I'm told by smart people that $$0.999999999\ldots=1$$ and I believe them, but is there a proof that explains why this is?
2026-04-01 16:49:52.1775062192
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Is it true that $0.999999999\ldots=1$?
60.6k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
16
There are 16 best solutions below
3
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Given (by long division):
$\frac{1}{3} = 0.\bar{3}$
Multiply by 3:
$3\times \left( \frac{1}{3} \right) = \left( 0.\bar{3} \right) \times 3$
Therefore:
$\frac{3}{3} = 0.\bar{9}$
QED.
18
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Indeed this is true. The underlying reason is that decimal numbers are not unique representations of the reals. (Technically, there does exist a bijection between the set of all decimal numbers and the reals, but it is not the natural/obvious one.)
Here's a very simple proof:
$$\begin{align} \frac13&=0.333\ldots&\hbox{(by long division)}\\ \implies0.333\ldots\times3&=0.999\ldots&\hbox{(multiplying each digit by $3$)} \end{align}$$
Then we already know $0.333\ldots\times3=1$ therefore $0.999\ldots=1$.
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.999... = 1 because .999... is a concise symbolic representation of "the limit of some variable as it approaches one." Therefore, .999... = 1 for the same reason the limit of x as x approaches 1 equals 1.
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You can visualise it by thinking about it in infinitesimals. The more $9's$ you have on the end of $0.999$, the closer you get to $1$. When you add an infinite number of $9's$ to the decimal expansion, you are infinitely close to $1$ (or an infinitesimal distance away).
And this isn't a rigorous proof, just an aid to visualisation of the result.
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What I really don't like about all the above answers, is the underlying assumption that $1/3=0.3333\ldots$ How do you know that? It seems to me like assuming the something which is already known.
A proof I really like is:
$$\begin{align} 0.9999\ldots × 10 &= 9.9999\ldots\\ 0.9999\ldots × (9+1) &= 9.9999\ldots\\ \text{by distribution rule: }\Space{15ex}{0ex}{0ex} \\ 0.9999\ldots × 9 + 0.9999\ldots × 1 &= 9.9999\ldots\\ 0.9999\ldots × 9 &= 9.9999\dots-0.9999\ldots\\ 0.9999\ldots × 9 &= 9\\ 0.9999\ldots &= 1 \end{align}$$
The only things I need to assume is, that $9.999\ldots - 0.999\ldots = 9$ and that $0.999\ldots × 10 = 9.999\ldots$ These seems to me intuitive enough to take for granted.
The proof is from an old high school level math book of the Open University in Israel.
7
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Suppose this was not the case, i.e. $0.9999... \neq 1$. Then $0.9999... < 1$ (I hope we agree on that). But between two distinct real numbers, there's always another one in between, say $x=\frac{0.9999... +1}{2}$, hence $0.9999... < x < 1$.
The decimal representation of $x$ must have a digit somewhere that is not $9$ (otherwise $x = 0.9999...$). But that means it's actually smaller, $x < 0.9999...$, contradicting the definition of $x$.
Thus, the assumption that there's a number between $0.9999...$ and $1$ is false, hence they're equal.
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\begin{align} x &= 0.999... \\ 10x &= 9.999... \\ &= 9 + 0.999... \\ &= 9 + x \\ 10x - x &= (9 + x) - x \\ (10 - 1)x &= 9 + (x - x) \\ 9x &= 9 \\ x &= 1 \end{align}
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One argument against this is that 0.99999999... is "somewhat" less than 1. How much exactly?
1 - 0.999999... = ε (0)
If the above is true, the following also must be true:
9 × (1 - 0.999999...) = ε × 9
Let's calculate:
0.999... ×
9 =
───────────
8.1
81
81
.
.
.
───────────
8.999...
Thus:
9 - 8.999999... = 9ε (1)
But:
8.999999... = 8 + 0.99999... (2)
Indeed:
8.00000000... +
0.99999999... =
────────────────
8.99999999...
Now let's see what we can deduce from (0), (1) and (2).
9 - 8.999999... = 9ε because of (2)
9 - 8.999999... = 9 - (8 + 0.99999...) = because of (1)
= 9 - 8 - (1 - ε) because of (0)
= 1 - 1 + ε
= ε.
Thus:
9ε = ε
8ε = 0
ε = 0
1 - 0.999999... = ε = 0
Quod erat demonstrandum. Pardon my unicode.
3
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Assuming:
- infinite decimals are series where the terms are the digits divided by the proper power of the base
- the infinite geometric series $a + a \cdot r + a \cdot r^2 + a \cdot r^3 + \cdots$ has sum $\dfrac{a}{1 - r}$ as long as $|r|<1$
$$0.99999\ldots = \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots$$
This is the infinite geometric series with first term $a = \frac{9}{10}$ and common ratio $r = \frac{1}{10}$, so it has sum $$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$
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If you take two real numbers x and y then there per definition of the real number z for which x < z < y or x > z > y is true.
