1 - PDF

# Number System ## Rab Nawaz Khan Jadoon Lecturer COMSATS Lahore Department of Computer Science DCS COMSATS Institute of Information Technology ## Computing for Management ## Common Number Systems | System | Base | Symbols | Used by humans? | Used in computers? | |---...

# Number System ## Rab Nawaz Khan Jadoon Lecturer COMSATS Lahore Department of Computer Science DCS COMSATS Institute of Information Technology ## Computing for Management ## Common Number Systems | System | Base | Symbols | Used by humans? | Used in computers? | |--------------|------|------------------------------|-----------------|--------------------| | Decimal | 10 | 0, 1, ..., 9 | Yes | No | | Binary | 2 | 0, 1 | No | Yes | | Octal | 8 | 0, 1, ..., 7 | No | No | | Hexadecimal | 16 | 0, 1, ..., 9, A, B, ..., F | No | Yes | ## Quantities/Counting ### (1 of 3) | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 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 | ### (2 of 3) | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 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 | ### (3 of 3) | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 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 | | 23 | 10111 | 27 | 17 | | ... | ... | ... | ... | ## Conversion Among Bases - The possibilities:  ## Quick Example $25_{10} = 11001_2 = 31_8 = 19_{16}$ This is a simple conversion from Decimal to Binary, Octal, and Hexadecimal demonstrating the concept of bases. ## Decimal to Decimal (just for fun)  It's a self-conversion which is used to demonstrate base and weight concepts. ## 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** $125_{10} = ?_2$ ``` 2 | 125 2 | 62 1 2 | 31 0 2 | 15 1 2 | 7 1 2 | 3 1 2 | 1 1 ``` Therefore, $125_{10} = 1111101_2$ ## Decimal to Binary Conversion Convert (0.6875)10 to binary. | Integer | Fraction | |---------|----------| | 1 | 0.3750 | | 0 | 0.7500 | | 1 | 0.5000 | | 1 | 0.0000 | Answer: (0.6875)10 = (0.1011)2 (41.6875)10 = (101001.1011)2 ## 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" $101011_2 =>$ ``` 1 x 2° = 1 1 x 2¹ = 2 0 x 2² = 0 1 x 2³ = 8 0 x 2⁴ = 0 1 x 2⁵ = 32 ``` Therefore, $101011_2 = 43_{10}$ ## 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** $724_8 =>$ ``` 4 x 8⁰ = 4 2 x 8¹ = 16 7 x 8² = 448 ``` Therefore, $724_8 = 468_{10}$ ## 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** $ABC_{16} =>$ ``` C x 16⁰ = 12 x 1 = 12 B x 16¹ = 11 x 16 = 176 A x 16² = 10 x 256 = 2560 ``` Therefore, $ABC_{16} = 2748_{10}$ ## Octal to Binary  - **Technique** - Convert each octal digit to a 3-bit equivalent binary representation - **Example** $705_8 = ?_2$ ``` 7 0 5 ↓ ↓ ↓ 111 000 101 ``` Therefore, $705_8 = 111000101_2$ ## Hexadecimal to Binary  - **Technique** - Convert each hexadecimal digit to a 4-bit equivalent binary representation. - **Example** $10AF_{16} = ?_2$ ``` 1 0 A F ↓ ↓ ↓ ↓ 0001 0000 1010 1111 ``` Therefore, $10AF_{16} = 0001000010101111_2$ ## Decimal to Octal  - **Technique** - Divide by 8 - Keep track of the remainder - **Example** $1234_{10} = ?_8$ ``` 8 | 1234 8 | 154 2 8 | 19 2 8 | 2 3 ``` Therefore, $1234_{10} = 2322_8$ ## Decimal to Octal Conversion Convert (0.513)10 to octal. | | | |----------|----------| | 0.513 x 8 | 4.104 | | 0.104 x 8 | 0.832 | | 0.832 x 8 | 6.656 | | 0.656 x 8 | 5.248 | | 0.248 x 8 | 1.984 | | 0.984 x 8 | 7.872 | Answer: (0.513)10 = (0.406517... )8 (153.513)10 = (231.406517)8 ## Decimal to Hexadecimal  - **Technique** - Divide by 16 - Keep track of the remainder - **Example** $1234_{10} = ?_{16}$ ``` 16 | 1234 16 | 77 2 16 | 4 13 = D ``` Therefore, $1234_{10} = 4D2_{16}$. ## Binary to Octal  - **Technique** - Group bits in threes, starting on right - Convert to octal digits - **Example** $1011010111_2 = ?