Exhaustive Study Notes on Number Systems, Base Conversions, and Binary Arithmetic

Overview of Number Systems & Hardware Logic

• Primary Objectives:

  • Understand the principles and foundational mathematics of number systems.
  • Understand why computer systems operate using the binary (Base-2) system.
  • Master positional notation across Base-2, Base-8, Base-10, and Base-16.
  • Master conversion methodologies among all four bases.

• Hardware Rationale for Binary Usage:

  • Modern computers are built from electronic switches.
  • Each physical switch operates in one of two discrete states: ON or OFF.
  • Numerical mapping of switch states:
  • Logic $1$ represents the ON state.
  • Logic $0$ represents the OFF state.
  • Electrical voltage levels in digital logic circuits:
  • Logic $0$ corresponds to $0\,\text{V}$.
  • Logic $1$ corresponds to either $+3.3\,\text{V}$ or $+5\,\text{V}$.

Taxonomy of Common Number Systems

• Characteristics of Standard Bases:

  • Binary: Base $2$, designated by symbol B. Uses digits $\{0, 1\}$. Utilized directly by computer hardware, not typically used by humans.
  • Octal: Base $8$, designated by symbol O. Uses digits $\{0, 1, 2, 3, 4, 5, 6, 7\}$. Serves as a human-readable shorthand representation for binary; not used directly by computers or in general human communication.
  • Decimal: Base $10$, designated by symbol D. Uses digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Used universally by humans, not used natively by computer hardware.
  • Hexadecimal: Base $16$, designated by symbol H. Uses digits $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F\}$, where letters represent values $10$ through $15$ ($A=10$, $B=11$, $C=12$, $D=13$, $E=14$, $F=15$). Serves as a compact human-readable shorthand for binary strings; not native to computer hardware.

• Counting Equivalence Across Number Systems ($0$ to $23$ Decimal):

  • Decimal $0_{10}$ = Binary $0_2$ = Octal $0_8$ = Hexadecimal $0_{16}$
  • Decimal $1_{10}$ = Binary $1_2$ = Octal $1_8$ = Hexadecimal $1_{16}$
  • Decimal $2_{10}$ = Binary $10_2$ = Octal $2_8$ = Hexadecimal $2_{16}$
  • Decimal $3_{10}$ = Binary $11_2$ = Octal $3_8$ = Hexadecimal $3_{16}$
  • Decimal $4_{10}$ = Binary $100_2$ = Octal $4_8$ = Hexadecimal $4_{16}$
  • Decimal $5_{10}$ = Binary $101_2$ = Octal $5_8$ = Hexadecimal $5_{16}$
  • Decimal $6_{10}$ = Binary $110_2$ = Octal $6_8$ = Hexadecimal $6_{16}$
  • Decimal $7_{10}$ = Binary $111_2$ = Octal $7_8$ = Hexadecimal $7_{16}$
  • Decimal $8_{10}$ = Binary $1000_2$ = Octal $10_8$ = Hexadecimal $8_{16}$
  • Decimal $9_{10}$ = Binary $1001_2$ = Octal $11_8$ = Hexadecimal $9_{16}$
  • Decimal $10_{10}$ = Binary $1010_2$ = Octal $12_8$ = Hexadecimal $A_{16}$
  • Decimal $11_{10}$ = Binary $1011_2$ = Octal $13_8$ = Hexadecimal $B_{16}$
  • Decimal $12_{10}$ = Binary $1100_2$ = Octal $14_8$ = Hexadecimal $C_{16}$
  • Decimal $13_{10}$ = Binary $1101_2$ = Octal $15_8$ = Hexadecimal $D_{16}$
  • Decimal $14_{10}$ = Binary $1110_2$ = Octal $16_8$ = Hexadecimal $E_{16}$
  • Decimal $15_{10}$ = Binary $1111_2$ = Octal $17_8$ = Hexadecimal $F_{16}$
  • Decimal $16_{10}$ = Binary $10000_2$ = Octal $20_8$ = Hexadecimal $10_{16}$
  • Decimal $17_{10}$ = Binary $10001_2$ = Octal $21_8$ = Hexadecimal $11_{16}$
  • Decimal $18_{10}$ = Binary $10010_2$ = Octal $22_8$ = Hexadecimal $12_{16}$
  • Decimal $19_{10}$ = Binary $10011_2$ = Octal $23_8$ = Hexadecimal $13_{16}$
  • Decimal $20_{10}$ = Binary $10100_2$ = Octal $24_8$ = Hexadecimal $14_{16}$
  • Decimal $21_{10}$ = Binary $10101_2$ = Octal $25_8$ = Hexadecimal $15_{16}$
  • Decimal $22_{10}$ = Binary $10110_2$ = Octal $26_8$ = Hexadecimal $16_{16}$
  • Decimal $23_{10}$ = Binary $10111_2$ = Octal $27_8$ = Hexadecimal $17_{16}$

