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So I recently learned that $0{.}9$ repeating is equal to $1$:

$$ x = 0{.}9\ldots\\ 10x = 9{.}9\ldots\\ 9x = 10x - x = 9{.}9\ldots - 0{.}9\ldots = 9\\ x = 9x/9 = 9/9 = 1\\ x = 1 $$

Or a simpler proof:

$$ x = 1/3 = 0{.}3\ldots\\ 3x = 3/3 = 0{.}9\ldots $$

Which I had no trouble stomaching, this also proves that two different numbers can be equal.

Here's my difficulty though, if the repeating decimal $0{.}9\ldots$ is equal to another, what is $0{.}1\ldots$, $0{.}2\ldots$, $0{.}3\ldots$ etc. up until $0{.}8\ldots$ equal to?

Asaf Karagila
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warspyking
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    Conventionally this is not showing "two different numbers can be equal" but that one number can have two different representations. The other repeating decimals you present do not have alternative representations; it's only the ones with repeating 9s. – Joffan Jan 09 '15 at 22:54
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    I am guessing that your mother calls you something other than "warspyking". Does that prove that two different people can be equal? – WillO Jan 09 '15 at 22:58
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    How did you prove they were different? While they may not look the same, they can represent the same value: $.5=\frac{1}{2}=\frac{50}{100}$ and a bunch of other ways to express that value. – JB King Jan 09 '15 at 23:52

4 Answers4

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The idea is the same, just solve for $x=0{.}\overline{8}$: $$10x=8{.}\overline{8}=8+x\implies 9x=8\implies x=\frac 89$$

Surely you can do the other fractions on your own, or just note that $0{.}\overline{1}=\frac19$ by the same argument, and therefore $0{.}\overline{2}=2\cdot\frac19=\frac29$ and so on.

Note that the "simpler" proof relies on the fact that $\frac13=0{.}\overline{3}$. If you don't understand why this is the case, then the proof is not simpler, since it sweeps the difficulty under rug, rather than dissolving it.

Asaf Karagila
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  • But wouldn't this be a circular proving the .3… is .3… which is .3… etc? Is there no "rounded" version of them through these calculations? – warspyking Jan 09 '15 at 23:01
  • What rounded version? Do you expect $0{.}\overline{3}$ to be equal to $0{.}34$ at some point? – Asaf Karagila Jan 09 '15 at 23:05
  • Something like that, why isn't it? – warspyking Jan 09 '15 at 23:10
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    Because if those were equal, multiply them by $100$ and you get that $34=33{.}\overline{3}$, which is clearly false. – Asaf Karagila Jan 09 '15 at 23:11
  • Ah, I never though of it that way. Thanks. – warspyking Jan 09 '15 at 23:15
  • Not you too, @AsafKaragila. Surely you know that you can't multiply an infinite series by a constant without properly defining the reals, defining the sum of a convergent series, and proving the theorem on term-by-term multiplication by a constant. And by the time you've done all that, you already understand why .888... = 8/9 and .999... = 1 since you've already proven a more mathematically sophisticated theorem. I'm a little dismayed to see you promoting this pseudo-argument, which I regard merely as a heuristic for high school students. Am I totally off base here, tilting at windmills? – user4894 Jan 09 '15 at 23:59
  • @user4894: No, I am well aware of the problem of talking about these things. But there is time and place for bringing up these issues. When you sit through your first calculus course, you don't expect the professor to begin by telling you what are the inference rules, and what are the axioms. You just do things, and they turn out alright, so you proceed. Later you learn about proofs, inferences, and so on, and you learn how to properly do what you've done informally. This is a crutch, a scaffold in the never ending tower of mathematics; as are these pseudo-arguments for high school students. – Asaf Karagila Jan 10 '15 at 00:59
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this also proves that two different numbers can be equal.

The point is that they are NOT different numbers, just two different representations.

You've already said that $0{,}\overline{3}$ (The overline is common notation for a repeating set of digits) is $\frac{1}{3}$.

To find out what the other repeating representations are equal to, use the same procedre, i.e. multiply by $10$ (or more generally $10^\text{length of period}$), subtract and divide.

You'll find that $0{,}\overline{x} = \frac{x}{9}$.

Henrik supports the community
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Your espression $0.3333...x$ is meaningless.

You can imitate the proof you gave like this: $$ x = 0.333..., \\ 10x = 3.333... \\ 10x-x = 3.000... = 3 \\ 9x = 3 \\ x = 3/9 = 1/3 $$

Your other questions can be answered similarly. $0.7777... = 7/9$ and so on.

GEdgar
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It's just that $1$ has $2$ different radix representations in base $10$: $\{1, 0.\bar9\}$. In fact, all integers have $2$ different radix representations in base $10$. The integer itself, and the previous integer $+ \space0.\bar 9$. $13$ can be written as either $13$ or $12.\bar 9$. These two representations point to the same value.

$0.\bar8$ is a radix representation for the rational $\dfrac89$.

You can show this:

$$0.\bar 8 = 8 \left(0.1 + 0.01 + 0.001 + \dots \right) = 8\sum_{k=1}^\infty \left(\frac{1}{10}\right)^k = \frac{\frac{8}{10}}{1 - \frac{1}{10}} = \frac89$$

user123
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