I list comments I have received so far to help me (and perhaps others) understand the difficulties people are having with this paper. My responses are in italics. I do not attribute these comments to specific names because most of these remarks are from emails, which often tend to include offhand remarks.

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Comment: Theorem 4 is true, but useless, because when $\sum f(n)$ is rational, a corresponding decomposition of $f$ can be obtained using $F(m):=\sum_m^\infty f(n)$ and  its translate by one.

My response: Correct, of course. But I had discussed this very issue in Remark 7 of paper.  [I have now added a line on p.8 of my paper to please read Remark 7, before raising this Comment] Theorem 4 and Theorem 6 would be unusable and would just be circular statements  unless you attach conditions (4), (5) mentioned on page 8 (and onwards).

Theorem 4 and Theorem 6 (with these two attached conditions) would be foundational theorems of mathematics (or foundational conjectures if you do not accept their proof with the two extra restrictions attached).

Do you not at least see that assuming Theorem 4 (or call it Conjecture 4 if you wish, in above form), the statement of Theorem 2 follows from it?

Comment: Your theorem 2 is a very interesting conjecture, but unfortunately your proof has a serious gap. In particular theorem 2 does not follow in any obvious way from theorem 4.

My response: Nothing  follows from theorem 4 unless we attach two additional conditions[see previous Comment].

Comment: The main problem is that just because the function f in theorem 4 is a quotient of polynomials, this does not imply that the functions you break it into in theorem 4 are also quotients of polynomials. I think it is unlikely that this problem can be fixed with the elementary methods you are using.

My response: It can. I insist that with the above attached conditions it can be proved that theorem 2 follows from theorem 4, and I have proved it in my paper. Please read p. 14 onwards.

Comment: Cannot  make sense of (4) and (5) [p. 8 and onwards in Remark 7] in this paper.

My response: Actually (4), (5) just require $\sum_{n=1} ^ {\infty}f(n)$ to be a "telescoping series" as explained in Part 7-1 of the paper. No great mystery there, college calculus books use the term "telescoping series." Just formalizing what such a series is.

Comment: For example, consider f(n) = 3/(n+1)^2 - 4/(2n+1)^2 = p(n)/q(n) where  p(n) = 8n^2 + 4n - 1,  q(n) = ((n+1)(2n+1))^2 Then the sum of f(n) for n = 1, 2, ... is the rational number 1 but I don't see how f(n) can be split into partial fractions that satisfy the condition in your Theorem 2 (unless I misunderstand it).

My response: The theorem 2 on the opening page of the website and in the emails I sent is not the actual theorem 2 but a simplified version, as I noted. The correct theorem 2 is in the downloadable paper available at the website . Your example is not a counter-example to that actual theorem 2.

Comment: sum
1/(n-1/24)^2+1/(n-23/24)^2
 +1/(n-5/24)^2+1/(n-19/24)^2
 +1/(n-7/24)^2+1/(n-17/24)^2
 +1/(n-11/24)^2+1/(n-13/24)^2
 -16/(n-1/6)^2-16/(n-5/6)^2
from n=1 to infinity,

the result is rational, namely 0

My response: Let us re-write as:

(24/(24n-1))^2+(24/(24n-23))^2
 +(24/(24n-5))^2+(24/(24n-19))^2
 +(24/(24n-7))^2+(24/(24n-17))^2
 +(24/(24n-11))^2+(24/(24n-13))^2
 -(24/(6n-1))^2-(24/(6n-5))^2
Then on the pair in the last row "increase relative occurrence" (similar to the
example on p.5 of paper). Replace n in the (6n-1) by 4n,4n-1,4n-2,4n-3 to get four
negatives which exactly cancel four of the positives in upper rows.
Do same for n in the (6n-5). Everything cancels as stated by theorem 2 and
$\lim_{n\to\infty}D(n)$ can be seen to converge to the rational number $0$.

Comment: Theorem is a statement you (you believe) have a proof of. This statement should be complete, and not modified by later remarks.

It can be seen that theorem 4 with its original statements (1), (2), (3) is easily proved but not useful. To make it usable one needs to add conditions (4), (5) to it -- and you are right in that with this modification (whose proof may presumably be questionable to some) it could convert to being  Conjecture 4 and cease to be a theorem. But in my paper, in violation of rules of exposition, I continue calling it theorem 4, even after these modifications, especially because the Theorem 4 with just the three conditions is an entirely different beast from the Modified Theorem 4 with the five conditions. I agree and apologize for that.

