Number and Numeration (Bases)

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Number bases (or numeration systems) allow us to represent numbers using different sets of digits. Our everyday number system is base 10 (decimal), but computers use base 2 (binary) and other bases are also important.

Base 10 (Decimal): Digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 The number 5723 means: $$5 \times 10^3 + 7 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 = 5000 + 700 + 20 + 3$$

Base 2 (Binary): Digits: 0, 1 only The number $1101_2$ means: $$1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 13_{10}$$

Base 8 (Octal): Digits: 0, 1, 2, 3, 4, 5, 6, 7 The number $527_8$ means: $$5 \times 8^2 + 2 \times 8^1 + 7 \times 8^0 = 320 + 16 + 7 = 343_{10}$$

Base 16 (Hexadecimal): Digits: 0-9, then A(10), B(11), C(12), D(13), E(14), F(15) The number $2AF_{16}$ means: $$2 \times 16^2 + 10 \times 16^1 + 15 \times 16^0 = 512 + 160 + 15 = 687_{10}$$

⚡ WAEC Tip: When converting from base $b$ to decimal, expand using powers of $b$. When converting from decimal to base $b$, use repeated division by $b$.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Converting Decimal to Other Bases:

Method: Repeated Division

Convert 27 to base 2: $$27 \div 2 = 13 \text{ remainder } 1$$ $$13 \div 2 = 6 \text{ remainder } 1$$ $$6 \div 2 = 3 \text{ remainder } 0$$ $$3 \div 2 = 1 \text{ remainder } 1$$ $$1 \div 2 = 0 \text{ remainder } 1$$

Reading remainders from bottom to top: $27 = 11011_2$

Convert 458 to base 8: $$458 \div 8 = 57 \text{ remainder } 2$$ $$57 \div 8 = 7 \text{ remainder } 1$$ $$7 \div 8 = 0 \text{ remainder } 7$$

Reading up: $458 = 712_8$

Convert 2543 to base 16: $$2543 \div 16 = 158 \text{ remainder } 15 \Rightarrow F$$ $$158 \div 16 = 9 \text{ remainder } 14 \Rightarrow E$$ $$9 \div 16 = 0 \text{ remainder } 9$$

Reading up: $2543 = 9EF_{16}$

Converting Between Bases:

Binary to Octal (and vice versa): Group binary digits in threes from the right: $$1101011_2 = \underbrace{001}{1} \underbrace{101}{5} \underbrace{011}_{3} = 153_8$$

Binary to Hexadecimal (and vice versa): Group binary digits in fours from the right: $$1101011_2 = \underbrace{0110}{6} \underbrace{1011}{B} = 6B_{16}$$

Octal to Hexadecimal: Use binary as an intermediary: $723_8 = 111\ 010\ 011_2 = 111010011_2 = \underbrace{0001}{1} \underbrace{1101}{D} \qdm 0011}{3} = 1D3{16}$

Addition in Different Bases:

In base $b$, digits range from 0 to $b-1$. When adding, carry when sum exceeds $b-1$.

Problem: Add $134_7 + 256_7$

$$134_7$$ $$+ 256_7$$ $$------$$ Units: $4 + 6 = 10_{10} = 13_7$ (write 3, carry 1) Sevens: $1 + 3 + 5 = 9_{10} = 12_7$ (write 2, carry 1) Forty-nines: $1 + 1 + 2 = 4$ (since $4 < 7$, write 4 directly) $$134_7 + 256_7 = 423_7$$

Subtraction in Bases:

Problem: Subtract $234_5 - 143_5$

$$234_5$$ $$- 143_5$$ $$------$$ Units: $4 - 3 = 1$ Fives: $3 - 4$: Can’t! Borrow 1 (which is 5 in base 5): $3 + 5 - 4 = 4$ Twenty-fives: $2 - 1 - 1$ (borrowed) = 0 $$234_5 - 143_5 = 041_5 = 41_5$$

⚡ Common Student Mistakes: Forgetting that in base $b$, you can only use digits 0 to $b-1$. Confusing the method for decimal-to-base conversion (it’s division, not multiplication). Forgetting to regroup/chunk properly when converting between binary and octal/hex.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for thorough preparation.

Computation in Bases:

Problem: Multiply $213_4 \times 3_4$

$$213_4$$ $$\times 3_4$$ $$------$$ Units: $3 \times 3 = 9_{10} = 21_4$ (write 1, carry 2) Fours: $3 \times 1 + 2 = 5_{10} = 11_4$ (write 1, carry 1) Sixteens: $3 \times 2 + 1 = 7_{10} = 13_4$ (write 3, carry 1) $$213_4 \times 3_4 = 1311_4$$

Problem: Divide $312_5$ by $4_5$

Convert to decimal: $312_5 = 3 \times 25 + 1 \times 5 + 2 = 75 + 5 + 2 = 82_{10}$ $4_5 = 4_{10}$ $82 \div 4 = 20$ remainder 2

Convert back: $20_{10} = 40_5$, $2_{10} = 2_5$ So $312_5 \div 4_5 = 40_5$ remainder $2_5$

Check: $40_5 \times 4_5 + 2_5 = 3 \times 40 + 2 = 120 + 2 = 122_5$ $312_5 - 122_5 = 190_5$… this is wrong.

