The Peano axiom of induction for natural numbers says that For any property $P(n)$ , if $P(0)$ holds, and that whenever $P(n)$ holds, $P(n++)$ holds, then $P(n)$ holds for all natural numbers.

Can this be replaced by

"For every natural number $n$ not equal to $0$, there exists another natural number $m$ such that $m++ = n$" ?

What are the problems associated with replacing this?

I am following Terence Tao's book Analysis I.

  • $\begingroup$ The axiom of induction offers a principle for proving some statement $P$ for the natural numbers. Your "axiom" does not even refer to some property $P$ to be proven over the natural numbers... How are the two connected? $\endgroup$
    – frabala
    Mar 17 '19 at 10:04
  • $\begingroup$ One of the reasons mentioned in the book to use the axiom of induction was to remove superfluous/extra elements. For example, the set Z = (0, 0++, 0++++, ........, x, x++, x++++.....) follows the previous axioms (where x is not a natural number). We could eliminate x and its increments by what I wrote in the suggestion. What I want to know is how does the induction axiom add more than just removing superfluous elements. $\endgroup$
    – Kosha Modi
    Mar 18 '19 at 10:13

Let $N=\{-1\}\times\mathbb N\cup\{1\}\times\mathbb Z$. If $(\pm1,m)\in N$, then let $(\pm1,m)++=(\pm1,m+1)$. Then all the Peano axioms (with the induction one replaced by yours) hold (assuming that $0=(-1,0)$), but what we have here is something which is different from the naturals. For instance, induction doesn't hold here.


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