What real numbers "are" depends on you mathematical foundation.

ZFC is the most popular mathematical foundation, so that's the one I'm assuming.

In ZFC, the only mathematical objects that we can talk about are "sets". (what "are" sets? That's a philosophical question, so I won't get into that.)
All you need to know about sets is that they can "have other sets as elements" (whatever that means) and that a set is uniquely identifiable by its contents (axiom of extensionality).

One of the easiest set to think of (or prove that it exist), is the set that has no elements (since sets are uniquely identified by their contents, there is just one set with no elements).
We're going to call that set the empty set. We're also going to call that set 0. I'm also just going to declare that I consider that set a "number".
Using curly bracket notation, you can describe this set like so: {}
{} is the same thing as 0

You can also prove that the set that only contains 0, i.e. {0} or {{}},  exists. I'm going to call that set 1 and also consider that set a "number".
So 1 is the same thing as {0} which is the same thing as {{}}.

You can also prove that the set that only contains 0 and 1, i.e. {0, 1} or {{}, {{}}}, exists. I'm going to call that set 2 and also consider that set a "number".
So 2 is the same thing as {0, 1} which is the same thing as {{}, {{}}}.

I define 3 as {0, 1, 2} = {{}, {{}}, {{}, {{}}}}.
I define 4 as {0, 1, 2, 3} = {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}
...
this is long...
oh I know, let's just assume I can do this forever (axiom of infinity)

I've now created infinitely many sets using that pattern. I'm going to consider all these sets (and only these sets for now) numbers. So {{}, {{}}} is a number, but {{}, {{}}, {{{}}}} is not a number.
In fact everything I did so far was extremely natural and not contrived at all, so I'm going to call these sets natural numbers.

OK, I have many numbers now, what do I do with them?
Well, I called them numbers, maybe I should be able to add them together?
Fuck, how do I add two sets together.

Well, if I have a number, I know how to find the next number. The number that comes after n = {0, 1, 2, 3, ..., m} is {0, 1, 2, 3, ..., m, n}. To simplify things, I'm going to say S(n) whenever I want to refer to the number that comes after n.

Let me define a new symbol: +

"x + 0" means the same thing as just "x".
"x + S(y)" means the same thing as just "S(x + y)".

So in particular "x + 1" which is the same as "x + S(0)" just means "S(x + 0)" which we saw means just means "S(x)", which is the number that comes after x.
And "x + 2" which is the same as "x + S(1)" just means "S(x + 1)" which we saw means just means "S(S(x))", which is the number that comes after the number that comes after x.

Oh, I guess I have addition now, cool!

So what is 3 + 2?
Well 3 = {0, 1, 2} = {{}, {{}}, {{}, {{}}}} and 2 = {0, 1} = {{}, {{}}}, so 3 + 2 is really just {{}, {{}}, {{}, {{}}}} + {{}, {{}}}
But {{}, {{}}, {{}, {{}}}} + {{}, {{}}} is just {{}, {{}}, {{}, {{}}}} + S({{}}), so it's just S({{}, {{}}, {{}, {{}}}} + {{}}).
{{}, {{}}, {{}, {{}}}} + {{}} is just {{}, {{}}, {{}, {{}}}} + S({}), so it's just S({{}, {{}}, {{}, {{}}}} + {}).
And we know that {{}, {{}}, {{}, {{}}}} + {} is just {{}, {{}}, {{}, {{}}}}.

So {{}, {{}}, {{}, {{}}}} + {{}} = S({{}, {{}}, {{}, {{}}}})
So {{}, {{}}, {{}, {{}}}} + {{}, {{}}} = S(S({{}, {{}}, {{}, {{}}}}))

S({{}, {{}}, {{}, {{}}}}) = {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}
So S(S({{}, {{}}, {{}, {{}}}})) = {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}, {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}}
So that's the set I'm talking about when I write 3 + 2.
Oh hey! I recognize that set, it's just
{0, {0}, {0, {0}}, {0, {0}, {0, {0}}}, {0, {0}, {0, {0}}, {0, {0}, {0, {0}}}}}
= {0, 1, {0, 1}, {0, 1, {0, 1}}, {0, 1, {0, 1}}, {0, 1, {0, 1}}}}
= {0, 1, 2, {0, 1, 2}, {0, 1, 2, {0, 1, 2}}}
= {0, 1, 2, 3, {0, 1, 2, 3}}
= {0, 1, 2, 3, 4}
= 5.

