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I have been taught that $0.\overline9=1$.

Now if we think of the finite series' in turn, each closer in turn to $1$ than the previous:

0.9

0.99

0.999 etc.

If we called the difference between this series and $1$ e then this difference approaches zero as the series is extended.

Next, I imagine constructing a rectangle of length l and height e. Clearly the height approaches zero as the length approaches 1. Would this mean that the limit of a two dimensional shape (a rectangle in this case) is a one dimensional shape (an interval)?

I would have though that for any e the shape would be a rectangle, but that a rectangle of zero width is a line segment.

it's a hire car baby
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Thomas
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    If you define the limit of a sequence of shapes in a suitable way, yes, then it would happen that the limit of a sequence of rectangles would be a line segment. But what does it have to do with $0.999\ldots=1$? Any sequence of rectangles with one dimension tending to zero (the other presumably a constant) would behave the same way, no? – Jyrki Lahtonen Jul 11 '16 at 10:13
  • I would have thought a rectangle was by definition 2 dimensions. – Thomas Jul 11 '16 at 10:16
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    So it looks like the property of a shape to be a rectangle is not (in math parlance) closed under taking the limit. This happens more often than not in math: the limit of a sequence of rational numbers need not be a rational number, the limit of a sequence of continuous functions need not be a continuous function,... – Jyrki Lahtonen Jul 11 '16 at 10:19

2 Answers2

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0.999... is defined as an infinite series. For example as: $$0 + 0.9 + 0.09 + 0.009 + \ldots$$ The value of such an expression is defined as the limit of the series. You are right, the limit of course is $1$. So the value itself simply is $1$.

If you have the infinite series, the difference to $1$ is always $0$. As the value of the series is always defined as the limit. For $e$ to be not $0$ you must make the series finite, i.e. stopping it. For example at $0.99$, then $e = 0.01$.

If you now construct your rectangle, it is $2$-dimensional as $e \ne 0$.

Of course, you can now study the behavior of your rectangle if $e$ approaches $0$. Then the rectangle approaches to your $1$-dimensional interval. But, which is important, it will never be the interval as the series must be finite to speak of an $e \ne 0$.

As said before, if you make the series infinite, then the series simply is the same as $1$.

Summarized, your rectangle approaches the $1$-dimensional interval but it will never reach it. That is why we speak of the term $limit$.

$$\lim\limits_{e \to 0} rectangle = interval$$ $$rectangle \ne interval$$

Zabuzard
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  • Thanks for that. It clarifies my question, but I am still struggling with the idea that 0.9999... is equal to 1. If so, then if I start with a rectangle of height e, while I am dealing with 0.9999... I have some height, thus two dimensions. When I am dealing with 1.0, e must be zero and this would fail my understanding of a rectangle. So I feel there is an important difference between 0.9999... and 1. – Thomas Jul 11 '16 at 10:27
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    There is no difference between $0.999\ldots$ and $1$. The value of an infinite series is defined as the limit. The value of a finite sequence is the value of it when you sum it up. Thus, you only have an $e \ne 0$ if you use a finite series, for example $0.999$ instead of $0.999\ldots$. An example: $\sum\limits_{n = 0}^\infty n := \lim\limits_{k \to \infty} \sum\limits_{n = 0}^k n$. You may take a look at: https://en.wikipedia.org/wiki/Series_(mathematics)#Definition – Zabuzard Jul 11 '16 at 10:30
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    @Thomas you're conflating the limit of the decreasing width of your rectangle with the limit of its length as that increases. The quickest and clearest way to see why $0.999..=1$, is to subtract $0.999...$ from $1$ and write down the difference. When you've finished writing it down, tell us what the magnitude of the last digit is. – it's a hire car baby Jul 11 '16 at 10:42
  • The following approach might be good to understand. Are you with me if I say $0.999\ldots$ is defined as $\sum\limits_{k = 1}^\infty 9 \cdot 10^{-k}$ and not as $\sum\limits_{k = 1}^m 9 \cdot 10^{-k}$, where $m \in \mathbb{N}$. The difference is that the first series is called an infinite series and in the second series we have a fixed $m$. Yes, the $m$ can be veeeery large, but it is fixed. Okay, an infinite series is defined as the limit: $\sum\limits_{k = 1}^\infty 9 \cdot 10^{-k} := \lim\limits_{m \to \infty} \sum\limits_{k = 1}^m 9 \cdot 10^{-k}$. And the limit of this series is $1$. – Zabuzard Jul 11 '16 at 14:49
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This is an insightful question because you have struck upon a property of integral calculus. Your example demonstrates how, by taking infinite limits, we can create transformations from an area to a line.

In the words of Jyrki Lahtonen above, "The property of a shape to be a rectangle is not closed under taking the limit." What this means is that the taking of limits of a rectangle's properties has the power to transcend whether it is a rectangle or not.

However in your example this capacity to become "not a rectangle" does not derive from the $0.999...$ approaching $1$, it derives from the width approaching zero.

You should take care to define $l$ in your numerical format. l can be written as a function of the number of 9's in your sequence, as follows: $$l(m)=\sum_{n=1}^{m}{9\times10^{-n}}$$

And you are defining $e=1-l$

To state that $0.\overline9=1$ is equivalent to stating $$\lim_{m\to\infty} l(m) = 1$$

The area of your your rectangle is $e\times l$

And you are stating that $$\lim_{m\to\infty}( e(m) \times l(m) )= 0$$

This is clear because the limit of $e$ is zero, the limit of $l$ is $1$ so the limit of their product is $0$.

Therefore the limit of the area of a rectangle, as its width approaches zero; $m\to\infty$ is zero.

Which as you correctly postulated, implies that a line segment can be thought of as the limit of a rectangle as its width approaches zero.

it's a hire car baby
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