A community for students.
Here's the question you clicked on:
 0 viewing
math_proof
 2 years ago
Peano's axioms prove
math_proof
 2 years ago
Peano's axioms prove

This Question is Closed

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0Starting from Peano's axioms prove that if \[n \in \mathbb{Z}^+ and n \ne1\] then n is a successor, i.e. s(a)=n for some \[a \in \mathbb{Z}^+ \]

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0[Let A=Im(s) U {1} and prove that A=Z+]

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1I think the first could happen through induction. Base case: If \(n=2\), then \(s(1)=2\), so we're good. Now we assume up to some \(k<n\). Now we need to show it's true for \(k+1\). But since \(s(k)=k+1\) for \(k\in\mathbb{Z}^+\), we're done. I've got to go for a bit now. I'll come back to the second part later.

math_proof
 2 years ago
Best ResponseYou've already chosen the best response.0which second part? whats in thee brackets is the hint

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1If that's the hint, then we're completely done. For \(n\in\mathbb{Z}_{\ge 2}\), we have that \(n\in \text{im}(s)\), and since \(\text{im}(s)\subseteq \mathbb{Z}^+\) by definition, we've proved that \(A=\mathbb{Z}_{\ge2}\), so \(A\cup \{1\}=\mathbb{Z}^+\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.