In my book, it is mentioned that the Quotient Field of the Integral Domain of Integers is the Field of Rational Numbers.
However, I have a confusion- the Quotient Field of $\mathbb{Z}$ would be the set of equivalence classes of ordered pairs $(a,b)$ (in $\mathbb{Z}$ where $b$ is non zero) under the relation that $(a,b)\sim(c,d)$ if and only if $ad = bc$.
In that case, wouldn't the Quotient Field of $\mathbb{Z}$ actually be the set of equivalence classes of every fraction $a/b$, where $b$ is non zero? I mean to say that, rather than individual rational numbers $(a/b)$ being elements of the quotient field, shouldn't an individual element of the quotient field be all fractions that simplify to $a/b$?
In that case, the quotient field will not be $\mathbb{Q}$, since its elements would be sets rather than numbers.
What is a rational number? Is there a difference between $\frac{2}{3}$ and $\frac{4}{6}$? I guess you would say no to that.
But now we run into the problem on how to make sense of this non-difference. One solution is to only consider reduced fractions. But then the symbol $\frac{4}{6}$ is not allowed any more, which I would say is inconvenient. Another solution is to say that a rational number is the collection of all those fractions $\frac{n}{d}$, which represent the same number. This is to say in mathematical terms, the symbol $\frac{m}{n}$ denotes the equivalence class of all fractions, which represent the same number. This "represent the same number" has to be made precise in the form of an equivalence relation, which should not use the term fraction. This is precisely what the relation $(m,n) \sim (a,b) \Leftrightarrow mb=na$ is for. Using fractions as notation for equivalence classes this translates into $\frac{m}{n} = \frac{a}{b} \Leftrightarrow mb=na$, which you will agree is the defining property of fractions.