So I tried looking around for this question, but I didn't find much of anything – mostly unrelated-but-similarly-worded stuff. So either I suck at Googling or whatever but I'll get to the point.
So far in my coursework, it seems like we've mostly taken for granted that $(\mathbb{R},+,\cdot)$ is a field. I'm not doubting that much, that would seem silly. However, my question is: how would one prove this? In particular, how would one prove that $(\mathbb{R},+)$ and $(\mathbb{R}\setminus \{0\}, \cdot)$ are closed under their respective operations?
I understand the definition of closure, but to say "a real number plus/times a real number is a real number" seems oddly circular since, without demonstrating that, it essentially invokes the assumptions we're trying to prove. Obviously, there's something "more" to the definition of "real number" that would make proving this possible.
Though I'm not sure what property would be used for this.
One thought I dwelled on for a while was instead looking at what the real numbers are not. For example, they are numbers lacking those "imaginary" components you see in their higher-dimensional generalizations – the complex numbers ($i$), quaternions ($i,j,k$), and so on. But that didn't seem quite "right" to me? Like I'm not sure if it's really wrong, it just irked me in some way. Like it's simple enough to say "a real number is any complex number with a zero imaginary component," take two real numbers, show their imaginary parts sum/multiply to zero, and thus we have a real number.
Maybe it's just a personal issue? Like I said – I'm not saying it's inherently wrong (it might be, though, I don't know – if it is, I would like to know why). Maybe it's just the whole idea of "defining a number by what's it's not" that bugs me. Like I said, I'm not really sure, and I think I'm rambling/unclear enough as it is, so I'll get straight to the point.
In short, how does one properly demonstrate, if not in the above way,
$$a,b \in \mathbb{R} \Rightarrow (a+b)\in \mathbb{R}$$
$$a,b\in \mathbb{R \setminus \{0\}} \Rightarrow (a\cdot b) \in \mathbb{R \setminus \{0\}}$$
(And again, I'm not at all doubting that these are true. I'm just curious as to how one would demonstrate these facts in the most appropriate manner since I don't believe it's come up in my coursework and I've been curious on how one would prove it for a few days now.)
Best Answer
As I seem to be so fond of saying, it all comes down to definitions. Specifically, before you can answer the question "Why are the real numbers closed under addition and multiplication?", one needs to answer the question "What is a real number?" There are lots of ways of doing this–I'll outline two common approaches.
Axiomatic Approachs
The real numbers can be defined axiomatically, i.e. we simply list the properties that we want the real numbers to have, then work with those properties directly. The "usual" axiomatization is to define the real numbers to be the (unique, up to isomorphism) totally ordered, Dedekind-complete field. Basically, if we let $\mathbb{R}$ denote the set of real numbers, then $\mathbb{R}$ is defined by the properties that:
If $\mathbb{R}$ is defined in this manner, then the answer to your question is trivial: $\mathbb{R}$ is a field because we have defined it to be so, and fields are closed under addition and multiplication. These properties are "baked in" to the definition of the real numbers.
It is worth noting that other axiomatizations are possible. For example, Tarski's axiomitization, which defines $\mathbb{R}$ as a linearly ordered abelian group with some other nice properties. Under this axiomatization, $\mathbb{R}$ is, by definition, closed under addition, but no multiplication operation is defined a priori. I am not as familiar with this construction, but the above cited Wikipedia article suggests that Tarski was able to define a multiplication operation and show that it behaved as expected, making $\mathbb{R}$ into a field. I suspect that one could also go the route of defining the group $\mathbb{R}$ in terms of the Tarski axioms, then obtaining the field $\mathbb{R}$ as the field of fractions.
A More Constructive Approach
(In the interest of brevity, the following discussion is very loosey-goosey. It can mostly be made rigorous, but I'm not sure that it would be terribly useful to do so.)
The downside of a axiomatic approach is that you are kind of asserting the existence of of a mathematical object which as the properties that you want. It might be more satisfying to build the real numbers from more basic building blocks. We still need some axioms as a starting point, be we can start from a much more primitive place, say the Peano axioms or the ZFC axioms for set theory.
