- #1
Andraz Cepic
- 31
- 3
So I was just writing a proof that every natural number is either even or odd. I went in two directions and both require that 1 is odd, in fact I think that 1 must always be odd for every such proof as the nature of naturals is inductive from 1.
I am using the version where 1 is the smallest number of natural numbers. Using the one with 0 shouldn't be a problem.
Now by the definition of oddness, a number n in naturals is odd iff n = 2k + 1 for some k in naturals.
Thereby 1 "could" be 1 = 2*0 + 1, however 0 does not exist from Peano's axioms, thus such an operation is not defined in naturals, moreover using n = 2k - 1 as the definition of oddness is also not sufficient, as the natural numbers have no notion of -1.
Therefore, my question is what am I missing here, as I do not yet know almost any abstract algebra to fully understand such rigor here(I'll be a math freshman starting this October), but this question is getting me nervous :D. Could it be that with these axioms we cannot define evenness or oddness of the members of the set they describe?
I am using the version where 1 is the smallest number of natural numbers. Using the one with 0 shouldn't be a problem.
Now by the definition of oddness, a number n in naturals is odd iff n = 2k + 1 for some k in naturals.
Thereby 1 "could" be 1 = 2*0 + 1, however 0 does not exist from Peano's axioms, thus such an operation is not defined in naturals, moreover using n = 2k - 1 as the definition of oddness is also not sufficient, as the natural numbers have no notion of -1.
Therefore, my question is what am I missing here, as I do not yet know almost any abstract algebra to fully understand such rigor here(I'll be a math freshman starting this October), but this question is getting me nervous :D. Could it be that with these axioms we cannot define evenness or oddness of the members of the set they describe?