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| ▲ | Sharlin 19 minutes ago | parent | next [-] |
| Subtraction is not closed over positive integers, which is untidy. The point of Tarski’s conjecture was to propose a minimal number of axioms and operations, AFAICS they define the standard semiring of positive integers (with the natural definition of exponentiation added). (Edit: positive integers aren’t exactly a semiring because 0 is excluded, although some authors do define a semiring without the requirement of an additive identity element.) |
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| ▲ | munchler 10 minutes ago | parent [-] | | Well, yes, but negative numbers are also well known to every high school math student. | | |
| ▲ | Sharlin 5 minutes ago | parent [-] | | Sure. But "High School Algebra (Excluding Subtraction) Problem" isn’t as catchy a name. | | |
| ▲ | brookst a minute ago | parent [-] | | They subtracted the subtraction exclusion in the name of simplicity? |
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| ▲ | stevefan1999 19 minutes ago | parent | prev | next [-] |
| I'm not sure, but maybe it is due to that the expression a - b can be replaced as a + (-b)? Similarly, I think a * b and a / b can be replaced with the same trick, but then I realized it may not work on non-abelian, or where multiplicative inverse is not available... |
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| ▲ | Sharlin 17 minutes ago | parent [-] | | We’re in the semiring of positive integers, so there are no additive (or multiplicative) inverses. |
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| ▲ | woadwarrior01 12 minutes ago | parent | prev | next [-] |
| Because subtraction is not a total operation on positive integers. Negative numbers leave the domain. |
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| ▲ | Transformanshen 14 minutes ago | parent | prev [-] |
| The subtraction point is interesting but I don't think it makes the result disappointing. The whole point of Tarski's problem is what follows from that very restricted set of elementary identities so finding the exact minimum countermodel under those rules still seems like a pretty satisfying result. |
