I am currently studying first-order logic and I am struggling on a problem.
We work on a first-order language with non-logical symbols of arithmetics and the axioms of arithmetic.
We define a non-standard model $\mathfrak{M}$ of $T$ in which there exists $a \in |\mathfrak{M}|$ such that $\mathfrak{M}(S^n0)<\mathfrak{M}(a)$, for all $n$. (such an $a$ is called non-standard number).
I have to prove that for any $\mathfrak{M}$, a non-standard model of $T$, and any non-standard number $b \in |\mathfrak{M}|$, there exists an $a <\mathfrak{M}(b)$ such that for each $p$ prime, $\mathfrak{M}(S^p0)$ divides $a$.
I don't know were to start, I know we can build a non-standard model by expanding $\mathcal{L}$ with a constant $c$ and then we consider $\Sigma=\{0<c,\text{S}0<c,\text{SS}0<c,\cdots\}.$ I thought adding a new constraint with a constant $d$ such as it can be divided by all $p$ prime, and adding another constraint $d<c$. And conclude with the completeness theorem.
But I am not sure this method can be applied to my problem.
Can anyone help me?
Thanks a lot!
Welcome to Math.SE!
Instead of constructing a new model $\mathcal{M}(c,d)$, the problem asks you to work in the model $\mathcal{M}(b)$ that you are given.
The following three statements follow from the axioms of arithmetic, and therefore they hold in all models of arithmetic, including $\mathcal{M}(b)$.
This follows from the principle of induction. In the base case, $n=0$ we can choose $m = 1$. In the inductive case, $n=k+1$, and by the inductive hypothesis there is a number $m'$ that is divisible by every number less than or equal to $k$. But then the number $nm'$ is divisible by every number less than $n$ (since all such numbers divide $m'$), and clearly by $n$ itself as well. So we can prove the inductive case by setting $m=nm'$. By the principle of induction, we conclude that for every $n$, we can find a number $m'$ that is divisible by every number less than or equal to $n$.
Using the previous statement, we see that arithmetic proves the following as well:
For every $n$, there is a unique least number $Q_n$ that is divisible by every number less than or equal to $n$, (least meaning that for any other number $P_n$ with this property, the inequality $Q_n \leq P_n$ holds).
For all $n > 1$, there is a unique largest number $m$ such that $Q_m<n$.
We call $x \in \mathcal{M}(b)$ standard if $x \leq \mathcal{M}(b)(S^ko)$ for some $k \in \mathbb{N}$. Notice that if $x$ is standard, then so is $Q_x$, since e.g. $Q_x \leq \mathcal{M}(b)(S^{k!}o)$.
Now consider $b \in \mathcal{M}(b)$. Since the third statement above holds in $\mathcal{M}(b)$, we can find a largest $m \in \mathcal{M}(b)$ such that $Q_m < b$. If this $m$ was standard, then so would be $m+1$, and by the observation above $Q_{m+1}$ as well. But $b$ is not standard, so $Q_{m+1} < b$ would hold, contradicting that $m$ was the largest number such that $Q_m < b$. Therefore $m$ is not standard, and so $m$ is larger than all $\mathcal{M}(b)(S^po)$ with $p \in \mathbb{N}$. But $Q_m$ is divisible by all the numbers less than $m$, so we can set $a=Q_m$ to get $a<b$ that is divisible by all the numbers of the form $\mathcal{M}(b)(S^po)$ with $p \in \mathbb{N}$. This was to be shown.