Question: Let I := [0, 1] and let f : I R be defined by f (x) := x for x rational, and f(x) :=

Let I := [0, 1] and let f : I → R be defined by f (x) := x for x rational, and f(x) := 1 - x for x irrational. Show that f is injective on I and that f (f(x)) = x for all x ∈ I. (Hence f is its own inverse function!) Show that f is continuous only at the point x = 1/2.

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If x I is rat i onal then fx x is also rat i onal so ffx fx x if y I is irrational then fy ... View full answer

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