Confusing concerning the nature of $1=0.999...?$

Question

Let that

$\frac{1}{3}=0.333...$, then multiply by $3$:

$\frac{1}{3}\times 3=1\tag1$

$0.333...\times3=0.999...\tag2$

$1-0.999...=0.000...0001$

Here it seems that $1 \ne 0.999...$

So what is the correct view concerning $1$ and $0.999...$

Are they equal or not ?

Are you sure $1-0.999... = 0.000...1$ ?? How many $0$ are there?? — JoseSquare, Jun 12 '19 at 18:16
related: https://math.stackexchange.com/questions/11/is-it-true-that-0-999999999-dots-1 — Henry, Jun 12 '19 at 18:17
$1 - 0.999... = 1 - \sum_{i = 1}^{\infty} 9\cdot 10^{-i} = 1 - 1 = 0.$ — Alexander Geldhof, Jun 12 '19 at 18:18
Do you already know what a sequence and its limit is? If not, it is hard to really explain why $0.9999999\ldots =1$. — maxmilgram, Jun 12 '19 at 18:20

Bernard · Accepted Answer · 2019-06-12T21:59:52.933

One can prove there exists a correspondence between real numbers and infinite decimal expansions. In this correspondence, rational numbers correspond to ultimately periodic infinite decimal expansions. The correspondence is bijective, except for decimal numbers, which correspond either to infinite decimal expansions which end in an infinity of $9$s (period $1$), or to the more usual form, obtained from the previous one as follows: that last digit $d<9$ is replaced with the digit $d+1$, and the following $9$s are replaced with $0$s. Of course, usually, one does not write the final $0$s.

Confusing concerning the nature of $1=0.999...?$

1 Answers1