Positional Weight & Base Mechanics

• Positional Weight Structure (Base-10 Model):

  • Digits located to the left of the radix (decimal) point represent values between zero and nine multiplied by increasing non-negative powers of ten ($10^0, 10^1, 10^2, 10^3, \dots$).
  • Digits located to the right of the radix point represent values between zero and nine multiplied by increasing negative powers of ten ($10^{-1}, 10^{-2}, 10^{-3}, \dots$).
  • Place values and positional weights:
  • Thousands: $10^3 = 1000$
  • Hundreds: $10^2 = 100$
  • Tens: $10^1 = 10$
  • Ones: $10^0 = 1$
  • Tenths: $10^{-1} = 0.1$
  • Hundredths: $10^{-2} = 0.01$
  • Thousandths: $10^{-3} = 0.001$

• Fully Expanded Decimal Examples:

  • Example 1 ($725.194_{10}$):
  • $7 \times 10^2 + 2 \times 10^1 + 5 \times 10^0 + 1 \times 10^{-1} + 9 \times 10^{-2} + 4 \times 10^{-3}$
  • $= 7 \times 100 + 2 \times 10 + 5 \times 1 + 1 \times 0.1 + 9 \times 0.01 + 4 \times 0.001$
  • $= 700 + 20 + 5 + 0.1 + 0.09 + 0.004 = 725.194_{10}$
  • Example 2 ($125_{10}$):
  • $1 \times 10^2 + 2 \times 10^1 + 5 \times 10^0 = 100 + 20 + 5 = 125_{10}$

Detailed Base Conversions & Worked Examples

• Binary to Decimal Conversion:

  • Method: Multiply each binary digit by $2^n$, where $n$ is the position weight starting from $0$ on the rightmost bit, and calculate the sum of all terms.
  • Example 1 ($10011_2$):
  • Term 0: $1 \times 2^0 = 1$
  • Term 1: $1 \times 2^1 = 2$
  • Term 2: $0 \times 2^2 = 0$
  • Term 3: $0 \times 2^3 = 0$
  • Term 4: $1 \times 2^4 = 16$
  • Sum: $16 + 0 + 0 + 2 + 1 = 19_{10}$
  • Example 2 ($101011_2$):
  • Bit 0 (weight $0$): $1 \times 2^0 = 1$
  • Bit 1 (weight $1$): $1 \times 2^1 = 2$
  • Bit 2 (weight $2$): $0 \times 2^2 = 0$
  • Bit 3 (weight $3$): $1 \times 2^3 = 8$
  • Bit 4 (weight $4$): $0 \times 2^4 = 0$
  • Bit 5 (weight $5$): $1 \times 2^5 = 32$
  • Sum: $32 + 0 + 8 + 0 + 2 + 1 = 43_{10}$
  • Example 3 ($11001010_2$):
  • Expansion: $1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0$
  • Sum: $128 + 64 + 0 + 0 + 8 + 0 + 2 + 0 = 202_{10}$

• Octal to Decimal Conversion:

  • Method: Multiply each octal digit by $8^n$, where $n$ is the position weight starting from $0$ on the rightmost digit, and add all term values.
  • Positional Placeholders for Octal System:
  • For an octal value like $242_8$:
    • "Sixty-Fours" position ($8^2$): $8^2 \times 2 = 64 \times 2 = 128$
    • "Eights" position ($8^1$): $8^1 \times 4 = 8 \times 4 = 32$
    • "Ones" position ($8^0$): $8^0 \times 2 = 1 \times 2 = 2$
  • Example ($724_8$):
  • Digit 4 (weight $0$): $4 \times 8^0 = 4 \times 1 = 4$
  • Digit 2 (weight $1$): $2 \times 8^1 = 2 \times 8 = 16$
  • Digit 7 (weight $2$): $7 \times 8^2 = 7 \times 64 = 448$
  • Sum: $448 + 16 + 4 = 468_{10}$

• Hexadecimal to Decimal Conversion:

