The Riemann Hypothesis is the number one mathematical challenge of today. We offer a geometric solution to the problem that is derived from 4th dimensional mathematics, which proves the conjecture to be true.

## Overview

Our solution is derived from the system of 4th dimensional mathematics, developed by Colin Power. Through the process of Simultaneous Equation, we solve for (x + iy) where x and y hold the values -1, +1, -∞, +∞ that forms a comparison chart, from which we arrive at four zero solutions.

Next we solve for k for each of the solutions for (x + iy) and the same for 1/k. We take the minimum range of +1 and the maximum range as +∞-1, at the infinite set does not include zero. From this we create two sets of nine solutions, with the pair of zero points for each set (4 zero solutions; Summery Table X.Y). Next we take these results and solve for k=+1 and k= +∞-1. And then the same for its reciprocal.

From the Table Summery K we find that all solutions result in only six states. For k=1 we find three states ∞-1, 0 , ∞+1. And for k=+∞-1 we find the other three results -1, 0, +1. As k is in its infinite state, we can adjust this result to add or subtract additional whole number infinite sets from k= +∞-1 to obtain the three results, +∞-2, +∞-1, +∞. We can use the results in a 4th dimensional calculator to view the zeta function.

However, to complete the solution we solve for f and formulate a comparison table that can reveal the scope of the proposed equation. We do this by performing calculations of all possible outcomes for the unique solutions for f(x +iy), where f=1, and then again for f=∞.

This produces another nine solutions for each state. From this we can draw up a Comparison Table FK. We find a two zero solutions where F=1 and K=1, where the first is derived from the zero solution that appears at the centre of the table, whereas the second solution is found where F and K both equal ∞+1.

THE

# Concept

## 4th Dimensional Solution

The concept of ∞+1 in 4^{th} Dimensional mathematics causes the number line to rotate in numerical space, which is driven by the squaring of ZERO at the centre of the number line. (see rotational Squaring). As infinity can not exceed itself on the linear plain, it must move dimension in order to compensate for this extra number density.

Using this solution we can enter the values for k into the 4D calculator, and examine the wave function. Where k=1 we enter the into the time function, as either -1, 0 or +1, we notice that only the values +1 and -1 produce a mathematically acceptable output, whereas the zero state produces the output x/0 = DIV0! This in normal mathematics has no solution, but in 4D maths produces the division of the number line into 2 infinite sets; positive, and negative. For this reason the domain of the complex plain can only be extended into the negative by the bending of numerical space, through the process of analytic continuation.

In 4D maths we can use the secondary *equation** **line* in order to achieve the same result, which sets the critical strip from -1 – 0 – +1, instead of 0 – 0.5 – 1, as we do not need to **bend** number space which distorts the numerical results. In 4D mathematics we can perform the equivalent of analytic continuation by **folding** number space over zero, which maintains the proportion of the infinite densities aleph 0.5 for reciprocal space and Aleph 1 for whole number space.

In this way we can explain the nature of the critical strip as the number ZERO is off-set by 0.5 through the bending of number space, which collapse of the infinite density of whole numbers, aleph 1.

Therefore we find that the collapse of Aleph 1 (whole number space), through the bending of numerical space that creates the non-trivial zeros along the critical strip, instead of at the zero line. This is achieves by inverting the negative values of i into positive values, which is what the equation that produces the curvature is expressing. N= 1 to ∞, whereas the second side of the equation is the reciprocal of n^{s}, which is an expression of reciprocal number law, n / n² = 1/n , on the complex plain.

## Bending the Square

The number 2²= 4. We can see that in the above images the number 4 is denotes the point where the red rectangle meets the yellow one. Yet on the complex plain this numerical space is also represent as a square divided into 4 smaller squares. Half of the square lies in the the whole number plain whereas the other half in the reciprocal space between 0 and 1. As these spaces have an infinite density ratio of 0.5:1 (whole space is half as dense as reciprocal space), when the square is folded through the circle, the density of whole numbers is wrapped into the reciprocal space. Therefore instead of ending up at the number zero, it ends at the number 0.5, forming the critical strip.

## Number compression of the Hexagonal plain

The complex plain it based on the square, which in 4D mathematics comes out of the notion of ZERO². The second number plain based in the triangle is not represented on the complex plane, but is represent in 4D mathematics derived from the notion of ZERO³. These produces the two types regular 2D tapestries (formed only of 2 alternating colours), from the square and the triangle.

