# $0.00\dots 1=0$ claim

I was reading this webpage about a claim to do with $0 \times \infty$. In it he states in red that $0.00\dots 1=0$ for the same reasons that $0.9999\dots= 1$. I fail to see his logic and his supposed proof that he provides (also in red) is just a Wikipedia article proving $0.9999\dots = 1$. Can anyone explain whats going on? Is this guy talking rubbish or is there some truth to his argument?

• $0.00\dots1$ makes no sense, it is not a number. – Andrés E. Caicedo Aug 3 '13 at 20:23
• .99999... represents the real number that a particular infinite sum converges to. 0.000...1 is not a real number. There is no last decimal place. – William Aug 3 '13 at 20:23
• All part of "A Guide for the Survival of Humankind and Helping the World, Society, and Yourself". Which could make one feel pessimistic about said survival(s)... – Did Aug 3 '13 at 20:27
• Maybe he meant $\lim\limits_{n\to\infty}10^{-n}$ – llllllllllllllllllllllllllllll Aug 3 '13 at 20:30
• The thesis is not sustainable as it stands. I believe it is possible to index the positions of an extended decimal expansion with general ordinals, and make some sense of the idea - but that requires a sophistication and level of understanding beyond what is evident here. We can choose "numbers" with whatever properties we like, in the end. The Real Numbers have proven to have useful and unique properties - which is why we study and use them so much. What is proposed is not possible with Real Numbers. We go beyond when there is some object in view which we cannot otherwise attain. – Mark Bennet Aug 3 '13 at 20:44

## 2 Answers

The only truth is that if you interpret his "number" the only way possible,

$$\lim_{n \to \infty} \frac1{10^n}$$

then indeed this limit is $0$. But his notation is completely whack in my opinion, because you can't write a bunch of zeros and then literally write a $1$ at position "infinity"

• +1 for correctness, as well as the phrase "his notation is completely whack" – Zev Chonoles Aug 3 '13 at 20:35
• Hang on, you can write a formal string of infinitely many zeros with a 1 at position $\omega$, where the string is indexed by the ordinal number $\omega + 1$. However, it isn't obvious how to interpret this string as a decimal that evaluates to a number. – Chris Culter Aug 3 '13 at 20:37
• Sure, we could extend decimal expansions beyond being indexed by positive integers, but then the only conclusion I can see is that the contribution of every new "decimal place" beyond the finite positive integer index range would be 0, so he might as well have written $0.0000....999...$ = 0 as well. – user2566092 Aug 3 '13 at 22:18

In math, you need to be much more precise otherwise you end up in a conundrum such as this one.

In order to resolve the question we need to allow something that is not usual, which is extending the positional decimal system. In the standard decimal notation, there is no infinity out of the box! Period. So let us do something about it first.

If you want to have an infinite positional system you need to define the mapping $\{0,1,2,3,4,5,6,7,8,9\} \to \mathbb{N}$, first. Then, you create the mapping of elements $\{0,1,2,3,4,5,6,7,8,9\} \to \mathbb{N}$ to the real numbers from $0$ to $1$, and then extend this on using ...$-1+m$, $1+m$, $2+m$... to all reals, where $m$ is the member of the above mapping.

This mapping is simply saying that each possible position has one of the ten digits. Instantly, each position gets one of the ten digits. There is no limiting sequence like in classical case $0.999...$ ${0.9,0.99,0.999,...}$ No. For that case, we simply have the element $\{9\} \to \mathbb{N}$

This infinite positional system can be split further in two subsets. One, the Approximants, include all those 'finite' decimal notations like $0.1=0.100000...$ $0.12=0.1200000...$ that ends with an infinite number of trailing zeros, and another one with truly within infinite positional system like $0.03030303...$ that never ends with an infinite number of trailing zeros, there is always some digit from $1$ to $9$ that interrupts the tail. We will call this second one the Complete map.

The Approximants do not represent $\pi$ for example, not even $\frac{1}{3}$. The other group is complete and is a representation of the set of real numbers between $0$ and $1$.

Your case is asking to do this: Take $0.1$ and start adding all more and more zeros $0.001$, $0.0001$, $0.00001$... Each of these fits into the first category of the Approximants. And now you ask what is $0.0...01$, with an infinite number of zeros to the left of $1$?

This is not a representation of a real number. Notice that there is a mapping between the Complete map and the region $(0,1]$. ($0$ does not belong to it as we ask to have at least one other digit that interrupts the zero tail.) It is not difficult to prove that $(\{9\} \to \mathbb{N}) \to 1$, if you consider multiplication for example of numbers in the region $(0,1]$. $(\{9\} \to \mathbb{N})$ acts as the identity element. It is not really equal to $1$, unless we extend the notion of equivalence, which we do in math quite often.

Your newly devised element $0.0...1$ does not belong to the mapping $\{0,1,2,3,4,5,6,7,8,9\} \to \mathbb{N}$, and there is no real number between $0$ and $1$ that does not, for a very simple reason that we cannot define the position of that single $1$. Notice that in the above mapping we always know the position of each digit, it is always one natural number. (It does not belong to the Approximants either because the infinite tail of zeros is broken at least once.)

To make something out of your case, let us try to find some other fitting. For example, your new element may correspond to something like $\{0,1,2,3,4,5,6,7,8,9\}^2 \to \mathbb{N}^2$. Digit $1$ starts after all natural number positions are filled in. For this reason you need to write it less confusing as $0.0...;10...$.

This notation and the system that comes from $\{0,1,2,3,4,5,6,7,8,9\}^2 \to \mathbb{N}^2$ belongs to the set of hyperreal numbers, that is qualitatively different from the set of real numbers.

In order to compare a hyperreal and real number you need the function known as the standard part $\mathrm{st}(x)$. If you do all the math it is indeed $\mathrm{st}(0.0...;10...)=0$.

This is to say that $0.0...;10...$ in your notation $0.0...1$ represents an infinitesimal number.

But you are correct, there is a way of having $0.9999...$ representing $1$ and there is a way of $0.0...1$ representing $0$. However, these two are very different paths and quote: "the same reason" in their argumentation is very far from the truth.