As adults, people many people feel discouraged for their lack of skill in Mathematics. This is unfortunate as Mathematics is there to empower people regardless of their skills and asks only two things in return Guessing and consistency. Math is the great teacher and is there to give people the opportunity to discover things by themselves.

Math is not mainly about proofs. Proofs are the finished product, the cake, Math is about linguistic consistency. Learning to judge cakes is not the same as baking them. Just like with any other language the younger you are, the easier it's for you to pick it up and the more relative time you can devote to master it. If not introduced at a young age there are still ways to do beautiful and useful math even to an advanced degree. For this, you need to be able to see a clear path to achieve through feedback.

How to guess? by playing

The idea of fairness is something important in a game, for this you need rules. The simpler and more consistent the rules the more approachable and reliable the game (A lot of us have experienced rules that change in board games that depend on who are you playing against like with UNO, right? it can be frustrating)

Wild guessing is no different from being lost, thus having a grasp of at least something to base your guess on is called an educated guess, and you must try to improve it for the next guess.

I will give you 7 steps.

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Let's start with a plane. This plane is a surface that is ideally flat.

(I'm talking about true level here )

This plane is so big that is indistinguishable from infinite. This plane is also super slim and rigid. You can't change it or bend it, is so slim you can't be bothered to measure it, but thick enough you can observe where it is.

If you can reference your position relative to that plane, you have cut space. (e.g.Up from this plane, down from this plane)
Every time you have a new plane you are cutting space into (divisions, fractions, parts)

Here's the game: y̲o̲u̲ ̲c̲u̲t̲ ̲t̲h̲e̲ ̲s̲p̲a̲c̲e̲ ̲w̲i̲t̲h̲ ̲5 ̲p̲l̲a̲n̲e̲s̲. ̲l̲i̲k̲e̲ ̲i̲s̲ ̲a̲ ̲c̲a̲k̲e̲.

• How many parts?
  One person can start guessing. Guessing is not difficult, but the real problem is hard. Mainly for a reason, most problems are ill-posed. One might think it understands the problem but most of the time nature gives even the question incomplete. (Step 1: 𝔦𝔰 𝔱𝔥𝔢 𝔮𝔲𝔢𝔰𝔱𝔦𝔬𝔫 𝔠𝔬𝔪𝔭𝔩𝔢𝔱𝔢?)
  
  Why 5 planes? why not 4 or 6? This is where simplifying the actual question comes into place, the reason we try to use an easier more simple version of the question.
  The simplest model to work with must be 1 plane, so let's use it.
  The above is a picture of the horizon. Is not the actual horizon, just a representation of that space. The picture is a meta-plane itself, at this point of view you are the water-air interface. The union is the plane. Now, let's keep track of this.
  

Planes	Parts
1	2

Now with two planes, (do they intersect?) If they don't, you get 3 parts. One of the parts is shared by both divisions. that's not difficult, is even easy to see that 5 non-intersecting divisions will give us 6 parts like in a musical pentagram. but what if they do intersect? it should look something like this.

Planes	Parts
2	4

With three, one immediately notices that avoiding all intersections is impossible as each plane is almost infinitely big. There was only 1 scenario where intersections where avoided, all parallels. (𝔱𝔥𝔢 𝔮𝔲𝔢𝔰𝔱𝔦𝔬𝔫 𝔴𝔞𝔰 𝔦𝔫𝔠𝔬𝔪𝔭𝔩𝔢𝔱𝔢𝔩𝔶 𝔰𝔱𝔞𝔱𝔢𝔡) If they are not parallel and in my sight, they will intersect for this problem.

• Is this an exception or a rule?

You could try to pass them all through the same point, in this case with the 5 planes you get 10 parts, like a wheel. we can see there's a minimum number of intersections and a maximum number of intersections and that's where the answer will lie. An important observation is that since the number of positions a plane can ocupate can go from 0 to infinity, even the slightest shift in orientation changes the parallel state. This indicates that the cases where there's a limited number of intersections are the exception and not the norm. (Step 2: 𝔤𝔬𝔦𝔫𝔤 𝔱𝔬 𝔱𝔥𝔢 𝔢𝔵𝔱𝔯𝔢𝔪𝔢 𝔠𝔞𝔰𝔢𝔰?)

I didn't expose the really simplest case, the extreme case, 0 planes.