When we consider the nature of the hexagonal plane, we find that 6 triangles join to form a hexagon.

This tapestry is created automatically from equal interlocking of two circles of the same dimension.

This creates a geometric form called the Vesica Piscis. Once formed, more circles of the same

dimension can be centred on the two nodes created to initiate the tapestry. The Vesica Piscis has a

ratio of 1:√3, which is reflected throughout the number plain.

The function f=s=(s² ) squares number space by rotating the x axis through 180° and 90° rotation on

the i plain. This alters the the number density relationship of the whole numbers (Aleph 1) to the

reciprocal space (aleph 0.5). However, instead of the points crossing at √3, (as per the triangular

number plain) the number lines cross at the point of i2. This maintains a 90° angle at all intersecting

points, instead of the 60° of the triangle, found on the 0 3 plain. We can examine this transformation

from the perspective of the Silver and Golden ratio.

#### When we superimpose the silver mean and √2 fractal, the bending of the complex plain through analytic continuation matches the fractal pattern.

## The Geometric Solution

Here we can see that the rotation of the plain move +2i to the number -2. Whereas the value -1 is

rotate double the distance creating the offset in infinite density. The square (Diagonal √2) exactly half

density of the rectangle (2 Squares Diagonal √5). When the √5 triangle rotates, to lie along the same

orientation as the √2 of the square, its corner point defines the halfway mark between 0 and +1, which

creates the critical strip A. For this reason all non-trivial zeroes will fall on this line, just as any

reciprocal square number series will devolve to zero. ON the negative side of the number plain we

notice a +2i arcs through number space to arrive at the -1 position on the number line. Normally this

rotation should find a position at at n= -2. On the square plain this offset forms a second critical strip

(B) from the diagonal of the square in the negative plain, from a diagonal ratio of -√2:√8, which is a

function of the silver ratio. However, on the positive plain the natural squaring function is not present,

and √2:√8 is replace with √5:2, which is a function of the golden ratio. Such as

When we examine the difference between the silver and golden ratio equations, both exhibit the

function of ± 1. However, when the number √2 is replace by √5, the whole equation must be halved to

maintain this ± 1. From the we can calculate the height triangle formed where √5 triangle touches the

critical strip A as:

#### √(((√5−√2)²−0.5²) = 0.652261205

To this value we can add 1 to ascertain the distance from the zero point of the critical strip,and then

multiply by √3 (the ratio of the triangular plain.)

#### 1.652261205 × √3 = 2.861800354

The number is also very close to √8, producing a squared value of 8.189901266 (≈8.19), whereas

multiplying the value by √3 results in a value just short of the number 5.

** (2.861800354×√3 =**

**4.956783614 ≈ 4.95678)**

We notice that 8.19 is a number form of the digits 81 (9 2 ) and 9, whereas the number sequence

formed after 2.8 is (6180) which is close to the golden ratio. Whereas the number (28= 7×4) which is

the number proceeding it multiplied by 10, (base 10, see subtraction of 7 below.)

When considered from the perspective on the 0 2 (square) plain, a rectangle of sides 2:1 ratio a triangle

with diagonal of √5, instead of √2 as is true for the square of 1.

Next we can subtract √3+1 from 2.861800354 to find the difference = 0.129749546, which has a reciprocal value of 7.707156062. And then we find that √(7.707156062 – 7) = 0.500069696 = the critical strip. The number 7 is subtracted to **de-π** the value as 22/7 is close to π, and due to the nature of number e which demonstrates the difference in line length between the circle and hexagon in the triangular number plain.

(see ((2π−6)×(2.25×e) = 1.731999 ≈ √3).

Here the value 2.25 is the relationship density x2 plus 0.5/2=0.25.

This function is most easily expressed in geometric terms as:

Here we see the reciprocal square number law has a limit of 1>∞ =2. The squares above 1² are

reflected above and below, mirroring the density. The space left is formed of all reciprocal square

numbers, from ¼ >∞ = 2, that have a diagonal of √0.5, which is also reflected on either side of the central line. As we compress the scale on the right to form the critical strip, so there is an expansion

of number density along the x axis, (negative plain of the zeta function).

If we make a final comparison of the rotation of the line on the triangular plain we find that two unit

rotations will produce a point on the critical strip that is at n=√3.

