Skip to content
Q13Section II (Mathematics)13 of 190

Thomae's function: Riemann integrable, continuous at irrationals, discontinuous at rationals

Suppose f(x)f(x) is defined on [0,1][0,1] as follows: f(x)=0f(x)=0 if xx irrational, 1/q1/q if xx rational and x=p/qx=p/q with (p,q)=1(p,q)=1. Then f is:

Options

Pick an option to check your answer.