How would you convince the child-me this is false: $0.\bar{9} = 1 \implies \bar{9} = - 1$?

Question

So I remember as a child when I was taught: $ . \bar9 =1 $ The proof was taught as:

$$x = 0.\bar{9} \\ 10x = 9.\bar{9} \\ 10x - x = 9.\bar{9} - 0.\bar{9} \\ 9x = 9 \\ x = 1 \\ \therefore 0.\bar{9} = 1$$

I was found the whole thing quite counter-intuitive and created my own "analog proof" of why it "must" be an absurd statement:

Let us "assume" $\bar 9$ can exist the same way we assumed $. \bar 9$ can exist.

$$x = \bar{9} \\ \frac{x}{10} = \bar9.{9} \\ x - \frac{x}{10} = \bar{9} - \bar{9}.9 \\ .9 x = -.9 \\ x = -1 \\ \therefore \bar{9} = - 1$$

Hence, by the same set of logic if $. \bar 9 = 1$ then $\bar 9 = -1$. I remember the maths teacher being really frustrated with me because of these kind of "stunts." I kind of sympathize with him that it would be really difficult to explain to a child without using "radius of convergence", etc.

Question

Is it possible to make sense to the "child version" of me without using the words "radius of convergence"?

Answer 1

Note first that $\bar{9}$ doesn't contain a decimal point, so is presumably intended as an "integer" with infinitely many 9s going to the left. In other words, it's $9+90+900+\cdots$, which anyone can see is infinite. (@MattiP. has pointed out division by $10$ might not turn the $9$ into $0.9$, if we never travel infinitely far to the right to reach it, but let's put the aside for now.) Both arguments try to compute the sum of a geometric series, i.e. of terms that multiply by the same factor, be it $1/10$ or $10$. What we've shown is that if a finite value for such a sum exists, it'll be computable in a certain way.

So what's the difference between the two sums? Well, you have to check whether adding successive terms gradually brings us closer to the intended result. $0.9,\,0.99,\,0.999,\,\cdots$ get $10$ times closer at each step to $1$, whereas $9,\,99,\,999,\,\cdots$ get $10$ times further at each step from $-1$. So $1$ is a limit of the first sequence, but -1 isn't a limit of the second.

Or is it? If your definition of the distance from a limit isn't the modulus of a difference, you'll identify different sequences as convergent. You might find this interesting.

Answer 2

Let us "assume" $\bar 9$ can exist the same way we assumed $. \bar 9$ can exist.

In mathematics, we don't just assume things. We define things that are useful and necessary so that things we do in mathematics make sense.

We define infinite decimal notation because without it, we are not capable of writing $\frac13$ in decimal form. There is a mathematical necessity to introduce infinite places after the decimal point. Therefore, if we are able to make sense of a world with infinite places after the decimal, we should. And it turns out that indeed, there is a way to define infinite places after the decimal so that it all makes sense. *

There is no such necessity on the left side of the decimal point. There is no need to define numbers that have infinitely many spaces before the decimal point. Also, if we did define them, as you showed above, this would lead to strange conclusions such as $\overline 9 = -1$. Therefore, we don't have a good way of using expressions such as $\overline 9$, and we don't use them.

* The fact that our definition of infinite decimal places works is something a 5 year old will simply have to believe.