I Proof that 1 is an odd number using Peano Axioms of naturals

[Andraz Cepic]

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?

 

[andrewkirk]

If your natural numbers start at 1 rather than 0, I don't think you'd be able to succeed by defining an odd number as $2k+1$, for the reason you pointed out.

However the following problem:

moreover using n = 2k - 1 as the definition of oddness is also not sufficient, as the natural numbers have no notion of -1.

is easily fixed.

Instead of defining an odd number to be a number $n$ for which there exists a number $k$ such that $n=2k-1$ which, as you point out, is undefined, define it to be a number $n$ for which there exists a number $k$ such that $n+1=2k$. Then $n=1$ is odd with $k=1$.

 

[PeroK]

I think you're getting nervous unnecessarily. In maths things are what they are defined to be. If we take $\mathbb{N}$ not to include $0$ and define odd and even as $2n+1$ and $2n$ respectively, then this doesn't cover the number $1$.#

Note that if the definition of odd is $2n+1$ for some $n \in \mathbb{N}$, then nothing you can do is going to prove that $1$ is odd from that; precisely because $1$ fails that definition.

We see that $2, 4, 6 \dots$ are even and that $3, 5, 7 \dots$ are odd.

We would then have to define $1$ to be odd. Later, when we have $0$ we can redefine odd and even and take care of $1$ then.

Note that an inductive argument can just as easily start with $3$ as with $1$. You could prove that $3, 5, 7 \dots$ are odd without saying anything about $1$.

 

[Andraz Cepic]

I think you're getting nervous unnecessarily. In maths things are what they are defined to be. If we take $\mathbb{N}$ not to include $0$ and define odd and even as $2n+1$ and $2n$ respectively, then this doesn't cover the number $1$.#

Note that if the definition of odd is $2n+1$ for some $n \in \mathbb{N}$, then nothing you can do is going to prove that $1$ is odd from that; precisely because $1$ fails that definition.

We see that $2, 4, 6 \dots$ are even and that $3, 5, 7 \dots$ are odd.

We would then have to define $1$ to be odd. Later, when we have $0$ we can redefine odd and even and take care of $1$ then.

Note that an inductive argument can just as easily start with $3$ as with $1$. You could prove that $3, 5, 7 \dots$ are odd without saying anything about $1$.

Hmm so I was right to hypothesise that we cannot prove this theorem with naturals starting at 1 as 1 fails the definitions of oddness and evenness, except of course we define oddness of n as n+1=2k for some k, then 1 satsfies it perfectly. I did not even think about this solution!

Thank you :)