# Indeed, 0.999…= 1

Standard

There is a video by Vi Hart titled “Why Every Proof that .999… = 1 is Wrong”

It is true that few real numbers (with cardinality same as that of rational numbers) can’t have unique decimal representation, like, 0.493499999… = 0.4935000…. So, the point made in this video is that you can’t prove two different representations of a number to be equal.

But why so? As pointed out in my earlier post, this has something to do with the way we construct numbers. Such ambiguity in representation holds irrespective of representation system (binary or decimal)

In decimal representation of real numbers  we subdivide intervals into ten equal subintervals. Thus, given $x\in [0,1]$, if we subdivide $[0,1]$ into ten equal subintervals, then $x$ belongs to a subinterval $\left[\frac{b_1}{10}, \frac{b_1 + 1}{10}\right]$ for some integer $b_1$ in $\{0,1,\ldots ,9\}$. We obtain a sequence $\{b_n\}_n$ of integers with $0 \leq b_n \leq 9$ for all $n \in \mathbb{N}$ such that $x$ satisfies

$\frac{b_1}{10}+ \frac{b_2}{10^2}+ \ldots + \frac{b_n}{10^n}\leq x \leq \frac{b_1}{10} + \frac{b_2}{10^2} + \ldots + \frac{b_n+1}{10^n}$

In this case we say that $x$ has a decimal representation given by

$x = 0.b_1 b_2\ldots b_n \ldots$

The decimal representation of $x\in [0,1]$ is unique except when $x$  is a subdivision point at some stage, which is when $x=\frac{m}{10^n}$ for some $m,n \in \mathbb{N}; 1 \leq m \leq 10^n$. We may also assume that $m$ is not divisible by 10.

When $x$ is a subdivision point at the $n^{th}$ stage, one choice for $b_n$ corresponds to selecting the left subinterval, which causes all subsequent digits to be 9, and the other choice corresponds to selecting the right subinterval, which causes all subsequent digits to be 0.

For example, $x = \frac{1}{2} = \frac{5}{10} = 0.4999\ldots = 0.5000\ldots$ unlike $x = \frac{1}{3} = 0.333...$

A nice exposition is available on Wikipedia: https://en.wikipedia.org/wiki/0.999…