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math_proof Group TitleBest ResponseYou've already chosen the best response.0
Starting 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 \[a \in \mathbb{Z}^+ \]
 one year ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
I 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.
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
which second part? whats in thee brackets is the hint
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
If 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 \(A\cup \{1\}=\mathbb{Z}^+\).
 one year ago
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