Planes	Parts
0	1

In the case of 3 planes, with 1 intersection you get 6 parts, with two points of the intersection you get
the maximum number of intersections, in this case, two you get.

Planes	Parts
3	8

In the next case. At this point, we have been probably practicing the next step, so let's try 4 planes, how many parts?

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At this point, people can make educated guesses. Even if they do not provide the maximum amount of information, they have a basis and that basis is observation. When you observe, you can find a pattern and if you are brave enough you say it generalizes to all the cases.

Some might say the next number of parts is 16. That's is an educated guess, but it needs something else, that is the point of consistency in the language of math, where the natives recognize those that are not versed in the langue, this part of the grammar is doubt in the form of ｔｅｓｔ.

Planes	Parts
4	16 ?

• What have I observed?

Is difficult to imagine 4 planes that intersect at random, this is a subproblem and as such it's to be addressed by using a simpler version since it this approach has brought us so far.

This self-reference to the problem previous stages in the problem or self-comparison is what it means to observe. This is a type of counting. You are checking the trail of crumbles has not moved (Step 3: 𝔬𝔟𝔰𝔢𝔯𝔳𝔢: 𝔣𝔦𝔫𝔡𝔦𝔫𝔤 𝔢𝔪𝔢𝔯𝔤𝔢𝔫𝔱 𝔰𝔲𝔟-𝔭𝔯𝔬𝔟𝔩𝔢𝔪𝔰 𝔞𝔫𝔡 𝔠𝔬𝔪𝔭𝔞𝔯𝔦𝔫𝔤 𝔱𝔥𝔢𝔪 𝔱𝔬 𝔱𝔥𝔢 𝔟𝔦𝔤 𝔭𝔯𝔬𝔟𝔩𝔢𝔪)

Going back to the case of 3 planes, there's an interesting phenomenon when there are two points of intersection. While almost all parts until now have been extending infinitely, here appears a part that is not like this. Shaped like a triangle.

Remember our perspective is hiding the extension of this triangle formed by the 3 planes and two points of intersection, if we were to isolate a section of it to look from the side it would look something like this open on top and bottom. . One way to counter this is by using this isolated figures to derive the not observable parts. We have 1 part, delimited by the triangle, at the each vertex we have one part and at each edge, we have 1 part. 1+3+3=7

This that we

Now we can clear something, due to our perspective, we said points of intersection, now we do realize that each point was actually an infinitely extending line, an edge.

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If we limit this visualization, in 4 planes with 3 points, or rather edges of the intersection, then we get something like this. 4 planes that include bounded bodies.

This tetrahedic bounded figures are also isolations of the infinitely extending figure.
Now by counting the figures, we have 1 fully closed tetrahedron at the center. 4 almost closed, but lacking a closed base, tetrahedrons at each vertex. 4 faces each one delimited by an infinitely extending triangle and edges that represent the meeting of two planes. For a total of 15 parts.

Planes	intersections	Parts
4	3	15

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If a point is intersected by all four planes, the result is 2⁴, meaning that for one of the extreme cases, the answer will be 2ⁿ.
This is what it would look like.

It seems we can generalize from this. Let, P= planes, I=Intersections, s= parts

s=2^p - (p - i)

Nonetheless there's one more interesting case. All five planes, sharing all points at the same time. This is 5 planes would behave like a single plane, giving place to two parts.

s=2; p=5; i=∞ ?

The algebra seems to help us here. We discover quickly that when a plane occupies completely the space of another place the intersection is negative. Like a mirror image. (e.g. for 4 planes all occupying the same space, 2²-(2-{-2}), of course, there other ways to fulfill this)

This creates a new symmetry. Sounds familiar?

The integers line can be seen as the mathematical representation of these two lines.

In the case of more than 2 planes occupying the same place the numbers go out of the space itself, I will explain this special case in a future post. What is really important is that the special cases can be dismissed because they lack a pattern and be taken as a dead end.

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For the majority of random cases, we can see that there's an alternative. By following the derivational nature of intersections we see that just like planes divide space, planes are divided by edges or lines and that lines are divided by points. This is what it means to observe.

Planes	space/planes	planes/lines	lines/points
0	1	1	1
1	2	2	2
2	4	4	3
3	8	7	4
4	15	?	5
5	?	?	6
n	?	?	n+1

• How can I complete my observation?

