# 2018 Group 6

## Contents

**Origami-Constructible Real Numbers**

In geometry and algebra, a real number α is constructible if and only if, given a unit length, a line segment of length |α| can be constructed from the unit length in a ﬁnite number of steps using only a ruler (i.e. a straight unmarked edge) and pair of compasses.[math]^{[1]}[/math]

We start with an informal definition of origami and how this can be applied to mathematics: origami constructions are made up of series of folds in the paper, when the paper is folded and unfolded, it leaves a crease which acts as a line. In such origami constructions, a point only exists if it is the intersection of two creases.

## History

### Origin of Origami Algebra

The Japanese art of paper folding has an impressive history, despite its foundations in geometry and algebra only being explored over the past few hundred years. The foundations for constructions are primarily classical, dating back to Pythagorus, Euclid, and Pappus.[math]^{[2]}[/math] In 1893, Indian mathematician T. Sundara Rao published "Geometric Exercises in Paper Folding" which used paper folding to demonstrate proofs of geometrical constructions.[math]^{[3]}[/math] This work was inspired by the use of origami in the kindergarten system.

The axioms that describe the operations made when folding a piece of paper, were first discovered by Jacques Justin in 1986.[math]^{[4]}[/math] Axioms 1 to to 6 were then rediscovered by Japanese-Italian mathematician Humiaki Huzita and reported at *the First International Conference on Origami in Education and Therapy in 1991*.[math]^{[5]}[/math] Axioms 1 to 5 were rediscovered by Auckly and Cleveland in 1995.[math]^{[6]}[/math] Axiom 7 was rediscovered by Koshiro Hatori in 2001.[math]^{[7]}[/math]

### The Seven Axioms

We shall describe a set of axioms for paper folding which will be used to describe diﬀerent subﬁelds of the real numbers, similar to the way that ruler and compass constructions are used to build ﬁelds. There are 6 axioms that were outlined in 1992 by Italian-Japanese mathematician Humiaki Huzita[math]^{[5]}[/math]:

**O1.**Given points [math]P_1[/math] and [math]P_2[/math], we can fold a line that goes through both of them.**O2.**Given points [math]P_1[/math] and [math]P_2[/math], we can fold [math]P_1[/math] onto [math]P_2[/math] (i.e. ﬁnd the perpendicular bisector of segment [math]P_1P_2[/math]).**O3.**Given two lines [math]L_1[/math] and [math]L_2[/math], we can fold [math]L_1[/math] onto [math]L_2[/math] (i.e. bisect the angle between them).**O4.**Given a point [math]P[/math] and a line [math]L[/math], we can fold a line perpendicular to [math]L[/math] that goes through [math]P[/math].**O5.**Given two points [math]P_1[/math] and [math]P_2[/math] and a line [math]L[/math], we can fold [math]P_1[/math] onto [math]L[/math] with a line that goes through [math]P_2[/math].**O6.**Given two points [math]P_1[/math] and [math]P_2[/math] and two lines [math]L_1[/math] and [math]L_2[/math], we can fold [math]P_1[/math] onto [math]L_1[/math] and [math]P_2[/math] onto [math]L_2[/math] with a single line.

A 7th axiom was introduced in 2002 by Hatori[math]^{[7]}[/math]:

**O7.**Given a point [math]P_1[/math] and two lines [math]L_1[/math] and [math]L_2[/math], we can make a fold perpendicular to [math]L_2[/math] that places [math]P_1[/math] onto line [math]L_1[/math]

## Origami Field Theory

### Definitions

Recall the definition of a * constructible real number*:
Given a unit length, a real number α is constructible if a line segment of length |α| can be constructed from the unit length in a ﬁnite number of steps using only a ruler (i.e. a straight unmarked edge) and pair of compasses.[math]^{[1]}[/math]
Then we have the set K = {x ∈ [math]\mathbb{R}[/math]| x is constructible}

