As a postscript, we ought to ask if the same reasoning applies to rationals; if it does, that would contradict our earlier argument that rational numbers can only have a handful of 2-good solutions. To show that there is no contradiction, note that in the rational case, s can conceivably reach zero (i.e., a / b can be an exact approximation). Next, rewrite r as p / q in the left-hand portion of the earlier inequality:
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Ray Tracing/Path Tracing