What is a rational number? What is the quotient of two integers?

Question

How can we form an idea of the result of the operation of dividing a by b with a and b integers and b not equal to zero?

Answer 1

Let $(a,b),(c,d)$ be two ordered pairs with $a,b,c,d\in\mathbf{Z}$ and $bd\neq 0$. Define the equivalence relation $(a,b)\sim (c,d)$ if $ad=bc$. Then, we can define the rationals $\mathbf{Q}$ to be the set of equivalence classes with $+$ and $\times$ such that $$\begin{align*}(a,b)+(c,d) &=(ad+bc,bd) \\ (a,b)\times (c,d) &= (ac,bd).\end{align*}$$ Now just show this construction satisfies field axioms. Hope this answers your question.

Answer 2

What are the rational numbers? The first thing you need to get rid of is your conception that a rational number $\frac{a}{b}$ means a divided by b. While this is true in practice, that's not how we define the rational numbers.

Take the set of integers, $\mathbb{Z}$. Take the set of nonzero integers, $\mathbb{Z}\setminus\{0\}$.

Now take the cartesian product of these sets: $\mathbb{Z} \times \mathbb{Z}\setminus\{0\}$, which is the set of all ordered pairs $(m,n)$ such that $m,n \in \mathbb{Z}, n \neq 0$. We may also write $\mathbb{Z} \times \mathbb{Z}\setminus\{0\} = \{(m,n) \mid m,n \in \mathbb{Z}, n \neq 0 \}$.

Then you write that two pairs $(m_1,n_1),(m_2,n_2)$ are equivalent, denoted $(m_1,n_1)\sim(m_2,n_2)$, iff $m_1n_2 - m_2n_1 = 0$. This relation $\sim$ is an equivalence relation, and allows us to say that fractions like $\frac12$ and $\frac24$ are equivalent.

Then $\mathbb{Q}$, the set of rational numbers, is the set of all equivalence classes given by $\sim$ on the set $\mathbb{Z} \times \mathbb{Z}\setminus\{0\}$. So, formally, if you denote the class of fractions equivalent to $\frac12$ by $C\left(\frac12\right)$, you can think of $\mathbb{Q}$ like this:

$$\mathbb{Q} = \{C\left(\frac12\right),C\left(\frac13\right),\ldots,C\left(\frac57\right),\ldots,C\left(\frac{11}{19}\right)\} $$

Where you note that every equivalence class of fractions is represented by its irreducible form (when the denominator and numerator are coprime).

Answer 3

Not sure if I understand your question right, but you can just take an axiomatic approach: First of all rational numbers are the set

$\mathbb Q = \{ \frac a b : a \in \mathbb Z, b \in \mathbb N \}$

and we define the two operations $+$ and $\cdot$ by

$+ : \mathbb Q \times \mathbb Q \to \mathbb Q: (\frac a b, \frac c d) \mapsto \frac {ad + bc}{bd} $

$\cdot : \mathbb Q \times \mathbb Q \to \mathbb Q: (\frac a b, \frac c d) \mapsto \frac {ac}{bd} $

Having defined this just by axioms one can prove that $(\mathbb Q, +, \cdot)$ is a field any many other properties. More over this definition is compatible with addition and multiplication in $\mathbb Z$.

Or did you want some "deeper" insight into the nature of rational numbers?