Useless Not-Even-CP Information #2: 0.999 repeating = 1 thing and real numbers

Правка en2, от greateric, 2026-06-07 22:16:50

(No, these will not turn into a daily occurrence, but if you all like hearing about vaguely CP-related things then I'll keep writing them occasionally. Anyway.)

You've probably heard a "proof" that $$$0.99\overline{9} = 1$$$ that looks something like this:

  • $$$\frac{1}{3} = 0.33\overline{3},$$$ (This first step is already non-rigorous, because you're already using circular reasoning to say "yeah, if a repeating decimal is not an approximation, then it is not an approximation")
  • $$$1 = \frac{3}{3} = 0.99\overline{9}.$$$

Or this:

  • $$$x = 0.99\overline{9},$$$ (Okay, this is fine so far...)
  • $$$10x = 9.99\overline{9},$$$ (How are you allowed to multiply these?)
  • $$$9x = 9.99\overline{9} - 0.99\overline{9} = 9,$$$ (Is this true? Sure, the limiting behavior does approach $$$9$$$, but that's different from being identically equal.)
  • $$$x = 1$$$.

So let's formalize this, starting with:

What is a real number?

Before we do this, let's lay out what we're allowed to take for granted. Everything else we have to prove from first principles.

  • Integers and rational numbers exist.
  • We can perform basic arithmetic (addition, multiplication, division, etc.) on integers and rationals.

There are two equivalent definitions.

Dedekind Cut

A real number $$$r$$$ is defined by its set $$$A$$$ of rationals, where:

  • $$$A$$$ consists of all rational numbers strictly less than $$$r$$$.
  • The above implies that $$$A$$$ has to be "closed downwards": if $$$x \in A$$$, and $$$y \lt x$$$, then $$$y$$$ must also be in $$$A$$$.
  • There's a bonus requirement that $$$A$$$ does not contain a maximum element (it's to prevent there from being 2 representations for a rational number), but that's not important to us here.

Two real numbers $$$x$$$ and $$$y$$$ are equal if and only if their $$$A$$$-sets $$$A_x$$$ and $$$A_y$$$ are the same.

For example, $$$\sqrt{2}$$$ can be defined as $$$A_{\sqrt{2}}$$$ containing every rational number $$$q$$$ s.t. $$$q^2 \lt 2$$$.

Infinite Cauchy Sequences

We define a real number as the number that an infinite Cauchy sequence of rationals approximates.

Loosely speaking, a sequence $$$a_1, a_2, ...$$$ is Cauchy if it converges. More formally speaking, given any $$$\epsilon \gt 0$$$, we can find a critical point $$$c$$$ such that for all $$$i \gt c, j \gt c$$$, we have $$$|a_i - a_j| \lt \epsilon$$$. In words, given any $$$\epsilon$$$, past a certain point $$$c$$$, all the elements of the sequence are close to each other within $$$\epsilon$$$.

Then, a real number can be identified by a Cauchy sequence.

The same real number can be identified by multiple Cauchy sequences — two Cauchy sequences $$$a_1, a_2, ...$$$ and $$$b_1, b_2, ...$$$ identify the same real number if and only if $$$\lim_{i \rightarrow \infty} a_i - b_i = 0$$$.

For example, $$$\sqrt{2}$$$ can be defined by a Cauchy sequence of its decimal approximation: $$$1.4, 1.41, 1.414, 1.4142, ...$$$ (Technically, a decimal approximation is a nebulous term. More formally, we'll say that $$$a_i = \frac{k}{10^i}$$$, where $$$k$$$ is the highest number where $$$a_i^2 \lt 2$$$ still holds,)

You can notice that $$$1.45, 1.415, 1.4145, 1.41425, ...$$$ is a different Cauchy sequence but it intuitively identifies the same $$$\sqrt{2}$$$, and formally you can see that the limit $$$a_i - b_i$$$ becomes arbitrarily small.

Defining a repeating decimal

Let's formally define a repeating decimal $$$0.\overline{d_1 d_2 d_3 ... d_k}$$$.

Let's also define the sequence $$$a$$$ as $$$0.d_1, 0.d_1 d_2, 0.d_1 d_2 d_3, ...$$$

Under Dedekind cuts

We will construct the set $$$A$$$ as follows: $$$q$$$ is in $$$A$$$ if and only if there exists some $$$i$$$ where $$$q \lt a_i$$$. In other words, there must exist a certain point where the successive finite decimal representations outnumber us.

Let's take $$$2/3 = 0.\overline{6}$$$. The rational number $$$0.662$$$ is in $$$A$$$ because $$$0.662 \lt a_3 = 0.666$$$.

Under Cauchy sequences

This one is easier. It's literally just the sequence $$$a$$$ of the successive finite decimal representations.

Finally showing $$$0.\overline{9} = 1$$$

Dedekind cuts

We'll show that $$$A_{0.\overline{9}}$$$ and $$$A_{1}$$$ are the same set by showing both are subsets of each other.

$$$q \in A_{0.\overline{9}} \rightarrow q \in A_{1}$$$

  • $$$a_i \lt 1$$$ for all $$$a_i$$$ in the finite representations of $$$0.\overline{9}$$$, so if $$$q \in A_{0.\overline{9}}$$$, then for some $$$i$$$, we have $$$q \lt a_i \lt 1$$$ and $$$q \in A_{1}$$$.

$$$q \in A_{1} \rightarrow q \in A_{0.\overline{9}}$$$

  • Take any $$$q \lt 1$$$. It must have a difference $$$d = 1 - q \gt 0$$$. Then, $$$q \lt a_i$$$ is equivalent to $$$d \gt 1 - a_i$$$. We have $$$a_i = 0.999...9_{\textit{i nines}}$$$, or $$$1 - a_i = 10^{-i}$$$, so we only need to show that there exists some $$$i$$$ where $$$d \gt 10^{-i}$$$. Clearly, $$$10^{-i}$$$ can be arbitrarily small, so choosing a high enough $$$i$$$ proves that $$$q \in A_{0.\overline{9}}$$$.

Therefore, both $$$A$$$-sets are the same, and these are the same real number.

Cauchy sequences

The sequence for $$$0.\overline{9}$$$ is $$$a_i = 1 - 10^{-i}$$$: $$$0.9, 0.99, 0.999, ...$$$

The sequence for $$$1$$$ is $$$b_i = 1$$$: $$$1, 1, 1, ...$$$

So we have $$$|a_i - b_i| = 10^{-i}$$$, and this clearly approaches $$$0$$$ as $$$i \rightarrow \infty$$$.

Therefore, these are the same real number.

Bonus: defining arithmetic with real numbers

Dedekind cuts

Given $$$x: A_a$$$ and $$$y: A_b$$$, define $$$A_{x+y}$$$ as the set containing all $$$q$$$ where $$$q \le cd, c \in A_a, d \in A_b$$$.

Multiplication can be defined the same way.

Cauchy sequences

Given $$$x: [x_1, x_2, ...]$$$ and $$$y: [y_1, y_2, ...]$$$, define $$$z = x+y$$$ as $$$[z_i = x_i + y_i]$$$.

Multiplication can again be defined the same way.

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en2 Английский greateric 2026-06-07 22:16:50 4 fix wrong statement
en1 Английский greateric 2026-06-07 22:12:44 5805 Initial revision (published)