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Firstly I'm aware of the proofs/reasons regarding .999...=1. I'm not asking for anyone to reference or reiterate them but rather to look at my proof in isolation and help me understand my own mistakes and fallacies.

Another disclaimer I suppose; it's difficult to call this 'proof' my 'own' as it's extremely obvious and simple. Nonetheless I can't find the inconsistency in it.

A 'proof' .999... =/= 1

Suppose .999... = 1 then there is some number x = 1 / .999... and x should obviously be 1. If we take the values for x in the equation x = 1 / .999.. starting with x = 1 / .9 and as x approaches 1 / .999... we get a value for x where x =/= 1. x equals 1.01 (with a repeat bar over the zero and then an additional repeat bar over the zero and one together - I don't know how to write a double repeat bar like that) I'm also going to call the value .01 repeating bar over zero with an additional repeat bar over both the zero and one ε for simplicity's sake.

Reason: 1 / .9 = 1.1 (repeating) 1 / .99 = 1.01 (repeating) 1 / .999 = 1. 001 (repeating) etc.

When you reach an infinite series of .9's the zeros from the value of 1 / .999... becomes infinite before reaching the first 1 then when you reach that first 1 another series of infinite zeroes occurs before another 1 and then that pattern repeats infinitely.

I then find the value for x for x = 1 / .999... is actually x = 1 + ε (the value I defined as the .01 double repeating bars I referred to earlier)

Subtracting 1 from both sides from the equation 1 + ε = 1 / .999... gives the literal value for what I'm referring to as the infinitesimal; ε = .01 (repeating bar over zero then additional repeat bar over both zero and one)

So then the equation ends up not being .999... = 1 but rather .999... + ε = 1.

If I do some basic checks I find ε = 1 - .999... to be true and I have to show I can derive .999... + ε = 1 from 1 + ε = 1 / .999... and due to ε + ε = ε (#) I can show that 1*.999.. + ε*.999... = 1 = .999... + ε

(#) I'm suggesting that in the infinite series ε 'adding' any more of the exact same ε to that infinite series doesn't change the 'value' of that infinite series, you still end up with the same infinite number of the exact same steps. Thus ε * .999... = ε as well as ε * n = ε

So there you have it, 'my' 'proof' .999... + ε = 1 and that ε = .01 repeating bar over zero with an additional repeating bar over both the zero and one.

I fail to find where it's incorrect, any help would be appreciated! And again I'm aware of the reasons for .999...=1 so please don't just simply reiterate them, I'd like to know where I went wrong in the logic of what I wrote, thank you!

Georg Anatoly
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    .01 repeating bar over zero That's just $,0,$ plain and simple. There are no additional decimals after an infinity of $0$s. – dxiv Feb 27 '18 at 04:55
  • You do realize the concept of having infinite zeroes before a one doesn't make any sense? The fact that you have infinite zeroes implies the fact there will never be a one because there is no real end to place the one. Thus $0.{\overline 0}1 = 0$. – Andrew Li Feb 27 '18 at 04:58
  • Please use MathJax to format your posts. – gen-ℤ ready to perish Feb 27 '18 at 05:03
  • Why wouldn't it make sense? The one is still there, there's just an infinite number of zeroes before it. That's like arguing because .9 repeating infinitely doesn't stop the number 1 doesn't exist or that at some point in the series infinitely repeating nines it magically turns the entire infinite string of nines into the number 1. That at best is arbitrary. Let me put it this way, in the number line there is an infinite series of digits between pi and the next number, just because pi literally doesn't end doesn't mean pi's value changes to the next value. – Georg Anatoly Feb 28 '18 at 23:00

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The very assertion that:

$$\dfrac{1}{0.\bar 9} = 1 + 0.\bar 01$$

Proves that $1 = 0.\bar 9$. This is because:

$$0.\bar 01 = 0$$

Conceptually speaking, there is no way to stick a $1$ at the end of an infinite number of $0$s because there is no end to an infinite number of $0$s. There will just be more $0$s—by definition an infinite number of them, before you ever reach a one. You can't write an infinite amount of $0$s then write a one—then you wouldn't have an infinite amount of $0$s! Thus there is no "final" $1$ so $0.\bar 01$ is just an infinite number of $0$s which is $0$.

The same exactly logic applies to the problem of $1-0.\bar 9$. You end up with your result of $0.\bar 01$ again, which by the same reasoning is $0$. Thus you can again conclude that since the difference between $1$ and $0.\bar 9$ is $0$ that $1 = 0.\bar 9$.

Andrew Li
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  • I really need to learn how to type using correct formatting for that I apologize. But anyway: 0.01 (point zero repeating) =/= 0 it equals 0.01 (zero repeating). The fact that there are an infinite number of zeroes before the 1 has nothing to do with whether or not that one exists. It just means you'll never count enough zeroes to get there. It's still there, and it still has a value giving its value to that endless string of zeroes making 0.01 (point zero repeating) infinitesimally greater than zero. – Georg Anatoly Feb 28 '18 at 23:24
  • @GeorgAnatoly It makes no sense to say a number with an infinite zero then a one exists. That's because you're suggesting that the infinity has an end then you have a one, which is not possible infinity has no end. – Andrew Li Feb 28 '18 at 23:30
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There is not just one point at which you make some kind of error. You're exploring an idea and trying some notation out and it seems ok here and there. But the question is really one that starts even sooner.

What do you mean by $0.999\cdots$?

If you resolve this in one way, you may find that your epsilon vanishes and there's no difference between $1$ and $0.\overline{9}$. For instance you can ask "What digit is at the $n$th place?" Well, it's a 9. So if you are subtracting $1-0.\overline{9}$ you feel you should have to put a $1$ somewhere, but no place is quite right, because you've always got another $9$ to go before you can put the $1$ down.

In haste, you conclude, "At the end of an infinity of $0$s is a $1$ so $1-0.\overline{9} = 0.000\cdots 1$." But if you didn't say what you meant by $0.999\cdots$ then you're no better off here, because we now need to know what you mean by $0.\overline{0}$ before we can think about what you might mean by $0.\overline{0}1$.

Perhaps you could say that $1 - \sum_{n=1}^N 9\cdot 10^{-n} = 10^{-N}$. And now what you want is a number $\epsilon$ that satisfies $\epsilon < 10^{-1}$ and $\epsilon < 10^{-2}$ and $\epsilon < 10^{-3}$ and... and $\epsilon < 10^{-N}$ for any $N\in \mathbb{N}$ but also satisfies $\epsilon > 0$.

In the real numbers such an epsilon does not exist: the real number is the limit of the sequence of descending powers of ten and that number is identically 0.

But what if it did exist? Then, if you're very careful, you would have the hyperreals.

law-of-fives
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