Question
 If f(x) is continuous for all real values of
x and f(x) takes on only rational values, then if f(1)=1, the value of f(0) isSolution
Property: - Let f:[0,1]→ R be continuous such that f(x)∈Q for any x∈[0,1] then f (x) is constant. Proof Suppose f isn't constant. Then for some a,b∈[0,1], f(a)≠f(b); Without Loss of Generality, f(a)
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