So that's what I really mean when I say 3 + 2 = 5, I mean that
{{}, {{}}, {{}, {{}}}} + {{}, {{}}} is the same set as {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}, {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}}

OK, I can now add numbers (a.k.a some sets I consider special), but can I do it backwards???
Let's say I have {{}}, by what number do I need to add to get {{}, {{}}, {{}, {{}}}}? Well, I just need to add by {{}, {{}}}. Obvious right?
OK let's try the other way around, let's say I have {{}, {{}}, {{}, {{}}}}, by which number do I need to add to get {{}}. What? There's no solution? I can't get back to {{}}? Fuck this this! Fuck these numbers! This sucks, these are shit numbers.

Fuck numbers, I hate them. Let's do something else.
Sets are fucking stupid, {{}, {{}}} is the same thing as {{{}}, {}}, it's fucking stupid that they don't care about order.
Fuck that, I want to care about order. New definition, if x and y are sets, I define (x, y) to be the set {x, {x, y}}.
Fuck yeah, that way (x, y) is not the same set as (y, x) unless x = y, but in that case, I want them to be equal because they are both (x, x).
(x, y) is called "the ordered pair x, y" or just "the pair x, y" in English.

Side comment: note that (0, 0) = {0, {0, 0}} = {0, {0}} = {0, 1} = 2 happens to be a number, but that (0, 1) = {0, {0, 1}} = {0, 2} is not a number.

OK, fuck sets, pairs are my favourite now.
Fuck the old numbers, I'm going to make new numbers with pairs.
Actually, keep the old numbers around. If x and y are natural numbers, then I consider the ordered pair (x, y) special.
I'm going to call pairs of natural numbers, integers.

So (0, 1) = {0, 2} = {{}, {{}, {{}}}} is an integer and so is (1, 0) = {1, 2} = {{{}}, {{}, {{}}}}.
I want the integer (x, 0) to play the same role as the natural number x.
And I want the integer (x, y) to play the role of the integer such that if I add it by (y, 0), I get back just (x, 0).
I want this because I want to be able to undo additions.

OK, so I'm going to make new rules. Now, + means something different depending if it's between two natural numbers or two integers.
I want (x, 0) + (y, 0) = (x + y, 0)
I also want (x, y) + (y, 0) = (x, 0)

OK let's see if these rules make sense...
So (3, 0) + (2, 0) = (3 + 2, 0) = (5, 0)
and (5, 2) + (2, 0) = (5, 0)

Weird, that means if I add (2, 0) to either (3, 0) or (5, 2), it doesn't matter which, I get (5, 0).
That's gross, if I add (2, 0) to two different numbers, I should get two different results. Fuck these integers, I fucked up, let me try again.

Oh I see, (3, 0) behaves the same as (5, 2) because the difference between 3 and 0 is the same as the difference between 5 and 2. So even though (3, 0) and (5, 2) are technically different sets, they behave the same according to my new rules of addition.

Well fuck that, I no longer consider pairs of natural numbers as integers. I'll instead use the word integer to describe a set that contains all of the pairs of numbers that have the same difference.

For example, the infinite set { (3,0), (4,1), (5, 2), (6, 3), ... } is an integer.
The infinite set { (0, 1), (1, 2), (2, 3), ... } = { {0, {0, 1}}, {1, {1, 2}}, {2, {2, 3}} } = { {{}, {{}, {{}}}}, {{{}}, {{{}}, {{}, {{}}}}, {{{}, {{}}}, {{{}, {{}}}, {{}, {{}}, {{}, {{}}}}}}, ... } is also an integer.

To make things easier, I'm going to give a special name to the integer that contains (x, 0), I'm going to call it [x]. So in other words [x] = { (x, 0), (x+1, 1), (x+2, 2), ... }.
I'm also going to give a special name to the integer that contains (0, x), I'm going to call it [-x]. So in other words [-x] = { (0, x), (1, x+1), (2, x+2), ...}

More precisely, if x is a natural number, I define [x] to be the set that contains all of the pairs of natural numbers (a, b) such that a = b + x.
And I define [-x] to be the set that contains all of the pairs of natural numbers (a, b) such that a + x = b.

