IEEE 754 floating-point standard — significant bits, special values

WHAT is being stored?

The 32-bit (single precision) and 64-bit (double precision) layouts:

Format	Total bits	Sign $s$	Exponent $E$	Fraction $f$	Bias
single	32	1	8	23	127
double	64	1	11	52	1023

Figure — IEEE 754 floating-point standard — significant bits, special values

HOW the bits become a number (derivation from scratch)

Step 1 — Normalize in binary. Any nonzero real in scientific binary looks like $x = \pm 1.b_1b_2b_3\ldots \times 2^{E_{\text{real}}}.$ Why this step? In base 2 the leading digit of a normalized number is always 1 (it can't be 0, or it wouldn't be the leading digit). So we get that bit for free — it is implicit (the "hidden bit").

Step 2 — Store only the fraction. Since the leading $1$ is implied, we store just $f = b_1b_2\ldots$ So the significand is $m = 1.f = 1 + \sum_{i=1}^{p} b_i\,2^{-i}.$ Why this step? Storing the hidden $1$ would waste a bit and buy a free bit of precision: 23 stored fraction bits behave like 24 significant bits.

Step 3 — Bias the exponent. Exponents can be negative ( $2^{-5}$ ) or positive ( $2^{+5}$ ). Instead of a sign for the exponent, we store $E = E_{\text{real}} + \text{bias}, \qquad \text{bias}=2^{k-1}-1$ where $k$ = number of exponent bits. Why? A plain unsigned $E$ then lets simple integer comparison sort floats correctly, and reserves the all-0 and all-1 patterns for special meanings.

Significant bits & precision

Special values (the reserved exponent patterns)

The all-0 and all-1 exponent fields are not ordinary numbers:

$E$ field	fraction $f$	meaning
all 0	$0$	$\pm 0$ (signed zero)
all 0	$\neq 0$	subnormal (denormal) number, no hidden 1: $x=(-1)^s\,(0.f)\,2^{1-\text{bias}}$
$1\ldots254$ (single)	any	normal number (formula above)
all 1	$0$	$\pm\infty$
all 1	$\neq 0$	NaN (Not a Number)

Worked examples

Common mistakes (steel-manned)

Recall Feynman: explain to a 12-year-old

Imagine you can only write 7 digits on a tiny card, but you also get a little "shift the dot" note. So you write $6022$ and a note "move the dot 23 places right" to mean a huge number, or "move it left" for a tiny one. The card is the mantissa, the note is the exponent, and a $+/-$ box is the sign. Because the card has only a few digits, some numbers like $0.1$ can't fit perfectly — you get the closest card you can write. Two special notes are saved: one says "this is so big it's basically infinity," and one says "this isn't a real number at all" (like dividing by zero).

Flashcards

IEEE 754 single-precision bit split (sign/exp/frac)?

1 / 8 / 23 bits

IEEE 754 double-precision bit split?

1 / 11 / 52 bits

Value formula for a normal float?

(-1)^s (1.f)_2 \times 2^{E-\text{bias}}

Bias for single and double precision?

127 and 1023 (i.e.

2^{k-1}-1

)

What is the "hidden bit"?

The implicit leading 1 of a normalized significand, not stored

Effective significant bits in single precision?

24 (23 stored + 1 hidden) ≈ 7 decimal digits

Definition of machine epsilon?

Gap between 1.0 and next float,

\varepsilon=2^{-(p-1)}

Machine epsilon (single, double)?

2^{-23}\approx1.19e{-7}

2^{-52}\approx2.22e{-16}

How is

\pm\infty

encoded?

Exponent all 1s, fraction = 0

How is NaN encoded?

Exponent all 1s, fraction

\neq 0

How is a subnormal encoded and valued?

Exp all 0s,

f\neq0

(-1)^s(0.f)\,2^{1-\text{bias}}

Why do subnormals exist?

Gradual underflow — fill the gap between smallest normal and 0

Quick test for NaN in code?

x != x is true only for NaN

Why isn't 0.1 exact?

Its binary expansion repeats forever; 5 is not a power of 2

Decimal digits from p bits?

p\log_{10}2

Connections

Floating-point round-off error
Numerical stability and conditioning
Binary number representation
Two's complement vs biased exponent
Catastrophic cancellation
Machine epsilon and unit in the last place

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, computer ke paas sirf 32 ya 64 bits hote hain, lekin real numbers infinite hote hain. Toh IEEE 754 ek smart jugaad hai: number ko binary scientific notation me likho — $(-1)^s \times m \times 2^e$ . Yaani ek sign bit, ek exponent (kitna shift karna hai), aur ek mantissa (significant digits). Single precision me split hota hai 1 + 8 + 23, aur double me 1 + 11 + 52.

Sabse mazedaar trick hai hidden bit: normalized binary number ka pehla digit hamesha 1 hota hai, toh use store karne ki zaroorat hi nahi — automatic maan lete hain. Isliye 23 stored bits actually 24 significant bits ki tarah kaam karte hain, jo lagbhag 7 decimal digits ke barabar hai. Exponent ko bias ke saath store karte hain (single me 127 add karke), taaki negative exponents bhi unsigned form me aa jaayen aur comparison aasaan ho.

Special values bhi yaad rakho: exponent agar all zeros ho toh number ya to zero hai ya subnormal (bahut chhote numbers, gradual underflow ke liye). Exponent all ones ho toh ya $\pm\infty$ (jaise $1/0$ ) ya NaN (jaise $0/0$ ya $\sqrt{-1}$ ). Ek important baat: NaN kabhi khud ke barabar nahi hota, isliye x != x se NaN check karte hain.

Kyun important hai? Kyunki rounding ki wajah se $0.1 + 0.2 \ne 0.3$ ho jaata hai computer me — $0.1$ binary me exactly store hi nahi hota. Numerical methods me yahi chhoti-chhoti errors add hoke bada blunder ban sakti hain, isliye floats ko == se compare mat karo, hamesha tolerance ke saath compare karo.

Test yourself — Numerical Methods

Connections