Number Systems Revision Notes: Binary, Octal, Hexadecimal Conversions and Complements
Complete revision notes on number systems for Computer Science teacher exams like BPSC TRE: conversions, binary arithmetic, 1's and 2's complement, with solved examples and exam shortcuts.
CodeOrbit Learn TeamPublished 3 min read
Number systems questions are quick marks in any Computer Science paper, if you know the methods well. These notes cover every conversion, binary arithmetic and complements with solved examples.
The four number systems
System | Base | Digits used | Example |
|---|---|---|---|
Binary | 2 | 0, 1 | 1011₂ |
Octal | 8 | 0–7 | 17₈ |
Decimal | 10 | 0–9 | 25₁₀ |
Hexadecimal | 16 | 0–9, A–F (A=10 … F=15) | 2F₁₆ |
The value of a digit = digit × base^position, positions counted from 0 at the right.
Any base → Decimal
Multiply each digit by its place value and add.
Example: 1011₂ = 1×2³ + 0×2² + 1×2¹ + 1×2⁰ = 8 + 0 + 2 + 1 = 11₁₀
Example: 2F₁₆ = 2×16 + 15×1 = 47₁₀
Fractions: places after the point use negative powers. 0.101₂ = 1×2⁻¹ + 0×2⁻² + 1×2⁻³ = 0.5 + 0.125 = 0.625₁₀
Decimal → Any base
Integer part: divide repeatedly by the base and read the remainders bottom to top.
Example: 25₁₀ to binary
Division | Quotient | Remainder |
|---|---|---|
25 ÷ 2 | 12 | 1 |
12 ÷ 2 | 6 | 0 |
6 ÷ 2 | 3 | 0 |
3 ÷ 2 | 1 | 1 |
1 ÷ 2 | 0 | 1 |
Reading upward: 11001₂
Fraction part: multiply repeatedly by the base and read the integer parts top to bottom.
Example: 0.375₁₀ to binary: 0.375×2 = 0.75 → 0; 0.75×2 = 1.5 → 1; 0.5×2 = 1.0 → 1. Answer: 0.011₂
Binary ↔ Octal ↔ Hexadecimal (the grouping shortcut)
Because 8 = 2³ and 16 = 2⁴:
- Binary → Octal: group bits in 3s from the point.
- Binary → Hex: group bits in 4s from the point.
Example: 1101011₂
- Octal: 1 | 101 | 011 → 1 5 3 → 153₈
- Hex: 110 | 1011 → 6 B → 6B₁₆
Octal → Hex: go through binary. 153₈ → 001 101 011 → 1101011 → 6B₁₆
Binary arithmetic
Addition rules: 0+0=0, 0+1=1, 1+1=10 (write 0, carry 1), 1+1+1=11 (write 1, carry 1).
1 0 1 1 (11)
+ 0 1 1 1 ( 7)
---------
1 0 0 1 0 (18)Subtraction is usually done with 2's complement (below).
1's and 2's complement
- 1's complement: flip every bit. 1's complement of 01011 is 10100.
- 2's complement: 1's complement + 1. 2's complement of 01011 is 10100 + 1 = 10101.
Shortcut for 2's complement: from the right, copy bits up to and including the first 1, then flip the rest. 01011000 → 10101000.
Subtraction using 2's complement
To compute A − B: add A to the 2's complement of B. If there is a carry out, drop it; the result is positive.
Example: 9 − 5 with 4 bits: 1001 + (2's complement of 0101 = 1011) = 1 0100 → drop carry → 0100 = 4 ✔
Range of signed numbers
With n bits in 2's complement, the range is to . For 8 bits: −128 to +127.
Codes you should know
Code | Key idea |
|---|---|
BCD (8421) | Each decimal digit written as 4 bits: 25 → 0010 0101 |
Excess-3 | BCD + 3 for each digit: 2 → 0101 |
Gray code | Successive numbers differ in only one bit |
ASCII | 7-bit character code: 'A' = 65, 'a' = 97, '0' = 48 |
Binary → Gray: the first bit stays the same; each next Gray bit = XOR of the current and previous binary bits. 1011 → 1, 1⊕0=1, 0⊕1=1, 1⊕1=0 → 1110.
Quick practice
- Convert 156₁₀ to hexadecimal.
- Convert 3A₁₆ to binary and octal.
- Find the 2's complement of 10110100.
- What is the range of a 6-bit signed number in 2's complement?
Answers: 1) 9C₁₆ (156 = 9×16 + 12). 2) 0011 1010₂ = 72₈. 3) 01001100. 4) −32 to +31.
Tags:#Computer Science#Number Systems#BPSC TRE#Digital Logic
Frequently asked questions
Why do computers use 2's complement?
It gives a single representation of zero and lets the same adder circuit do both addition and subtraction.
How many bits are needed to store a number N?
The smallest n with 2ⁿ > N. For example, 100 needs 7 bits because 2⁷ = 128.
What is the base of the number system used in BCD?
BCD stores decimal (base-10) digits, each using 4 binary bits.
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