COMP101: Introduction to Computing - Number Systems

Course Overview and Institutional Integration

• Course Information:

  • Course Code & Title: COMP101: Introduction to Computing
  • Module: Module 2: Number System (Weeks 4 and 5)
  • Institution: Kolehiyo ng Lungsod ng Dasmariñas (KLD)
  • Academic Department: Institute of Foundational Studies
  • Address: Brgy. Burol Main, City of Dasmariñas, Cavite, Philippines 4114
  • Website: www.kld.edu.ph

• Course Outcomes & Intended Learning Outcomes:

  • Course Outcome (CO4): Determine and transform number system conversions.
  • Intended Learning Outcome (LO1): Learn the different types of number systems.

• SDG and KLD REGAL100 Integration:

  • SDG 4: Quality Education
  • SDG 8: Decent Work and Economic Growth
  • SDG 9: Industry, Innovation, and Infrastructure
  • REGAL 100: Information Literacy Program

Introduction to Number Systems

• Definition and Context:

  • A number system is a structured way to represent or express numbers.
  • Numbers represented in any specific number system category can be converted into any other system.
  • Standard mathematical number systems include whole numbers and real numbers, but computer architecture utilizes specialized digital number systems.

• Rationale for Digital Systems:

  • In digital hardware, instructions and data are transmitted using electric signals.
  • Signal variations are controlled by varying signal voltage.
  • Implementing ten distinct voltage levels to process the decimal system directly in digital equipment is technically complex and unreliable.
  • Consequently, digital systems rely on alternative positional number systems that are simpler to implement using binary voltage logic.

• Four Core Computer Number Systems:

  • Decimal Number System
  • Binary Number System
  • Octal Number System
  • Hexadecimal Number System

Characteristics of Core Number Systems

• Decimal Number System:

  • Base / Radix: $10$
  • Digits Used: $10$ digits ($0, 1, 2, 3, 4, 5, 6, 7, 8, 9$)
  • Positional weight increases to the left of the decimal point, representing units, tens, hundreds, thousands, and so on.
  • Every position corresponds to an increasing power of the base $10$.
  • Digit values are strictly position-dependent. For example, considering the value of the digit $7$ across three different numbers:
  • In $734$, the value of $7$ is $7$ hundreds = $700$ = $7 \times 100$ = $7 \times 10^2$
  • In $971$, the value of $7$ is $7$ tens = $70$ = $7 \times 10$ = $7 \times 10^1$
  • In $207$, the value of $7$ is $7$ units = $7$ = $7 \times 1$ = $7 \times 10^0$

• Binary Number System:

  • Base / Radix: $2$
  • Digits Used: $2$ digits ($0$ and $1$)
  • Represents the hardware switch states: off ($0$) and on ($1$).
  • Bit Definition: A single binary digit is called a bit.
  • Positional values are represented by increasing powers of $2$.
  • Key Terms:
  • Least Significant Bit (LSB): The rightmost digit in a binary number.
  • Most Significant Bit (MSB): The leftmost digit in a binary number.
  • Decimal Evaluation Example:
  • $11010_2 = 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$
  • $11010_2 = 16 + 8 + 0 + 2 + 0 = 26_{10}$

• Computer Memory Capacity Conversion Table:

  • $1\text{ byte (B)} = 8\text{ bits}$
  • $1\text{ Kilobyte (KB)} = 1024\text{ bytes}$
  • $1\text{ Megabyte (MB)} = 1024\text{ KB}$
  • $1\text{ Gigabyte (GB)} = 1024\text{ MB}$
  • $1\text{ Terabyte (TB)} = 1024\text{ GB}$
  • $1\text{ Exabyte (EB)} = 1024\text{ PB}$
  • $1\text{ Zettabyte} = 1024\text{ EB}$
  • $1\text{ Yottabyte (YB)} = 1024\text{ ZB}$

• Octal Number System:

  • Base / Radix: $8$
  • Digits Used: $8$ digits ($0, 1, 2, 3, 4, 5, 6, 7$)
  • Widely used in computer applications and electronic systems.