For those who are interested we can square these numbers to get the ratio 3+49 = 52. As a function

of π this becomes expressed as a ratio 3:49 in the following equation:

(( π /( 3 / 49 )) + 1 ) 2 ≈ √2

(( π /( 3 / 49 )) + 1 ) x 7 ≈ 10

√2:10 scaling function

As the Value for +2i rotates to position n=-1, however +i3 will rotate to the value of -1², simply

because 3 x 4 = 12, which are the squared values for √3 and 2. As the function f(s)=(s² ) is a squaring

function, this compression ratio √3² : 2² maintains the perfect 90° angle of the zeta function as it

bends the number plain through the ratio π on the x plain to π/2 on the i plain.

π : π/2 = 2

( √3 / 2π )÷e ≈ 1

Where the result 1 in the second equation is equivalent to the +1 function in the √2:10 scaling

equation. In 4D mathematics this relationship is defined by the simultaneous equation:Simultaneous equations for infinite density function of e

**Function through 3**

((2π÷√3)÷e)÷((π×√3)÷e) = 1 / 1.5

((π÷√3)÷e)÷((2π×√3)÷e) = 1 / 6

((2π÷√3)÷e)÷((2π×√3)÷e) = 1 / 3

((√3×π)÷e)÷((π×√3)÷e) = 1

((2π÷√3)÷e)÷((π÷√3)÷e) = 2

((π×√3)÷e)÷((π÷√3)÷e) = 3

**Function through 2**

((√3÷2π)÷e)÷((2π×√3)÷e) = ¼

((√3÷π)÷e)÷((2π×√3)÷e) = ½

((√3÷2π)÷e)÷((2π÷√3)÷e) = ¾

((√3×2π)÷e)÷((2π×√3)÷e) = 1

((π×√3)÷e)÷((2π÷√3)÷e) = 1.5

((2π ×√3)÷e)÷((π ×√3)÷e) = 2

((2π×√3)÷e)÷((2π÷√3)÷e) = 3

### √12π ÷e ≈ 4

Notice that the zero point of the critical strip (n=0.5) is formed by an equation with a ratio of π : 2π. For

more information on the function of e can be demonstrated by the e-ratio calculator.

This function is more easily demonstrated by the following geometric presentation that shows how

the density of two squares with a surface of area of 1.5 are compressed into the triangular plain

through the number e.

Therefore 2.25 / √3 = 1.299038106 = √1.6875 ≈ 1.3

Notice that the square of 4 overlays the square of nine (bottom left), with equal density.

## Number unit compression

It we examine the analytic transformation of the zeta plane, we see the ‘red’ dot accumulates first at point n=1.5, which moves towards the √3 before collapsing into the number one. This is due to the infinite density of numbers on the hexagonal plane. The number e is the accumulation of all reciprocal factorials, which are a series of triangular numbers added together in time. I.e the collapsing of number space into a dot. This is what is happening when the complex plane is transformed. However, as the plane is bent instead of being folded, the numbers are compressed into the triangular plane, which is the genesis of the hexagon in a circle.

This is expressed in 4D mathematics by the equation:

((2π-6) × 2.25) × e = √3

Here, a circle with the radius of 1 has the 6 sides of the hexagon removed from its value. This is multiplied by 2.25 to form the ‘infinite density constant’ (unique to 4D maths). As the number e represents the infinite density of all numbers in reciprocal space compressed into a dot, the √3 represents the unit distance between the 2 tips of opposing triangles.

We can see from the image above that when rotated 180° the unit measure between the start and end point is double √3. Therefore the equation becomes:

((2π-6) × 4.5) × e = √12

((2π-6) × 3) × (e ÷2) = √3 / 2

However, in 4D mathematics whose axioms accepts the concept of negative squaring, due to the fact of the law of order (e.g: 2 x 3 is NOT 3 x 2 in 4D Maths), the 4 functions of squaring ( ++, – – , + – , and – + ), means that the above equation becomes expressed in terms of the number 9.

((2π-6) × 9) × e = √24

This is the ‘real’ equation in 4D mathematical space, that will never be expressed on a complex number plane that is missing ¾ of all numbers.

