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Proof that a guard digit bound the error of subtraction

I was reading What Every Computer Scientist Should Know About Floating-Point Arithmetic, which is extremely interesting. But I have some troubles understanding the proof of Theorem 9 (page 33).

First a pretty trivial question. When the formula $ (15)$ say: $ $ y – \bar{y} \lt (\beta – 1)(\beta^{-p} + \dots + \beta^{-p-k})$ $ Shouldn’t it be $ \le$ instead of $ \lt$ , or did I miss something?

More importantly, I do not understand why it say that if $ x-\bar{y} \lt 1$ , then $ \delta = 0$ . How can there be no rounding error?

It is then said that: $ $ x – y \ge 1.0 – 0.\overbrace{0 \dots 0}^k\overbrace{\rho \dots \rho}^k \; \textrm{with} \; \rho = \beta – 1$ $

Why is that so? Can’t the difference be arbitrarily small or even $ 0$ ?

And overall, correct me if I’m wrong, but the guard digit only save the day if the input floats $ x$ and $ y$ are the exact numbers we want to subtract. If they are the exactly-rounded result of another computation, then we might still have a catastrophic cancellation. The guard digit would only put the computed value within the correct range.

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