Comment:   Philosophy: a possibly vague statement, for which you have reasons to expect something similar is true.

My response:  I agree that Part 7-2 of Remark 7 which proves (or aims to prove) why (4), (5) can be added could be considered “philosophical” by some. One can change Theorem 4 to Conjecture 4 with these conditions, if they feel I have not proved in Remark 7 how I can add (4), (5).

Comment: Examples can be nice. Proofs are crucial. Don't mix them.

My response: I believe that the statement of Theorem 2 follows reasonably easily from Modified Theorem 4 (or Conjecture 4 as discussed above). On p. 16 through 23, in the proof of Theorem 2, I do use some examples; but they are used to illustrate the mechanism I will be adopting and I believe I have proved that Theorem 2 follows.

Comment: It is not easy to check a proof.

My response:  I agree. However, the methods used in the proof of Theorem 2 from Modified Theorem 4 (or Conjecture 4) are essentially elementary and thus it should be not be as time consuming to check it.

Comment: Following true formula is 2-adic convergence, i.e. in the field of 2-adic numbers, not in ${\Bbb R}$, but should it matter, and, if yes, why: $\sum 2^n/n=0$.

My response:  I wrote a line on the last page of my paper stating that I believe the cancellation principles of the paper should be applicable to other places in abstract algebra, giving similar results. Questions relating to such extensions could be premature because it is not clear what mathematicians have to say about the statements in the paper as they pertain to just rational numbers such as whether they believe/agree that (a) the statement of Theorem 2 is true? (b) statement of Theorem 4 (modified to include two conditions as above) is true? (c) Theorem 2 follows from Theorem 4? (d) What about the statements of the other Theorems mentioned? (e) And if the proof of Theorem 4 (modified to include two conditions as above) is accepted then there is nothing left to argue about!

Comment: Your theorems [4 and 6, with the two added conditions] are virtually a tautology.

My response: As long it is not a tautology why should it matter if it is almost one? From this "almost a tautology" we can derive theorems 2 and 8 and Abel would have had a new method to attack series such as the one mentioned on p. 26 of the paper.

Comment: Fail to give a mathematical proof.

My response: Okay, but could you explain where you feel I failed. Is it in the proof of theorem 4/6 (with the two added conditions) or is it in how theorem 2 follows from these or is it in both? Theorem 2 follows naturally and in a standard mathematical manner from theorem 4/6, and if needed I can revise and re-write any part of that proof to satisfy any objections. As for the proof of theorem 4/6 (with the two added conditions), I would agree that Part 7-2 of Remark 7 does not look like mathematical proofs one generally reads, but that does not make it not a mathematical proof.

Comment: Your paper seems to be a political statement against the rest of the world stating that people like Euler, Lindemann and everyone else who have worked or now work on questions of irrational and transcendental numbers were just being (to put it plainly) stupid and that you have superseded their thinking on the subject. This is not quite the proper tone with which to approach mathematics, and there are certain statements "against the world" which need not have been included in the paper.

My response: Can you tell me which specific statements? I do not mention Euler in the main body of the paper, his name only happens to be in the title of one of the citations; the paper does not mention Lindemann at all (though theorem 2 does imply that pi is transcendental). The only name I mention in the body of the paper is  Abel, and I was not criticizing Abel for  anything.

However, I believe your statements may be correct. But in your proof you use a different sort of logic which seems like you are posing puzzles.

My response: As I agreed above, there is one part of my paper which does not look like mathematical proofs one generally reads.

Comment: I am not able to understand what you have written, and I never will be able to understand it because you write in a very vague way.

My response: I don't think that is a fair way to dismiss the whole paper without any specific criticism whatsoever. Remark 7-2 could be considered different from “normal mathematics exposition” but the rest of the paper is not.

Comment: I dismissed your text when I did not find in it anything of the  required depth, and when I saw you had tweaked statements just to cover counterexamples which were given to you.

My response: You seem unhappy at the simplicity of my discoveries. Have you considered the possibility that simple and basic natural truths which would change the course of these seemingly formidable unsolved  problems were not seen --  it is human arrogance for mathematicians to demand a solution along the lines of the “deep” thoughts they have had on the matter. As for tweaked statements -- as I state in the acknowledgement at the end of my paper -- I did that once (and only once) because,  unlike you perhaps, I sometimes make mistakes that need correction as one mathematician helped me do.

