User:Trovatore/Arithmetic and exponentiation in various numerical structures

Many times, usually in the context of discussions about the expression $0^{0}$, I have wanted to express certain ideas that I think are relevant, but it would be too long-winded inline, and also I didn't want to spend the time. So I'm hoping that pointers to this essay may prove useful.

Opening remarks

It is often claimed that the natural numbers $\mathbf {N}$ form a subset of the integers $\mathbf {Z}$, which in turn form a subset of the rational numbers $\mathbf {Q}$, which in their turn form a subset of the real numbers $\mathbf {R}$, and the reals a subset of the complex numbers $\mathbf {C}$. In symbols this would be $\mathbf {N} \subseteq \mathbf {Z} \subseteq \mathbf {Q} \subseteq \mathbf {R} \subseteq \mathbf {C}$.

For most purposes this is true.

However, it's worth noting that in any of the most common implementations of mathematics in set theory, it is not literally true. In a typical implementation, the natural numbers are identical to the finite ordinal numbers.

Then the integers are represented ordered pairs of natural numbers, for example $(n,m)$, intended to be understood as $n-m$, up to the equivalence relation $(n_{0},m_{0})\sim (n_{1},m_{1})\iff n_{0}+m_{1}=n_{1}+m_{0}$. (The universe of the structure $\mathbf {Z}$ is then the quotient of $\mathbf {N} \times \mathbf {N}$ by $\sim$.)

Note that this already breaks the subset relationship from the first paragraph. For example, the natural number 0 is the finite ordinal number 0, which is literally simply the empty set. However, the integer zero is an infinite set, namely $\{(0,0),(1,1),(2,2),\ldots \}$. Therefore the natural number 0 is not an integer, so $\mathbf {N} \not \subseteq \mathbf {Z}$.