If 9.999... = 10, then is there a general proof for any number that has infinite trailing 9s?

Question

I've read about $9.999...=10$, and I would say that I understand it. However, I am looking to apply that proof to all real numbers with trailing 9s. For example:

$72.999...=73$

I have the following potential proof for this particular scenario:

$73/3=x$

$x=24.333...$

$3x=72.999...$

$72.999...=73$

This works, but how would I create a general proof for all numbers that have an infinite trail of 9s?

Answer 1

Since $$0.999\ldots =1$$ we add the integer $n$ to both sides to get $$n.999\ldots =n+1.$$

Answer 2

Let $x = k.999999\ldots $ for any integer $k$.

Compute $10 x - x$.

Answer 3

First of all, assume that $x\in(0,1)$. Then $x=0.d_1d_2\ldots d_n99999\ldots$, with $d_n\ne9$. Therefore $10^nx=a+0.999\ldots$, where $a=10^{n-1}a_1+10^{n-2}a_2+\cdots+a_0$. So, $10^nx=a+1$, and therefore $x=0.d_1d_2\ldots d_{n-1}(d_n+1)$.

In the general case, $x$ can be written as $m+x'$, with $m\in\Bbb Z$ and where $x'\in(0,1)$ has trailing $9$'s. So, the previous argument aples to $x'$.