Number Systems (Binary and Hexadecimal)

Context: FIT1047_MOC · positional notation in bases $2$, $10$, $16$ · the Assignment 1 hand skill · basis of Signed Integer Representation (Two’s Complement) and Floating Point Numbers (IEEE 754)

Quick Revision

• 🎯 Objective: each digit is worth (digit) × (base)${}^{\text{position}}$ ➔ convert any direction between binary/decimal/hex + add binary by hand.
• 📦 Core Components: weights table (binary→dec) ➔ repeated division (dec→binary) ➔ 4-bits-per-hex-digit (binary↔hex).
• ⚡ Key Constraint: $n$ bits represent $2^n$ values ($0$ to $2^n-1$) — the counting question examiners love.

📝 Core

• Positional notation ➔ $2396_{10}=2\times 10^3+3\times 10^2+9\times 10^1+6\times 10^0$; same idea in base 2: $2396_{10}=100101011100_2$.
• Binary → decimal ➔ write weights $2^7\dots 2^0$ over the bits, add the weights under the 1s.
• Decimal → binary ➔ repeated division by 2, remainders read bottom-up (standard method for the A1 skill; slides teach the weights direction — drill both).
• Capacity ➔ $n$ bits ⟹ $2^n$ values: 3 bits → $\mathrm{0..7}$, 5 bits → $\mathrm{0..31}$, 8 bits → $\mathrm{0..255}$.
• Binary addition ➔ long addition with carries: $1+1=10_2$ (write 0 carry 1).
• Hex ➔ base 16, digits $\mathrm{0..9},A..F$; one hex digit = exactly 4 bits ➔ group binary in 4s from the right: $100101011100_2=95C_{16}$.

⚖️ Core Decision Matrix

Conversion	Method	Micro-example
binary → decimal	weights table, sum the 1-positions	$11011101_2=128+64+16+8+4+1=221$
decimal → binary	÷2, remainders bottom-up	$13=1101_2$ ($13\to 6r1,6\to 3r0,3\to 1r1,1\to 0r1$)
binary → hex	group 4 bits from right → digit each	$100101011100_2\to 100101011100\to 95C_{16}$
hex → binary	each digit → 4 bits	$A{7}_{16}\to 10100111_2$
decimal ↔ hex	via binary (fastest by hand)	$221\to 11011101_2\to D{D}_{16}$

📊 Exam Execution Trace

Manual Execution Trace — binary addition $1001_2+0101_2$

Step	Column ($2^k$)	Bits + carry	Write	Carry
1	$2^0$	$1+1$	$0$	$1$
2	$2^1$	$0+0+1$	$1$	$0$
3	$2^2$	$0+1$	$1$	$0$
4	$2^3$	$1+0$	$1$	$0$
Result: $1110_2=14_{10}$ ✓ ($9+5$).

✍️ Practice

(a) Convert $174_{10}$ to binary and hex. (b) How many different values can 6 bits represent, and what is the largest? (c) Add $10111_2+1101_2$.

Answer

• (a) $174=10101110_2=A{E}_{16}$ (check: $128+32+8+4+2=174$).
• (b) $2^6=64$ values, largest $=2^6-1=63$.
• (c) $10111_2+1101_2=100100_2$ ($23+13=36$ ✓).
• Key move: always verify by converting back to decimal.

⚠️ Common Mistakes

• 💡 $2^n$ values vs largest value ➔ $n$ bits give $2^n$ values but the max is $2^n-1$ (zero occupies a slot).
• 💡 Group hex from the RIGHT ➔ padding goes on the left; grouping from the left mangles the number.
• 💡 Write the base subscript ➔ $101$ is ambiguous; $101_2\ne 101_{10}\ne 101_{16}$ — unlabeled bases lose marks.