December 27, 2014
Hiroshi Michiwaki
Basis for distance and angle, and division by zero
Theorem 1: When using radius set R and arc length set A, and angle set Θ is defined bydivision A/R,holds.
Theorem 2: The center of a rotating body does not rotate.
Proof: All sets X include the empty set ∅, in other words,𝑋⊇∅(1)

Therefore, radius set R is
R = {r} ∪ ∅R

(2)

if element r of R is used as {r ∈ R|0 < r}. However, ∃∅R∀{r ∉ ∅R|0 < 𝑟}. Similarly, the arc
length set A is
A = {a} ∪ ∅A(3)if element a of A is used as {a ∈ A|0 < a ≤ 2𝜋𝑟}. Here, ∃∅A∀a{a ∉ ∅A|0 < a ≤ 2πr}, and π isthe ratio of the circumference of a circle to its diameter. Here, if the radius set R is fixed,
considering the mapping f that maps the arc length set A to angle set Θ,
f : A → Θ (4)it clearly makes a bijection. This is because the arc length set A satisfies Equation (3). On theother hand, if the element θ of the angle set Θ is {θ ∈Θ|0 < θ ≤ 2𝜋}, from Equation (1),Θ = {θ} ∪ ∅Θ(5)Here, ∃∅Θ∀θ{θ ∉ ∅Θ|0 < 𝜃 ≤ 2π}. At this time,
f:a→θ(6)is clearly a bijection. On the other hand, if the arc length set A is an empty set ∅A, then theangle set Θ will clearly also be an empty set ∅Θ, and the opposite also holds, as bijectionf :∅A → ∅Θ (7)is established. Therefore,f : {a}∪∅A →{θ}∪∅Θ∴ 𝑓: A → Θ (8)makes a bijection.
On the other hand, if the arc length set A is fixed, if we consider mapping 𝑔 that maps theradius set R to the angle set Θ,𝑔: R  Θ(9)this mapping is clearly not a bijection. We will show this below.
First, the radius set R satisfies Equation (2). On the other hand, assuming Equation (5), theangle set Θ
𝑔: r → θ(10)is clearly a bijection. However, when the arc length set A is fixed by the empty set ∅A, fromEquation (7), the angle set Θ is also an empty set ∅Θ; therefore, if we assume that𝑔: ∅R → ∅Θ (11)holds, on the other hand, the inverse mapping (𝑔 − 1) of mapping 𝑔,𝑔 − 1: ∅Θ → ∅R(12)generally does not hold. In other words, Equation (11) is an injection. Therefore, Equation (9)is not a bijection.
Here, it is assumed that Equation (11) holds. Thus, let us show that Equation (11) does hold.
If a straight line that makes the radius set R is on the baseline B, the reference point O isshared. Thus, based on the definition, 𝛩 = ∅Θ = 0 is clear. Here, if the radius R is 0, the radius
set is clearly R = ∅R = 0. Here, empty set ∅ = 0 is handled by the standard definition of ageneral set.
Even if the radius set R is 0; i.e., the empty set ∅R, the radius set R shares the reference point
O on baseline L. At this time, clearly the arc length set A becomes the empty set ∅A. In otherwords, for mapping h,h: ∅R → ∅A

(13)

h−1: ∅A → ∅R(14)holds, butdoes not. On the other hand, from the relationship of the injectivity of Equation (13) and thebijectivity of Equation (8),∅R → ∅A → ∅𝛩(15)holds. Therefore, the injectivity of Equation (11) holds.
Now, based on the definition, the angle set Θ is stipulated asbased on the ratio of the radius set R and the arc length set A; therefore, if we apply Equation(15) to Equation (16),is obtained. As such, Theorem 1 is proven.
If radius R is constant, and both sides of Equation (16) are differentiated by time t, this can beexpressed as

Here, the left side is the angular velocity ω, while the differential form on the right side is thetangential velocity v. Thus, if we substitute these expressions, the above equation becomesEquation (19) expresses the relationship between the radius R of a rotating body, tangential
velocity v, and angular velocity ω, but the relationship in Equation (17) is clearly angularvelocity ω = 0 when R = 0, showing that the center of a rotating body does not rotate. As such,Theorem 2 is proven.