9 ms·
But 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.
by Biganon 2y ago
But 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.