We can form a similar definition for * origami-constructible numbers*: An element x ∈ [math]\mathbb{R}[/math] ≥ 0 is said to be origami constructible if, starting with the unit segment and using only axioms

*O1*to

*O6*, it is possible to construct a segment of length x. By deﬁnition, an element x ∈ [math]\mathbb{R}[/math] < 0 is origami constructible if and only if so is −x. [math]^{[8]}[/math] Then we have the set O = {x ∈ [math]\mathbb{R}[/math]| x is origami constructible}

We can extend this further by defining an origami pair[math]^{[9]}[/math]:

* [math]\{P,L\}[/math] is an origami pair if [math]P[/math] is a set of points in [math]\mathbb{R^2}[/math] and [math]L[/math] is a collection of lines in [math]\mathbb{R^2}[/math] satisfying:*

**a)** The point of intersection of any two non-parallel lines in [math]L[/math] is a point in [math]P[/math]

**b)** Given any two distinct points in [math]P[/math], there is a line [math]L[/math] going through them.

**c)** Given any two distinct points in [math]P[/math], the perpendicular bisector of the line segment with given end points is a line in [math]L[/math]

**d)** If [math]L_1[/math] and [math]L_2[/math] are lines in [math]L[/math], then the line which is equidistant from [math]L_1[/math] and [math]L_2[/math] is in [math]L[/math].

**e)** If [math]L_1[/math] and [math]L_2[/math] are lines in [math]L[/math], then there exists a line [math]L_3[/math] in [math]L[/math] such that [math]L_3[/math] is the mirror reflection of [math]L_2[/math] about [math]L_1[/math]

### Lemmas

There are many lemmas that apply to the field of origami constructible numbers, but we have chosen the most relevant to mention below:

**Lemma 1***It is possible to construct a line parallel to a given line through any given point using origami.*[math]^{[10]}[/math]

__Proof__:

Consider the point [math]P_1[/math] and line [math]L_1[/math], we can fold a line parallel to [math]L_1[/math] that goes through [math]P_1[/math]. We obtain this from two applications of *O4*, the first gives us a line perpendicular to [math]L_1[/math], which we label [math]L_2[/math] the second application gives us a line [math]L_3[/math] that is perpendicular to [math]L_2[/math] and therefore parallel to [math]L_1[/math] passing through [math]P_1[/math].

We mention a second lemma which will be applied later in order to construct cubic roots:

**Lemma 2***Origami can solve any cubic equation.*[math]^{[11]}[/math]

Italian mathematician *Margherita Beloch* proved this in the 1930s.

__Proof__:

we construct the points [math]P_1[/math] = (a, 1) and [math]P_2[/math] = (c, b). Considering the real axis as the x-axis and the imaginary axis to be the y-axis, we also construct the two lines [math]L_1[/math] and [math]L_2[/math], defined by y = −1 and x = −c, respectively. We then apply O6 to [math]P_1[/math], [math]P_2[/math], [math]L_1[/math], and [math]L_2[/math] to obtain a new line [math]L_3[/math]. This new line is defined by the equation y = mx + z (since we know it is not vertical), where m is a solution to the original cubic equation.

This was realised to simply be an application of | Lill's method for finding real roots of a polynomial.

In general, we can make some conclusions about the field of origami numbers, the first being that [math]\mathbb{Q}[/math] ⊆ O ⊆ [math]\mathbb{C}[/math]. This means that, for any subfield k of O, if an extension K/k has degree 2 or 3, then K is also a subfield of O.[math]^{[11]}[/math]

## Constructions

Similar to the field of constructible numbers, we can show that the following properties still hold within the field of origami-constructible numbers: for any origami-constructible numbers α and β, α−β and αβ are origami-constructible and α[math]^{-1}[/math] is origami-constructible if α does not equal 0.

Using the lemmas outlined in the previous section, we can perform the following constructions in order to demonstrate that the above properties hold.