For x = 0.99999... and y = 1 you can't find a z and therefore 0.99999... = 1.
8
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Okay, I burned a lot of reputation points (at least for me) on MathOverflow to gain clarity on how to give some intuition into this problem, so hopefully, this answer will be at least be somewhat illuminating.
To gain a deeper understanding of what is going on, first we need to answer the question, "What is a number?"
There are a lot of ways to define numbers, but in general numbers are thought of as symbols that represent sets.
This is easy for things like natural numbers. So 10 would correspond to the set with ten things -- like a bag of ten stones. Pretty straightforward.
The tricky part is that when we consider ten a subset of the real numbers, we actually redefine it. This is not emphasized even in higher mathematics classes, like real analysis; it just happens when we define real numbers.
So what is 10 when constructed in the real numbers? Well, at least with the Dedekind cut version of the real numbers, all real numbers correspond to a set with an infinite amount of elements. This makes 10 under the hood look drastically different, although in practice it operates exactly the same.
So let's return to the question: Why is 10 the same as 9.99999? Because the real numbers have this completely surprising quality, where there is no next real number. So when you have two real numbers that are as close together as possible, they are the same. I can't think of any physical object that has this quality, but it's how the real numbers work (makes "real" seem ironic).
With integers (bag of stones version) this is not the same. When you have two integers as close to each other as possible they are still different, and they are distant one apart.
Put another way, 10 bags of stones are not the same as 9.9999999 but 10 the natural number, where natural numbers are a subset of the real numbers is.
The bottom line is that the real numbers have these tricky edge cases that are hard to understand intuitively. Don't worry, your intuition is not really failing you. :)
I didn't feel confident answering until I got this Terence Tao link:
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Both proposed number expressions represent the same Dedekind cut of the set of rational numbers, i.e. the same real number.
The answer is "true".
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Let $R$ be any ring containing an element $x$ such that $1-10x=0$. Suppose further that $R$ contains the formal power series $\sum_{i=1}^\infty x^i$.
Formally this means there is a ring homomorphism $f\colon S\to R$, from a subring $S\subseteq\mathbb{Z}[[t]]$ containing $\sum_{i=1}^\infty t^i$, such that $f(t)=x$.
Then: $$1=\sum_{i=1}^\infty 9x^i.$$
Proof: We have $$0=(1-10x)\left(1+\sum_{i=1}^\infty x^i\right)=1-\sum_{i=1}^\infty 9x^i.$$
Apology: I looked through the previous $31$ answers and did not see any that made clear that this result can be proved algebraically (independently of any topology, analysis or order structure). I think this is a legitimate contribution to the discussion, as once someone has understood why the identity follows from the definitions, they tend to wonder if they could not have defined the reals "better", to avoid it (or perhaps that is what they were wondering in the first place, without realising).
1
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One missing link in other answers.
If $0.9999....$ and $1$ are to represent real numbers then they have to follow the properties of the set of real numbers.
One of them is saying that real numbers are densely ordered, meaning there is always another real number between two different real numbers.
If $0.9999...$ and $1$ are different, there would have to be another real number between the two. However you cannot change any of the digits of either of the two to record this number.
Because of that, $0.9999...$ and $1$ must be considered the same element of the set of real numbers.
0
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We can do this by computing the square root of $0.99...$ by long division method.
Now, we will firstly get a zero and decimal point in the quotient as $0<0.99...\leq 1$. Then $9$ as the divisor for the first pair of $9$'s, then $189$ for the next pair of $9$'s, $1989$ further and so on($19989$, $199989$, ...). The quotient is appended with a $9$ at each step and we can see a pattern in the dividends, divisors and the remainders so as to assure that the quotient will have only $9$'s after the decimal point. Though I'm not sure how to prove that there will be no digit except $9$, maybe observing the pattern is enough. (Inputs on this proof will be appreciated.)
Further we get $\sqrt{0.999...}=0.999...$, moreover $0<0.99...\leq 1$.
Now, if $x\in\mathbb{R},\ x=\sqrt{x} \implies x =0\ \text{or}\ x=1$
Thus we get $0.999...=1$
What does it mean when you refer to $.99999\ldots$? Symbols don't mean anything in particular until you've defined what you mean by them.
In this case the definition is that you are taking the limit of $.9$, $.99$, $.999$, $.9999$, etc. What does it mean to say that limit is $1$? Well, it means that no matter how small a number $x$ you pick, I can show you a point in that sequence such that all further numbers in the sequence are within distance $x$ of $1$. But certainly whatever number you choose your number is bigger than $10^{-k}$ for some $k$. So I can just pick my point to be the $k$th spot in the sequence.
A more intuitive way of explaining the above argument is that the reason $.99999\ldots = 1$ is that their difference is zero. So let's subtract $1.0000\ldots -.99999\ldots = .00000\ldots = 0$. That is,
$1.0 -.9 = .1$
$1.00-.99 = .01$
$1.000-.999=.001$,
$\ldots$
$1.000\ldots -.99999\ldots = .000\ldots = 0$