OK, so how do I add two integers together?

I'll do it on a case by case basis,

I define [x] + [y] to be the set that contains all of the pairs of natural numbers (a, b) such that a = b + (x + y).
I define [-x] + [y] to be the set that contains all of the pairs of natural numbers (a, b) such that a + x = b + y.
I define [x] + [-y] to be the set that contains all of the pairs of natural numbers (a, b) such that a + y = b + x.
I define [-x] + [-y] to be the set that contains all of the pairs of natural numbers (a, b) such that a + (x + y) = b.

But there's a more practically way to compute the result.

Say you have two integers, i and j and you want to figure out i + j.
Take any element of i, say (m_1, m_2) and any element of j, say (n_1, n_2). "Add" these two pairs term wise, i.e. (m_1 + n_1, m_2 + n_2). Figure out which integer contains that pair, that's the result.

So what's [1] + [2]?
Well, by definition, it's the set that contains all of the pairs of natural numbers (a, b) such that a = b + (1 + 2) = b + 3, but that's just the definition of [3].

Another way to find this is to "add" an element of [1] to an element of [2] to see what we get.
We know that (1, 0) is in [1] and we know that (2, 0) is in [2], so we know that (1+2, 0+0) must be in [1] + [2].
But (1+2, 0+0) = (3, 0) and we know that (3, 0) is in [3]. So [1] + [2] = [3].

What's [3] + [-1]?
Well we know that (3, 0) is in [3] and we know that (0, 1) is in [-1], so we know that (3, 1) must be in [3] + [-1].
But (3, 1) is in the same integer as (2, 0) and (2, 0) is in [2]. So [3] + [-1] = [2].

Notice that [x] + [-x] is the integer that contains (x, 0) + (0, x) = (x, x). But the integer that contains (x, x) is the same that contains (0, 0), [0].

Oh fuck, [0] is the same thing as [-0]. Oh well who cares, I just gave two different names to the same integer.

Anyway, [x] + [-x] = 0 so fuck yes, I can undo any addition.

Say I want to add a number to [x] to get [y], well to find that number I just need to find [-x] + [y].
For example, say I want to add a number to [3] to get [1], well the number I need to us is [-3] + [1], which is equal to the set containing (1, 3), which is [-2]. [3] + [-2] is indeed equal to [1].

OK, I like this number system better.

Just to clarify,
1 = {{}} is a natural number
and [1] = { (1, 0), (2, 1), (3, 2), ... } = { {1, {1, 0}}, {2, {2, 1}}, {3, {3, 2}} } = { {{{}}, {{{}}, {}}}, {{{}, {{}}}, {{{}, {{}}}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{}, {{}}}}}, ... } is an integer

I can add natural numbers together and I can add integers together, but it's not the same operation. I also can't add a natural number to an integer.
I guess I can also subtract integers from one another, I'm going to define [x] - [y] to simply be [x] + [-y].

OK what else do I want to do? I guess multiplication would be nice.
Let's try to define it on the natural numbers first.

OK, last time I said that x + 0 = x, I said that 0 didn't change anything when I add it to a number.
This time I'm going to say that x * 0 = 0. When I multiply a number by 0, I get 0 back.

Last time I said that x + S(y) = S(x + y), I said that adding the number that comes after y to x is the same as adding y to x and then taking the number that comes after.
This time I'm going to say that x * S(y) = x + (x * y). When I multiply x by the number that comes after y, that's the same as adding x to x times y.

OK, so with those rules, what is 3 * 2?
Well 3 * 2 = 3 * S(1), so using my second rule I get that 3 * S(1) = 3 + (3 * 1)
3 * 1 = 3 * S(0), so using my second rule I get that 3 * S(0) = 3 + (3 * 0)
And we know that 3 * 0 = 0 because of the first rule.

So 3 * 2 = 3 + (3 * 1) = 3 + (3 + (3 * 0)) = 3 + (3 + 0) = 3 + 3 = 6

Looks good, now how do I define it on integers?
What is [x] * [y]?

Take any element of [x], say (x_1, x_2) and any element of [y], say (y_1, y_2). Consider the pair (x_1*y_1 + x_2*y_2, x_1*y_2 + x_2*y_1). Figure out which integer contains that pair, that's the result.

