April 17, 2017
Hiroshi Michiwaki
The Density Principle Solved with a Graph, and Division by Zero
Theorem It may be beneficial to graph the density principle examined in the Division by Zero paper#39 (“The Density Principle and Division by Zero with the Resulting Remainder”) and use the graphto demonstrate the principle. To this end, a coordinate system is considered with area along the x-axis
and mass along the y-axis. In this coordinate system, let the maximum of the x-axis be 𝑠₀, which is thesurface area of A₀, the bottom surface of a given cylinder. If some point along the x-axis between theorigin and 𝑠₀ (the bottom surface area of the cylinder) is the target surface area 𝑠, the mass
corresponding to that area, that is, the mass 𝑀in sitting on top of the target surface A, is considered.
The mass on top of A₀ is the total mass 𝑀. As the slope σ of the graph of the mass 𝑀in with respect tothe target surface area 𝑠 is a function of 𝑠, it is expressed as σ = σ(𝑠). If now the cylinder has auniform mass distribution ρ, then at every point from where the area is greater than 0 up to 𝑠₀ (thebottom surface area of the cylinder), the target surface area 𝑠 and the partial mass atop it, that is, themass 𝑀in sitting on top of the target surface, is proportional with constant σ.

Fig 1. Density principle and division by zero: σ(𝑠) = const > 0
It is now assumed that starting from the state where the target surface area 𝑠 is at its maximum, equal
to the bottom surface area of the cylinder 𝑠₀, it is gradually shrunk. After shrinking it to a suitable
extent, we obtain what is shown in Fig. 1, where the mass 𝑀in sitting on the target surface area 𝑠 canbe determined by the linear equation 𝑀in=σ𝑠, and the total mass 𝑀 can be determined by 𝑀=σ𝑠+ 𝑀out,
a constant linear function of 𝑠.
To simplify the equations discussed so far, they will be reexamined by letting the bottom surface areaof the cylinder 𝑠₀ = 100, the total mass 𝑀 = 100, and the linear slope σ = 1 (𝑠 > 0). When the target
surface area 𝑠 is at its maximum, i.e., equal to the bottom surface area of the cylinder 𝑠₀, the ratio of
the total mass 𝑀 to the target surface area 𝑠, by way of compound numbers, iswhere the leading number 1 is the value of σ. Then, if 𝑠 is shrunk by only 1/100,

Using compound numbers as in the previous expression, it can be seen that the ratio reaches a valueslightly larger than 1. The leading number 1 is the value of σ as well.
As these formulations are rather counter-intuitive, the relations among the divided, divisor, quotient,and remainder will now be reorganized. Here, the initial numerator corresponds to the divisor, thedenominator to the divisor, the leading number to the quotient, and the subsequent numerator to theremainder. The divisor does not change; thus, this relationship is expressed as

In this sequence of calculations, the first term in the expression following the arrow ⇒ corresponds tothe size of the mass 𝑀in sitting on the area, which is represented by the vertical blue line in Fig. 1(parallel to the mass axis), and the second term corresponds to the mass 𝑀out sitting outside the area,that is, the remainder, which is represented by the vertical green line (parallel to the mass axis).
Thus, if 𝑠 = 0, thenAs is clearly seen in Fig. 2, the target surface area 𝑠 is 0; thus, 𝑀in, the mass on it, is also 0. Moreover,𝑀out, the mass outside it, accounts for the entire total mass 𝑀 and is thus 100.

Fig. 2. Density principle and division by zero: σ(𝑠) = 0
Furthermore, if the leading number is 1 in (3) as well, then

However, although ∞ × 0 implies that there is not even a single ∞, it is undefined as its magnitude isstill unclear. That is, (4) is absurd, and the equation does not hold true. This will happen even if theleading number is changed to any real number other than 0. This shows thatis absurd. Consequently, it is clear that 100 = □ × 0 + 100 is true if and only if □ = 0, yielding 100 = 0
× 0 + 100. Based on this, it is clear that

What follows is an alternative proof for this.
For two real numbers 𝑎, b ≥ 0, let

and

Then,

Hence, (6) is obtained:

Furthermore, from the above hypothesis, the following holds for two real numbers 𝑎, b ≥ 0 ∧ 𝑎 ≥ 𝑛𝑏
(𝑛 ∈ N):Thus, if 𝑏 = 0, thenThis shows that when (3) is written in the form of a compound fraction,
dividend 100 converted dividend[A1] 0 + remainder 100
remainder 100== quotient 0
divisor 0
divisor 0
divisor 0⇒ dividend 100 = quotient 0 divisor 0 + remainder 100(3')

the quotient is uniquely determined to be 0, based on (6). Consequently, (3) holds true. Generally, thefollowing is true:= quotient 0 ⋯ remainder 𝑎 ∧ dividend 𝑎 = quotient 0 × divisor 0 + remainder 𝑎(12)
It should be noted that the converted dividend in (3′) is the dividend obtained after the original
dividend splits into the sum of a secondary dividend and the remainder. When the divisor is not 0, thedividend can be split arbitrarily; however, when the divisor is 0, the only way to split the dividend sothat the number can be turned into a compound fraction results in a converted dividend of 0. That is,when the divisor is 0, the quotient is uniquely 0, and in terms of reducible set theory, the remainder is
uniquely equal to the dividend. This may be redundant, but if 𝑏 = 0 when 𝑎 ≠ 0, then 𝑥 ≠ 0, and thus

