Let $\mathbb G(q,l)$ to be the set of all floating-point numbers in the base $q$ and with a mantissa of a length $l$. In the lecture notes I am trying to learn from, the following is said to be obvious (and consequently has no proof):
$$ \forall x \in \mathbb G(q,l), \forall y \in \mathbb G(q,l)\setminus\{x\}: |x-y| \ge |x|q^{-l} $$
What I have tried so far:
Let $x=(-1)^m a \cdot q^\alpha$ and $y=(-1)^n b \cdot q^\beta \in \mathbb G(q,l)$ such that $x \ne y$, then
\begin{align} |x-y| &\ge |x|-|y|
\\ &=|(-1)^m a \cdot q^\alpha|-|(-1)^n b \cdot q^\beta|
\\ &=a \cdot q^\alpha – b \cdot q^\beta
\\ &\ge …?
\end{align}
Best Answer
Since it's true for $x=0$, we may assume $x \ne 0$. By symmetry, assume $x > 0$. Let $x = a \cdot q^\beta$ with $a$ an integer, $q^{\ell-1} \le a \le q^{\ell} - 1$. Let $y_1$ and $y_2$ be the greatest floating-point number $< x$ and the least floating-point number $>x$ respectively. If $q^{\ell-1} < a$, then $y_1 = (a-1) \cdot q^\beta$ and $|x - y_1| = q^\beta$ while $x q^{-\ell} = a \cdot q^{\beta - \ell} < q^{\beta}$. Similarly, if $a < q^{\ell}-1$, $|x - y_2| = q^\beta$. I'll leave it to you to try $|x - y_1|$ with $a = q^{\ell-1}$ and $|x - y_2|$ with $a = q^{\ell}-1$.