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Questions tagged [peano-axioms]

For questions on Peano axioms, a set of axioms for the natural numbers.

17 votes
1 answer
815 views

Is every model of modular arithmetic either even or odd?

Modular Arithmetic (MA) has the same axioms as first order Peano Axioms (PA) except $\forall x (Sx \ne 0)$ is replaced with $\exists x(Sx = 0)$. (http://en.wikipedia.org/wiki/Peano_axioms#First-...
Russell Easterly's user avatar
0 votes
1 answer
798 views

Peano's Postulates Proofs

How can I prove the following two questions: Prove using Peano's Postulates for the Natural Numbers that if a and b are two natural numbers such that a + b = a, then b must be 0? Prove using Peano's ...
Jakemmarsh's user avatar
4 votes
5 answers
748 views

Peano postulates

I'm looking for a set containing an element 0 and a successor function s that satisfies the first two Peano postulates (s is injective and 0 is not in its image), but not the third (the one about ...
Toby Carter's user avatar
4 votes
2 answers
1k views

Induction as Peano Axiom

Let P be some proposition. If we have that $P(0)$ is true and that if $P(n)$ is true, then $P(S(n))$ is true, where $S(n)$ is the successor of natural number $n$. Then we have that $P(n)$ is true for ...
Praslow D.'s user avatar
18 votes
6 answers
5k views

Why do we take the axiom of induction for natural numbers (Peano arithmetic)?

More precisely, when we define the set of natural numbers $\mathbb{N}$ using the Peano axioms, we assume the following: $0\in\mathbb{N}$ $\forall n\in\mathbb{N} (S(n)\in\mathbb{N})$ $\forall n\in\...
russell11's user avatar
  • 245
26 votes
4 answers
10k views

Why is Peano arithmetic undecidable?

I read that Presburger arithmetic is decidable while Peano arithmetic is undecidable. Peano arithmetic extends Presburger arithmetic just with the addition of the multiplication operator. Can someone ...
chinu's user avatar
  • 685
28 votes
1 answer
2k views

Is there a natural model of Peano Arithmetic where Goodstein's theorem fails?

Goodstein's Theorem is the statement that every Goodstein sequence eventually hits 0. It is known to be independent of Peano Arithemtic (PA), and in fact, was the first such purely number theoretic ...
Jason DeVito - on hiatus's user avatar

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