Considering 𝑥 = 1 yields

This shows that a ratio (comparison) with 0 only makes it possible to compare the two qualitative
states of “being” and “non-being.” This shows that a given size cannot be quantitatively measured
using the size itself as reference, because the dividend, which is the reference size, is 0. Furthermore,when there is a difference in the size of the numerator, no conclusion can be drawn. Nevertheless, itdoes show that there is a state of “being,” which is distinct from the state of “non-being.”
The notation of compound numbers is as follows. Fractional equalities will be denoted using three
real numbers 𝑎, b, as

and 𝑎 is called the leading number, 𝑏 the divisor, 𝑎𝑏+𝑐 the dividend, and 𝑐 the remainder. Then, if theleading number 𝑎 is called the general quotient, the general quotient 𝑎 that minimizes the remainder 𝑐is equivalent to the quotient 𝐴 of the previous division. That is,in which caseand the following relation between the set {𝑎} of general quotients 𝑎 and quotient 𝐴, as well as
between the set {𝑐} of remainders 𝑐 and 𝐶, is true:

Using the above notation for compound numbers, (7) can be written as

Then, Equation (8) will be establishedwhereupon if 𝑏 = 0,and thus (9)holds true. In this case, by (9), Equation (19) is uniquely expressed as

Moreover, by the definition of compound numbers, the leading number 0 in (20) is an element in theset of general quotients; moreover, it is the largest in the set, as (20) is uniquely true. It is also
equivalent to the quotient of the previous division. Consequently,

is uniquely true.
A coordinate system with area along the x-axis and mass along the y-axis will again be considered. In
this coordinate system, let the maximum of the x-axis be 𝑠₀, which is the surface area of A₀, thebottom surface of a given cylinder. Let A be the pressure receptor of a flat and perfectly rigid body,with a surface area 𝑠 at some point along the x-axis between the origin and 𝑠₀ (the bottom surface areaof the cylinder) such that A receives the total mass 𝑀on of a cylinder with a uniform base. That is, 𝑀on,the total mass acting on pressure receptor A, is considered. Of course, the mass on the underside of
the cylinder A₀ is equal to the mass 𝑀. Then, the slope σ of the graph of the mass 𝑀on with respect tothe surface area of the pressure receptor 𝑠 is expressed as a function of 𝑠, σ = σ(𝑠). Now, if thecylinder has a uniform mass distribution ρ and is a perfectly rigid body, then at every point from
where the area is greater than 0 up to 𝑠₀ (the bottom surface area of the cylinder), the target surface
area 𝑠 and the total mass 𝑀 of the cylinder sitting atop it (the mass 𝑀on acting on the pressure
receptor) are in a linear relationship with the size of σ(𝑠), that is, the slope of the graph σ, as itchanges.
It is now assumed that starting from the state where the surface area of the pressure receptor 𝑠 is at its
maximum, equal to the bottom surface area of the cylinder 𝑠₀, 𝑠 gradually shrinks. After it shrinks toa suitable extent, we obtain what is shown in Fig. 3, where the mass 𝑀on on the surface area of thepressure receptor 𝑠 remains constant and can be found by the linear equation 𝑀on = σ(𝑠)𝑠. That is, σ(𝑠)is inversely proportional to 𝑠. Of course, the total mass 𝑀 is expressed as the sum of 𝑀on, the mass on
the surface area of the pressure receptor A, and 𝑀off, the mass not on the surface area of the pressure
receptor, yieldingThis implies that the total mass 𝑀 can be expressed as

Fig. 3. Density principle and division by zero: [MATH]
To simplify the equations discussed so far, they will be reexamined by letting the bottom surface areaof the cylinder 𝑠₀= 100, the total mass 𝑀 = 100, and the linear slope σ(s) ≧ 1 (s > 0). When the target
surface area 𝑠 is equal to the bottom surface area of the cylinder 𝑠₀, to determine the ratio of the total
mass 𝑀 to the surface area of the pressure receptor 𝑠, if the dividend is split into the sum of two
numbers to minimize the remainder, it can be seen that, by way of compound numbers, it iswhere the leading number 1 is the value of σ. Then, if 𝑠 is shrunk by only 1/100,

Using compound numbers as in the previous expression, it can be seen that the ratio reaches a valueslightly larger than 1. The leading number 1.0̇1̇ is here the value of σ as well.
As before, the relations among the divided, divisor, quotient, and remainder below will be organized.
Here, the initial numerator corresponds to the divisor, the denominator to the divisor, the leading
number to the quotient, and the subsequent numerator to the remainder. The divisor does not change;
thus, this relationship is expressed asAccording to the density principle 0/0 = 0 as well as (21),is uniquely true. This also takes the exact same form as (3), which expresses Fig. 2 numerically. The
graph in Fig. 4 is similarly equivalent to Fig. 2 as well. The only difference between these twosystems is that the former has a surface outside the target to support the mass sitting outside it,whereas the latter has no surface outside the pressure receptor to support the mass sitting outside it.
This implies that in the former, the cylinder does not undergo any changes even when the target
surface area is 0, whereas in the latter, when the surface area of the pressure receptor is 0, there ceases
to be any surface capable of supporting any of the cylinder’s mass, causing the cylinder to fall.

