Use that the rationals are dense in by chusing for every x in the domain, a sequence of rational numbers in the domain converging to x. Since f is continuous, , and since the LHS is zero, the RHS is too.
I have heard of the topic of `denseness', but we have not formally discussed this concept in my analysis class. Is there another way of framing your reply, or perhaps could you restate the reply while including the definition of denseness???
(a) For every real number x, there exists a natural number n such that n > x.
(b) Similarly, for every real number y, there exists a natural number m such that 1/m < y.
(a) Should be trivial. (b) Follows from (a) (raise both sides of the inequality to the (-1), and of course reverse the inequality).
So, if I have this correctly (in order to prove there exists a rational number between x & y):
Let and consider We must find a rational number inside this interval. By the Archimedean Theorem, We then find an Now, we have that thus proving an
Is that correct? If so, I am uncertain of how to prove this for
By (a) we know that an m exists such that m > nx, which implies that x < m/n, the left half of the inequality. By (b) we know that an n exists such that 1/n < y/m, which implies that m/n < y, the right half of the inequality. Thus, there exists a rational number q = m/n x < m/n < y for any two arbitrary real numbers x and y.
For the irrationals, choose an irrational number w, and add w to x, m/n, y. Then, prove that m/n + w is no longer rational.
As for a formal proof, sorry.
Or, maybe more understandable.
Suppose that for some . and let . Since is continuous we should be able to find some such that . But, no matter what you pick there will be some such that and so