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

• I
• Andraz Cepic
In summary, the conversation discusses the proof that every natural number is either even or odd. The speaker goes in two directions, both requiring that 1 is odd. However, the problem arises when trying to define oddness using the number 1, as it fails the definition. The solution proposed is to define oddness as n+1=2k for some k, which would make 1 satisfy the definition and solve the problem. It is also noted that an inductive argument can start with 3 instead of 1.
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?

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##.

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##.

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 :)

PeroK

## What are the Peano Axioms of naturals?

The Peano Axioms are a set of axioms that define the natural numbers, which are used as the building blocks of mathematics. They include the axioms of zero, successor, induction, and ordering.

## Why is it important to prove that 1 is an odd number using Peano Axioms?

Proving that 1 is an odd number using Peano Axioms is important because it provides a solid foundation for understanding the properties of odd numbers and their relationship to the natural numbers. This proof also demonstrates the power and usefulness of the Peano Axioms in mathematical reasoning.

## What is the definition of an odd number according to Peano Axioms?

According to Peano Axioms, an odd number is any natural number that can be expressed as the successor of another natural number, followed by an additional increment of 1. In other words, an odd number is always one more than an even number.

## How does the Peano Axioms proof show that 1 is an odd number?

The Peano Axioms proof starts with the definition of an odd number as the successor of a natural number, followed by an additional increment of 1. By substituting 0 for the natural number, we can see that 1 is the successor of 0, making it an odd number.

## Can the Peano Axioms proof be extended to show that all natural numbers are either odd or even?

Yes, the Peano Axioms proof can be extended to show that all natural numbers are either odd or even. This can be done by using the axiom of induction, which states that if a property holds for the natural number 0 and the successor of any natural number, then it holds for all natural numbers. By applying this axiom, we can prove that every natural number is either 0 (an even number) or the successor of an even number (an odd number).

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