How would that work? For starters, there are no even numbers and addition is a ternary operation. So the natural numbers here are 1, 3, 5, 7, and so on.
There are five basic operations: addition (5 + 3 + 1 = 9), subtraction (5 – 3 – 1 = 1), middle (5 + 3 – 1 = 7, or 5 – 3 + 1 = 3. There are multiple ways of writing it.), multiplication (5 * 3 = 15), and division (51 / 3 = 17). Note that subtraction and middle could also be written using negative numbers.
A big issue you may notice is that we can’t just pick an arbitrary predicate and count how many numbers satisfy it. This could probably be fixed with ternary logic, but I’ll just ignore it and move in a different direction:
Adjusting the field axioms to this type of number system (we will call them threelds), we get:
1) a + b + c = a + c + b = c + a + b. (Commutativity of addition)
3) a + -a + b = b (Inverse of addition)
5) a + b + (c + d + e) = a + (b + c + d) + e = (a + b + c) + d + e (Associativity of addition)
7) 1 * x = x (Identity)
9) x * y = y * x (Commutativity of multiplication)
11) x * (y * z) = (x * y) * z (Associativity of multiplication)
13) (a + b + c) * x = a * x + b * x + c * x (Distributivity)
15) x * 1/x = 1. (Inverse of multiplication. Note that this applies to all numbers due to the nonexistence of zero.)
Here’s a proof that the only threeld of odd size is the threeld of order 1:
Because of the commutativity and associativity of addition, the sum of all elements in a set makes sense if that set has an odd size.
Consider the sum S of the set of all elements in the threeld. Multiplication is a group, and thus multiplication by any element permutes them. Thus, S is preserved when multiplying by any element. That is, for any elements a and b, a * S = b * S. Dividing by S on both sides, we get that a = b for any elements a and b, and thus that there is only one element.
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