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

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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?
 

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  • #2
gb7nash
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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...
 
  • #3
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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...
 

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