1. ## Continuity

Determine the values where the function is continuous.

f(x)={x for rational numbers
{1-x for irrational numbers

I drew a sketch of what the graph would look like, and it seems that the function is not continuous anywhere. The two "lines" cross at (.5, .5), but .5 is rational, not irrational, so it doesn't satisfy the second half of the function. Is this logic correct?

2. ## Re: Continuity

Originally Posted by lovesmath
Determine the values where the function is continuous.
f(x)={x for rational numbers
{1-x for irrational numbers
Suppose that $\alpha>0$ is irrational and $\beta=\frac{\alpha}{2}$.
Then $\beta$ is irrational.
Thus $f(.5+\beta)=.5-\beta~\&~f(.5-\beta)=.5+\beta$.
Now if $|.5-x|<\beta$ what about $|f(.5)-f(x)|~?$

3. ## Re: Continuity

f(x) is continuous at x = 1/2 if lim f(x) as x→1/2 =1/2 and f(1/2) = 1/2.
So show that, given ε, δ exixts st│f(x)-1/2│ < ε if │x-1/2│ < δ

x rational: │f(x)-1/2│= │x-1/2│ so δ=ε
x irrational: │f(x)-1/2│= │1-x-1/2│=│x-1/2│so δ=ε

So │f(x)-1/2│ < ε if │x-1/2│< ε and lim exists. Also, f(1/2) = 1/2