This is about the proof of Theorem A on pg. 235 of Knuth's "The Art of Computer Programming" Vol. 2, 3rd Ed.
Background: By "normalized floating point number" Knuth means a number of the form \begin{equation} \pm (0.a_1 \ldots a_p)_b \times b^e \qquad \text{(1)} \end{equation} where the base $b$ is an integer at least 2, the precision $p$ is an integer at least 1, and $a_1 \in \{1, \ldots p-1\}$. The exponent $e$ is an integer in some range - in fact it seems to me that by $e$ Knuth might mean here what he called earlier $e-q$, where the integer $q$ is a so-called "excess". Throughout we are assuming that all exponents remain within that range, i.e. there is no exponent underflow nor overflow. The notation $\text{round}(x)$ means the closest normalized number to $x$, where we make no assumption about how ties are broken. Given normalized floating point numbers $u$ and $v$, the notation $u \oplus v$ stands for $\text{round}(u+v)$ and $u \ominus v$ stands for $\text{round}(u-v)$. The system described is a precursor of the IEEE-754 standard for floating-point numbers and arithmetic.
Knuth writes:
Theorem A. Let $u$ and $v$ be normalized floating point numbers. Then \begin{equation} ((u \oplus v) \ominus u) + ((u \oplus v) \ominus ((u \oplus v) \ominus u)) = u \oplus v, \qquad \text{(40)} \end{equation} provided that no exponent overflow or underflow occurs.
He introduces the notation
\begin{align} u' &= (u \oplus v) \ominus v , & v' &= (u \oplus v) \ominus u; \\ u'' &= (u \oplus v) \ominus v', & v'' &= (u \oplus v) \ominus u'. \end{align}
The proof of Theorem A then begins with a definition and a Lemma, the proof of which is an exercise. I quote:
Proof. Let us say that $t$ is a tail of $x$ modulo $b^e$ if \begin{equation} t \equiv x \text{ (modulo $b^e$) }, \qquad |t| \leq \tfrac{1}{2}b^e; \end{equation}
I take this to mean, "Given real numbers $x$ and $t$ and an integer $e$, $t$ is a tail of $x$ modulo $b^e$ iff..."
thus, $x-\text{round}(x)$ is always a tail of $x$. The proof of Theorem A rests largely on the following simple fact proved in exercise 11:
Lemma T. If $t$ is a tail of the floating point number $x$, then $x \ominus t = x - t$.
I assume here that $t$, like $x$, is intended to be a floating-point number, otherwise it cannot be an operand for the operator $\ominus$, as I understand it. I have written a proof of Lemma T which I omit here.
Let $w = u \oplus v$. Theorem A holds trivially when $w=0$. By multiplying all variables by a suitable power of $b$, we may assume without loss of generality that $e_w=p$.
I suppose that $e_w$ here means the $e$ in the representation of $w$ as in (1).
Then $u+v = w + r$, where $r$ is a tail of $u+v$ modulo 1. Furthermore $u' = \text{round}(w-v) = \text{round}(u-r) = u-r-t$, where $t$ is a tail of $u-r$ modulo $b^e$ and $e=e_{u'}-p$.
If $e \leq 0$, then $t \equiv u-r \equiv -v$ (modulo $b^e$), hence $t$ is a tail of $-v$ and $v'' = \text{round}(w-u') = \text{round}(v+t) = v+t$; this proves (40). If $e>0$, then $|u-r| \geq b^p - \tfrac{1}{2}$; and since $|r| \leq \tfrac{1}{2}$, we have $|u| \geq b^p-1$. It follows that $u$ is an integer, so $r$ is a tail of $v$ modulo 1. If $u'=u$, then $t=-r$ is a tail of $-v$. Otherwise the relation $\text{round}(u-r) \neq u$ implies that $|u|=b^p-1$, $|r|=\tfrac{1}{2}$, $|u'|=b^p$, $t=r$; again $t$ is a tail of $-v$.
Which is the end of the proof of Theorem A. I understand it up until "If $e>0$, then $|u-r| \geq b^p - \tfrac{1}{2}$". Instead, I have the weaker inequality $|u-r| \geq b^p - \tfrac{1}{2}b$. My logic being as follows: We have $u-r=u'+t$, whence $|u-r| \geq | |u'| - |t| | = |u'| - |t|$. Since \begin{equation} u' = \pm (0.a_1 \ldots a_p)_b \times b^{e_{u'}}, \quad 1 \leq a_1 < b, \end{equation} we have $|u'| \geq (0.1)_b \times b^{e_{u'}} = b^{e_{u'}-1} = b^{e+p-1}$. We also have that $|t| \leq \tfrac{1}{2}b^e$, so $|u-r| \geq b^{e+p-1} - \tfrac{1}{2}b^e = b^e(b^{p-1}-\tfrac{1}{2})$.
But $e>0$ is an integer, therefore $e \geq 1$, so $|u-r| \geq b^p - \tfrac{1}{2}b$. My question is, how to prove the stronger inequality $|u-r| \geq b^p - \tfrac{1}{2}$ that appears in the proof of Theorem A ?
