Let be the set of natural numbers and be the set of positive rational numbers. For every positive integer , there exists a unique real number such that . The question is does there exists an such that be rational? In other words, does there exist a positive rational such that ? At present, we don’t even know if is always rational, let alone irrational or transcendental. Investigation seems to suggest that for any . However this has not been completely established. In this post, we shall look at some on the results in this line of research.
Expanding the Riemann zeta function about gives
where the constants
are known as Stieltjes constants. Inverting the above series we obtain:
Theorem 1. Let and ; then,
Proof. The proof follows by inverting the Stieltjes series expansion of .
We can rewrite the Stieltjes series expansion of the Riemann zeta function in the following form which we shall use regularly in our analysis:
We want to find an interval where the function is strictly decreasing. Clearly is continuous and strictly decreasing in the interval . The trivial zeros of occur at where . In other words, the trivial zeros of occur at and all these zeros lie in the interval . Hence in this interval, will have infinitely many points of inflexion. Excluding the interval , we can divide the real line in to two parts and ; and in each of these two intervals, in strictly decreasing. For positive rationals , the scope of our current study is the interval .
Lemma 2. For every finite real , there exists a unique real number , such that
Proof. Let lemma follows form the fact that in our interval of interest , is strictly decreasing.
Theorem 3. Let be as defined in Lemma 2. The asymptotic expansion of is
Proof. This is obtained by inverting the above Stieltjes series expansion of .
The next result gives us a method to compute numerically. We have
Lemma 4. Let be defined recursively as
where . Then .
Marek Wolf, has sent me the computed values of for . Some of these numbers were used in the proof of the following theorem.
Theorem 5. If then .
Proof. We have . There are exactly twenty three positive fractions with such that
For each of these fractions, we find (by actual computation) that there exists a integer such that
Since is strictly increasing, it implies that if then the denominator of has to be . Hence the theorem follows.
If we continue working with the idea given the proof of Theorem 5, we can prove for some large natural number . But this method gives no hope of proving that . Therefore we must look for some other ideas.
Theorem 6. Let , and such that
If where is any non-negative integer; then .
Some applications of Theorem 6 are given below.
Example 7. If and then at least one of the two numbers
is not a integer.
Example 8. If then at least one of the three numbers
is not a integer.