Formal construction of $\mathbb Q$: interpretation and equality of elements Formally the rational numbers are defined as $\mathbb Z \times \mathbb Z / \{0\}$, where $(m_1, n_1)$ and $(m_2, n_2)$ being equivalent if $m_1n_2 = m_2 n_1$.
This set equipped with $+$ and $\times$ as defined in the Wiki (http://en.wikipedia.org/wiki/Rational_number) form a field, that is the field of quotients of $\mathbb Z$.
How can one say for $n \in \mathbb Z$ that $n = (n,1)$ ? This equality is true only up to isomorphism ?
Also, how can one say $(m_1, n_1) = \text {some decimal expansion} : c_n c_{n-1} \ldots c_0. c_{-1}c_{n-2}\ldots$ ?
I mean $(m_1, n_1)$ is an element of a set but in normal use we set it equal to a real number ?
Also the result of $m_1 / n_1$ is understood to be a real number ? Not just the integer part ?
How just one decide that $1/3 = 0.333333\ldots$ and not some other real number ?
 A: Let $f: \Bbb{Z} \to \Bbb{Q}: \ z \to z/1$.  Show that this is an injective ring homomorphism.  Note that $z/1$ refers to an equivalence class of objects, for instance when you're working in arithmetic with:
$$
\dfrac{3}{5} =  \dfrac{21}{35} = \dfrac{3\cdot 7}{5\cdot 7}
$$
In particular, notice the equals.  So you've already worked with such a structure.
If $\frac{a}{b}$ is a non-integer rational and your base expansions (eg. $0.333333\dots$) are in base $c$ (eg. base $c = 10$, or "decimal"). Then $x = \frac{a}{b} = \lfloor \dfrac{a}{b} \rfloor + \dfrac{a \pmod b}{b} = I_1 + \dfrac{r_1}{b} = I_1 + \dfrac{1}{c}(\lfloor \dfrac{c \cdot r_1}{b} \rfloor+ \dfrac{c\cdot r_1 \pmod b}{b}) = \dots $. Thus $\frac{a}{b}$ produces an infinite base $c$ expansion if the sequence 
$$r_1 = a \pmod b, \\ 
r_2 = c\cdot r_1 \pmod b, \\
r_3 = c \cdot r_2 \pmod b, \\ \dots
$$ reaches zero at some finite point in time.  However if $c^k r_1 \equiv c^k a = 0 \pmod b$ and $c$ is a unit $\pmod b$, in other words $\gcd(c,b) = 1$, then the above can be multiplied by $c^{-k}$ and so $a = 0\pmod b$.  But we already said $\frac{a}{b}$ was non-integer.  Therefore whenever the bottom of the fraction $b$ is relatively prime to the base of the expansion representation $c$, you'll get an infinite base-$c$ expansion.  Prove that for rationals this expansion always starts repeating a sequence after enough places passed the decimal.  This should be true since there are a finite number of remainders $\pmod b$.   Another important thing that's sort of related is that this sort of base expansion representation is not uniquely determined, for instance $1 = .9999999\dots$.
