Math is beautiful

If you’ve read my posts on the integers and the rationals you’ll see that the naturals aren’t really a subset of the integers, which are not really a subset of the rationals. We can identify certain natural numbers with certain integers, sure, but it’s definitely not true that natural 0, {}, equals integer 0, {{{0}}, {{1}}, {{2}}, {{3}}, …}.

Unless…

The integers

We could create a new set. This will be the actual set of the integers, Z. This new set will include both the original set N of the natural numbers and the set of the equivalence classes of the integers - let’s call it B. Then Z = N ∪ B.

In N we have an equality = which is an equivalence relation between two natural numbers and goes something like this:

∀a, b (a = b ↔ [∀c (c ∈ a ↔ c ∈ b)])

Two natural numbers a and b are equal if and only if all members of a are also members of b and vice-versa. That’s just the Axiom of Extensionality, really, and it’s valid for any sets. Since the construction of all the naturals is constant and so is that of the integers, that equality is good enough for them. But not for us. So now we’ll define a new equivalence relation in Z: [=] (I put it in italics to highlight that this is a special relation whose strict symbol is [=]). It goes like this:

∀a, b (a [=] b ↔ (a, b ∈ Z) ∧ [(a = b) ∨ (a ∈ N ∧ b ∈ B ∧ ∀c ∈ N [(c + a, c) ∈ b]) ∨ (b ∈ N ∧ a ∈ B ∧ ∀c ∈ N [(c + b, c) ∈ a])])

Under this equivalence relation, a natural number x equals an Integer y if and only if all members of that integer are ordered pairs of the form (k + x, k); a natural number [=]s another if they’re equal; same goes for two integers.

So, you see, with this set and this equivalence relation we can properly say that N is a proper subset of Z!

Now we need to also define our new operations. The easiest way to define them is to make all results be integers instead of natural numbers - that way we won’t get any confusions!

Addition

∀a, b, c ∈ Z (a [+] b [=] c ↔ ∀x, y ∈ B (x [=] a ∧ y [=] b ∧ x + y = c))

In this definition, [+] is the new addition we’re defining on Z, [=] is the equivalence relation we explained above, + is integer addition on B and = is integer equality on B.

Subtraction

∀a, b, c ∈ Z (a [-] b [=] c ↔ ∀x, y ∈ B (x [=] a ∧ y [=] b ∧ x - y = c))

Once again we’re defining [-] using integer - and = on B (remember that integer subtraction is defined as addition to the inverse of the second term.)

Multiplication

∀a, b, c ∈ Z (a [*] b [=] c ↔ ∀x, y ∈ B (x [=] a ∧ y [=] b ∧ x * y = c))

Once again, pretty easy, with integer multiplication.

Division

Aaahh… now things get complicated. However, we can just use this and replace the -s with [-]s and the +s with [+]s and the =s with [=]s et voilá.

Inequality

∀a, b ∈ Z (a [≤] b ↔ ∀x, y ∈ B (x [=] a ∧ y [=] b ∧ x ≤ y))

In this case, ≤ is the inequality defined in B and [≤] is the one we want to define in Z. We can define the others based on this one. For every a and b in Z:

a [<] b ↔ a [≤] b ∧ ¬(a [=] b)

a [≥] b ↔ b [≤] a

a [>] b ↔ b [<] a

And we’re done with the inequalities.

Set membership

We can even invent a new kind of set membership, [∈]! It goes like this:

∀s (a [∈] s ↔ ∃b ∈ s (a [=] b))

That is, if we have any integer or natural number b that’s a member of some set s then all numbers that are equivalent to b under [=] possess the [∈] relation with s.

So whenever we talk about the set Z of integers, that’s the one we’re talking about, and the operations we just defined are the canonical operations we’d expect. So we can indeed add an integer and a natural number and we’ll get an integer. And we can say that the set of natural numbers is a proper subset of the integers. We really have managed to make this set behave exactly as we were taught the set of integers should behave.

The rationals

Now we do the same thing for rationals. Of course things get more complicated, but not too much so. Let’s call the set of the equivalence classes of the rationals… I dunno, M perhaps. And the set we’re trying to build is Q = B ∪ M. Anyway, whenever we do not mark the operations below, we mean the operations on the rationals.

