:It seems obvious that one cannot both " accept " the hypothesis that all P's are Q and also reject the contrapositive, i . e . that all non-Q's are non-P . Yet it is easy to see that on the Neyman-Pearson theory of testing, a test of " All P's are Q " is " not " necessarily a test of " All non-Q's are non-P " or vice versa.
42.
I'd appreciate a nudge in the right direction, as I've been stuck on this for a while and I think it's probably a very short argument, just one I'm not getting-I believe I've proved X mn " c is irreducible implies both X m " c and X n " c are irreducible via the contrapositive, but I didn't use the tower law, or the fact that m, n are coprime : simply write Y = X m ( assuming X n-c reducible ), then X mn-c = Y n-c is reducible.
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