2.1 Bits, Bytes & Positional Number Conversions

Theory: Bits, Bytes & Positional Number Conversions

Everything a computer stores (numbers, text, pictures, instructions) is ultimately a pattern of bits. Before we can understand how a CPU adds, compares or stores numbers, we need to be fluent at moving between the number systems humans like (decimal) and the ones hardware likes (binary, and its shorthands octal and hexadecimal). This page builds that fluency from scratch (Null & Lobur, Ch. 2).

Bits, Nibbles, Bytes and Words

Unit	Size	Notes
Bit (binary digit)	1 bit	The most basic unit of information: $0$ or $1$. Physically a state of a circuit: “off/on”, “low/high voltage”.
Nibble	4 bits	Half a byte. Exactly one hexadecimal digit.
Byte	8 bits	The smallest addressable unit in most machines (byte-addressable memory). Two hex digits.
Word	a contiguous group of bytes	Machine dependent; 16, 32 and 64 bits are most common. The “natural” size the CPU processes at once.

Byte-addressable vs word-addressable memory. An address names one storage location.

• In a byte-addressable system every byte has its own address, so a 32-bit word occupies 4 consecutive addresses.
• In a word-addressable system the word is the smallest addressable unit: each address names a whole word, and individual bytes inside it cannot be addressed directly.

The number of address bits needed is $\log_2(\text{number of addressable units})$.

Mini-example

A memory holds $2^{24}$ bytes and uses 32-bit (4-byte) words.

• Byte-addressable: $2^{24}$ addresses, so 24 address bits.
• Word-addressable: $2^{24} / 2^{2} = 2^{22}$ words, so 22 address bits.

Positional Numbering Systems

In a positional (weighted) number system with base (or radix) $b$, each digit is multiplied by a power of $b$ that depends on its position. Positions to the left of the radix point have weights $b^0, b^1, b^2, \dots$ (from right to left); positions to the right have weights $b^{-1}, b^{-2}, \dots$

$$(d_{n-1} \cdots d_1 d_0 \,.\, d_{-1} d_{-2} \cdots)_b = \sum_{i} d_i \times b^{\,i}, \qquad 0 \le d_i \le b-1$$

A base-$b$ system uses exactly $b$ distinct digit symbols, $0$ to $b-1$.

Base	Name	Digits
2	binary	0, 1
8	octal	0–7
10	decimal	0–9
16	hexadecimal	0–9, A(10), B(11), C(12), D(13), E(14), F(15)

Any integer quantity can be represented exactly in any base. The two most important radices in computer science are binary (what the hardware uses) and hexadecimal (a compact, human-readable shorthand for binary).

Mini-example

$$241_{10} = 2 \times 10^2 + 4 \times 10^1 + 1 \times 10^0, \qquad 1001_2 = 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 9_{10}$$

Converting Any Base → Decimal (Expansion)

Procedure: write each digit times its positional weight and add everything up in decimal. This works for fractional digits too (negative powers).

Mini-example

$$10212_3 = 1(81) + 0(27) + 2(9) + 1(3) + 2(1) = 81 + 0 + 18 + 3 + 2 = 104_{10}$$ $$101.11_2 = 4 + 0 + 1 + \tfrac12 + \tfrac14 = 5.75_{10}$$

Converting Decimal → Base $b$ (Whole Numbers)

There are two equivalent methods.

Method 1: Subtraction (of powers).

1. Find the largest power $b^k$ that is $\le$ the number (use $k = \lfloor \log_b N \rfloor$).
2. See how many times $b^k$ fits (that count is the digit for position $k$, between $0$ and $b-1$); subtract.
3. Repeat with $b^{k-1}, \dots, b^0$. Every power down to $b^0$ gets a digit, even if it is $0$.

Method 2: Division-remainder (the one used most in practice).

1. Divide the number by $b$. Write down the remainder (it is a digit, $0$ to $b-1$).
2. Replace the number by the quotient and repeat until the quotient is $0$.
3. Read the remainders from bottom to top (the first remainder is the least significant digit).

Mini-example (division-remainder, $104_{10}$ to base 3)

Division	Quotient	Remainder
$104 \div 3$	34	2 (LSD)
$34 \div 3$	11	1
$11 \div 3$	3	2
$3 \div 3$	1	0
$1 \div 3$	0	1 (MSD)

Reading bottom to top: $104_{10} = 10212_3$.

Why does it work? Dividing by $b$ peels off the last base-$b$ digit as the remainder, exactly like dividing a decimal number by 10 gives its last digit as the remainder ($241 \div 10 = 24$ r $1$).

