Discussion
Let \(f : \mathbf{R} \to Q\) be a non-constant increasing function. Then :
Answer: A
The answer is that \(f\) is constant on some interval \([a, b]\).
Because \(f\) takes only rational values, its range is countable. For each value \(q\) in the range, the set \(f^{-1}(q)\) is an interval (if \(x \lt y \lt z\) and \(f(x) = f(z) = q\), monotonicity forces \(f(y) = q\)). So R is a union of countably many such intervals. If all of them were single points, R would be countable, which it is not. Hence some \(f^{-1}(q)\) is a non-degenerate interval, and \(f\) is constant on a closed interval \([a, b]\) inside it.
This makes D false, and B false too, since a function constant on an interval is not one-one. C does not hold in general: the floor function \(f(x) = \lfloor x \rfloor\) is increasing, non-constant and rational valued, yet it never takes the value 1/2.
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