**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 . *

**Click
to go to last comment**

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.*

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 ${4\over 7}$ the``son 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.*