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

  • #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?
 
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  • #2
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:
Andraz Cepic said:
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##.
 
  • #3
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##.
 
  • #4
PeroK said:
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 :)
 
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