Equality

Rational equality is {=} and it goes like this:

∀a, b (a {=} b ↔ (a, b ∈ Q) ∧ [(a = b) ∨ (a [=] b) ∨ ∃c, x ∈ B (((c [=] a) ∧ (b ∈ M) ∧ ((c*x, x) ∈ b)) ∨ ((c [=] b) ∧ (a ∈ M) ∧ ((c*x, x) ∈ a)))])

This complicated definition will say that a {=} b if at least one of the following is true:

a = b under the general definition of equality in set theory;

a [=] b, that is, they’re both integers and are equal;

there is some pure integer c such that either a [=] c, b is a rational and is the representation of a or b [=] c, a is a rational and is the representation of b.

These three cases cover everything: if a and b are both naturals, integers or rationals, the equality holds; if a and b are members of Z that would be equivalent under [=], the equality holds; if either a or b is a member of Z and either b or a is a member of M (the set of the equivalence classes of rationals) then we find a member c of B (which is a proper integer) that’s equivalent to our member of Z and check whether the other one is also the rational representation of the former.

Addition

∀a, b, c ∈ Q (a {+} b {=} c ↔ ∀x, y ∈ M (x {=} a ∧ y {=} b ∧ x + y = c))

In this case, {+} is the new operation of addition we want to define and + is the operation as it is on the rationals.

Subtraction

∀a, b, c ∈ Q (a {-} b {=} c ↔ ∀x, y ∈ M (x {=} a ∧ y {=} b ∧ x - y = c))

Multiplication

∀a, b, c ∈ Q (a {*} b {=} c ↔ ∀x, y ∈ M (x {=} a ∧ y {=} b ∧ x * y = c))

Division

And now we got to the meat of rationals. Since we can this time get the results of all divisions without bothering with silly remainders and whatnot, we can have a proper definition of rational division:

∀a, b, c ∈ Q (a {/} b {=} c ↔ ∀x, y ∈ M (x {=} a ∧ y {=} b ∧ x / y = c))

We can divide any rationals we want, and now we found an operation that works even when those rationals are also naturals or integers.

Inequality

∀a, b ∈ Q (a {≤} b ↔ ∀x, y ∈ M (x {=} a ∧ y {=} b ∧ x ≤ y))

But wait! We haven’t defined the ≤ operation on the equivalence classes of the rationals yet! How does it work?

It’s a bit more complicated, too. It goes like this:

∀a, b ∈ M (a ≤ b ↔ ∀w, x, y, z ∈ B (x > 0 ∧ z > 0 ∧ [(w, x)] ∈ a ∧ [(y, z)] ∈ b → wz ≤ xy))

That is, we find a representation of the rationals a and b as their equivalence class such that the second term of the ordered pair is positive and the inequality on the integers wz ≤ xy follows.

With this we can define <, ≥, >, {<}, {≥} and {>} in the usual way.

You should pay attention to what kinds of mathematical objects we’re dealing with. Since w, x, y and z are proper integers, their comparison is the integer one; since a and b are proper rationals, their comparison is the one we want to define.

Set membership

∀s (a {∈} s ↔ ∃b ∈ s (a {=} b))

It’s the same thing as before: a new kind of set membership that automatically “includes" in all sets that have some rational q all sets equivalent to it under {=}.

So what?

Now that we’ve built those sets, all the properties we’d expect them to possess are there. With this formalisation we don’t need to worry about the set of natural numbers not really being a subset of the integers and so on.

When we talk about equality, inequality or the operations on those sets, we’ll be talking about the ones defined above. If I say n ≤ z where n is a natural number and z is an integer, I’ll just be talking about the relation [≤] which is perfectly analysable and perfectly convenient.

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Trigonometry!

Multiplying both sides of an equality by a negative changes its direction

That is, if you have a ≤ b and multiply both sides by some c < 0 you’ll get ac ≥ bc. Let’s prove that.

Under their equivalence class formulation, standard formulation and intuitive formulation we get the same properties of multiplication. From this point on, whenever I talk about the set of integers, I mean the set Z with the [] relations as defined in the last of those three posts. If you haven’t read it, it’s a formulation which I invented that has all the usual relations and operations working as they intuitively should, with the very cool fact that, if N is the set of natural numbers, then N ⊆ Z.