From the Peano axioms, we get the natural numbers immediately, so let's start there. From this point of view, $\mathbb{N}$ is a fairly simple object: about the only thing we get from the definition is a successor operation and zero, which gives us an order on $\mathbb{N}$, but not a lot else. However, from these, we can build up $\mathbb{N}$ as a monoid (there is an additive structure with identity, but no inverses).
From the natural numbers, we can define the integers: first, we define an equivalence relation on $\mathbb{N}\times \mathbb{N}$ by declaring that $(a,b)\sim (c,d)$ if $a+d = c+b$. The integers are then elements of the set $\mathbb{N}\times\mathbb{N}/\sim$, i.e. equivalence classes modulo this relation. The basic idea is that $(a,b)$ represents the number $a-b$ (so $(0,b)$ is the integer $-b$, for example). Since subtraction and negative numbers are not defined by default, everything gets done in terms of addition. When the dust settles, we get a group $(\mathbb{Z},+)$. This group has all the nice additive properties which we expect from the integers, so we can now forget that the integers are "really" equivalence classes of ordered pairs of natural numbers. Note that $\mathbb{N}$ embeds nicely into $\mathbb{Z}$, and that the order on $\mathbb{N}$ can be extended to $\mathbb{Z}$. So, now we have an ordered group. Hoo-ray!
On the integers, it is now possible to define a multiplication operation which is compatible with the addition operation and the order relation. For example, we can define multiplication here as repeated addition, then use some fancy induction. The problem with the integers is that most integers don't have multiplicative inverses, so we need to patch that up. We do this in much the same way that we got the integers. This time, we define an equivalence relation on pairs of integers by declaring that $(p,q)\sim (r,s)$ if $ps = qr$, then mod out by this relation. This gives us the set $\mathbb{Q} := \mathbb{Z}\times\mathbb{Z}/\sim$, i.e. the field of fractions of $\mathbb{Z}$. Again, $\mathbb{Z}$ can be embedded into the rationals in a way that preserves all of the operations as well as the order.
Finally, to get the real numbers, we want some kind of notion of completeness. The usual constructions are via Dedekind cuts or Cauchy sequences. At the end of the day, the two constructions can be shown to be equivalent, but, as a guy who does analysis on metric spaces, I am more comfortable with the construction via Cauchy sequences.
Since this is the heart of your question, I'll be a little more detailed here.
Let $|\cdot| : \mathbb{Q} \to \mathbb{Q}_{\ge 0}$ denote the usual absolute value on $\mathbb{Q}$. What we would really like to do is use the absolute value to define a metric on $\mathbb{Q}$, but we don't have the real numbers in hand yet, so this would be a little circular. However, the absolute value does define a valuation on $\mathbb{Q}$, which is morally the same thing as a norm which can induce the moral equivalent of a metric.
We would like to live a world where Cauchy sequences converge to limits. However, there are examples of Cauchy sequences in $\mathbb{Q}$ which do not converge in $\mathbb{Q}$. Consider, for example, the sequence of partial sums $$ S_N := \sum_{n=0}^{N} \frac{1}{n!}, $$ which is Cauchy but has no rational limit. To rectify this situation, we once again build an equivalence relation.
Now we declare an equivalence relation on the set of sequences. Specifically, we say that $(a_n) \sim (b_n)$ if the sequence $(a_n - b_n)$ is a null sequence. It is a little bit of work to show that this is an equivalence relation, but it works out. Hoo-ray! We can now define $\mathbb{R}$ to be $\mathscr{C}(\mathbb{Q})/\sim$, where $\mathscr{C}(\mathbb{Q})$ denotes the set of Cauchy sequences.
A real number is then an equivalence class of Cauchy sequences. To finish the construction, we need an order relation, as well as addition and multiplication operations. Since your question is about the additive and multiplicative closure of $\mathbb{R}$, I'll ignore the order relation and note only that addition and multiplication are defined pointwise, i.e.
To show that the reals are closed, it is necessary to show that $(a_n + b_n)$ and $(a_n \cdot b_n)$ are both Cauchy. This isn't too bad to do, so I'll leave it as an exercise. Once the exercise is completed, the question is answered. :)