Integers, rationals and reals as sets? Natural numbers can be represented as pure sets by defining them to contain every number that is smaller than them. Arithmetic can be performed on them using the Peano axioms. Are there any similar definitions for integers, rationals and reals?

For example, I could define a rational to be an ordered pair of dividend and divisor. But that would leave the two rationals $\frac{1}{2}$ and $\frac{2}{4}$ not equal to each other, and it would be based on ordered things rather than pure sets.
Yes the procedure you outline is correct, but you must define equivalence classes,
$(a,b)\sim (b,c)$ if $ac=bd$. Thus a rational number is an equivalence class of pairs.
One then defines the arithmetic operations addition and multiplications and show that they are invariant under equivalence. The same can be done to define the integers from the natural numbers here the relation is $(n,m)\sim (r,s)$ if $n+s=r+m$.
The reals are then defined as Dedekind cuts of the rationals.
Yes.


Define the ordered pair $(a, b)$ as the set $\{a, \{a, b\}\}$
Define the integers as $(\{1, 0\} \times \mathbb N) \setminus (0, 0)$: ordered pairs where the first element represents the sign and the second represents the magnitude, excluding negative $0$. In the first slot, $1$ represents positive, $0$ represents negative.
Define the rational number $\frac{p}{q}$ as $\{(a, b): aq=bp\}$: the set of all ordered pairs that reduce to the same thing $(p, q)$ does in lowest terms. (That's not quite what this definition says, but it's easier to say and amounts to the same thing.)
Define real numbers as sets of rationals for which if $\frac{p}{q}$ is in the set, then all rationals lower than $\frac{p}{q}$ are in the set. (Also, the empty set and $\mathbb Q$ are specifically excluded from being real numbers.)