Conversion between decimal and binary systems

Conversion between decimal and binary systems

1. Decimal to Binary (Integer Part) — Repeated Division by 2

Divide the number repeatedly by 2, note each remainder (always 0 or 1), and stop when the quotient becomes 0. Read the remainders from bottom to top.

Example: (45)₁₀ = (?)₂

Step	Division	Quotient	Remainder
1	45 ÷ 2	22	1 (LSB)
2	22 ÷ 2	11	0
3	11 ÷ 2	5	1
4	5 ÷ 2	2	1
5	2 ÷ 2	1	0
6	1 ÷ 2	0	1 (MSB)

Reading upwards: 101101₂. Check: $32+0+8+4+0+1 = 45$ ✓

2. Binary to Decimal (Integer Part) — Positional Expansion

Multiply each bit by its positional weight $2^{position}$ and add, from the right (position 0).

Example: (101101)₂ = (?)₁₀

Position	5	4	3	2	1	0
Bit	1	0	1	1	0	1
Weight	32	16	8	4	2	1
Product	32	0	8	4	0	1

$32 + 0 + 8 + 4 + 0 + 1 =$ (45)₁₀

For long patterns the doubling method is faster: start with 0 and for each bit (left to right) do double and add the bit.
1 → 1; 1 → 3; 0 → 6; 1 → 13; 1 → 27; 0 → 54; 1 → 109 (so 1101101₂ = 109₁₀).

3. Fractional Part — Repeated Multiplication by 2

Multiply the fraction by 2; the integer part produced (0 or 1) is the next binary digit; keep the fractional part and repeat. Read the digits top to bottom.

Example: (0.625)₁₀ = (?)₂

Step	Multiply	Integer part	Fraction left
1	0.625 × 2 = 1.25	1	0.25
2	0.25 × 2 = 0.50	0	0.50
3	0.50 × 2 = 1.00	1	0.00 → stop

So 0.625₁₀ = 0.101₂. Verify: $0.5 + 0 + 0.125 = 0.625$ ✓

A fraction whose denominator is not a power of 2 never terminates — e.g. 0.1₁₀ = 0.0001100110011…₂. This is exactly why 0.1 + 0.2 is not exactly 0.3 in floating-point arithmetic.

4. Whole Number with a Fraction

Convert the two parts separately and join them: 45.625₁₀ = 101101.101₂

5. Quick Method for Small Integers — 8-4-2-1 Weights

Subtract the largest power of 2 that fits, repeatedly, and put a 1 in the corresponding position:

Weight	64	32	16	8	4	2	1
Bit	1	0	0	1	1	0	1

6. Range and Precision Points

• $n$ bits give $2^{n}$ patterns: unsigned range $0$ to $2^{n} - 1$ (8 bits → 0-255; 16 bits → 0-65535).
• Each extra fractional bit halves the resolution ($2^{-k}$): 8 fractional bits give a resolution of $1/256 ≈ 0.0039$.
• Conversion errors come only from the fractional part (truncation) — integer conversions are always exact.

7. Practice Numericals

Problem	Answer
(156)₁₀ → binary	10011100
(1000)₁₀ → binary	1111101000
(1101011)₂ → decimal	107
(0.40625)₁₀ → binary	0.01101
(25.75)₁₀ → binary	11001.11
(11111111)₂ → decimal	255 (= 2⁸ − 1)

8. Exam Points

• Convert a given decimal integer and fraction to binary (show the table) — 5 marks, almost certain.
• Convert a binary number to decimal by positional expansion — 3 marks.
• With a worked example, explain why 0.1 cannot be represented exactly — 3-5 marks.
• Find the range of numbers representable in n bits (unsigned) — 3 marks.
• State the number of bits required for a given range of values — 2-3 marks.