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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 \(A\cup \{1\}=\mathbb{Z}^+\).
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