Comment: I don't think it would be of any use for me to try to test your proof if you don't first give a self-contained statement of what you claim,  and then what you believe to be a proof of it.

My response: I believe I have done that in the paper and clarified further in Comments on my website, and the elementary nature of the paper makes these exposition issues not that big a deal. Please forget these somewhat bureaucratic objections.

The Main theorems 4/6 do not have any restriction that f be a quotient of polynomials but state that if f(n) is some mathematical expression/formula that takes rational number values for all n and gives a convergent infinite series, then the series will converge to a rational number if and only if the series is a (truly) telescoping series (conditions (4), (5) of Remark 7 of the paper enforce the "truly"). This is the simple truth about infinite series that mathematicians did not realize.

Comment: The question of when $\sum a_n=$ rational, if true, should be provable by  having a closed'' expression for the sum $S_N$ up to $N$ is interesting.\hfill\break

You should decide whether your statements are to be considered as theorems (you claim to have a proof) or conjectures: evaluation criteria differ.

My response: They are theorems. Using the result of the general statement  ---  that $\sum a_n$ will converge to a rational number iff there is a closed'' expression for the sum $S_N$ up to $N$ --- theorem $2$ applies it to $a_n$ being a quotient of polynomials and makes a statement that predicts whether $\sum a_n$ will be rational or irrational. The first thing is to test this hypothesis on known series converging to rational numbers. There are some who believe that such simple truths cannot be true, {\it but I think they are (perhaps unconsciously) basing this on their feeling that if these were true why did no one else see them}? For example, about a month ago, one mathematician who posted on a public board that he believes my statements are false found what he suggested was a counter-example and I showed it was not, see Comments section of my website about his long $a_n$ involving the number $24$ in denominator  \hfil\break

On a philosophical level why should this simple statement be true --- because it all ties in with Georg Cantor's showing that almost all real numbers are irrational; see page 8 of my paper. (Here, the term almost all'' means that all reals that are not irrational have measure zero). Infinite series have to have a very special property to converge to a rational number and my statements imply that almost all infinite series converging will converge to irrational. Algebraic numbers also need a very special property, and we can see that from my note on transcendental numbers on page $5$ of the paper. So this is an independent way to come to Cantor's realizations. \hfil\break

There are two proofs at issue. The proof of the wider statement and (assuming that) the proof of Theorem $2$. Theorem $2$ follows naturally and in a standard mathematical manner from the wider statement, and if needed I can revise and re-write any part of that proof to satisfy any objections. As for the proof of wider statements ( theorems 4/6 with the two added conditions), I would agree that Part $7-2$ of Remark $7$ does not look like mathematical proofs one generally reads, but that does not make it {\it not a mathematical proof}. \hfil\break

I am a great admirer of Cantor (of course, his ideas are now universally acknowledged to be revolutionary) and it was very tragic that Leopold Kronecker blocked and dismissed Cantor, saying his work was simply not mathematics'' (which is believed to have played a central role in Cantor ending up in a mental asylum). But Kronecker was, in a sense, absolutely correct when he said irrational and transcendental numbers do not exist" and are the work of man.'' {\it In a deeper sense} this is a central property of irrational numbers which can be exploited to lead to new truths which my paper mentions; {\it Part $7-2$ of Remark $7$ may be subtle and but is solid logic}. \hfil\break

Comment: Why should the case of $\sum a_n=$ rational be so different from the case of $\sum a_n=$ rational power of $\pi$?

My response: Because anything which is not rational is {\it just a name given to the sum of a convergent series} --- that other series may sum to a new irrational number which is the rational power of what this series sums to, or that $\pi$ may have other properties that are interesting does not change this fact --- they are all just names.  Consider that square root of $2$'' is just a name of the convergent infinite series which has the property that its square is $2$ --- one could alternatively call it the son of $2$.'' Of course one could call ${4\over 7}$ theson of ${16\over 49}$" but it will still be ${4\over 7}$ and we can ask why a series converges to this particular number -- because we have something more than a name. Arguing that the name $\pi$'' came first and it was later that mathematicians found a series that equalled it does not change anything. Please read Remark $7-2$ of my paper for the full discussion, an excerpt from which I attach below: Suppose [the infinite series] does not go to a rational number (but converges). Then we can just say that let us call the irrational number it goes to by some name, say$qi$''. If we make such an assignment then it makes no sense to have an investigation of what special properties the terms of the series must have so as to go to the number $qi$''.'' \hfil\break