The hexagon divides 2π into 6 parts. When multiplied by 9 we get 1.5 hexagons, which will rotate

around the circle 1.5 times to end at the negative side of the number line. The square on the other

hand, will perform the same rotation (1.5) in six steps. Therefore 6 squares x 1.5 = 9. However, as the

zeta function bends the square number plane (complex plane) in both, the up and down direction from the real number line, the number e is produced at half density.

## ((2π-6) × 9) × (e ÷2) = √12

Furthermore, the rotation of number compression is accomplished through a single rotation of 180°,

which is 2 steps on the square plane and 3 steps on the triangular plane. Therefore, we get:

## ((2π-6) × 3) × (e ÷2) = √3 / 2

In 4D mathematics the number √3 /2 is found as the diagonal of a cube side length 0.5. Normally, this

describes a hyper-cubic number space, a concept that is beyond a capabilities of the complex plane to express, as it excludes the function of negative squaring. The result is that the bending of number space of the zeta function can only accommodate numbers >+1, and no negative numbers.

In 4D mathematics the zeta function is depicted as a cube of number space that becomes flattened

into a square and off-set when expressed on the complex plane.

In the next section we explain how 4D space exposes the correct views of number space beyond the number line, and the methods employed to calculate the infinite densities of all real numbers greater than 1.

## Appendix

### Solve for x + iy

**Solution A – All Whole Numbers**

**X = +∞, Y =+∞**

√-1 * +∞ = -∞ = (-Y∞) + (+X∞) = 0

**X = -∞, Y =-∞**

√-1 * -∞ = +∞ = (+Y∞) + (-X∞) = 0

**Solution B – Square Space**

**X = -∞, Y =+∞**

√-1 * +∞ = -∞ = (-Y∞) + (-X∞) = 2∞

**X = +∞, Y =-∞**

√-1 * -∞ = +∞ = (+Y∞) + (+X∞) = 2∞ +n * +n = +n

**Solution C – Reciprocal Space**

**X = +1, Y =+1**

√-1 * +1 = -1 = (-Y1) + (+X1) = 0

**X = -1, Y =-1**

√-1 * -1 = +1 = (+Y1) + (-X1) = 0

**Solution D – Reciprocal Square Space**

**X = -1, Y =+1**

√-1 * +1 = -1 = (-Y1) + (-X1) = -2

**X = +1, Y =-1**

√-1 * -1 = +1 = (+Y1) + (+X1) = +2

**Solution E – Negative integers**

**X = +1, Y =+∞**

√-1 * +∞ = -∞ = (-Y∞) + (+X1) = -∞+1

**X = +∞, Y = +1**

√-1 * +1 = -1 = (-Y1) + (-X∞) = -∞-1

**Solution F – Positive integers**

**X = +1, Y =-∞**

√-1 * -∞ = +∞ = (+Y∞) + (+X1) = +∞+1

**X = +∞, Y = -1**

√-1 * -1 = +1 = (-Y1) + (+X∞) = +∞-1

**Solution G – Integers Stop Counting**

**X = -1, Y = -∞**

√-1 * -∞ = +∞ = (+Y∞) + (-X1) = +∞-1

**X = -∞, Y = -1**

√-1 * -1 = +1 = (+Y1) + (-X∞) = -∞-1

##### Unique solutions

+2∞ +∞+1 +∞-1 +2 0, 0, 0, 0 -2-∞+1 -∞-1 -2∞

Note: there are 4 ZERO solutions. This is to be expected as:

Y+1 = X-1

Y-1 = X=+1

Y+∞ = X-∞

Y-∞ = X+∞

These are the four ‘REAL’ solutions in 4D mathematics, with the other 8 solutions being a 4 the dimensional

identity, where 2 and -2 perform the function of ‘Rotational Squaring’, (Yellow square’), that require a number

twice the length of the squared value, which is the function of the orange square in 4D mathematics.

The functions of the eight states of ±∞ ±1 (Red and Green Squares) are the function of infinity that generates the functional algorithm +1, -1, +1, -1, which moves the infinite number line onto the diagonal, as a result of the combine infinite densities of a ‘squared’ pair of number lines. We term this function ZERO² in 4D Mathematics.

This function is demonstrated geometrically as the relationship to the sliver mean the defines π and the √2:1

ratio that forms the fractal of a square rotated inside on another. Due to its fractal nature, this relationship is a

scalable phenomena that is distributed throughout the square number plain. The Sliver mean maintains the

infinite density of the number line (2π/4= ½π), as it is rotated through ¼ of the number plain.

