2
$\begingroup$

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"?

$\endgroup$
  • 7
    $\begingroup$ Are you absulutely sure that $$ x = \bar{9} \qquad \Rightarrow \qquad \frac{x}{10} = \bar{9}.9 $$ ? I think this would require some deeper contemplation. Think about it: What is infinity divided by ten? $\endgroup$ – Matti P. Sep 17 at 7:29
  • $\begingroup$ You seem to have a geometric series $9+9\times 10+9\times 10^2+\cdots = \frac{9}{1-10}=-1$. That is no better or worse than saying $1+2+4+8+\cdots =-1$ but in decimal rather than binary $\endgroup$ – Henry Sep 17 at 7:39
  • 2
    $\begingroup$ @MattiP. I know .. which is precisely why I wrote "assume" $\bar 9$ can exist" I mean at that age we aren't explained why $.\bar 9$ can exist ... I think the answers reflect my point. $\endgroup$ – More Anonymous Sep 17 at 7:43
  • $\begingroup$ @Henry I know ... But the point of the "education" tag is to "educate" a student in way he will understand. $\endgroup$ – More Anonymous Sep 17 at 7:44
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I feel you've sneaked in partial summation (though still intuitive) $\endgroup$ – More Anonymous Sep 17 at 7:34
  • 1
    $\begingroup$ @MoreAnonymous I've didn't sneak it in, I rather explicitly said we need to consider it. I saw no rule against using it. Sure, it's one extra concept to explain, but for an ELI5 audience it's much, much friendlier than a radius of convergence. $\endgroup$ – J.G. Sep 17 at 7:52
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ While I do understand you saying it for the benefit of the student "The fact that our definition of infinite decimal places works is something a 5 year old will simply have to believe" ... I always found that way of mathematics dry and boring .,. I always enjoyed it when I could understand the motivations of the mathematician $\endgroup$ – More Anonymous Sep 17 at 7:49
  • $\begingroup$ @MoreAnonymous I mean, if the student wants, we can go into detail how we define infinite decimals, but there is no going around the fact that what we are dealing with are convergent series. Which is exactly what the other answer here did, as he started talking about limits. $\endgroup$ – 5xum Sep 17 at 7:57
  • $\begingroup$ yes ... I'm not sure where you'd start with such a student (who was me) :P Either way I feel I would have been satisfied with J.G method of bringing in partial summation as an intuitive concept $\endgroup$ – More Anonymous Sep 17 at 8:00
  • $\begingroup$ @MoreAnonymous It's a good approach. Just remember that each student needs a different approach. While J.G. provides a nice reason from within mathematics, some students aren't happy with that. My reasoning is more from outside mathematics, answering the more meta-answer of "why do infinite decimals exist to the right, but not to the left of the decimal point". The answer is because (1) they are useful and (2) we know how to make them work. $\endgroup$ – 5xum Sep 17 at 8:05
  • $\begingroup$ I remember asking my math teacher if something has $1$ unit of distance what do we measure in the experiment ? Let's just say he wasn't pleased with the question in this context. I agree this has it's applications ... Unfortunately my teacher wasn't great ... Maybe you can update your answer in light (1) ... Also J.G. has a way to make (2) work (in his last line) ... Like I said I always disliked maths without context (the context doesn't have to be physical world) $\endgroup$ – More Anonymous Sep 17 at 8:52
-1
$\begingroup$

Bro, that's called Real Analysis like thing kicks in!

Let me show you something.

$$Let \, x= 1 + 2 + 3 + \dots \tag{1}$$$$ \Rightarrow x+0 = 1+2+3+\dots \tag{2} \\ (2) - (1) \\ 0 = 1 + 1 + 1 +\dots \\ \boxed{0=+\infty} $$

This is what happens when you do elementary operations to divergent series. You've to ensure that a given series is Convergent, before adding, subtracting, substituting or canceling!

The first one $0.\bar 9$ is convergent because it is bounded between $1 \,\& \,0 $ (Bolanzo-Weierstrass Theorem) .

Second one isn't.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.