3 ms·
The Dirac solution doesn’t involve gamma, just N square roots and 2 logarithms.
by vlovich123 2y ago
The Dirac solution doesn’t involve gamma, just N square roots and 2 logarithms.
- Biganon 2y agoBut the square root has a hidden operand, 2. We don't write it because by convention the default root is 2, but that still feel like cheating to me.
- necovek 2y agoThis is why people have brought up the successor function too: A + B, is, by definition B applications of Succ() on A: A + B = Succ(Succ(Succ(...Succ(A)))) So using your own argument, we could say that using '+' is simply a convention on how we can write down the above — if we insist on spelling "conventional" things out, we must be able to use the underlying elementary function[]. Or isn't a factorial n! really n(n-1)...21, so all those numbers spelled out? The mathematical root probably first appeared as a square root and was later extended to support other exponents. But is there any fun in this? As noted elsewhere, the game is in finding the rules, and a solution within those rules. []Since all the natural numbers other than 1 are defined using a Succ() functions, there's a trivial solution. But if we only limit ourselves to this most elementary operation, we can't get a 1 because that's an axiom in itself ("There exists 1" or "There is a set of cardinality 1").
- orthoxerox 2y agoI think the correct definition of A + B is A + 0 = A (or A + 1 = Succ(A), if you insist 0∉ℕ) A + Succ(B) = Succ(A) + B
- necovek 2y agoNot sure what you mean with "correct", because "correct" is an equivalence class of slightly different definitions? Eg. https://en.wikipedia.org/wiki/Peano_axioms https://en.wikipedia.org/wiki/Peano_axioms has, instead of both yours and mine: A + 0 = A A + Succ(B) = Succ(A + B) They would all be proven in the same manner, though some might be slightly stronger in relation to commutation, making some proofs easier off the bat.