#497 Four touching spheres form a tetrahedron – Thank you for the inspiration, Nanodots! – A new minimal geometric composition each day

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### Putting the sets together

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 ZN ∪ 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!

∀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.

∀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 ∈ (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.

There are still caveats. There will always be caveats. One example is that whenever we make a set with rationals, for example, we’ll need to get not only the pure rationals but also the integers and naturals that are equivalent to them under {=}. The relations [∈] and {∈} aren’t really the same thing as ∈, but we can use them in a mostly intuitive way and you can also easily see that a ∈ b → a [∈] b and also that a ∈ b → a {∈} b (because both [=] and {=} are equivalence relations) so if the set we’re building includes all “versions” of the same number under [=] or {=} then we’re good.

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I think by now you guys can tell that I am fond of “colorful mathematics”.

(Source: quantumaniac)

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Created by a team of designers, ‘Pasta by Design’ book reveals the hidden mathematical beauty of pasta: its geometrical shapes and surfaces are explained by mathematical formulae, drawings and illustrations

(Source: imathematicus)

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Mandelbrot island - Mandelbrot set rendered as an island with Terragen, a fractal-based landscape generator.

Fathom the Universe

Source:http://commons.wikimedia.org/wiki/File:Mandelbrot_island.jpg

(via visualizingmath)

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3D Laser Cut Paper - Geometric Art by Eric Standley

Eric Standley is an artist and educator currently living and working in Virginia. In his incredible series of 3D laser cut paper art, Standley’s work is found at the intersection of art, technology, history and mathematics.

His vector drawings were initially inspired by the geometry in Gothic and Islamic architectural ornamentation. The pieces are painstakingly assembled from laser-cut paper, layered to create elaborate 3-D works of art. Often these works are created using well over 100 layers of paper and can take months of planning and drawing. The result is so intricately detailed that the pieces must be viewed from multiple perspectives to be fully appreciated.

Standley uses an array of colors woven together with mathematical precision to create his art, combining 12th century architectural aesthetics with contemporary technology. In designing his pieces, Standley envisions three to seven layers of paper at one time, picturing how they will build upon one another. Source.

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The Prime Pages (prime number research, records and resources)

A very, very interesting website dedicated to primes!

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(Source: facuontivero, via mathmajik)

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Probability. Probability is the study of events that cannot be determined completely, involving some degree of randomness.

Sorry for the shortage of posts. I finally found a question on StackExchange that I can answer and wasn’t already answered! I thought I’d share it with you :)

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