• Hexadecimal Number System:

  • Base / Radix: $16$
  • Digits/Symbols Used: $16$ symbols ($0, 1, 2, 3, 4, 5, 6, 7, 8, 9, \text{A}, \text{B}, \text{C}, \text{D}, \text{E}, \text{F}$)
  • Each position represents a power of $16$.
  • Because the base exceeds $10$, standard numeric digits are supplemented by alphabetic letters:
  • $\text{A} = 10$
  • $\text{B} = 11$
  • $\text{C} = 12$
  • $\text{D} = 13$
  • $\text{E} = 14$
  • $\text{F} = 15$
  • The inclusion of letters $\text{A}$ through $\text{F}$ is purely conventional.

Decimal Conversion Methods

• Decimal to Binary Conversion:

  • Procedure: Repeatedly divide the decimal integer by $2$, record the remainder at each step, and continue until the quotient becomes $0$. Write the remainders in backward order (from bottom to top, LSB to MSB).
  • Example: Convert $25_{10}$ to binary.
  • $25 / 2 = 12$ with remainder $1$
  • $12 / 2 = 6$ with remainder $0$
  • $6 / 2 = 3$ with remainder $0$
  • $3 / 2 = 1$ with remainder $1$
  • $1 / 2 = 0$ with remainder $1$
  • Reading remainders backward yields: $25_{10} = 11001_2$

• Decimal to Octal Conversion:

  • Procedure: Repeatedly divide the decimal integer by $8$, record the remainder at each step, and continue until the quotient is $0$. Write remainders in backward order.
  • Example: Convert $128_{10}$ to octal.
  • $128 / 8 = 16$ with remainder $0$
  • $16 / 8 = 2$ with remainder $0$
  • $2 / 8 = 0$ with remainder $2$
  • Reading remainders backward yields: $128_{10} = 200_8$

• Octal to Decimal Conversion:

  • Procedure: Multiply each digit of the octal number by decreasing powers of base $8$, starting from MSB to LSB, and sum all products together.
  • Example 1: Convert $22_8$ to decimal.
  • $22_8 = 2 \times 8^1 + 2 \times 8^0 = 16 + 2 = 18_{10}$
  • Example 2: Convert $215_8$ to decimal.
  • $215_8 = 2 \times 8^2 + 1 \times 8^1 + 5 \times 8^0$
  • $215_8 = 2 \times 64 + 1 \times 8 + 5 \times 1 = 128 + 8 + 5 = 141_{10}$

• Decimal to Hexadecimal Conversion:

