Binary Numbers and Place Value
On this page 11 headings
Start here One light bulb can show only two things: off or on. Two bulbs can show four patterns. Three bulbs, eight. With enough bulbs you can show any number. A computer is just a huge row of “bulbs” (bits).
One bit gives only 2 values (0 and 1). To show more, put bits side by side:
| Bits | Patterns | Count |
|---|---|---|
| 1 | 0, 1 | 2 |
| 2 | 00, 01, 10, 11 | 4 |
| 3 | 000, 001, 010, 011, 100, 101, 110, 111 | 8 |
| n | … | 2ⁿ |
Rule: n bits give 2ⁿ different values.
You already know this. In 329, each place is worth 10 times the place to its right:
329 = 3×100 + 2×10 + 9×1 = 3×10² + 2×10¹ + 9×10⁰
Binary works the same way, but each place is worth 2 times the place to its right:
101 in binary = 1×4 + 0×2 + 1×1 = 5
Binary place values (right to left): 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024…
Tip: write the base as a small subscript when it could be confusing: 101₂ = 5₁₀.
- A digital system uses a limited set of values.
- A binary system uses two: 0 and 1, held as low and high voltages.
- Groups of bits represent bigger numbers, letters, colours, anything.
In one line: n bits give 2ⁿ values, and binary place values double from right to left (1, 2, 4, 8, …), just as decimal places multiply by 10. Exam words: bit, binary (base 2), decimal (base 10), place value, positional number system, 2ⁿ, least significant bit (LSB, rightmost), most significant bit (MSB, leftmost).
Steps: answer = 2ⁿ. Example: 5 bits → 2⁵ = 32 values (0 to 31). Trap: the biggest number is 2ⁿ − 1, not 2ⁿ, because we start from 0.
Steps: find the smallest n with . Example: 26 letters. 2⁴ = 16 (too small), 2⁵ = 32 ≥ 26 → 5 bits. Trap: needs 5 bits, but needs 6.
You’ll see this again in sequential circuits: “A system has 64 states. Are 6 flip-flops enough?” Each flip-flop stores 1 bit, so n flip-flops give 2ⁿ states. 2⁶ = 64 → yes. 65 states would need 7.
Steps:
- Write the place values over each bit, starting with 1 on the right.
- Add the place values where the bit is 1.
Example: 110101₂
| Place value | 32 | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|---|
| Bit | 1 | 1 | 0 | 1 | 0 | 1 |
32 + 16 + 4 + 1 = 53
Method A, keep dividing by 2:
- Divide by 2 and write down the remainder (0 or 1).
- Divide the answer by 2 again, and keep going until you reach 0.
- Read the remainders from bottom to top.
Example: 13
| Divide | Answer | Remainder |
|---|---|---|
| 13 ÷ 2 | 6 | 1 |
| 6 ÷ 2 | 3 | 0 |
| 3 ÷ 2 | 1 | 1 |
| 1 ÷ 2 | 0 | 1 |
Read bottom to top: 1101₂. Check: 8 + 4 + 0 + 1 = 13 ✓
Method B, take away the biggest place values: 13 → largest place value ≤ 13 is 8 (left 5) → 4 fits (left 1) → 2 doesn’t fit → 1 fits (left 0). So the places are 8, 4, 1 → 1101₂.
Trap: in Method A, reading the remainders top to bottom gives the number backwards.