Trying to be Edgy

Maths and making stuff! I spent some time this weekend folding up the Dodecahedron pictured above. Here is the link for Instructions on how to build your own Origami Dodecahedron. Coming from MathCraftNZ, great site with some high quality and fun math related builds.

A Dodecahedron is a member in the platonic solid family. A platonic solid is a 3D shape where each face is the same regular polygon and the same number of polygons meet at each vertex. A dodecahedron has 12 faces, 20 vertices and 30 edges. The 12 faces are in the shape of a pentagon, and 3 pentagons meet at each vertex.

There are only five platonic solids:

  1. Tetrahedron
  2. Cube
  3. Octahedron
  4. Dodecahedron
  5. Icosahedron

A proof, of the fact that there can only be five platonic solids is offered by Euler’s Formula (noting this is a different formula than Euler’s Formula for Complex Numbers). Euler’s Formula states:
F + V – E = 2 
F is the Number of Faces
V is the Number of Vertices
E is the Number of Edges.

An easy example is the cube. For the cube: F = 6. V = 8. E = 12.
F + V – E = 2
6 + 8 – 12 = 2

Let’s say that we wanted to build a platonic solid out of a polygon with N number of edges. Then NF = 2E, because in the fully assembled 3D shape, every edge is shared by two faces, meaning each edge is counted twice. R number of edges are going to meet at each of the vertices as we build up this shape, giving, RV = 2E.

capture
Exploded Cube

If we put the two equations together, rearrange and substitute back into Euler’s Formula:
NF = 2E so F = 2E/N
RV = 2E so V = 2E/R
F + V – E = 2
(2E/N) + (2E/R)  – E = 2
(1/N) + (1/R)  – (1/2) = (1/E)
Looking at the last term, we note that the number of Edges can’t be zero, which yields the equation that proves there are only five platonic solids.
(1/N) + (1/R)  – (1/2) > 0
(1/N) + (1/R)  > (1/2)
Plugging in any positive integer values for N and R shows that only five combinations of N and R are possible! Those combinations of N (Number of Edges) and R (Number of Edges that meet at each of the vertices), are the combinations for the Tetrahedron, Cube, Octahedron, Dodecahedron, and Icosahedron.

No coffee required to make these regular solids! Thanks for reading! Feel free to share pictures of your own Dodecahedron builds, comments or questions.

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One, Plus, Minus and Infinity

… It never ends.

The Question: What does 1 – 1 + 1 – 1 + 1 … equal? Write down your answer and save it for later, I’m about to bend your brain.

1 – 1 + 1 – 1 + 1 – 1… = ?     [1]

For those of you not familiar with mathematical notation the … at the end of [1] means that the pattern continues until infinity. You are subtracting and adding 1’s an infinite number of times.

The Answer(s):

Case 1:
1 – 1 + 1 – 1 + 1 – 1 … = 0
Consider grouping the numbers in the following manner:
(1 – 1) + (1 – 1) + (1 – 1) … = 0
0 + 0 + 0 … = 0

Case 2:
1 – 1 + 1 – 1 + 1 – 1 … = 1
Consider grouping the numbers in the following manner:
1 – (1 – 1) – (1 – 1) – … = 1
1 – (0) – (0) – … = 1

Case 2a:
1 – 1 + 1 – 1 + 1 – 1 … = 1
Consider grouping the numbers the following manner:
1+(-1+1)+(-1+1) … = 1
1+0 + 0 + … =1

So what’s the deal with the grouping? Isn’t throwing the parenthesis into the equation breaking some rule? The short answer, no. The way that I’m inserting the parenthesis into the equation is making use of the Associative Property of Addition: a + (b + c) = (a + b) + c.
By this property we see that the order the addition takes place in doesn’t matter.

Taking a closer look at Case 2, one might say that I’m “changing” the order of the subtraction, and we know that the order of subtraction does matter since subtraction is not associative. There is NOT a change in the order of subtraction here. The negative gets distributed across the parenthesis first (thank you order of operations): 1–(1 – 1)=1 +(-1+1).
Case 2a doesn’t break any rules either and makes more sense when you think of the fact that subtraction is really the same as adding a negative:
1 -1 = 1 + -1.

