5 Basics of Computers · 5.1 Computer basics: number systems, codes, logic, algorithms and spreadsheets
5.1.2 Number systems: binary, octal, hexadecimal and conversion to and from decimal
A number system is fixed by its base (radix): computers work in binary (base 2), and octal (base 8) and hexadecimal (base 16) are short ways of writing binary; any base converts to decimal by adding digit × place value, and decimal converts to any base by repeated division (whole part) and repeated multiplication (fraction).
- APPSC Paper I
- Advanced level
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Core facts
- Base (radix) = number of distinct digits. Digits run from 0 to base − 1.
| System | Base | Digits | Example |
|---|---|---|---|
| Binary | 2 | 0, 1 | 101101₂ |
| Octal | 8 | 0 to 7 | 55₈ |
| Decimal | 10 | 0 to 9 | 45₁₀ |
| Hexadecimal | 16 | 0 to 9, A to F | 2D₁₆ |
- Hex letters: A = 10, B = 11, C = 12, D = 13, E = 14, F = 15.
- Positional value: each place is a power of the base. In binary, from the right: 1, 2, 4, 8, 16, 32, 64, 128, ...; after the point: ½, ¼, ⅛, ...
- Any base → decimal: multiply each digit by its place value and add.
- Decimal (whole part) → base b: divide by b repeatedly; read the remainders from bottom to top.
- Decimal (fraction) → base b: multiply by b repeatedly; read the integer parts from top to bottom.
- Binary ↔ octal: groups of 3 bits (8 = 2³). Binary ↔ hex: groups of 4 bits (16 = 2⁴). Group from the point outwards; pad with zeros.
- Octal ↔ hex: go through binary.
- MSB = most significant (leftmost) bit; LSB = least significant (rightmost) bit.
- n bits give 2ⁿ patterns; the largest unsigned value is 2ⁿ − 1 (8 bits: 0 to 255).
- Useful pairs: 2¹⁰ = 1,024; 255 = 11111111₂ = 377₈ = FF₁₆.
Example 1. Convert 45₁₀ to binary. Solution. 45 ÷ 2 = 22 r 1; 22 ÷ 2 = 11 r 0; 11 ÷ 2 = 5 r 1; 5 ÷ 2 = 2 r 1; 2 ÷ 2 = 1 r 0; 1 ÷ 2 = 0 r 1. Read upwards. Answer: 101101₂
Example 2. Convert 11010₂ to decimal. Solution. 16 + 8 + 0 + 2 + 0. Answer: 26
Explained
1. Decimal to octal and hexadecimal.
Example 3. Convert 156₁₀ to octal and to hexadecimal. Solution. Octal: 156 ÷ 8 = 19 r 4; 19 ÷ 8 = 2 r 3; 2 ÷ 8 = 0 r 2 → 234₈. Hex: 156 ÷ 16 = 9 r 12 (C); 9 ÷ 16 = 0 r 9 → 9C₁₆. Check: 2 × 64 + 3 × 8 + 4 = 156; 9 × 16 + 12 = 156. Answer: 234₈ and 9C₁₆
2. Octal and hexadecimal to decimal.
Example 4. Convert 754₈ and 3F7₁₆ to decimal. Solution. 7 × 64 + 5 × 8 + 4 = 448 + 40 + 4 = 492. 3 × 256 + 15 × 16 + 7 = 768 + 240 + 7 = 1,015. Answer: 492 and 1,015
3. Binary to octal and hex by grouping.
Example 5. Write 1101011₂ in octal, hexadecimal and decimal. Solution. Groups of 3 from the right: 1 | 101 | 011 → 153₈. Groups of 4: 110 | 1011 → 6 | B → 6B₁₆. Decimal: 64 + 32 + 8 + 2 + 1 = 107.
Example 6. Convert A5₁₆ to octal. (a) 245 (b) 165 (c) 255 (d) 225 Solution. A5 = 1010 0101 → regroup in threes: 10 | 100 | 101 → 2 | 4 | 5. Answer: (a) 245₈ (A5₁₆ = 245₈ = 165 in decimal; (b) is the decimal value, a trap).
