# Peano axioms proof

Given that $$K$$ is an ordered field satisfying the least upper bound property and $$1$$ as the multiplicative identity, the set of natural numbers $$\mathbb{N}_K$$ in $$K$$ is defined as: $$1 \in \mathbb{N}_K$$ and $$N + 1 \in \mathbb{N}_K$$ if $$N \in \mathbb{N}_K$$. In this case, $$N = 1 + ... + 1 \}$$ $$N$$ times. Show that $$\mathbb{N}_K$$ satisfies the Peano axioms.

I am new to analysis and was not really taught about the Peano axioms and can't seem to find anything on the web about exactly what I need to prove in this problem. Can someone please briefly list and/or describe the axioms that need to be shown in order to complete the above proof? Any assistance is much appreciated.

• You can find the list here. To show that any of those axioms are satisfied by $\mathbb{N}_K$ you'll want to use the fact that $K$ is a completely ordered field, i.e., going back to the field axioms, etc.. Jan 28, 2021 at 23:37
• @Hayden It starts with "$0$ is a natural number" however in the book I use, it suggests that they start from $1$? Jan 28, 2021 at 23:56
• Yeah, whether $0$ is considered a natural number is a bit up to personal preference. Within the first-order axiomatization, replace the first axiom with $\forall x ~(S(x) \neq 1)$, the third axiom with $\forall x ~(x+1 = S(x))$, the fifth axiom with $\forall x ~(x \cdot 1 = x)$, and in the induction axiom schema, replace $0$ with $1$. Although I haven't checked the details carefully, any model of this modified system of axioms should arise as the set of all nonzero elements of a model of the original system, and vice versa. Jan 29, 2021 at 3:01

Suppose you are allowed to sum numbers, and if you are in the natural numbers, this is quite natural. And define the operation $$S(x) = x+1$$ to be it, then we have to believe:
1. $$S(x) = S(y)$$, meaning that our $$x$$ equals $$y$$, so $$S$$ is injective;
2. 1 is the only element, that doesn't succed any other, then there is no $$x$$ in natural numbers such that $$S(x) = 1$$;
3. Suppose there is a subset of natural numbers with 1 inside it, then if in this subset there is another, then it's because $$S(x) \in X$$, and so, this subset is the natural numbers set.
• First, I want to know why $N= 1+1+1+ ... +1$ in this case above as you described? Jan 29, 2021 at 0:17