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The Field of Origami-Constructible Numbers

The field of origami-constructible numbers is the field of all numbers that can be constructed on a sheet of paper by the intersection of two lines (or creases) caused by folding the paper given, that the paper has the two points 0 and 1 marked on it, where the paper represents the complex plane with the real axis going through the marked points and imaginary axis being perpendicular to it and passing through point 0. This seemingly elementary origami can be used to demonstrate all Euclidean constructions (these are defined as being straight-edge and compass constructions that are permissible by Euclid's 5 axioms) as well as some constructions that can't be achieved using a straight-edge and compass alone such as trisecting an arbitrary angle, constructing ${\sqrt[{3}]{2}}$ and constructing regular Heptagons and nonagons.

Euclidean Constructions

Constructions comes from a need to demonstrate and therefore create certain objects for our proofs. In times of Plato, geometric constructions were restricted to the use of only a straight-edge and compass and no markings could be placed on the straight-edge to be used to make measurements. Because of how critical and prominent these geometric constructions were in Euclid’s Elements, these constructions are sometimes known as Euclidean constructions. Such constructions lay at the heart of the three classical geometric problems: (1) squaring the circle, (2) doubling the cube, and (3) trisecting an arbitrary angle. It took until the eighteenth and nineteenth century for the problems to be proved to be impossible.

Near the beginning of the first book of the Elements, Euclid gives five postulates (axioms) for plane geometry, in terms of constructions:

1. To draw a straight line from any point to any point.

2. To produce a finite straight line continuously in a straight line.

3. To draw a circle with any centre and radius.

4. That all right-angles are equal to one another.

5. The parallel postulate: If a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than two right angles.

Huzita-Hatori Axioms

The field of all numbers can be constructed by a series of folds in the paper to create different lines. All these lines will be straight and the types of lines that we are able to create are defined by the Huzita-Hatori axioms. At the First International Meeting of Origami Science and Technology, Humiaki Huzita and Benedetto Scimemi presented a series of papers. One of which identified six uniquely different ways by which one could create a single crease by aligning one or more combinations of points and lines on a sheet of paper. Those six operations became known as the Huzita axioms. In 2002, Japanese had, through his own study, discovered a type of single-fold alignment that was not detailed in any of the six Huzita axioms. Thus, the 7 Huzita-Hatori axioms came into being (though some believe Huzita-Justin to be a more appropriate name as the seventh axiom is largely credited to Jacques Justin. In 1989, Justin published a paper in which he specified 7 possible combinations of alignments — which turned out to be the 6 Huzita axioms plus Hatori’s 7th.)

• Axiom 1. Given points $p_{1}$ and $p_{2}$, we can fold a line that goes through both.

  

  In a parameterised form, the equation for the line passing through the two points $p_{1}$ and $p_{2}$ would be:

  $F(\lambda )=p_{1}+\lambda (p_{2}-p_{1})$

• Axiom 2. Given points $p_{1}$ and $p_{2}$, we can fold $p_{1}$ onto $p_{2}$.

  

  This is equivalent to finding a perpendicular bisector to the midpoint of the line connecting $p_{1}$ and $p_{2}$. The parametric equation for this fold would therefore be:$F(\lambda )=p_{m}+\lambda \cdot V$. With $p_{m}$ being the midpoint between $p_{1}$ and $p_{2}$ and $V$ being the vector perpendicular to the line segment $p_{1}p_{2}$.

• Axiom 3. Given two lines $l_{1}$ and $l_{2}$, we can fold $l_{1}$ onto $l_{2}$ (i.e. bisect the angle between them).

This is equivalent to finding a bisector of the angle between $l_{1}$ and $l_{2}$. Let $p_{1}$ and $p_{2}$ be any two points on $l_{1}$, and let $q_{1}$ and $q_{2}$ be any two points on $l_{2}$.

We let $u$ and $v$ be the unit direction vectors of the lines $l_{1}$ and $l_{2}$ respectively. That is:

$u=(p_{2}-p_{1})/\left\vert (p_{2}-p_{1})\right\vert$

$v=(q_{2}-q_{1})/\left\vert (q_{2}-q_{1})\right\vert$.

If the two lines are parallel, they will have no point of intersection. The fold must be the line midway between the two and parallel to them. Should the two lines not be parallel, the point of intersection will be:

$p_{I}=p_{1}+s_{I}\cdot u$

Where:

$s_{I}=-{\frac {v^{\perp }\cdot (p_{1}-q_{1})}{v\cdot u}}$

The direction of one of the bisectors is:

$w={\frac {\left\vert u\right\vert v+\left\vert v\right\vert u}{\left\vert u\right\vert +\left\vert v\right\vert }}$

and, therefore, the parametric form of the equation of the fold will be:

$F(\lambda )=p_{I}+\lambda \cdot w$.

