Show that the set of positive elements in $\mathbb{Z}$ can be identified with $\mathbb{I(N)}$

Question

I have an exercise in my book that states

Show that the set of positive elements in $\mathbb{Z}$ can be identified with $I\mathbb{(N)}$

we finally define a mapping $I$ from $\mathbb{N}$ to $\mathbb{Z}$ by $$ I(a) := [(a+1,1)]$$

The mapping $I$ defined above is an order preserving injection, i.e. it is injective and $$I(a) < I(b) \textit{ iff } a<b, \ a,b \in \mathbb{N} , and $$ $$I(a \cdot b) = I(a) \cdot I(b), \ a, b \in \mathbb{N} $$

What those "identifie" means? English is not my native language. Should I show that $I$ is bijective?

Answer 1

You are supposed to show that there is no essential difference between $\Bbb N$ and $I(\Bbb N)$, that is:

• The map $I\colon \Bbb N\to I(\Bbb N)$ is bijective (of course, injective suffices)
• All essential properties defined for $\Bbb N$ (such as addition, multiplication, order) are transported by $I$ to the extended corresponding properties in $\Bbb Z$.

It seems that you are almost done, only $I(a+b)=I(a)+I(b)$ seems to be missing.

Once you have shown this, the a priory existing necessity to distinguish between $a$ and $I(a)$ or between addition/multiplication/comparison of naturals and addition/multiplication/comparison of their images in $\Bbb Z$ becomes moot. This justifies to say that $42$ is an element of $\Bbb Z$, for example, when "really" only $[43,1]\in\Bbb Z$. To make $\Bbb N $a real subset of $\Bbb Z$, we should instead replace $\Bbb Z$ with $(\Bbb Z\setminus I(\Bbb N)\cup \Bbb N$ and accordingly "glue" together the two additions/multiplications/comparisons - but who cares when the additions/multiplications/comparisons are in fact "the same" as far as anybody could ever be interested? In fact, once the construction of $\Bbb Z$ from $\Bbb N$ is completed, we may abstract from all those equivalence classes and simply pretend that there is a set $\Bbb Z$ which has $\Bbb N$ as subset and such that $\Bbb Z$ is a ring containing $\Bbb N$ as a sub-semiring and not containing any proper subring ...