  • Method: Multiply each hexadecimal digit by $16^n$, where $n$ is the position weight starting from $0$ on the rightmost digit, and sum the results. Replace letter symbols with decimal values ($A=10, B=11, C=12, D=13, E=14, F=15$).
  • Example ($ABC_{16}$):
  • Digit $C$ ($12_{10}$, weight $0$): $C \times 16^0 = 12 \times 1 = 12$
  • Digit $B$ ($11_{10}$, weight $1$): $B \times 16^1 = 11 \times 16 = 176$
  • Digit $A$ ($10_{10}$, weight $2$): $A \times 16^2 = 10 \times 256 = 2560$
  • Sum: $2560 + 176 + 12 = 2748_{10}$

• Decimal to Binary Conversion:

  • Method (Repeated Division by $2$):
  • Divide the decimal integer by 2$.\n * Record the remainder (0$or$1$) on the right. The first remainder generated is bit$0 (Least Significant Bit - LSB).\n * Divide the resulting quotient by 2 again, recording the remainder.\n * Repeat until the quotient reaches 0$.
  • Write the binary string starting from the last remainder generated (Most Significant Bit - MSB) on the left to the first remainder (LSB) on the right.
  • Example 1 (Convert $125_{10}$ to binary):
  • $125 \div 2 = 62\text{ remainder } 1$ (LSB)
  • $62 \div 2 = 31\text{ remainder } 0$
  • $31 \div 2 = 15\text{ remainder } 1$
  • $15 \div 2 = 7\text{ remainder } 1$
  • $7 \div 2 = 3\text{ remainder } 1$
  • $3 \div 2 = 1\text{ remainder } 1$
  • $1 \div 2 = 0\text{ remainder } 1$ (MSB)
  • Result: $125_{10} = 1111101_2$
  • Example 2 (Convert $333_{10}$ to binary):
  • $333 \div 2 = 166\text{ remainder } 1$ (LSB)
  • $166 \div 2 = 83\text{ remainder } 0$
  • $83 \div 2 = 41\text{ remainder } 1$
  • $41 \div 2 = 20\text{ remainder } 1$
  • $20 \div 2 = 10\text{ remainder } 0$
  • $10 \div 2 = 5\text{ remainder } 0$
  • $5 \div 2 = 2\text{ remainder } 1$
  • $2 \div 2 = 1\text{ remainder } 0$
  • $1 \div 2 = 0\text{ remainder } 1$ (MSB)
  • Result: $333_{10} = 101001101_2$

• Octal to Binary Conversion:

  • Method: Convert each octal digit directly into its equivalent 3-bit binary group.
  • Example (Convert $705_8$ to binary):
  • Digit $7_8 \rightarrow 111_2$
  • Digit $0_8 \rightarrow 000_2$
  • Digit $5_8 \rightarrow 101_2$
  • Result: $705_8 = 111000101_2$

• Hexadecimal to Binary Conversion:

  • Method: Convert each hexadecimal digit directly into its equivalent 4-bit binary group.
  • Example (Convert $10AF_{16}$ to binary):
  • Digit $1_{16} \rightarrow 0001_2$
  • Digit $0_{16} \rightarrow 0000_2$
  • Digit $A_{16} \rightarrow 1010_2$
  • Digit $F_{16} \rightarrow 1111_2$
  • Result: $10AF_{16} = 0001000010101111_2$

• Decimal to Octal Conversion:

  • Method (Repeated Division by $8$):
  • Divide the decimal integer by $8$ and record the remainder.
  • Continue dividing quotients by $8$ until the quotient is 0$.\n * Read remainders from bottom (MSB) to top (LSB).\n * Example 1 (Convert 1234_{10} to octal):\n * 1234 \div 8 = 154\text{ remainder } 2\n * 154 \div 8 = 19\text{ remainder } 2\n * 19 \div 8 = 2\text{ remainder } 3\n * 2 \div 8 = 0\text{ remainder } 2\n * Result: 1234_{10} = 2322_8\n * Example 2 (Convert 44978_{10} to octal):\n * 44978 \div 8 = 5622\text{ remainder } 2\n * 5622 \div 8 = 702\text{ remainder } 6\n * 702 \div 8 = 87\text{ remainder } 6\n * 87 \div 8 = 10\text{ remainder } 7\n * 10 \div 8 = 1\text{ remainder } 2\n * 1 \div 8 = 0\text{ remainder } 1\n * Result: 44978_{10} = 127662_8\n\n* Decimal to Hexadecimal Conversion:\n * Method (Repeated Division by 16):\n * Divide the decimal number by 16 and record the remainder.\n * Replace remainders 10$through$15$with corresponding symbols$A$through$F$.
  • Repeat until quotient is $0$. Read remainders from bottom to top.
  • Example (Convert $1234_{10}$ to hexadecimal):
  • $1234 \div 16 = 77\text{ remainder } 2$
  • $77 \div 16 = 4\text{ remainder } 13\text{ (symbol } D\text{)}$
  • $4 \div 16 = 0\text{ remainder } 4$
  • Result: $1234_{10} = 4D2_{16}$

