&gt;&gt; The really interesting thing about this proof is that it doesn&#x27;t use divisibility, just mathematical induction in its &quot;Z is well-ordered&quot; form.<p>&gt;That&#x27;s quite a bold statement. It would mean that this proof should generalize to other well-ordered sets that do not provide any notion of divisibility.<p>It doesn&#x27;t, for the somewhat obvious reason that divisibility makes sense in any set that has a notion of &#x27;multiplication&#x27;. The proof itself doesn&#x27;t use any statements of the form &#x27;x is divisible by y&#x27; though, not even implicitly as far as I can tell.<p>Even fraction laws don&#x27;t really count as &#x27;using&#x27; divisibility. Since fractions can be defined in a way that doesn&#x27;t directly use divisibility (although the concepts are obviously related).<p>Being well ordered is a far more restrictive property than &#x27;having a notion of divisibility&#x27;.<p>&gt; Moreover, these fractions are not even fractions in Z, but in R, or at least in Z[sqrt(m)]. So there&#x27;s still a lot of specialized structure involved in this proof.<p>According to the proof the fractions are in Q not R.