Binary number system and bit - byte concepts: Worked examples

The scenario matrix

Every binary/byte question is really one of these case classes. We work at least one example of each.

Cell	Case class	What makes it tricky	Covered by
A	Small binary → decimal	line up column weights	Ex 1
B	Full byte → decimal (all 1s)	the max value $255$	Ex 2
C	Zero / degenerate input	$0$ and leading zeros	Ex 3
D	Decimal → binary (odd)	remainders bottom-up	Ex 4
E	Decimal → binary (exact power of 2)	a single lone $1$	Ex 5
F	"How many bits fit this number?"	$2^{n} > N$ , off-by-one	Ex 6
G	Real-world storage word problem	KiB vs kB units	Ex 7
H	Exam twist: max-value trap	$2^{n}$ vs $2^{n} - 1$	Ex 8

Figure — Binary number system and bit - byte concepts

Keep that picture in your head — each bit is a switch, each column doubles. Every example below is just this ruler used forwards or backwards.

Cell A — small binary → decimal

Cell B — the full byte (all ones)

Cell C — zero and degenerate inputs

Cell D — decimal → binary, odd number

Worked example Convert

$3 7_{10}$ to binary (repeated division by 2) Forecast: $37$ is between $32$ and $64$ , so it needs a $32$ -column → 6 bits. Guess the pattern.

Divide by 2 repeatedly, record each remainder:

step	divide	quotient	remainder (bit)
1	$37/2$	$18$	$1$
2	$18/2$	$9$	$0$
3	$9/2$	$4$	$1$
4	$4/2$	$2$	$0$
5	$2/2$	$1$	$0$
6	$1/2$	$0$	$1$
Why this step? Each division asks "how many pairs, and is one left over?" The remainder is the current lowest bit.

Read remainders bottom → top: $10010 1_{2}$ . Why bottom-up? The first remainder is the $2^{0}$ (rightmost) bit; the last is the most significant. Read downward and you'd reverse the number.

Verify: $10010 1_{2} = 32 + 0 + 0 + 4 + 0 + 1 = 37$ . ✓

Cell E — decimal → binary, exact power of two

Cell F — "how many bits do I need?"

Cell G — real-world storage word problem

Cell H — the exam twist (max-value trap)

Case coverage check

Recall Did every matrix cell get worked?

A small→dec ::: Ex 1 ( $110 1_{2} = 13$ ) B all-ones byte ::: Ex 2 ( $1111111 1_{2} = 255$ ) C zero / leading zeros ::: Ex 3 ( $0 = 0$ , $0010 1_{2} = 5$ ) D decimal→binary odd ::: Ex 4 ( $37 = 10010 1_{2}$ ) E exact power of two ::: Ex 5 ( $32 = 10000 0_{2}$ ) F bits needed ::: Ex 6 ( $500 \to 9$ bits) G real-world units ::: Ex 7 ( $3$ MiB $= 25, 165, 824$ bits, $85$ photos) H max-value trap ::: Ex 8 (4 bits max $= 15$ )

Connections

Binary number system and bit - byte concepts — the parent rules these examples exercise.
Hexadecimal number system — group the 8 bits of Ex 2 into two nibbles for a hex shortcut.
Two's complement — how the same bit patterns encode negative numbers.
Logic Gates — gates act on exactly the $0/1$ bits we manipulated here.
Memory addressing — why the byte in Ex 7 is the addressable unit.