4. Fractions.
Example 7. Convert 0.625₁₀ to binary. Solution. 0.625 × 2 = 1.25 → 1; 0.25 × 2 = 0.5 → 0; 0.5 × 2 = 1.0 → 1. Read downwards. Answer: 0.101₂
Example 8. Convert 25.375₁₀ to binary. Solution. 25 = 11001; 0.375 × 2 = 0.75 → 0; 0.75 × 2 = 1.5 → 1; 0.5 × 2 = 1.0 → 1, so 0.011. Answer: 11001.011₂
Example 9. Convert 1101.11₂ and 1A.8₁₆ to decimal. Solution. 8 + 4 + 1 + ½ + ¼ = 13.75. 1 × 16 + 10 + 8/16 = 26.5. Answer: 13.75 and 26.5
Summary of methods
| From → to | Method |
|---|---|
| Base b → decimal | Σ digit × b^position |
| Decimal → base b | ÷ b (whole), × b (fraction) |
| Binary → octal / hex | groups of 3 / 4 bits |
| Octal ↔ hex | via binary |
Deeper
- Not every decimal fraction ends in binary. 0.1₁₀ = 0.000110011...₂ (the block 0011 repeats), which is why 0.1 + 0.2 is not exactly 0.3 in most programs. A fraction ends in binary only if its denominator (in lowest terms) is a power of 2.
- Bits needed for a decimal number N: ⌊log₂ N⌋ + 1. 100 needs 7 bits (1100100); 1,000 needs 10 bits.
- Validity of digits: a numeral is valid in base b only if every digit is less than b. "789" is not an octal number; "1234" needs base 5 or more.
- Why octal and hex? They compress long bit strings: one hex digit = 4 bits, two hex digits = one byte. Memory addresses, colour codes (#FF0000) and machine code listings use hex.
- Shifting: appending a 0 to a binary number doubles it (1011 → 10110, 11 → 22); dropping the last bit halves it (whole part only).
Example 10. If 121 in base b equals 49 in decimal, find b. (a) 5 (b) 6 (c) 7 (d) 8 Solution. b² + 2b + 1 = 49, so (b + 1)² = 49 and b = 6. Answer: (b) 6
Example 11. Convert 2F₁₆ to octal. Solution. 2F = 0010 1111 = 101 111 → 57₈. Check: 2 × 16 + 15 = 47 = 5 × 8 + 7. Answer: 57₈
Example 12. Arrange in ascending order: 110011₂, 62₈, 33₁₆. Solution. 110011₂ = 51; 62₈ = 50; 33₁₆ = 51. Answer: 62₈ < 110011₂ = 33₁₆
Common traps
- ✗ "Read the remainders from top to bottom." ✓ For whole numbers read bottom to top (last remainder is the MSB).
- ✗ "Read fractional digits bottom to top." ✓ For fractions read the integer parts top to bottom.
- ✗ "Group bits from the left." ✓ Group from the binary point outwards (right to left for the whole part), padding with zeros.
- ✗ "Octal uses digits 0 to 8." ✓ Octal uses 0 to 7; hex uses 0 to 9 and A to F.
- ✗ "Hex F = 16." ✓ F = 15; 10₁₆ = 16.
- ✗ "8 bits can hold up to 256." ✓ 8 bits give 256 patterns, so the largest unsigned value is 255.
- ✗ "Every decimal fraction converts exactly." ✓ 0.1₁₀ recurs in binary.
Quick revision
- Binary 2, octal 8, decimal 10, hex 16; digits 0 to base − 1.
- A to F = 10 to 15.
- To decimal: Σ digit × place value.
- Whole part: ÷ base, remainders bottom up.
- Fraction: × base, integer parts top down.
- 3 bits = 1 octal digit; 4 bits = 1 hex digit.
- n bits → 2ⁿ patterns, max 2ⁿ − 1.
- 45 = 101101₂ = 55₈ = 2D₁₆.
- 255 = 11111111₂ = 377₈ = FF₁₆.
- 0.625 = 0.101₂; 0.1 recurs in binary.
Practise this topic. Questions on Number systems & conversions, with answers explained.
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