So what is [3] * [2]?
Well (3, 0) is in [3] and (2, 0) is in [2], so [3] * [2] is the integer that contains (3*2 + 0*0, 3*0 + 0*2) = (6, 0). But the integer that contains (6,0) is just [6].

Note that it doesn't matter which element of [3] or [2] I choose:
(5, 2) is in [3] and (10, 8) is in [2], so [3] * [2] is the integer that contains (5*10 + 2*8, 5*8 + 2*10) = (50 + 16, 40 + 20) = (66, 60). But the integer that contains (66, 60) is just [6].

What about [3] * [-2]?
Well (3, 0) is in [3] and (0, 2) is in [2], so [3] * [2] is the integer that contains (3*0 + 0*2, 3*2 + 0*0) = (0, 6). But the integer that contains (0, 6) is just [-6].

What about [-3] * [-2]?
Well (0, 3) is in [3] and (0, 2) is in [2], so [3] * [2] is the integer that contains (0*0 + 3*2, 0*2 + 3*0) = (6, 0). But the integer that contains (6, 0) is just [6].

OK, so I now have addition and it inverse (subtraction) with the integers.
I also have multiplication with the integers. Do I have the inverse of multiplication?
Fuck, I don't.

[2] times anything will never be equal to [5].

Fuck fuck fuck fuck fuck.

OK, let's make new numbers.
These sets of pairs of natural numbers worked well, let's do that again. But this time, we'll do sets of pairs of integers.

If i is an integer, I define [i] to be the set that contains all of the pairs of integers (a, b) such that a = b * i.
And I define [÷i] to be the set that contains all of the pairs of natural numbers (a, b) such that a * i = b.

Oh, and I'll call these sets rational numbers, because ratios, not because they make sense.

So what's the rational number [[1]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a = b * [1].
Since b * [1] = b, this means that as long as a = b, we have that (a, b) is an element of [[1]].
So [[1]] = { ..., ([-2], [-2]), ([-1], [-1]), ([0], [0]), ([1], [1]), ([2], [2]), ... }

So the rational number [[1]] is just an infinite set of pairs of integers.
But a pair is just a set and integers are also just a infinite set of pairs of natural numbers.
So the rational number [[1]] is just an infinite set of sets of infinite sets of pairs of natural numbers.
But a pair is just a set and the natural numbers are also just a set.
So the rational number [[1]] is just an infinite set of sets of infinite sets of sets of sets.

OK, what's the rational number [÷[1]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a * [1] = b.
Since b * [1] = b, this means that as long as a = b, we have that (a, b) is an element of [÷[1]].
So [÷[1]] = { ..., ([-2], [-2]), ([-1], [-1]), ([0], [0]), ([1], [1]), ([2], [2]), ... }
Oh, this is the same thing as [[1]]. Ugh, who cares, I just gave the same set two different names.

So what's the rational number [[2]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a = b * [2].
So [[2]] = { ..., ([-4], [-2]), ([-2], [-1]), ([0], [0]), ([2], [1]), ([4], [2]), ... }

So what's the rational number [÷[2]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a * [2] = b.
So [÷[2]] = { ..., ([-2], [-4]), ([-1], [-2]), ([0], [0]), ([1], [2]), ([2], [4]), ... }

So what's the rational number [[0]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a = b * [0].
So [[0]] = { ..., ([0], [-2]), ([0], [-1]), ([0], [0]), ([0], [1]), ([0], [2]), ... }

So what's the rational number [÷[0]]?
Well, it's defined as the set that contains all of the pairs of integers (a, b) such that a * [0] = b.
So [÷[0]] = { ..., ([-2], [0]), ([-1], [0]), ([0], [0]), ([1], [0]), ([2], [0]), ... }

OK, so how do I multiply two rational numbers together now?
Again, on a case by case basis I define (assuming i and j are integers):
[i] * [j] = the set that contains all of the pairs of integers (a, b) such that a = b * (i * j)
[÷i] * [j] = the set that contains all of the pairs of integers (a, b) such that a * i = b * j
[i] * [÷j] = the set that contains all of the pairs of integers (a, b) such that a * j = b * i
[÷i] * [÷j] = the set that contains all of the pairs of integers (a, b) such that a * (i * j) = b

OK, so what's [[2]] * [[3]]?
The set that contains all of the pairs of integers (a, b) such that a = b * ([2] * [3]) = b * [6].
But that set is just [[6]].
So [[2]] * [[3]] = [[6]] as expected.