So is the answer 1 or 0? In case your brain wasn’t hurting yet:

Case 3:
1 – 1 + 1 – 1 + 1 – 1 … = ½
Consider setting the series equal to a variable S and then subtract S from 1.
S = 1 – 1 + 1 – 1 + 1 – 1 …
1 – S = 1 – (1 + 1 – 1 + 1 – 1) …
Order of Operations: Distributing the Negative
1 – S = 1 – 1 + 1 – 1 + 1 – 1 …
On the left hand side we now see we ended up with what we started with, so we can go ahead and solve for S.
1 – S = S
S = ½
1 – 1 + 1 – 1 + 1 – 1 … = ½

The Explanation: This infinite series is called Grandi’s Series, and the reason you can get different answers for the sum is because it is a divergent series. A divergent series means that the infinite sequence of the partial sums of the series does not have a finite limit.

grandis-series
Grandi’s Series in Sigma Notation

We see that Grandi’s Series doesn’t converge to the value 1 or 0 as the number of terms in the sequence gets infinitely big, as we would expect of a convergent series. Instead, it can be both 0 and 1 and in a sense it oscillates between 0 and 1. The average of those oscillations is ½, which is why that number shows up in Case 3.

Brain Chow:
What happens if the we turned Grandi’s Series into a partial sum where n is now finite and odd?
What happens if the we turned Grandi’s Series into a partial sum where n is now finite and even?
Is infinity even or odd?

I’d tell you a joke about infinity but it never ends. As always thanks for reading! Thoughts, comments, feedback, and responses to the brain chow questions are always welcome.

The Seven Bridges of Königsberg

All these bridges and I can’t get over it.

In 1735 the city of Königsberg, Prussia  (now Kalinigrad, Russia) was set on both sides of the Pregel River. The river divided Königsberg into four regions with seven bridges connecting the regions.

The Problem: Can you find a path that allows you to walk around the city, crossing each of the seven bridges only once?

The Seven Bridges of Königsberg
Figure 1: The Seven Bridges         Source:Wikipedia

[Spoiler Alert] The Answer:  It is impossible. There is no path in Königsberg that allows you to walk around the city using each bridge only once.

The Proof: You can simply show that one path doesn’t work, with some pencil and paper. However, showing that all paths do not work requires a bit more mathematical rigor. In 1736, Leonhard Euler provided a solution to the seven bridges problem and a general solution for any number of bridges with any number of regions.

Looking at Figure 2, consider each of the four regions of Königsberg a Vertex (v) and each bridge an Edge (E). Counting the number of Edges(E) that touch a Vertex(v) provides the Degree (deg(v)). In Königsberg, we see that degree of every vertex is odd.
deg(B,C,D) = 3     deg(A) = 5

In order to to cross each bridge once and only once, the number of vertices of an odd degree must be zero or two. If there are two vertices with an odd degree, then those vertices are your starting and ending points. These are the conditions that make up an Euler Path, which is a continuous path that passes through every edge once and only once.

Euler's Representation
Figure 2: Euler’s Representation      Source: Weber University

The Seven Bridges of Königsberg do not meet the conditions for a Euler Path. However, if one of the bridges is removed (consider the removal of bridge b in Figure 2), then the degree of vertices A and B become even (deg(A,B) = 2). Vertices C and D remain odd, leaving these as your starting and ending points for the Euler Path.

“…this type of solution bears little relationship to mathematics, and I do not understand why you expect a mathematician to produce it, rather than anyone else, for the solution is based on reason alone, and its discovery does not depend on any mathematical principle. Because of this, I do not know why even questions which bear so little relationship to mathematics are solved more quickly by mathematicians than by others.” – Leonhard Euler, 1736

Euler should have been one to know mathematicians love a good puzzle, no matter how simple. Even though Euler found the problem trivial, he provided a general solution to the problem for any number of bridges or regions and kicked off an entirely new branch of mathematics called Graph Theory.

The Seven Bridges of Königsberg no longer exist today. Two of the seven bridges were lost in the bombing of Königsberg in WWII and two more were replaced by a highway, leaving only three bridges and an Eulerian path that is now possible.

Alright so edges, vertices, degrees, graphs and a solution to a trivial question leaves us with the humble beginnings of Graph Theory. Why do we care? Graph Theory makes up many modern day technologies. Providing your GPS with the ability to find the shortest path home, ranking the relevance of the links in your Google searches, even making up the topology of the circuits in just about every electronic device.

All these bridges and I still can’t get over it. Thanks for reading! If you’re interested in learning more about Eulerian paths or Graph Theory I’ve provided some links below with the reference material that I used to put this together and resources for going further.

Mathematical Association of America
Google and Pregel
NRICH
Graph Theory