Certainly! We can prove the existence and uniqueness of an integer $n$ such that $n \leq r < n + 1$ for a rational number $r$ without relying on the Archimedean property.

Proof:

Step 1: Existence

Let $r$ be a rational number, so $r$ can be written as $r = \frac{a}{b}$, where $a$ and $b$ are integers, and $b > 0$.

We need to show that there exists an integer $n$ such that $n \leq r < n + 1$. To find such an $n$, we first multiply the inequality by $b$ (since $b > 0$) to eliminate the fraction:

$b \cdot n \leq a < b \cdot (n + 1)$

This simplifies to:

$bn \leq a < bn + b$

Now, let's define $n$ as:

$n = \left\lfloor \frac{a}{b} \right\rfloor$

where $\left\lfloor \frac{a}{b} \right\rfloor$ is the floor of $\frac{a}{b}$, which is the largest integer less than or equal to $\frac{a}{b}$.

By definition, we have:

$n \leq \frac{a}{b} < n + 1$

This implies:

$n \leq r < n + 1$

Thus, we have found an integer $n$ that satisfies the inequality.

Step 2: Uniqueness

Suppose there exist two different integers $n_1$ and $n_2$ such that both satisfy $n_1 \leq r < n_1 + 1$ and $n_2 \leq r < n_2 + 1$. Without loss of generality, assume $n_1 < n_2$.

Then, we would have:

$n_1 < n_2 \leq r < n_1 + 1$

But this implies:

$n_1 + 1 > n_2$

Since $n_2$ is an integer and $n_2 > n_1$, we must have $n_2 = n_1 + 1$. However, this leads to a contradiction because $r$ cannot be both $\geq n_2$ and $< n_1 + 1 = n_2$.

Therefore, $n_1$ must be equal to $n_2$, proving that $n$ is unique.

Conclusion:

For any rational number $r = \frac{a}{b}$, there exists a unique integer $n$ such that $n \leq r < n + 1$, without relying on the Archimedean property of real numbers.

Would you like any further explanations or have any questions? Here are some related questions you might consider:

1. How does the proof change if $r$ is an irrational number?
2. What is the significance of the floor function in number theory?
3. How can this concept be applied to solve problems in modular arithmetic?
4. Is there a similar approach to finding $n$ for complex numbers?
5. How does this proof relate to the concept of divisibility?

Tip: When proving mathematical properties, it's often useful to try different approaches, like avoiding reliance on more abstract concepts, to deepen your understanding.