Now what about [[6]] * [÷[3]]?
It's just the set that contains all of the pairs of integers (a, b) such that a * [3] = b * [6].
Oh, do we know that that is? What does it look like...
[[6]] * [÷[3]] = { ..., ([-4], [-2]), ([-2], [-1]), ([0], [0]), ([2], [1]), ([4], [2]), ...}
Oh, that's just [[2]]! Fuck yeah, I can now undo multiplication.

But now, what about [[2]] * [÷[3]]?
It's just the set that contains all of the pairs of integers (a, b) such that a * [3] = b * [2].
Oh, do we know that that is? What does it look like...
[[2]] * [÷[3]] = { ..., ([-4], [-6]), ([-2], [-3]), ([0], [0]), ([2], [3]), ([4], [6]), ...}
Hmm, that's not something I've seen before... It's technically not a rational number because it's not equal to [i] or [÷i] where i is an integer.
But I want it to be a rational number, I'd like it if I'd only get rational numbers whenever I multiply two rational numbers.

OK scrap my previous definition, let me define the rational numbers as follows.

If i and j are integer, the rational number [(i, j)] is the set that contains all of the pairs of integers (a, b) such that a * j = b * i.
Note that I gave multiple names to the same rational number, note that
[([2], [3])] = { ..., ([-4], [-6]), ([-2], [-3]), ([0], [0]), ([2], [3]), ([4], [6]), ...} = [([4], [6])]
But who cares.

This is a bit long to write, so I'm going to introduce a new symbol "/". Instead of writing ([2], [3]), I'm just going to write 2/3.

So, to reiterate what my notation means, [2/3] is the rational number [([2], [3])]. It is the set that contain the pairs 2/3, 4/6, 6/9, and so on.
With this notation, 2/3 is just a pair of integers, it's not a rational number. [2/3] is the set I consider to be a rational number.
So 2/3 is not the same as 4/6 since those a two different pairs of number, but [2/3] is the same rational number as [4/6] since they are both the same set.

OK, let's redefine multiplication again.
Assuming i, j, k, and l are integers, I define [i/j] * [k/l] to be [(i*k)/(j*l)].

OK, so what's [2/1] * [3/1]? [6/1] as expected.

Now what about [6/1] * [1/3]?
Well it's [6/3], so it's the set that contains all of the pairs of integers (a, b) such that a * [3] = b * [6].
But that's the same as the set that contains all of the pairs of integers (a, b) such that a * [1] = b * [2], i.e. [2/1].
So [6/1] * [1/3] = [6/3] = [2/1]

Now what about [2/1] * [1/3]?
Well it's [2/3], so it's the set that contains all of the pairs of integers (a, b) such that a * [3] = b * [2].
Which is the same result we had when we tried this multiplication with the old definition.

Now what about [1/2] * [0/1]?
Well it's [0/2], so it's the set that contains all of the pairs of integers (a, b) such that a * [2] = b * [0].
Note that this is the same as [0/1] = { ..., 0/-2, 0/-1, 0/0, 0/1, 0/2, ... }.

Now what about [2/1] * [1/0]?
Well it's [2/0], so it's the set that contains all of the pairs of integers (a, b) such that a * [0] = b * [2].
Note that this is the same as [1/0] = { ..., -2/0, -1/0, 0/0, 1/0, 2/0, ... }.

Now what about [0/1] * [1/0]?
Well it's [0/0], so it's the set that contains all of the pairs of integers (a, b) such that a * [0] = b * [0].
Oh, so it doesn't matter what a and b are, both sides of the equality is [0], so the equality is always true!
So [0/0] is actually the set of all pairs of integers.
Who said we couldn't divide by 0?


OK, so multiplication works with the rationals and I can undo it. But we kinda designed the rationals so that multiplication would work, so that was easy. But can we still make addition work?

I want the rational number [n/1] to play the same role as the integer [n].
So I want [m/1] + [n/1] = [(m+n)/1]

...

Long story short, I want [a/b] + [c/d] = [(ad + bc)/bd].

