Show that any model of $\Delta$ is a Nonstandard Model of Arithmetic I was hoping that someone could help check my proof. I originally thought I was spot of with my proof, but my professor suggested that my method was wrong. So, I went to check the hint in the back of the book, but the hint seemed like it was just my solution. He may have misunderstood me or I could have not said what I mean. Hence, I was hoping someone here could help me check it. Any help at all would be much appreciated! I will state the problem, then show my proof.
By true arithmetic we mean the set Γ of all sentences of the language of arithmetic that are true in the standard interpretation. By a nonstandard model of arithmetic we mean a model of this $\Gamma$  that is not isomorphic to the standard interpretation. Let $\Delta$ be the set of the sentences obtained by adding the sentences $c\not = \boldsymbol 0 $, $c\not = \boldsymbol 1 $, $c\not = \boldsymbol 2 $, and so on, to $\Gamma$. Show that any model of $\Delta$ would give us a nonstandard model of arithmetic. 
Proof: Let $N$ a model of $\Gamma$. Moreover, let $M$ be the model of $\Delta$. Now, suppose, for the sake of contradiction, that $N\cong M$. 
Then, we can say that there is a bijection $\alpha :|N|\to |M|$ such that for all $c$ we have $\alpha (c_{N})=c_{M}$.
Now, if $c_{N}=\boldsymbol 2_{N}$, then $\alpha (\boldsymbol 2_{N})=c_{M}=\boldsymbol2_{M}$. However, we know that $c_{M} \not =\boldsymbol2_{M}$. 
Thus, we have reached a contradiction and our interpretations are not isomorphic.
If I am wrong where did I mess up, and how bad is it?
Thanks again!
 A: If you assume $N\models \Gamma$ and $M\models \Delta$, then you cannot conclude that $N\not\cong M$, since of course $M\models \Gamma$ because $\Delta\supseteq \Gamma$.
Instead, the question is asking you to show that, if $M\models\Delta$, then $M\not\cong N$, where $N$ is the standard model of $\Gamma$ - that is, $N=(\mathbb{N}, +, \times, 0, <)$.
Except this isn't quite right! $\Gamma$ is a set of sentences in the language $\{+, \times, <, 0, c\}$, whereas the standard model $N$ of arithmetic is only a $\{+, \times, <, 0\}$-structure. What you're really being asked to show is that, for any $a\in N$, we have $$(\mathbb{N}, +, \times, 0, <, a)\not\cong M.$$
You've started on the right direction - you've shown that $a$ cannot be 2. The same argument will show that $a\not = n$ for any standard natural number $n$; but these are all the elements of $N$! So you're essentially already done.
Note that the nonstandard model of arithmetic arising from $M$ is the reduct of $M$ to the smaller language $\{+, \times, 0, <\}$ of arithmetic.
