Question

Let \(f(x)=x\) if \(x\) is rational, f(x)=0 if \(x\) is irrational. Show that \(f\) is continuous at \(x=0\) and nowhere else.

Answer 1

ok so I know that\[\forall x: 0\leq |f(x)|\leq|x|\]

Answer 2

so that shows that f is cts at x=0

Answer 3

how do I show the second part?

Answer 4

so for \(a\neq 0\)
if a is rational I have \(f(a)=a\neq 0\)
if a is irrational I have \(f(a)=0\)

Answer 5

are there like an infinite number of irrational numbers that can get arbitrarily close to a?

Answer 6

ok so if there are, and I think there are. then I know that this is true. how would I formally show this? obviously you would have an infinite number of "holes" in the graph. could someone walk me through the epsilon delta?