0.999... = 1? What is the secret?

Question

If x = 0.999... (a) -> 10x=9.999... -> 10x-x=(9.999...)-(0.999...) -> 9x = 9 -> x = 9/9 = 1 (b)

x = 0.999 (a) = 1 (b) ?

So... what is the right explanation for this occur? I know that exists a tiny error somewhere... I think that the subtraction of the repeating decimal is incorrect, because the principle of infinity. That's it? Am I right?

There is no error. Although you have to be careful with addition of infinite series it is handled properly here. — Graham Kemp, Oct 11 '16 at 00:09
If 0.999... = 1, why there is a need to sum 0.999... + 0.001... to match 1? — Marco, Oct 11 '16 at 00:12
@GrahamKemp well not sure Macro gets that, so you have to explain him not me — reuns, Oct 11 '16 at 00:13
@Marco it is still wrong. Just do the addition by hand. What happens after the first $1$? — user251257, Oct 11 '16 at 00:15
The easiest explanation I've seen for this is as follows: $\frac{1}{3}=.333...$.$\frac{1}{3}\times3=1=.999...$ — AfronPie, Oct 11 '16 at 00:17
@GrahamKemp I think you mean $9/10^k$ as it's incorrect the way you wrote it. — Logan Clark, Oct 11 '16 at 00:19
@Marco If you want to keep the nice properties of algebra that we have (without working from surreals), it must be that $.999... = 1$. — Logan Clark, Oct 11 '16 at 00:22
I would like to point out that $0.000...1$ makes no sense as a number. For a decimal expansion of a given number, if there is a $1$, it must occur somewhere (and "at the infinityth place" is not somewhere). — D_S, Oct 11 '16 at 00:28

score 2 · Answer 1 · answered Oct 11 '16 at 00:22

2

There is no error in your logic. Infinity does weird things, and one thing it does is that the infinitely long decimal expression $0.999...$ turns out to be equal to 1. The implication is that there are actually many real numbers with two equivalent decimal expressions for them (so $0.5829999... = 0.583$, for example).

There's a Wikipedia article that goes into ridiculously deep detail on the matter.

answered Oct 11 '16 at 00:22

ConMan

24,300

10(10/10.001...)-(10/10.001...)= 10x-x=9x 9x = (10/10.001...)/9= x = [(10/10.001...)/9]/9 x = 0,0123443074226321
I messed up on something?
– Marco Oct 11 '16 at 00:27
That is to explain that 9x (x=(10/10.001...) is equal to 0.999..., so x is different than 1. – Marco Oct 11 '16 at 00:32
Where is that .001 coming from? It does not exist. – Graham Kemp Oct 11 '16 at 00:33
I mean 10,001... is equal to 10,000000000000000000000000000000000000000000000000000000000000001 – Marco Oct 11 '16 at 00:36
@Marco then you mean 10,000...01, whatever that means – Ben Grossmann Oct 11 '16 at 00:38
Yes, thanks @Omnomnomnom – Marco Oct 11 '16 at 00:40
@Marco or do you mean 10,000...001? Are those different? And what would you make of 10,000...00999...? – Ben Grossmann Oct 11 '16 at 00:42
I think 10,000...01 is the same that 10,000...001, but i did not know why did you ask that. – Marco Oct 11 '16 at 00:45
1

In the real numbers, there's no such thing as the last decimal place, so there's nowhere to put that 1 you're trying to put. The limit of the sequence (1, 0.1, 0.01, 0.001, 0.0001, ...) is exactly 0. – ConMan Oct 11 '16 at 03:27

score 2 · Answer 2 · answered Oct 11 '16 at 00:26

A priori, real numbers have nothing to do with decimals. If you want to see why this is, look up Peano arithmetic and the precise definition of real numbers.

It just so happens that every real number can be written out in decimal form. However, the decimal expansion of a given real number need not be unique. The real numbers corresponding to the decimal forms

$$1.0000 \cdots$$

$$0.9999 \cdots$$

are the same, as you have observed.

score 2 · Answer 3 · answered Oct 11 '16 at 00:31

$$\begin{align} x ~=~& 0.\dot{\overline 9}\\[1ex] ~=~& \lim\limits_{n\to\infty}\sum_{k=1}^n \tfrac 9{10^k} \\[2ex] 10 x ~=~& \lim\limits_{n\to\infty}\sum_{k=1}^n \tfrac {9}{10^{k-1}}\\[1ex] ~=~& 9+ \lim\limits_{n\to\infty}\sum_{k=2}^n \tfrac {9}{10^{k-1}}\\[1ex] ~=~& 9+ \lim\limits_{n\to\infty}\sum_{j=1}^{n-1} \tfrac {9}{10^{j}} \tag{$\star$} \\[1ex] ~=~& 9+ x\\[2ex] 9x~=~&9 \\[1ex] x~=~&1 \end{align}$$

$(\star)$: Assuming the limit of $F(n)$ as $n\to\infty$ exists, then $F(n-1)$ has the same limit.

Red Banana · Accepted Answer · 2018-02-22T13:59:56.147

I'll recommend you the best article I know in this subject. Fred Richman's: Is $0.999\dots=1$? [MAA: Mathematics Magazine, $12/1999$].

From the beginning: "Arguing whether $0.999\dots=1$ is a popular sport on the newsgroup sci.math. It seems that people are often too quick to dismiss the idea that these numbers might be different."

The article is interesting because it actually discusses why it is and why it couldn't be. What are the needed tools for it to be. Also, there are amazing references.

EDIT: I also recommend Courant's Differential and Integral Calculus (Chapter 1, Section2). There is a great explanation in there which makes the subject appear less magic that it seems.

0.999... = 1? What is the secret?

4 Answers4