If and consider the open interval:
This interval contains a rational because:
and the endpoints of are irrationals.
considering the cases prirme and composite.
Let's take function given by a condition:
Prove the existence of the derivative of in all points .
So, I am aware that if there was standing in the formula instead of , the derivative wouldn't exist (that would simply be the Thomae's function). The thing I couldn't figure out is, why would the replacement change anything and where should I start the proof?
No, they are specifically talking about which is almost exactly the same function except that the function in the article is defined to be 1 at x= 0, while the function here is not defined at x= 0. Since the conclusion is "there exist am uncountable set or irrational numbers, dense in the real numbers, on which is not continuous, and 0, of course, is rational, that would not affect the proof.
Thanks for all the answers.
So, basically speaking, the theorem I stated was entirely false.
What does Fernando's reasoning prove then? (I am going to look into it later but was unable to decipher it quickly.)
And where the heck is the square of Thomae's function differentiable?