https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation

For example, Grandi’s series is the sequence:
1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,…
evaluates by Cesaro summation (and other methods for evaluating divergent series) to its mean value 1/2.

Hypothesis 1. Every real number can be expressed as the evaluation of a Cesaro summable sequence of adjacent integers.
Hypothesis 2. Every Cesaro summable sequence of adjacent integers can be evaluated as a real number.

In other words, this is a new way to define the real numbers. As the mean of a sequence of adjacent integers.

Example 1. Take a square mat and inscribe a circle on it. Toss a sand grain at random U(0,1;0,1) on it. The probability of the sand grain falling within the circle is pi/4 = 0.785398163… Construct a sequence, every time the sand grain lands within the circle write 1 and every time the sand grain land outside the circle write 0. The mean of the resulting sequence will tend to a value of pi/4.

eg. 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, …

We can do better. Rather then tossing the sand grain at random, use a dithering algorithm to minimise the error at each step of the sequence. This gives a unique sequence for all irrational real numbers between 0 and 1, and for all rational numbers p/q where p and q are coprime and q is odd. When q is even, we get two sequences, one where ½ is rounded down to 0, and the other where ½ is rounded up to 1.

pi/4 = 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, …

You can see at once that the mean of this sequence is between 3/4 and 4/5, like pi/4.

.thankS

This is no use unless I can prove closure under +, -, *, /.

Clearly I can’t use term by term, because 0 / 0 is undefined.

Let’s use the definitions of +, -, *, / from Conway’s surreal numbers and see if that works.

pi / 4 ∈ { 0 | 1 } which is to be interpreted as pi / 4 sits on the open interval (0,1).

Using Conway’s method.
pi/2 = pi/4 + pi/4 = { 0 + pi/4 | 1 + pi/4}

This is correct, pi/2 > pi/4 and pi/2 < 1 + pi/4. And I’ve retained the unit interval.

But the interval is no longer between two adjacent integers. Hmm.

Addition goes by shift and add. Subtraction by subtract and shift.

½ = 1, 0, 1, 0, 1, 0, 1, 0, …
½ = 0, 1, 0, 1, 0, 1, 0, 1, …
½ + ½ = 1 = 1, 1, 1, 1, 1, 1, 1, 1, …

1/3 = 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, …
1/3 = 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, …
1/3 + 1/3 = 2/3 = 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, …

pi/4 = 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, …
pi/4 = 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, …
pi/4 +pi/4 = pi/2 = 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, …

That’s plus and minus taken care of (sufficiently)

Now to consider multiply and divide.

To be continued.

mollwollfumble said:

Addition goes by shift and add. Subtraction by subtract and shift.

½ = 1, 0, 1, 0, 1, 0, 1, 0, …
½ = 0, 1, 0, 1, 0, 1, 0, 1, …
½ + ½ = 1 = 1, 1, 1, 1, 1, 1, 1, 1, …

1/3 = 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, …
1/3 = 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, …
1/3 + 1/3 = 2/3 = 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, …

pi/4 = 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, …
pi/4 = 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, …
pi/4 +pi/4 = pi/2 = 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, …

That’s plus and minus taken care of (sufficiently)

Now to consider multiply and divide.

To be continued.

This isn’t pretty enough for me. It can be moved back to two adjacent integers using a dithering algorithm. But I’d rather skip the sequence of fractions step entirely.