Range and Spacing of Floating-Point Numbers

Only a finite set of numbers can be represented in a floating-point system. This section answers three questions:

1. How many numbers can be represented?
2. What are the smallest and largest representable values?
3. How are the numbers spaced on the number line?

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Working Example Setup

We use Convention 1 (Standard Form) with:

$\beta = 2,\quad m = 3,\quad e_{\min} = -1,\quad e_{\max} = 2$

Step 1: List the Distinct Mantissas

In Convention 1, $d_1 = 1$ is fixed; $d_2$ and $d_3$ each range over $\{0, 1\}$.
This gives $2^2 = \mathbf{4}$ distinct mantissas:

Mantissa	Decimal value
$0.100$	$\frac{1}{2} = 0.5$
$0.101$	$\frac{1}{2} + \frac{1}{8} = \frac{5}{8} = 0.625$
$0.110$	$\frac{1}{2} + \frac{1}{4} = \frac{3}{4} = 0.75$
$0.111$	$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} = 0.875$

Step 2: List the Possible Exponents

$e \in \{-1,\; 0,\; 1,\; 2\}$ → $\mathbf{4}$ exponent values.

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Counting Representable Numbers

$\text{Non-negative numbers} = 4 \text{ mantissas} \times 4 \text{ exponents} = \mathbf{16}$

Including the sign bit: $16 \times 2 = \mathbf{32}$ representable numbers (zero is a special case in Convention 1 — it cannot be stored exactly).

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Smallest and Largest Values

Smallest positive number: smallest mantissa, smallest exponent:

$$X_{\min} = (0.100)_2 \times 2^{-1} = \frac{1}{2} \times \frac{1}{2} = \boxed{\frac{1}{4}}$$

Largest positive number: largest mantissa, largest exponent:

$$X_{\max} = (0.111)_2 \times 2^{2} = \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right) \times 4 = \frac{7}{8} \times 4 = \boxed{\frac{7}{2} = 3.5}$$

Considering the sign, the full representable range is $\left[-\dfrac{7}{2},\; \dfrac{7}{2}\right]$.

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Spacing Between Adjacent Numbers

For a fixed exponent $e$, moving from one mantissa to the next changes only the last digit $d_m$, contributing $\beta^{-m} \times \beta^e = \beta^{e-m}$.

For our example ($m = 3$, $\beta = 2$), the spacing at exponent $e$ is $2^{e-3}$:

Exponent $e$	Numbers (positive)	Spacing $= 2^{e-3}$
$-1$	$\frac{1}{4},\; \frac{5}{16},\; \frac{3}{8},\; \frac{7}{16}$	$\frac{1}{16}$
$0$	$\frac{1}{2},\; \frac{5}{8},\; \frac{3}{4},\; \frac{7}{8}$	$\frac{1}{8}$
$1$	$1,\; \frac{5}{4},\; \frac{3}{2},\; \frac{7}{4}$	$\frac{1}{4}$
$2$	$2,\; \frac{5}{2},\; 3,\; \frac{7}{2}$	$\frac{1}{2}$

Verify for $e = -1$:

$\frac{5}{16} - \frac{1}{4} = \frac{5}{16} - \frac{4}{16} = \frac{1}{16} \checkmark$

Visualising the Distribution

Notice how the numbers become more spread out as the exponent grows. Near zero, numbers are densely packed; near the maximum, they are far apart.

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The Key Insight: Non-Uniform Spacing

Caution

Floating-point numbers are not evenly spaced on the number line. The absolute gap between adjacent numbers grows proportionally with the exponent. However, the relative gap (gap ÷ value) stays approximately constant, which is about $\beta^{1-m}$ for Convention 1. This is why floating-point is described as a relative error system: you get roughly the same number of significant digits regardless of the magnitude of the number you are representing.

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