Continuity proved by differentiation

[losin]

f: (0,+inf)->R and

f(x) is

0 if x is irrational

1/n if x is rational (n is positive integer)


For each rational and irrational, i want to show continuity/discontinuity of f

Intuitively, i think at each rational f is discontinuous, and at each irrational f is continuous,

but i cannot figure out how should i apply differentiability to this problem...

 

[ice109]

you haven't defined n?

 

[HallsofIvy]

I strongly suspect that this is supposed to be one of the "Dirichlet" examples:

f(x)= 0 if x is irrational, f(x)= 1/n, if x is rational where n is the denominator of x expressed as a fraction in lowest terms. It can be shown that $lim_{x\to a} f(x)= 0$ for all x so, yes, it is continuous for all irrationals. It is not defined at x= 0 but if you define f(0)= 0, it is continuous at x= 0 and discontinuous for all other rationals.

There is no way to "apply differentiability" to this problem, the function is not differentiable.

 

[losin]

for f(x)=1/n when x is rational, n is random

so differentiability is not applicable?

since differentiability implies continuity, i tried to use that method..

 

[sutupidmath]

for f(x)=1/n when x is rational, n is random

so differentiability is not applicable?

since differentiability implies continuity, i tried to use that method..

A function can fail to be differentiable but be continuous. On this note, there are functions that are continuous everywhere but differentiable nowhere. So showing that a function is not differentiable, doesn't tell you anything about continuity. It works the other way around. That is, since if a function f is differentiable then it is continuous, it means that if it is not continuous then it is not differentiable.

...and what do you mean 'n' is random? random what?

 

[losin]

n is a random positive integer.

and how should i show discontinuity when x is rational?

when i prove 'f is continuous when x is irrational', does it follows that

'for rationals, f is not continuous'...?