Prove f(x) Continuous at x=1/2 for x Rational, Irrational

  • Thread starter Thread starter ashok vardhan
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 3K views
ashok vardhan
Messages
18
Reaction score
0
f[x]=x,x is rational, 1-x, x is irrational. prove that f(x) is only continuous at x=1/2.
 
on Phys.org
please give a proper mathematical solution to this
 
Why? It looks like homework to me. You need to try yourself, first, and show us what you have.

Here's a nudge: for any x, there exist a sequence of rational numbers converging to x and there exist a sequence or irrational numbers converging to x.
 
i have already solved the problem in the way you suggested.but i have a problem in solving it using epsilon and those things
 
f(x+y)=f(x)+f(y).f is continuous at x=0.prove that f(kx)=kf(x). i have proved it for k os an integer.for k a rational number i assumed it to be of p/q.and i can't proceed further to prove this. would you like to help in this
 
Is this a completely new problem? f(x+ y) is definitely NOT equal to f(x)+ f(y) for the problem you gave before. For example, [itex]1+ \sqrt{2}[/itex] is irrational and so [itex]f(1+ \sqrt{2})= 1- (1+ \sqrt{2})= -\sqrt{2}[/itex] but since 1 is rational and [itex]\sqrt{2}[/itex] is irrational, [itex]f(1)+ f(\sqrt{2})= 1+ (1- \sqrt{2})= 2- \sqrt{2}[/itex].

You didn't say anything about using [itex]\epsilon[/itex] and [itex]\delta[/itex] in your first post. If you are not allowed to use "[itex]\lim_{x\to a} f(x)= L[/itex] if and only if, for any sequence [itex]\{x_n\}[/itex] that converges to a, the sequence [itex]\{f(x_n)\}[/itex] converges to L", then copy the proof of that theorem, for this particular function.