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Where 1 <= x <= 4: f(x) = (x-1)/9 + 1, maps bijectively from [1,4] to [1,4/3] g(x) = (x-1)/9 + 4/3, maps bijectively from [1,4] to [4/3,5/3] h(x) = (x-1)/9 + 5/3 maps bijectively from [1,4] to [5/3,2] Therefore, [1,2] must contain three times as many numbers as [1,4], right? It doesn't work like that. |