  • Procedure: Repeatedly divide the decimal integer by $16$, record the remainder, and write the remainders backward when quotient reaches 0$.\n * **Example:** Convert 128_{10} to hexadecimal.\n * 128 / 16 = 8$with remainder$0\n * 8 / 16 = 0$with remainder$8\n * Reading remainders backward yields: 128_{10} = 80_{16}\n\n* **Hexadecimal to Decimal Conversion:**\n * **Procedure:** Multiply each hexadecimal digit by powers of 16$starting from the units place ($16^0) at the LSB and increase powers moving left to the MSB, then sum the terms.\n * **Example:** Convert 7 ext{CF}{16} to decimal.\n * Hexadecimal symbol values: 7 = 7$,$ ext{C} = 12$,$ ext{F} = 15\n * 7 ext{CF}{16} = (7 imes 16^2) + (12 imes 16^1) + (15 imes 16^0)\n * 7 ext{CF}{16} = (7 imes 256) + (12 imes 16) + (15 imes 1)\n * 7 ext{CF}{16} = 1792 + 192 + 15 = 1999_{10}\n\n# Binary Conversion Methods\n\n* **Binary to Decimal Conversion:**\n * **Procedure:** Multiply each binary digit starting from MSB to LSB by decreasing powers of the base 2, then sum all terms.\n * **Example:** Convert 1101_2 to decimal.\n * 1101_2 = 1 imes 2^3 + 1 imes 2^2 + 0 imes 2^1 + 1 imes 2^0\n * 1101_2 = 8 + 4 + 0 + 1 = 13_{10}\n\n* **Binary to Octal Conversion Methods:**\n * Direct one-step manual conversion is not typically used; two main systematic methods exist.\n * **Method 1 (Indirect via Decimal):**\n * Step 1: Identify the binary number.\n * Step 2: Convert binary to decimal by multiplying each digit by 2^{n-1}$(where$n is digit position from the right).\n * Step 3: Compute the equivalent decimal sum.\n * Step 4: Divide the resulting decimal number by 8$.
  • Step 5: Note the remainder.
  • Step 6: Repeat division with the quotient until quotient equals 0$.\n * Step 7: Write the remainders in reverse order.\n * Step 8: The resulting digits form the target octal number.\n * **Example:** Convert 1011101_2 to octal.\n * Binary to Decimal: 1011101_2 = (1 imes 2^6) + (0 imes 2^5) + (1 imes 2^4) + (1 imes 2^3) + (1 imes 2^2) + (0 imes 2^1) + (1 imes 2^0)\n * 1011101_2 = 64 + 0 + 16 + 8 + 4 + 0 + 1 = 93_{10}\n * Decimal to Octal (Divide 93$by$8):\n * 93 / 8 = 11$remainder$5\n * 11 / 8 = 1$remainder$3\n * 1 / 8 = 0$remainder$1\n * Collecting remainders in reverse order gives: 135_8\n * Therefore, 1011101_2 = 135_8\n * **Method 2 (Direct Grouping):**\n * Step 1: Identify the binary number (base 2$digits$0$or$1).\n * Step 2: Group binary digits into sets of three starting from the right side.\n * Step 3: Add leading zeros to the left if the leftmost group contains fewer than three digits.\n * Step 4: Map each 3-bit group to its octal equivalent digit.\n * Step 5: Concatenate the resulting octal digits.\n * **Example:** Convert 01110101_2 to octal.\n * Grouping into 3-bit sets: 001 / 110 / 101\n * Converting groups: 001_2 = 1$,$110_2 = 6$,$101_2 = 5\n * Output octal value: 165_8\n\n# Octal to Binary Conversion\n\n* **Procedural Steps (Two-Step Process):**\n * **Step 1: Octal to Decimal Conversion**\n * Count the total number of digits (n).\n * Multiply each digit in the n^ ext{th}$position from the right by$8^{n-1}.\n * If a fractional component exists, multiply fractional digits in the m^ ext{th}$position after the decimal point by$8^{-m}.\n * Sum all products to obtain the equivalent decimal number.\n * **Step 2: Decimal to Binary Conversion**\n * Divide the resulting decimal number by 2$.
  • Note down the remainder.
  • Continue dividing the quotient by $2$ until the quotient reaches 0$.\n * Write remainders in reverse order to produce the equivalent binary number.\n\n* **Demonstration Example:** Convert 41_8 to binary.\n * Step 1 (Octal to Decimal):\n * 41_8 = (4 imes 8^1) + (1 imes 8^0) = 4 imes 8 + 1 imes 1 = 32 + 1 = 33_{10}\n * Step 2 (Decimal to Binary):\n * 33 / 2 = 16$remainder$1\n * 16 / 2 = 8$remainder$0\n * 8 / 2 = 4$remainder$0\n * 4 / 2 = 2$remainder$0\n * 2 / 2 = 1$remainder$0\n * 1 / 2 = 0$remainder$1\n * Writing remainders backward yields: 100001_2\n * Therefore, 41_8 = 100001_2\n\n# Binary to Hexadecimal Conversion\n\n* **Purpose:** Provides a compact way to group and represent large binary strings.\n\n* **Indirect Method (Binary to Decimal to Hexadecimal):**\n * **Example:** Convert 1101010_2 to hexadecimal.