To take the next step — and we’re close to the finish line! — note that the proof doesn’t put any constraint on the upper value of K. If we choose some definite K1, the proof establishes the existence of a single 2-good pair, which we can label a1 and b1. If we choose K2, it proves the existence of a pair we’ll call a2 and b2; that pair may or may not produce a functionally different approximation of r. Maybe there’s just a single solution that repeats for every K?
Many early-stage crowdfunders left empty-handed as Tilray acquires beer company’s UK and Irish assets
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