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I am looking for a simple way to describe the underlying set of the real numbers without getting into cauchy sequences or dedekind cuts. Furthermore, I want the description to not rely on some notion of equivalence (like how one can use the notion of coprime to give the rationals unique representatives). I think the following works:

$\mathbb{R} = \{\text{All decimal expansions | does not end in repeating 9s}\}$.

My question is am I forgetting about something or does this do the trick?

Edit: Indeed, this is not closed under the usual definition of $+$ and $\cdot$ but we can redefine these operations to "round up" when necessary. I can verify the field axioms on my own time; I am merely seeing if anyone can spot a subtlety that I missed or if my description truly does give unique representatives of the reals.

Edit 2: (Some context) The posters have given some great ways to define the reals if you want to spend a decent amount of time on it. I want to give the reals as a set of decimal expansions without any fancy notion of equivalence so that I can move on and do linear algebra. For example, if I were teaching a first year course, the field axioms would take a whole lecture and be super boring and Cauchy sequences/dekekind cuts would have students lining up to drop the course.

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    $\begingroup$ As usual, Tim Gowers has interesting things to say: dpmms.cam.ac.uk/~wtg10/decimals.html $\endgroup$ – Hans Lundmark Oct 3 '14 at 6:44
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    $\begingroup$ I wanted to point out that it seems contradictory to forbid the use of equivalences but ask for something 'quick and painless'. $\endgroup$ – Hurkyl Oct 3 '14 at 9:30
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    $\begingroup$ Aside: are you sure that the "usual definition of $+$ and $\cdot$" that you have in mind is actually well-defined on infinite decimal expansions? $\endgroup$ – Hurkyl Oct 3 '14 at 9:42
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    $\begingroup$ "I want to give the reals as a set of decimal expansions without any fancy notion of equivalence so that I can move on and do linear algebra." Why do you need to define the reals as a set of decimal expansions before doing linear algebra? $\endgroup$ – JiK Oct 3 '14 at 19:53
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    $\begingroup$ If I was teaching a first year course, I'd assume the student has a sufficiently good intuitive understanding of the real numbers and run with that, leaving the explicit study of foundations to a later course. Even in the later courses, it's probably more appropriate for the real analysis course to do that exercise. And if you wanted to do foundations and still be self-contained and aren't doing any analysis, all you really need to state is the real closed field axioms (because you want to talk about positive numbers and factor polynomials). $\endgroup$ – Hurkyl Oct 4 '14 at 4:08
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It will do the trick, but it will be messy and painful to prove the field axioms. Moreover, the definition in this way is somewhat arbitrary. Why choose base $10$? There is a more natural way to do it which is the Bourbaki definition of the reals. It relies on the completion construction of a uniform space. The completion of a uniform space is precisely the set of minimal Cauchy filters (so in particular, there is no equivalence involved). The reals are the completion of the rationals, and the rationals can be given the structure of a uniform space (basically, since it is topological group), thus the reals can be defined to have underlying set the set of all minimal Cauchy filters in $\mathbb Q$.

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    $\begingroup$ Let $10 = n$ for any $n$. This construction isn't actually that bad and it illustrates the idea that the intuition (so what if your intuition is base 10?) is totally on track and can be made rigorous. If I recall from Spivak, the trick is to prove completeness first and use that everywhere, or something. $\endgroup$ – djechlin Oct 3 '14 at 9:20
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    $\begingroup$ I don't know what "$10=n$ for any $n$" means. $\endgroup$ – Ittay Weiss Oct 3 '14 at 10:34
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Your definition is not closed under the arithmetic operations, at least with their usual definitions on decimal expansions. For instance, take the decimal expansion $1/9 = 0.111...$ and multiply by 9 to obtain $0.999...$, which is not an element of your set.

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    $\begingroup$ obviously, the arithmetic operation will take this into account. This is not the real issue with this definition. In particular, this definition can be used to define the reals. The details of the proofs though are almost horrific. $\endgroup$ – Ittay Weiss Oct 3 '14 at 6:23
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    $\begingroup$ True, you could define the operations in such a way as to avoid this. But you're essentially encoding the equivalence relation into the operations; maybe you get to escape using the phrase "equivalence relation," but that's about it. I agree with your answer as well! $\endgroup$ – Dylan Murphy Oct 3 '14 at 6:26
  • $\begingroup$ The equivalence relation you refer to is a totally different one than the one using, e.g., Cauchy sequences. Primarily, you don't need the axiom of choice to choose one representative from each equivalence class, since there is a canonical one. With Cauchy sequences there is no canonical choice of representative. With the uniform completion, you can first look at all Cauchy filters, and define an equivalence relation. The nice thing is that with the filters there is a minimality condition that chooses for you a unique representative from each equivalence class. $\endgroup$ – Ittay Weiss Oct 3 '14 at 6:41
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If you can read German, here is a detailed development of the approach Gowers is advocating at the link given in Hans Lundmark's comment:

http://www.math.ethz.ch/~blatter/Dualbrueche_2.pdf

But note the following: Whichever approach you take, the amount of work to be done in order to verify all the details is about the same.

