Floating-Point Arithmetic and Loss of Significance

When we perform arithmetic on floating-point numbers, each individual operation can introduce a small rounding error. Most of the time these errors stay small. But when subtracting two nearly equal numbers, the errors grow drastically. This is called Loss of Significance (or catastrophic cancellation).

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Rounding in Arithmetic Operations

Setup

Use Convention 1 with $\beta = 2$, $m = 3$, $e_{\min}=-1$, $e_{\max}=2$.

Let: $x = \frac{5}{8} = (0.101)_2 \times 2^0 \qquad y = \frac{7}{8} = (0.111)_2 \times 2^0$

Both are already exact floating-point numbers, so $fl(x)=x$ and $fl(y)=y$.

Computing $x \times y$

The exact product:

$x \times y = \frac{5}{8} \times \frac{7}{8} = \frac{35}{64}$

Converting to binary:

$\frac{35}{64} = (0.100011)_2 \times 2^0$

This has 6 mantissa digits, but our system only holds $m = 3$. We must round.

The 4th bit (the bit immediately past $d_3$) is 0, so we truncate:

$fl(x \times y) = (0.100)_2 \times 2^0 = \frac{32}{64} = \frac{1}{2}$

Rounding error:

$\left|\frac{35}{64} - \frac{32}{64}\right| = \frac{3}{64} \approx 0.047$

This is a normal, bounded rounding error, therefore $\delta \le \xi_M$ holds.

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Where Loss of Significance Comes From

Suppose $x$ and $y$ are not exact: $fl(x) = x(1+\delta_1)$ and $fl(y) = y(1+\delta_2)$.

Computing $fl(x) - fl(y)$:

$$fl(x) - fl(y) = x(1+\delta_1) - y(1+\delta_2) = (x-y) + x\delta_1 - y\delta_2 = (x-y)\!\left(1 + \frac{x\delta_1 - y\delta_2}{x - y}\right)$$

The relative error in the result is:

$\frac{x\delta_1 - y\delta_2}{x - y}$

Danger

If $x \approx y$, the denominator $x - y$ is tiny. Even though $x\delta_1 - y\delta_2$ is small, dividing by a tiny denominator makes the relative error enormous. This is Loss of Significance.

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Classic Example: Quadratic Roots

Solve $x^2 - 56x + 1 = 0$ on a toy computer that rounds to 4 significant figures.

Exact roots (7 s.f.)

$x = \frac{56 \pm \sqrt{56^2 - 4}}{2} = 28 \pm \sqrt{783}$

$x_1 = 28 + \sqrt{783} \approx 55.9822 \qquad x_2 = 28 - \sqrt{783} \approx 0.017862$

With 4 significant figures

The toy computer computes $\sqrt{783} \approx 27.98$ (4 s.f.).

$x_1 = 28 + 27.98 = \mathbf{55.98} \qquad \text{✓ accurate}$

$x_2 = 28 - 27.98 = \mathbf{0.02000} \qquad \text{✗ wrong! (actual: 0.017862)}$

For $x_2$, the two nearly equal numbers $28$ and $27.98$ cancel, leaving only 1 significant figure in the result, which isa severe Loss of Significance.

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Avoiding Loss of Significance

The Workaround: Use Vieta’s Formulas

For $x^2 - 56x + 1 = 0$, Vieta’s formulas give:

$x_1 \cdot x_2 = \frac{c}{a} = \frac{1}{1} = 1 \qquad x_1 + x_2 = \frac{b}{a} = 56$

Strategy: compute the root that is far from zero (no cancellation) using the standard formula, then obtain the other root by division.

Step 1: compute $x_1$ (addition, no cancellation):

$x_1 = 28 + 27.98 = 55.98$

Step 2: compute $x_2$ from the product relation:

$x_1 \cdot x_2 = 1 \;\Longrightarrow\; x_2 = \frac{1}{x_1} = \frac{1}{55.98} = \mathbf{0.01786}$

This perfectly recovers the accurate value of $x_2 \approx 0.017862$ to 4 significant figures.

Caution

General Rule: If two roots $x_1, x_2$ of $x^2 - bx + c = 0$ satisfy $x_1 x_2 = c$, compute the numerically stable root first (the one that does not subtract nearly equal quantities), then find the other as $c / x_1$. This completely avoids the cancellation.

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Summary of Conditions That Cause Loss of Significance

Operation	When dangerous	Why
$x - y$	$x \approx y$	Leading significant digits cancel
$x + y$ with opposite signs	$\|x\| \approx \|y\|$	Same as subtraction
Quadratic formula $\dfrac{-b - \sqrt{b^2-4ac}}{2a}$	$b^2 \gg 4ac$	$\sqrt{b^2-4ac} \approx \|b\|$

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