Minimality in the construction of integers and rationals (Analysis I, Theorem 9.1 & 9.2). The book Analysis I by Herbert Amann & Joachim Escher can be found here on page 85 & 86.
When constructing $\mathbb{Z}$ from $\mathbb{N}$ the author defines an equivalence relation on $\mathbb{N}^2$, $$(m,n)\sim(m',n'):\Leftrightarrow m+n'=m'+n,$$
and let $\mathbb{Z}=\mathbb{N}^2/\sim$.
Similarly for rationals, an equivalence relation on $\mathbb{Z}\times\mathbb{Z}^{\times},$ $$(a,b)\sim(a',b'):\Leftrightarrow ab'=a'b,$$
and let $\mathbb{Q}=\mathbb{Z}\times\mathbb{Z}^{\times}/\sim$.
In both of the proofs, the author then writes that the sets ($\mathbb{Z}$ and $\mathbb{Q}$ respectively) constructed are "by construction" minimal. Why is that true?
 A: From the book Analysis I by Herbert Amann & Joachim Escher:
9.1 Theorem There is a smallest domain (commutative ring without zero divisors) with unity, $\mathbb{Z}$, such that $\mathbb{N} \subset \mathbb{Z}$ and the ring operations on $\mathbb{Z}$ restrict to the usual operations on $\mathbb{N}$. This ring is unique up to isomorphism and is called the ring of integers. 
We argue, not following the book, that $\mathbb{Z}$ is the smallest ring containing $\mathbb{N}$ by making several observations, leaving the details to the OP:
Assume that $\mathbb{N}$ is contained in $R$. Suppose in $R$ there exist an element $g$ such that $g + 1 = 0$. Then for $m \in \mathbb{N}$,
$mg + m = m (g+1) = m\text{*} 0 = 0$, so $mg$ is the additive inverse of $m$.
The set $Z = \{mg\;|\; m \gt 0 \} \cup \mathbb{N}$ must be contained in $R$.
But this set can be identified with the set $\mathbb{Z}$.
There is only one way to define addition and multiplication on $Z$ (it is generated by 'adding' $g$ to the natural numbers), so it can be algebraically identified with the construction of $\mathbb{Z}$.

9.2 Theorem There is, up to isomorphism, a unique smallest field $\mathbb Q$, which contains $\mathbb{Z}$ as a subring. 
In this second part we want to show that $\mathbb Q$  is contained in any field that contains (via an injective ring morphism) $\mathbb Z$.
In the first part above, we 'threw in' the additive inverses to get $\mathbb Z$. In the same way, we now want to 'throw' $F = \{2^{-1},3^{-1},4^{-1},5^{-1},\dots, n^{-1},\dots\}$ into $\mathbb Z$ and 'generate' $\mathbb Q$.
We leave it to the OP to show that each block (element of $\mathbb Q = \frac{\mathbb Z \times \mathbb Z^{\text{x}}}{\text{~}}$) has a representative that is '$\mathbb Z$ generated' by an element in $F$.
The remaining details are also left as an exercise.
A: Ok, I think I got it. 
They do there one of the classical constructions both of integral domain $\;\Bbb Z\;$ and of the field $\;\Bbb Q\;$. The minimality of $\;\Bbb Z\;$ follows from the fact that any domain containing $\;\Bbb N\;$ has to contain the constructed $\;\Bbb Z\;$ (as the operations defined on $\;\Bbb Z\;$ must restrict to the ones already given in $\;\Bbb N\,$ !). This is what the authors do when they write on page $85$, after $\;(9.4)\;$, "Now let $\;R\supset\Bbb N\;$ be some commutative ring...", and then they pass to prove that $\;\Bbb Z\;$ can be embedded within $\;R\;$.
The minimality for $\;\Bbb Q\;$ follows from the one of $\;\Bbb Z\;$, and the end of the proof, after  $\;(9.5)\;$ , is very similar to the one above.
A: The minimality it is because the numbers added are just the needed to pass from a ring to a field, i.e. you need from $\Bbb Z$ just the multiplicative inverses to transform it in the field $\Bbb Q$. 
By the other side: if you quit just a number of $\Bbb Q$ then it is not a field anymore. In the case of $\Bbb N$ to $\Bbb Z$ is the same: we construct the needed numbers to transform $\Bbb N$ in a ring ($\Bbb Z$), i.e. the additive inverses.
The proof of minimality is that if you quit a number you lose the algebraic structure. By example: taking some $x\in\Bbb Z_{<0}$ then the natural number $-x$ lost it additive inverse. Then $\Bbb Z$ is minimal.
For the case of $\Bbb Q$ if you quit some $p/q$ then you must quit too $-p/q$, $q/p$ and $-q/p$. But then the number $p\cdot q^{-1}$ doesnt exist, so this altered $\Bbb Q$ is not a field anymore. Then you can try to quit $q^{-1}$ too but then $q$ doesnt have anymore inverse.
To prove this rigorously we can do something based in the above to construct a proof by contradiction. You can prove that doesnt exists some $r\in\Bbb Q$ such that $r\neq p/q$ for any pair $p,q\in\Bbb Z$, but by construction such $r$ cannot exists.
And you can prove that you are unable to quit any $p/q$ to maintain a field based in the operations of addition and multiplication over $\Bbb Z$. Notice that if you quit some $p/q$ this implies that you must quit too $q^{-1}$ for some $q\in\Bbb Z$ because $p/q=p\cdot q^{-1}$.