### Addition

We can apply the previous lemma to outline how to find the sum of two origami numbers.[math]^{[10]}[/math]

The following steps outline the addition operation for origami numbers:

**1.** We start out with three points [math]P_1[/math],[math]P_2[/math],and [math]0[/math] as shown in *figure 1*

**2.** We use *O1* to fold a line that goes through [math]0[/math] and [math]P_1[/math] which we label [math]L_1[/math] as in *figure 2*

**3.** We repeat *O1* to fold a second line that goes through [math]0[/math] and [math]P_2[/math] which we label [math]L_2[/math] just like *figure 3*

**4.** Consider the point [math]P_2[/math] and line [math]L_1[/math], we can fold a line parallel to [math]L_1[/math] that goes through [math]P_2[/math] using Lemma 1, as demonstrated in *figure 4*

**5.** Consider the point [math]P_1[/math] and line [math]L_2[/math], we can repeat the procedure in step **(4)** with the respective sides in order to obtain a line [math]L_4[/math] which is parallel to [math]L_2[/math] and passes through [math]P_1[/math] as done in *figure 5*

**6.** The intersection of the lines [math]L_3[/math] and [math]L_4[/math] gives us the point [math]P_3[/math] that is equal to [math]P_1[/math] + [math]P_2[/math] as can be seen in *figure 6*

### Multiplication

In fact, it is also possible to find the product of two origami numbers.[math]^{[10]}[/math]

The following steps outline how the multiplication operation for origami numbers:

**1.** Just as in the proof for constructible numbers (refer to lecture notes) we use the properties of similar triangles to multiply a number [math]P_1[/math] by a real number [math]r[/math]

**2.** We start out with four points [math]0[/math], [math]1[/math], [math]P_1[/math] and [math]r[/math] as shown in *figure 1* and *figure 2*

**3.** Apply *O1* to fold a line that goes through [math]0[/math] and [math]P_1[/math] which we label [math]L_1[/math] as in *figure 3*

**4.** Apply *O1* again to fold a second line that goes through [math]1[/math] and [math]P_1[/math] which we denote as [math]L_2[/math] just as in *figure 4*

**5.** Consider the point [math]r[/math] and line [math]L_2[/math], we can fold a line parallel to [math]L_2[/math] that goes through [math]r[/math], we denote this as [math]L_3[/math] as demonstrated in *figure 5*

**6.** Then the intersection of lines [math]L_1[/math] and [math]L_3[/math] gives us the point [math]P_2[/math] that is equal to [math]r[/math][math]P_1[/math] as can be seen in *figure 6*

Since this method is used primarily for complex origami-constructible numbers, we make the following adjustments since we are focusing on the field of origami-constructible real numbers:

add [math]i[/math] to [math]P_1[/math], multiply by [math]r[/math], and project this product onto the real axis with an application of *O4*

### Inverse

Similar to the field of Constructible numbers, we can find the inverse of origami-constructible numbers.[math]^{[10]}[/math]

**1.** We use the properties of similar triangles to divide a number [math]P_1[/math] by a real number [math]r[/math]

**2.** We start out with four points [math]0[/math], [math]1[/math], [math]P_1[/math] and [math]r[/math] as shown in *figure 1*

**3.** Apply *O1* to fold a line that goes through [math]0[/math] and [math]P_1[/math] which we label [math]L_1[/math] just as in *figure 2*

**4.** Apply *O1* again to fold a second line that goes through [math]r[/math] and [math]P_1[/math] which we denote as [math]L_2[/math] which can be seen in *figure 3*

**5.** Consider the point [math]1[/math] and line [math]L_2[/math], we can fold a line parallel to [math]L_2[/math] that goes through [math]1[/math], we denote this as [math]L_3[/math] as in *figure 4*

**6.** Then the intersection of lines [math]L_1[/math] and [math]L_3[/math] gives us the point [math]P_2[/math] that is equal to [math]1[/math]/[math]P_1[/math] as shown in *figure 4*

Since this method is used primarily for complex origami-constructible numbers, we make the same adjustment as described for multiplication of real origami numbers.