Let me redo in base 5: $312_5 \div 4_5$: 4 goes into 31 (base 5) = 3 times (3 × 4 = 12_5 = 7₁₀) $31_5 - 12_5 = 31_5 - 12_5 = (3×5 + 1) - (1×5 + 2) = 16 - 7 = 9_{10} = 14_5$ Bring down 2: $14_5 \div 4_5 = 2$ remainder… $14_5 - 12_5 = 2_5$

So answer: $32_5$ remainder $2_5$

Fractional Bases:

Problem: Convert $0.625_{10}$ to base 2.

$$0.625 \times 2 = 1.25 \quad \Rightarrow \text{first digit } 1, \text{ carry } 0.25$$ $$0.25 \times 2 = 0.5 \quad \Rightarrow \text{second digit } 0, \text{ carry } 0.5$$ $$0.5 \times 2 = 1.0 \quad \Rightarrow \text{third digit } 1, \text{ carry } 0$$

So $0.625_{10} = 0.101_2$

Problem: Convert $0.3_{10}$ to base 2.

$$0.3 \times 2 = 0.6 \quad \Rightarrow 0, \text{ carry } 0.6$$ $$0.6 \times 2 = 1.2 \quad \Rightarrow 1, \text{ carry } 0.2$$ $$0.2 \times 2 = 0.4 \quad \Rightarrow 0, \text{ carry } 0.4$$ $$0.4 \times 2 = 0.8 \quad \Rightarrow 0, \text{ carry } 0.8$$ $$0.8 \times 2 = 1.6 \quad \Rightarrow 1, \text{ carry } 0.6$$

The pattern repeats. $0.3_{10} = 0.0100110011…_2$ (recurring)

Number Base Systems — Complement Arithmetic:

Uses of Complements:

• Simplified subtraction
• Representation of negative numbers

Two’s Complement in Binary: For n bits, the two’s complement of a number is $2^n - \text{number}$

Example: Find 8-bit two’s complement of $27$: $2^8 - 27 = 256 - 27 = 229 = 11100101_2$

To subtract 27 from 50 in binary using complements:

• 50 = $00110010_2$
• -27 = $11100101_2$ (two’s complement)
• Add: $00110010 + 11100101 = 100010111_2$
• Discard overflow bit (9th bit): $00010111_2 = 23_{10}$

Uses of Different Bases:

Computing:

• Binary (base 2): Fundamental to digital circuits (0 = off, 1 = on)
• Octal (base 8): Used as shorthand for binary (3 bits per octal digit)
• Hexadecimal (base 16): Used in computing (4 bits per hex digit), memory addresses

Everyday life:

• Base 60: Time (60 seconds/minutes), angles (60 minutes)
• Base 12: Months, hours, dozens
• Base 7: Days of the week
• Base 360: Degrees in a circle

Modular Arithmetic:

Problem: What day of the week will it be 100 days from Monday?

$100 \div 7 = 14$ remainder 2 So 2 days after Monday = Wednesday

Problem: Find the remainder when $3^{100}$ is divided by 7.

$3^1 \bmod 7 = 3$ $3^2 \bmod 7 = 9 \bmod 7 = 2$ $3^3 \bmod 7 = 6$ $3^4 \bmod 7 = 18 \bmod 7 = 4$ $3^5 \bmod 7 = 12 \bmod 7 = 5$ $3^6 \bmod 7 = 15 \bmod 7 = 1$ $3^7 \bmod 7 = 3$ (pattern repeats every 6)

$100 \div 6 = 16$ remainder 4 So $3^{100} \bmod 7 = 3^4 \bmod 7 = 4$

Converting Fractional Numbers Between Bases:

Problem: Convert $0.36_{10}$ to base 5.

$$0.36 \times 5 = 1.80 \quad \Rightarrow 1, \text{ carry } 0.80$$ $$0.80 \times 5 = 4.00 \quad \Rightarrow 4, \text{ carry } 0$$

So $0.36_{10} = 0.14_5$

Check: $1/5 + 4/25 = 0.2 + 0.16 = 0.36$ ✓

Base 3 Arithmetic:

Digits: 0, 1, 2

Problem: Convert $212_3$ to decimal. $$212_3 = 2 \times 9 + 1 \times 3 + 2 = 18 + 3 + 2 = 23_{10}$$

Problem: Convert $47_{10}$ to base 3. $$47 \div 3 = 15 \text{ remainder } 2$$ $$15 \div 3 = 5 \text{ remainder } 0$$ $$5 \div 3 = 1 \text{ remainder } 2$$ $$1 \div 3 = 0 \text{ remainder } 1$$ $$47_{10} = 1202_3$$

Pattern Recognition in Bases:

The expansion of $\frac{1}{3}$ in base 10 is $0.\overline{3}$.

In base 3: $\frac{1}{3} = 0.1_3$ (terminating because 3 divides $3^1$)

The expansion of $\frac{1}{6}$ in base 10 is $0.1\overline{6}$.

In base 6: $\frac{1}{6} = 0.1_6$ (terminating because 6 = 2 × 3, and base 6 has factors 2 and 3)

⚡ WAEC Examination Patterns: Convert numbers between bases (decimal to binary/octal/hex and vice versa). Add and subtract numbers in different bases. Multiply numbers in bases. Convert fractional numbers between bases. Use complements for subtraction. Solve problems involving modular arithmetic.