Post #1819350

P.S. Just to note this; people must have been doing it for decades. Consider the following well-known identities for rational p and q: (sqrt x) ≡ (expt x 1/2) (* (expt x p) (expt x q)) ≡ (expt x (+ p q)) With an evaluator that applies the respective transformations, perhaps with a bit of lazy evaluation thrown in, we "automatically" get (* (sqrt 2) (sqrt 2)) => 2 (exactly). @screwlisp @simon_brooke