• Axiom 4. Given a point $p$ and a line $l$, we can fold a line perpendicular to $l$ that goes through $p$.

  

  This is equivalent to finding a perpendicular to $l_{1}$ that passes through $p_{1}$. If we work out a vector $v$ that is perpendicular to the line $l_{1}$, the parametric equation of the fold will be:

$F(\lambda )=p_{1}+\lambda \cdot v$

• Axiom 5. Given two points $p_{1}$ and $p_{2}$ and a line $l$, we can fold $p_{1}$ onto $l$ with a line that goes through $p_{2}$.

This is equivalent to finding the intersection of a line with a circle. This means that it may have 0, 1, or 2 solutions. The line is defined by $l_{1}$, and the circle has its centre at $p_{2}$, and a radius equal to the distance between $p_{1}$ and $p_{2}$. If the line does not meet the circle at any point, there are no solutions. If the line is tangent to the circle, there is one solution, and if the line intersects the circle in two places, there are two solutions.

The parametric equation of the folds are:

$F_{1}(\lambda )=p_{1}+1/2(d_{1}-p_{1})+\lambda (d_{1}-p_{1})^{\perp }$

$F_{2}(\lambda )=p_{1}+1/2(d_{2}-p_{1})+\lambda (d_{2}-p_{1})^{\perp }$

where $d_{1}$and $d_{2}$are solutions of the quadratic equation, using the combination of the general equation of the circle and the parametric form of line $l_{1}$.

• Axiom 6. Given two points $p_{1}$ and $p_{2}$ and two lines $l_{1}$ and $l_{2}$, we can fold $p_{1}$ onto $l_{1}$ and $p_{2}$ onto $l_{2}$ with a single line.

  

  This is equivalent to finding a line simultaneously tangent to two parabolas, and can be considered equivalent to solving a third-degree equation as there are in general three solutions. The two parabolas have foci at $p_{1}$ and $p_{2}$, respectively, with directrices (lines used to define a curve or surface defined) by $l_{1}$ and $l_{2}$, respectively. This fold is called the Beloch fold after Margharita P. Beloch, who in 1936 showed using it that origami can be used to solve general cubic equations.

• Axiom 7. Given one point $p$ and two lines $l_{1}$ and $l_{2}$, we can fold $p$ onto $l_{1}$ and is perpendicular to $l_{2}$.

  

  We have seen that Axiom 5 can have 0, 1, or 2 solutions and Axiom 6 can have 0, 1, 2, or 3 solutions. Because of this, the geometries of origami are considered stronger than the geometries of straight-edge and compass, where the maximum number of solutions an axiom can have is 2. Thus, compass and straightedge geometry solves second-degree equations, while so-called origami geometry, can solve third-degree equations, and solve problems such as angle trisection and doubling of the cube. It is worth noting, though, that the construction of the fold by Axiom 6 requires the "sliding" of the paper, which is not allowed in classical straight-edge and compass constructions

Field operations

The field of all origami-constructible numbers forms a subfield of the field Complex numbers, $\mathbb {C}$. For all $x$ and $y$ that exist within the field then $x-y$ and $xy$ also exist and $x^{-1}$ exists if $x\neq 0$ .

Elementary Operations

There are two elementary operations that are combinations of multiple axioms that help us to form create the more well known operations of addition, multiplication and inversion.

E1. Given a point $p$ and a line $l$, fold a line parallel to $l$ that goes through $p$.

This can be done by two uses of A4. The first to construct a line through $p$ perpendicular to $l$ and the second to create a new line through $p$ perpendicular to the first line created and hence parallel to the original line.

E2. Given a point $p$ and a line $l$, reflect $p$ across $l$.

This can be done by using A4 to construct a line through $p$ perpendicular to $l$ . Then by picking a point on $l$ that is not on the new create line (such a point an be created by the intersection of $l$ and a line passing through 0 or 1). Now using A5 on this new point, $p$ and the new line to fold $p$ across $l$ and mark the line going through $p$ and the new created point. This marks the new point which is the reflection of $p$ across $l$.

Multiplication

To Multiply by a real number can be done by using A1 on points 0 and $p_{1}$ and on 1 and $p_{1}$ to create two new lines, $l_{1}$ and $l_{2}$ respectively. Then by applying E1 to $l_{2}$ and the number to multiply by $m$ This will create a new line that will intersect $l_{1}$ at the point $mp_{1}$.

To multiply by $i$ we use A1 on 0 and $p_{1}$ to create line $l_{1}$ then use A4 on 0 and $l_{1}$ to create line $l_{2}$ then use A2 on $l_{1}$ and $l_{2}$ to create line $l_{3}$. Finally using E2 on $p_{1}$ and $l_{3}$ will create the point $ip_{1}$.