Here observation starts to bear fruits, by comparing the data we arrive at a conclusion. (Step 4: 𝔣𝔦𝔫𝔡 𝔞 𝔭𝔞𝔱𝔱𝔢𝔯𝔫)
By following the derivational relationships we found while counting, we see that we can relate the columns and rows to obtain new consistent information. By adding the value of the number immediately to the right of itself, we can find the number in the column, immediately below it.

Planes	space/planes	planes/lines	lines/points
0	1	1	1
1	2	2	2
2	4	4	3
3	8	7	4
4	15	11?	5
5	26?	16?	6
n	?	?	n+1

There are still two squares empty. The ones with the rule.

• What is the mathematical algorithm?

(Step 5: 𝔤𝔢𝔫𝔢𝔯𝔞𝔩𝔦𝔷𝔢) We found our answer, but we still don't have a way to test it. Our guess is a good guess, is a guess by induction. The product of observation. We passed from wild guesses, to educated guesses to a guess by induction.

If we check, by adding a line to our initial restricted perspective, we get

3 closed triangles, 3 open parts at the vertex and 5 open parts at the edges. 3+3+5=11

This means, now our induction is quite strong as it has evidence even if circumstantial.

Planes	space/planes	planes/lines	lines/points
n	?		n+1

(Step 6: 𝔱𝔢𝔰𝔱 ) In mathematics when you use numbers derived from others in a sequence, most of the time one refers to this as combinatorics. By combinatorial analysis:

When you cut space to produce the maximum number of parts, the nth numbers must also be added in the sequence. If p(n) is the number of divisions, then n is the last cut and p(n-1) is the cut before the last one.

Thus p(n)=n+(n-1)+(n-2)+...+1+p(0)

Being f(0)=1 as is the piece before any cut is made.

Rearrenging, p(n)=1+(1+2+3+...+n)
Simplified as an aritmetic progression, is equal to

The rationale is this. Each number in the sequence equals a triangular number plus 1

You can express it as a binomial coefficient.

This is part of a problem known as The Lazy caterer

(step 7: 𝔟𝔶 𝔞𝔫𝔞𝔩𝔬𝔤𝔶), we can extend this to generalize the next case and prove it.

Planes	space/planes	planes/lines	lines/points
n			n+1

This is the result. If n! expresses a factorial, and we write the binomial coefficients by then,

First the conditions:
(1) We must avoid parallel planes
(2) No parallel intersecting lines
(3) There's no point common to four or more lines

This problem is known as the cake numbers problem. A proof goes like this.

Let C_k(n) be the maximum number of k-dimensional regions with n-slices of (k-1) dimensional planes. We want to show that

We have by recursion: . If we assume the left/right argument works for any dimension, we get the general recursion .

With this recursion, we induct on k. For the base case, we can see . Now assume that . From the recursion, we have

and by telescoping

Thus, from the induction hypothesis:

using the Hockey-Stick identity, then

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Again, the instructions for induction compiled are:

1. Complete the question

2. Go to the extreme cases

3. Observe

🅄 3.a Find the emergent sub-problem and compare to the great problem (this is actually part of completing the question)

4. Find a pattern

5. Generalize

6. Test

7. Construct an Analogy

8. Derive first principles

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Some common questions about this subject.

• Can people see higher dimensions?

R - Probably no. Most of what we know are transformations or translations into our spatial language, as we are not natives to the language of higher dimensions. We can use other properties to represent such changes, like motion and color. The whole field of data visualization is proof of this.

The closest thing to this is people with partial dimensional negligence, that actually have less information than most of us and savants or artists that can see fractals.

• I've heard that reasoning from first principles is better than to reason from analogy, what does this mean?

R - Remember the cake analogy. First principles are already proven concepts. The result of previous inductive processes, some of them tested by centuries of scientific production. When someone reasons from first principles, this means deductively, what that person is doing is taking proven tools for implementation. Bringing into existence things that are merely conceptual. You already have the question fully stated, while with induction you didn't know that you didn't know the question.

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References:

• Let Us Teach Guessing, George Pólya. 1966. Washington, DC: MAA Video.
• Sloane, N. J. A. Sequence A000125/M1100 in "The On-Line Encyclopedia of Integer Sequences."
• Yaglom, A. M. and Yaglom, I. M. Challenging Mathematical Problems with Elementary Solutions, Vol. 1. New York: Dover, 1987.

Images: Google images labeled for reuse or constructed by me. LaTex transcribe to images using https://www.codecogs.com/latex/eqneditor.php