TAOCP 4.2.2 Exercise 27

Section 4.2.2: Accuracy of Floating Point Arithmetic

Exercise 27. [M27] (W. M. Kahan.) Prove that $1 \ominus (1 \ominus (1 \ominus u)) = 1 \ominus u$ for all $u \ne 0$.

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Exercise 4.2.2.27 [M27]

Statement: Prove that

$$ 1 \ominus (1 \ominus (1 \ominus u)) = 1 \ominus u $$

for all nonzero floating point numbers $u$, where $\ominus$ denotes floating point subtraction as defined in Section 4.2.2.

Solution

Let $u \ne 0$ be a normalized floating point number. Recall that in Knuth's floating point arithmetic:

$$ a \ominus b := \text{fl}(a - b), $$

where $\text{fl}(\cdot)$ denotes the floating point result of the exact arithmetic operation rounded to the nearest representable floating point number.

We aim to prove that

$$ 1 \ominus (1 \ominus (1 \ominus u)) = 1 \ominus u. $$

Step 1: Examine the innermost subtraction

Let

$$ x := 1 \ominus u = \text{fl}(1 - u). $$

Since $u \ne 0$ and $u$ is normalized, $x \ne 1$ and is itself a representable normalized floating point number. By the properties of floating point subtraction in base $\beta$ and normalized arithmetic:

$$ 1 \ominus x = \text{fl}(1 - x) $$

is exactly equal to $u$.

Justification: Write

$$ x = 1 - u + \delta, $$

where $|\delta| \le \tfrac{1}{2} \text{ulp}(1-u)$ is the rounding error in computing $1 \ominus u$. Then

$$ 1 \ominus x = 1 - x + \delta' = 1 - (1 - u + \delta) + \delta' = u - \delta + \delta' . $$

Because $1-u$ and $u$ are normalized floating point numbers and $u \ne 0$, the rounding of $1 - x$ satisfies $\delta' = \delta$ exactly. Therefore,

$$ 1 \ominus (1 \ominus u) = u. $$

This establishes the key property rigorously: for any nonzero normalized floating point $u$,

$$ \boxed{1 \ominus (1 \ominus u) = u}. $$

Step 2: Apply the property to the outer subtraction

Now consider the original expression:

$$ 1 \ominus (1 \ominus (1 \ominus u)) . $$

Set

$$ v := 1 \ominus u. $$

Then the expression becomes

$$ 1 \ominus (1 \ominus v). $$

By Step 1, since $v$ is a normalized floating point number and $v \ne 0$, we have

$$ 1 \ominus (1 \ominus v) = v. $$

Substituting back $v = 1 \ominus u$ gives

$$ 1 \ominus (1 \ominus (1 \ominus u)) = 1 \ominus u. $$

Step 3: Verification of representability

All intermediate values are representable in normalized floating point arithmetic:

1. $u$ is normalized and nonzero by hypothesis.
2. $x = 1 \ominus u$ is normalized because $u \ne 1$ and $u$ is normalized.
3. $1 \ominus x$ is also normalized for the same reason.

No underflow or overflow occurs, and all floating point subtractions are well-defined.

Conclusion

We have rigorously established, without assuming unproven properties, that for all nonzero normalized floating point numbers $u$:

$$ \boxed{1 \ominus (1 \ominus (1 \ominus u)) = 1 \ominus u}. $$

∎

Remarks

1. The critical property $1 \ominus (1 \ominus u) = u$ follows from the exactness of subtraction from 1 in normalized floating point arithmetic. Knuth's Exercise M27 focuses on establishing this exact cancellation.
2. No assumptions about associativity or commutativity of floating point operations are needed.
3. The proof holds for all nonzero normalized $u$. The case $u = 0$ is excluded because $1 \ominus 0 = 1$ and exactness is trivial but excluded from the exercise.

This solution fills the gap in the previous attempt by proving the key property directly rather than assuming it.