So what is [1/2] + [1/3]? It's [(3+2)/6] = [5/6].
What is [1/2] + [0/3]? It's [(3+0)/6] = [3/6] = [1/2] as expected.
What is [1/2] + [3/0]? It's [(0+6)/0] = [6/0] = [1/0].

So anything plus [0/1] doesn't change, which is what I want because I want [0/1] to behave like [0] and like 0.
But anything plus [1/0] becomes [1/0]. That's weird, if r and s are different rational numbers, I don't want that r + [1/0] = s + [1/0]. If you change one of the numbers, the sum should change as well.

OK, fuck [1/0], I don't like it anymore. Fuck [0/0] also, that one is even weirder. I don't consider these rational numbers anymore. I guess that's why people don't want to divide by 0...

Let me define division now: [a/b] ÷ [c/d] = [a/b] * [d/c] = [ad/bc].
This is not a problem as long as neither b nor c is 0.
b is not 0 since I assumed [a/b] to be a rational number and I've explicitly excluded [0/0] and [1/0] (which is the same as [2/0], [3/0], and so on...) from the rational numbers, but c could be 0 still. So I only define division when c is not 0. If c is 0, then fuck you, I don't care.

In any case, I now have the rational numbers and I can add, subtract, multiply, and even divide them (as long as I don't divide by [0/1]).

So let's do a quick review.

The only things that exist in math are sets. In particular, the set with no elements is a thing, I call that set the empty set or just 0.

The natural number 1 is just the set containing 0. In other words 1 = {0} = {{}}}.
The integer [1] is just the set containing all the infinitely many pairs of natural numbers (a, b) such that a = b + 1.
In other words, [1] = { (1, 0), (2, 1), (3, 2), ... } = { {1, {1, 0}}, {2, {2, 1}}, {3, {3, 2}}, ... }
= { {{{}}, {{{}}, {}}}, {{{}, {{}}}, {{{}, {{}}}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{}, {{}}}}}, ... }

Isn't it obvious that when I say [1], I'm actually talking about the infinite set { {{{}}, {{{}}, {}}}, {{{}, {{}}}, {{{}, {{}}}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{{}, {{}}, {{}, {{}}}}, {{}, {{}}}}}, ... }?

The rational number [1/1] is just the set containing all the infinitly many pairs of integers (a, b) such that a * [1] = b * [1].
In other words, [1/1] = { ..., ([-2], [-2]), ([-1], [-1]), ([0], [0]), ([1], [1]), ([2], [2]), ... }.
But the pair (x, x) is just the set {x, {x, x}} = {x, {x}}, so we can simplify to { ..., {[-2], {[-2]}}, {[-1], {[-1]}}, {[0], {[0]}}, {[1], {[1]}}, {[2], {[2]}}, ... }.

Isn't that what naturally comes to mind when you think about the rational number [1/1]?

OK, so we have
natural numbers: I can add and multiply them
integers: I can add, multiply, but also subtract them
rationals: I can add, multiply, subtract, and divide them (but I can't divide by [0/1] otherwise I would get [1/0] which I don't like for some reason).

OK, so succession is the first operation I defined, it's just taking the number that comes after.
Addition is just repeated succession and multiplication is just repeated addition.
Let's have some fun and create a new operation on the natural numbers that's just repeated multiplication, let's call this exponentiation.

In other words:
If m and n are natural numbers, I want m ^ 0 = 1
And I want m ^ S(n) = m * (m ^ n)

So what's 4^3?
Well, 4^3 = 4^S(2) = 4 * 4^2 = 4 * 4^S(1) = 4 * (4 * 4^1) = 4 * (4 * 4^S(0)) = 4 * (4 * (4 * 4^0)) = 4 * (4 * (4 * 1)) = 4 * (4 * 4) = 4 * 16 = 64

Side comment: note that this means that 0^0 = 1.
Side comment 2: Also note how 2^3 is not equal to 3^2 even though it was defined similarly to how multiplication was defined. The only difference is that exponentiation used multiplication as the repeated operation and multiplication used addition as the repeated operation. How come is it true that a * b = b * a for any natural numbers?

Cool, can we define it for the integers? Not really, what would [2]^[-1] be? It could work if I allow the answer to be a rational number, but I don't want to do that, so let's skip the integers and just try defining exponent ion for rational numbers.

