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I am currently learning about fractions, and there is something that I am finding it hard to make sense of. When a fraction it added to the right of the decimal point, the number becomes slightly bigger right? So for example if I add enough fractions to the 0.99999999, will this number eventually become equal to 1?

So will this number eventually become equal to 1 (if I added enough 9): 0.9999999999999999999999999999999999999999999999999999999999999999999999...9

Steve
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  • Have you learned about geometric series? – user119615 Jan 30 '15 at 04:06
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    No finite sequence of $9$s is $1$. Infinite $9$s is, by definition, the limit of those finite sequences of $9$s, and that limit is $1$. – Thomas Andrews Jan 30 '15 at 04:08
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    I don't think this is a duplicate; it is not asking about the infinite series, and it is based off an entirely different observation (i.e. that adding more digits to a decimal makes it bigger). The answer to this question is surely different than the answer to the other one. Indeed, this question has an answer without necessarily appealing to limits or infinity or anything. – Milo Brandt Jan 30 '15 at 04:14
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    For $n$ nines, the value is $1-\frac{1}{10^n}$, which is not $1$. The limit of $1-\frac{1}{10^n}$ is one. – Thomas Andrews Jan 30 '15 at 04:16
  • The method of "proof" that was in the (now removed) answer is as unsatisfying as having a bag of crisps become stuck in a vending machine. – Gahawar Jan 30 '15 at 04:22
  • Yeah, the $0.99999\cdots$ question is so common here, there was a knee-jerk response by voters. Too many beginners want to "prove" that $1\neq 0.9999\cdots$. – Thomas Andrews Jan 30 '15 at 04:22
  • No reason to be insulting, @Gahawar . – Thomas Andrews Jan 30 '15 at 04:23
  • Usually when saying that something (for example, a Turing machine halting) will happen "eventually", we mean within a finite number of steps. That gives this question a different answer than if we supposed an infinite number of $9$s. – David K Jan 30 '15 at 04:24
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    This also brings up a new question to me. How do we reopen a question that's marked duplicate incorrectly. – The Artist Jan 30 '15 at 04:26
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    (Some) people can vote to reopen. – Thomas Andrews Jan 30 '15 at 04:27
  • @ThomasAndrews I did not intend to insult. I just experience the same feeling when I come across both circumstances (That is, I frown slightly). Certainly, there are much better ways to answer the issue at hand. Regardless, $0.\overline{9}$ is a bad form in my opinion. – Gahawar Jan 30 '15 at 04:29
  • @ThomasAndrews I see, and what's the threshold of up votes needed? (btw I have already upvoted) – The Artist Jan 30 '15 at 04:30
  • @TheArtist We have three votes to reopen at this moment. Just need two more. (Upvotes are a different thing. You need more reputation to vote to close or reopen.) – David K Jan 30 '15 at 04:32
  • @DavidK Oh I see. Thanks. – The Artist Jan 30 '15 at 04:34
  • It's not upvotes, it is reopen votes. – Thomas Andrews Jan 30 '15 at 04:35
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    @Gahawar It's not clear what you mean by "bad form." It has a well-defined meaning that we can prove is equal to $1$. Sorry, but math has you beat here. – Thomas Andrews Jan 30 '15 at 04:36
  • @ThomasAndrews A "bad form" in the sense that it often leads to the confusion of many a student. – Gahawar Jan 30 '15 at 04:39
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    Lots of things lead to confusion of students. Too bad. If you are going to teach that $\frac{1}{3}=0.\overline{3}$, you have to teach $1=0.\overline{9}$. Stuff being hard doesn't mean we can avoid it. Starting from here lets us later introduce infinite geometric series, infinite non-repeating decimal expansions, etc. But you have to start somewhere. @Gahawar – Thomas Andrews Jan 30 '15 at 04:42
  • @ThomasAndrews Touché – Gahawar Jan 30 '15 at 04:45
  • @Meelo Yes, this is exactly what I mean. And it does not have to be just 0.99999999, it could be 0.55555555, so the idea is that with each new digit the number is becoming bigger. – Steve Jan 30 '15 at 04:50
  • This is a duplicate! – RE60K Jan 30 '15 at 05:04
  • @ADG: Since the OP seems to be talking about finitely many 9's, this does not seem to be a duplicate. – Rory Daulton Jan 30 '15 at 06:48

2 Answers2

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Yes and no. When you are adding more $9$'s to the end of the number, you are adding $0.9$, $0.09$, or $9(0.1)^n$, where $n$ is the $n^{th}$ digit. So if we look at it with a geometric series, we find that adding $n \,9's=\sum_{k=1}^{n} 9\frac{1}{10}^k=\sum_{k= 0}^{n-1} \frac{9}{10} \left(\frac{1}{10}\right)^{k}= \dfrac{9}{10}\dfrac{1-\left(\frac{1}{10}\right)^{n-1}}{1-\frac{1}{10}}$ Let's see if this can equal $1$ for any $n$:

$$1=1-\frac{1}{10}^{n-1}\implies 0=-\frac{1}{10}^{n-1}$$This is never true because an exponential is always positive. So no matter how many $9$'s you add, you will never reach $1$. However, if you add infinite $9$'s, we can take the limit as $n\to \infty$ and see that infinite 9's does equal 1.

ReliableMathBoy
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Teoc
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enter image description here Here is a kind of proof with handmade animation, Hope it will be useful.

enter image description hereAnother one is here too

Khosrotash
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