Notice that the diagonal in both cases is formed of 4 squares, which lie at the centre of a diagonal line between two infinite sets (x and i plain) . As one fractal expands from the centre (√2 compression), so the silver mean develops a diagonal of 6, with two of the squares across the diagonal rotated at 45° and expanded to the relative dimension of √2. In this way the circle ( π) is defined from the Center point of the silver mean fractal. Note that in 4D mathematics we normally define π from the ‘radius’ of the circle, which has double the the value compared to mathematical π. This gives the quarter arc a value of ½ π.

At the top of the above diagram we see the fractal relationship between the two forms. On the left, the Silver

mean fractal projects the diagonal (blue line) line 1.5 units of the √2 fractal’s side-length. Notice the 1.5 2 =

2.25, which appears in the number equation, therefore ((2π-6) × 1.5 2 ) × e = √3. Whereas when the √2 fractal is

superimposed over the quarter square of the silver mean fractal, it produces a diagonal (yellow line) that

defines the bottom left corner square of a second √2 fractal placed next to it, which defines its relative unit

length.

K^{(iy + x)} = +1^{(iy + x)} > +∞^{(iy + x)}

**Solve for K = +1 ^{(iy + x)}**

+1^{+}^{2}^{∞} = +1

+1^{+}^{∞+1} = +1

+1^{+}^{∞ -1} = +1

+1^{+2} = +1

**+1 ^{0 } = 0**

+1^{-2} = -1

+1^{–}^{∞ +1} = -1

+1^{–}^{∞ -1} = -1

+1^{–}^{2}^{∞} = -1

**S****olve for K = +****∞-****1**^{(iy + x)}

+∞-1^{+}^{2}^{∞} = (+2∞-2)^{2}

+∞-1^{+}^{∞+1} = (+∞-1)^{2}

+∞-1^{+}^{∞ -1} = (+∞+1)^{2}

+∞-1^{+2} = (+∞-1)^{2}

**+****∞-****1**^{0}** = 0**

+∞-1^{-2 } = (-∞+1)^{2}

+∞-1^{–}^{∞ +1} = (-∞-1)^{2}

+∞–1^{–}^{∞ -1} = (-∞+1)^{2}

+∞-1^{–}^{2}^{∞} = (-2∞+2)^{2}

# Solve for K * 1/K^{(iy + x)}

**K=+1 ^{(iy + x)}**

1 * (1 / +1) = +1/+1 = +1

1 * (1 / +1) = +1/+1 = +1

1 * (1 / +1) = +1/+1 = +1

1 * (1 / 0) = +1/0 = 0

1 * (1 / –1) = +1/-1 = –1

1 * (1 / –1) = +1/-1 = –1

1 * (1 / –1) = +1/-1 = –1

**K=+∞-1 ^{(iy + x)}**

+∞-1 * (1 / (+∞-1)^{2}) = (+∞+1)

+∞-1 * (1 / (+∞+1)^{2}) = (+∞–1)

+∞-1 * (1 / (+∞-1)^{2}) = (+∞-1)

+∞-1 * (1 / (∞-1^0) = 0

+∞-1 * (1/ (-∞+1)^{2}) = (-∞-1)

+∞-1 * (1/ (-∞-1)^{2}) = (-∞-1)

+∞-1 * (1/ (-∞+1)^{2}) = (-∞+1)

First we solve for K in the infinite producing potential extremes. K can equal any value from +1 to ∞-1 (as the number Zero is excluded from the set). We then solve for both types of K with to the power of (x + iy), and then do the same for the reciprocal. In this way we establish the relationship of K to 1/K which is called the ‘Bounce Point of the number in 4D Mathematics. The next step is to find the instances when F is equal to K, however, from the above table we are able to establish the 4th Dimensional proportions to solve the problem using a simple 4D calculator. See video Example.

