r/numbertheory

MyImprovedArithmetic

I’ve improved arithmetic.

I’m sure everyone knows what an abacus looks like. Here’s a question for everyone: Show me where zero (0) is on an abacus. It isn’t there. And that’s exactly what I’m going to talk about: emptiness.

The main flaw in modern arithmetic is that it counts emptiness. So I fixed that. It’s very simple, based on how a computer works—or more precisely, a processor. For a processor, 0 or 1 isn’t emptiness; it’s a value. But emptiness is present; it’s NULL. And emptiness is present in life. One more example before I move on to my arithmetic. A little problem for Pinocchio, just slightly modified. Pinocchio had an apple on his plate. Pinocchio wasn’t greedy and gave the apple to Artemon. How many apples are left on Pinocchio’s plate? Everyone will say the answer is zero apples. But that’s not the correct answer. A void remains. Because one could answer that there are zero pears or something else left. This is the first flaw in modern arithmetic, which requires that we not divide by zero. The second flaw is that in decimal arithmetic, in the ones place, we can only count up to nine, but it should be up to ten.

Now I’m correcting traditional arithmetic with my own. So, for me, 0 is emptiness. And you can count up to ten objects by adding the “Ten” symbol. Of course, you could invent a new symbol, but it isn’t on the keyboard yet. So I chose the Latin “Ten”—X.

Let's start counting: 1, 2, 3, 4, 5, 6, 7, 8, 9, X (or 10), 11, 12, 13, 14, 15, 16, 17, 18, 19, 1X (that is, the word “Twenty,” or 20, where we carry the ten from the ones place to the tens place), 21, 22, 23, 24, 25, 26, 27, 28, 29, 2X (that is, “Thirty,” or 30). The rest is clear. But in my arithmetic, there is a void or emptiness, which is 0, “Zero”. We do not perform any arithmetic operations with zero, it is only used as a statement. That is, it can be 0 (emptiness), or emptiness was filled with anything. Example: 1 - 1 = 0; we got emptiness X - 9 - 1 = 0; we got emptiness 0 + 1 = 1; we filled emptiness with a thing

A void can be present in both single-digit and multi-digit numbers. I’ve already shown a single-digit number. 1 - 1 = 0; Example of a multi-digit number: 25 - 5 = 20; 255 - 50 = 205; The void can be replaced with a value: 20 => 1X, meaning from the two tens, carry the one from the second digit to the first digit. The reverse operation is also possible: 1X => 20.

Thus, the error is corrected. Try multiplying or dividing in a column; it all works. Just remember, operations with 0 are not performed, because it is not the item for calculation. We can only make emptiness or fill emptiness.

Since e is generally proved to be irrational by contradiction, I wanted to write a proof that directly shows it cannot be rational. When I presented this to Claude it took some cajoling for it to say the proof was correct, and it was unable to find a similar/direct proof, so if my logic isn't clear or has errors I would appreciate any critiques and am interested if anyone has encountered a direct proof like this one.

I don't know if it's an existing work, but this identity is independently developed by me.

I've rewritten this StackEchange posting from a few years ago, making the results more rigorous (although it's certainly not 100% rigorous yet). As explained there, the starting point is the idea that the sum of a series, regardless of whether it is convergent or divergent, should be taken to be the sum of the partial sum and the remainder term.

In case of a convergent series, the remainder term tends to zero in the limit of the truncation point to infinity, which allows us to compute the sum of such a series without having to consider the remainder term. In case of a divergent series, we then do need to consider the remainder term.

While the remainder term looks like something that is completely arbitrary, I show in section 3 of the stackexchange posting that the remainder term for the rescaled summand is related to that of the original summand, see eq. (3.11). I derived this for the convergent case, but by invoking analytic continuation, I argue that this should be generally valid.

If we're summing f(k) from k = p to infinity, we can consider summing f(k/N). The remainder term for truncating at the argument of the summand of x is denoted by R(x,N). This means that the index value at which we're truncating is N x. We then do have invoked analytic continuation to any real or complex values for x.

Eq (3.11) then says that:

R(x,1/N) = sum from k = 1 to N of R(x + k/N -1)

Where the remainder term in the summation without the second argument is the original remainder term with N = 1.

I then show in section 4 that this relation directly implies the value of the sum over all positive integers.

More powerful summation methods are derived in section 5 from (3.11) by considering the limit of N to infinity. One result is eq. (5.5) which gives the sum X of a divergent series in terms of an integral over the partial sum S(t):

X = Constant term in the large-x expansion of Integral from x -1 to x of S(t) dt

And another result is eq. (5.6) which gives the prescription of how to correctly use regularization to compute the value of divergent series. We're then summing a summand f(k) that leads to a convergent summation with value X, and they both depend on another parameter. By doing some manipulations involving that parameter, be it analytic continuation, or series expansions or something else, one formally gets to the desired divergent sum.

However, eq.(5.6) tells us that to get to the correct value of the divergent sum, one has to also consider the integral of f(t) from x to infinity, do whatever is done to the regularized series to this integral, extract the constant term of the large-x expansion from this and subtract that from the result of the manipulations to the regularized sum.

In section 6 I give some examples of computations involving (5.5) and (5.6). And I've given more examples of how doing the regularization correctly resolves ambiguities in other postings. See e.g. this MathOverflow posting and in this posting I show how it eliminates an ambiguity with choosing the branch of a logarithm.