October 10, 2017
Hiroshi Michiwaki
Super-composite Numbers as a Class of Nonnegative Integers and Division by Zero
Definition When the number of nonnegative integers that can evenly divide a given nonnegative
integer is 1, that nonnegative integer is called an identity element; when it is 2, a prime number; andwhen it is 3 or more, a composite number. In particular, when it is 3 or more and the nonnegative
integer cannot be divided by 2, it is called an odd composite number; and if it can be divided by 2, it
is called an even composite number. Furthermore, a nonnegative integer that can be evenly divided by
all nonnegative integers is called a super-composite number (or super-even composite number). Then,the following theorem is true.
Theorem 0 is a super-composite number.
Proof When 0 is divided by 0, according to the fundamental theorem of reducible set-theoreticaldivision by zero

(where ···𝑎 denotes the remainder of 𝑎), if 𝑎 = 0 in (1), then ···𝑎=0; thus, 0 is clearly evenly divisible
by 0.
If 𝑛 is any natural number, it is self-evident that

Based on the above, 0 is evenly divisible by all nonnegative integers. Consequently, 0 is a supercomposite number according to Definition. QED
Equation (1) implies that dividing 1 by 0 yields a remainder of 1. Thus, 1 is not divisible by 0, only by
1, and is an identity element according to Definition. Furthermore, dividing a positive prime integer
by 0 yields the integer as a remainder; being indivisible by only two numbers, 1 and itself, it is also aprime number according to the definition. Numbers other than 0, 1, and prime numbers yield the same
number when divided by 0 but are evenly divisible by 1 as well as 2 or more different prime numbers.
This implies that they are divisible by 3 or more nonnegative integers and are thus compositenumbers.
Note Let it be assumed that all nonnegative integers are expressed as the product of the highest powerof the different prime factors by which they are divisible. Considering the range of natural numbers, if
𝑁 is any natural number, 𝑝j the 𝑗-th prime in ascending order, and 𝑛j the largest nonnegative integer
satisfying 𝑛j ≥ 0 that is the highest power of the prime 𝑝j by which 0 is evenly divisible, a natural
number 𝑁 satisfies

as well asprovided that 𝑀 is a sufficiently large constant.
It is now assumed that 𝑁 is extended from natural numbers to nonnegative integers, that is, the zeroelement is added to the set 𝑁.
Based on the above theorem, 0 is a super-composite number that has all prime numbers 𝑝j as factors;
thus, 𝑛j, i.e., the highest power of prime factors 𝑝j, satisfies 𝑛j > 0. That is,is true. Here, ∞̇ is a point beyond infinity, or true infinity, and not the limit of the increase sequence.
Likewise, 𝑝j is not an extreme quantity called “infinity,” but rather a truly infinite quantity. As supercomposite numbers are also divisible by the composite number 𝑝jnj, which consists of the power of
each prime number 𝑝j, 𝑛j, i.e., the highest power of prime factors 𝑝j, has a truly infinite size.
Consequently, the super-composite number 0 can be expressed asThat is, ∞̇=0 is considered to be true.
This would also seem to make it obvious, in a sense, that nonnegative integers except 0 are notdivisible by 0. The reason for this is that although 0 has all nonnegative integers as factors (i.e., it has
an infinite number of divisors), all nonnegative integers except 0 (i.e., natural numbers) have only afinite number of divisors. Thus, there is no reason they should be divisible by 0, which has an infinite
number of divisors.