**F= +1**

+1 * +2∞ = +2∞

+1 * +∞ +1 = +∞ +1

+1 * +∞ = +∞

+1 * +2 = +2

+1 * 0 = 0

+1 * -2 = -2

+1 * –∞ +1 = –∞ +1

+1 * –∞ -1 = –∞ -1

+1 * –2∞ = –2∞

**F= ∞**

+∞ * +2∞ = +2∞^{2}

+∞ * +∞ +1 = +∞ +1^{2}

+∞ * +∞ = +∞^{2}

+∞ * +2 = +2∞

+∞ * 0 = 0

+∞ * -2 = -2∞

+∞ * –∞ +1 = –∞ +1^{2}

+∞ * –∞ -1 = –∞ -1^{2}

+∞ * –2∞ = –2∞^{2}

If f may be in any state from 1 to infinity, there will be a specific point where it will fall into equilibrium

with k, so that when multiplied by its reciprocal to the power (x + iy) the result will be equal to f * (x +

iy). This produces 2 solutions, one for f=+1 where f = R, and another f=∞ Where f = R.

The table below show the above to value sets combined in a 4 tables, with the order maintained.

Table 1: f=+1 and k=+1 (top left), show a zero solution where +∞+1 is found, which is the 1:1 solution.

Table 2: f=+1 and k=+∞ (top right) shows a Zero solution where both f=+∞ and k+∞.

The light blue squares on these table show where each on appears reflected over two axis.

Table 3: f=+∞ and k=+1 (bottom left), show the zero 2 solution, which have no Zero solution, but

instead exhibits +∞+1 solution, as a result of the squaring of zero, which produces the number cross

in space, equivalent to the complex plane but in 2D space. This is because if f moves to infinity, then

normally k will also need to change in order to remain in equilibrium with infinity. The significance of

+∞+1 the reduction of the infinite density of infinity, by increasing the whole number line by one unit

beyond infinity. This expansion can only be compensated for by the expansion of 1D linear number

space into the 2D number plane, of which there are exactly two options, the square and the triangle.*

Finally, table 4 (bottom right), show the values when f=+∞ and k=+∞, which is the maximum values

for each function. Here, we find the ratio ∞:∞ 2 where the number ∞ replaces the position of ZERO on

the table above, as k:f fall into an x 2 relationship.

Notice, that on the bottom table the f numbers (green row) are squared from the numbers above. The

table above is the linear solution, and the pair of table below is the 0 2 ‘Number Cross’ solution.

As conversational mathematics has excluded the possibility of ‘negative’ square numbers, we have

maintained f in the positive. However, in 4D mathematics, the negative function as positive and

negative orders do matter in 4D. See ‘Order Matters’ Principles of 4D mathematics.

*For more information, see, 2 types of regular 2D space, Squaring triangles and squares

This table represents a 4D blueprint of the solutions to the Riemann Hypothesis. From this table we can incorporate k=1 and k=∞ into a 4th dimensional calculator, and solve for each state.

THE

# Conclusion

## How Does this Solve the Riemann Hypothesis?

We can solve the problem thorugh simple geometry, whereby the Golden and Silver Ratio are employed to demonstrate how the complex number plain can be folded, and yet still offer an analytic continuation that is geometrically coherent. The geometric solution is explained in more detail in terms of infinite density function and 4D mathematics which offers a novel new kind of numerical space that does not express √-1 in the traditional sense.

## The complex plain VS infinite density function

In 4D maths we employ the conept of infinite dencity function, which explores the nature of infinity from the perspective of a unit length. When a line is increase in length to its infinite density decreases relative to the numerical space between zero and one. Root number are geometrically positioned at a 45° angle to a square, defining the density function between to number lines placed at 90° to each other. The complex plain places the root value of -1 at a 90° angle, which in turn leads to the ‘unit circle, which is unable to represent reciprocal numbers in the correct infinite density (Aleph 0.5). Therefore the critical strip will always act as a zero point instead of the zero line.

#### Carry On Learning

###### This article is part of our new theory, ‘Maths of Infinity‘

###### Read the main article or browse more interesting post from the list below

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## YOUR QUESTIONS ANSWERED

Got a Question? Then leave a comment below.

###### Question?

I did not quite get the part where you say “double the result we create the whole number value for the next cardinal step”?

###### ANSWER?

Beginning with 1/1=1, we double it to get 2. Next we add 1/2 + 2/2= 1.5. double it to get 3, the next cardinal number. Then 1/3 + 2/3 + 3/3 = 2, which when doubled creates the next cardinal number 4, and so on.

###### Question?

Isn’t the result of 1/1 the same as 2/2?

###### ANSWER?

Yes it is. However the calculation that generates the result is different. When we perform mathematical calculations we like to include the quality of the equation itself. In this way we can establish a pattern that becomes predictable, i.e: an ordered set.

## STILL Looking for Answers?

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