Related Concepts: 02 Base Complements & Non-Decimal Arithmetic | 03 The 8-Bit Two’s Complement Problem | 01 BCD & Weighted Digital Codes

1.01 Advanced Base Conversions

Concept Overview: Base Conversions

Converting a number from decimal to any target base {radix $r$} requires splitting the number into two distinct components: the integer part and the fractional part.

• Integer Part: Computed via successive division by the target base $r$, recording the remainders from bottom (MSB) to top (LSB).
• Fractional Part: Computed via successive multiplication by the target base $r$, recording the integer carries generated from top to bottom.

graph TD
    Input[Decimal Number N] --> Split{Split Number}
    
    Split -->|Integer Part| IntDiv[Successive Division by Base r]
    IntDiv --> CollectRem[Collect Remainders Bottom to Top MSB to LSB]
    
    Split -->|Fractional Part| FracMult[Successive Multiplication by Base r]
    FracMult --> CollectInt[Collect Integer Carries Top to Bottom]
    
    CollectRem & CollectInt --> Combine[Combined Base-r Result]

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1. Decimal to “Unusual” Bases (Base-3, Base-4, Base-7, Base-12)

While standard conversions to binary, octal, and hexadecimal are common, term exams heavily test conversions to non-standard bases {such as base-3, base-4, base-7, and base-12}.

Major Exam Problem: Convert $249.6_{10}$ to Base-4 and Base-7 (2015, 2016, 2022, 2023, 2025)

Part A: Converting $249.6_{10}$ to Base-4

Step 1: Integer Part ($249_{10}$) — Successive Division by 4

• $249÷4=62\text{remainder }1\text{(LSB)}$
• $62÷4=15\text{remainder }2$
• $15÷4=3\text{remainder }3$
• $3÷4=0\text{remainder }3\text{(MSB)}$

$\text{Integer Result: }(3321{)}_4$

Step 2: Fractional Part ($0.6_{10}$) — Successive Multiplication by 4

• $0.6\times 4=2.4\to \text{integer }2$
• $0.4\times 4=1.6\to \text{integer }1$
• $0.6\times 4=2.4\to \text{integer }2$
• $0.4\times 4=1.6\to \text{integer }1$

$\text{Fractional Result: }(.2121\dots {)}_4$

$\text{Final Base-4 Result: }(3321.2121\dots {)}_4$

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Part B: Converting $249.6_{10}$ to Base-7

Step 1: Integer Part ($249_{10}$) — Successive Division by 7

• $249÷7=35\text{remainder }4\text{(LSB)}$
• $35÷7=5\text{remainder }0$
• $5÷7=0\text{remainder }5\text{(MSB)}$

$\text{Integer Result: }(504{)}_7$

Step 2: Fractional Part ($0.6_{10}$) — Successive Multiplication by 7

• $0.6\times 7=4.2\to \text{integer }4$
• $0.2\times 7=1.4\to \text{integer }1$
• $0.4\times 7=2.8\to \text{integer }2$
• $0.8\times 7=5.6\to \text{integer }5$

$\text{Fractional Result: }(.4125\dots {)}_7$

$\text{Final Base-7 Result: }(504.4125\dots {)}_7$

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2. Hexadecimal Fractions to Decimal, Octal, and Binary

Converting between Hexadecimal ($2^4=16$), Octal ($2^3=8$), and Binary ($2^1=2$) is accomplished by grouping binary bit patterns directly.

Major Exam Problem: Convert $2AC5.D_{16}$ to Binary, Octal, and Decimal (2015, 2017, 2025)

graph LR
    Hex[Hexadecimal 2AC5.D] -->|4-Bit Expansion| Binary[Binary 0010 1010 1100 0101 . 1101]
    Binary -->|3-Bit Grouping| Octal[Octal 25305.64]
    Hex -->|Polynomial Expansion| Decimal[Decimal 10949.8125]

Step 1: Hexadecimal to Binary (4-Bit Expansion)

Expand each Hex digit into its 4-bit binary equivalent using positional weights ($8,4,2,1$):

• $2_{16}=0010_2$
• $A_{16}(10_{10})=1010_2$
• $C_{16}(12_{10})=1100_2$
• $5_{16}=0101_2$
• $D_{16}(13_{10})=1101_2$

$\text{Binary Result: }(0010101011000101.1101{)}_2$

Step 2: Binary to Octal (3-Bit Grouping)