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There are at least 3 sensible ways to do it, outlined in Spivak:

  1. Define a real number as Cauchy sequences of rational numbers, with the equivalence relation that two sequences are equivalent if their differences converge to 0. So technically a real number is an equivalence class of Cauchy sequences. Yes, this is quick and painless.
  2. Define a real number as a Dedekind cut. A Dedekind cut consists of two sets $(L, R)$ satisfying:

    • $L$ and $R$ form a nontrivial partition, meaning each is non-empty, $L \cap R = \emptyset$, and $L \cup R = \mathbb{Q}$.
    • Each element of $L$ is less than each element of $R$.
    • If $a \in L$, then all rational numbers less than $a$ are also in $L$. Similarly if $b \in R$ then all rational numbers greater than $b$ are in $R$.
    • $L$ has no greatest element. (The would-be greatest element is the real number being described by the cut. $R$ will have a least element if and only if the Dedekind cut corresponds to a rational number, namely that one.)

    Basically a Dedekind cut gets as close as it can to pinpointing a real number, without naming it (since it can't, since we're constructing it). I've personally liked this once, since it's elementary to learn but pulls at completeness the same way a Dedekind cut pulls for a real number.

  3. Your approach. The salvage for closure is to allow repeated 9s, and create an equivalence relation. It's exactly how clock arithmetic works: 7+7 would violate closure on a clock (i.e. $\mathbb{Z}/ 12\mathbb{Z}$) except we know $14 \cong 2$ so it's fine. The rigorous way to do this is to think of numbers on the clock as the sets $\{0, 12, 24, \ldots\}, \{1, 13, 25, \ldots\}, \ldots$ and define addition on those. Yes, this is "no frills."
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  • $\begingroup$ OP was specifically looking for a construction other than Cauchy sequences or Dedekind cuts, and one that avoids an equivalence relation. For approach three, there really is no need to introduce an equivalence relation for which every equivalence class has at most 2 elements, and of these a canonical choice can be made. So, the technical difficulties of this construction aside, that would be what OP was looking for. $\endgroup$ – Ittay Weiss Oct 3 '14 at 10:37
  • $\begingroup$ I think that if you omitted the paragraph with the first number bullet point, the rest of your answer would be better. That's because I feel that the definition by Dedekind cuts is simpler because a computable Cauchy sequence doesn't necessarily approach a computable number and it's harder to prove that that definition of a real number works. $\endgroup$ – Timothy Dec 9 '18 at 19:43
  • $\begingroup$ @Timothy in that case (empty, R) is a Dedekind cut and it ruins the bijection we want to set up between cuts and real numbers. Unless it corresponds to infinity but that's not what we're trying to do. I think (4) is only needed to ensure uniqueness of cuts -> reals. Without it, there's two ways to refer to rational points. $\endgroup$ – djechlin Dec 10 '18 at 15:28
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There's no universal agreement on what a real number is so I will construct a set with operations +, $\times$, and $\leq$ that can be proven to be a complete ordered field but won't bother to give a proof that it's a complete ordered field and will call those the real numbers. I don't know of any really simple way of describing what a real number is so I will give as simple a description as I can. The following is just my description of the whole thing, not a universal one so if you use it for teaching, you should make it clear that you are using your own definitions and explaining how you define them. Maybe if you do a really good job of researching how to teach and teach less material so that you can move onto new material at a slower rate, your students will learn it really well. For example, the course should not teach a not very useful concept and how to make calculations about it because that might make there not be enough time to teach more useful material without teaching it too quickly such as the definition of a real number that I will later give that I think should be taught in full to make other course work later easier and another course could teach people background information from which they will later be able to easily learn so many different previously hard to learn math concepts instead of spending so much time teaching only one of them and then from that, they will be able to figure out how to write a formal proof in ZF that there is a complete ordered field which is unique up to isomorphism when they are asked to do so on a test and should only get full marks if they write the formal proof properly so that a future job can get better people which are those who have shown that they have the ability to figure out how to write a complete formal proof the way they were told to and not just an intuitive proof that they cannot figure out how to break down into a complete formal proof. Maybe you could join a research group and give your ideas on how a text book for that course should be written.

