Let be continuous on and suppose that for each rational number x in .Prove that
Ok, Do I need to do a proof by contradiction to prove that . What it comes down to is proving that for each irrational number , right???
Otherwise I don't really see how to do it, I mean even if I do by contradiction the way I thought about doing it was maybe say something like Assume that
I am not sure... any suggestion?
So from the problem we know the behavior of the function when x is rational so that is why we just have to look at the function when x is irrational, correct?
I don't understand that part, how come you if you have x as an irrational number you have to have a sequence of rational numbers so that ??? Can you explain a little?
I think I get the second part better but lel me check if I understand it well, so since we know that the sequence converges to then the and is that correct??
Ok, I understand what you mean. I think I am starting to understand better but I am still a little confused about one little part of your explanation (sorry for asking so many question).
So you said
That part I follow and understand it no problem now.
The only thing that I am not sure I understand is
because the problem states that for each rational number x in , however in your thing you said that being an irrational number which is the opposity of what is stated in the problem isn't it??? Or am I missing something? Or is it because since is rational then
according to the problem?? So since , then more or less???
I am sorry to ask that many question, I am juste tying to be sure I understand it well and that I don't get this answer for the wrong reason.
Thanks for all your help.