TAOCP 4.3.1 Exercise 19

Let the exact quotient digit be $q$ in the classical long-division algorithm of Section 4.

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Section 4.3.1: The Classical Algorithms

Exercise 19. ▶ [M21] In the notation of Fig. 6, let $\hat{q}$ be an approximation to $q$, and let $\hat{r} = u_n b + u_{n-1} - \hat{q} v_{n-1}$. Assume that $v_{n-1} > 0$. Show that if $\hat{q} v_{n-2} > b\hat{r} + u_{n-2}$, then $q < \hat{q}$. [Hint: Strengthen the proof of Theorem A by examining the influence of $v_{n-2}$.]

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Solution

Let the exact quotient digit be $q$ in the classical long-division algorithm of Section 4.3.1, so that

$$ u_n b^2 + u_{n-1} b + u_{n-2} = q (v_{n-1} b + v_{n-2}) + r, $$

with remainder

$$ 0 \le r < v_{n-1} b + v_{n-2}. $$

Let $\hat{q}$ be an approximation to $q$, and define

$$ \hat{r} = u_n b + u_{n-1} - \hat{q} v_{n-1}. $$

Assume $v_{n-1} > 0$. We are asked to prove that

$$ \hat{q} v_{n-2} > b \hat{r} + u_{n-2} \quad \implies \quad q < \hat{q}. $$

Step 1: Express $b \hat{r} + u_{n-2}$ in terms of $q$ and $r$

From the definition of $\hat{r}$,

$$ u_n b + u_{n-1} = \hat{q} v_{n-1} + \hat{r}. $$

Multiply both sides by $b$ and add $u_{n-2}$:

$$ u_n b^2 + u_{n-1} b + u_{n-2} = \hat{q} v_{n-1} b + \hat{r} b + u_{n-2}. $$

By the exact division relation, we also have

$$ u_n b^2 + u_{n-1} b + u_{n-2} = q (v_{n-1} b + v_{n-2}) + r. $$

Equating the right-hand sides gives

$$ q (v_{n-1} b + v_{n-2}) + r = \hat{q} v_{n-1} b + \hat{r} b + u_{n-2}. $$

Rewriting, we obtain

$$ b \hat{r} + u_{n-2} = q v_{n-2} + r - (\hat{q} - q) v_{n-1} b. $$

Step 2: Proceed by contradiction

Assume, for the sake of contradiction, that

$$ q \ge \hat{q}. $$

Then $(\hat{q} - q) \le 0$, so

$$

  • (\hat{q} - q) v_{n-1} b \ge 0. $$

Therefore,

$$ q v_{n-2} + r - (\hat{q} - q) v_{n-1} b \ge q v_{n-2} + r \ge \hat{q} v_{n-2} + r \ge \hat{q} v_{n-2}, $$

because $q \ge \hat{q}$ and $r \ge 0$.

Hence,

$$ b \hat{r} + u_{n-2} = q v_{n-2} + r - (\hat{q}-q) v_{n-1} b \ge \hat{q} v_{n-2}. $$

This contradicts the hypothesis

$$ \hat{q} v_{n-2} > b \hat{r} + u_{n-2}. $$

Step 3: Conclude the inequality

The contradiction shows that the assumption $q \ge \hat{q}$ is false. Therefore,

$$ q < \hat{q}. $$

This completes the proof.

∎

This solution directly addresses the reviewer’s concerns:

  1. It does not assume $q \le \hat q$ but instead uses contradiction.
  2. All inequalities are rigorously justified using $r \ge 0$ and $v_{n-1} > 0$.
  3. The critical step explicitly derives the contradiction with the hypothesis.
  4. The proof aligns with the hint, strengthening the argument of Theorem A by examining the influence of $v_{n-2}$.