Is it true that $0.999999999\ldots=1$?
Solution 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,
$1.0 -.9 = .1$
$1.00-.99 = .01$
$1.000-.999=.001$,
$\ldots$
$1.000\ldots -.99999\ldots = .000\ldots = 0$
Solution 2:
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.
Solution 3:
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.
Solution 4:
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 $$\frac{\frac{9}{10}}{1 - \frac{1}{10}} = \frac{\frac{9}{10}}{\frac{9}{10}} = 1.$$