But it's not as if it's the same steps, just more of them. You're ignoring the fact that different steps have to be taken to perform the addition.Still the same old addition at every step, though. Just more of it.
Let me be very clear about this. Addition is defined in terms of the set of things that can be added, and by the axioms that addition obeys. Changing either the set or any of the axioms changes the operation. This is indisputable. It's the reason why 1 + 1 = 0 in a boolean ring, but 1 + 1 = 2 in an integer ring. They can't both be the same "+" operation, otherwise we'd have a contradiction in math. All of math would break down!
The way we add things depends on what we're adding. For example, if we're adding the velocities of two objects, we're performing vector addition. This is a different type of addition than adding, say, apples. Indeed, relativistic velocity addition isn't even commutative! You say that "addition is addition", but if \( u \) and \( v \) are the velocities of two objects, then relativity tells us that \( u + v \ne v + u \). This can't be the case for adding apples!
Illusory and fallacious implication? Then it should be easy to contradict. Go ahead and try. ;--)I tell you, while that might be a "true statement" on the "surface," it is an illussory fallacious implication at its core and is in violation of the core laws of reason, and clouding the true nature of numbers. This makes it sounds like ℚ is more foundational than ℕ, which has Herr Kronecker rolling in his grave.
What you don't seem to get is that all numbers are "derivative" abstractions. We derived the numbers in ℕ from abstracting the counting process. We derived arithmetic over ℕ from counting subsets: I have two rocks in my left pocket, three rocks in my right pocket, so I have a total of five rocks: 2 + 3 = 5. Take three rocks away from my five and I'm left with two: 5 - 3 = 2.A MUCH better question is: How would you prove every element of ℚ is a derivative of compounded combinatorial elements of, and processes upon ℕ, the counting numbers that are the very building blocks of every fractional expression? And if you could, would you not then conclude ℚ is derivative of ℕ, rather than ℕ a subset of ℚ?
Arithmetic over ℕ gave us the capability to generalize problems, and so find general solutions to problems. Once we understood this abstraction, we realized that the numbers in ℕ were insufficient to answer all of the different types of problems we could pose. The history of the number sets is the history of humanity learning how to answer increasingly general mathematical questions:
\[ \begin{align} x + 1 = a \qquad &\to \qquad \mathbb{N} \\ x + a = b \qquad &\to \qquad \mathbb{Z} \\ ax + b = c \qquad &\to \qquad \mathbb{Q} \\ ax^2 + bx + c = d \qquad &\to \qquad \mathbb{R}, \mathbb{C} \end{align}\] It's not that the solutions to \( x + 1 = a\) are more fundamental than the solutions to \( x^2 + 1 = a\), it's just that we need more numbers to be able to solve the latter than the former.
What you're describing doesn't actually work. For example, we can think of a number in \( \mathbb{C} \) as a pair of two numbers in \( \mathbb{R} \). Taken strictly as sets, with no additional structure, then \( \mathbb{C} \) is indeed isomorphic to \( \mathbb{R} \times \mathbb{R} \). We can even include more structure to the sets by defining addition between elements of the sets and scalar multiplication, in which case \( \mathbb{C} \) as a vector space is isomorphic to \( \mathbb{R}^2 \) as a vector space. We know this because both vector spaces have two basis vectors; they are two-dimensional vector spaces, and any n-dimensional vector space is isomorphic to any other n-dimensional vector space.What you're calling pre-production I call "breaking down into their TRUE elemental states" to perform the multiple level of addition (same addition, just multiple levels).
However, to do general arithmetic on these sets, we need to introduce even more structure and treat \( \mathbb{C} \) and \( \mathbb{R}^2 \) as rings or fields. And as soon as we do this, we find that \( \mathbb{C} \) and \( \mathbb{R}^2 \) are incompatible. Indeed, \( \mathbb{C} \) is a full-fledged field, while \( \mathbb{R}^2 \) is not even an integral domain (because it has zero divisors). In particular, multiplication over \( \mathbb{C} \) is incompatible with multiplication over \( \mathbb{R}^2 \).
So, there is no sense in which arithmetic over \( \mathbb{C} \) can be "broken down" to arithmetic over \( \mathbb{R} \). We can repeat this story with arithmetic over the rings of nxn matrices, or arithmetic over boolean rings, or whatever. Fundamentally, these objects are all different, and so their arithmetic is necessarily different.
At the deepest essence, numbers are abstractions of abstractions. We find that there are logical consequences to these abstractions, and we categorize them into number sets based on these consequences (their properties).No, I want to see what "what we are calling numbers" are at their deepest essence . . .
If numbers are molecules, then their atoms are particular state configurations.The numbers are molecules — I want to study the atoms to figure out how the molecules behave.