Converting Decimal Fractions → Base $b$ (Repeated Multiplication)

For the part after the radix point, use the target radix as a multiplier:

1. Multiply the fraction by $b$.
2. The integer part of the product is the next digit (going left to right after the point).
3. Keep only the fractional part and repeat.
4. Stop when the fractional part becomes 0, or when you have produced the required number of digits.
5. Read the digits from top to bottom (the first digit produced is the one right after the point).

For a mixed number (like $2.8125$), convert the integer part by division-remainder and the fraction by repeated multiplication, then join them.

Mini-example ($0.8125_{10}$ to binary)

Step	Multiply	Product	Digit
1	$0.8125 \times 2$	1.6250	1
2	$0.6250 \times 2$	1.2500	1
3	$0.2500 \times 2$	0.5000	0
4	$0.5000 \times 2$	1.0000	1 (fraction is now 0, stop)

Top to bottom: $0.8125_{10} = 0.1101_2$. Since $2_{10} = 10_2$, we get $2.8125_{10} = 10.1101_2$.

Non-terminating fractions. A decimal fraction terminates in base 2 only if, in lowest terms, its denominator is a power of 2. Something as innocent as $0.1_{10}$ becomes the repeating pattern $0.0\overline{0011}_2 = 0.000110011001\ldots_2$, so it cannot be stored exactly in any finite number of bits. You must stop after a fixed number of bits, which introduces a (small) error. This is one root cause of floating-point errors (section 2.5).

Binary ↔ Hexadecimal ↔ Octal (Grouping)

Because $16 = 2^4$ and $8 = 2^3$, every hex digit corresponds to exactly 4 bits and every octal digit to exactly 3 bits. No arithmetic is needed, just grouping.

Binary → hex (or octal):

1. Starting at the radix point, split the bits into groups of 4 (for hex) or 3 (for octal): moving left for the integer part and right for the fraction part.
2. Pad the outermost groups with zeros (leading zeros on the left, trailing zeros on the right of the fraction).
3. Replace each group by its digit.

Hex (or octal) → binary: replace each digit by its 4-bit (or 3-bit) pattern.

Hex ↔ octal: go through binary.

Hex	Binary	Hex	Binary	Hex	Binary	Hex	Binary
0	0000	4	0100	8	1000	C	1100
1	0001	5	0101	9	1001	D	1101
2	0010	6	0110	A	1010	E	1110
3	0011	7	0111	B	1011	F	1111

Mini-example

$$1011\,1110_2 = \text{BE}_{16}, \qquad 10\,111\,110_2 = 010\,111\,110_2 = 276_8$$

Check: $\text{BE}_{16} = 11 \times 16 + 14 = 190$ and $276_8 = 2 \times 64 + 7 \times 8 + 6 = 190$ ✓.

How Many Digits? (Logarithms)

• With $n$ base-$b$ digits you can write $b^n$ different values, $0$ to $b^n - 1$. So $n$ bits hold unsigned values $0$ to $2^n - 1$.
• The highest power of $b$ needed for a number $N$ is $k = \lfloor \log_b N \rfloor$ (this is the place to start the subtraction method).
• The number of digits needed is $\lfloor \log_b N \rfloor + 1$.
• On a calculator, $\log_b N = \dfrac{\ln N}{\ln b} = \dfrac{\log_{10} N}{\log_{10} b}$.

Mini-example

$\log_3 104 = \dfrac{\ln 104}{\ln 3} \approx 4.228$, so the highest power is $3^4 = 81$ and $104$ needs $4 + 1 = 5$ base-3 digits (indeed $10212_3$ has 5 digits).

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Worked Examples

Exercise 1

Convert $0011010100011011_2$ to hexadecimal, octal and decimal.

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Solution

Hexadecimal: group into 4s from the right.

$$0011\ 0101\ 0001\ 1011 \;\to\; 3\ \ 5\ \ 1\ \ \text{B}$$

So $0011010100011011_2 = \text{351B}_{16}$.

Octal: group into 3s from the right. 16 bits do not divide evenly into 3s, so pad the leftmost group with leading zeros:

$$\underbrace{0}_{\to\, 000}\ 011\ 010\ 100\ 011\ 011 \;\to\; 0\ 3\ 2\ 4\ 3\ 3$$

So the value is $32433_8$.

Decimal: expand the hex form (fewer terms than expanding 16 bits):

$$\text{351B}_{16} = 3 \times 16^3 + 5 \times 16^2 + 1 \times 16 + 11 = 12288 + 1280 + 16 + 11 = 13595_{10}$$

Check via octal: $3 \times 8^4 + 2 \times 8^3 + 4 \times 8^2 + 3 \times 8 + 3 = 12288 + 1024 + 256 + 24 + 3 = 13595$ ✓.