_8$ ``` 1 011 010 111 ↓ ↓ ↓ ↓ 1 3 2 7 ``` Therefore, $1011010111_2 = 1327_8$ ## Binary to Hexadecimal  - **Technique** - Group bits in fours, starting on right - Convert to hexadecimal digits - **Example** $1010111011_2 = ?_{16}$ ``` 10 1011 1011 ↓ ↓ ↓ 2 B B ``` Therefore, $1010111011_2 = 2BB_{16}$ ## Octal to Hexadecimal  - **Technique** - Use binary as an intermediary - **Example** $1076_8 = ?_{16}$ ``` 1 0 7 6 ↓ ↓ ↓ ↓ 001 000 111 110 ↓ ↓ ↓ ↓ 2 3 E ``` Therefore, $1076_8 = 23E_{16}$ ## Hexadecimal to Octal  - **Technique** - Use binary as an intermediary - **Example** $1F0C_{16} = ?_8$ ``` 1 F 0 C ↓ ↓ ↓ ↓ 0001 1111 0000 1100 ↓ ↓ ↓ ↓ 1 7 4 14 ``` Therefore, $1F0C_{16} = 17414_8$ ## Exercise - Convert ... | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 33 | | | | | | 1110101| | | | | | 703 | | | | | | 1AF | ## Exercise - Convert ... **Answer** | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 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 | μ | .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  This is a screenshot of a hard drive's properties, showing the used space. The example illustrates how to convert bytes to Gigabytes using the concept of Base 2 powers. ## Review - Multiplying Powers - For common bases, add powers $a^b * a^c = a^{b+c}$ $2^6 * 2^{10} = 2^{16} = 65,536$ or... $2^6 * 2^{10} = 64 * 2^{10}= 64k$ ## Binary Addition (1 of 2) - **Two 1-bit values** | A | B | A + B | |---|---|-------| | 0 | 0 | 0 | | 0 | 1 | 1 | | 1 | 0 | 1 | | 1 | 1 | 10 | "two" ## Binary Addition (2 of 2) - **Two n-bit values** - Add individual bits - Propagate carries - E.g., ``` 1 1 10101 + 11001 ----- 101110 ``` This is a binary addition example illustrating how to add two binary numbers. ## Multiplication (1 of 3) - **Decimal (just for fun)** ``` 35 x 105 ---- 175 000 35 ---- 3675 ``` This is a simple decimal multiplication to show how to multiply numbers in decimal representation. ## Multiplication (2 of 3) - **Binary, two 1-bit values** | A | B | A x B | |---|---|-------| | 0 | 0 | 0 | | 0 | 1 | 0 | | 1 | 0 | 0 | | 1 | 1 | 1 | ## Multiplication (3 of 3) - **Binary, two n-bit values** - As with decimal values -E.g., ``` 1110 x 1011 ----- 1110 1110 0000 1110 ----- 10011010 ``` This is a binary multiplication example showcasing how to multiply two binary numbers. ## Fractions - **Decimal to decimal (just for fun)** $3.14 => 4 * 10^{-2} = 0.04$ $1* 10^{-1} = 0.1$ $3 * 10^{0} = \frac{3}{3.14}$ - **Binary to decimal** $10.1011 => 1 * 2^{-4} = 0.0625$ $1 * 2^{-3} = 0.125$ $0 * 2^{-2} = 0.0$ $1 * 2^{-1} = 0.5$ $0 * 2^{0} = 0.0$ $1 * 2^{1} = 2.0$ Therefore, $10.1011 = 2.6875$ ## Decimal to binary $3.14579$ ``` x.14579 2 0.29158 2 0.58316 2 1.16632 2 0.33264 2 0.66528 2 1.33056 2 etc. ``` Therefore, $3.14579 = 11.001001...$ ## Exercise - Convert ... | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 29.8 | | | | | | 101.1101| | | | | | 3.07 | | | | | | C.82 | ## Exercise - Convert ... **Answer** | Decimal | Binary | Octal | Hexadecimal | |---------|--------|-------|-------------| | 29.8 | 11101.110011... | 35.63... | 1D.CC... | | 5.8125 | 101.1101 | 5.64 | 5.D | | 3.109375| 11.000111 | 3.07 | 3.1C | | 12.5078125| 1100.10000010 | 14.404 | C.82 | ## Thanks for the tolerance <br> This document provides an overview of different number systems and their conversion methods. It also includes exercises for practicing those conversions. This should help you understand how numbers are represented and manipulated in different bases, which is essential for working with computers and digital systems.