(related to Proposition: Algebraic Structure of Rational Numbers Together with Addition and Multiplication)
This proof is a direct application of the theorem construction of fields from integral domains \(( * )\), in which we will construct the field \((\mathbb Q, +,\cdot)\) from the integral domain \((\mathbb Z, +,\cdot)\).
In the proof of the theorem \(( * )\), a field \((F,\ast,\circ)\) is constructed from an integral domain \((R, +,\cdot )\), as the set of all equivalence classes \([a,b]\), which are represented by ordered pairs \((a,b),(c,d)\in R\times (R\setminus\{0\})\) together with the following equivalence relation:
$$(a,b)\sim (c,d)\Leftrightarrow ad=bc.$$
In this proof, the operations \(\ast\) and \(\circ\) are defined as follows: $$\begin{array}{ccl} \lbrack a,b\rbrack \ast\lbrack c,d\rbrack &:=&\lbrack ad + cb,~bd\rbrack ,\\ \lbrack a,b\rbrack \circ\lbrack c,d\rbrack &:=&\lbrack ac,~bd\rbrack .\\ \end{array} $$
For the special case of the integral domain \((R, + ,\cdot) =(\mathbb Z, + ,\cdot )\), the equivalence relation is defined for two ratios
$$\frac ab\sim \frac cd\Leftrightarrow ad=bc,~~~~~~~~~~a,c\in\mathbb Z,~~b,d\in\mathbb Z\setminus\{0\}$$
and the field operations in \((\mathbb Q,\ast,\circ)\), "addition \(\ast\)" and the "multiplication \(\circ\)" can be now written as $$\begin{array}{ccccc} \frac ab \ast \frac cd &:=&\frac ab + \frac cd &=&\frac {a\cdot d + c\cdot b}{b\cdot d},\\ \frac ab \circ\frac cd &:=&\frac ab \cdot \frac cd &=&\frac {a\cdot c}{b\cdot d}.\\ \end{array} $$ Thereby, we have replaced the notation \((\mathbb Q,\ast,\circ)\) by the more common notation \((\mathbb Q, + ,\cdot)\).
Furthermore, according to the proof of \(( * )\), the integral domain \((\mathbb Z, +,\cdot )\) is isomorphic to a proper subset \(S\subset\mathbb Q\) with (using our new notation) $$\mathbb Z\simeq S:=\left\{\frac {a\cdot x}x~|~a\in\mathbb Z,~x\in\mathbb Z\setminus\{0\}\right\}.$$ Please note, that since \((\mathbb Z, +,\cdot )\) is an integral domain, we can even write \(\frac {a\cdot x}x=\frac {a\cdot \cancel x}{\cancel x}=\frac a1\). Thus, \((\mathbb Z, +, \cdot)\) can be regarded as a proper subset of \((\mathbb Q, +,\cdot)\) (i.e. each integer is a special case of a fraction), preserving the effect of the operations \( + \) and \(\cdot\), defined for fractions.
We also learn from the proof of theorem \(( * )\):
This completes the proof.
