This question already has an answer here:

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 xy 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 xz because y would fit in between them meaning that, y = x and y = z but xz. 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.


marked as duplicate by Don Thousand, Holo, Ennar, Namaste, Lord Shark the Unknown Oct 20 '18 at 6:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 7
    $\begingroup$ Point is, $0.0...1$ can't be a number, because that simply violates a lot of the fundamental tenants of the real numbers. There are extended number systems that attempt to create well defined notions of numbers as such, but your construction certainly isn't one. There are so many flaws with what you've presented that it's not really worth discussing. I would check out the link I've tagged. It discusses all of the things I've mentioned in this comment. $\endgroup$ – Don Thousand Oct 19 '18 at 22:15
  • 1
    $\begingroup$ Like @RushabhMehta said: 0.0...1 is not a number, 0.999... is also not a number, it is informal way to write $0.\overline9=\sum_{i=1}^\infty 10^{-i}9=0.9+0.09+...$. The problem with your way is that your definition of = is wrong, without getting too deep into this in math A=B means "A and B are one and the same" so $c\ne a=b=c$ is not possible, ever. The system you are thinking about is infinitesimal numbers, such system is nealy never used nowadays. Also note that if your numbers exists than 0.0...2/2=0.0...1 so 0.0...1/2=0.0...1 thus 0.0...1=0 $\endgroup$ – Holo Oct 19 '18 at 22:29
  • 4
    $\begingroup$ @RelhokSacul Please don't say to someone that only tries to help "you are missing the point of math entirely" and try to understand what they are trying to say. RushabhMehta is not wrong in what he is saying, your construction simply does not make sense mathematically, before $i$ existed people didn't just one day thought "ah maybe we can define this" they defined this number in a way it won't destroy previously defined mathematics, by extending the field. There are a lot of things you need to make sure that works, for example how do you define multiplication? how do you solve $0.0...1x=1$ $\endgroup$ – Holo Oct 19 '18 at 22:39
  • 1
    $\begingroup$ @Relhok Sacul, "the only reason it can't work is because it violates a lot of preexisting notions"... what better reason do you need? $\endgroup$ – Ennar Oct 19 '18 at 22:54
  • 1
    $\begingroup$ what you says about numbers like $0.0\ldots1$ can be (possibly) understood as some kind of infinitesimals. However you need a theory to make it sense. $\endgroup$ – Masacroso Oct 19 '18 at 23:31

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.


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

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


Not the answer you're looking for? Browse other questions tagged or ask your own question.