Yes, the statement is true. Here's why:

Proof:

Given $r$ is a rational number, we can express $r$ as $r = \frac{a}{b}$ where $a$ and $b$ are integers, and $b \neq 0$.

Now, by the Archimedean property of the real numbers (which also holds for rational numbers), for any real number $r$, there exists a unique integer $n$ such that:

$n \leq r < n + 1$

This integer $n$ is called the floor of $r$, often denoted as $\lfloor r \rfloor$. The floor function $\lfloor r \rfloor$ is defined as the greatest integer less than or equal to $r$.

Uniqueness:

The integer $n$ is unique because the floor function $\lfloor r \rfloor$ always rounds $r$ down to the nearest integer. If there were two such integers $n_1$ and $n_2$ that satisfy $n_1 \leq r < n_1 + 1$ and $n_2 \leq r < n_2 + 1$, then $n_1$ and $n_2$ must be the same integer. Therefore, $n$ is unique.

Conclusion:

For any rational number $r$, there exists a unique integer $n$ such that $n \leq r < n + 1$.

Would you like more details or have any questions about this proof? Here are some related questions you might consider:

1. How does the Archimedean property apply to irrational numbers?
2. What is the difference between the floor and ceiling functions?
3. How can you prove that the floor function is always an integer?
4. Can this property be extended to real numbers, and if so, how?
5. How does this concept relate to rounding in arithmetic?

Tip: The floor function $\lfloor r \rfloor$ is very useful in algorithms and computer science, especially in cases where you need to determine the largest integer less than or equal to a given number.