A super weird quality of numbers that is hard to explain.

Question

First off, I know that 0.9… = 1 and I'm not trying to prove or debate that, but my discussion about it is necessary for understanding the question. I was talking to my brother about whether 0.9… equaled one and I was using the “No holes in the number line proof” to get my point across when he said that 0.9… + (0.0…1)/2 would fit in between 0.9… and 1. Now I didn't know whether 0.0…1 was a valid number or not but I guess technically had to (if it's not feel free to tell me but for the rest of this we'll be pretending it is, think of it as a fun what if situation). I told him that it couldn't be divided because it's equal to zero, nothing could fit in between it and zero. He said (0.0…1)/2 could. I told him that assuming he could, it would equal 0.0…05 and infinity plus one equals infinity meaning that it could be simplified down to 0.0…5 and that 0.0…5 doesn't fit in between 0 and 0.0…1 meaning that 0.0…1 = 0 so 0.9… + 0.0…1 = 1 could be simplified down to 0.9… + 0 = 1, 0.9… = 1. He told me that infinity plus one does not equal infinity and I gave up. All that said and done, I was still left thinking about the numbers 0.0…1 through 0.0…9 and I realized they are quite weird. If we define x ≠ y as x having space in between it and y on the number line, and x = y as x having no space in between it and y on the number line you will see where the weirdness starts. For just one example, if x = 0.0…1 and y = 0.0…2 then x = y. Now if z = 0.0…3 then y = z. But the thing is x ≠ z because y would fit in between them meaning that, y = x and y = z but x ≠ z. To keep this from breaking the transitive property of equality y would have to be both equal to and not equal to both x and z at the same time. That is why I have dubbed these numbers “Quantum Numbers”. Does anyone know whether or not anyone has published anything on them, I would be interested to know, and if you have anything to add please do.

Answer 1

In the numerator of your added quantity (the thing divided by $2$), how many zeroes are meant to be before the final $1$? If it is infinitely many, then that "decimal" doesn't make sense because what power of $1/10$ goes with the final $1$? If it is a finite number of zeroes the argument doesn't work anyway.

Answer 2

The thing is, writing $0.999\cdots$ or $0.\overline 9$ means $\sum_{n=1}^\infty \dfrac{9}{10^n}$, which, in turn, equals $\displaystyle\lim_{n \to \infty}\left(\sum_{k=1}^n \dfrac{9}{10^k}\right)$.

At no time is $$0.\underbrace{999\dots999}_{\text{n nines}}=\sum_{k=1}^n \dfrac{9}{10^k}$$ ever actually equal to $1$. But the difference between that sum and $1$

$$1-0.\underbrace{999\dots999}_{\text{n nines}} = 0.\underbrace{000\dots00}_{n-1\text{ zeros}}1=\dfrac{1}{10^n}$$

can be made smaller than any given positive number just by making $n$ large enough.

You wrote $.00\dots 01$ but you didn't define what you mean by it. Well, the only definition I can think of is

$$.00\dots 01 = \lim_{n \to \infty} 0.\underbrace{000\dots00}_{n-1\text{ zeros}}1 = \lim_{n \to \infty}\dfrac{1}{10^n} = 0 $$