According to standard mathematics, the Natural Numbers are given.
Moreover, they are given as a (completed) Infinite Set. This set is commonly
denoted as:
$$
\mathbb{N} = \left\{ 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,\,
\dots \right\}
$$
Theorem. The set of all natural numbers is a completed infinity.
Proof. A set is infinite (i.e. a completed infinity) if there
exists a bijection between that set and a proper subset of itself. Now consider
the even naturals. They are a proper subset of the naturals and a bijection can
be defined between the former and the latter. The "numerosity" of the evens is equal to the "numerosity" of the naturals. There are "as many" evens as there are naturals. As follows:
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 ... | | | | | | | | | | | | | | | | | | | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...Galileo's Paradox. The cardinality of the squares equals the cardinality of the naturals. There are "as many" squares as there are naturals. Proof. There exists a bijection between the naturals and the squares. The numerosity of the squares equals the numerosity of the naturals:
1 2 3 4 5 6 7 8 9 10 11 12 13 .. | | | | | | | | | | | | | 1 4 9 16 25 36 49 64 81 100 121 144 169 ..The cardinality of the powers of $7$ equals the cardinality of the naturals. There are "as many" powers of $7$ as there are naturals. Proof. There exists a bijection between the naturals and these powers:
1 2 3 4 5 6 7 8 9 10 .. | | | | | | | | | | 7 49 343 2401 16807 117649 823543 5764801 40353607 282475249 ..In general. Let there be defined a function $\;A(n) : \mathbb{N} \rightarrow \mathbb{N}$ . Assume that $\,A(n)\,$ is a sequence which is monotonically increasing with $\;n\;$ . Then we have the following bijection:
1 2 3 4 5 6 7 8 9 .. | | | | | | | | | A(1) A(2) A(3) A(4) A(5) A(6) A(7) A(8) A(9) ..It can be concluded that the cardinality of the set $\;\left\{n\in\mathbb{N}:A(n)\right\}\;$ is $\;\aleph_0\;$, which is the cardinality of the naturals. Examples are:
- $A(n) = 2\cdot n\;$ (the even numbers)
- $A(n) = n^2\;$ (Galileo's paradox)
- $A(n) = 7^n\;$ (powers of seven)
1 2 3 4 5 6 7 8 9 10 .. 48 49 50 .. 343 .. 2401 .. 0 0 0 0 0 0 1 1 1 1 1 2 2 3 4 : D(n) A(0) A(1) A(2) A(3) A(4)Let $D(n)$ be the number of $A(m)$ values (count) less than or equal to $n$, where $(m,n)$ are natural numbers. Then we have the following theorem, tentatively called the Inverse Function Rule: $$ \lim_{n\rightarrow \infty} \frac{D(n)}{ A^{-1}(n) } = 1 $$ Proof. $A(n)$ is monotonically increasing with $n$ , therefore the inverse sequence $A^{-1}(n)$ exists in the first place. And it is monotonically increasing as well. Furthermore, we see that $\,D(n)\,$ is $\,m\,$ for $\,A(m) \le n < A(m+1)$ .
Consequently: $\;A(D(n)) \le n < A(D(n)+1)\;$ , hence $\;D(n) \le A^{-1}(n) < D(n) + 1\;$ , therefore $\;A^{-1}(n) - 1 < D(n) \le A^{-1}(n)\;$ .
Divide by $\;A^{-1}(n)\;$ to get: $\;1 - 1/A^{-1}(n) < D(n) / A^{-1}(n) \le 1\;$ . For $\,n\rightarrow \infty\,$ now the theorem follows, because $\;A^{-1}(n) \rightarrow \infty\;$ . Written as an asymptotic equality: $\,D(n) \approx A^{-1}(n) $ .
Examples. Revealing the Double Think.
- $A(n) = 2.n \;\Longrightarrow\;\lim_{n\rightarrow\infty} D(n)/[\,n/2\,] = 1 \;\Longrightarrow\; D(n) \approx n/2$
The numerosity of the evens is not equal but half the numerosity of the naturals - $A(n) = n^2 \;\Longrightarrow\;\lim_{n\rightarrow\infty} D(n)/\sqrt{n} = 1\;\Longrightarrow\; D(n) \approx \sqrt{n}$
The numerosity of the squares is the square root of the numerosity of the naturals - $A(n) = 7^n \;\Longrightarrow\;\lim_{n\rightarrow\infty} D(n)/[\,\ln(n)/\ln(7)\,] = 1\;\Longrightarrow\; D(n) \approx \log_7(n)$
The numerosity of the powers of 7 is the 7-logarithm of the numerosity of the naturals
Wouldn't even dare to ask about the more important thing, namely what is the "correct" way to no double think about numerosities: cardinalities or via the "inverse function rule" ?