Help with natural number problems Suppose that $\mathbb{F}$ is an ordered field with identities $0$ and $1^{\star}$ (in this problem, $1 \in \mathbb{N}$ )
Define inductively $f: \mathbb{N} \rightarrow \mathbb{F}$ by:
$$f(1) = 1^{\star}\qquad f(n + 1) = f(n) + 1^{\star}$$
(Imprecisely, $f(n) = \underbrace{1^{\star} + 1^{\star} + \cdots + 1^{\star}}_{\text{n times}}$)
c) Letting $\mathbb{N}^{\star} = \{f(n): n \in \mathbb{N}\}$, show that $f$ is a bijection from $\mathbb{N}$ to $\mathbb{N}^{\star}$. (Hint: Use this form $f(j + k) = f(j) + f(k)$ for $j,k\in\mathbb{N}$)
d) Show that $\mathbb{N}^{\star}$ is a "copy" of $\mathbb{N}$ in $\mathbb{F}$.  How can analogous $\mathbb{Z}^{\star}$ and $\mathbb{Q}^{\star}$ be defined, starting with $\mathbb{N}^{\star}$?
I don't need help for the first two parts of the problem since it's all just induction.  The first and second parts are easy.  I need help with third and fourth parts of the problems.
My Approach
Part(c) - I know that in order for $f$ to be a bijection, it needs to be 1-1 and onto.  If it's 1-1, then:
$$f(x) = f(y) \rightarrow x = y$$
I attempt to apply induction to show $f$ is $1$-$1$, but I'm stuck.
For the other part, I need to show that $f$ is onto.  That is: I need to prove that $f(y) = x$ for every $x$ and $y$, such that $y$ is the inverse of $f$.
Part(d) - My method would be to select $n \in \mathbb{N}$ and determine the range for each $n$.  That is:
$$f(1) = 1^{\star}$$
$$f(2) = f(1 + 1) = 1^{\star} + 1^{\star}$$
and so on.  But I think this may be wrong.
For the other part, I need to determine what are $\mathbb{Z}^{\star}$ and $\mathbb{Q}^{\star}$.  I can list all the elements from $\mathbb{Z}$ and $\mathbb{Q}$ and determine their equivalences in the field $\mathbb{F}$, but this seems to be the slow way to determine these sets.
Please help me.  My instructor is totally disorganized with his teaching.  He expects me to know how to solve these problems easily.
 A: c) By the definition of $\mathbb{N}^*$ as the image of $f$, you know that $f$ is onto. For one-to-one, if you would like to use induction, you can do the following. Let $x, y \in \mathbb{N}$ such that $x < y$. Define $n$ by $y = x+n$ and use induction on this $n$ to conclude that $f(x) \neq f(y)$. In fact, what you might want to prove here inductively is that $f(x+n) = f(x) + f(n)$ here to help you with part d. 
d) As mentioned in the previous part, show that $f(x+n) =f(x)+f(n)$ and I think from here you get that $\mathbb{N}^*$ is a copy of $\mathbb{N}$. For $\mathbb{Z}^*$ you may want to extend $f$ by $f(-n) = -f(n)$ when $n \in \mathbb{N}$ and $f(0) = 0^*$. And for $\mathbb{Q}^*$, you may try extending $f$ again by $f(m/n) = f(m)f(n)^{-1}$ whenever $m, n \in \mathbb{Z}$ $(n \neq 0)$.
Hope this helps.
A: For (c) I would prove by induction on $n$ that $$f(m+n)=f(m)+f(n)$$ for all $m,n\in\Bbb N$. The base case is built into your definition of $f$, which actually says that $f(m+1)=f(m)+f(1)$ for all $m\in\Bbb N$. You’ll want this result for (d), and once you verify that $f(n)\ne 0_{\Bbb F}$ for any $n\in\Bbb N$, you’ll be able to show very easily that $f$ is $1$-$1$.
Part (d) is a bit ambiguous. In my view showing that $\Bbb N^*$ is a ‘copy’ of $\Bbb N$ is a matter of showing that $f$ not only is a bijection but also respects the two field operations: $$f(m+n)=f(m)+f(n)$$ and $$f(mn)=f(m)f(n)$$ for all $m,n\in\Bbb N$. However, it’s possible that only the first of these is meant. For the rest, it’s not at all clear how you’re to interpret ‘starting with $\Bbb N^*$’. One possibility is that you’re to explain how to mimic the construction of $\Bbb Z$ from $\Bbb N$ and then of $\Bbb Q$ from $\Bbb Z$: $\Bbb Z^*$ would then be a certain set of equivalence classes of ordered pairs of elements of $\Bbb N^*$. Another, perhaps more likely, is that you’re to bypass all of this by using your knowledge of how things are supposed to work out, first extending $f$ to $g:\Bbb Z\to\Bbb F$ by setting
$$g(n)=\begin{cases}
f(n),&\text{if }n\in\Bbb N\\
0_{\Bbb F},&\text{if }n=0\\
-f(n),&\text{if }n<0\;.
\end{cases}$$
In somewhat similar fashion you can then extend $g$ to a function $h:\Bbb Q\to\Bbb F$ that ‘behaves properly’.
A: One of the things one learns in more advanced algebra classes is that there are such things as finite fields.  For these fields, there is a positive integer N (called the characteristic of the field) such that N times (any field element) is zero.  Classic example: Given a prime P, the set {0, 1, ..., P-1}, under addition and multiplication mod P, is a finite field.
Given that, any argument that your map f is one-to-one must fail.  Then the next question must be: Does the definition of an ordered field naturally exclude finite fields?  Any proof that f is one-to-one must involve the answer to this question.
