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@neauoire @eris Right, I guess what I'm trying to say is, that if someone where to define reduction as a mathematical function, they would have to define it like your operator. Meaning it would need something like this signature: Q -> (N, N) and not this one: Q -> Q. Because if they would choose the latter and define it such that reduce(4/6) = 2/3, they would just define the identity function, because 4/6 = 2/3. |
@TsRoe @eris the reduction of 4/6 is 2/3 yes, the special operator gives you the two whole numbers.