Exercise 2

Convert each number to decimal.

(a) $\text{2A3}_{16}$ (b) $1234_5$ (c) $1011.101_2$ (d) $\text{3D.8}_{16}$

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Solution

Multiply each digit by its positional weight and add.

(a) $\text{2A3}_{16} = 2 \times 256 + 10 \times 16 + 3 \times 1 = 512 + 160 + 3 = 675_{10}$

(b) $1234_5 = 1 \times 125 + 2 \times 25 + 3 \times 5 + 4 \times 1 = 125 + 50 + 15 + 4 = 194_{10}$

(c) Integer part $1011_2 = 8 + 0 + 2 + 1 = 11$. Fraction part:

$$.101_2 = 1 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} = 0.5 + 0 + 0.125 = 0.625$$

So $1011.101_2 = 11.625_{10}$.

(d) $\text{3D.8}_{16} = 3 \times 16 + 13 \times 1 + 8 \times 16^{-1} = 48 + 13 + 0.5 = 61.5_{10}$

Exercise 3

Convert $104_{10}$ to base 3 using (a) the subtraction method and (b) the division-remainder method. Verify your answer.

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Solution

(a) Subtraction. Powers of 3: $3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243$. The largest power $\le 104$ is $3^4 = 81$ (also $\lfloor \log_3 104 \rfloor = \lfloor 4.228 \rfloor = 4$).

Power	How many fit	Subtract	Remaining
$3^4 = 81$	1	$104 - 81$	23
$3^3 = 27$	0	$23 - 0$	23
$3^2 = 9$	2	$23 - 18$	5
$3^1 = 3$	1	$5 - 3$	2
$3^0 = 1$	2	$2 - 2$	0

Digits top to bottom: $10212_3$.

(b) Division-remainder.

Division	Quotient	Remainder
$104 \div 3$	34	2
$34 \div 3$	11	1
$11 \div 3$	3	2
$3 \div 3$	1	0
$1 \div 3$	0	1

Remainders bottom to top: $10212_3$. Both methods agree.

Verify: $1(81) + 0(27) + 2(9) + 1(3) + 2(1) = 81 + 18 + 3 + 2 = 104$ ✓.

Exercise 4

(a) Convert $147_{10}$ to binary using division-remainder, then write it in hexadecimal and octal.

(b) Convert $2.8125_{10}$ and $13.375_{10}$ to binary.

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Solution

(a) Divide repeatedly by 2:

Division	Quotient	Remainder
$147 \div 2$	73	1
$73 \div 2$	36	1
$36 \div 2$	18	0
$18 \div 2$	9	0
$9 \div 2$	4	1
$4 \div 2$	2	0
$2 \div 2$	1	0
$1 \div 2$	0	1

Bottom to top: $147_{10} = 10010011_2$.

• Hex: $1001\ 0011 \to 93_{16}$ (check: $9 \times 16 + 3 = 147$ ✓).
• Octal: $010\ 010\ 011 \to 223_8$ (check: $2 \times 64 + 2 \times 8 + 3 = 147$ ✓).

(b) Split each number into integer and fraction parts.

$2.8125$: integer $2 = 10_2$. Fraction $0.8125$:

Multiply	Product	Digit
$0.8125 \times 2$	1.625	1
$0.625 \times 2$	1.25	1
$0.25 \times 2$	0.5	0
$0.5 \times 2$	1.0	1

So $2.8125_{10} = 10.1101_2$. Check: $2 + 0.5 + 0.25 + 0.0625 = 2.8125$ ✓.

$13.375$: integer $13 = 1101_2$ ($13 \div 2 = 6$ r 1, $6 \div 2 = 3$ r 0, $3 \div 2 = 1$ r 1, $1 \div 2 = 0$ r 1). Fraction $0.375$:

Multiply	Product	Digit
$0.375 \times 2$	0.75	0
$0.75 \times 2$	1.5	1
$0.5 \times 2$	1.0	1

So $13.375_{10} = 1101.011_2$. Check: $8 + 4 + 1 + 0.25 + 0.125 = 13.375$ ✓.

Exercise 5

Convert $0.1_{10}$ and $0.6_{10}$ to binary, keeping 8 bits after the binary point. Do the conversions terminate? What decimal value does each 8-bit result actually represent?