### Haga's Theorem

Another interesting consequence of the field of origami-constructible numbers is Haga's Theorem. A retired professor of biology from Japan, came up with this method which allows us to fold any fraction we would like from a square piece of paper.[math]^{[12]}[/math]

Referring to the image on the left, fold the bottom left corner to a point a distance of k along the top edge

Then, the top-left triangle has sides of length k, x and 1 -x. So Pythagoras' theorem tells us that [math]x^2[/math] + [math]k^2[/math] = [math](1 – x)^2[/math]

Rearranging gives us x = (1 – [math]k^2[/math])/2 and by the similar triangles argument we have that y/(1 – k) = k/x

If we put these equations together we find y/(1 – k) = k/((1 – [math]k^2[/math])/2)

Which can be rearranged as y = (2(1-k)k)/(1 – [math]k^2[/math])

And since 1 – [math]k^2[/math] = (1 – k)(1 + k) this can be simplified to y/2 = k/(1 + k)

The image on the right hand side demonstrates how to fold 1/5 using Haga's Theorem:
1⁄5 can be generated with three folds; first halve a side, then use Haga's theorem twice to produce 2⁄3 and then finally 1⁄5.

## Solving the Greek Problems

It turns out that origami is much more powerful than straight-edge and compass creations because many things that cannot be created using straight-edge and compass can be constructed using origami. Origami allows us to:

- Construct [math]^3\sqrt{2}[/math]
- Trisect an arbitrary angle

These were previously impossible problems

### Doubling the Cube

Doubling the cube is equivalent to solving the cubic [math]x^3 − 2[/math] = 0 i.e. constructing [math]^3\sqrt{2}[/math].[math]^{[13]}[/math]

**1.** start with a square sheet of paper. We first fold the paper into thirds and define [math]L_1[/math], [math]L_2[/math], [math]P_1[/math] and [math]P_2[/math] as shown in *figure 1*

**2.** We now fold [math]P_1[/math] onto [math]L_1[/math] and [math]P_2[/math] onto [math]L_2[/math] using *O6*, mark the point where [math]P_2[/math] meets [math]L_2[/math] as in *figure 2*

**3.** Unfold and extend this point into the line [math]L_3[/math] as demonstrated in *figure 3*

**4.** Then the ratio of the lengths a and b in the diagram is precisely [math]^3\sqrt{2}[/math].

### Trisecting an Angle

This origami move is actually solving a cubic equation by finding a simultaneous tangent to 2 parabolas.[math]^{[13]}[/math]

We use the following steps to demonstrate how it is possible to trisect an angel using origami-constructible real numbers:

**1.** We start with a square piece of paper and fold a line [math]L_1[/math] to create any angle [math]θ[/math] as in *figure 1*

**2.** To trisect [math]θ[/math], we have to fold the paper into quarters from top to bottom and define [math]P_2[/math] as shown in *figure 2*

**3.** simultaneously fold [math]P_1[/math] onto [math]L_2[/math] and [math]P_2[/math] onto [math]L_1[/math] using *O6* and whilst the paper is still folded, extend [math]L_1[/math] by a new fold [math]L_3[/math], as shown in *figure 3*

**4.** If we now open the paper and extend [math]l_3[/math] to its full length, it will divide [math]θ[/math] in the ration 1:2, as shown in *figure 4*

**5.** Finally, use an application of *O3* to bisect the larger part of the angle which then splits [math]θ[/math] into three equal parts as required

### Squaring the Circle

The third Classical Greek Problem that is impossible with straight edge and compass, squaring the circle, is impossible even using Origami. This is because it involves constructing the transcendental ratio [math]\sqrt{π}[/math] which cannot be written as a root of a polynomial with ﬁnite number of terms.[math]^{[1]}[/math]

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