# Function proofs

1. Oct 2, 2008

### phyguy321

the function f(x) = 1 if x is rational
f(x) = 0 if x is irrational is not continuous for all real numbers, c

the function f(x) = x if x is rational
f(x) = 0 if x is irrationa is continuous at x=0 and not continuous for all other real numbers c

the function f(x) = 1/q if x is rational and x = p/q in lowest terms
f(x) = 0 if x is irrational
is continuous at c if c is irrational and not continuous at c if c is rational

2. Oct 2, 2008

### tiny-tim

Hi phyguy321;!

(have an epsilon: ε and a delta: δ )

Can you use an ε and δ method to prove that it is continuous at x = 0?

Have a go.

3. Oct 3, 2008

### phyguy321

I've got the proof for when f(x) is continuous at x=0 but I'm not sure how to prove that its discontinuous at every other value using delta epsilon. I understand it has to do with the density property of real numbers and that its full of holes be cause between every real number there exists rational and irrational numbers.

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4. Oct 4, 2008

### tiny-tim

Hi phyguy321!

Same method … start "if x ≠ 0, then for any ε < x, … "

5. Oct 4, 2008

### phyguy321

but how do i say if |x-c|<delta then |f(x) - f(not zero)|< epsilon?

6. Oct 5, 2008

### tiny-tim

But it's not continuous, so there isn't a δ.

You should be trying to prove that, no matter how small δ is, the neighbourhood will contain values that further away than ε.