0

Does the $0.\overline{9} = 1$ proof show that the gap between numbers adjacent numbers is $0$, and so are infinitesimal values equal to $0$ ? Would $10^{-\infty}=0$ ?

Edit: I feel like the smallest non-zero number would be $0.0...1$, which when subtracted from $1$ would 'appear' to be $0.999...$ , but it can't since $1 = 0.999...$

DeepSea
  • 77,651
  • 3
    "the gap between adjacent numbers" There is no such thing as "adjacent numbers." Given two real numbers, either they are equal or they have a number between them. – JMoravitz Feb 01 '18 at 02:10
  • $10^{-\infty}$ is not a valid expression. $\infty$ is not a real number. A related concept though that is valid to talk about is $\lim\limits_{x\to\infty}10^{-x}$ which is indeed zero. – JMoravitz Feb 01 '18 at 02:11
  • Read up about the property of Reals that they are Dense. – DynamoBlaze Feb 01 '18 at 02:13
  • But how can there be a number close to 1 than 0.999...? –  Feb 01 '18 at 02:14
  • 1
    "but how can there be a number close to 1 than 0.999..." It is unclear what exactly you are asking here since your grammar is so poor, but if you mean to ask "how can there be a number as close to $1$ as $0.999\dots$" or "how can there be a number closer to $1$ than $0.999\dots$" remember that $0.999\dots$ is $1$, it just just another way of writing it. When you have two unequal numbers, $a,b$ with $a<b$ then $a<\frac{a+b}{2}<b$ and $\frac{a+b}{2}$ is a number unequal to either and inbetween them. – JMoravitz Feb 01 '18 at 02:17
  • So does every number in between 0.999... and 1 also equal 1? –  Feb 01 '18 at 02:20
  • "So does every number inbetween $1$ and $1$ also equal $1$" Yes. – JMoravitz Feb 01 '18 at 02:20
  • I know that it does, I can't explain the question very well though, sorry. I feel like subtracting the smallest number from 1 would equal 0.999..., since the smallest number 'would' be 0.00....1, but 1 = 0.999... –  Feb 01 '18 at 02:25
  • 1
    See below. There is no such thing as the smallest number. Suppose that $\epsilon$ were such a "smallest number", but then $\frac{\epsilon}{2}<\epsilon$ contradicting that $\epsilon$ be the smallest. – JMoravitz Feb 01 '18 at 02:28
  • Stop! Just stop! There are no such things as "adjacent numbers" and there is no gap between them, there is no smallest positive number. Both $10^{-\infty}$ and $.000..... 0001$ are meaningless nonsense and don't mean anything. $.\overline{9999}$ is $1$. EXACTLY $1$. Not "infinitely close to $1$", not "as close as you like to $1$", but EXACTLY equal to $1$. $1 -.\overline{9999} = 0$, not something infinitely small but $0$. Because $1 -1=0$. And $.\overline{9999} = 1$ EXACTLY. Dont' ask "but what if" or "how about". Those are absolute facts. Get used to it. – fleablood Feb 01 '18 at 07:07
  • "I feel like the smallest non-zero number would be 0.0...1," Well, you feel wrong. There is no such thing as a smallest non-zero number. "which when subtracted from 1 would 'appear' to be 0.999..." you feel wrong. There is not smallest number so you can't subtract it. Anything that "appears" to be $0.9999....$ IS $1$. – fleablood Feb 01 '18 at 07:21

2 Answers2

1

You have: $0.9999...= \dfrac{9}{10}+\dfrac{9}{100}+\cdots= \dfrac{\frac{9}{10}}{1-\dfrac{1}{10}}=1$. For the other question $10^{-\infty} = \displaystyle \lim_{n \to -\infty} 10^n = 0$ ( you can prove this using definition ) .

DeepSea
  • 77,651
  • Would the bottom equation be equal to 0.0...1 if that number could exist? –  Feb 01 '18 at 02:16
  • I meant sort of like the smallest possible number. If you could minus that from 1 what would you get? Since the 'apparent' answer of 0.999... also equals 1. I know it probably doesn't have an answer but I couldn't wrap my head around it when I heard it. –  Feb 01 '18 at 02:23
  • @J.Doe "the smallest possible number." You are describing an infinitesimal. The only infinitesimal in standard real analysis is exact zero. Any other real number could be halved giving a smaller real number. – JMoravitz Feb 01 '18 at 02:27
  • That's why I was wondering if 10−∞=0 (since it would otherwise be infinitesimal) (and even though it is not a valid expression) –  Feb 01 '18 at 02:31
  • $10^{-\infty}$ is as you say, "not a valid expression." That means that anything you say about it is meaningless. It doesn't denote anything. There aren't any sensible questions you can ask about it. – saulspatz Feb 01 '18 at 02:51
  • But the answers says that its equal to that limit, which is equal to zero –  Feb 01 '18 at 03:01
  • @J.Doe $.000...1$ is not defined. Think of it this way. A decimal is a function that inputs a counting number $1, 2, 3, ...$ and outputs a digit. There is no end to the counting numbers so there is no "last decimal digit." – user4894 Feb 01 '18 at 07:08
  • Simply remove all concepts of "the smallest number or $0.0000 ....1$ or $.\overline{9999}$ being anything different than $1$ from your head. There is no such thing as "a smallest number" and $.\overline{9999}$ and $1$ are ONE thing. The exact same thing. $5 - 4 = .\overline{9999}$ and $.\overline{9999}\times 7 = 7$. $1 = .\overline{9999}$. PERIOD. And $1-0 = .\overline{9999}$ and $.\overline{9999} -1 =0$ and $1 - .\overline{9999} = 0$ and $.\overline{9999} -1 = 0$. Because statements are $1-0=1;1-1=0;1-1=0$ and $1-1 = 0$. – fleablood Feb 01 '18 at 07:16
0

The real numbers do not contain infinitesimals. In systems with infinitesimals, it is not true that $0.\overline 9 = 1$, using its usual definition as series limit. Indeed, in systems with infinitesimals, $0.\overline9$ is actually undefined, as the series won't converge.

If you define $0.\overline 9$ differently, then in those systems it may or may not be equal to $1$, depending on the definition.

However note that the real numbers are complete, that is, all Cauchy sequences converge in them, and therefore there is no need to introduce infinitesimal values. Indeed, when considering convergence they are actually harmful.

celtschk
  • 43,384