-2

As the title says. The proof for the last part is as follows: 1/3 = 0.333 with infinitely repeating threes. Multiplying both sides by 3, we have 1 = 0.333 with infinitely repeating threes * 3 = 0.9 with infinitely repeating 9s.

Edit:

The application for this question is with limits. Consider a function f(x) such that it defined as y = 1 for all x in [1,3].

lim f(x) x -> 1^- = DNE, since f(1^-) is not defined. But is this equation not the same as lim f(x) -> 0.99999 with infinite 9s? Since 0.99999 with infinite 9s = 1, then lim f(x) -> 0.99999 = f(1) = 1.

Jyrki Lahtonen
  • 133,153
Serendipitous Epiphany
  • 432
  • 3
  • 15
  • .... so what is your question? – Bram28 Aug 19 '17 at 16:52
  • 4
    Numbers don't move, so there is no "approaching". Your intuition is flawed here. – B. Goddard Aug 19 '17 at 16:52
  • There is no such thing as "the number that approaches one from the left". Numbers are things that are approached, they don't approach things. The infinite decimal expansion defines a sequence of rational numbers that approaches a thing. The thing it approaches is the real number represented by the infinite decimal expansion. The sequence $.9,.99,.999,\ldots$ has limit $1$, thus $1$ is the real number defined by the decimal expansion $.999\ldots$ – spaceisdarkgreen Aug 19 '17 at 17:06
  • Huh.... BECAUSE 0.999999..... "approaches" 1 from the left is WHY 0.99999..... = 1. I don't understand your question. Also $f(1^-)$ not being defined has nothing to do with anything. You arbitrarily cut the domain to be [1,3] for no reason whatsoever. You might as well have defined the domain to only be the irrational numbers or the rationals or whatever. Showing a lim for a function doesn't exist outside it's domain is irrelevent. And if g(x) =1 on a domain that includes $1^-$ then $g(1^-) = 1$. No biggee. EVERYTHING works! – fleablood Aug 19 '17 at 17:43
  • $\lim_{x\to0.999\ldots} f(x)$ is the same as $\lim_{x\to1} f(x).$ Why would either of these be the same as $\lim_{x\to1^-} f(x)?$ There is a number named $0.999\ldots$ (which also has another name, $1$) but there is no number named $1^-.$ – David K Aug 19 '17 at 18:23
  • @David K that helps a bit. Then I would like to clarify: what does 1^- represent? Is it the set of numbers less than one? – Serendipitous Epiphany Aug 19 '17 at 18:58
  • 1
    If we let $1^-$ be a set of numbers, then we have to define what it means for $x$ to approach a set, and it wouldn't work the way $x\to1^-$ is intended to work. There is not actually any mathematical object named $1^-$ in $\lim_{x\to1^-} f(x);$ instead, the superscript "$-$" tells us what kind of values of $x$ are allowed to occur in the definition of this limit, namely, only values less than $1.$ – David K Aug 19 '17 at 19:29

2 Answers2

0

Let $x=0.99999....$ So, $10x=9.99999.....$ Now by subtracting ,$9x=9$ Therefore $x=1$

Sufaid Saleel
  • 3,789
  • 2
    Why does the subtraction work as you presume, with an infinite decimal expansion? Specifically, how would you make a positional subtraction work for $0.\overline{26}-0.\overline{142857}$ ? –  Aug 19 '17 at 17:25
  • @G.Sassatelli with magic (actually calculus and limits) – Simply Beautiful Art Aug 19 '17 at 18:16
0

$$0.999999...=9 \sum_{n=1}^{ \infty} \frac{1}{10^n}=9 \times \frac{ \frac{1}{10}}{1- \frac{1}{10}}=1$$

Marios Gretsas
  • 21,906