Without using a cardinality argument, show that R is not ring-isomorphic to Q. (Where R is the set of all reals and Q is the set of rationals.)
I'll give it a shot. Assume is such an isomorphism. Say where . must satisfy
so
Thus, each integer, and in turn each rational, is mapped according to the identity mapping. Then where would the irrationals go? The homomorphism seems to be onto, but not one-to-one, so no bijection can exist.