• Binary to Octal Conversion:

  • Method: Group binary bits into sets of three starting from the rightmost bit (LSB). Convert each 3-bit group to its octal digit.
  • Example (Convert $1011010111_2$ to octal):
  • Bit grouping from right: $1 \, 011 \, 010 \, 111$
  • Group conversions:
    • $111_2 = 7_8$
    • $010_2 = 2_8$
    • $011_2 = 3_8$
    • $1_2 = 1_8$
  • Result: $1011010111_2 = 1327_8$

• Binary to Hexadecimal Conversion:

  • Method: Group binary bits into sets of four starting from the rightmost bit (LSB). Convert each 4-bit group to its hexadecimal digit.
  • Example (Convert $1010111011_2$ to hexadecimal):
  • Bit grouping from right: $10 \, 1011 \, 1011$
  • Group conversions:
    • $1011_2 = B_{16}$ ($11_{10}$)
    • $1011_2 = B_{16}$ ($11_{10}$)
    • $10_2 = 2_{16}$
  • Result: $1010111011_2 = 2BB_{16}$

• Octal to Hexadecimal Conversion:

  • Method: Use binary as an intermediary base:
  1. Convert octal digits into 3-bit binary groups.
  2. Regroup the resulting binary string into 4-bit groups starting from the right.
  3. Convert each 4-bit group into its hexadecimal digit.
  • Example (Convert $1076_8$ to hexadecimal):
  • Step 1 (Octal to 3-bit Binary):
    • $1_8 \rightarrow 001_2$
    • $0_8 \rightarrow 000_2$
    • $7_8 \rightarrow 111_2$
    • $6_8 \rightarrow 110_2$
    • Combined binary: $001000111110_2$
  • Step 2 (Regroup into 4-bit groups):
    • $0010 \, 0011 \, 1110$
  • Step 3 (4-bit Binary to Hex):
    • $1110_2 = E_{16}$
    • $0011_2 = 3_{16}$
    • $0010_2 = 2_{16}$
  • Result: $1076_8 = 23E_{16}$

• Hexadecimal to Octal Conversion:

  • Method: Use binary as an intermediary base:
  1. Convert hexadecimal digits into 4-bit binary groups.
  2. Regroup the resulting binary string into 3-bit groups starting from the right.
  3. Convert each 3-bit group into its octal digit.
  • Example (Convert $1F0C_{16}$ to octal):
  • Step 1 (Hex to 4-bit Binary):
    • $1_{16} \rightarrow 0001_2$
    • $F_{16} \rightarrow 1111_2$
    • $0_{16} \rightarrow 0000_2$
    • $C_{16} \rightarrow 1100_2$
    • Combined binary: $0001111100001100_2$
  • Step 2 (Regroup into 3-bit groups from right):
    • $001 \, 111 \, 100 \, 001 \, 100$
  • Step 3 (3-bit Binary to Octal):
    • $100_2 = 4_8$
    • $001_2 = 1_8$
    • $100_2 = 4_8$
    • $111_2 = 7_8$
    • $001_2 = 1_8$
  • Result: $1F0C_{16} = 17414_8$

• Comprehensive Conversion Reference Table:

  • Decimal $25_{10}$ | Binary $11001_2$ | Octal $31_8$ | Hexadecimal $19_{16}$
  • Decimal $33_{10}$ | Binary $100001_2$ | Octal $41_8$ | Hexadecimal $21_{10}$
  • Decimal $117_{10}$ | Binary $1110101_2$ | Octal $165_8$ | Hexadecimal $75_{16}$
  • Decimal $451_{10}$ | Binary $111000011_2$ | Octal $703_8$ | Hexadecimal $1C3_{16}$
  • Decimal $431_{10}$ | Binary $110101111_2$ | Octal $657_8$ | Hexadecimal $1AF_{16}$

Binary Arithmetic Operations

• Binary Addition Fundamentals:

  • Basic 1-bit Addition Rules:
  • $0 + 0 = 0$
  • $0 + 1 = 1$
  • $1 + 0 = 1$
  • $1 + 1 = 10_2$ ($0$ sum with a carry of $1$ to the next higher bit position)
  • Carry Comparison with Decimal:
  • Decimal (Base-10): $9 + 1 = 0\text{ carry } 1$
  • Binary (Base-2): $1 + 1 = 0\text{ carry } 1$
  • Procedure for Multi-bit ($n$-bit) Binary Addition:
  • Align numbers by position starting at LSB.
  • Add bits column by column.
  • Whenever a column sum is $2$ ($1+1$), record $0$ in that column and propagate a carry of $1$ to the left column.
  • Whenever a column sum is $3$ ($1+1+1$ from carry), record $1$ in that column and propagate a carry of $1$ to the left column.
  • Worked Example ($10101_2 + 11001_2$):
  • Value 1: $10101_2 = 21_{10}$
  • Value 2: $11001_2 = 25_{10}$
  • Column addition steps:
    • Column 0 (LSB): $1 + 1 = 0$, carry $1$
    • Column 1: $0 + 0 + 1 (\text{carry}) = 1$
    • Column 2: $1 + 0 = 1$
    • Column 3: $0 + 1 = 1$
    • Column 4: $1 + 1 = 0$, carry $1$
    • Column 5: $1$
  • Sum: $101110_2 = 46_{10}$

• Practice Arithmetic Exercises:

  • Addition Problems:
  1. $101_2 + 11_2 = 1000_2$
  2. $111_2 + 111_2 = 1110_2$
  3. $1010_2 + 1010_2 = 10100_2$
  4. $11101_2 + 1010_2 = 100111_2$
  5. $11111_2 + 11111_2 = 111110_2$
  • Subtraction Problems:
  1. $110_2 - 10_2 = 100_2$
  2. $101_2 - 11_2 = 10_2$
  3. $1001_2 - 11_2 = 110_2$
  4. $1101_2 - 11_2 = 1010_2$
  5. $10001_2 - 100_2 = 1101_2$

• Binary Multiplication Mechanics:

  • Decimal Comparison ($35 \times 105$):
  • $35 \times 5 = 175$
  • $35 \times 0 = 000$
  • $35 \times 100 = 3500$
  • Total Sum: $175 + 000 + 3500 = 3675$
  • Single-bit Binary Multiplication Rules:
  • $0 \times 0 = 0$
  • $0 \times 1 = 0$
  • $1 \times 0 = 0$
  • $1 \times 1 = 1$
  • Multi-bit Binary Multiplication Steps:
  • Multiply multiplicand by each bit of the multiplier to form partial products.
  • Shift each successive partial product one bit to the left.
  • Add all partial products together using binary addition.
  • Worked Example ($1110_2 \times 1011_2$):
  • Multiplicand: $1110_2$ ($14_{10}$)
  • Multiplier: $1011_2$ ($11_{10}$)
  • Partial Products:
    • Bit 0 ($1$): $1110$
    • Bit 1 ($1$): $11100$
    • Bit 2 ($0$): $000000$
    • Bit 3 ($1$): $1110000$
  • Summing Partial Products:
    • $1110_2 + 11100_2 = 101010_2$
    • $101010_2 + 1110000_2 = 10011010_2$ ($154_{10}$)

Fractional Number Representation & Conversions

• Fractional Expansion in Decimal:

  • Example ($3.14_{10}$):
  • $3 \times 10^0 = 3$
  • $1 \times 10^{-1} = 0.1$
  • $4 \times 10^{-2} = 0.04$
  • Sum: $3 + 0.1 + 0.04 = 3.14_{10}$

• Binary Fractional Conversion to Decimal:

  • Method:
  • For digits left of binary point, multiply by positive powers of two ($2^0, 2^1, 2^2, \dots$).
  • For digits right of binary point, multiply by negative powers of two ($2^{-1}, 2^{-2}, 2^{-3}, 2^{-4}, \dots$).
  • Negative powers of two equivalents:
    • $2^{-1} = \frac{1}{2} = 0.5$
    • $2^{-2} = \frac{1}{4} = 0.25$
    • $2^{-3} = \frac{1}{8} = 0.125$
    • $2^{-4} = \frac{1}{16} = 0.0625$
  • Worked Example (Convert $10.1011_2$ to decimal):
  • Integer Part ($10_2$):
    • $1 \times 2^1 = 2.0$
    • $0 \times 2^0 = 0.0$
  • Fractional Part ($.1011_2$):
    • $1 \times 2^{-1} = 0.5$
    • $0 \times 2^{-2} = 0.0$
    • $1 \times 2^{-3} = 0.125$
    • $1 \times 2^{-4} = 0.0625$
  • Sum of all terms:
    • $2.0 + 0.0 + 0.5 + 0.0 + 0.125 + 0.0625 = 2.6875_{10}$