So everything works exactly as it should. That means that:

∀a, b ∈ Z (a ≤ b ↔ ∃c ∈ N (a + c = b))

(You have no idea how happy it makes to finally be able to talk about N and Z in the same sentence without having to add a million excuses.)

Now suppose we take some number d < 0 and multiply both sides of the inequality by it. First things first:

∀d ∈ Z (d < 0 ↔ ∃k ∈ N* (d = -1*k))

(Yay! I can now multiply negative integers by natural numbers! That is so cool!)

So let’s see. Since a ≤ b means a + c = b then:

a + c = b [Premise]

d*(a + c) = d*b [Multiplying both sides of an equality by an integer maintains the equality]

d*a + d*c = d*b [Distributive property]

d*a - k*c = d*b [Replacing d with -k]

d*a = d*b + k*c [Adding k*c to both sides]

Now, let’s call d*a by a’, d*b by b’, and k*c by c’. Since both k and c are members of N it follows that so is c’. That is:

a’ = b’ + c’ ↔ a’ ≥ b’ ↔ d*a ≥ d*b

So if we multiply both sides of any inequality by a negative integer we get that the inequality is inverted.

A similar argument can be made based on the operations on the rationals.

Density

The rational numbers have a property called density. That means that between any two rationals there is one other rational, and therefore infinitely other rationals. More specifically, under ≤ and the other inequalities on Q there will always be a distinct rational number between two different rationals. Let’s just show that. If p = a/b and q = m/n and p < q then:

Let’s prove that. First:

Multiplying the right and the left ends by 2:

Subtracting a/b from both sides:

Which is what we started with. A similar argument can be made to show that that number is also less than m/n and therefore our triple inequality is true. And since there is one rational number between any two rational numbers it follows that there’s an infinity of them. In fact, a countable infinity.

The cardinality of the rationals

I have previously stated that the set of the rational numbers has the same cardinality as that of the integers and natural numbers. But I haven’t shown this. Now, the standard way of showing it is graphical and I don’t like it much. But I found a much cooler way of showing that. It goes like this.

I have previously talked about the base-b expansion of an integer. I said that every integer can be written uniquely as the sum of ki*b^n where b is the base, ki < b is an integer and n is an exponent. The number 14 in base-2 expansion (binary) is 1110 and in base-3 expansion is 112 and in base-16 (hexadecimal) expansion is E (because the symbols for hexadecimal expansion are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F). Now hold that thought for a minute.

Every rational number x = a/b can be written in the form of an equation bx = a. Not only that, but every rational number can be written in the form of an infinity of equations of that form. I’ll show that we can pair each such equation with a distinct integer.

First, let’s invent a base-13 alphabet for integer representation: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, x, = and -. In that case, the rational 5/4 is represented by the equation 4x = 5 which is paired with the number:

(That’s because the number 10 in base-13 representation equals x and the number 11 equals =.)

This way of representing formulae was discovered by Kurt Gödel and is used in his famous Incompleteness Theorem (of which I’ll speak in the future). Using that, not only can we link every rational to an integer, there are multiple integers linked to the same rational. For instance, 5/4 = 10/8 which can be represented by the equation 8x = 10:

We can see that there are multiple integers tied to the same rational. In order to avoid that kind of thing, however, let’s make a restriction in our representation: if bx = a is a rational’s representation then gcd(a, b) = 1 and b > 0.

So we created an injective function from the rationals to the integers. For instance, f(5/4) = 10,626, and we can find that value for every rational number. That means that |Q| ≤ |Z|. Now, you can intuitively see that there’s no way the set of the rationals has cardinality less than the set of integers. Indeed, because Z ⊆ Q we know that |Z| ≤ |Q|. That is, since Z is a subset of Q we have that it’s cardinality is less than or equal to Q's. Therefore, |Q| = |Z|, the two sets have exactly the same cardinality. Q.E.D.

Well-order of the rationals

According to the well-ordering axiom, every set has a well-order. If you don’t want to click the link, basically a well-order of a set A is a relation R on A such that all nonempty subsets S of A have a unique member s ∈ S with the property that, for all members a ∈ S, s R a. On the natural numbers, the typical example is the ≤ relation: every nonempty subset of the natural numbers has a least member under ≤.

Now, clearly ≤ is not a very good well-order of the rationals because they are dense. So suppose I take the open interval ]0, 1[ of the rationals. Then no matter what rational p I take in that interval, there will always be another one between 0 and p, so ≤ doesn’t well-order the rationals.

