Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Real analysis: show that a continuous function is defined for irrationals

  1. Nov 23, 2011 #1
    Let f be a continuous function defined on (a, b). Supposed f(x)=0 for all rational numbers x in (a, b). Prove that f(x)=0 on (a, b).

    i dont even know where to start...any tips just to point me in the right direction?
     
  2. jcsd
  3. Nov 23, 2011 #2

    gb7nash

    User Avatar
    Homework Helper

    There's a couple of ways you can approach this problem. Did you try proof by contradiction?

    Let f be a continuous function defined on (a, b). Assume f(x)=0 for all rational numbers x in (a, b) and assume that f(x)≠0 on (a, b), so...
     
  4. Nov 23, 2011 #3

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You'll need two things for this problem:

    1) If f is continuous and if [itex]x_n\rightarrow x[/itex], then [itex]f(x_n)\rightarrow f(x)[/itex].

    2) For every real number x there exists a sequence of rational numbers that converges to x. This is saying that [itex]\mathbb{Q}[/itex] is dense in [itex]\mathbb{R}[/itex]

    Try to do something with these things...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Real analysis: show that a continuous function is defined for irrationals
  1. Continuity - analysis (Replies: 12)

  2. Real Analysis (Replies: 0)

Loading...