User blog:Rgetar/Transfinite decimal fractions

Three months ago I thought: we can fill (-∞; +∞) with rational numbers, and it may seem that it is the most "dense" set of numbers within (-∞; +∞), because there are no "gaps" between rational numbers, that is there are rational numbers between any two different rational numbers. But it is not true: there are real numbers, and some of them are not rational. Set of rational numbers is countable, and set of real numbers has cardinality of continuum. And I asked myself: is set of real numbers the most "dense" in (-∞; +∞)? Or maybe there are numbers in (-∞; +∞), which are not real, and cardinality of set of them is larger than continuum? (Maybe, some of these numbers are located "between" reals, the same way as some of reals are located "between" rationals?)

I thought that real numbers are infinite decimal fractions (maybe, except such cases as 0.9999999..., which is usually considered equal to 1.0000000... = 1). That is a real number is an integer number, then decimal point, then 1st digit after point, then 2nd digit after point, then 3rd digit after point, then 4th digit after point, ... I thought: what if we introduce transfinite digits after point, that is ω-th digit after point, (ω + 1)-th digit after point, (ω + 2)-th digit after point, ...? Then real numbers are numbers with zero transfinite decimal fractional part, and we can fill (-∞; +∞) with set of numbers of any infinite cardinality. For example, for decimal fractions with digits up to Ω-th digit after point cardinality is Ω2, for decimal fractions with digits up to Ω2-th digit after point cardinality is Ω3, etc.

Example: a number, which integer part is 0, ω-th digit after point is 1, and any other digit after point is 0. It is larger than 0, and it is not a real number.

If we subtract this number from 1, we get 0.9999999..., but only up to ω-th digit after point. Any digit beginning from (ω + 1)-th is 0. Apparently, 0.9999999... = 1 only for 0.9999999... with any digit after point equal to 9, including transfinite.

Then I asked myself: what is ω-th digit of π? I thought: well, π is a real number, so ω-th and any after ω-th digit of π is 0.

But then I thought: what is ω-th digit of 1/3? Since 1/3 is also a real number, it is 0.3333333..., and any digit beginning from ω-th is 0.

But then I thought that must be


 * 1/3 * 3 = 1

However, if we multiply 0.3333333... with zero transfinite decimal fractional part by three, we get 0.9999999... with zero transfinite decimal fractional part, which is lesser than 1, since 1 is 0.9999999... with any digit after point equal to 9, including transfinite. So, 1/3 is 0.3333333... with any digit after point equal to 3, including transfinite.

So, 1/3 has non-zero transfinite decimal fractional part. What does it mean? Maybe, 1/3 is not a real number? I thought that it is unlikely. Because we can represent real number 1 as 0.9999999..., with any digit after point equal to 9, including transfinite. So I thought that some real numbers can contain non-zero transfinite decimal fractional part.

And 1/3 = 0.3333333... in decimal numeral system, but in ternary numeral system 1/3 = 0.1, that is 1/3 is a finite ternary fraction. But numbers should not depend on base of numeral system, since a numeral system is just a designation.

So I suggested that some real numbers have non-zero transfinite decimal fractional part.

But if it is so, then maybe ω-th digit of π is not 0.

Then I try to google it, since I thought that this idea is quite simple, and these numbers are already known.

I found something similar - surreal numbers. I think that transfinite decimal fractions might be a special case of surreal numbers, but I am not sure, since I have not studied them yet. I think that surreal numbers is a more general system. In particular, reals and ordinals are surreal numbers.

About surreal number I found
 * Wikipedia article
 * 2012 pdf (intruduction to surreal numbers)
 * 2019 pdf (another intruduction to surreal numbers)