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Define a number $n$ to be composite if it can be written as $a\cdot b$ for some $a,b$ where $a,b\neq 1$.

Define $p$ to be prime if $p=a\cdot b$ implies $a=1$ or $b=1$.

The theorem that every composite number has a prime factor seems to require a bit more induction than just everything is either 0 or a successor. So what is a model of Robinson Arithmetic where that theorem fails?

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up vote 11 down vote accepted

The nonnegative part of a discrete ordered ring always satisfies Robinson Arithmetic, so many examples can be found there. For a specific one, take the ring $R$ of formal Puiseux polynominals of the form $$a = a_0 + a_1 T^{q_1} + \cdots + a_k T^{q_k}$$ where $0 < q_1 < \cdots \lt q_k$ are rationals and $a_0,a_1,\ldots, a_k$ are integers. The order is given by $a \lt b$ if the most significant term of $b-a$ is positive (i.e. we think of $T$ as an infinitely large number). In this ring $T$ is composite (e.g. $T = T^{2/3}\cdot T^{1/3}$) but it has no prime factor.

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