Final Answer:
F is continuous on the irrationals but not on the rationals.
Step-by-step explanation:
The function F defined as
exhibits continuity on the irrationals and discontinuity on the rationals.
For irrational numbers in the interval (0, 1), the function behaves as the identity function,
, which is continuous. However, for rational numbers in the same interval, the function becomes
introducing a jump discontinuity at each rational point. This leads to F being continuous on the irrationals but not on the rationals.
The reason for this behavior lies in the fact that the rational numbers are dense in the real number line, creating a situation where the function jumps between two values at every rational point. The irrationals, on the other hand, do not exhibit such jumps, allowing for continuous behavior on that subset of the domain.