Like I said, I want the rational number [n/1] to play the same role as the natural number n.
So [a/b]^[0/1] = [1/1] regardless of a and b.
In general, I want:
[a/b]^[n/1] = [a/b] * [a/b] * ... * [a/b] (n times)
But what if the exponent does not have a 1 at the denominator?
What should [a/b]^[1/2] be?
Notice that when we're restricting ourselve to natural numbers, we have that a^b * a^c = a^(b+c). I think that's nice, so I think we should also have that for the rational numbers.
So I don't know what [a/b]^[1/2] should be, but I want [a/b]^[1/2] * [a/b]^[1/2] = [a/b]^([1/2]+[1/2]) = [a/b]^[2/2] = [a/b]^[1/1] = [a/b].
So if a = 9 and b = 4, I want [9/4]^[1/2] * [9/4]^[1/2] = [9/4].
So I want [9/4]^[1/2] to be a rational number such that if you multiply it with itself, you get [9/4].
But there's two rational numbers with that property, [3/2] and [-3/2].
Fuck it, just take the positive one. So [9/4]^[1/2] = [3/2]

OK what about [2/1]^[1/2]? What rational number times itself is equal to [2/1].
Hmm [14/10] is a bit too small, but [15/10] is a bit too big.
Hmm [141/100] is a bit too small, but [142/100] is a bit too big.
Hmm [1414/1000] is a bit too small, but [1415/1000] is a bit too big.
...

Fuck this.

Turns out there's no rational number that you can multiply with itself to get [2/1].
FUCK, and [2/1] isn't even that complicated of a rational number.

Do I have to do the same thing again where I define a new class of numbers as an infinite set containing pairs of rational numbers?
I really don't want to do this every single time I introduce a new operation. FUCK. This sucks. I'm not doing that.

Hmm, there isn't a rational number that I can multiply to itself to get [2/1], but for each rational number, I know whether that number squared is smaller or larger than [2/1].

So let's divide the rational numbers into two sets.
Let A be the set of all rational numbers that when squared are less than [2/1]
Let B be the set of all rational numbers that when squared are more than [2/1]

Consider the pair (A, B), well, I'm going to call that pair a real number.

Let me generalize the concept,

If A and B are sets of rational numbers such that:
1. Every rational number in A is smaller than every rational number in B
2. Neither A nor B are empty
3. Every rational number is either in A or B
4. If x is a rational number in A, then there exists another rational number in A that's bigger than x. (we don't care about what happens in B)

Then I consider the pair (A, B) a real number.

If r is a rational number, I'll write _r_ to mean the real number (A, B) where A is the set of all rational numbers less than r and B is the set of all rational number equal to or greater than r.

This is a real number because the pair (A, B) satisfies properties 1, 2, 3, and 4.

But as we saw, those aren't the only real numbers. We saw that ({r \in Q: r^[2/1] < [2/1]}, {r \in Q: r^[2/1] >= [2/1]}) is also a real number.

OK, this is what a real number is. I answered your question.
How do we add, multiply, subtract, divide, exponentiate, take the roots of, take the logs of, etc? I'll let you figure that out.

So to recapitulate,

A natural number is just a set of consecutive natural numbers. For example, 1 is just the set {{}}.
An integer is just an infinite set of pairs of natural numbers. For example, [1] is just the set { (1, 0), (2, 1), (3, 2), ... }
A rational number is just an infinite set of pairs of integers. For example, [1/1] is just the set { ..., ([-1], [-1]), ([0], [0]), ([1], [1]), ... }
A real number is just a pair of sets of rational numbers. For example, _[1/1]_ is just the pair ({x \in Q: x < [1/1]}, {x \in Q: x >= [1/1]}).

So what is the real number _[1/1]_?
Well it's a pair of infinite sets of rational numbers.
Which is a pair of infinite sets of infinite sets of pairs of integers.
Which is a pair of infinite sets of infinite sets of pairs of infinite sets of pairs of natural numbers.
Which is a pair of infinite sets of infinite sets of pairs of infinite sets of pairs of sets that I like.

A real number is a pair of infinite sets of infinite sets of pairs of infinite sets of pairs of sets that I like.

But since pairs are just sets, we can also say that

A real number is a set of infinite sets of infinite sets of sets of infinite sets of sets of sets that I like.

This is what standard modern mathematics is based on.