Say we have only defined the rational numbers. If we have two rational numbers, s and r, and we can show that $|s-r|<\epsilon$, for all positive rational $\epsilon$, is that enough to show that $s = r$? Is this a trivial question? or is there a definition of equality or an equivalence relation that would satisfy this property?
I tried to use the definition of equivalence on rationals. This was my attempt: Let $s = [(a, b)]$ and $r = [(c, d)]$ where $a,b,c,d \in \mathbb Z$ with the equivalence relation, $\sim$ defined as $(a_1, a_2) \sim (a_3, a_4)$ iff $a_1 a_4 - a_3 a_2 = 0$, subtraction, $-$ defined as $(a_1, a_2) - (a_3, a_4) = (a_1 a_4 - a_3 a_2, a_2 a_4)$ and "$<$" defined as $(a_1, a_2) < (a_3, a_4)$ iff $a_1 a_4 < a_3 a_2$. My goal was to express s and r as their respective equivalence classes, show that they were equivalent and use the transitivity of equivalence to finish the proof.
Let an arbitrary $\epsilon > 0$ and
$|s - r| < \epsilon$
$\lvert [(a,b)] - [(c,d)] \rvert < \epsilon$
$\lvert [(ad-bc,bd)] \rvert < \epsilon$
If we let $\epsilon = [(p, q)]$,
$\lvert [(ad-bc,bd)] \rvert < [(p, q)]$. WLOG, we can let s > r and thus, $[(ad-bc,bd)] < [(p, q)]$ $\implies q(ad - bc) < p(bd)$
I'm stuck. I cannot see anything that would imply that $ad - bc = 0$. Am I missing something?
If you already proved all the basic properties of addition and multiplication of rational numbers, I see no reason why you should ever go back to writing the elements of $\mathbb{Q}$ as equivalence classes of pairs. I mean, $\frac{a}{b}$ is the same thing as $[(a,b)]$, just a much more simple notation.
Anyway, $|s-r|$ is clearly a nonnegative rational number. Suppose it is positive, say $|s-r|=\frac{a}{b}$ with $a,b$ positive integers. But then for $\epsilon=\frac{a}{2b}$ we have $|s-r|>\epsilon$, contradicting the assumption.