First, we construct the integers with the operations of +, $\times$ and $\leq$. For each integer $x$ that's not a solution to $10 \times y = x$, we can invent a solution, $y$ to the equation. Each of the newly invented numbers $y$ is still not a solution to $10 \times z = y$ so for each of those numbers we can invent a solution. We can keep going for ever. There is only one intuitive way to define addition, multiplication, and inequality on the set of all numbers that can be constructed that way There is also only one intuitive way to define a notation for each of those numbers. Take any subset of that set that's nonempty and its complement is also nonempty and for every member of the set, all smaller members are in the set. Let's say a boundary number of that set is one where all smaller numbers are in the set and all larger numbers are not in the set. Some of those sets have boundary numbers and others don't. For each of those sets that doesn't have a boundary number, we can invent a number that's larger than all numbers in the set and smaller than all numbers in its complement and then call all the numbers we've constructed so far real numbers. Now, there's only one intuitive way to define inequality on that set and the notation for each number in that set and it does not invent a second notation for numbers I already invented a notation for such as 0.999... for 1. Rather we say the notation 0.999... is undefined. Now that the notation and inequality on that set has already been defined, there is exactly one way to define 1, 0, +, $\times$, and $\leq$ on that set that I will call $\mathbb{R}$ such that $(\mathbb{R}, 1, 0, +, \times, \leq)$ is a complete ordered field and + and $\times$ restricted to terminating decimals matches what I previously defined and the multiplicative identity 1 has the notation 1 and the additive identity 0 has the notation 0.

Now the way to add 2 notations is by determining which numbers they represent and figuring out the sum then writing the notation of the sum. It's probably better to start all over and do it how you would if you only began school now and try to figure out how you would do it if you didn't have the past experience of how you were taught to add base 10 notations in grade 1. After reading https://www.inc.com/bill-murphy-jr/science-says-were-sending-our-kids-to-school-much-too-early-and-that-can-hurt-th.html, I think that kids learn better from school if they start it when they're older. Maybe you were so young when you learned how to add that you misunderstood what they were trying to teach you and then had trouble breaking your old habits of thinking it's a universal rule. I once had an interview at Earls and they said they do their job a totally different way and new people tend to learn it better then people with previous experience because people with previous experience tend to have trouble breaking old habits.

That method of adding does not work on nonterminating real numbers when using my notation. The method was to work on the digits going from right to left but there's no last digit to start on. If we decide to instead use a notation that allows for a string of trailing 9's, then there is another simple way to do it that given any 2 notations can produce a notation that represents the sum or product of the two numbers those are notations for. I won't explain how to do multiplication with it but for addition, you can work from left to right. If the digits of one column add up to less than 9, the column to the left has no carry. If they add up to more than 9, the column to the left has a carry. If they add up to 9, the column to the left has a carry if and only if that column has a carry. For some pairs of positive real numbers, for some columns, to determine whether it has a carry, you have to first determine whether the one to the right has a carry but to do that, you first have to determine whether the one to the right has a carry and that will keep going for ever. If you are told the information that starting from a certain point, the digits in each column always add up to 9, then you know the notation for the sum. Now what about subtraction and division.

You may have learned about fractions in elementry school and misunderstood what you were getting taught and rather than understanding that a fraction is really one notation for division of integers which is defined for all pairs of real numbers whose denominator is nonzero, you instead thought of the rational numbers as being constructed from the integers and thought there were no numbers but rational numbers. Rather, division on $\mathbb{R}$ is defined in terms of multiplication on $\mathbb{R}$ and when you then restrict the already existing multiplication and division operations on $\mathbb{R}$ to $\mathbb{Q}$, it ends up agreeing with the way you learned how to multiply and divide fractions earlier and you must reject the assumption that the rational numbers are the only numbers unless you're fine with using only the rational numbers and saying there is no solution to the equation $x^2 = 2$. We can still define the property of being a rational number such that either its square is less than 2 or it is negative and then define $\sqrt{2}$ to mean that property.

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The simplest way: The real numbers are a set with operations +, -, *, / and relations < = > which follow the axioms of arithmetic, and where every nonempty set with an upper bound has a least upper bound.

Equivalence classes of Cauchy sequences are IMO the easiest way to prove that the real numbers exist, and they come kind of natural when you start talking about limits (just construct a sequence that should have the limit $2^{1/2}$ but doesn't in the rational numbers because the number doesn't exist).

(Going to decimals with unlimited number of digits will get you into huge trouble just defining the product of two numbers).

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    $\begingroup$ How does that answer OP's question? $\endgroup$ – Ittay Weiss Oct 3 '14 at 10:57

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