Floating Point Numbers (IEEE 754)

Context: FIT1047_MOC · binary scientific notation for numbers of varying scale · significand + exponent + sign · why $0.1$ is inexact

Quick Revision

• 🎯 Objective: represent $s\times 2^e$ with fixed-size fields ➔ IEEE 754 double: 52-bit significand + 11-bit exponent + 1 sign bit $=64$ bits.
• 📦 Core Components: binary point ➔ normalise to $1.xxxx\times 2^e$ ➔ store fraction + exponent (excess-$K$).
• ⚡ Key Constraint: finite bits ⟹ rounding — $0.1_{10}$ has no exact binary form; errors accumulate.

📝 Core

• Motivation ➔ integers can’t hold $c=3\times 10^5$ km/s and $t=15\times 10^{-5}$ s in one scheme (scaling is ad-hoc; rounding gives $t=0$) ➔ scientific notation makes it easy: $t\times c=45\times 10^{5-5}=45$ km.
• Binary fractions ➔ positions right of the point weigh $2^{-1},2^{-2},\dots$: $110.01_2=4+2+0.25=6.25_{10}$.
• Normalisation ➔ shift the point so the number starts "$1.$": $110.01_2=1.1001\times 2^2$ — exponent = shift count.
• Storage (IEEE 754 double) ➔ sign $1$ bit · exponent $11$ bits · significand $52$ bits; special FPU hardware computes on these.
• Exponent encoding ➔ negative exponents needed ➔ stored in excess-$K$ (biased), not 2’s complement (detail out of W1 scope).
• Precision trade ➔ 64-bit float vs 64-bit int: the float spends 12 bits on sign+exponent ⟹ large integers stored as floats lose exactness.
• The $0.1$ problem ➔ no finite sum of powers of $2$ equals $10^{-1}$; IEEE 754 stores $\approx 0.10000000000000000555\dots$ — tiny per-operation errors compound ⟹ never floats for money.

📊 Applied Exercise — encode $6.25_{10}$ (6-bit significand, 4-bit exponent, from lecture)

$$\begin{array}{rlrl}6.25_{10}& =110.01_2& & (4+2+\frac{1}{4})\\ & =1.1001\times 2^2& & \text{(normalise: point moves 2 left)}\\ s& =001001,e=2& & \text{(exponent stored in excess-}K\text{ in real IEEE 754)}\end{array}$$

Final Extracted Output: significand $001001$, exponent $2$, sign $0$.

✍️ Practice

Convert $10.75_{10}$ to binary, normalise it, and state the significand and exponent. Then explain in one sentence why a banking system must not store balances as floats.

Answer

• $10.75=1010.11_2=1.01011\times 2^3$ ⟹ significand $01011\dots$, exponent $3$.
• Money: decimal fractions like $0.10$ are inexact in binary ⟹ rounding errors accumulate across millions of transactions — use integer cents.
• Key move: normalise until exactly one $1$ precedes the point; the shift count IS the exponent.

⚠️ Common Mistakes

• 💡 “Looks exact” isn’t ➔ printed $0.1$ hides the stored $0.100000000000000005\dots$; equality tests on floats (a == 0.3) fail unpredictably.
• 💡 Floats don’t replace ints ➔ same 64 bits, but floats trade exactness for range; big integers silently round.
• 💡 Exponent isn’t 2’s complement ➔ IEEE 754 uses excess-$K$ bias — don’t decode exponent fields with the signed-integer rules.