[I have now added mention of Kronecker and square root of $2$'' to part $7 - 2$ of my paper]

Comment: Suppose that f(n) is a function which DOES satisfy your theorem 4.  I will
[attempt] to construct another function g from f which does not.  Note that
since f(n) is positive for all n, and is rational, there must be a number
k such that f(n) > exp(2,n) for all n > k.
This is because an exponential function decreases much faster than any
polynomial or rational function.

Now let g be defined as follows.  If n < 2k , then g(n) = f(n).
Otherwise if n is even,  g(n) = f(n) - exp(2,n)
and if n is odd, g(n) = f(n) + exp(2,n-1).
where exp(2,n) denotes
-n
2    (2 raised to power -n)

Thus g equals f plus small terms such that they cancel each other out
in successive steps.  Now g CANNOT satisfy the conditions of your theorem, it cannot be written in the form you want.  On the other hand the infinite sum of g(n) is also the sum
of f(n) which is rational by hypothesis.   Moreover, all values of g are
also rational since they differ from the values of f by a rational
number.

My response: Regarding your example, the key is that x^n=x^n/(1-x) –x^(n+1)/1-x=F(n)-F(n+1), (where ^ means power).

In your counterexample (for n>=2k) the part you add to f(n) is: -(1/2) ^n for n even and +(1/2)^(n-1) for n odd. In general we can replace ½ by x and talk of  –(x^n) for n even and +x^(n-1) for n odd (keeping |x|<1).

This part you add can be written as F(n)-F(n+1) where F(n) is defined  to be: -[x^n/(1-x)] for n even and +[x^(n-1)/(1-x)] for n odd. So the f (n) part (by hypothesis) and the part you add can each be written as F (n)-F(n+1) as stated by theorem 4 and thus their sum can be so written.  Thus g(n) does not form a counterproof.

I see what you were aiming to do and the (1/2)^n was just an unlucky choice in that it can be made to telescope. You  were heading in the direction of creatively constructing g(n) which will not satisfy conditions (4), (5) of thm 4.

So in general you were aiming for g(n)=f(n)+h(n) where you start with  f(n) that satisfy thm 4 but h(n) doesn’t. Your general idea was to add small (compared to f(n)) positive and negative pairs h(n) which have rational values and cancel each other and these h by themselves need not satisfy thm 4. The f(n) served the purpose of keeping g(n) positive, as required by thm 4.

Part 7-1 of the paper says that f(n) must be an expression/formula in n which is not a “conditional” function (such as being defined differently for even and odd n), which yours was. The f(n) that are naturally important in mathematics are of the type I restricted to. That saves me from finding F for every creative valid f that anyone comes up with. One can define your h(n) in all sorts of creative ways by saying it does this for such n and that for other n -- my paper wishes to not to address this endless creativity.

Again, though in part 7-1 of my paper I stated that I do not wish to address these, here is the resolution for the general case you were attempting, which case was: g(n)=f(n)-h(n) for n even and g(n)=f(n)+h(n-1) for n odd. By your hypothesis f(n) satisfies the telescoping requirement of thm 4 (we can find F such that f(n)=F(n)-F(n+1)) but h(n) does not have to and the infinite sum g(n) will still converge to the same rational number as sum of f(n) would have converged to. You claim that no G(n) exist that satisfy g(n)=G(n)-G(n+1) as predicted by thm 4 so this is a counter-example.

Here is the resolution:

for g(n) with n even: define G(n)= F(n) - ½ h(n) and for n even and G(n)= F(n) +½ h(n-1) for n odd.
for g(n) with n odd: define G(n)= F(n)+ ½ h(n-1) and for n odd and G(n) = F(n) -½ h(n-2) for n even.

(Also please note that instead of the ½’s we could use c and (1-c) for c
rational. For example,
for g(n) with n even: define G(n)= F(n) – (c) h(n) and for n even and G
(n)= F(n) +(1-c) h(n-1) for n odd).

Thus each of your g(n) can be written as G(n)-G(n+1) as required by thm 4.