Use universal instantiation or universal modus ponens to fill in a valid conclusion for the following argument. ∀ real numbers r, a, and b, if r is positive, then (ra)b = rab. r = 3, a = 1 2 , and b = 6 are particular real numbers such that r is positive. ∴ (−3)1/2 6 = (−3)(1/2)(6) 31/2 6 ≠ 3(1/2)(6) 31/2 6 = 3(1/2)(6) 1 2 3 6 = 13 26 (−3)1/2 −6 = (−3)(1/2)(−6)