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Catastrophic cancellation occurs when subtraction of nearly equal approximations removes their matching leading digits and exposes earlier roundoff error. The subtraction can itself be exact on the stored operands: the information may already have been lost while obtaining them. Goldberg distinguishes this from cancellation of accurately known quantities, which can be harmless.^[ref]

For $x>0$, consider the single real-valued functionThe rationalization of a square-root difference proves equality in exact arithmetic. The two evaluations use the same input $x$, but the direct version first forms a value near one and then subtracts one. The rearranged version keeps the small input in its numerator and divides by a denominator near two.

For binary64 x = 1e-16, the rounded sum 1.0 + x is one, so sqrt(1.0 + x) - 1.0 is zero. The rationalized evaluation returns approximately $5\times 10^{-17}$. A 100-digit Decimal calculation using the exact stored input gives relative errors of one and about $2.50\times 10^{-17}$ respectively. At x = 1e-12, the direct result is about $5.0004445029\times 10^{-13}$, with relative error $8.89\times 10^{-5}$; the rearranged result is about $4.99999999999875\times 10^{-13}$, with relative error $1.21\times 10^{-17}$.

The plot uses Decimal.from_float(x) to preserve the binary input exactly, computes $x/(\sqrt{1+x}+1)$ at 100 decimal digits, and verifies the reference at 160 digits. Exact conversion of a stored binary value is documented by Python.^[ref] Each error is $|\hat{f}-f|/|f|$, with the computed result converted exactly to Decimal before comparison. Hence the curves measure evaluation error for one common input, rather than a mixture of input conversion error and evaluation error. The plotted interval $10^{-16}\le x\le 10^{-1}$ is far above the subnormal number range.

The original problem is well conditioned with respect to relative changes in $x$. Using the derivative $f^{\prime }(x)=1/(2\sqrt{1+x})$ and rationalization of a square-root difference, its relative condition number isIt tends to one as $x$ tends to zero. The direct evaluation instead introduces the intermediate difference $s-1$ with $s=\sqrt{1+x}$; an error of order unit roundoff in $s$ is large relative to $f(x)\approx x/2$. This is a failure of the chosen algorithm's accuracy despite a small condition number for the same real-valued function. The rationalized evaluation avoids that intermediate catastrophic cancellation, retaining relative accuracy near the unit roundoff scale throughout the plotted interval. Recognizing this structure can improve a calculation far more than merely printing extra digits.

Use rationalization of a square-root difference to avoid subtracting two large, nearly equal quantities:Since $1/n\to 0$ and the square root is continuous at $1$, the denominator tends to $2$. The reciprocal limit law established above is applicable because this limit is nonzero. HenceThe limit is one half; rationalization exposes the finite difference hidden by cancellation.

Multiplying numerator and denominator of a fraction by the same nonzero expression can remove a square root from the denominator. For instance,when $a\ge 0$ and both denominators are nonzero. This uses the difference-of-squares identity $(\sqrt{a}+b)(\sqrt{a}-b)=a-b^2$. If $a=b^2$, the original expression may still be defined even though the rationalized quotient is not, so that case must be handled separately. The related rationalization of a square-root difference applies the same multiplication to a numerator.