Binary Numeral System and Base Conversion Principles

Overview of Numeral Systems and Decimal Fundamentals

• The decimal numeral system is a base-$10$ positional system where each digit represents a multiplier for a power of $10$.

• In a base-$10$ system, valid digits range from $0$ through $9$.

• Position values correspond to powers of $10$ starting from $10^0$ at the rightmost position ($\text{ones place}$), $10^1$ ($\text{tens place}$), $10^2$ ($\text{hundreds place}$), and continuing to higher powers.

• Detailed breakdown of decimal number $137_{10}$:

  • Digit $7$ occupies the $10^0$ position ($10^0 = 1$), representing $7 \times 10^0 = 7$

  • Digit $3$ occupies the $10^1$ position ($10^1 = 10$), representing $3 \times 10^1 = 30$

  • Digit $1$ occupies the $10^2$ position ($10^2 = 100$), representing $1 \times 10^2 = 100$

  • Polynomial expansion:     $137_{10} = (1 \times 10^2) + (3 \times 10^1) + (7 \times 10^0)$

  • Summation calculation:     $100 + 30 + 7 = 137$

The Binary Numeral System (Base 2)

• The binary system is a base-$2$ positional numeral system where all numbers are expressed as sums of powers of $2$.

• Valid digits in the binary system are strictly $0$ and $1$

• Binary corresponds directly to electronic digital systems and computer states:

  • State $1$ represents an "on" state

  • State $0$ represents an "off" state

• Subscript notation is used to distinguish binary numbers from decimal numbers, appending a subscript $2$ (for example, $1101_2$).

• Positional power values for base $2$ from right to left:

  • First position (rightmost): $2^0 = 1$

  • Second position: $2^1 = 2$

  • Third position: $2^2 = 4$

  • Fourth position: $2^3 = 8$

  • Fifth position: $2^4 = 16$

Converting Binary Numbers to Decimal

• Conversion procedure from base $2$ to base $10$:

  • Multiply each binary digit by its corresponding positional power of $2$

  • Add all resulting products together to produce the base-$10$ value

• Conversion of $1101_2$ to Decimal:

  • Positional alignment:

  • Rightmost digit $1$ is at position $2^0$

  • Second digit $0$ is at position $2^1$

  • Third digit $1$ is at position $2^2$

  • Fourth digit $1$ is at position $2^3$

  • Polynomial expansion:     $1101_2 = (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)$

  • Intermediate calculation:     $1101_2 = 8 + 4 + 0 + 1$

  • Final result:     $1101_2 = 13_{10}$

• Conversion of $111_2$ to Decimal:

  • Positional alignment:

  • Rightmost digit $1$ is at position $2^0$

  • Middle digit $1$ is at position $2^1$

  • Leftmost digit $1$ is at position $2^2$

  • Polynomial expansion:     $111_2 = (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)$

  • Intermediate calculation:     $111_2 = 4 + 2 + 1$

  • Final result:     $111_2 = 7_{10}$

Converting Decimal Numbers to Binary

• Method 1: Subtraction of Nearest Lower Powers of $2$

  • Identify the largest power of $2$ less than or equal to the target decimal value.

  • Subtract that power of $2$ from the decimal value to get the remainder.

  • Repeat this process for all remaining values.

  • Determine the total number of required positional places. To include a maximum power of $2^k$, $k + 1$ positions are required (from $2^0$ to $2^k$).

  • Assign $1$ to each included power position and $0$ to unused power positions.

  • Conversion Example for $18_{10}$:

  • Largest power of $2$ less than $18$ is $16$ ($2^4$)

  • Remainder calculation:       $18 - 16 = 2$

  • Remainder $2$ corresponds to $2^1$

  • Expression as sum of powers of $2$:       $18_{10} = 2^4 + 2^1$

  • Required positions: $5$ positions ($2^0$ through $2^4$)

  • Bit assignments:

    • Position $2^4$: $1$

    • Position $2^3$: $0$

    • Position $2^2$: $0$

    • Position $2^1$: $1$

    • Position $2^0$: $0$

  • Resulting binary representation:       $18_{10} = 10010_2$

  • Verification conversion back to decimal:       $(1 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0) = 16 + 2 = 18_{10}$

