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Sep 4, 2020 at 10:44 history edited JimN CC BY-SA 4.0
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Sep 4, 2020 at 10:40 comment added JimN Hmmm... you are right ... induction would be saying that a property holds for everything on the 1,2,4,8,16 .... path. I guess one actually has to find a different set for this, rather than a successfor function. Answer now edited to reflect this.
Sep 4, 2020 at 10:36 comment added Asaf Karagila Are you sure that induction holds? $A=\{2^n\mid n\in\Bbb N\cup\{0\}\}$ seems to satisfy $1\in A$, since $2^0=1$, and if $n\in A$, then $n=2^k$, so $s(n)=2^{k+1}\in A$. But very obviously, $A$ does not contain any odd numbers except $1$.
Sep 4, 2020 at 10:22 history answered JimN CC BY-SA 4.0