Group the binary bits into 3-bit clusters starting outward from the binary point:

• Integer Grouping (Left of point): $(010)(101)(011)(000)(101)⟹25305$
• Fractional Grouping (Right of point): $(110)(100)⟹.64\text{*(padded with two zeros at the end)*}$

$\text{Octal Result: }(25305.64{)}_8$

Step 3: Hexadecimal to Decimal (Polynomial Expansion)

Multiply each digit by its positional weight ($16^k$): $=(2\times 16^3)+(10\times 16^2)+(12\times 16^1)+(5\times 16^0)+(13\times 16^{-1})$ $=2(4096)+10(256)+12(16)+5(1)+13(0.0625)$ $=8192+2560+192+5+0.8125$

$\text{Decimal Result: }(10949.8125{)}_{10}$

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3. Fractional Binary Equivalence & Precision Analysis

Major Exam Word Problem: The "2/3" Fraction Analysis (2021, 2023)

Question: Calculate the binary equivalent of $2/3$ out to eight places. Convert the binary result back to decimal. How close is the result to $2/3$? Convert the binary result into hexadecimal, then convert that result to decimal. Is the answer the same?

Solution:

Part A: Binary Equivalent of $2/3$ to 8 Places

Decimal $2/3\approx 0.6666\dots$ Multiply the fraction $2/3$ by $2$ successively:

1. $\frac{2}{3}\times 2=\frac{4}{3}=1+\frac{1}{3}⟹a_{-1}=1$
2. $\frac{1}{3}\times 2=\frac{2}{3}=0+\frac{2}{3}⟹a_{-2}=0$
3. $\frac{2}{3}\times 2=\frac{4}{3}=1+\frac{1}{3}⟹a_{-3}=1$
4. $\frac{1}{3}\times 2=\frac{2}{3}=0+\frac{2}{3}⟹a_{-4}=0$

$\text{8-Bit Binary Fraction: }(0.10101010{)}_2$

Part B: Convert Binary back to Decimal & Quantify Error

$0.10101010_2=(1\times 2^{-1})+(0\times 2^{-2})+(1\times 2^{-3})+(0\times 2^{-4})+(1\times 2^{-5})+(0\times 2^{-6})+(1\times 2^{-7})+(0\times 2^{-8})$ $=\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}=0.5+0.125+0.03125+0.0078125=0.6640625_{10}$

• Precision Loss (Error Calculation): $\text{Exact Value}-\text{Approximated Value}=0.666666\cdots -0.6640625=\mathrm{0.00260416...}$

Part C: Convert Binary to Hexadecimal & Back to Decimal

• Group the 8-bit binary fraction into 4-bit nibbles: $(0.10101010{)}_2⟹(0.AA{)}_{16}\text{*(since }1010_2=10_{10}=A_{16}\text{)}$

• Convert Hexadecimal $(0.AA{)}_{16}$ to Decimal: $=(10\times 16^{-1})+(10\times 16^{-2})=\frac{10}{16}+\frac{10}{256}=0.625+0.0390625=0.6640625_{10}$

• Conclusion: Yes, the converted decimal answer ($0.6640625_{10}$) is identical! $■$

Concept Resolution: Why doesn't Binary-to-Hex conversion lose precision?

Converting between Binary, Octal, and Hexadecimal does not cause any precision loss. This is because their bases ($2^1$, $2^3$, and $2^4$) are exact powers of one another. We simply regroup bits (3-bit or 4-bit clusters) without truncation or approximation.

However, converting from a decimal fraction to binary can lose precision. A fraction that terminates in decimal (like $0.2_{10}=2/10$) does not terminate in binary ($0.00110011{\dots }_2$) because $10$ has prime factors of $2$ and $5$, whereas binary can only represent denominators that are powers of $2$. Truncating a repeating binary fraction to a finite number of bits (like 8 places) is what causes the precision loss.

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Past Year Questions (PYQs)

• [PYQ 2015, 2016, 2022, 2023, 2025]: Convert integer/fraction decimals to base-3, base-4, base-7, base-12.
• [PYQ 2015, 2017, 2025]: Convert hexadecimal with fractions to decimal, octal, binary.
• [PYQ 2021, 2023]: Calculate binary equivalent of fractions (e.g., $2/3$ out to 8 places) and verify accuracy through hex conversions.