Suppose we define addition on $\Bbb N^2$ as $(x_1,x_2)+(y_1,y_2)=(x_1+y_1,x_2+y_2)$. I need to prove that there is no way to define $0$ andd multiplication so that becomes a model of the Peano Axioms PA.
A hint says that every member of $\Bbb N$ is either even or odd. But I cannot find the connection between this fact and the question.
Could someone please tell me how can I see it from this hint? Or any other ways are also appreciated.
My version of PA is stated as follows:
I don't quite know what to do with the hint ... but here is a problem with trying to make $\mathbb{N}^2$ the domain of a model where $+$ is interpreted as given:
Using the Peano Axioms you can prove:
$$\forall x \forall y (\exists z \ x + z^+ = y \lor x = y \lor \exists z \ y + z^+ = x)$$
However, this will not be true for $(0,1)$ and $(1,0)$ (or any $(m,n)$ and $(n,m)$ with $m\not = n$), because with the $+$ as given, there is no $(m,n) \in \mathbb{N}^2$ such that $(0,1)+(m,n)=(1,0)$, we don't have that $(0,1) = (1,0)$, and there is no $(m,n) \in \mathbb{N}^2$ such that either $(1,0)+(m,n)=(0,1)$
So, you already cannot make a model with $\mathbb{N}$ as the domain and the $+$ as given, no matter how you would interpret $\overline{0}$ or $\cdot$.
UPDATE
OK, so here is something you can do with the hint: Given that a number $x$ is even if and only if there is some $y$ such that $y+y=x$, it follows that the only even numbers in your domain are $(2m,2n)$ with $m,n \in \mathbb{N}$. Defining a number $x$ to be odd if and only if there is some $y$ such that $x = (y + y) + \overline{0}^+$, you can use the Peano axioms (especially the induction axiom PA3) to prove that:
1) every number has to be odd or even, and
2) that any odd number plus odd number is an even number.
So, this means by 1) that $(0,1)$ and $(1,0)$ are odd, which by 2) means that $(0,1)+(1,0)=(1,1)$ should be even ... but clearly it is not.