Floating-Point Numbers

A floating-point number is scientific notation in base 2 with a fixed number of digits. Because most fractions do not end in binary (see Binary Fractions), nearly every real number has to be rounded to fit, and the pattern of that rounding explains a lot of computer arithmetic.

A (normalized) float with $p$ fraction bits is $$x = \pm(1.b_1b_2\ldots b_p)_2 \times 2^{e},$$ where the exponent $e$ lies in a fixed range. A double (IEEE 754 double precision) uses 1 sign bit, 11 exponent bits and $p = 52$ fraction bits: 64 bits in all, with 53 significant bits counting the leading 1, which is not stored.

Key idea

Between $2^{e}$ and $2^{e+1}$ the floats are evenly spaced, $2^{e - p}$ apart, and the spacing doubles at each power of 2. Machine epsilon $\varepsilon = 2^{-p}$ is the gap between $1$ and the next float; for doubles $\varepsilon = 2^{-52} \approx 2.2 \times 10^{-16}$. Rounding to the nearest float has relative error at most $u = \varepsilon/2$.

Why it works

Inside $[2^{e}, 2^{e+1})$ only the fraction bits change, and the last one is worth $2^{-p} \times 2^{e}$: that is the gap. For $x$ in that range the gap is at most $2^{-p} \cdot 2^{e} \le \varepsilon|x|$. Rounding to the nearest float moves $x$ by at most half a gap, so $$\frac{|\mathrm{fl}(x) - x|}{|x|} \le \frac{\varepsilon}{2} = u.$$ The absolute error grows with $|x|$, but the relative error never exceeds $u$: about 16 significant decimal digits for doubles.

A toy system with 3 fraction bits. The top strip shows every positive float; each colored band is one power-of-2 range, and the gaps double from one band to the next.

Worked example: Floats in a toy system

With $p = 3$ fraction bits, list the floats in $[1, 2)$ and find the gap in $[4, 8)$.

In $[1, 2)$: $(1.b_1b_2b_3)_2$ for the 8 choices of bits: $1, 1.125, 1.25, \ldots, 1.875$, a gap of $2^{-3} = 0.125$.
In $[4, 8)$ the exponent is $2$, so the gap is $2^{-3} \times 2^{2} = 0.5$: the floats there are $4, 4.5, 5, \ldots, 7.5$.
Machine epsilon for this system is $2^{-3} = 0.125$.

Answer: gap $0.125$ in $[1, 2)$, gap $0.5$ in $[4, 8)$

Worked example: Rounding and its error

In the same system, round $x = 2.7$ to the nearest float and find the relative error.

$2.7$ is in $[2, 4)$, where the gap is $2^{-3} \times 2 = 0.25$: the neighbors are $2.5$ and $2.75$.
$2.75$ is nearer: $\mathrm{fl}(2.7) = 2.75$, an error of $0.05$.
Relative error: $\frac{0.05}{2.7} \approx 0.0185$, below $u = \frac{\varepsilon}{2} = 0.0625$.

Answer: $2.75$, relative error about $0.0185$

Worked example: Why 0.1 + 0.2 is not 0.3

Explain what a computer does with $0.1 + 0.2$ in double precision.

$0.1$ is stored as $0.100\,000\,000\,000\,000\,005\,55\ldots$ and $0.2$ as $0.200\,000\,000\,000\,000\,011\,1\ldots$: each is rounded.
Their exact sum, $0.300\,000\,000\,000\,000\,016\,6\ldots$, is rounded again, to the double $0.300\,000\,000\,000\,000\,044\,4\ldots$
The double nearest $0.3$ is $0.299\,999\,999\,999\,999\,988\,9\ldots$, one gap lower. So the test "0.1 + 0.2 == 0.3" is false.
Compare floats with a tolerance instead: $|a - b| \le 10^{-12}\max(|a|, |b|)$, say.

Answer: Each value is rounded to the nearest double, and the roundings do not cancel.

The 64 bits of a double, the exact value it stores, and the 0.1 + 0.2 example.

Numbers worth knowing.
Double precision	Value
fraction bits $p$	$52$
machine epsilon $\varepsilon = 2^{-52}$	$\approx 2.22 \times 10^{-16}$
unit roundoff $u = 2^{-53}$	$\approx 1.11 \times 10^{-16}$
largest double	$\approx 1.8 \times 10^{308}$
smallest normal double	$\approx 2.2 \times 10^{-308}$

Common mistake

Expecting $0.1 + 0.2$ to equal $0.3$ exactly, or picturing floats as evenly spaced. Floats are dense near $0$ and sparse far out: near $10^{6}$ the gap between doubles is about $10^{-10}$, near $1$ it is about $10^{-16}$. What stays the same is the relative spacing.

Floating-Point Numbers

Practice

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Sources and further reading