Fig. 4. Density principle and division by zero: σ(𝑠)=0
So far, the density principle has been explained in terms of ratios and division by zero, using mass
surface density as an example. This could be called the ratio principle. Some examples will now be
used to deepen the understanding of the density principle (ratio principle) as seen in Figs. 3 and 4. The
density principle will be first considered followed by the ratio principle. For the former, adisappearing island is considered. This island has a total area of 100 ha and is home to 100 residents.
There are no births or deaths, nor is there immigration from other areas. However, there is one
problem: owing to global warming and sea level rise, the area of this island decreases by 1 ha every
year.
The current population density is 1 person/ha; in 20 years, it will be 1.25 people/ha. What will it be in
100 years?
Current population density:
[

]
[

]

= 1[people/ha]

[

]
[

]

= 1[people/ha] remainder 0 [people]

Population density in 20 years:
100[people]
100[people]0[people]== 1.25[people/ha]= 1.25[people/ha] remainder 0 [people]
100 − 1 × 20[ha]80[ha]
80[ha]

Population density in 50 years:
100[people]
100[people]0[people]== 2[people/ha]= 2[people/ha] remainder 0 [people]
100 − 1 × 50[ha]50[ha]
50[ha]

Population density in 90 years:
100[people]
100[people]0[people]== 10[people/ha]= 10[people/ha] remainder 0 [people]
100 − 1 × 90[ha]10[ha]
10[ha]

Population density in 99 years:
100[people]
100[people]0[people]== 100[people/ha]= 100[people/ha] remainder 0 [people]
100 − 1 × 99[ha]
1[ha]
1[ha]

That is, 100 people on 1 ha of land. What will happen 1 year after that? After 100 years, the island
will be completely submerged; hence, its surface area will be 0. However, as there are no births,
deaths, or immigration, the residents will have become sea-people floating in the sea. This is the sameas trying to determine the population density if there were no island to begin with. That is, as there isno “island population density” in the water, where there is no island, the population density is null φwhether people are there or not (in the sense of the population density of the island). This is the same
as being zero. Thus, the case where there is no island to begin with is mathematically equivalent to thecase where the island disappears. Consequently,
Population density in 100 years:
100[people]
100[people]
100[people]== 0[people/ha]= 0[people/ha] remainder 100 [people]
100 − 1 × 100[ha]
0[ha]
0[ha]

Moreover, for the ratio principle, the relationship between the mass and moving velocity of a neutrino
will be examined. Let the y-axis be the rest mass of the particle 𝑚₀, and the x-axis be the inverse of
the Lorentz factor corresponding to the particle’s velocity 𝑣, that is,(which will hereafter be referred to as the Lorentz quantity). Consequently, if the velocity 𝑣 is
assigned a value within the range 0 ≤ 𝑣 ≤ 𝑐, where 𝑐 is the speed of light, the value of the x-axis will
be between 0 and 1. If it is assumed that this graph passes through the origin, its slope represents (𝑣),the mass of the particle at velocity 𝑣, as Fig. 5 shows. That is,Fig. 5. Relationship between a neutrino’s velocity 𝑣 and its mass (𝑣) (0 ≤ 𝑣 < 𝑐)When the particle’s velocity 𝑣 is equal to the speed of light 𝑐, the Lorentz quantity is 0, the graph isequal to the y-axis, and its height is equal to the rest mass 𝑚₀. What is (𝑐), the mass of the particle
when its velocity 𝑣 is the speed of light 𝑐?
For particle velocity 𝑣 in the range of 0 ≤ 𝑣 <𝑐,relativistic mass 𝑚(𝑣) ==non − relativistic mass 𝑚𝑣
Lorentz quantity 1 −𝑐non − relativistic mass 𝑚 + non − relativistic mass 0𝑣
Lorentz quantity 1 −𝑐non − relativistic mass 0= non − relativistic mass 𝑚(𝑣)_𝑣
Lorentz quantity 1 −𝑐= non − relativistic mass 𝑚(𝑣) remainder non − relativistic mass 0

Therefore, if 𝑣=𝑐,
relativistic mass 𝑚(𝑐) =non − relativistic mass 𝑚𝑐𝑐non − relativistic mass 0 + non − relativistic mass 𝑚=
Lorentz quantity 0
non − relativistic mass 𝑚= relativistic mass 0_
Lorentz quantity 0= non − relativistic mass 𝑚(𝑐) = 0 remainder non − relativistic mass 𝑚
Lorentz quantity 1 −Consequently, the increase in mass (𝑐) of a particle due to the relativistic effect disappears when theparticle moves at the speed of light 𝑐, that is, its relativistic mass (𝑐) is 0, and its rest mass 𝑚₀ is
measured as the non-relativistic mass.