Can anything? Is it possible to find a well-order of the rationals that can be described? Turns out it is.

We just proved that the rationals have the same cardinality as the integers which means they have the same cardinality as the natural numbers. So if we find a bijection f(n) between the rationals and the naturals we can say that f(n0) W f(n1) ↔ n0 ≤ n1 where W is the well-order we’re looking for in the rationals.

The above is the most famous example of a bijection between naturals and rationals. It’s even on Wikipedia n stuff. Basically we order p/q such that p1/q1 W p2/q2 ↔ (p1 + q1 < p2 + q2) ∨ (p1 + q1 = p2 + q2 ∧ p1 ≤ p2). There are two caveats: we only take the canonical representation of each rational number (the canonical representation of 0 being 0/1); and the < and ≤ relations described above aren’t the regular integer ones but rather the ones of the well-order of the integers.

Now, it may not be very pleasant because it’s not very formulaic, but if you want a good formula, you can find a bijection h between Z and Q such as we defined above and then use the usual well-order of Z to find the least member of a given set in Q.

Of course, this is much less satisfying than the obvious well-order of the integers and natural numbers. We can’t really see much here. But that’s the price we pay for having a dense set.

These remarks make some interesting observations in reference to a previous post. The intuition here is correct, and the motion displayed in the 2D case can be thought of as a special case of a certain 3D motion when viewed from a particular perspective. Let’s experiment with this idea here.

Its important to note that two properties are at play that make this phenomenon work the way it does. This first of these, as mentioned in the response, is that the white circles do not move back and forth along their respective lines at the same speed. Each white circle actually slows down as it approaches the edge of the black circle, but speeds up as it passes through the center. More specifically, how the velocity of the white circle changes with time can be described as a sine wave. Compare the motion of the white and blue circles in the animation below. The white circle moves back and forth while maintaining the same speed throughout, whereas the blue circle’s motion is sinusoidal.

The other property that the circles must exhibit in order to create the outline of a rolling circle is to reach the edge of the black circle in their respective motions at just the right time. This is accomplished by  spacing them out just right. Consider the case where each of the circles all start at the outer edge at the same time. The resulting motion would just look like this:

Here, the circles velocities do not very sinusoidally and each moves with constant speed. By having each circle reach the outer edge at different times, the following motion results:

Notice how in this case the white circles do not form the outline of a perfect circle. Now, by keeping the circles spaced out just right, and also having the speeds vary sinusoidally we arrive at the desired effect that we saw in the original post:

Let’s  try to generalize this idea into three dimensions. Instead of circles we’ll use spheres, and instead of circles moving back and forth along a straight line we’ll have the spheres move in a circular orbit. In the animation below, we see a sphere that is orbiting along the outer shell of a larger sphere in a circular path along the outer shell of a larger sphere. Note that the sphere is moving with constant speed. The plane in which the sphere is orbiting is displayed along with a bounding box for a better sense of depth in 3D.

Now we add some more orbiting spheres that also move in circular paths, where each of the paths lies in evenly spaced planes that are all perpendicular to a plane (not shown) that would be parallel to the front-most face of the bounding box.

In the above animation there are 8 spheres that are orbiting, and for each of the 4 planes there is a pair of spheres that orbit within the plane. Let’s now view this same scene looking head-on through the front face of the bounding box:

From this perspective the planes now look like lines since we’re viewing them along their edges. Here, we are taking into account the proper perspective in 3D, and you can tell by noticing that the sizes of the spheres change as they move closer or further away. We can ignore the perspective, and view the scene as projected into 2D  via an orthographic projection yielding the following animation:

It can now be seen that this effect results in exactly what we observed in the 2D case! Something interesting happened though. We  did start with 8 spheres originally but in this view half of them get obscured by the other half since one is in front of the other. If we really wanted to see 8 spheres moving around, we would actually need to work with 16 spheres in total. The following animations use 16 spheres in the same three perspectives we just experimented with, but this time the planes have been removed:

The relationship here between the linear sinusoidal motion in the 2D case and the circular orbit with constant speed in the 3D case can be seen as a generalization of an analogous relationship between the sin function and the unit circle. We wont invest in the details here, but just take a look at this animation for an idea (taken from Wikipedia):