• Method 2: Systematic Successive Division by $2$

  • Continuously divide the decimal integer by $2$

  • Record the quotient and remainder ($0$ or $1$) for each division step

  • Repeat until the integer quotient reaches $0$

  • The remainders represent the coefficients for powers of $2$, where the first division remainder corresponds to $2^0$

  • Construct the binary number by reading remainders from bottom to top (last division step to first division step)

  • Systematic Division for $137_{10}$:

  • Step 1:       $137 \div 2 = 68 \quad \text{remainder } 1 \quad (2^0 \text{ place})$

  • Step 2:       $68 \div 2 = 34 \quad \text{remainder } 0 \quad (2^1 \text{ place})$

  • Step 3:       $34 \div 2 = 17 \quad \text{remainder } 0 \quad (2^2 \text{ place})$

  • Step 4:       $17 \div 2 = 8 \quad \text{remainder } 1 \quad (2^3 \text{ place})$

  • Step 5:       $8 \div 2 = 4 \quad \text{remainder } 0 \quad (2^4 \text{ place})$

  • Step 6:       $4 \div 2 = 2 \quad \text{remainder } 0 \quad (2^5 \text{ place})$

  • Step 7:       $2 \div 2 = 1 \quad \text{remainder } 0 \quad (2^6 \text{ place})$

  • Step 8:       $1 \div 2 = 0 \quad \text{remainder } 1 \quad (2^7 \text{ place})$

  • Reading remainders bottom-to-top yields:       $137_{10} = 10001001_2$

  • Systematic Division for $25_{10}$:

  • Bounds evaluation: $25$ lies between $16$ ($2^4$) and $32$ ($2^5$)

  • Subtraction breakdown:       $25 = 16 + 8 + 1 = 2^4 + 2^3 + 2^0$

  • Step 1:       $25 \div 2 = 12 \quad \text{remainder } 1 \quad (2^0 \text{ place})$

  • Step 2:       $12 \div 2 = 6 \quad \text{remainder } 0 \quad (2^1 \text{ place})$

  • Step 3:       $6 \div 2 = 3 \quad \text{remainder } 0 \quad (2^2 \text{ place})$

  • Step 4:       $3 \div 2 = 1 \quad \text{remainder } 1 \quad (2^3 \text{ place})$

  • Step 5:       $1 \div 2 = 0 \quad \text{remainder } 1 \quad (2^4 \text{ place})$

  • Reading remainders bottom-to-top yields:       $25_{10} = 11001_2$

  • Verification check:       $(1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 16 + 8 + 0 + 0 + 1 = 25_{10}$

Alternative Numeral Systems and High-Base Representations

• Ternary Numeral System (Base 3)

  • System using base $3$ positional notation

  • Valid digits are $0$, $1$, and $2$

  • Successive positions represent powers of $3$ ($3^0, 3^1, 3^2, \dots$)

• Hexadecimal Numeral System (Base 16)

  • System using base $16$ positional notation

  • Requires $16$ distinct digit values ($0$ through $15$)

  • To represent values $10$ through $15$ as single symbols without ambiguity, standard alphabetic letters are used:

  • $0$ through $9$ represent numeric values $0$ through $9$

  • $A$ represents value $10$

  • $B$ represents value $11$

  • $C$ represents value $12$

  • $D$ represents value $13$

  • $E$ represents value $14$

  • $F$ represents value $15$

• Conversion of Decimal $25_{10}$ to Hexadecimal (Base 16)

  • Power breakdown:     $25_{10} = 16 + 9$

  • Expressed as powers of $16$:     $25_{10} = (1 \times 16^1) + (9 \times 16^0)$

  • Placeholders for hexadecimal positions:

  • Coefficient for $16^1$ position is $1$

  • Coefficient for $16^0$ position is $9$

  • Resulting hexadecimal representation:     $25_{10} = 19_{16}$

Universal Core Concept of Base Conversions

• For any numeral system with base $b$:

  • Numbers are written as sequences of positional digits corresponding to integer powers of $b$ ($b^0, b^1, b^2, b^3, \dots$)

  • Conversion from base $b$ to base $10$ involves expanding into a positional sum of powers of $b$

  • Conversion from base $10$ to base $b$ involves repeated division by $b$, recording remainders at each step to determine digit coefficients from right to left ($b^0$ to $b^n$)