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Solution

$0.1_{10}$:

Step	Multiply	Product	Digit
1	$0.1 \times 2$	0.2	0
2	$0.2 \times 2$	0.4	0
3	$0.4 \times 2$	0.8	0
4	$0.8 \times 2$	1.6	1
5	$0.6 \times 2$	1.2	1
6	$0.2 \times 2$	0.4	0
7	$0.4 \times 2$	0.8	0
8	$0.8 \times 2$	1.6	1

At step 6 the fraction $0.2$ has appeared before (after step 1), so the pattern repeats forever: $0.1_{10} = 0.0\overline{0011}_2 = 0.000110011001\ldots_2$. It never terminates.

Kept to 8 bits: $0.00011001_2 = \dfrac{25}{256} = 0.09765625$. The error is $0.1 - 0.09765625 = 0.00234375$.

$0.6_{10}$:

Step	Multiply	Product	Digit
1	$0.6 \times 2$	1.2	1
2	$0.2 \times 2$	0.4	0
3	$0.4 \times 2$	0.8	0
4	$0.8 \times 2$	1.6	1
5	$0.6 \times 2$	1.2	1
6	$0.2 \times 2$	0.4	0
7	$0.4 \times 2$	0.8	0
8	$0.8 \times 2$	1.6	1

The fraction $0.6$ returns at step 5, so $0.6_{10} = 0.\overline{1001}_2$, also non-terminating.

Kept to 8 bits: $0.10011001_2 = \dfrac{153}{256} = 0.59765625$.

Both fail to terminate because $0.1 = \tfrac{1}{10}$ and $0.6 = \tfrac{3}{5}$ have a factor 5 in the denominator, which is not a power of 2. By contrast $0.8125 = \tfrac{13}{16}$ and $0.375 = \tfrac{3}{8}$ terminate.

Exercise 6

Convert $\text{7A9.C}_{16}$ to binary and to octal. Verify by converting both to decimal.

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Solution

Hex → binary: replace each hex digit by 4 bits.

$$7 \to 0111,\quad \text{A} \to 1010,\quad 9 \to 1001,\quad \text{C} \to 1100$$$$\text{7A9.C}_{16} = 0111\,1010\,1001.1100_2 = 11110101001.11_2$$

(drop the leading 0 and the trailing 0s).

Binary → octal: regroup in 3s outward from the point. Integer part, from the right: $11\,110\,101\,001 \to 011\,110\,101\,001$. Fraction part, from the left: $.11 \to .110$.

$$011\ 110\ 101\ 001\ .\ 110 \;\to\; 3\ 6\ 5\ 1\ .\ 6$$

So $\text{7A9.C}_{16} = 3651.6_8$.

Verify:

$$\text{7A9.C}_{16} = 7(256) + 10(16) + 9 + 12/16 = 1792 + 160 + 9 + 0.75 = 1961.75$$$$3651.6_8 = 3(512) + 6(64) + 5(8) + 1 + 6/8 = 1536 + 384 + 40 + 1 + 0.75 = 1961.75 \ ✓$$

Grouping always starts at the radix point. Grouping the fraction from the right-hand end (e.g. turning $.11$ into $.011 = .3_8$) gives the wrong answer.

Exercise 7

(a) How many bits are needed to represent $1000_{10}$ in unsigned binary?

(b) What is the highest power of 3 needed to write $104_{10}$ in base 3, and how many base-3 digits does it have?

(c) What range of unsigned values fits in 12 bits? How many hex digits does a 16-bit word take?

(d) A machine has $2^{24}$ bytes of memory and 32-bit words. How many address bits are needed if memory is (i) byte-addressable, (ii) word-addressable?

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Solution

(a) $\log_2 1000 \approx 9.97$, so the highest power is $2^9 = 512$ and the number of bits is $\lfloor 9.97 \rfloor + 1 = 10$. Check: $2^9 = 512 \le 1000 \lt 1024 = 2^{10}$, so 9 bits (max 511) are too few and 10 bits (max 1023) suffice. Indeed $1000_{10} = 1111101000_2$ (10 bits).

(b) $\log_3 104 = \dfrac{\ln 104}{\ln 3} \approx \dfrac{4.644}{1.099} \approx 4.228$. The highest power is $3^{\lfloor 4.228 \rfloor} = 3^4 = 81$, and the number of digits is $4 + 1 = 5$ (matches $10212_3$).

(c) 12 bits give $2^{12} = 4096$ patterns: unsigned values $0$ to $4095$. A 16-bit word is $16 / 4 = 4$ hex digits (each hex digit = one nibble).

(d)

• (i) Byte-addressable: one address per byte, $2^{24}$ addresses → 24 bits.
• (ii) Word-addressable: each 32-bit word is 4 bytes $= 2^2$ bytes, so there are $2^{24}/2^2 = 2^{22}$ words → 22 bits.