Question

define function y = f(x) on open interval (0,1): if x is irrational, let f(x)= 0 if x is rational write x as fraction in lowest terms and left f(x)= denominator. is function continuous at x = 1/2 and x=(sqrt2)/2? explain why

Answer 1

@asnaseer continuity problem

Answer 2

@SmoothMath continuity problem

Answer 3

what does "left f(x)=denominator"
mean?

Answer 4

It's never continuous. The rational numbers and irrational numbers are both infinitely dense.

Answer 5

But denominator means it's the bottom of the fraction.
So for example, f(6/8) = 4 because the denominator of that fraction in reduced form is 4.

Answer 6

This function is intentionally designed to be the furthest from continuous it could be. Thing is effing crazy.

Answer 7

define a function y=f(x) on the open interval (0,1) in this way: if x is irrational let f(x)=0, IF X IS RATIONAL, WRITE X AS A FRACTION IN LOWEST TERMS AND LET F(X)=DENOMINATOR. FOR EXAMPLE F(radical 2 over 2 = 0 and f (9/15 =5. Is this function continuous at x=1/2? why? is this function continuous at x = radical 2 over 2? why?