Addition

Addition can be performed using E1 and A1. starting with the two points to be added: $p_{1}$ and $p_{2}$ and the point 0 then use A1 to create lines through $p_{1}$ and 0 and $p_{2}$ and 0. Then by using E1 twice to find the lines parallel to these lines that pass through the other point. This will create a parallelogram with points $0,p_{1},p_{2}$ and a forth point. this forth point is the addition of $p_{1}$ and $p_{2}$.

This method only works if $p_{1}$, $p_{2}$ and 0 are not all on the same line. If this is the case a new method must be used. Using A1 to create a line through all three points and using A2 to fold $p_{1}$ onto $p_{2}$ so a new point $p_{3}$ lies over 0. Using A4 on 0 and the line with the paper still folded creates a new point This will mark $p_{3}$, the sum of the two points.

Inversion

The process to invert a number is very similar to multiplication. To start with two lines can be made using A1 on $p_{1}$ and 0 and A1 on $p_{1}$ and $m$ Then a third line is made using E1 on 1 and the second created line. This line intersecting with the first will mark the inverse.

This can be done for complex numbers in the same manner as multiplication.

Other Constructions

Doubling the Cube

Doubling the cube can be done in origami by creating the length ${\sqrt[{3}]{2}}$. This can be done by taking a piece of square paper and folding it into 3 equal horizontal strips. A bottom corner is then moved onto the top edge where the mark at at the intersection of the bottom edge and one of the fold creases is place directly over the other crease. The point now marked at the top edge will be ${\sqrt[{3}]{2}}$ times further from one corner than the other.

Trisecting the Angle

Trisecting the angle can also be done by creating two parallel folds $p$ and $q$ with the distances $p$ to $q$ and $q$ to a edge of the angle being equal and the angle is on the bottom left corner. The left edge of the fold $p$ is the placed over the other edge of the angle with the bottom left corner being placed on the fold $q$. The points marked by the new positions of the bottom left corner and the left edge of $q$ can then be used to trisect the angle by creating folds between them and the angle.

Solving Polynomials

There are a few methods for solving polynomials using origami. A method of solving cubic equations such as $x^{3}+ax^{2}+bx+c$ is to construct the points $(a,1)$ and $(c,b)$and the lines $y=-1$ and $x=-c$. Using Axiom 6 will create a new line $y=mx+z$ where $m$ is the solution to the original equation.

This method can also be used to find cube roots by solving the equation $x^{3}-r=0$.

A method of solving polynomials to any degree called Lill's Method was adapted to solve cubic equations in origami by Margharita P. Beloch. This method uses connecting line segments with lengths of the coefficients of the polynomial and lines at certain $\theta$ to find lengths equal to the roots of the polynomial. This method can be used to find the roots of polynomials of any degree using origami if simultaneous folds are allowed.

Power of Origami and Neusis

The advantage origami techniques have over regular compass and straight edge constructions comes from the ability to use Neusis construction as a method of solving problems. Neusis construction consists of fitting a line element of given length $\alpha$ in between two given lines ($l$and $m$), in such a way that the line element, or its extension, passes through a given point $P$. That is, one end of the line element has to lie on $l$, the other end on $m$, while the line element is "inclined" towards P. This extra ability is the reason for all constructions available to origami, and not otherwise, including the neusis construction. This allows the construction of a common tangent line of two parabolas, trisecting angles, doubling the cube and many other constructions.

Other constructible numbers

Negative numbers can be constructed. Negating a number is equivalent to reflecting it across 0. We can do this using A1, A4 and E2 in order. Square and cube roots are also constructible. Square roots can be constructed by using circle geometry to create a point where a circle centred at $(0,(r-1)/2)$ with radius $(r+1)/2$ intersects the real axis at point ${\sqrt {r}}$. Cube roots can be solved by solving cubic equations as mentioned above.

Another origami-constructibe number is ${\sqrt {1+x^{2}}}$ assuming that $x$ itself if constructible. This can be constructed as a right angle triangle with 1 and $x$ as the sides of the triangle and ${\sqrt {1+x^{2}}}$ as the hypotenuse.

Any $n$ sided regular polygon is constructible if and only if $n=2^{r}3^{s}p_{1}...p_{m}$ where $0\leq r,s,m\in \mathbb {Z}$ and $p_{i}$ are distinct primes of the form $p_{i}=2^{ai}3^{bi}+1$.

Extensions of the Field

The field of origami constructible numbers could possibly be extended in many ways. The most common method of doing so is to introduce new rules to allow the creation of new numbers. These include allowing the use of a compass. This can allow for the creation of circles and other curves but does not actually extend the field (no numbers that can be created by allowing the use of a compass can't be created without it).

Other possible ways of extending the field are to allow the use of tracing paper and allow cutting.