If and consider the open interval:
This interval contains a rational because:
and the endpoints of are irrationals.
Now analize:
considering the cases prirme and composite.
Regards.
Fernando Revilla
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?
If and consider the open interval:
This interval contains a rational because:
and the endpoints of are irrationals.
Now analize:
considering the cases prirme and composite.
Regards.
Fernando Revilla
Hmm, this article thinks otherwise. Apparently, it was published in American Mathematical Monthly in 2009.
I can't see most of characters in the article, but surely they are no talking exactly about the same function.
Regards.
Fernando Revilla
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. I'm afraid I need a new computer. :-)
Regards.
---
Fernando Revilla
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?