\n * Convert Binary to Decimal:\n * 1101010_2 = 1 imes 2^6 + 1 imes 2^5 + 0 imes 2^4 + 1 imes 2^3 + 0 imes 2^2 + 1 imes 2^1 + 0 imes 2^0\n * 1101010_2 = 64 + 32 + 0 + 8 + 0 + 2 + 0 = 106_{10}\n * Convert Decimal to Hexadecimal:\n * 106_{10} = 6 imes 16^1 + 10 imes 16^0 = 6 ext{A}{16}\n * Answer: 6 ext{A}{16}\n\n* **Direct Grouping Method:**\n * Because base 16$consists of$16$symbols ($0-9$and$ ext{A}- ext{F}$), every single hexadecimal digit maps directly to a unique 4-bit binary sequence ($2^4 = 16).\n * **Algorithm Steps:**\n * Group binary digits into sets of four.\n * For integer parts, start grouping from right to left.\n * For fractional parts, start grouping from left to right.\n * Pad with leading 0$s on the far left (MSB) for integers or trailing$0s on the far right (LSB) for fractions to complete any incomplete 4-bit set.\n * Replace each 4-bit set with its corresponding hexadecimal digit.\n * **Example 1 (Integer Number):** Convert 1010101101001_2 to hexadecimal.\n * Grouping from right: 1 imes 0101 imes 0110 imes 1001\n * Add leading zeros to complete 4-bit groups: 0001 imes 0101 imes 0110 imes 1001\n * Map each group: 0001_2 = 1$,$0101_2 = 5$,$0110_2 = 6$,$1001_2 = 9\n * Result: 1569_{16}\n * **Example 2 (Fractional Number):** Convert 001100101.110111_2 to hexadecimal.\n * Integer grouping (right to left from point): 001100101 ightarrow 0 imes 0110 imes 0101 ightarrow 0110 imes 0101 = 6 imes 5\n * Fractional grouping (left to right from point): .110111 ightarrow .1101 imes 1100 = ext{D} imes ext{C}\n * Result: 65. ext{DC}_{16}\n\n# Hexadecimal to Binary Conversion\n\n* **Indirect Conversion Steps (Hexadecimal to Decimal to Binary):**\n * Step 1: Take the given hexadecimal number.\n * Step 2: Determine the total number of digits (n).\n * Step 3: Multiply each digit by 16^{n-1}$based on its positional index$n from the right.\n * Step 4: Sum all multiplied products together.\n * Step 5: Obtain the equivalent decimal integer.\n * Step 6: Divide the decimal number by 2$.
  • Step 7: Record the remainder.
  • Step 8: Repeat division with the quotient until quotient equals 0$.\n * Step 9: Write remainders in reverse order.\n * Step 10: Output the derived binary sequence.\n\n# Inter-System Conversions (Octal and Hexadecimal)\n\n* **Octal to Hexadecimal Conversion Methods:**\n * Direct single-step mapping between octal and hexadecimal is unavailable; two indirect pathways are applied.\n * **Method A (Via Binary Representation):**\n * Step 1: Express every individual octal digit as a 3-bit binary pattern.\n * Step 2: Regroup the continuous binary pattern into 4-bit groups (from right to left).\n * Step 3: Convert each 4-bit binary group into its corresponding hexadecimal digit.\n * Reference values: 5_8 = 101_2$,$6_8 = 110_2\n * **Example:** Convert 56_8 to hexadecimal.\n * Step 1: Convert 56_8$to binary:$5 ightarrow 101$,$6 ightarrow 110 ightarrow 101110_2\n * Step 2: Regroup into 4-bit sets from right: 10 imes 1110 ightarrow 0010 imes 1110\n * Step 3: Map to hex: 0010_2 = 2$,$1110_2 = 14 = ext{E}\n * Answer: 2 ext{E}_{16}\n * **Method B (Via Decimal Representation):**\n * Step 1: Convert octal to decimal.\n * Step 2: Convert decimal to hexadecimal via repeated division by 16$.
  • Example: Convert $56_8$ to hexadecimal.
    • Step 1 (Octal to Decimal): $5 \times 8^1 + 6 \times 8^0 = 40 + 6 = 46_{10}$
    • Step 2 (Decimal to Hexadecimal):
    • $46 / 16 = 2$ remainder $14$ ($\text{E}$)
    • $2 / 16 = 0$ remainder $2$
    • Reading remainders backward yields: $2\text{E}_{16}$

• Hexadecimal to Octal Conversion:

  • Procedure: Cannot be converted directly. Convert the hexadecimal number to its equivalent decimal number, then convert the decimal number to octal.
  • Step-by-Step Algorithm:
  • Step 1: Take the hexadecimal input and count its digits.
  • Step 2: Multiply each digit in position $n$ from the right by $16^{n-1}$.
  • Step 3: Sum all terms to compute the equivalent decimal number.
  • Step 4: Divide the resulting decimal number by 8$.\n * Step 5: Note down the remainder.\n * Step 6: Repeat division with the quotient until quotient equals 0$.
  • Step 7: Write all remainders in reverse order.
  • Step 8: The resulting sequence is the equivalent octal number.