21sir pls explain why it is sufficient to prove that
|1/q- 1/b| is non zero
and why this is so true
Find all local strict maximum of the function
F(x) = 0, x irrational
= 1/q,x= p/q in lowest terms
-
UP 0 DOWN 0 1 3
3 Answers
62all rational numbers will be points of maximums....
look at all the numbers of the set
Nq={n/b: n is an integer and b is any natural number less than q}
If we can prove that the min of |1/q-1/b|>0 over the set above then we are done...
this is true because this set will be finite..
21
341If x0 it is a maximum, then you will find a deleted neighbourhood A of x0 such that \forall \ x \in A, f(x)<f(x_0)
For integers this is obvious
Let us examine whether such a scenario is violated for rational points that are not integers.
Since, the next higher value that f can take is \frac{1}{q-1}, we need to examine if every neighbourhood of f contains a point x such thatf(x) = \frac{1}{q-1}
So x has to be of the form \frac{z}{q-1}
But, \left| \frac{p}{q} - \frac{z}{q-1} \right| = \left| \frac{p(q-1) - qz}{q(q-1)} \right| \ge \frac{1}{q(q-1)}
(since the numerator is an integer). You can see this bound is true for any natural number less than q
So, if you chose \epsilon < \frac{1}{q(q-1)}, then the condition for maximum is satisfied.