This is probably very simple but I'm not sure if the last step is right. Let A be a dense set in the reals and f(x)=0 for all x in A. If f is continuous, prove that f(x)=0 for all x. Let a be in real number. By definition, for all [tex]\epsilon > 0[/tex] there exists [tex]\delta > 0[/tex] such that [tex]|x-a|< \delta \Rightarrow |f(x)-f(a)|< \epsilon [/tex]. But in any open interval there lies element of A, so in particular: for all x in A such that [tex]|x-a|< \delta [/tex] we have [tex]|f(a)|< \epsilon [/tex]. Here's where I deduce f(a)=0 and this is the step I'm not sure about. Does this make sense?